INSTRUCTOR’S SOLUTIONS MANUAL Richard N. Aufmann Palomar College Vernon C. Barker Palomar College Richard D. Nation Palo...
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INSTRUCTOR’S SOLUTIONS MANUAL Richard N. Aufmann Palomar College Vernon C. Barker Palomar College Richard D. Nation Palomar College Christine S. Verity
COLLEGE ALGEBRA AND TRIGONOMETRY SIXTH EDITION
Aufmann/Barker/Nation
HOUGHTON MIFFLIN COMPANY BOSTON NEW YORK
Editor-in-Chief/Publisher: Richard Stratton Sponsoring Editor: Molly Taylor Marketing Manager: Jennifer Jones Editorial Associate: Andrew Lipsett Marketing Associate: Mary Legere Editorial Assistant: Anthony D’Aries
Copyright © 2008 by Houghton Mifflin Company. All rights reserved. Houghton Mifflin Company hereby grants you permission to reproduce the Houghton Mifflin material contained in this work in classroom quantities, solely for use with the accompanying Houghton Mifflin textbook. All reproductions must include the Houghton Mifflin copyright notice, and no fee may be collected except to cover the cost of duplication. If you wish to make any other use of this material, including reproducing or transmitting the material or portions thereof in any form or by any electronic or mechanical means including any information storage or retrieval system, you must obtain prior written permission from Houghton Mifflin Company, unless such use is expressly permitted by federal copyright law. If you wish to reproduce material acknowledging a rights holder other than Houghton Mifflin Company, you must obtain permission from the rights holder. Address inquiries to College Permissions, Houghton Mifflin Company, 222 Berkeley Street, Boston, MA 02116-3764. Printed in the U.S.A. ISBN 13: 978-0-618-82508-0 ISBN 10: 0-618-82520-7 1 2 3 4 5 6 7 8 9-XX-11 10 09 08 07
Contents Chapter P
Preliminary Concepts
Chapter 1
Equations and Inequalities
36
Chapter 2
Functions and Graphs
98
Chapter 3
Polynomial and Rational Functions
177
Chapter 4
Exponential and Logarithmic Functions
238
Chapter 5
Trigonometric Functions
303
Chapter 6
Trigonometric Identities and Equations
361
Chapter 7
Applications of Trigonometry
436
Chapter 8
Topics in Analytic Geometry
496
Chapter 9
Systems of Equations and Inequalities
598
Chapter 10
Matrices
691
Chapter 11
Sequences, Series, and Probability
762
1
Projects
818
Additional College Trigonometry Solutions
871
Correlation Chart
Correlation Chart College Trigonometry, 6e Solutions/Projects Chapter 1 Section 1.1 Section 1.2* Section 1.3* Section 1.4* Section 1.5* Section 1.6* Section 1.7 Exploring Concepts with Technology Chapter 1 Assessing Concepts Chapter 1 Review Exercises Chapter 1 Quantitative Reasoning Chapter 1 Test Chapter 2 Chapter 2 up through Chapter 2 Test Cumulative Review Exercises Chapter 3 Chapter 3 up through Chapter 3 Test Cumulative Review Exercises Chapter 4 Section 4.1 Section 4.2 Section 4.3 Exploring Concepts with Technology Chapter 4 Review Exercises Chapter 4 Assessing Concepts Chapter 4 Test Cumulative Review Exercises Chapter 5 Section 5.1 Exercises 1-62 Section 5.1* Exercises 63-72 Section 5.2 Section 5.3 Exploring Concepts with Technology Chapter 5 Assessing Concepts Chapter 5 Review Exercises Chapter 5 Quantitative Reasoning Chapter 5 Test Cumulative Review Exercises Chapter 6 Sections 6.1-6.6 Section 6.7, Exercises 1-18 Section 6.7, Exercises 19-40 Chapter 6 Connecting Concepts Chapter 6 Review Exercises Chapter 6 Test Cumulative Review Exercises Chapter 7 Chapter 7 up through Chapter 7 Test Cumulative Review Exercises
College Algebra and Trigonometry, 6e Solutions/Projects See Additional College Trigonometry Solutions Section 2.1 Section 2.2 Section 2.5 Section 2.6 Section 4.1 Section 2.7 Chapter 2 Exploring Concepts with Technology See Additional College Trigonometry Solutions
Chapter 5 up through Chapter 5 Test See Additional College Trigonometry Solutions Chapter 6 up through Chapter 6 Test See Additional College Trigonometry Solutions Section 7.1 Section 7.2 Section 7.3 Chapter 7 Exploring Concepts with Technology Exercises 1-45 Chapter 7 Review Exercises See Additional College Trigonometry Solutions
Section P.6 Exercises 1-62 See Additional College Trigonometry Solutions Section 7.4 Section 7.5
See Additional College Trigonometry Solutions
Chapters 8.1-8.6 Section 8.7, Exercises 1-18
See Additional College Trigonometry Solutions
Section 4.2-4.7 up through Chapter 4 Test See Additional College Trigonometry Solutions
* See the Additional College Trigonometry Solutions for solutions to the Prepare for Section exercises.
Solutions
Chapter P
Preliminary Concepts Section P.1 1.
− 15 : rational, real; 0: integer, rational, real; –44: integer, rational, real; π : irrational, real; 3.14: rational, real; 5.05005000500005…: irrational, real;
2.
81 = 9 : integer, rational, prime, real; 53: integer, rational, prime, real
5 : irrational, real; 5 : rational, real; 31: integer, rational, prime, real; −2 1 : rational, real; 4.235653907493: rational, real; 7 7 2
51: integer, rational, real; 0.888… = 0.8 = 8 : rational, real 9 3.
Let x = 1, 2, 3, 4. Then {2x | x is a positive integer} = {2, 4, 6, 8}
4.
Let x = 0, 1, 2, 3. (We could have used x = –3, –2, –1, 0.) Then {|x| | x is an integer} = {0, 1, 2, 3}
5.
Let x = 1, 2, 3, 4. (Recall 0 is not a natural number.) Then {y | y = 2x + 1, x is a natural number} = {3, 5, 7, 9}
6.
Let x = 0, 1, 2, 3. (We could have used x = −3, −2, −1, 0.) Then {y | y = x2 − 1} = {−1, 0, 3, 8}
7.
Let x = 0, 1, 2, 3. (We could have used x = −3, −2, −1, 0.) Then {z | z = |x|, x is an integer}= {0, 1, 2, 3}
8.
Let x = −1, −2, −3, −4. Then {z | z = |x| − x, x is a negative integer}= {2, 4, 6, 8}
9.
A ∪ B = {−3, −2, −1, 0, 1, 2, 3, 4, 6}
11.
A ∩ C = {0, 1, 2, 3}
14.
(A ∩ C) = {0, 1, 2, 3} B ∪ (A ∩ C) = {−2, 0, 1, 2, 3, 4, 6}
16.
(A ∩ B) ∪ (A ∩ C) = {−2, 0, 2} ∪ {0, 1, 2, 3} = {−2, 0, 1, 2, 3}
17.
(B ∪ C) ∩ (B ∪ D) = {−2, 0, 1, 2, 3, 4, 5, 6} ∩ {−3, −2, −1, 0, 1, 2, 3, 4, 6} = {−2, 0, 1, 2, 3, 4, 6}
18.
(A ∩ C) ∪ (B ∩ D) = {0, 1, 2, 3} ∪ ∅ = {0, 1, 2, 3}
19.
10. 12.
C ∩ D = {1, 3}
21.
23.
{x | −3 < x < 3}
24.
26.
28.
{x | −5 ≤ x ≤ −1}
{x | x ≥ 2}
{x | x < 4} 27.
( −∞, − 1)
(3, 5) 29.
[−2, ∞) 30.
[0, 1]
[−1, 5)
B∩D=∅
(B ∪ C) = {−2, 0, 1, 2, 3, 4, 5, 6} D ∩ (B ∪ C) = {1, 3}
{x |1 ≤ x ≤ 5}
{x | −2 < x < 3}
25.
13. 15.
20.
22.
C ∪ D = {−3, −1, 0, 1, 2, 3, 4, 5, 6}
(−4, 5]
31.
–5
32.
–(4)2 = –16
33.
3(4) = 12
34.
|−3| − |−7| = 3 − 7 = −4
35.
π2 + 10
36.
10 – π2
Copyright © Houghton Mifflin Company. All rights reserved.
2
Chapter P: Preliminary Concepts
37.
x−4 + x+5 = 4− x+ x+5=9
38.
39.
2 x − x − 1 = 2 x − (1 − x ) = 2x − 1 + x = 3x − 1
40.
|x + 1| + |x – 3| = (x + 1) + (x – 3) = 2x – 2
41.
|x–3|
45.
m−n
46.
p −8
47.
a−4 <5
48.
z −5 > 7
49.
x+2 >4
50.
y+3 >6
42.
|a––2|=|a+2|
x + 6 + x − 2 = x + 6 + x − 2 = 2x + 4
43.
44.
x − −2 = 4 x+2 =4
51.
52.
53.
54.
55.
56.
57.
58.
59.
60.
61.
62.
63.
64.
65.
66.
z −5 =1
67.
−( −2)3 = −( −8) = 8
68.
−( −2)2 = −(4) = −4
69.
2(3)( −2)( −1) = 12
70.
−3(3)( −1) = 9
71.
−2(3)2 ( −2)2 = −2(9)(4) = −72
72.
2( −2)3 ( −1)2 = 2( −8)(1) = −16
73.
3( −2) − ( −1)[3 − ( −2)]2 = 3( −2) − ( −1)[3 + 2]2 = (3)( −2) − ( −1)[5]2 = (3)( −2) − ( −1)(25) = −6 + 25 = 19
74.
[ −1 − 2( −2)]2 − 3( −1)3 = ( −1 + 4)2 − 3( −1)3 = (3)2 − 3( −1) = 9 + 3 = 12
75.
32 + ( −2)2 9 + 4 13 = = = 13 3 + ( −2) 1 1
76.
2(3)( −2)2 ( −1)4
77.
3( −2) 2( −1) −6 −2 − = − = − 2 − 1 = −3 3 3 −2 −2
78.
(3 + 1)2 (3 − 1)2 = 42 ⋅ 22 = 16 ⋅ 4 = 64
79.
( ab2 )c = a (b2 c ) Associative property of multiplication
80.
2 x − 3 y = −3 y + 2 x Commutative property of addition
81.
4(2a − b) = 8a − 4b Distributive property
82.
6 + (7 + a) = 6 + (a + 7) Commutative property of addition
83.
(3x ) y = y (3x ) Commutative property of multiplication
84.
4ab + 0 = 4ab Identity property of addition
[ −2 − ( −1)]4
=
2(3)(4)(1) ( −2 + 1) 4
Copyright © Houghton Mifflin Company. All rights reserved.
= 24 4 = 24 ( −1)
Section P.1
3
85.
1 ⋅ (4 x ) = 4 x Identity property of multiplication
86.
7(a + b) = 7(b + a) Commutative property of addition
87.
x2 + 1 = x2 + 1 Reflexive property of equality
88.
If a + b = 2, then 2 = a + b Symmetric property of equality
89.
If 2x + 1 = y and 3x – 2 = y, then 2x+ 1 = 3x – 2 Transitive property of equality
90.
4x + 2y = 7 and x= 3, then 4(3) = 2y = 7 Substitution property of equality
91.
4⋅ 1 =1 4 Inverse property of multiplication
92.
ab + ( −ab) = 0 Inverse property of addition
93.
3(2 x )
−2(4 y ) −8 y
95.
3(2 + x )
3x−1x 4 2 3x−2x 4 4 1x 4
99.
94.
6x
97.
2a + 5a 3 6 4a + 5a 6 6 9a 6 3a 2
101.
5 − 3(4 x − 2 y )
98.
5(4 r − 7t ) − 2(10r + 3t ) 20r − 35t − 20r − 6t 20r − 20r − 35t − 6t −41t
107.
Area = 1 bh = 1 (3 in)(4 in) = 6 in 2 2 2
≈ 66 Heart rate is about 66 beats per minute.
4 + 2(2a − 3) 4 + 4a − 6 4a − 2
3(2a − 4b) − 4( a − 3b) 6a − 12b − 4a + 12b 6a − 4a − 12b + 12b 2a
105.
5a − 2[3 − 2(4a + 3)] 5a − 2(3 − 8a − 6) 5a − 2( −8a − 3) 5a + 16a + 6 21a + 6
106.
6 + 3[2 x − 4(3x − 2)] 6 + 3(2 x − 12 x + 8) 6 + 3( −10 x + 8) 6 − 30 x + 24 30 x + 30
= −0.5(110) + 120(110) − 2000 = −0.5(12100) + 120(110) − 2000 = −6050 + 13200 − 2000 = 5150 The profit for selling 110 bicycles is $5150.
= 65 + 53 41
100.
103.
= −0.5(110)2 + 120(110) − 2000
111. Heart rate = 65 + 53 4t + 1 53 = 65 + 4(10) + 1
2 + 3(2 x − 5) 2 + 6 x − 15 6 x − 13
7 − 2(5n − 8m) 7 − 10n + 16m 16m − 10n + 7
109. Profit = = −0.5 x 2 + 120 x − 2000 2
−2(4 + y ) −2 y − 8
3x + 6
102.
5 − 12 x + 6 y −12 x + 6 y + 5 104.
96.
108. V = lwh = (40 ft)(30 ft)(12 ft) = 14,400 ft 3 110. Circulation = n 2 − n + 1 = 122 − 12 + 1 144 − 12 + 1 = 133 ≈ 11.5 The circulation of the magazine after 12 months is approximately 11.5 thousand or 11,500 subscriptions.
112. BMI = 7052w h 705(160) 112800 = = ≈ 23 4900 (70)2 The body mass index (BMI) of a person who weighs 160 pounds and is 5 feet 10 inches (70 inches) tall is about 23.
Copyright © Houghton Mifflin Company. All rights reserved.
4
Chapter P: Preliminary Concepts
113. Height = −16t 2 + 80t + 4 = −16(2)2 + 80(2) + 4 = −16(4) + 80(2) + 4 = −64 + 160 + 4 = 100 After 2 seconds, the ball will have a height of 100 feet.
114. Concentration = 50t t +1 50(24) = 24 + 1 1200 = = 48 25 After 24 minutes, the concentration will be 48 grams per liter.
.......................................................
Connecting Concepts
115. For any set A, A ∪ A = A . 116. For any set A, A∩ A = A .
117. For any set A, A ∩ ∅ = ∅.
119. If A and B are two sets and a A ∪ B = A, then all elements of B are contained in A. So B is a subset of A.
120. If A and B are two sets and a A ∩ B = B, then all elements of B are contained in A. So B is a subset of A.
121. No. (8 ÷ 4) ÷ 2 = 2 ÷ 2 = 1 8 ÷ (4 ÷ 2) = 8 ÷ 2 = 4
122. No. 5–3=2 3 – 5 = –2
123. All but the multiplicative inverse property
124. All
125.
x+7 x+7 = x+7 = x+7 = x+7 = x + x −1 x − ( x − 1) 1 x + x −1
126.
x+3 x+3 x+3 x+3 = = = = x+3 1 1 1 1 1 1 1⎞ ⎛ 1⎞ ⎛ + x− + x+ −⎜ x − ⎟ + ⎜ x + ⎟ x− + x+ 2 2 2 2 2 2 2⎠ ⎝ 2⎠ ⎝
127. |x – 2| < |x – 6|
128. |x – a| < |x – b|
131. 2 < |x – 4| < 7
132. b < |x – a| < c
129. |x – 3| > |x + 7|
....................................................... PS1. 22 ⋅ 23 = 4 ⋅ 8 = 32 3
Alternate method: 2 ⋅ 2 = 2
2 +3
5
= 2 = 32
130. |x| > |x – 5|
Prepare for Section P.2 PS2.
2
118. For any set A, A ∪ ∅ = A.
43 = 43−5 = 4 −2 = 1 = 1 45 42 16 3 Alternate method: 45 = 51−3 = 12 = 1 16 4 4 4
PS3. (23 ) 2 = 82 = 64
PS4. 3.14(105 ) = 3.14(100,000) = 314,000
Alternate method: (23 )2 = 23(2) = 26 = 64
Alternate method: To multiply by 105 , move the decimal point 5 places to the right. 000. = 314,000 Thus, 3.14(105 ) = 314 → → → → →
PS5. False
34 ⋅ 32 = 36 , not 96 .
PS6. False
(3 + 4)2 = 72 = 49 but 32 + 42 = 9 + 16 = 25.
Copyright © Houghton Mifflin Company. All rights reserved.
Section P.2
5
Section P.2 0
1.
−53 = −(53 ) = −125
2.
( −5)3 = −125
3.
⎛ 2⎞ =1 ⎜ ⎟ ⎝ 3⎠
4.
−60 = −(60 ) = −1
5.
4 −2 = 12 = 1 16 4
6.
3−4 = 14 = 1 81 3
7.
1 = 25 = 32 2 −5
8.
1 = 33 = 27 3−3
9.
2 −3 = ⎛ 2 ⎞ ⎜ ⎟ 6 −3 ⎝ 6 ⎠
10.
4 −2 = 23 = 8 = 1 2 −3 42 16 2
11.
−2 x 0 = −2
12.
x0 = 1 4 4
−3
= ⎛⎜ 1 ⎞⎟ ⎝ 3⎠
−3
3
= ⎛⎜ 3 ⎞⎟ = 33 = 27 ⎝1⎠
13.
2 x −4 = 2 ( x −4 ) = 24 x
14.
3 y −2 = 3 ( y −2 ) = 32 y
15.
5 = 5z 6 z −6
16.
8 = 8z5 x −5
17.
( x 3 y 2 )( xy 5 ) = x 3+1 y 2+5 = x 4 y 7
18.
(uv 6 )(u 2 v ) = u1+ 2 v 6+1 = u 3v 7
19.
( −2ab4 )( −3a 2 b5 ) = ( −2)( −3)a1+ 2 b4+5 = 6a 3b9
20.
(9 xy 2 )( −2 x 2 y 5 ) = (9)( −2) x1+ 2 y 2+5 = −18 x 3 y 7
21.
16a 7 = 16 a 7 −1 = 8a 6 2a 2
22.
24 z 8 = 24 z 8−3 = −8 z 5 −3z 3 −3
23.
6a 4 = 6 a 4 −8 = 3 a −4 = 3 4 8a 8 8 4a 4
24.
12 x 3 = 12 x 3− 4 = 3 x −1 = 3 4 4x 16 x 4 16
25.
12 x 3 y 4
2 y2 = 12 x 3−5 y 4 −2 = 2 x −2 y 2 = 2 18 3 18 x y 3x
26.
5v 4 w −3 = 5 v 4 −8 w −3 = 1 v −4 w −3 = 1 10 2 10v8 2 v 4 w3
27.
36a −2 b3 = 36 a −2 −1b3− 4 = 12a −3b −1 = 12 3 a 3b 3ab4
28.
−48ab10 = −48 a1− 4 b10−3 = 3 a −3b7 = 3b7 2 2a 3 −32a 4 b3 −32
29.
( −2m 3n 2 )( −3mn 2 )2 = ( −2m3n 2 )(9m 2 n 4 )
30.
(2a 3b2 )3 ( −4a 4 b2 ) = (8a 9 b6 )( −4a 4 b2 )
5 2
31.
= ( −2)(9)m 3+ 2 n 2 + 4
= (8)( −4)a 9 + 4 b6+ 2
= −18m5n 6
= −32a13b8
( x −2 y )2 ( xy ) −2 = ( x −4 y 2 )( x −2 y −2 )
32.
( x −1 y 2 ) −3 ( x 2 y −4 ) −3 = ( x 3 y −6 )( x −6 y12 )
= x −4 − 2 y 2 − 2
= x 3−6 y −6+12
= x −6 y 0 = 16 x
= x −3 y 6 =
y6 x3
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6
33.
35.
Chapter P: Preliminary Concepts 2
⎛ 3a 2 b3 ⎞ 9a 4 b6 ⎜ 4 4⎟ = 36a 8b8 ⎝ 6a b ⎠ = 9 a 4 −8 b 6 −8 36 = 1 a −4 b −2 4 = 14 2 4a b
34.
( −4 x 2 y 3 ) 2
36.
2 3
(2 xy )
=
9
= 8c 125
16 x 4 y 6 3 6
( −3a 2 b3 )2
4 6 = 9a 3b 12 ( −2ab ) −8a b = − 9 a 4 −3b6−12 8 = − 9 ab −6 8 = − 9a6 8b 4 3
8x y
= 2 x 4 −3 y 6−6 = 2x
37.
3
3 6 9 ⎛ 2ab2 c3 ⎞ = 8a b 3c 6 ⎜ 2 ⎟ ⎝ 5ab ⎠ 125a b = 8 a 3−3b6−6 c9 125 = 8 a 0 b0 c 9 125
2
⎛ a −2 b ⎞ a −4 b2 ⎜ 3 −4 ⎟ = 6 −8 a b ⎝a b ⎠
38.
= a −4 −6 b2 −( −8)
⎛ x −3 y −4 ⎞ ⎜⎜ x −2 y ⎟⎟ ⎝ ⎠
−2
=
x 6 y8 x 4 y −2
= x 6− 4 y 8−( −2)
= a −4 −6 b2 +8
= x 6 − 4 y 8+ 2
= a −10 b10
= x 2 y10
10 = b10 a
39.
2,011,000,000,000 = 2.011 × 1012
40.
49,100,000,000 = 4.91 × 1010
41.
0.000000000562 = 5.62 × 10−10
42.
0.000000402 = 4.02 × 10−7
43.
3.14 × 107 = 31,400,000
44.
4.03 × 109 = 4,030,000,000
45.
−2.3 × 10−6 = −0.0000023
46.
6.14 × 10−8 = 0.0000000614
47.
(3 × 1012 )(9 × 10−5 ) = (3)(9) × 1012 −5
48.
(8.9 × 10−5 )(3.4 × 10−6 ) = (8.9)(3.4) × 10−5−6
49.
= 27 × 107
= 30.26 × 10−11
= 2.7 × 108
= 3.026 × 10−10
9 × 10−3 = 9 × 10−3−8 6 × 108 6
50.
2.5 × 108 = 2.5 × 108−10 5 5 × 1010 = 0.5 × 10−2
= 1.5 × 10−11
= 5 × 10−3
51.
(3.2 × 10−11 )(2.7 × 1018 ) 1.2 × 10
−5
=
(3.2)(2.7) × 10−11+18−( −5) 1.2
52.
(6.9 × 1027 )(8.2 × 10−13 ) 4.1 × 1015
=
(6.9)(8.2) × 1027 −13−15 4.1
= 7.2 × 10−11+18+5
= 13.8 × 10−1
= 7.2 × 1012
= 1.38 × 100
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Section P.2
53.
7
(4.0 × 10−9 )(8.4 × 105 ) −6
18
(3.0 × 10 )(1.4 × 10 )
=
(4.0)(8.4) × 10−9 + 5+ 6−18 (3.0)(1.4)
= 8 × 10
54.
(7.2 × 108 )(3.9 × 10−7 ) (2.6 × 10
−10
−8
)(1.8 × 10 )
−16
=
(7.2)(3.9) × 108−7 −( −10)−( −8) (2.6)(1.8)
= 6 × 108−7 +10+8 = 6 × 1019
3
56.
−163/ 2 = − 16 = −43 = −64
3
57.
−642 / 3 = − 3 64 = −42 = −16
59.
1 = 1 = 1 = 1 9 −3/ 2 = 3/ 9 2 ( 9 )3 33 27
60.
32 −3/ 5 =
55.
43/ 2 = 4 = 23 = 8
58.
1254 / 3 = 3 125 = 54 = 625
61.
⎛4⎞ ⎜ ⎟ ⎝9⎠
63.
⎛1⎞ ⎜ ⎟ ⎝8⎠
65.
(4a 2 / 3b1/ 2 )(2a1/ 3b3 / 2 ) = (4)(2)a 2 / 3+1/ 3b1/ 2 + 3/ 2 = 8a 3 / 3b4 / 2 = 8ab2
66.
(6a 3/ 5b1/ 4 )( −3a1/ 5b3/ 4 ) = (6)( −3)a 3 / 5+1/ 5b1/ 4 +3/ 4 = −18a 4 / 5b
67.
( −3x 2 / 3 )(4 x1/ 4 ) = ( −3)(4) x 2 / 3+1/ 4 = −12 x8 /12 + 3/12 = −12 x11/12
68.
( −5 x1/ 3 )( −4 x1/ 2 ) = ( −5)( −4) x1/ 3+1/ 2 = 20 x 2 / 6+ 3/ 6 = 20 x 5 / 6
69.
(81x8 y12 )1/ 4 = 811/ 4 x8 / 4 y12 / 4 = 4 81x 2 y 3 = 3x 2 y 3
70.
(27 x 3 y 6 )2 / 3 = 272 / 3 x 6 / 3 y12 / 3 = 3 27 x 2 y 4 = 32 x 2 y 4 = 9 x 2 y 4
71.
16 z 3/ 5 = 16 z 3/ 5−1/ 5 = 4 z 2 / 5 12 3 12 z1/ 5
73.
(2 x 2 / 3 y1/ 2 )(3x1/ 6 y1/ 3 ) = (2)(3) x 2 / 3+1/ 6 y1/ 2+1/ 3 = 6 x 5 / 6 y 5 / 6
74.
x1/ 3 y 5 / 6
4
1/ 2
= 4= 4=2 9 9 3
−4 / 3
4
= 84 / 3 = 3 8 = 24 = 16
⎛ 16 ⎞ ⎜ ⎟ ⎝ 25 ⎠
64.
⎛ 8 ⎞ ⎜ ⎟ ⎝ 27 ⎠
3
1 = 1 = 1 =1 323/ 5 ( 5 32 )3 23 8
3
3 ⎛ ⎞ = ⎜ 16 ⎟ = ⎛⎜ 4 ⎞⎟ = 43 = 64 125 5 ⎝ 25 ⎠ ⎝ 5 ⎠
−2 / 3
= ⎜⎛ 27 ⎟⎞ ⎝ 8 ⎠
2/3
2
2
2 ⎛3 ⎞ ⎛ ⎞ = ⎜ 3 27 ⎟ = ⎜ 327 ⎟ = ⎜⎛ 3 ⎟⎞ = 9 4 ⎝2⎠ ⎝ 8 ⎠ ⎝ 8 ⎠
2
x
2 / 3 1/ 6
y
72.
= x1/ 3− 2 / 3 y 5 / 6−1/ 6 = x −1/ 3 y 4 / 6 =
x1/ 3
9a 3/ 4 b = 9a 3 / 4 − 2 / 3b1− 2 = 3a 9 /12 −8 /12 b −1 = 3a1/12 3 b 3a 2 / 3b2
76.
12 x1/ 6 y1/ 4 16 x
3 / 4 1/ 2
y
=
12 x1/ 6−3/ 4 y1/ 4 −1/ 2 3x 2 /12 −9 /12 y1/ 4 − 2 / 4 3x −7 /12 y −1/ 4 3 = = = 7 /12 16 4 4 4x y1/ 4
45 = 32 ⋅ 5 = 3 5
77. 3
3
6a 2 / 3 = 6a 2 / 3−1/ 3 = 2a1/ 3 = 2a1/ 3 9 3 3 9a1/ 3
y2 / 3
75.
80.
3/ 2
62.
2
135 = 33 ⋅ 5 = 33 5
78. 81.
3
3
75 = 52 ⋅ 3 = 5 3
79.
3
24 = 23 ⋅ 3 = 2 3 3
−135 = 3 ( −3)3 ⋅ 5
82.
3
−250 = 3 ( −5)3 ⋅ 2
= −3 3 5
Copyright © Houghton Mifflin Company. All rights reserved.
= −5 3 2
8
Chapter P: Preliminary Concepts
83.
24 x 2 y 3 = 22 x 2 y 2 ⋅ 6 y = 2 xy 6 y
84.
16a 3 y 7 = 3 23 a 3 y 6 ⋅ 3 2 y = 2ay 2 3 2 y
86.
18 x 2 y 5 = 32 x 2 y 4 ⋅ 2 y = 3 x y 2 2 y 3
3
3
3
54m 2 n 7 = 33 n 6 ⋅ 2m 2 n = 3n 2 2m 2 n
85.
3
87.
2 32 − 3 98 = 2 16 ⋅ 2 − 3 49 ⋅ 2 = 2(4) 2 − 3(7) 2 = 8 2 − 21 2 = −13 2
88.
5 3 32 + 2 3 108 = 5 3 8 ⋅ 4 + 2 3 27 ⋅ 4 = 5 3 8 ⋅ 3 4 + 2 3 27 ⋅ 3 4 = 5 23 ⋅ 3 4 + 2 33 ⋅ 3 4 = 5(2) 3 4 + 2(3) 3 4
3
3
= 10 3 4 + 6 3 4 = 16 3 4 89.
4
4
−8 4 48 + 2 4 243 = −8 4 16 ⋅ 3 + 2 4 81 ⋅ 3 = −8 4 16 ⋅ 4 3 + 2 4 81 ⋅ 4 3 = −8 24 ⋅ 4 3 + 2 34 ⋅ 4 3 = −8(2) 4 3 + 2(3) 4 3 = −16 4 3 + 6 4 3 = −10 4 3
90.
3
3
3
3
2 3 40 − 33 135 = 2 3 8 ⋅ 5 − 33 27 ⋅ 5 = 2 3 8 ⋅ 3 5 − 33 27 ⋅ 3 5 = 2 23 ⋅ 3 5 − 3 33 ⋅ 3 5 = 2 23 ⋅ 3 5 − 3 33 ⋅ 3 5 = 2(2) 3 5 − 3(3) 3 5 = 4 3 5 − 9 3 5 = −5 3 5
91.
4 3 32 y 4 + 3 y 3 108 y = 4 3 8 y 3 ⋅ 4 y + 3 y 3 27 ⋅ 4 y = 4 3 8 y 3 ⋅ 3 4 y + 3 y 3 27 ⋅ 3 4 y 3
= 4 3 23 y 3 ⋅ 3 4 y + 3 y 33 ⋅ 3 4 y = 4(2 y ) 3 4 y + 3 y (3) 3 4 y = 8 y 3 4 y + 9 y 3 4 y = 17 y 3 4 y 92.
3
3
3
3
3
3
−3x 54 x 4 + 2 16 x 7 = −3x 27 x 3 ⋅ 2 x + 2 8 x 6 ⋅ 2 x = −3x 27 x 3 ⋅ 3 2 x + 2 8 x 6 ⋅ 3 2 x 3
3
= −3x 33 x 3 ⋅ 3 2 x + 2 23 x 6 ⋅ 3 2 x = −3x (3x ) 3 2 x + 2(2 x 2 ) 3 2 x = −9 x 2 3 2 x + 4 x 2 3 2 x = −5 x 2 3 2 x 93.
3
x 3 8 x 3 y 4 − 4 y 3 64 x 6 y = x 3 8 x 3 y 3 ⋅ y − 4 y 3 64 x 6 ⋅ y = x 3 8 x 3 y 3 ⋅ 3 y − 4 y 64 x 6 ⋅ 3 y 3
= x 3 23 x 3 y 3 ⋅ 3 y − 4 y 43 x 6 ⋅ 3 y = x (2 xy ) 3 y − 4 y (4 x 2 ) 3 y = 2 x 2 y 3 y − 16 x 2 y 3 y = −14 x 2 y 3 y 94.
4 a 5b − a 2 ab = 4 a 4 ⋅ ab − a 2 ab = 4 a 4 ⋅ ab − a 2 ab = 4a 2 ab − a 2 ab = 3a 2 ab
95.
( 5 + 3)( 5 + 4) = 5 + 4 5 + 3 5 + (3)(4) = 5 + 7 5 + 12 = 17 + 7 5
96.
( 7 + 2)( 7 − 5) = 7 − 5 7 + 2 7 + (2)( −5) = 7 − 3 7 − 10 = −3 − 3 7
97.
( 2 − 3)( 2 + 3) = 2 + 3 2 − 3 2 + ( −3)(3) = 2 − 9 = −7
98.
(2 7 + 3)(2 7 − 3) = (2 7)2 − 3(2 7) + 3(2 7) + (3)( −3) = 4(7) − 6 7 + 6 7 − 9 = 28 − 9 = 19
99.
(3 z − 2)(4 z + 3) = (3)(4) z + 3(3 z ) − 2(4 z ) + ( −2)(3) = 12 z + 9 z − 8 z − 6 = 12 z + z − 6
100.
(4 a − b )(3 a + 2 b ) = (4 a )(3 a ) + (4 a ) ⋅ (2 b ) − b (3 a ) − b (2 b )
2
2
2
2
2
= (4)(3) a + (4)(2)( a )( b ) − 3( a )( b ) − 2 b
2
= 12a + 8 ab − 3 ab − 2b = 12a + 5 ab − 2b 101.
2
( x + 2)2 = x + 2( x )(2) + 22 = x + 4 x + 4
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Section P.2
9 2
102.
(3 5 y − 4)2 = 32 5 y + 2(3 5 y )( −4) + ( −4)2 = 9(5 y ) − 24 5 y + 16 = 45 y − 24 5 y + 16
103.
( x − 3 + 2)2 = x − 3 + 2( x − 3)(2) + 22 = x − 3 + 4 x − 3 + 4 = x + 4 x − 3 + 1
104.
( 2 x + 1 − 3)2 = 2 x + 1 + 2( 2 x + 1)( −3) + ( −3)2 = 2 x + 1 − 6 2 x + 1 + 9 = 2 x − 6 2 x + 1 + 10
2
2
3x = 3 x ⋅ 3 = 3 x 3 = 3 x 3 = 3 x 3 = x 3 2 3 3 3 3 3 3
105.
2 = 2 ⋅ 2=2 2=2 2= 2 2= 2 2 2 2 2 2 2 2
107.
5 = 5 = 5 ⋅ 2 = 5 ⋅ 2 = 5 ⋅ 2 = 10 = 10 18 2 6 2 ⋅ 32 2 ⋅ 32 22 ⋅ 32 22 ⋅ 32 2 ⋅ 3
108.
7 = 7 = 7 ⋅ 2 ⋅ 5 = 7 ⋅ 2 ⋅ 5 = 7 ⋅ 2 ⋅ 5 = 7 ⋅ 2 ⋅ 5 = 70 40 20 23 ⋅ 5 23 ⋅ 5 2 ⋅ 5 24 ⋅ 52 22 ⋅ 5 24 ⋅ 52 3
3
109.
3 = 3 ⋅ 22 = 3 2 2 = 33 4 3 3 3 2 2 3 2 3 22 2
110.
2 = 2 = 2 ⋅ 3 2 = 23 2 = 23 2 = 2 3 2 = 3 2 3 2 2 4 3 22 3 2 2 3 2 3 23 2
111.
4 3
112.
106.
8x
2
=
4 3
3 2
2 x
=
4 3
2 x
2
=
3 3 3 = 2 = 2 ⋅3x =2 x =2 x 3 2 3 2 3 3 x x x x x 2 x
4
3
2
3 3 3 4 2 3 4 2 3 4 4 4 2 = 2 = 2 ⋅ 2 y = 2 2 y = 2 4y = 2 4y = 4y 4 4y 4 22 y 4 22 y 4 22 y 3 4 24 y 4 2y 2y y
113.
3( 3 − 4) 3( 3 − 4) 3( 3 − 4) 3 3 − 12 3 = 3 ⋅ 3−4 = = = = = − 3 3 − 12 2 2 3 16 13 13 − − 3+4 3 + 4 3 − 4 ( 3 + 4)( 3 − 4) 3 −4
114.
2( 5 + 2) 2( 5 + 2) 2( 5 + 2) 2 5 + 4 2 = 2 ⋅ 5+2 = = = = =2 5+4 2 5−4 1 5−2 5 − 2 5 + 2 ( 5 − 2)( 5 + 2) 5 − 22 3
115.
3( 5 − 1) 3( 5 − 1) 6 6 6 = = = 3 = 3 ⋅ 5 −1 = = 2 2 5 + 2 2( 5 + 1) 2 ( 5 + 1) 5 +1 5 + 1 5 − 1 ( 5 + 1)( 5 − 1) 5 −1 =
116.
−7(3 2 + 5) −7(3 2 + 5) −7(3 2 + 5) −7(3 2 + 5) −7 = −7 ⋅ 3 2 + 5 = = = = (9)(2) − 25 18 − 25 3 2 − 5 3 2 − 5 3 2 + 5 (3 2 − 5)(3 2 + 5) (3 2)2 − 52 =
117.
3( 5 − 1) 3 5 − 3 = 5 −1 4
−7(3 2 + 5) −7 (3 2 + 5) = =3 2 +5 −7 −7
3( 5 − x ) 3 3 = ⋅ 5− x = = 3 5−3 x = 3 5 −3 x 5− x 5+ x 5 + x 5 − x ( 5 + x )( 5 − x ) ( 5)2 − ( x ) 2
Copyright © Houghton Mifflin Company. All rights reserved.
10
118.
Chapter P: Preliminary Concepts
5 = y− 3
5 ⋅ y− 3
y+ 3
=
y+ 3
5( y + 3) ( y − 3)( y + 3)
=
5 y +5 3 2
( y ) − ( 3)
119.
$8.1 × 1012 = $8.1 × 1012 −8 per person ≈ $2.72 × 104 per person 2.98 × 108 people 2.98
120.
4.7 × 1021 bacteria ⋅
670 femtograms 1 × 10−15 grams ⋅ 1 bacteria 1 femtogram
121.
−15 4.7 × 1021 ⋅ 670 ⋅ 1 × 10 = (4.7)(6.70)(1) × 1021+ 2 −15 1 1 = 3.149 × 109
2
=
5 y +5 3 y−3
1 seed ⋅ 1 ounce 3.2 × 10−8 ounce package 1 1 ⋅1 = × 108 −8 3.2 3.2 × 10
= 0.3125 × 108
= 3.13 × 107 seeds per package
They weigh 3.149 × 109 grams. 122.
800 nm ⋅ 1 × 10−9 m = (8 × 102 )(1 × 10−9 ) = 8 × 10−7 m 1 1 nm 1 frequency = wavelength 1 = = 1.25 × 108 cycles per second 8 × 10−7
123.
Red shift =
λr − λs λs −7
× 10 = 5.13 × 10 − 5.06 5.06 × 10−7 =
−7
(5.13 − 5.06) × 10−7 5.06 × 10−7 −7
= 0.07 ⋅ 10−7 5.06 10 = 1.38 × 10−2 124.
5.2 AU = 5.2(9.3 × 107 miles)
125.
= (5.2)(9.3) × 107 miles = 48.36 × 107 miles = 4.84 × 108 miles
1 sec ⋅ 1.5 × 1011 m ⋅ 1 min 60 sec 3 × 108 m 11
1 ⋅ 1.5 × 1011 ⋅ 1 = (1)(1.5)(1) × 10 60 3 × 108 3(60) × 108 = 1.5 × 1011−8 180 ≈ 0.008 × 103 ≈ 8 minutes
126.
1 gram ⋅ 1 atom 6.023 × 1023 atoms 1 ⋅ 1 = 1 × 10−23 6.023 6.023 × 1023 ≈ 0.1660302175 × 10−23
≈ 1.66 × 10−24
One hydrogen atom weighs approximately 1.66 × 10−24 gram. 127. Evaluate R when x = 20.
R = 1250 x (2 −0.007 x ) = 1250(20)(2 −0.007(20) ) = 25,000(2 −0.14 ) ≈ 25,000(0.907519) ≈ 22,688 When the company sells 20 thousand phones, the revenue is $22,688.
Copyright © Houghton Mifflin Company. All rights reserved.
Section P.2
128. a. b.
11
Evaluate A when t = 4. A = 2(10−0.0078t ) = 2(10−0.0078(4) ) = 2(10−0.0312 ) ≈ 2(.930679) ≈ 1.86 Four hours after taking 2 milligrams of digoxin, about 1.86 milligrams remain in the patient’s blood. Determine the sum of the amounts of medication remaining at 6:00 PM from each of the two doses. For the 1:00 PM dose, use t = 5; for the 5:00 PM dose, use t = 1. A = 2(10−0.0078(5) ) + 2(10−0.0078(1) ) = 2(10−0.039 ) + 2(10−0.0078 ) ≈ 2(.9141) + 2(0.9822) ≈ 1.828 + 1.964 ≈ 3.79 At 6:00 PM, the amount of digoxin remaining in the patient’s blood is 3.79 milligrams.
129. Evaluate P when n = 50. P = 6.5(20.016354 n ) = 6.5(20.016354(45) ) = 6.5(20.73593 ) ≈ 6.5(1.665) ≈ 10.8
In 2050, the world’s population will be approximately 10.8 billion. 130. One hour is 60 minutes. Evaluate P when t = 60. P = 90 − 3t 2 / 3 = 90 − 3(60)2 / 3 ≈ 90 − 3(15.3261) ≈ 44.02
Thus, the percent of the students who remembered the number after 1 hour was 44.02%. 131. a.
Evaluate P when d = 10. P = 102 − d / 40 = 102 −10 / 40 = 102 −0.25 = 101.75 ≈ 56 The amount of light that will pass to a depth of 10 feet below the ocean’s surface is about 56%.
b.
Evaluate P when d = 25. P = 102 − d / 40 = 102 − 25 / 40 = 102−0.625 = 101.375 ≈ 24 The amount of light that will pass to a depth of 25 feet below the ocean’s surface is about 24%.
.......................................................
Connecting Concepts
x x y 132. If 2 x = y , then 2 x − 4 = 2 x ⋅ 2 −4 = 2 4 = 2 4 = 4 2 2 2
133. No, if a and b are nonzero numbers and a < b, then the statement a −1 < b −1 is not a true statement. Let a = 2 and b = 3. Then a < b, but a −1 = 2 −1 = 1 and b −1 = 3−1 = 1 . 1 > 1 so a −1 > b −1. 3 2 3 2 134.
450 ⋅ 5100 = 450 ⋅ 550 ⋅ 550 = (4 ⋅ 5 ⋅ 5)50 = 10050 = (102 )50 = 10100 10100 is 1 followed by 100 zeros, thus has 101 digits.
135.
a 2 / 5a p = a 2 2+ p=2 5 p =2− 2 5 8 p= 5
137.
x −3 / 4 = x 4 x3 p − 3 − 3p = 4 4 −3 p = 19 4 p = − 19 12
136.
138.
b −3/ 4 b2 p = b3 − 3 + 2p = 3 4 2p = 3+ 3 4 15 ⎛ p = ⎜ ⎞⎟ 1 ⎝ 4 ⎠2 p = 15 8
( x 4 x 2 p )1/ 2 = x (4 + 2 p) 1 = 1
2 2 + p =1 p = −1
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12
Chapter P: Preliminary Concepts 2 2 4 + h − 2 = 4 + h − 2 ⋅ 4 + h + 2 = ( 4 + h − 2)( 4 + h + 2) = ( 4 + h ) − 2 = 4 + h − 4 h h 4+h +2 h ( 4 + h + 2) h( 4 + h + 2) h ( 4 + h + 2)
139.
=
h 1 h = = 4+h +2 h( 4 + h + 2) h ( 4 + h + 2)
2 2 9 + h − 3 = 9 + h − 3 ⋅ 9 + h + 3 = ( 9 + h − 3)( 9 + h + 3) = ( 9 + h ) − 3 = 9 + h − 9 h h h( 9 + h + 3) h( 9 + h + 3) h( 9 + h + 3) 9+h +3
140.
=
h h 1 = = h( 9 + h + 3) h ( 9 + h + 3) 9+h +3
141.
2 2 2 2 2 n 2 + 1 − n = n 2 + 1 − n ⋅ n 2 + 1 + n = ( n + 1 − n )( n + 1 + n ) = ( n + 1) − n = n 2 + 1 − n 2 = 1 1 1 n2 + 1 + n n2 + 1 + n n2 + 1 + n n2 + 1 + n n2 + 1 + n
142.
2 2 2 2 2 n 2 + n − n = n 2 + n − n ⋅ n 2 + n + n = ( n + n − n )( n + n + n ) = ( n + n ) − n = n 2 + n − n 2 = n 2 2 2 2 2 1 1 n +n +n n +n +n n +n +n n +n +n n +n +n
143.
(
2
2
)
2
= 2
( 2 )( 2 )
= 2
( 2 )2
2
= 2 =2
.......................................................
Prepare for Section P.3
PS1. −3(2a − 4b) −6a + 12b
PS2. 5 − 2(2 x − 7) 5 − 4 x + 14 −4 x + 19
PS3. 2 x 2 + 3x − 5 + x 2 − 6 x − 1
PS4. 4 x 2 − 6 x − 1 − 5 x 2 + x
2 x 2 + x 2 + 3x − 6 x − 5 − 1
4 x 2 − 5x 2 − 6 x + x − 1
(2 + 1) x 2 + (3 − 6) x − (5 + 1)
(4 − 5) x 2 + ( −6 + 1) x − 1
3x 2 − 3x − 6
− x 2 − 5x − 1
PS6. 12 + 15 = 27 4 4 = 6 3 ≠ 18 4 False.
?
PS5.
4 − 3x − 2 x 2 =− 2 x 2 − 4 x + 4 ?
−2 x 2 − 3x + 4 =− 2 x 2 − 4 x + 4 False.
Section P.3 1.
D
2.
E
3.
H
4.
F
5.
6.
I
7.
B
8.
C
9.
J
10. A
11.
a. b. c. d. e.
x2 + 2x − 7 2 1, 2, −7 1 x2, 2x , − 7
12.
a. b. c. d. e.
−12x4 − 3x2 − 11 4 −12, − 3, − 11 −12 −12x4, − 3x2, − 11
13.
a. b. c. d. e.
Copyright © Houghton Mifflin Company. All rights reserved.
x3 − 1 3 1, −1 1 x3, − 1
G
Section P.3
14.
a. b. c. d. e.
17.
13
4x2 − 2x + 7 2 4, −2, 7 4 4x2, −2x, 7
2x4 + 3x3 + 4x2 + 5 4 2, 3, 4, 5 2 2x4, 3x3, 4x2 , 5
15.
a. b. c. d. e.
3
18.
20.
6
21.
23.
(3x2 + 4x + 5) + (2x2 + 7x – 2) = 5x2 + 11x + 3
25.
(4w3 – 2w + 7) + (5w3 + 8w2 – 1) = 9w3 + 8w2 – 2w + 6
26.
(5x4 – 3x2 + 9) + (3x3 – 2x2 – 7x + 3) = 5x4 + 3x3 – 5x2 – 7x + 12
27.
(r2 – 2r – 5) – (3r2 – 5r + 7) = r2 – 2r – 5 – 3r2 + 5r – 7 = –2r2 + 3r – 12
28.
(7s2 – 4s + 11) – (–2s2 + 11s – 9) = 7s2 – 4s + 11 + 2s2 – 11s + 9 = 9s2 – 15s + 20
29.
(u3 – 3u2 – 4u + 8) – (u3 – 2u + 4) = u3 – 3u2 – 4u + 8 – u3 + 2u – 4 = –3u2 – 2u + 4
30.
(5v4 – 3v2 + 9) – (6v4 + 11v2 – 10) = 5v4 – 3v2 + 9 – 6v4 – 11v2 + 10 = –v4 – 14v2 + 19
31.
(2x2 + 7x – 8) (4x – 5) = 8x3 – 10x2 + 28x2 – 35x – 32x + 40 = 8x3 + 18x2 – 67x + 40
32.
(3x2 – 8x – 5) (5x – 7) = 15x3 – 21x2 – 40x2 + 56x – 25x + 35 = 15x3 – 61x2 + 31x + 35
33.
16.
a. b. c. d. e.
3
19.
5
2
22.
4
3x 2 − 2 x + 5
24.
−5x3 + 3x2 + 7x − 1 3 −5, 3, 7, −1 −5 −5x3, 3x2 , 7x, − 1
(5y2 – 7y + 3) + (2y2 + 8y + 1) = 7y2 + y + 4
34.
2y3 − 3 y + 4
2 x2 − 5x + 2
2 y2 − 5 y + 7
+ 6 x 2 − 4 x + 10
+ 14y3
− 21 y + 28
3
2
4
3
2
4 y5
4
3
2
4 y 5 − 10 y 4 + 8 y 3 + 23 y 2 − 41 y + 28
− 15 x + 10 x − 25 x + 6 x − 4 x + 10 x
− 10 y
6 x − 19 x + 26 x − 29 x + 10
4
2
+ 15 y − 20 y − 6 y3 + 8 y 2
35.
(2x + 4) (5x + 1) = 10x2 + 2x + 20x + 4 = 10x2 + 22x + 4
36.
(5x − 3) (2x + 7) = 10x2 + 35x − 6x − 21 = 10x2 + 29x − 21
37.
(y + 2) (y + 1) = y2 + y + 2y + 2 = y2 + 3y + 2
38.
(y + 5) (y + 3) = y2 + 3y + 5y + 15 = y2 + 8y + 15
39.
(4z − 3) (z − 4) = 4z2 − 16z − 3z + 12 = 4z2 – 19z + 12
40.
(5z − 6) (z − 1) = 5z2 − 5z − 6z + 6 = 5z2 – 11z + 6
41.
(a + 6) (a − 3) = a2 − 3a + 6a – 18 = a2 + 3a − 18
42.
(a − 10) (a + 4) = a2 + 4a − 10a − 40 = a2 – 6a − 40
43.
(5x − 11y) (2x − 7y) = 10x2 − 35xy − 22xy + 77y2 = 10x2 – 57xy + 77y2
44.
(3a − 5b) (4a − 7b) = 12a2 − 21ab − 20ab + 35b2 = 12a2 – 41ab + 35b2
45.
(9x + 5y) (2x + 5y) = 18x2 + 45xy + 10xy + 25y2 = 18x2 + 55xy + 25y2
46.
(3x − 7z) (5x − 7z) = 15x2 − 21xz − 35xz + 49z2 = 15x2 – 56xz + 49z2
47.
(3p + 5q) (2p − 7q) = 6p2 − 21pq + 10pq – 35q2 = 6p2 – 11pq − 35q2
48.
(2r − 11s) (5r + 8s) = 10r2 + 16rs – 55rs – 88s2 = 10r2 – 39rs − 88s2
49.
(4d − 1)2 − (2d − 3)2 = (16d 2 − 8d + 1) – (4d 2 – 12d + 9) = 16d 2 − 8d + 1 – 4d 2 + 12d − 9 = 12d 2 + 4d − 8
50.
(5c − 8)2 − (2c − 5)2 = (25c2 − 80c + 64) – (4c2 – 20c + 25) = 25c2 − 80c + 64 – 4c2 + 20c − 25 = 21c2 − 60c + 39 Copyright © Houghton Mifflin Company. All rights reserved.
14
Chapter P: Preliminary Concepts
51.
52.
r 2 − rs + s 2 r+s
r 2 + rs + s 2 r−s
+ r 2 s − rs 2 + s3
− r 2 s − rs 2 − s3
r 3 − r 2 s + rs 2 r3
53.
r 3 + r 2 s + rs 2 + s3
r3
(3c – 2)(4c + 1)(5c – 2) = (12c2 – 5c – 2)(5c – 2)
54.
− s3
(4d – 5)(2d – 1)(3d – 4) = (8d2 – 14d + 5)(3d – 4)
12c 2 − 5c − 2
8d 2 − 14d + 5
5c − 2
3d − 4
2
2
− 24c + 10c + 4 3
− 32d + 56d − 20
2
3
24d − 42d 2 + 15d
60c − 25c − 10c 60c3 − 49c 2
24d 3 − 74d 2 + 71d − 20
+4
55.
(3x + 5)(3x – 5) = 9x2 – 25
56.
(4x2 – 3y)(4x2 + 3y) = 16x4 – 9y2
57.
(3x2 – y) 2 = 9x4 – 6x2y + y2
58.
(6x + 7y)2 = 36x2 + 84xy + 49y2
59.
(4w + z) 2 = 16w2 + 8wz + z2
60.
(3 x − 5 y 2 )2 = 9 x 2 − 30 xy 2 + 25 y 4
61.
[(x + 5) + y][(x + 5) – y] = (x + 5)2 – y2 = x2 + 10x + 25 – y2
62.
[(x – 2y) + 7][(x – 2y) – 7] = (x – 2y)2 – 49 = x2 – 4xy + 4y2 – 49
63.
x2 + 7x – 1 = 32 + 7(3) – 1 = 9 + 21 – 1 = 29
64.
x2 – 8x + 2 = 42 – 8(4) + 2 = 16 − 32 + 2 = −14
65.
−x2 + 5x – 3 = −(−2)2 + 5(−2) – 3 = −4 − 10 – 3 = −17
66.
−x2 – 5x + 4 = −(−5)2 – 5(−5) + 4 = −25 + 25 + 4 = 4
67.
3x3 − 2x2 – x + 3 = 3(−1) 3 − 2(−1)2 – (−1) + 3 = 3(−1) – 2(1) + 1 + 3 = −3 – 2 +1 + 3 = −1
68.
5x3 − x2 + 5x − 3 = 5(−1)3 − (−1)2 + 5 (−1) − 3 = 5(−1) – (1) – 5 − 3 = −5 – 1 – 5 – 3 = −14
69.
1 – x5 = 1 – (−2)5 = 1 – (−32) = 1 + 32 = 33
71. a. b.
73. a.
b.
70.
Substitute the given value of v into 0.016v2. Then simplify 0.016 v2 0.016(10)2 = 1.6 The air resistance is 1.6 pounds. 0.016 v2 0.016(15)2 = 3.6 The air resistance is 3.6 pounds.
72.
Substitute the given value of h and r into π r2h. Then simplify π r2h π (3)2 (8) = 72π The volume is 72π in3 . π r2h π (5)2 (12) = 300π The volume 300π cm3.
74.
1 – x3 – x5 = 1 – 23 − 25 = 1 – 8 – 32 = −39
a. b.
a.
b.
Substitute the given value of v into 0.015v2 + v + 10. Then simplify. 0.015v2 + v + 10 0.015(30)2 + 30 + 10 = 53.5 The safe distance is 53.5 feet. 0.015v2 + v + 10 0.015(55)2 + 55 + 10 = 110.375 The safe distance is 110.375 feet. Substitute the given value of v into −0.02v2 + 1.5v + 2. Then simplify. 0.02v2 + 1.5v + 2 −0.02 (45)2 + 1.5(45) + 2 = 29 The fuel efficiency is 29 miles per gallon. 0.02v2 + 1.5v + 2 −0.02 (60)2 + 1.5(60) + 2 = 20 The fuel efficiency is 20 miles per gallon.
Copyright © Houghton Mifflin Company. All rights reserved.
Section P.3
15
75.
Substitute the given value of v into 0.005x2 – 0.32x + 12. a. 0.005x2 – 0.32x + 12 0.005(20)2 – 0.32(20) + 12 = 7.6 The reaction time is 7.6 hundredths of a second or 0.076 seconds. b. 0.005x2 – 0.32x + 12 0.005(50)2 – 0.32(50) + 12 = 8.5 The reaction time is 8.5 hundredths of a second or 0.085 seconds.
76.
1 3 1 2 1 1 1 1 n − n + n = (21)3− (21)2 + (21) = 1330 committees 3 6 3 6 2 2
77.
1 2 1 1 1 n − n= (150)2 − (150) = 11,175 chess matches 2 2 2 2
78.
a. b. c.
4.3 × 10−6 (1000)2 − 2.1 × 10−4 (1000) = 4.09 sec 4.3 × 10−6 (5000)2 − 2.1 × 10−4 (5000) = 106.45 sec 4.3 × 10−6 (10,000)2 − 2.1 × 10−4 (10,000) = 427.9 sec
79.
a. b.
1.9 × 10−6 (4000)2 − 3.9 × 10−3 (4000) = 14.8 sec 1.9 × 10−6 (8000)2 − 3.9 × 10−3 (8000) = 90.4 sec
80.
a.
velocity = 6r 2 − 10r 3
81.
Evaluate −16t 2 + 4.7881t + 6 when t = 0.5 height = −16t 2 + 4.7881t + 6
= −10r 3 + 6r 2 m/s
= −16(0.5)2 + 4.7881(0.5)t + 6 = 4.39 Yes. The ball is approximately 4.4 feet high when it crosses home plate.
b.
Evaluate −10r 3 + 6r 2 when r = 0.35. velocity = −10(0.35)3 + 6(0.35)2 = −10(0.042875) + 6(0.1225) = −0.42875 + 0.735 = 0.30625 The velocity of the air in a cough when the radius of the trachea is 0.35 centimeters is 0.31 m/s.
82.
a.
Evaluate 0.0002t 3 − 0.0114t 2 + 0.0158t + 104 when t = 0
b.
Evaluate 0.0002t 3 − 0.0114t 2 + 0.0158t + 104 when t = 25
Temp = 0.0002t 3 − 0.0114t 2 + 0.0158t + 104
Temp = 0.0002t 3 − 0.0114t 2 + 0.0158t + 104
= 0.0002(0)3 − 0.0114(0)2 + 0.0158(0) + 104
= 0.0002(25)3 − 0.0114(25)2 + 0.0158(25) + 104
= 104 The patient’s temperature was 104°F before taking the medication.
83.
= 100.395 The patient’s temperature was 100.4°F 25 minutes after taking the medication.
211 ⋅ 36 ⋅ 53 ⋅ 72 ⋅ 11 ⋅ 13 = 2 ⋅ 3 ⋅ (22 ) ⋅ 5 ⋅ (2 ⋅ 3) ⋅ 7 ⋅ (23 ) ⋅ (32 ) ⋅ (2 ⋅ 5) ⋅ 11 ⋅ (22 ⋅ 3) ⋅ 13 ⋅ (2 ⋅ 7) ⋅ (3 ⋅ 5) = 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 ⋅ 11 ⋅ 12 ⋅ 13 ⋅ 14 ⋅ 15 = 15! n = 15
....................................................... 3
3
2
3
3
2
2
3
3
2
3
(a − b) = a − 3a b + 3ab − b
3
2
(a + b) = a + 3a b + 3ab + b
87.
(y + 2) = y + 3y (2) + 3y(2) + 2 = y + 6y + 12y + 8
3
2
85.
84.
2
3
Connecting Concepts 86.
3
3
2
(x – 1) = x – 3x + 3x − 1
Copyright © Houghton Mifflin Company. All rights reserved.
16
Chapter P: Preliminary Concepts 3
3
2
3
3
2
2
3
3
2
2
88.
(2x – 3y) = (2x) + 3(2x) (−3y) + 3(2x)(−3y) + (−3y) = 8x – 36x y + 54xy – 27y
89.
(3x + 5y) = (3x) + 3(3x) 5y + 3(3x)(5y) + (5y) = 27x + 135x y + 225xy + 125y
2
3
3
2
2
3
3
.......................................................
PS2. ( −12 x 4 )3x 2 = ( −12)(3) x 4 + 2 = −36 x 6
PS1. 6 x 3 = 3x 3−1 = 3x 2 2x
x 6 = ( x 2 )3
PS3. a.
Prepare for Section P.4
b.
x 6 = ( x 3 )2
PS5. −3(5a − ?) = −15a + 21 = −3(5a − 7) Thus, ? = 7
PS4. 6a 3b4 ⋅ ? = 18a 3b7 = 6a 3b4 (3b3 )
Thus, ? = 3b3 PS6. 2 x (3x − ?) = 6 x 2 − 2 x = 2 x (3x − 1) Thus, ? = 1
Section P.4 2
2
1.
5x + 20 = 5(x + 4)
3.
−15x − 12x = −3x (5x + 4)
5.
l0x y + 6xy − 14xy = 2xy(5x + 3 − 7y)
7.
(x − 3)(a + b) + (x − 3)(a + 2b) = (x − 3)(a + b + a + 2b) = (x − 3)(2a + 3b)
8.
(x − 4)(2a − b) + (x + 4)(2a − b) = (2a − b)(x − 4 + x + 4) = (2a − b)(2x)
9.
x + 7x + 12 = (x + 3)(x + 4)
11.
a – 10a – 24 = (a – 12)(a + 2)
13.
6x + 25x + 4 = (6x + 1)(x + 4)
15.
51x – 5x − 4 = (17x + 4)(3x – 1)
17.
6x + xy – 40y = (3x + 8y)(2x – 5y)
19.
x + 6x + 5 = (x + 5)(x + 1)
21.
6x + 23x + 15 = (6x + 5)(x + 3)
23.
b – 4ac = 26 − 4(8)(15) = 196 = 14 The trinomial is factorable over the integers
25.
b – 4ac = (−5) – 4(4)(6) = −71 The trinomial is not factorable over the integers.
27.
b – 4ac = (−14) – 4(6)(5) = 76 The trinomial is not factorable over the integers.
2
2
2
2
2
2
2
2
4
2
4
2
2
2
2
2
2
2
2
2
2
2
2
2
2.
8x + 12x − 40 = 4(2x + 3x − 10)
4.
−6y – 54y = −6y (y + 9)
6.
6a b – 12a b + 72ab = 6ab(a b − 2a + 12b )
2
3 2
2
3
2
2
2
10.
x + 9x + 20 = (x + 4)(x + 5)
12.
b + 12b – 28 = (b + 14)(b – 2)
14.
8a – 26a + 15 = (4a – 3)(2a – 5)
16.
57y + y − 6 = (19y – 6)(3y + 1)
18.
8x + 10xy – 25y = (4x – 5y)(2x + 5y)
20.
x + 11x + 18 = (x + 9)(x + 2)
22.
9x + 10x + 1 = (9x + 1)(x + 1)
24.
b – 4ac = 8 – 4(16)(−35) = 2304 = 48 The trinomial is factorable over the integers.
26.
b – 4ac = 8 – 4(6)(− 3) = 136 The trinomial is not factorable over the integers.
28.
b – 4ac = (–4) – 4(10)(–5) = 216 The trinomial is not factorable over the integers.
2
2
2
2
4
2
2
4
2
2
2
2
2
2
2
2
2
2
2
2
Copyright © Houghton Mifflin Company. All rights reserved.
Section P.4
17
29.
x4 – x2 – 6 = (x2 – 3)(x2 + 2)
30.
x4 + 3x2 + 2 = (x2 + 1)(x2 + 2)
31.
x2y2 – 2xy – 8 = (xy – 4)(xy + 2)
32.
2x2y2 + xy – 1 = (2xy – 1)(xy + 1)
33.
3x4 + 11x2 – 4 = (3x2 – 1)(x2 + 4)
34.
2x4 + 3x2 – 9 = (2x2 – 3)(x2 + 3)
35.
3x6 + 2x3 – 8 = (3x3 – 4)(x3 + 2)
36.
8x6 – 10x3 – 3 = (4x3 + 1)(2x3 – 3)
37.
x – 9 = (x –3)(x + 3)
38.
x – 64 = (x – 8)(x + 8)
39.
4a – 49 = (2a – 7)(2a + 7)
40.
81b2 – 16c2 = (9b – 4c)(9b + 4c)
41.
1 – 100x2 = (1 − 10x)(1 + 10x)
42.
1 − 121y2 = (1 − 11y)(1 + 11y)
43.
x − 9 = (x2 − 3)(x2 + 3)
44.
y − 196 = (y2 – 14)(y2 + 14)
45.
(x + 5) − 4 = (x + 5 − 2)(x + 5 + 2) = (x + 3)(x + 7)
46.
(x − 3) − 16 = (x − 3 − 4)(x − 3 + 4) = (x − 7)(x + 1)
47.
x + 10x + 25 = (x + 5)
48.
y + 6y + 9 = (y + 3)
49.
a − 14a + 49-= (a − 7)
50.
b − 24b + 144 = (b − 12)
51.
4x + 12x + 9 = (2x + 3)
52.
25y + 40y + 16 = (5y + 4)
53.
z + 4 z w + 4w = (z + 2w )
54.
9x − 30x y + 25y
55.
x − 8 = (x – 2 )(x + 2x + 4)
56.
b + 64 = (b + 4)(b − 4b + 16)
57.
8x − 27y = (2x – 3y)(4x + 6xy + 9y )
58.
64u −27v = (4u – 3v)(16u + 12uv + 9v )
59.
8 − x = (2 − x )(4 + 2x + x )
60.
1+y
61.
(x − 2) − 1 = [(x − 2) − 1] [(x − 2) + (x − 2) + 1] = (x − 3)(x − 4x + 4 + x − 2 + 1) = (x − 3)(x − 3x + 3)
62.
(y + 3) + 8 = ((y + 3) + 2)((y + 3) − 2(y + 3) + 4) = (y + 5)(y + 6y + 9 – 2y − 6 + 4) = (y + 5)(y + 4y + 7)
63.
3x + x + 6x + 2 = x (3x + 1) + 2(3x + 1) = (3x + 1)(x + 2)
64.
18w + 15w + 12w + 10 = 3w (6w + 5) + 2(6w + 5) = (6w + 5)(3w + 2)
65.
ax – ax + bx – b = ax(x – 1) + b(x – 1) = (x – 1)(ax + b)
66.
a y – ay + ac – cy = ay (a – y) + c(a – y) = (a – y)(ay + c)
67.
6w + 4w – 15w – 10 = 2w (3w + 2) – 5(3w + 2) = (3w + 2)(2w – 5)
68.
10z − l5z – 4z + 6 = 5z (2z – 3) – 2(2z – 3) = (2z – 3)(5z – 2)
69.
18x – 2 = 2(9x – 1) = 2(3x – 1)(3x + 1)
70.
4bx + 32b = 4b(x + 8) = 4b(x + 2)(x – 2x + 4)
71.
16x – 1 = (4x – 1)(4x + 1) = (2x – 1)(2x + 1)(4x + 1)
2
2
4
2
2
2
2
2
2
2
4
2 2
4
3
2
22
2
3
3
2
6
3
2
2
2
4
3
2
2
3
2
2
2
2
3
2
4
2
2
2
2
2
2
2
4
2 2
3
4
2
2
3
3
12
2
4
4
2
2
2
2
2
2 2
3
3
2
3
2
2
2
2
2
2
2
3 4
2
2
3
2
2
2
8
= (1 + y )(1 – y + y )
2
2
22
= (3x − 5y )
2
Copyright © Houghton Mifflin Company. All rights reserved.
2
18
Chapter P: Preliminary Concepts 4
2
2
2
72.
81y – 16 = (9y – 4)(9y + 4) = (3y – 2)(3y + 2)(9y + 4)
73.
12ax – 23axy + 10ay = a(12x – 23xy + 10y ) = a(3x – 2y)(4x – 5y)
74.
6ax – 19axy – 20ay
75.
3bx + 4bx – 36x – 4b = bx (3x + 4) – b(3x + 4) = (3x + 4)(bx – b) = b(3x + 4)(x – 1) = b(3x + 4)(x – 1)(x + 1)
76.
2 x6 − 2 = 2( x6 − 1) = 2(x – 1)(x + 1) = 2(x – 1)(x + x + 1)(x + 1)(x – x + 1)
77.
72bx + 24bxy + 2by = 2b(36x + 12xy + y ) = 2b(6x + y)
78.
64y – 16y z + yz = y(64y – 16yz + z ) = y(8y – z)
79.
(w – 5) + 8 = [(w – 5) + 2][(w – 5)2 – 2(w – 5) + 4] = (w – 3)(w – 10w + 25 – 2w + 10 + 4) = (w−3)(w – 12w + 39)
80.
5xy + 20y – 15x – 60 = 5(xy + 4y – 3x – 12) = 5[y(x + 4) – 3(x + 4)] = 5(x + 4)(y − 3)
81.
x + 6xy + 9y – 1 = (x + 3y) – 1 = (x + 3y – 1)(x + 3y + 1)
82.
4y – 4yz + z – 9 = (2y – z) – 9 = (2y – z – 3)(2y – z + 3)
83.
8x + 3x – 4 is not factorable over the integers.
85.
5x(2x – 5) − (2x – 5) = (2x – 5) [5x − (2x – 5)] = (2x − 5) (5x − 2x + 5) = (2x − 5) (3x + 5)
86.
6x(3x + 1) − (3x + l) = (3x + 1) [6x − (3x + 1)] = (3x + 1) (6x – 3x – 1) = (3x + 1) (3x – 1)
87.
4x2 + 2x – y – y = 4x – y + 2x – y = (2x – y)(2x + y) + (2x – y) = (2x – y)(2x + y + l)
88.
a +a+b−b
2
2
2 3
2
2
2
2
= a(6x – 19xy – 20y ) = a(6x + 5y)(x – 4y)
2
2
2
3
2
3
2
3
2
2
2
2
2
2
2
2
2
2
2
2
3
2
2
2
2
2
2
2
2
2
84.
2
16x + 81 is not factorable over the integers.
2
3
2
2
2
3
4
3
3
3
2
2
2
2
2
2
2
= a − b + a + b = (a − b)(a + b) + (a + b) = (a + b)(a – b + 1)
....................................................... 2
2
2
89.
x + kx + 16 = (x + 4) = x + 8x + 16, thus k = 8
90.
36x + kx + 100y2 = (6x + 10y)
91.
x + 16x + k = (x +
92.
x – 14xy + ky = (x −
93.
x
94.
x
2
2 2
2
− 1 = (x
4n
− 2x
2n
2n
– 1)(x
+ 1= (x
2
2
= 36x + 120x + 100y2, thus k = 120
2
k ) = x + 2x k + k ⇒ 16x = 2x k ⇒ 8 = k ⇒ k = 64
2
4n
Connecting Concepts
2n
2n
k y)
2
2
= x – 2xy k + ky
2
⇒ −14xy = −2xy k ⇒ 7 =
k ⇒
n n 2n + 1) = (x − 1)(x + 1)(x + 1)
– 1)(x
2n
n
n
n
n
n
2 n
2
– 1) = (x – 1)(x + 1)(x – 1)(x + 1) = (x – 1) (x + 1)
Copyright © Houghton Mifflin Company. All rights reserved.
k = 49
Section P.5
19 2
2
2
2
95.
A = π R − π r = π (R − r ) = π (R – r)(R + r)
97.
A = (2r) − π r = r (4 − π )
2
2
2
2
2
2
2
96.
A = π r + (2r) = π r + 4r = r2 ( π + 4)
98.
A = x − y = (x – y)(x + y)
2
2
....................................................... PS1. 1 +
Prepare for Section P.5 −1 PS2. ⎛ w ⎞ ⎛ y ⎞ ⎜ ⎟ ⎜ ⎟ ⎝x⎠ ⎝z⎠
1 = 1+ 1 ⋅⎛ 3⎞ ⎜ ⎟ 2−1 2 − 1 ⎝ 3⎠ 3 3 1⋅ 3 =1+ 2⋅3− 1 ⋅3 3 =1+ 3 = 1+ 3 6 −1 5 = 1 3 or 8 5 5
−1
⎛ = ⎛⎜ x ⎞⎟ ⎜ ⎝ w ⎠⎝
z⎞ ⎟ y⎠
= xz wy
PS3. x2 + 2x – 3 = (x + 3)(x – 1) x2 + 7x + 12 = (x + 4)(x + 3) The common factor is x + 3.
PS4. (2 x − 3)(3x + 2) − (2 x − 3)( x + 2) = (2 x − 3)[(3x + 2) − ( x + 2)] = (2 x − 3)(2 x ) = 2 x (2 x − 3)
PS5. x2 – 5x – 6 = (x – 6)(x + 1)
PS6. x3 – 64 = (x – 4)(x2 + 4x + 16)
Section P.5 1.
x 2 − x − 20 ( x + 4)( x − 5) x + 4 = = 3 x − 15 3( x − 5) 3 x3 − 9 x
3. 3
2
x + x − 6x
4.
x3 + 125
2 x3 − 50 x 5.
a3 + 8 a2 − 4
6.
7.
=
=
=
x ( x 2 + x − 6)
2 x( x 2 − 25)
2 x 2 − 5 x − 12 2
2x + 5x + 3
=
(2 x + 3)( x − 4) x − 4 = (2 x + 3)( x + 1) x + 1
x ( x − 3)( x + 3) x − 3 = x( x + 3)( x − 2) x − 2
=
( x + 5)( x 2 − 5 x + 25) x 2 − 5 x + 25 = 2 x( x − 5)( x + 5) 2 x( x − 5)
( a + 2)( a 2 − 2a + 4) a 2 − 2a + 4 = a−2 ( a − 2)( a + 2)
y 3 − 27 y2 + 3y + 9 ( y − 3)( y 2 + 3 y + 9) ( y − 3)( y 2 + 3 y + 9) = = =− − ( y − 8)( y − 3) y −8 − y 2 + 11 y − 24 − ( y 2 − 11 y + 24) x 2 + 3x − 40
=
( x − 5)( x + 8) x+8 =− − ( x − 5)( x + 2) x+2
2 x3 − 6 x 2 + 5 x − 15 9 − x2
9.
=
( x + 5)( x 2 − 5 x + 25)
− ( x 2 − 3 x − 10) 8.
x ( x 2 − 9)
2.
4 y 3 − 8 y 2 + 7 y − 14 − y 2 − 5 y + 14
=
=
2 x 2 ( x − 3) + 5( x − 3) − ( x 2 − 9)
=
4 y 2 ( y − 2) + 7( y − 2) − ( y 2 + 5 y − 14)
( x − 3)(2 x 2 + 5) 2 x2 + 5 =− − ( x − 3)( x + 3) x+3
=
( y − 2)(4 y 2 + 7) 4 y2 + 7 =− − ( y + 7)( y − 2) y+7
Copyright © Houghton Mifflin Company. All rights reserved.
20
10.
Chapter P: Preliminary Concepts
x3 − x 2 + x 3
x +1
x( x 2 − x + 1)
=
2
( x + 1)( x − x + 1)
=
x x +1
12.
4 3 4 ⎛ 12 x 2 y ⎞⎛ 25 x 2 z 3 ⎞ ⎟ = − 12 ⋅ 25 x yz = − 4 x ⎟⎜ − ⎜ ⎜ 5 z 4 ⎟⎜ 15 y 2 ⎟ yz 5 ⋅ 15 y 2 z 4 ⎠ ⎠⎝ ⎝
14.
⎛ 4r 2 s ⎞ ⎜ ⎟ ⎜ 3t 3 ⎟ ⎝ ⎠
15.
x 2 + x 3 x 2 + 19 x + 28 x( x + 1) (3x + 7)( x + 4) x(3x + 7) = ⋅ ⋅ = 2x + 3 2 x + 3 ( x + 4)( x + 1) 2x + 3 x2 + 5x + 4
−1
2
x + 7 x + 12 17.
18.
x 2 − 4 x − 21
⋅
2
x − 4x
y3 − 8
⋅
y2 + 3y 3
2
12 y 2 + 28 y + 15 2
6 y + 35 y + 25 z 2 − 81 z 2 − 16 21.
a2 + 9 2
a − 64 22.
⎛ 6 p2 ⎞ ⎜ ⎟ ⎜ 5q 2 ⎟ ⎝ ⎠
−1
2
2 2 ⎛ 2p ⎞ ⎜ ⎟ = 5q ⋅ 4 p = 10 ⎜ 3q 2 ⎟ 6 p 2 9q 4 27 q 2 ⎝ ⎠
=
( x − 4)( x + 4) ( x + 3)( x − 7) x − 7 ⋅ = x( x − 4) x ( x + 3)( x + 4)
⋅
y + y − 6 y + 2y + 4y
20.
13.
2 x 2 + 16 x + 30 3( x − 15) 2( x 2 + 8 x + 15) 3( x − 15) 2( x + 3)( x + 5) x+3 ⋅ = = ⋅ = 2 2 + + − + + 6 x 9 3 ( 2 x 3 ) 2 ( x 5 )( x 5 ) 3 ( 2 x 3 ) 2 x+3 2 x − 50 2( x − 25) 3x − 15
2
19.
⎛ 4a ⎞⎛ 6b ⎞ 24ab 8 ⎟⎜ ⎟ = − ⎜⎜ − =− 2 ⎟⎜ 4 ⎟ 4 2 3a b a3b ⎝ 3b ⎠⎝ a ⎠
3 3 2 ⎛ 6rs3 ⎞ ⎜ ⎟ = 3t ⋅ 6rs = 9s t ⎜ 5t 2 ⎟ 4r 2 s 5t 2 10r ⎝ ⎠
x 2 − 16
16.
11.
÷
÷
÷
2
3 y + 11 y − 20
z 2 + 5 z − 36
=
2
a + 5a − 24
4x − 9 y
÷
=
=
a2 + 9 ( a − 3)( a + 8) a2 + 9 (a − 3)(a + 8) 1 ⋅ = ⋅ = 2 (a − 8)(a + 8) a (a − 3) + 9(a − 3) (a − 8)(a + 8) (a − 3)(a 2 + 9) a − 8
3x 2 − xy − 2 y 2 2
(6 y + 5)(2 y + 3) (3 y − 4)( y + 5) (2 y + 3)(3 y − 4) ⋅ = (6 y + 5)( y + 5) (2 y − 3)( y + 1) (2 y − 3)( y + 1)
( z − 9)( z + 9) ( z + 9)( z − 4) ( z − 9)( z + 9)( z + 9) ( z − 9)( z + 9)2 ⋅ = = ( z − 4)( z + 4) ( z − 5)( z + 4) ( z + 4)( z − 5)( z + 4) ( z + 4) 2 ( z − 5)
a3 − 3a 2 + 9a − 27
2
y ( y + 3) ( y − 2)( y 2 + 2 y + 4) =1 ⋅ ( y − 2)( y + 3) y ( y 2 + 2 y + 4)
2 y2 − y − 3
z 2 − z − 20
6 x 2 + 13xy + 6 y 2 2
=
2 x + xy − 3 y
2
=
(3 x + 2 y )(2 x + 3 y ) (2 x + 3 y )( x − y ) 2 x + 3 y ⋅ = (2 x − 3 y )(2 x + 3 y ) (3 x + 2 y )( x − y ) 2 x − 3 y 2 s + 5t −2s + 3t 2 s + 5t − 2 s + 3t 8t + = = =2 4t 4t 4t 4t
23.
p + 5 2 p − 7 p + 5 + 2p − 7 3p − 2 + = = r r r r
25.
7x 8 x 2 − 32 x 8 x( x − 4) x x( x + 3) + 7 x( x − 5) x 2 + 3x + 7 x 2 − 35 x + = = = = x−5 x+3 ( x − 5)( x + 3) ( x − 5)( x + 3) ( x − 5)( x + 3) ( x − 5)( x + 3)
26.
2x 5x 2 x( x − 7) + 5 x(3 x + 1) 2 x 2 − 14 x + 15 x 2 + 5 x 17 x 2 − 9 x x(17 x − 9) + = = = = 3x + 1 x − 7 (3 x + 1)( x − 7) (3x + 1)( x − 7) (3 x + 1)( x − 7) (3x + 1)( x − 7)
27.
5 y − 7 2 y − 3 (5 y − 7) − (2 y − 3) 5 y − 7 − 2 y + 3 3 y − 4 − = = = y+4 y+4 y+4 y+4 y+4
24.
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Section P.5
21
28.
6 x − 5 3 x − 8 (6 x − 5) − (3 x − 8) 6 x − 5 − 3x + 8 3x + 3 3( x + 1) = = = = − x −3 x −3 x−3 x−3 x−3 x−3
29.
4z 5z 4 z ( z − 5) + 5 z (2 z − 3) 4 z 2 − 20 z + 10 z 2 − 15 z 14 z 2 − 35 z 7 z (2 z − 5) + = = = = 2z − 3 z − 5 (2 z − 3)( z − 5) (2 z − 3)( z − 5) (2 z − 3)( z − 5) ( 2 z − 3)( z − 5)
30.
3 y − 1 2 y − 5 (3 y − 1)( y − 3) − (2 y − 5)(3 y + 1) (3 y 2 − 10 y + 3) − (6 y 2 − 13 y − 5) − = = 3y + 1 y − 3 (3 y + 1)( y − 3) (3 y + 1)( y − 3) =
31.
3 y 2 − 10 y + 3 − 6 y 2 + 13 y + 5 −3 y 2 + 3 y + 8 = (3 y + 1)( y − 3) (3 y + 1)( y − 3)
x ( x + 4) − (3x − 1)( x − 3) x − x 3x − 1 3x − 1 = − = ( x − 3)( x + 3)( x + 4) x 2 − 9 x 2 + 7 x + 12 ( x − 3)( x + 3) ( x + 3)( x + 4)
=
32.
2 ( x 2 + 4 x ) − (3x 2 − 10 x + 3) x 2 + 4 x − 3x 2 + 10 x − 3 = = −2 x + 14 x − 3 ( x − 3)( x + 3)( x + 4) ( x − 3)( x + 3)( x + 4) ( x − 3)( x + 3)( x + 4)
( m − n )( m − n ) + (3m − 5n )( m − 3n ) 3m − 5n m−n m−n + 2 3m − 5n 2 = + = 2 m n m n m n m n ( 2 )( 3 ) ( )( 2 ) ( m + 2n )( m − 3n )( m − n ) + − − + m − mn − 6n m + mn − 2n 2
2 2 2 2 2 2 = m − 2mn + n + 3m − 14mn + 15n = 4m − 16mn + 16n ( m + 2n )( m − 3n )( m − n ) ( m + 2n )( m − 3n )( m − n )
=
33.
4( m 2 − 4mn + 4n 2 ) 4( m − 2n ) 2 = ( m + 2n )( m − 3n )( m − n ) ( m + 2n )( m − 3n )( m − n )
1 + 2 ⋅ 3x 2 + 11x − 4 = 1 + 2 ⋅ (3x − 1)( x + 4) = 1 + 2( x + 4) = 1( x − 5) + x[2( x + 4)] x 3x − 1 x −5 x 3x − 1 x x−5 x ( x − 5) ( x − 5) 2 2 (2 x − 1)( x + 5) = x − 5 + 2 x + 8x = 2 x + 9 x − 5 = x ( x − 5) x ( x − 5) x ( x − 5)
34.
2 2 − 3 ⋅ y − 1 = 2 − 3 ⋅ ( y − 1)( y + 1) = 2 − 3( y − 1) = 2( y + 4) − 3( y − 1) y = 2( y + 4) − y[3( y − 1)] y y +1 y + 4 y y +1 y+4 y y+4 y ( y + 4) y ( y + 4)
=
35.
2 y + 8 − 3 y 2 + 3 y −3 y 2 + 5 y + 8 3y2 − 5 y − 8 (3 y − 8)( y + 1) = =− =− y ( y + 4) y ( y + 4) y ( y + 4) y ( y + 4)
q +1 2q q + 5 q +1 2q q − 3 q + 1 2q ( q + 1)( q + 5) − 2q (q − 3) ÷ − = − ⋅ = − = (q − 3)(q + 5) q −3 q −3 q −3 q −3 q −3 q+5 q−3 q+5 =
36.
q 2 +6q + 5 − 2q 2 + 6q − q 2 +12q + 5 = (q − 3)(q + 5) (q − 3)(q + 5)
p p p+2 p p ( p + 3)( p − 4) p p( p + 3) p ( p + 2) + p ( p + 3)( p + 5) + = + ⋅ = + = ÷ p + 5 p − 4 p 2 − p − 12 p + 5 p − 4 p+2 p+5 p+2 ( p + 5)( p + 2)
=
p 2 + 2 p + p( p 2 + 8 p + 15) p 2 + 2 p + p3 + 8 p 2 + 15 p = ( p + 5)( p + 2) ( p + 5)( p + 2)
=
p3 + 9 p 2 + 17 p p ( p 2 + 9 p + 17) = ( p + 5)( p + 2) ( p + 5)( p + 2)
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22
37.
Chapter P: Preliminary Concepts
1 1 1 1 + 1 + 1 = + + x 2 + 7 x + 12 x 2 − 9 x 2 − 16 ( x + 3)( x + 4) ( x − 3)( x + 3) ( x − 4)( x + 4) 1( x − 3)( x − 4) + 1( x − 4)( x + 4) + 1( x − 3)( x + 3) = ( x + 3)( x + 4)( x − 3)( x − 4) 2 2 2 3x 2 − 7 x − 13 = x − 7 x + 12 + x − 16 + x − 9 = ( x + 3)( x + 4)( x − 3)( x − 4) ( x + 3)( x + 4)( x − 3)( x − 4)
38.
2 5 2 3 5 + 3 − = + − a 2 − 3a + 2 a 2 − 1 a 2 + 3a − 10 ( a − 1)( a − 2) ( a − 1)( a + 1) ( a − 2)( a + 5) 2( a + 1)( a + 5) + 3( a − 2)( a + 5) − 5( a − 1)( a + 1) = ( a − 1)( a − 2)( a + 1)( a + 5) =
2( a 2 + 6a + 5) + 3( a 2 + 3a − 10) − 5( a 2 − 1) ( a − 1)( a − 2)( a + 1)( a + 5)
2 2 2 = 2a + 12a + 10 + 3a + 9a − 30 − 5a + 5 ( a − 1)( a − 2)( a + 1)( a + 5)
=
3(7a − 5) 21a − 15 = ( a − 1)( a − 2)( a + 1)( a + 5) ( a − 1)( a − 2)( a + 1)( a + 5)
39.
1 ⎞ ⎛ x 2 ⎞⎛ 3 x 1 ⎞ ⎛ x + 2 ⎞⎛ 3x − 1 ⎞ ( x + 2)(3 x − 1) ⎛ 2 ⎞⎛ ⎟= ⎟⎜ ⎜1 + ⎟⎜ 3 − ⎟ = ⎜ + ⎟⎜ − ⎟ = ⎜ x ⎠⎝ x ⎠ ⎝ x x ⎠⎝ x x ⎠ ⎝ x ⎠⎝ x ⎠ ⎝ x2
40.
1 ⎞⎛ 2 ⎞ ⎛ 4 z 1 ⎞⎛ 4 z 2 ⎞ ⎛ 4 z − 1 ⎞⎛ 4 z + 2 ⎞ (4 z − 1)(4 z + 2) (4 z − 1)2(2 z + 1) 2(4 z − 1)(2 z + 1) ⎛ = = + ⎟=⎜ − ⎟⎜ ⎟= ⎟⎜ ⎜ 4 − ⎟⎜ 4 + ⎟ = ⎜ z z⎠ ⎝ z z ⎠⎝ z z ⎠ ⎝ z ⎠⎝ z ⎠ ⎠⎝ ⎝ z2 z2 z2
41.
44.
45.
1 ⎛⎜ 4 + 1 ⎞⎟ x 4x + 1 x⎠ x =⎝ = 1 1⎞ x −1 ⎛ 1− ⎜1 − ⎟ x x x ⎠ ⎝ 4+
42.
2 ⎛⎜ 3 − a =⎝ 3 ⎛ 5+ ⎜5 + a ⎝ 3−
2⎞ ⎟a 3a − 2 a⎠ = 3⎞ 5a + 3 ⎟a a⎠
43.
x −2 y = y−x
2 2 2 2 2 2 3( x − 3) 2 3+ 3+ 3+ 3+ 3+ + x−3 = x−3 = x−3 = x−3 = x−3 = x−3 = x−3 x−3 1 1 1 2x + 1 4(2 x + 1) x x x 4+ 4+ 4+ 4 +1÷ 4 + 1⋅ 4+ + 1 2x 1 2x + 1 + + + +1 2 1 2 1 2 1 2 x x x x x + 2+ x x x x 3( x − 3) + 2 3x − 9 + 2 3x − 7 3x − 7 9 x + 4 3x − 7 2 x + 1 (3x − 7)(2 x + 1) − 3 x = = x−3 = x−3 = ÷ = ⋅ = 4(2 x + 1) + x 8 x + 4 + x 9 x + 4 x − 3 2 x + 1 x − 3 9 x + 4 ( x − 3)(9 x + 4) 2x + 1 2x + 1 2x + 1 3+
1 1 1 1 1 1 5( x + 2) 1 5− 5− 5− 5− 5− − x+2 = x+2 = x+2 = x+2 = x+2 = x+2 = x+2 x+2 3 3 3 3x 1( x + 3) 3x x+3 x 1+ 1+ 1+ 1+ 3÷ 1 + 3⋅ 1+ + 3 1( x ) 3 x+3 3 3 3 3 x x x x + + + +3 1+ + x x x x 5( x + 2) − 1 5 x + 10 − 1 5 x + 9 x+2 x + 2 = x + 2 = 5 x + 9 ÷ 4 x + 3 = 5 x + 9 ⋅ x + 3 = (5 x + 9)( x + 3) = = 1( x + 3) + 3x x + 3 + 3x 4 x + 3 x + 2 x+3 x + 2 4 x + 3 ( x + 2)(4 x + 3) x+3 x+3 x+3 5−
Copyright © Houghton Mifflin Company. All rights reserved.
⎛x ⎞ ⎜⎜ − 2 ⎟⎟ y y ⎝ ⎠ = x − 2y ( y − x )y y ( y − x)
Section P.5
46.
23
1( x + h )2 1 − 1( x + h )2 1 − 1( x 2 + 2 xh + h 2 ) 1 − x 2 − 2 xh − h 2 1 1 1 − − ( x + h )2 ( x + h )2 ( x + h )2 ( x + h )2 ( x + h)2 ( x + h )2 = = = = h h h h h 2 2 = 1 − x − 2 xh2 − h ( x + h)
1+
48.
r−
r 1 r+ 3
=r−
=
1−
51.
2−
54.
55.
56.
( x + h)
h
h( x + h)
r r 3r + 1 ⎞ 3 ⎞ 3r ⎛ ⎛ =r− = r − ⎜r ÷ ⎟ = r −⎜r ⋅ ⎟=r− 3r 1 3r + 1 r r 3 3 1 3 + +1 ⎝ ⎠ ⎝ ⎠ + 3 3 3
r (3r + 1) r (3r + 1) − 3r 3r 2 + r − 3r 3r 2 − 2r r (3r − 2) 3r − = = = = 3r + 1 3r + 1 3r + 1 3r + 1 3r + 1 3r + 1
⎛ 1 ⎞ ⎜⎜1 − ⎟⎟ x2 ⎠ ⋅ x =⎝ 1 ⎛ 1⎞ 1+ ⎜1 + ⎟ x x⎠ ⎝
49.
53.
1
⎛ 1 ⎞ 1 ⎜1 + ⎟ b − 2 = ⎝ b − 2 ⎠ ⋅ (b − 2)(b + 3) = 1(b − 2)(b + 3) + 1(b + 3) = b2 + b − 6 + b + 3 = b2 + 2b − 3 = (b + 3)(b − 1) (b − 2)(b + 2) ⎛ 1 − 1 ⎞ (b − 2)(b + 3) 1(b − 2)(b + 3) − 1( b − 2) b2 + b − 6 − b + 2 b2 − 4 1− 1 ⎜ ⎟ b + 3 ⎝ b + 3⎠
47.
52.
2 2 2 2 ÷ h = 1 − x − 2 xh2 − h ⋅ 1 = 1 − x − 2 xh2− h
1
2
x2
x2 − 1
( x − 1)( x + 1) x − 1 = = = 2 2 x( x + 1) x x x +x
50.
1 1 1 b+a ab ab = = = 1÷ = 1⋅ = 1 1 1b 1a b+a ab b+a b+a + + a b ab ab ab
(m) m m m m m =2− =2− =2− =2− m =2− ⋅ = 2 − m2 1 1 1( ) 1 1 m m m m m − − + − 1 m − ⎛ 1⎞ m 1− 1+ + ⎜ ⎟ m m m −m m m ⎝m⎠
⎛ x + h +1 − x ⎞ ⎜ ⎟ x + 1 ⎠ ⋅ ( x + h )( x + 1 = ( x + h + 1)( x + 1) − x ( x + h ) = x 2 + x + xh + h + x + 1 − x 2 − xh = 2 x + h + 1 ⎝ x+h (h) ( x + h )( x + 1) h ( x + h )( x + 1) h( x + h )( x + 1) h( x + h )( x + 1) ⎛1 x−4⎞ ⎜ − ⎟ 2 2 ⎝ x x + 1 ⎠ ⋅ x( x + 1) = x + 1 − x( x − 4) = x + 1 − x + 4 x = − x + 5 x + 1 x x( x + 1) x( x) x2 x2 x +1
⎛ 2 3y − 2 ⎞ ⎜⎜ − ⎟ y − 1 ⎟⎠ y ( y − 1) 2( y − 1) − y (3 y − 2) 2 y − 2 − 3 y 2 + 2 y − 3 y 2 + 4 y − 2 ⎝y ⋅ = = = y ( y − 1) y( y) ⎛ y ⎞ y2 y2 ⎜⎜ ⎟⎟ ⎝ y −1 ⎠ 2 ⎞ ⎛ 1 − ⎟ ⎜ x − 1 − 2x − 6 − x−7 + − 1 ⎠ ( x + 3)( x − 1) 1( x − 1) − 2( x + 3) 3 x x ⎝ ⋅ = = = 3 ⎞ ( x + 3)( x − 1) x( x + 3) + 3( x − 1) x 2 + 3 x + 3 x − 3 x 2 + 6 x − 3 ⎛ x + ⎟ ⎜ ⎝ x −1 x + 3 ⎠
x+2 x+2 + 1 + 1 2 − + x x x + 1 ( x − 1)( x + 1)(2 x + 1) ( x + 2)(2 x + 1) + 1( x − 1)(2 x + 1) ( 1)( 1) + x 1 x −1 = ⋅ = 1 1 ( x − 1)( x + 1)(2 x + 1) x x x ( x + 1) + 1( x + 1)(2 x + 1) + + 2 x 2 − x − 1 x − 1 (2 x + 1)( x − 1) x − 1 2
2 2 2 (2 x + 1) = 2 x 2+ 5 x + 2 +22 x − x − 1 = 4 x2 + 4 x + 1 = x + x + 2 x + 3x + 1 3x + 4 x + 1 (3x + 1)( x + 1)
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24
Chapter P: Preliminary Concepts
x 2 + 3 x − 10
57.
( x − 2)( x + 5) x+5 x+5 x+5 x + 5 2x − 3 2x − 3 ( x − 2)( x + 3) ÷ = = x+3 = = ⋅ = ( x + 5)( x − 6) x+5 x + 3 2x − 3 x + 3 x + 5 x+3 (2 x − 3)( x − 6) 2 x − 3
x2 + x − 6 x 2 − x − 30 2 x 2 − 15 x + 18
58.
59.
2 y 2 + 11 y + 15
(2 y + 5)( y + 3) 2y + 5 2y + 5 2y + 5 2y + 5 y − 7 y−7 ( y + 3)( y − 7) ÷ = = = = ⋅ =1 2 y y y (3 − 2)(2 + 5) 2 + 5 y−7 y−7 y − 7 2y + 5 6 y + 11 y − 10 (3 y − 2)( y − 7) y−7 3 y 2 − 23 y + 14 y 2 − 4 y − 21
1 1 1b 1a b+a + + a −1 + b −1 a b ab ab b+a = = = ab = a−b a−b a−b a−b ab
60.
e
61.
−2
−f ef
−1
1 2 = e
a −1b − ab −1 a 2 + b2
63.
(a + b )
− 2 −1
a.
ef
1 f
1f =
2
e f
− ef
1e2 2
e f
b+a 1 a+b ⋅ = ab a − b ab(a − b)
f − e2 =
e2 f ef
=
f − e2 e2 f
÷ a 2 + b2
b2 − a 2 b2 − a 2 1 (b − a)(b + a ) ⋅ = = 2 2 ab a +b ab(a 2 + b 2 ) ab(a 2 + b 2 )
⎛ 1 ⎞ ⎟⎟ = ⎜⎜ a + b2 ⎠ ⎝
−1
⎛ ab 2 1 ⎞⎟ =⎜ + ⎜ b2 b 2 ⎟⎠ ⎝
−1
⎛ ab 2 + 1 ⎞ ⎟ =⎜ ⎜ b2 ⎟ ⎝ ⎠
−1
=
2 2 290 = = 2÷ 1 1 110 + 180 180(110) + 180 110 180(110) = 2⋅
f − e2 1 f − e2 ⋅ = e2 f ef e3 f 2
÷ ef =
(b)b a (a ) b 2 a 2 b 2 − a 2 b a − − − b2 − a 2 b)a b( a) ab ab ( = a b = = = ab = ab a 2 + b2 a2 + b2 a 2 + b2 a 2 + b2 =
62.
−
÷ ( a − b) =
b2 ab 2 + 1
b.
2v v 2v v 2 2 = = 1 2 = 1 2 1 + 1 v2 + v1 v2 + v1 v1 + v2 v1 v2 v1v2
b.
(v + v ) 2 c2 (v + v ) v1 + v2 = 1 2 ⋅ c2 = 2 1 2 vv vv c + v1v2 1 + 1 22 ⎜⎛ 1 + 1 2 ⎟⎞ c c c2 ⎠ ⎝
66.
1 − 1 = x+2− x = 2 x x + 2 x ( x + 2) x ( x + 2)
(180)(110) ≈ 136.55 mph (to the nearest hundredth) 290
8 8 v1 + v2 = 1.2 × 10 8+ 2.4 × 10 8 ≈ 3.4 × 108 v1v2 (1.2 × 10 )(2.4 × 10 ) 1+ 2 1+ c (6.7 × 108 )2
64.
a.
65.
1 + 1 = x + 1 + x = 2x + 1 x x + 1 x ( x + 1) x ( x + 1)
67.
1 + 1 + 1 = x ( x + 2) + ( x − 2)( x + 2) + x ( x − 2) = x 2 + 2 x + x 2 − 4 + x 2 − 2 x = 3x 2 − 4 x−2 x x+2 x ( x − 2)( x + 2) x ( x − 2)( x + 2) x ( x − 2)( x + 2)
Copyright © Houghton Mifflin Company. All rights reserved.
Section P.5
68.
25
x 2 ( x + 2)2 + ( x − 2)2 ( x + 2) 2 + x 2 ( x − 2)2 1 1 + 1 + = ( x − 2 )2 x 2 ( x + 2 )2 x 2 ( x − 2)2 ( x + 2)2 =
x 2 ( x 2 + 4 x + 4) + ( x 2 − 4 x + 4)( x 2 + 4 x + 4) + x 2 ( x 2 − 4 x + 4) x 2 ( x − 2)2 ( x + 2)2
4 3 2 4 2 4 3 2 4 16 = x + 4 x + 4 x +2 x − 8 x2 + 16 +2 x − 4 x + 4 x = 2 3x + x ( x − 2) ( x + 2) x ( x − 2)2 ( x + 2)2
.......................................................
Connecting Concepts
69.
( x + 5) − x( x + 5) −1 x + 5 ( x + 5) 2 − x x 2 + 10 x + 25 − x x 2 + 9 x + 25 = = = ⋅ x+5 x+5 ( x + 5) 2 ( x + 5) 2 ( x + 5) 2
70.
( y + 2) + y 2 ( y + 2)−1 y + 2 ( y + 2) 2 + y 2 y 2 + 4 y + 4 + y 2 2 y 2 + 4 y + 4 2( y 2 + 2 y + 2) = = = = ⋅ y+2 y+2 ( y + 2)2 ( y + 2) 2 ( y + 2)2 ( y + 2) 2
71.
72.
73.
74.
1 − 4 xy 1 − 4y 2 x (1 − 4 xy ) x x = = ⋅ x2 = −1 −1 (1 − 2 xy )(1 + 2 xy ) ( x − 2 y )( x + 2 y ) ⎜⎛ 1 − 2 y ⎟⎞ ⎜⎛ 1 + 2 y ⎟⎞ ⎛ 1 − 2 xy ⎞ ⎛ 1 + 2 xy ⎞ x ⎝x ⎠⎝ x ⎠ ⎜⎝ x ⎟⎠ ⎜⎝ x ⎟⎠ x −1 − 4 y
x+ y ⋅ x− y
1 − x+ y x = ⋅ x −1 + y −1 x − y 1 + x x −1 − y −1
1 ⎡ ⎢1 − (1 + i ) n R⎢ ⎢ i ⎢ ⎣⎢
1 y xy x + y y − x ⋅ = ⋅ = −1 1 xy x − y y + x y
⎡ (1 + i ) n − 1 ⎤ ⎤ ⎢ ⎥ ⎥ n n ⎡ ⎤ ⎥ = R ⎢ (1 + i ) ⎥ = R ⎢ (1 + i ) − 1 ⎥ ⎢ ⎥ ⎥ i ⎢⎣ i (1 + i ) n ⎥⎦ ⎢ ⎥ ⎥ ⎢ ⎥ ⎦⎥ ⎣ ⎦
1 1 RR R R1R2 R3 ⋅ 1 2 3 = = 1 1 1 ⎛ ⎞ R R R R R R1R3 + R1R2 + 1 1 1 1 2 3 2 3 + + ⎜ ⎟ + + ⎟ R1 R2 R3 ⎜ R R R 2 3⎠ ⎝ 1
....................................................... PS1. (2 − 3x )(4 − 5 x ) = 8 − 10 x − 12 x + 15 x 2
Prepare for Section P.6 PS2. (2 − 5 x )2 = 22 + 2(2)( −5 x ) + ( −5 x ) 2
= 15 x 2 − 22 x + 8
= 4 − 20 x + 25 x 2 = 25 x 2 − 20 x + 4
PS3. PS4.
96 = 16 ⋅ 6 = 4 6
( 2 + 3 5 )( 3 − 4 5 ) = 6 − 8
5 + 9 5 − 12 ( 5 ) = 6 + 5 − 60 = −54 + 5 2
PS5. 5 + 2 = 5 + 2 ⋅ 3 + 2 = 15 + 8 2 + 2 = 17 + 8 2 9−2 7 3− 2 3− 2 3+ 2
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26
Chapter P: Preliminary Concepts
PS6. a.
81 − x 2 is a difference of perfect squares with integer coefficients, which does factor over the integers.
b.
81 − x 2 = (9 − x )(9 + x )
9 + z 2 is a sum of perfect squares with integer coefficients. If there is no common factor, sums of perfect squares with integer coefficients are not factorable over the integers.
Section P.6 1.
−81 = i 81 = 9i
2.
−64 = i 64 = 8i
3.
−98 = i 98 = 7i 2
4.
−27 = i 27 = 3i 3
5.
16 + −81 = 4 + i 81 = 4 + 9i
6.
25 + −9 = 5 + i 9 = 5 + 3i
7.
5 + −49 = 5 + i 49 = 5 + 7i
8.
6 − −1 = 6 − i 1 = 6 − i
9.
8 − −18 = 8 − i 18 = 8 − 3i 2
10.
11 + −48 = 11 + i 48 = 11 + 4i 3
11.
(5 + 2i ) + (6 − 7i ) = 5 + 2i + 6 − 7i = (5 + 6) + (2i − 7i ) = 11 − 5i
12.
(4 − 8i ) + (5 + 3i ) = 4 − 8i + 5 + 3i = (4 + 5) + ( −8i + 3i ) = 9 − 5i
13.
( −2 − 4i ) − (5 − 8i ) = −2 − 4i − 5 + 8i = ( −2 − 5) + ( −4i + 8i ) = −7 + 4i
14.
(3 − 5i ) − (8 − 2i ) = 3 − 5i − 8 + 2i = (3 − 8) + ( −5i + 2i ) = −5 − 3i
15.
(1 − 3i ) + (7 − 2i ) = 1 − 3i + 7 − 2i = (1 + 7) + ( −3i − 2i ) = 8 − 5i
16.
(2 − 6i ) + (4 − 7i ) = 2 − 6i + 4 − 7i = (2 + 4) + ( −6i − 7i ) = 6 − 13i
17.
( −3 − 5i ) − (7 − 5i ) = −3 − 5i − 7 + 5i = ( −3 − 7) + ( −5i + 5i ) = −10
18.
(5 − 3i ) − (2 + 9i ) = 5 − 3i − 2 − 9i = (5 − 2) + ( −3i − 9i ) = 3 − 12i
19.
8i − (2 − 8i ) = 8i − 2 + 8i = −2 + (8i + 8i ) = −2 + 16i
22.
( −3i )(2i ) = −6i 2 = −6( −1) =6
24.
20.
23.
5i ⋅ 8i = 40i 2 = 40( −1) = −40
−50 ⋅ −2 = i 50 ⋅ i 2 = 5i 2 ⋅ i 2
25.
3(2 + 5i ) − 2(3 − 2i ) = 6 + 15i − 6 + 4i = (6 − 6) + (15i + 4i ) = 19i
27.
(4 + 2i )(3 − 4i ) = 4(3 − 4i ) + (2i )(3 − 4i )
2
= 6i ( 3) = 6( −1)(3) = −18
26.
21.
= 5i 2 ( 2)2 = 5( −1)(2) = −10
−12 ⋅ −27 = i 12 ⋅ i 27 = 2i 3 ⋅ 3i 3 2
3 − (4 − 5i ) = 3 − 4 + 5i = (3 − 4) + 5i = −1 + 5i
3i (2 + 5i ) + 2i (3 − 4i ) = 6i + 15i 2 + 6i − 8i 2 = 6i + 15( −1) + 6i − 8( −1) = 6i − 15 + 6i + 8 = ( −15 + 8) + (6i + 6i ) = −7 + 12i
= 12 − 16i + 6i − 8i 2 = 12 − 16i + 6i − 8( −1) = 12 − 16i + 6i + 8 = (12 + 8) + ( −16i + 6i ) = 20 − 10i
Copyright © Houghton Mifflin Company. All rights reserved.
Section P.6
28.
27
(6 + 5i )(2 − 5i ) = 6(2 − 5i ) + 5i (2 − 5i )
29.
( −3 − 4i )(2 + 7i ) = −3(2 + 7i ) − 4i (2 + 7i )
2
= −6 − 21i − 8i − 28i 2 = −6 − 21i − 8i − 28( −1) = −6 − 21i − 8i + 28 = ( −6 + 28) + ( −21i − 8i ) = 22 − 29i
= 12 − 30i + 10i − 25i = 12 − 30i + 10i − 25( −1) = 12 − 30i + 10i + 25 = (12 + 25) + ( −30i + 10i ) = 37 − 20i 30.
( −5 − i )(2 + 3i ) = −5(2 + 3i ) − i (2 + 3i )
31.
(4 − 5i )(4 + 5i ) = 4(4 + 5i ) − 5i (4 + 5i )
2
= 16 + 20i − 20i − 25i 2 = 16 + 20i − 20i − 25( −1) = 16 + 20i − 20i + 25 = (16 + 25) + (20i − 20i ) = 41
= −10 − 15i − 2i − 3i = −10 − 15i − 2i − 3( −1) = −10 − 15i − 2i + 3 = ( −10 + 3) + ( −15i − 2i ) = −7 − 17i 32.
(3 + 7i )(3 − 7i ) = 3(3 − 7i ) + 7i (3 − 7i )
33.
= 9 − 21i + 21i − 49i 2 = 9 − 21i + 21i − 49( −1) = 9 − 21i + 21i + 49 = (9 + 49) + ( −21i + 21i ) = 58
34.
(3 + −4 )(2 − −9 ) = (3 + i 4 )(2 − i 9) = (3 + 2i )(2 − 3i ) = 3(2 − 3i ) + 2i (2 − 3i ) = 6 − 9i + 4i − 6i 2 = 6 − 9i + 4i − 6( −1) = 6 − 9i + 4i + 6 = (6 + 6) + ( −9i + 4i ) = 12 − 5i
(5 + 2 −16)(1 − −25) = (5 + 2i 16)(1 − i 25) = [5 + 2i (4)][1 − i (5)] = (5 + 8i )(1 − 5i ) = 5(1 − 5i ) + 8i (1 − 5i ) = 5 − 25i + 8i − 40i 2 = 5 − 25i + 8i − 40( −1) = 5 − 25i + 8i + 40 = (5 + 40) + ( −25i + 8i ) = 45 − 17i
35.
(3 + 2 −18)(2 + 2 −50) = (3 + 2i 18)(2 + 2i 50) = [3 + 2i (3 2 )][2 + 2i (5 2 )] = (3 + 6i 2 )(2 + 10i 2) = 3(2 + 10i 2 ) + 6i 2(2 + 10i 2 ) = 6 + 30i 2 + 12i 2 + 60i 2 ( 2)2 = 6 + 30i 2 + 12i 2 + 60( −1)(2) = 6 + 30i 2 + 12i 2 − 120 = (6 − 120) + (30i 2 + 12i 2 ) = −114 + 42i 2
36.
(5 − 3 −48)(2 − 4 −27 ) = (5 − 3i 48)(2 − 4i 27 ) = [5 − 3i (4 3)][2 − 4i (3 3)] = (5 − 12i 3)(2 − 12i 3) = 5(2 − 12i 3) − 12i 3(2 − 12i 3) = 10 − 60i 3 − 24i 3 + 144i 2 ( 3) 2 = 10 − 60i 3 − 24i 3 + 144( −1)(3) = 10 − 60i 3 − 24i 3 − 432 = (10 − 432) + ( −60i 3 − 24i 3) = −422 − 84i 3 4
−8 = − 8 = −4 = −4 ⋅ i = −4i = −4i = 4i −1 2i 2i i i i i2
37.
6 = 6 ⋅ i = 6i = 6i = −6i i i i i 2 −1
39.
6 + 3i = 6 + 3i ⋅ i = 6i + 3i 2 = 6i + 3( −1) = 6i − 3 = 3 − 6i i i i −1 −1 i2
40.
4 − 8i = 4(1 − 2i ) = 4 (1 − 2i ) = 1 − 2i = 1 − 2i ⋅ i = i − 2i 2 = i − 2( −1) = i + 2 = −2 − i 4i 4i 4i i i i −1 −1 i2
41.
1(7 − 2i ) 1 = 1 ⋅ 7 − 2i = = 7 − 2i = 7 − 2i = 7 − 2i = 7 − 2i = 7 − 2 i 7 + 2i 7 + 2i 7 − 2i (7 + 2i )(7 − 2i ) 49 − 4i 2 49 − 4( −1) 49 + 4 53 53 53
38.
Copyright © Houghton Mifflin Company. All rights reserved.
28
Chapter P: Preliminary Concepts
42.
5(3 − 4i ) 5 = 5 ⋅ 3 − 4i = = 15 − 20i = 15 − 20i = 15 − 20i = 15 − 20i = 15 − 20 i = 3 − 4 i 3 + 4i 3 + 4i 3 − 4i (3 + 4i )(3 − 4i ) 9 − 16i 2 9 − 16( −1) 9 + 16 25 25 25 5 5
43.
2i = 2i ⋅ 1 − i = 2i (1 − i ) = 2i − 2i 2 = 2i − 2( −1) = 2i + 2 = 2 + 2i = 2 + 2 i = 1 + i 1 + i 1 + i 1 − i (1 + i )(1 − i ) 1 − ( −1) 1+1 2 2 2 1 − i2
44.
5i = 5i ⋅ 2 + 3i = 5i (2 + 3i ) = 10i + 15i 2 = 10i + 15( −1) = 10i − 15 = −15 + 10i = − 15 + 10 i 2 − 3i 2 − 3i 2 + 3i (2 − 3i )(2 + 3i ) 4 − 9( −1) 4+9 13 13 13 4 − 9i 2
45.
5 − i = 5 − i ⋅ 4 − 5i = (5 − i )(4 − 5i ) = 5(4 − 5i ) − i (4 − 5i ) = 20 − 25i − 4i + 5i 2 4 + 5i 4 + 5i 4 − 5i (4 + 5i )(4 − 5i ) 4(4 − 5i ) + 5i (4 − 5i ) 16 − 20i + 20i − 25i 2 20 − 25i − 4i + 5( −1) 20 − 25i − 4i − 5 (20 − 5) + ( −25i − 4i ) 15 − 29i 15 29 = = = = = − i 16 − 25( −1) 16 + 25 16 + 25 41 41 41
46.
4 + i = 4 + i ⋅ 3 − 5i = (4 + i )(3 − 5i ) = 4(3 − 5i ) + i (3 − 5i ) = 12 − 20i + 3i − 5i 2 = 12 − 20i + 3i − 5( −1) 3 + 5i 3 + 5i 3 − 5i (3 + 5i )(3 − 5i ) 9 − 25( −1) 9 − 25i 2 9 − 25i 2 12 − 20i + 3i − 5( −1) 12 − 20i + 3i + 5 (12 + 5) + ( −20i + 3i ) 17 − 17i 17 17 = = = = = − i = 1 − 1i 9 − 25( −1) 9 + 25 34 34 34 34 2 2
47.
2 32 + 2(3)(2i ) + (2i )2 9 + 12i + 4i 2 9 + 12i + 4( −1) 3 + 2i = 3 + 2i ⋅ 3 + 2i = (3 + 2i ) = = = 3 − 2i 3 − 2i 3 + 2i (3 − 2i )(3 + 2i ) 9 − 4( −1) 32 − (2i )2 9 − 4i 2 = 9 + 12i − 4 = 5 + 12i = 5 + 12 i 9+4 13 13 13
48.
8 − i = 8 − i ⋅ 2 − 3i = (8 − i )(2 − 3i ) = 8(2 − 3i ) − i (2 − 3i ) = 16 − 24i − 2i + 3i 2 = 16 − 24i − 2i + 3( −1) 2 + 3i 2 + 3i 2 − 3i (2 + 3i )(2 − 3i ) 4 − 9( −1) 22 − (3i )2 4 − 9i 2 = 16 − 24i − 2i − 3 = 13 − 26i = 13 − 26i = 1 − 2i 4+9 13 13 13
49.
−7 + 26i = −7 + 26i ⋅ 4 − 3i = ( −7 + 26i )(4 − 3i ) = −7(4 − 3i ) + 26i (4 − 3i ) = −28 + 21i + 104i − 78i 2 4 + 3i 4 + 3i 4 − 3i (4 + 3i )(4 − 3i ) 42 − (3i )2 16 − 9i 2 =
50.
−4 − 39i = −4 − 39i ⋅ 5 + 2i = ( −4 − 39i )(5 + 2i ) = −4(5 + 2i ) − 39i (5 + 2i ) = −20 − 8i − 195i − 78i 2 = 5 + 2i 5 − 2i 5 − 2i 5 + 2i (5 − 2i )(5 + 2i ) 5 + 2i 52 − (2i )2 25 − 4i 2 =
51.
−28 + 21i + 104i − 78( −1) −28 + 21i + 104i + 78 50 + 125i 50 125 = = = + i = 2 + 5i 16 − 9( −1) 16 + 9 25 25 25
−20 − 8i − 195i − 78( −1) −20 − 8i − 195i + 78 58 − 203i 58 203i = = = − = 2 − 7i 25 − 4( −1) 25 + 4 29 29 29
(3 − 5i )2 = 33 + 2(3)( −5i ) + ( −5i )2
52.
(2 + 4i )2 = 22 + 2(2)(4i ) + (4i )2
= 9 − 30i + 25i 2
= 4 + 16i + 16i 2
= 9 − 30i + 25( −1)
= 4 + 16i + 16( −1)
= 9 − 30i − 25
= 4 + 16i − 16
= −16 − 30i
= −12 + 16i
Copyright © Houghton Mifflin Company. All rights reserved.
Section P.6
53.
55.
29
(1 + 2i )3 = (1 + 2i )(1 + 2i )2
54.
= (1 + 2i )[12 + 2(1)(2i ) + (2i ) 2 ]
= (2 − i )[22 + 2(2)( −i ) + ( −i )2 ]
= (1 + 2i )[1 + 4i + 4i 2 ]
= (2 − i )[4 − 4i + i 2 ]
= (1 + 2i )[1 + 4i + 4( −1)]
= (2 − i )[4 − 4i − 1]
= (1 + 2i )[1 + 4i − 4]
= (2 − i )(3 − 4i )
= (1 + 2i )( −3 + 4i )
= 2(3 − 4i ) − i (3 − 4i )
= 1( −3 + 4i ) + 2i ( −3 + 4i )
= 6 − 8i − 3i + 4i 2
= −3 + 4i − 6i + 8i 2
= 6 − 8i − 3i + 4( −1)
= −3 + 4i − 6i − 8
= 6 − 8i − 3i − 4
= −11 − 2i
= 2 − 11i
Use the Powers of i Theorem. The remainder of 15 ÷ 4 is 3.
56.
i15 = i 3 = −i 57.
(2 − i )3 = (2 − i )(2 − i )2
Use the Powers of i Theorem. The remainder of 66 ÷ 4 is 2. i 66 = i 2 = −1
Use the Powers of i Theorem. The remainder of 40 ÷4 is 0.
58.
−i 40 = −(i 0 ) = −1
Use the Powers of i Theorem. The remainder of 51 ÷4 is 3. −i 51 = −(i 3 ) = −( −i ) = i
59.
Use the Powers of i Theorem. The remainder of 25 ÷4 is 1. 1 = 1 = 1 ⋅ i = i = i = −i i 25 i i i i 2 −1
60.
Use the Powers of i Theorem. The remainder of 83 ÷4 is 3. 1 = 1 = 1 ⋅i = i = i =i i83 i 3 i 3 i i 4 1
61.
Use the Powers of i Theorem. The remainder of 34 ÷4 is 2. 1 = 1 = 1 = −1 i −34 = 34 i i 2 −1
62.
Use the Powers of i Theorem. The remainder of 52 ÷4 is 0. 1 = 1 = 1 =1 i −52 = 52 i i0 1
63.
Use a = 3, b = −3, c = 3.
64.
Use a = 2, b = 4, c = 4. 2 −b + b2 − 4ac = −(4) + (4) − 4(2)(4) 2a 2(2)
2
−b + b2 − 4ac = −( −3) + ( −3) − 4(3)(3) 2a 2(3)
= −4 + 16 − 32 = −4 + −16 4 4 = −4 + i 16 = −4 + 4i 4 4 4 4 i − = + = −1 + i 4 4
= 3 + 9 − 36 = 3 + −27 6 6 = 3 + i 27 = 3 + 3i 3 6 6 3 3 3 1 = + i = + 3i 6 6 2 2 65.
Use a = 2 b = 6, c = 6.
66. 2
−b + b2 − 4ac = −(6) + (6) − 4(2)(6) 2a 2(2) = −6 + 36 − 48 = −6 + −12 4 4 − + 6 2 3 i − + 6 12 i = = 4 4 = −6 + 2i 3 = − 3 + 3 i 4 4 2 2
Use a =2, b = 1, c = 3. 2 −b + b2 − 4ac = −(1) + (1) − 4(2)(3) 2a 2(2)
= −1 + 1 − 24 4 1 − + −23 = −1 + i 23 = 4 4 = − 1 + 23 i 4 4
Copyright © Houghton Mifflin Company. All rights reserved.
30
67.
Chapter P: Preliminary Concepts
Use a = 4, b = −4, c = 2.
68. 2
Use a = 3, b = −2, c = 4. 2 −b + b2 − 4ac = −( −2) + ( −2) − 4(3)(4) 2a 2(3)
−b + b − 4ac = −( −4) + ( −4) − 4(4)(2) 2a 2(4) 2
= 2 + 4 − 48 = 2 + −44 6 6 + + i i 2 44 2 2 11 = = 6 6 = 2 + 2i 11 = 1 + 11 i 6 6 3 3
= 4 + 16 − 32 = 4 + −16 8 8 + 4 i 16 + 4 4 i = = 8 8 i 4 4 1 1 = + = + i 8 8 2 2
.......................................................
Connecting Concepts
69.
x 2 + 16 = x 2 + 42 = ( x + 4i )( x − 4i )
70.
x 2 + 9 = x 2 + 32 = ( x + 3i )( x − 3i )
71.
z 2 + 25 = z 2 + 52 = ( z + 5i )( z − 5i )
72.
z 2 + 64 = z 2 + 82 = ( z + 8i )( z − 8i )
73.
4 x 2 + 81 = (2 x )2 + 92 = (2 x + 9i )(2 x − 9i )
74.
9 x 2 + 1 = (3x )2 + 12 = (3x + i )(3x − i )
75.
If x = 1 + 2i, then x 2 − 2 x + 5 = (1 + 2i )2 − 2(1 + 2i ) + 5 = 1 + 4i + 4i 2 − 2 − 4i + 5 = 1 + 4i + 4(−1) − 2 − 4i + 5 = 1 + 4i − 4 − 2 − 4i + 5 = (1 − 4 − 2 + 5) + (4i − 4i ) = 0
76.
If x = 1 − 2i, then x 2 − 2 x + 5 = (1 − 2i )2 − 2(1 − 2i ) + 5 = 1 − 4i + 4i 2 − 2 + 4i + 5 = 1 − 4i + 4( −1) − 2 + 4i + 5 = 1 − 4i − 4 − 2 + 4i + 5 = (1 − 4 − 2 + 5) + ( −4i + 4i ) = 0
77.
Verify that ( −1 + i 3)3 = 8 . ( −1 + i 3)3 = ( −1 + i 3)( −1 + i 3)2 = ( −1 + i 3)[( −1)2 + 2( −1)(i 3) + (i 3)2 ] = ( −1 + i 3)[1 − 2i 3 + 3i 2 ] = ( −1 + i 3)[1 − 2i 3 + 3( −1)] = ( −1 + i 3)[1 − 2i 3 − 3] = ( −1 + i 3)( −2 − 2i 3) = −1( −2 − 2i 3) + i 3( −2 − 2i 3) = 2 + 2i 3 − 2i 3 − 2i 2 ( 3)2 = 2 + 2i 3 − 2i 3 − 2( −1)(3) = 2 + 2i 3 − 2i 3 + 6 = (2 + 6) + (2i 3 − 2i 3) =8
Verify that ( −1 − i 3)3 = 8 . ( −1 − i 3)3 = ( −1 − i 3)( −1 − i 3)2 = ( −1 − i 3)[( −1)2 + 2( −1)( −i 3) + ( −i 3)2 ] = ( −1 − i 3)[1 + 2i 3 + 3i 2 ] = ( −1 − i 3)[1 + 2i 3 + 3( −1)] = ( −1 − i 3)[1 + 2i 3 − 3] = ( −1 − i 3)( −2 + 2i 3) = −1( −2 + 2i 3) − i 3( −2 + 2i 3) = 2 − 2i 3 + 2i 3 − 2i 2 ( 3)2 = 2 − 2i 3 + 2i 3 − 2( −1)(3) = 2 − 2i 3 + 2i 3 + 6 = (2 + 6) + ( −2i 3 + 2i 3) =8 78.
⎡
⎤
2
Verify that ⎢ 2 (1 + i ) ⎥ = i . ⎣ 2 ⎦ 2
2
⎡ 2 ⎤ 2 2 2 2 1 1 1 ⎢ 2 (1 + i ) ⎥ = 2 (1 + i ) = 4 (1 + 2i + i ) = 2 [1 + 2i + ( −1)] = 2 (1 + 2i − 1) = 2 (2i ) = i 2 ⎣ ⎦ 79.
i + i 2 + i 3 + i 4 + ... + i 28 = 7(i + i 2 + i 3 + i 4 ) = 7(i + ( −1) + ( −i ) + 1) = 7(0) = 0
80.
i + i 2 + i 3 + i 4 + ... + i100 = 25(i + i 2 + i 3 + i 4 ) = 25(i + ( −1) + ( −i ) + 1) = 25(0) = 0
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Review
31
.......................................................
Exploring Concepts with Technology
Can You Trust Your Calculator? 1.
Iteration 1 3 5 10 15 19 20
4p – 3p2 1.25 0.95703125 0.8198110957 0.3846309658 0.5610061236 1.218765181 0.4188950245
p + 3p(1 – p) 1.25 0.95703125 0.8198110957 0.3846309658 0.5610061236 1.218765181 0.4188950251
.......................................................
Assessing Concepts
1.
True
2.
False
3.
Any real number a between 0 and 1
4.
Any negative integer n
5.
−3x 2 + 6 x − 7
6.
b
7.
c
8.
a
9.
e
10.
c
11.
b
12.
e
.......................................................
Chapter Review
1.
Integer, rational number, real number, prime number [P.1]
2.
Irrational number, real number [P.1]
3.
Rational number, real number [P.1]
4.
Rational number, real number [P.1]
5.
A ∪ B = {1, 2, 3, 5, 7, 11} [P.1]
6.
A ∩ B = {5} [P.1]
7.
Distributive property [P.1]
8.
Commutative property of addition [P.1]
9.
Associative property of multiplication [P.1]
10.
Closure property of addition [P.1]
11.
Identity property of addition [P.1]
12.
Identity property of multiplication [P.1]
13.
Symmetric property of equality [P.1]
14.
Transitive property of equality [P.1]
15.
−4 < x < 2 [P.1]
17.
[−3, 2) [P.1]
16.
(−4, 2] 19.
| 7 | = 7 [P.1]
23.
| −3 – 14 | = 17 [P.1]
26.
( 22 ⋅ 3−2 )2 −1 3
3 2
20.
x < −1
or x > 3 [P.1]
18.
(−∞, −1] ∪ (3, ∞ )
−3 ≤ x < 2
|2 – π | = −(2 – π) = π – 2, 21. because π > 2 [P.1]
| 4 – π | = 4 − π, [P.1] because 4 > π
24.
5 − ( − 2 ) = 5 + 2 [P.1]
(−1, ∞) [P.1] x > −1
25.
22.
|−11| =11 [P.1]
−52 + (−11) = −25 − 11 = −36 [P.1]
4 −4 = 2−13 3 = 24 −33−4 −( −1) = 24 −33−4 +1 = 213−3 = 23 = 2 [P.1] 27 3 2 3
Copyright © Houghton Mifflin Company. All rights reserved.
32
Chapter P: Preliminary Concepts 2
2
27.
(3 x 2 y )(2 x3 y ) 2 = 3 x 2 y ⋅ 4 x 6 y 2 = 12 x8 y 3 [P.2]
28.
2 8 4⎞ ⎛ ⎛ 2a 2b3c − 2 ⎞ ⎟ = ⎜ 2ab ⎟ = 4a b [P.2] ⎜ − 2 1 ⎜ 3c ⎟ ⎟ ⎜ 3ab 9c 4 ⎠ ⎝ ⎠ ⎝
29.
251 / 2 = 25 = 5 [P.2]
30.
− 27 2 / 3 = − 3 27
31.
x 2 / 3 ⋅ x3 / 4 = x 2 / 3 + 3 / 4 = x8 /12 + 9 /12 = x17 /12 [P.2]
32.
⎛ 8 x5 / 4 ⎞ ⎜ 1/ 2 ⎟ ⎝ x ⎠
33.
⎛ x2 y ⎞ ⎜⎜ x1/ 2 y −3 ⎟⎟ ⎝ ⎠
34.
( x1/ 2 − y1/ 2 )( x1/ 2 + y1/ 2 ) = x − y
2/3
= (8 x 5 / 4 − 1/ 2 )
1/ 2
2/3
= (8 x 5 / 4 −
= ( x 2 − 1/ 2 y1 − ( −3) )
1/ 2
)
2/4 2/3
= ( x4 / 2
12a3b = 4a 2 ⋅ 3ab = 2a 3ab [P.2]
38.
18 x3 y 5 = 9 x 2 y 4 ⋅ 2 xy = 3 xy 2 2 xy [P.2]
39.
54 xy 3 = 10 x
40.
−
24 xyz 3 15 z
5y
42.
3 9y
=
6
9 y2 ⋅ 3y
=−
5y
⋅
8 xy 5z 3
=−
3
3y2
3 y 3 3y2
3 2
5
=
2 2 xy z 5z
=
⋅
)
3y 3y 5
⋅
2/3
− 1/ 2 1 + 3 1/ 2
y
[P.2]
36.
27 y 3 = 5
= (8 x 3 / 4 )
5 5
=
( )2 = −(3)2 = −9 [P.2]
= 82 / 3 x (3/ 4)(2 / 3) = (23 )2 / 3 x (3/ 4)(2 / 3) = 22 x1 / 2 = 4 x1 / 2 [P.2]
= ( x3 / 2 y 4 )
1/ 2
= x (3/ 2)(1/ 2) y 4(1/ 2) = x 3/ 4 y 2 [P.2]
35.
48a 2b7 = 16a 2b6 ⋅ 3b = 4ab3 3b [P.2]
37.
72 x 2 y = 36 x 2 ⋅ 2 y = 6 x 2 y [P.2]
3 y 15 y [P.2] 5
2 10 xyz 5z [P.2] =− 5z 5z 2
5 y3 3 y 2 53 3 y 2 [P.2] = 3y 3
− 250 xy 6 = 3 − 125 y 6 ⋅ 2 x = −5 y 2 3 2 x [P.2]
7x
41.
=
7x 3
⋅
3 2
2 x
3 2
=
3
2 x2
43.
3
− 135 x 2 y 7 = 3 − 27 y 6 ⋅ 5 x 2 y = −3 y 2 3 5 x 2 y [P.2]
45.
620,000 = 6.2 × 10 [P.2]
2x2
2 x
5
44.
3
46.
0.0000017 = 1.7 × 10
49.
(2a + 3a – 7) + (−3a – 5a + 6) =[2a + (−3a2)] + [3a + (−5a)] + [(−7) + 6] = −a – 2a − 1 [P.3]
50.
(5b 2 − 11) − (3b 2 − 8b − 3) = 5b 2 − 11 − 3b 2 + 8b + 3 [P.3]
−6
2
2
47.
[P.2]
4
3.5 × 10 = 35,000 [P.2]
2
= 2b 2 + 8b − 8
7 x 3 4 x 73 4 x = [P.2] 2x 2
48.
4.31 × 10−7 = 0.000000431 [P.2]
2
51.
2 x 2 + 3x − 5 [P.3] 3x2 − 2 x + 4 + 8 x 2 + 12 x − 20 − 4 x3 − 6 x 2 + 10 x 6 x 4 + 9 x3 − 15 x 2 6 x 4 + 5 x3 − 13 x 2 + 22 x − 20
52.
(3 y − 5)3 = (3 y − 5)2 (3 y − 5) = (9 y 2 − 30 y + 25)(3 y − 5) = 27 y 3 − 45 y 2 − 90 y 2 +150 y + 75 y −125 = 27 y 3 −135 y 2 + 225 y −125 [P.3]
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Review
33
53.
3x 2 + 30 x + 75 = 3( x 2 + 10 x + 25) = 3( x + 5) 2 [P.4]
54.
25 x 2 − 30 xy + 9 y 2 = (5 x − 3 y ) 2 [P.4]
55.
20a 2 − 4b 2 = 4(5a 2 − b 2 ) [P.4]
56.
16a3 + 250 = 2(8a3 + 125) = 2(2a + 5)( 4a 2 − 10a + 25) [P.4]
57.
6 x 2 − 19 x + 10
=
2
2 x + 3 x − 20 58.
4 x3 − 25 x
=
4
8 x + 125 x
(3 x − 2)(2 x − 5) 3x − 2 = [P.5] (2 x − 5)( x + 4) x+4
x(4 x 2 − 25) 3
x(8 x + 125)
=
x(2 x − 5)(2 x + 5) 2
x(2 x + 5)(4 x − 10 x + 25)
=
2x − 5 2
4 x − 10 x + 25
[P.5]
59.
10 x 2 + 13 x − 3 6 x 2 + 5 x + 1 (2 x + 3)(5 x − 1) (2 x + 1)(3 x + 1) 2 x + 3 = ⋅ = ⋅ [P.5] 6 x 2 − 13 x − 5 10 x 2 + 3 x − 1 (2 x − 5)(3 x + 1) (2 x + 1)(5 x − 1) 2 x − 5
60.
15 x 2 + 11x − 12 25 x 2 − 9
61.
÷
3x 2 + 13 x + 12 10 x 2 + 11x + 3
=
15 x 2 + 11x − 12 10 x 2 + 11x + 3 (5 x − 3)(3x + 4) (5 x + 3)(2 x + 1) 2 x + 1 ⋅ = ⋅ = [P.5] x+3 25 x 2 − 9 3 x 2 + 13 x + 12 (5 x − 3)(5 x + 3) (3 x + 4)( x + 3)
x ( x + 4) + 2 x ( x + 3) x 2 + 4 x + 2 x 2 + 6 x 2x 2x x + x = + = = 2 x − 9 x + x − 12 ( x − 3)( x + 3) ( x + 4)( x − 3) ( x − 3)( x + 3)( x + 4) ( x − 3)( x + 3)( x + 4) x (3x + 10) 3x 2 + 10 x = = ( x − 3)( x + 3)( x + 4) ( x − 3)( x + 3)( x + 4) 2
62.
63.
3x (2 x − 1) − x ( x + 4) x x 3x 3x − = − = [P.5] x 2 + 7 x + 12 2 x 2 + 5 x − 3 ( x + 3)( x + 4) (2 x − 1)( x + 3) ( x + 3)( x + 4)(2 x − 1) 2 2 x (5 x − 7) 5x 2 − 7 x = 6 x − 3x − x − 4 x = = ( x + 3)( x + 4)(2 x − 1) ( x + 3)( x + 4)(2 x − 1) ( x + 3)( x + 4)(2 x − 1) ⎛ 2+ 1 ⎜2 + x−5 = ⎝ ⎛3− 3− 2 x − 5 ⎜⎝
64.
1
2+
65.
[P.5]
3 1+ 4 x
=
1 ⎞ ⎟ x − 5 ⎠ ⋅ x − 5 = 2( x − 5) + 1 = 2 x − 10 + 1 = 2 x − 9 [P.5] 2 ⎞ x − 5 3( x − 5) − 2 3x − 15 − 2 3 x − 17 ⎟ x −5⎠
1
2+
3 x+4 x x
=
1 2+
3 x+4 x
=
x+4 x + 4 = x + 4 [P.5] 1 1 1 = = ⋅ x+4 = = 3x x + 4 2( x + 4) + 3x 2 x + 8 + 3x 5 x + 8 x x 4 ⎛ + ⎞ ⎛ ⎞ 2 + 2 + ⎜3÷ ⎟ 2 + ⎜ 3⋅ ⎟ x+4 x ⎠ ⎝ ⎝ x+4⎠
5 + −64 = 5 + 8i [P.6]
66.
2 − −18 = 2 − i 18
[P.6]
= 2 −i 9⋅2 = 2 − 3i 2 67.
(2 − 3i ) + (4 + 2i ) = 2 − 3i + 4 + 2i [P.6] = (2 + 4) + ( −3i + 2i ) =6−i
68.
(4 + 7i ) − (6 − 3i ) = 4 + 7i − 6 + 3i [P.6] = (4 − 6) + (7i + 3i ) = −2 + 10i
69.
2i (3 − 4i ) = 6i − 8i 2
70.
(4 − 3i )(2 + 7i ) = 4(2 + 7i ) − 3i (2 + 7i ) [P.6]
= 6i − 8( −1) = 6i + 8 = 8 + 6i
[P.6]
= 8 + 28i − 6i − 21i 2 = 8 + 22i − 21( −1) = 8 + 22i + 21 = 29 + 22i
Copyright © Houghton Mifflin Company. All rights reserved.
34
71.
Chapter P: Preliminary Concepts
72.
(3 + i )2 = 32 + 2(3)(i ) + i 2 [P.6] = 9 + 6i + ( −1) = 8 + 6i
Use the Powers of i Theorem. [P.6] The remainder of 345 ÷ 4 is 1 i 345 = i1 = i
73.
4 − 6i = 2(2 − 3i ) = 2 (2 − 3i ) = 2 − 3i = 2 − 3i ⋅ i = 2i − 3i 2 = 2i − 3( −1) = 2i + 3 = −3 − 2i [P.6] −1 −1 i i i 2i 2i 2i i2
74.
2 − 5i = 2 − 5i ⋅ 3 − 4i = (2 − 5i )(3 − 4i ) = 2(3 − 4i ) − 5i (3 − 4i ) = 6 − 8i − 15i + 20i 2 = 6 − 8i − 15i + 20( −1) [P.6] 3 + 4i 3 + 4i 3 − 4i (3 + 4i )(3 − 4i ) 9 − 16( −1) (3)2 − (4i )2 9 − 16i 2 − − − − − i i i 6 8 15 20 14 23 14 23 = = =− − i 9 + 16 25 25 25
....................................................... QR1. Evaluate z =
λ0 − λs when λ0 = 390.5 × 10–9 λs
Quantitative Reasoning QR2. Evaluate z =
and λs = 375.4 × 10–9
λ0 − λs when λ0 = 412.3 × 10–9 λs
and λs = 401.5 × 10–9
−9
× 10 z = 390.5 × 10 − 375.4 375.4 × 10−9
−9
−9 = 15.1 × 10 −9 ≈ 0.040 375.4 × 10
−9 × 10−9 = 10.8 × 10−9 ≈ 0.027 z = 412.3 × 10 − 401.5 −9 401.5 × 10 401.5 × 10−9
2 2 ⎡ ⎤ ⎡ ⎤ QR3. Evaluate v = c ⎢ ( z + 1) − 1 ⎥ when c = 3 × 105 and z = 0.032. QR4. Evaluate v = c ⎢ ( z + 1) − 1 ⎥ when c = 3 × 105 and z = 0.041. 2 2 ⎣ ( z + 1) + 1 ⎦ ⎣ ( z + 1) + 1 ⎦
⎡ (0.032 + 1)2 − 1 ⎤ 5 ⎡ 0.065024 ⎤ v = 3 × 105 ⎢ ≈ 9446 ⎥ = 3 × 10 ⎢ 2 ⎣ 2.065024 ⎥⎦ (0.032 + 1) + 1 ⎣ ⎦ The relative speed is 9446 kilometers per second.
⎡ (0.041 + 1)2 − 1 ⎤ 5 ⎡ 0.083681 ⎤ v = 3 × 105 ⎢ ≈ 12,048 ⎥ = 3 × 10 ⎢ 2 ⎣ 2.083681 ⎥⎦ (0.041 + 1) + 1 ⎣ ⎦ The relative speed is 12,048 kilometers per second.
....................................................... 1. 3.
2. A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} [P.1]
Distributive property [P.1] −12 − (−5) = −12 + 5 = −7 = 7 [P.1]
2
5.
Chapter Test
2 3 1 2 6 ( 2a −1bc −2 ) 22 a −2 b2 c −4 = = 2 ⋅ 2 3⋅ 32 ⋅ 4b c 3 ba a c ( 3−1 b )( 2−1 ac −2 ) ( 3−1 b )( 2−3 a 3c −6 )
4. a ( −2 x 0 y −2 )
2
( −3x 2 y −1 )
−2
= ( 4 y −4 )( 3−2 x −4 y 2 ) =
4 [P.2] 9 x4 y2
5 2 2 = 2 ⋅ 35⋅ bc = 96bc [P.2] 5 a a
6.
0.00137 = 1.37 × 10−3 [P.2]
7.
x1 / 3 y −3 / 4
x5 / 6 = x1 / 3 − ( −1 / 2) y −3 / 4 − 3 / 2 = x1 / 3 +1 / 2 y −3 / 4 − 3 / 2 = x 2 / 6 +3 / 6 y −3 / 4 −6 / 4 = x5 / 6 y −9 / 4 = [P.2] x −1 / 2 y3 / 2 y9 / 4
8.
3x3 81xy 4 − 2 y 3 3 x 4 y = 3 x3 27 y 3 ⋅ 3 xy − 2 y 3 x3 ⋅ 3 xy = 3 x ⋅ 3 y 3 3 xy − 2 y ⋅ x3 3 xy = 9 xy3 3 xy − 2 xy3 3 xy = 7 xy3 3 xy [P.2]
x
9. 4
2 x3
=
x 4
2 x3
⋅
4 3
2 x
4 3
2 x
4
=
x 23 x 4 4 4
2 x
=
x4 8x 4 8x = [P.2] 2x 2
10.
3 = x +2
3 x −2 3 x −6 [P.2] ⋅ = x−4 x +2 x −2
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Chapter Test
35
11.
( x − 2 y )( x 2 − 2 x + y ) = x3 − 2 x 2 + xy − 2 x 2 y + 4 xy − 2 y 2 = x3 − 2 x 2 + 5 xy − 2 x 2 y − 2 y 2 [P.3]
12.
If y = −3, 3 y 3 − 2 y 2 − y + 2 = 3(−3)3 − 2(−3) 2 − (−3) + 2 = 3(−27) − 2(9) + 3 + 2 = −81 − 18 + 3 + 2 = −94 [P.4]
13.
7 x 2 + 34 x − 5 = (7 x − 1)( x + 5) [P.4]
14.
3ax − 12bx − 2a + 8b = (3ax − 12bx) − (2a − 8b) = 3x(a − 4b) − 2(a − 4b) = (a − 4b)(3x − 2) [P.4]
15.
16 x 4 − 2 xy 3 = 2 x (8 x 3 − y 3 ) = 2 x (2 x − y )(4 x 2 + 2 xy + y 2 ) [P.4]
16.
x 2 − 2 x − 15 = ( x − 5)( x + 3) = ⎛ x − 5 ⎞⎛ x + 3 ⎞ = −1 ⋅ ⎛ x + 3 ⎞ = − x + 3 [P.5] ⎜ ⎟⎜ ⎟ ⎜ ⎟ (5 − x )(5 + x ) ⎝ 5 − x ⎠⎝ x + 5 ⎠ x+5 ⎝ x + 5⎠ 25 − x 2
17.
x ( x − 3) − 2( x + 3) 2 x x [P.5] − 2 2 = − = x + x − 6 x − 5 x + 6 ( x − 2)( x + 3) ( x − 2)( x − 3) ( x − 2)( x + 3) ( x − 3) 2 ( x − 6)( x + 1) x 2 − 5x − 6 = x − 3x − 2 x − 6 = = ( x − 2)( x + 3) ( x − 3) ( x − 2)( x + 3) ( x − 3) ( x − 2)( x + 3) ( x − 3)
18.
2
2 x ( x + 2) 2 x 2 + 3x − 2 ÷ 2 x 2 − 7 x + 3 = 2 x 2 + 3x − 2 ⋅ x 3 − 3x 2 = (2 x − 1)( x + 2) ⋅ x ( x − 3) [P.5] = 2 3 2 2 2 (2 x − 1)( x − 3) x ( x − 3) x −3 2x − 7x + 3 x − 3x x − 3x x − 3x
19.
2 3 ⋅ a 2 − b2 − 5 = 3 ⋅ ( a − b)( a + b) − 5 = 3( a − b) − 5 = 3a ( a − b) − 5(2a − b) = 3a − 3ab − 10a + 5b [P.5] a + b 2a − b a a + b a a (2a − b) 2a − b 2a − b a a ( 2a − b)
20.
x−
21.
7 + −20 = 7 + 2i 5 [P.6]
23.
(2 + 5i )(1 − 4i ) = 2(1 − 4i ) + 5i (1 − 4i ) [P.6]
x = x − x = x − x = x − x ÷ 2x + 1 = x − x ⋅ 2 = x − 2x 2x + 1 2x + 1 2 2x + 1 2x + 1 x+1 2 2 2 2 [P.5] 2 2 2 x (2 x + 1) x (2 x − 1) 2 2 + 2 2 + − 2 2 − x x x x x x x x x = − = − = = = 2x + 1 2x + 1 2x + 1 2x + 1 2x + 1 2x + 1 2x + 1 22.
[P.6] (4 − 3i ) − (2 − 5i ) = 4 − 3i − 2 + 5i = (4 − 2) + ( −3i + 5i ) = 2 + 2i
= 2 − 8i + 5i − 20i 2 = 2 − 8i + 5i − 20( −1) = 2 − 8i + 5i + 20 = (2 + 20) + ( −8i + 5i ) = 22 − 3i 24.
3 + 4i = 3 + 4i ⋅ 5 + i = (3 + 4i )(5 + i ) = 3(5 + i ) + 4i (5 + i ) = 15 + 3i + 20i + 4i 2 = 15 + 3i + 20i + 4( −1) [P.6] 5−i 5 − i 5 + i (5 − i )(5 + i ) 25 − ( −1) 52 − i 2 25 − i 2 (15 − 4) + (3i + 20i ) 11 + 23i 11 23 = 15 + 3i + 20i − 4 = = = + i 25 + 1 26 26 26 26
25.
Use the Powers of i Theorem. [P.6] The remainder of 97 ÷ 4 is 1.
i 97 = i1 = i
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 1
Equations and Inequalities Section 1.1 1.
5.
8.
2 x + 10 = 40 2 x = 40 − 10 2 x = 30 x = 15
−3 y + 20 = 2
2.
y=6
2( x − 3) − 5 = 4( x − 5) 2 x − 6 − 5 = 4 x − 20 2 x − 11 = 4 x − 20 2 x − 4 x = −20 + 11 −2 x = −9 x=9 2
6.
6(5s − 11) − 12(2 s + 5) = 0 30s − 66 − 24 s − 60 = 0 6s − 126 = 0
9.
6s = 126 s = 126 6 s = 21
11.
3.
−3 y = 2 − 20 −3 y = −18
5 x + 2 = 2 x − 10 5 x − 2 x = −10 − 2 3x = −12 x = −4
5( x − 4) − 7 = −2( x − 3) 5 x − 20 − 7 = −2 x + 6 5 x − 27 = −2 x + 6 5 x + 2 x = 6 + 27 7 x = 33 x = 33 7
7.
3x+1 = 2 4 2 3 12 ⋅ ⎛⎜ 3 x + 1 ⎞⎟ = 12 ⋅ ⎛⎜ 2 ⎞⎟ 2⎠ ⎝4 ⎝ 3⎠ 9x + 6 = 8 9x = 8 − 6 9x = 2 x=2 9
10.
1 x + 7 − 1 x = 19 2 4 2 4 ⋅ ⎛⎜ 1 x + 7 − 1 x ⎞⎟ = 4 ⋅ ⎛⎜ 19 ⎞⎟ 4 ⎠ ⎝2 ⎝ 2⎠ 2 x + 28 − x = 38 x + 28 = 38 x = 38 − 28 x = 10
13.
4.
4 x − 11 = 7 x + 20 4 x − 7 x = 20 + 11 −3x = 31 x = − 31 3
4(2 r − 17) + 5(3r − 8) = 0 8r − 68 + 15r − 40 = 0 23r − 108 = 0 23r = 108 r = 108 23
x −5= 1 4 2 4 ⋅ ⎛⎜ x − 5 ⎞⎟ = 4 ⎛⎜ 1 ⎞⎟ ⎝4 ⎠ ⎝2⎠ x − 20 = 2 x = 2 + 20 x = 22
0.2 x + 0.4 = 3.6 0.2 x = 3.6 − 0.4
2 x −5= 1 x −3 3 2 2 ⎛ ⎞ 6 ⋅ ⎜ x − 5 ⎟ = 6 ⋅ ⎛⎜ 1 x − 3 ⎞⎟ ⎝2 ⎠ ⎝3 ⎠ 4 x − 30 = 3x − 18 4 x − 3x = −18 + 30 x = 12
12.
14.
0.04 x − 0.2 = 0.07 0.04 x = 0.07 + 0.2 0.04 x = 0.27 x = 6.75
15.
x + 0.08(60) = 0.20(60 + x ) x + 4.8 = 12 + 0.20 x x − 0.20 x = 12 − 4.8 0.80 x = 7.2 x=9
16.
6(t + 1.5) = 12t 6t + 9 = 12t 6t − 12t = −9 −6t = −9 t=3 2
17.
3( x + 5)( x − 1) = (3x + 4)( x − 2)
18.
5( x + 4)( x − 4) = ( x − 3)(5 x + 4)
19.
5[ x − (4 x − 5)] = 3 − 2 x 5( x − 4 x + 5) = 3 − 2 x 5( −3x + 5) = 3 − 2 x −15 x + 25 = 3 − 2 x −15 x + 2 x = 3 − 25 −13x = −22 x = 22 13
3 ( x 2 + 4 x − 5) = 3 x 2 − 2 x − 8
5 ( x 2 − 16 ) = 5 x 2 − 11x − 12
3x 2 + 12 x − 15 = 3x 2 − 2 x − 8
5 x 2 − 80 = 5 x 2 − 11x − 12
12 x + 2 x = −8 + 15
11x = −12 + 80
14 x = 7
11x = 68
x=1 2
x = 68 11
0.2 x = 3.2 x = 16
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Section 1.1
37
20.
6[3 y − 2( y − 1)] − 2 + 7 y = 0 6(3 y − 2 y + 2) − 2 + 7 y = 0 18 y − 12 y + 12 − 2 + 7 y = 0 13 y + 10 = 0 13 y = −10 y = − 10 13
21.
40 − 3x = 6 x + 7 5 8 ⎛ ⎞ ⎛ 6x + 7 ⎞ 40 3 x − 40 ⋅ ⎜ ⎟ = 40 ⋅ ⎜ ⎟ ⎝ 5 ⎠ ⎝ 8 ⎠ 8(40 − 3x ) = 5(6 x + 7) 320 − 24 x = 30 x + 35 −24 x − 30 x = 35 − 320 −54 x = −285 x = 95 18
22.
12 + x = 5 x − 7 + 2 3 −4 ⎛ 5x − 7 + 2 ⎞ 12 x + ⎛ ⎞ 12 ⋅ ⎜ ⎟ = 12 ⋅ ⎜ ⎟ ⎝ −4 ⎠ ⎝ 3 ⎠ −3(12 + x ) = 4(5 x − 7) + 24 −36 − 3x = 20 x − 28 + 24 −36 − 3x = 20 x − 4 −3x − 20 x = −4 + 36 −23x = 32 x = − 32 23
23.
−3( x − 5) = −3x + 15
24.
2 x + 1 = 6x + 1 3 3 1 ⎛ ⎞ ⎛ 3 ⋅ ⎜ 2 x + ⎟ = 3 ⋅ ⎜ 6 x + 1 ⎟⎞ 3⎠ ⎝ ⎝ 3 ⎠ 6x + 1 = 6x + 1 Identity
25.
2 x + 7 = 3( x − 1) 2 x + 7 = 3x − 3 2 x − 3x = −3 − 7 − x = −10 x = 10 Conditional equation
4[2 x − 5( x − 3)] = 6 4[2 x − 5 x + 15] = 6 4[ −3x + 15] = 6 −12 x + 60 = 6 −12 x = −54 x=9 2 Conditional equation
27.
4x + 8 = x + 8 4 4 x + 8 = 4( x + 8)
28.
3[ x − (4 x − 1)] = −3(2 x − 5)
3[ x − 2( x − 5)] − 1 = −3x + 29 3[ x − 2 x + 10] − 1 = −3x + 29 3[ − x + 10] − 1 = −3x + 29 −3x + 30 − 1 = −3x + 29 −3x + 29 = −3x + 29 Identity
30.
−3x + 15 = −3x + 15 Identity
26.
29.
32.
3[ x − 4 x + 1] = −6 x + 15 3[ −3x + 1] = −6 x + 15
4 x + 8 = 4 x + 32
−9 x + 3 = −6 x + 15 −3x = 12
8 = 32 Contradiction
33.
3( x − 4) + 7 = 3x − 5 3x − 12 + 7 = 3x − 5
x = −4 Conditional equation
4[3( x − 5) + 7] = 12 x − 32 4[3x − 15 + 7] = 12 x − 32 4[3x − 8] = 12 x − 32 12 x − 32 = 12 x − 32 Identity
34.
x =4 x=4
31.
2x − 8 = −x + 9 3x = 17 x = 17 3 Conditional equation
x =7 x=7
or x = –4
or
x = −7
3x − 5 = 3x − 5 Identity 35.
36.
x −5 = 2 x−5 = 2
x − 5 = −2
or
x=7
38.
x −8 = 3
x=3
39.
2 x = 24 x = 12
or
2 x − 3 = −21 2 x = −18 x = −9
x=5
40.
2 x + 6 = 10 2 x + 6 = 10 2x = 4 x=2
or
2 x − 5 = 11 2 x − 5 = 11 2 x = 16 x =8
x − 8 = −3
or
x = 11
2 x − 3 = 21 2 x − 3 = 21
37.
x −8 = 3
or
2x − 5 = −11 2x = −6 x = −3
2 x + 14 = 60
2x + 6 = −10
2 x + 14 = 60
2x = −16 x = −8
2 x = 46 x = 23
Copyright © Houghton Mifflin Company. All rights reserved.
or
2 x + 14 = −60 2 x = −74 x = −37
38
41.
43.
Chapter 1: Equations and Inequalities
x−4 =8 2 x−4 =8 2 x − 4 = 8(2)
42. x − 4 = −8 2 x − 4 = −8(2)
x+3 =6 4 x+3=6 4 x + 3 = 6(4)
x − 4 = 16
x − 4 = −16
x + 3 = 24
x + 3 = −24
x = 20
x = −12
x = 21
x = −27
or
44.
2 x + 5 = −8 2x + 5 ≥ 0
−17 ≥ 0 Contradiction. There is no solution. 46.
2 x + 3 + 4 = 34 2 x + 3 = 30 x + 3 = 15 x + 3 = 15 x = 12
47.
or
x = a+b 2
or
3 x − 5 − 16 = 2 3 x − 5 = 18 x−5 =6 x −5=6 x = 11
x + 3 = −15 x = −18
48.
2 x − a = b, b > 0
2x − a = b 2x = a + b
4 x − 1 = −17 4 x −1 ≥ 0
−8 ≥ 0 Contradiction. There is no solution. 45.
x + 3 = −6 4 x + 3 = −6(4)
or
2x − a = − b 2x = a − b
x − 5 = −6 x = −1
3 x − d = c, c > 0 x−d = c 3 x−d = c 3
x = a−b 2
or
or
x=d+c 3
x−d=−c 3
x=d −c 3
49.
1.6 x + 1.87 = Revenue 1.6 x + 1.87 = 10 1.6 x = 8.13 x = 8.13 ≈ 5 1.6 2000 + 5 = 2005 The revenue first exceeded $10 billion in 2005.
50.
93.8 x + 542.8 = Number of megawatts (MW) 93.8 x + 542.8 = 1200 93.8 x = 657.2 x = 657.2 ≈ 7 93.8 2000 + 7 = 2007 The energy will exceeded 1200 MW in 2007.
51.
d = 210 − 50t
52.
m = − 1 s − 55 + 25 2 22 = − 1 s − 55 + 25 2 −3 = − 1 s − 55 2 6 = s − 55
60 = 210 − 50t or − 60 = 210 − 50t −150 = −50t − 270 = −50t t =3 t = 5.4 5.4 hours = 5 hours 24 minutes Ruben will be exactly 60 miles from Barstow after 3 hours and after 5 hours and 24 minutes.
53.
45x + 550 = Cost 45 x + 550 = 3800 45 x = 3250 x ≈ 72 Rounded to the nearest yard, 72 sq yards can be carpeted for $3800.
−6 = s − 55 6 = s − 55 or 49 = s 61 = s Kate can drive at 49 mph or 61 mph to obtain gas mileage of exactly 22 miles per gallon. 54.
1.75x + 8.00 = Retail price 1.75 x + 8.00 = 156.75 1.75 x = 148.75 x = 85 The wholesale price of the coat is $85.00.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.1
55.
39
100 −
42,000 t = Percent remaining 500,000
42,000 t 500,000 21 100 − t 250 21 − t 250 t
100 −
56.
2650 − 475t 2650 − 475t − 475t t 3.5 hours
58.
max = 0.85(220 − a ) 153 = 0.85(220 − a )
= 25 = 25
= miles remaining = 1000 = −1650 ≈ 3.5 to the nearest tenth
= −75
≈ 892.857142857 seconds ⎛ 1 min ⎞ 892.857142857 sec ⋅ ⎜ ⎟ ≈ 15 min ⎝ 60 sec ⎠ 57.
max = 0.85(220 − a ) min=0.65(220 − a ) = 0.65(220 − 25) = 0.85(220 − 25) = 0.65(195) = 0.85(195) = 126.75 = 165.75
153 = 187 − 0.85a − 34 = −0.85a 40 = a
The maximum exercise heart rate for a person who is 25 years of age is 166 beats per minute (to the nearest beat).
A person should have a maximum exercise heart rate of 153 beats per minute at age 40.
The minimum exercise heart rate for a person who is 25 years of age is 127 beats per minute (to the nearest beat). 59.
ax + b = c ax = c − b
60.
x = c−b, a ≠ 0 a
61.
if x + 4 ≥ 0 x ≥ −4 x+4= x+4 an identity
63.
62.
x+4 = x+4
ax + b = cx + d ax − cx = d − b x(a − c) = d − b x = d −b, a −c ≠ 0 a−c x −1 = x −1 if x − 1 ≥ 0
if x + 4 < 0 x < −4
x ≥1 x −1 = x −1
x + 4 = − ( x + 4)
x + 4 = −x − 4 2 x = −8 x = −4
an identity
if x − 1 < 0 x <1 x − 1 = −( x − 1) x −1 = −x + 1 2x = 2 x =1
The case x + 4 < 0 has no solution since there is no real number x such that x < − 4 and x = − 4.
The case x −1 < 0 has no solution since there is no real number x such that x < 1 and x = 1.
{x | x ≥ −4}
{x | x ≥ 1} 64.
x + 7 = −( x + 7)
if x + 7 ≥ 0 x ≥ −7 x + 7 = −( x + 7) x + 7 = −x − 7 2 x = −14 x = −7
{x | x ≤ −7}
if x + 7 < 0 x < −7 x+7 = x+7 an identity
x − 3 = −( x − 3)
if x − 3 ≥ 0 x≥3 x − 3 = −( x − 3) x − 3 = −x + 3 2x = 6 x=3
{x | x ≤ 3}
Copyright © Houghton Mifflin Company. All rights reserved.
if x − 3 < 0 x<3 x−3 = x−3 an identity
40
65.
Chapter 1: Equations and Inequalities
66.
2x + 7 = 2x + 7
if 2 x + 7 ≥ 0
if 3x − 11 ≥ 0 3x ≥ 11 x ≥ 11 3 3x − 11 = −3x + 11 6 x = 22 x = 11 3
if 2 x + 7 < 0
7 x≥− 2 2x + 7 = 2x + 7 an identity
3x − 11 = −3 x + 11
7 x<− 2 2 x + 7 = −( 2 x + 7)
2 x + 7 = −2 x − 7 4 x = −14 7 x=− 2
if 3x − 11 < 0 3x < 11 x < 11 3 3x − 11 = −( −3x + 11) an identity
{x | x ≤ 113 }
The case 2x + 7 < 0 has no solution since there is no real number x such that x < − 7 and x = − 7 . 2 2 7⎫ ⎧ ⎨x | x ≥ − ⎬ 2⎭ ⎩
....................................................... PS1. 32 − x
Prepare for Section 1.2 PS2.
32 − 8 1 = 23 1 2 2 PS3. 2l + 2w = 2(l + w) Distributive property
PS5.
1 bh 2 1⋅2⋅4 = 4 2 3 5 15
PS4. ⎛⎜ 1 b ⎞⎟ h = 1 bh 2 ⎝2 ⎠ Associative property of multiplication
2 x + 1 x = 6 x + 5 x = 11 x 5 3 15 15 15
PS6.
1 = 1 1 = = 1 = ab 1 +1 1⋅b+ 1⋅a b + a b+a a+b a b a b b a ab ab ab
Section 1.2 1.
V = 1 π r2h 3 3 V = π r2h 3V = h π r2
5.
F=
Gm1m2 d
2.
P = S − Sdt Sdt = S − P t= S−P Sd
6.
2
Fd 2 = Gm1m2 Fd = m 1 Gm2 y − y1 = m( x − x1 ) y − y1 = mx − mx1 y − y1 + mx1 = mx y − y1 + mx1 =x m
A = 1 h (b1 + b2 ) 2 2 A = hb1 + hb2
I = Pr t I =t Pr
A =P 1 + rt
a1 1− r S (1 − r ) = a1 S − Sr = a1 S − a1 = Sr S − a1 =r S S=
an = a1 + ( n − 1)d
7.
an − a1 = ( n − 1)d an − a1 =d n −1
2 A − hb2 = b1 h
9.
A = P + Pr t A = P(1 + rt )
4.
2 A − hb2 = hb1
2
8.
3.
10.
P1V1 P2V 2 = T1 T2 P1V1T2 = P2V 2T1 P1V1T2 = V2 P2T1
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.2
41
11.
qb rating = 100 [0.05(61.3 − 30) + 0.25(6.39 − 3) + 0.2(3.29) + (2.375 − 0.25(4.78))] 6 ≈ 70.8
12.
qb rating = 100 [0.05(61.3 − 30) + 0.25(6.64 − 3) + 0.2(3.36) + (2.375 − 0.25(2.3))] 6 ≈ 82.5
13.
SMOG = 42 + 3 = 9.5
17.
14.
Let x = the number 1 1 1 x + x = x −5 5 4 2 1 ⎞ ⎛1 ⎛1 ⎞ 20⎜ x + x ⎟ = 20⎜ x − 5 ⎟ 5 4 2 ⎝ ⎠ ⎝ ⎠ 4 x + 5 x = 10 x − 100 9 x = 10 x − 100 100 = x
15.
SMOG = 105 + 3 = 13.2 18.
GFI = 0.4(14.8 + 15.1)
The original fraction is
x−4 . x
= 13.2
W
P = 2 L + 2W 174 = 2(2W − 3) + 2W 174 = 4W − 6 + 2W 180 = 6W W = 30 ft L = 2W − 3 = 2(30) − 3 = 60 − 3 = 57 ft
x + 10 = 5 x − 50 − 4 x = −60
The original fraction is
15 − 4 11 = . 15 15
21.
22.
110 = 2 L + 2 ⎡ 1 L + 1⎤ ⎣2 ⎦ 110 = 2 L + L + 2 108 = 3L L = 36 m W = 1 L + 1 = 1 (36) + 1 2
3x + 3 x + x = 84 7 x = 84 x = 12 3 x = 3(12) = 36 The shortest side is 12 cm. The longer sides are each 36 cm.
2 x 3
2 x 3
3x x
P = 2 L + 2W
2W − 3
19.
x − 4 + 14 =5 x − 10 x + 10 =5 x − 10 x + 10 = 5( x − 10)
3x
GFI = 0.4(18.8 + 14.2)
≈ 12.0
x = 15
20.
16.
x
2 x + 2 x + x = 161 3 3 2 x + 2 x + 3x = 483 7 x = 483 x = 69 2 x = 2 (69) = 46 3 3 The longest side is 69 miles. The two shorter sides are each 46 miles.
2
= 18 + 1 = 19 m
Copyright © Houghton Mifflin Company. All rights reserved.
42
23.
Chapter 1: Equations and Inequalities
24.
d = 6t d = 2(160 − t)
d = 15t d = 10(7.5 − t)
Let t = the time to run to the end of the track. Let 160 – t = the time in seconds to jog back. 6t = 2(160 − t ) 6t = 320 − 2t 8t = 320 t = 40 d = 6(40) = 240 meters
Let t = the time (in hours) to travel to the island. Let 7.5 – t = the time (in hours) to return. d = 15t , and d = 10(7.5 − t ) Thus, 15t = 10(7.5 − t ) 15t = 75 − 10t 25t = 75 t = 3 hours d = 15(3) = 45 nautical miles
25.
26.
d = 240(t + 3)
Let t = time (in hours) of the first plane. Let t – 1 = time (in hours) of the second plane.
d = 600t
Let t = the time (in hours) of the second plane. Let t + 3 = the time (in hours) of the first plane. d = 240(t + 3) d = 600t 240(t + 3) = 600t 240t + 720 = 600t 720 = 360t 2=t t = 2 hours
Los Angeles 660( t − 1)
Chicago 540 t
1800 miles
660(t − 1) + 540t = 1800 660t − 660 + 540t = 1800 1200t = 2460 t = 2.05 hours Thus, the planes pass each other 2.05 hours after the first plane leaves Chicago. The distance the first plane is from Chicago is 540(2.05) = 1107 miles.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.2
27.
43
Let x + y + z = the distance to Jon’s house, where x = the 28. distance uphill, y = the distance down hill, and z = the distance on level ground. Note that x, the distance uphill on the way to Jon’s house equals the distance down hill on the way home, and that z, the distance down hill on the way to Jon’s house equals the distance uphill on the way home. Also note that y is the distance on level ground on the way to Jon’s house and on the way home. d rt = d ⇒ t = r
rate
To Jon’s house
Back home
Up
4
Level
6
Down
12
Up
12
Level
6
Down
4
time x 4 y 6 z 12 x 12 y 6 z 4
30 seconds = 0.5 minutes = 1
500 meters = a
2
km.
rate Faster car
x
Slower car
80
0. 5 1 hour = hour. 60 120
time 1 120 1 120
distance
x ⎛⎜ 1 ⎞⎟ − 80 ⎛⎜ 1 ⎞⎟ = 1 ⎝ 120 ⎠ ⎝ 120 ⎠ 2
x
⎡ ⎤ 120 ⎢ x ⎛⎜ 1 ⎞⎟ − 80 ⎛⎜ 1 ⎞⎟ ⎥ = 120 ⎛⎜ 1 ⎞⎟ 120 120 ⎝2⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ x − 80 = 60 x = 140 140 km/h
y z x y z
x + y + z + x + y + z =1 4 6 12 12 6 4 y y ⎛ ⎞ 12 ⎜ x + + z + x + + z ⎟ = 12(1) ⎝ 4 6 12 12 6 4 ⎠ 3x + 2 y + z + x + 2 y + 3z = 12 4 x + 4 y + 4 z = 12 4( x + y + z ) = 12 x+ y+z=3 The distance to Jon’s house is 3 miles.
29.
Let x = the score on the next test. 80 + 82 + 94 + 71 + x = 85 5 327 + x = 85 5 327 + x = 425 x = 98
30.
A score of 98 will produce an average of 85.
31.
Let x = the number of sunglasses. Profit = Revenue − Cost 17,884 = 29.99 x − 8.95 x 17,884 = 21.04 x x = 850 The manufacturer must sell 850 sunglasses to make a profit of $17,884.
Let x = the score on the final examination. 90 + 74 + 82 + 90 + 2 x = 85 6 336 + 2 x = 85 6 336 + 2 x = 510 2 x = 174 x = 87 A score of 87 on the final examination score will produce an average of 85.
32.
Let x = the number of glasses of orange juice. Profit = Revenue − Cost 2337 = 0.75 x − 0.18 x 2337 = 0.57 x 4100 = x The owner must sell 4100 glasses of orange juice to make a profit of $2337.
Copyright © Houghton Mifflin Company. All rights reserved.
44
Chapter 1: Equations and Inequalities
33.
Let x = cost last year. x − 0.20 x = 750 0.80 x = 750 x = 937.50 The cost of a computer last year was $937.50.
34.
35.
Let x = amount invested at 8%. (14,000 − x) = amount invested at 6.5%. 0.08 x + 0.065(14,000 − x) = 1024 0.08 x + 910 − 0.065 x = 1024 0.015 x = 114 x = 7600 14,000 − x = 6400
36.
39.
5.5%
2500
8% 7%
x
7%
7500 − x
$6000 was invested at 5%. $1500 was invested at 7%. 38.
0.068
4600
x
0.09
x
2500 + x
0.08
4600 + x
0.055(2500) + 0.08 x = 0.07(2500 + x ) 137.5 + 0.08 x = 175 + 0.07 x 0.01x = 37.5 x = 3700
0.068(4600) + 0.09 x = 0.08(4600 + x ) 312.8 + 0.09 x = 368 + 0.08 x 0.01x = 55.2 x = 5520
$3750 additional investment.
$5520 additional investment
1.00
x
0.45 0.50
40.
0.40
x
200 − x
0.24
4
200
0.30
4+x
0.40 x + 0.24(4) = 0.30(4 + x ) 0.40 x + 0.96 = 1.2 + 0.30 x 0.10 x = 0.24 x = 2.4
x + 0.45(200 − x ) = 0.50(200) x + 90 − 0.45 x = 100 0.55 x = 10 x = 18 2 g pure silver 11 41.
5%
0.05 x + 0.07(7500 − x) = 405 0.05 x + 525 − 0.07 x = 405 − 0.02 x = −120 x = 6000 7500 − x = 1500
$7600 was invested at 8%. $6400 was invested at 6.5%.
37.
Let x = cost last year. x + 0.04 x = 26 1.04 x = 26 x = 25 The cost of the subscription last year was $25.
0
x
0.12 0.20
2.4 liters of 40% sulfuric acid 42.
0.25
6
160
0.25
x
160 − x
1.00
x
0.33
6
0.12(160) − 0 = 0.20(160 − x) 19.2 = 32 − 0.20 x 0.20 x = 12.8
x = 64 64 liters of water
0.25(6) − 0.25 x + x = 0.33(6) 1.5 + 0.75 x = 1.98 0.75 x = 0.48 x = 0.64 0.64 liter of water should be replaced. Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.2
43.
45
44.
14
x
25
3000 − x
14 x + 25( 3000 − x) = 61,800 14 x + 75,000 − 25 x = 61,800 − 11x = −13,200 x = 1200 3000 − x = 1800
$12 $9 $10
x
0.10
2x
0.25
255 − 3x
0.05 x + 0.10(2 x) + 0.25(255 − 3x) = 41.25 0.05 x + 0.20 x + 63.75 − 0.75 x = 41.25 − 0.50 x = −22.50 x = 45
1200 tickets at $14 each 1800 tickets at $25 each
45.
0.05
x = 45 nickels 2 x = 90 dimes 255 − 3 x = 120 quarters 46.
X 20 − x 20
8
x (gold)
5
42 − x (silver)
8 x + 5(42 − x) = 246 8 x + 210 − 5 x = 246 3x = 36
12 x + 9(20 − x) = 10(20) 12 x + 180 − 9 x = 200
x = 12
3 x = 20 x=6
2 3
12 gold coins 42 – 12 = 30 silver coins
2 lb of $12 coffee 3 2 1 20 − 6 = 13 lb of $9 coffee 3 3 6
47.
1.00
48.
x
1.00
4
14 7 or 24 12
15
14 7 or 24 12
x
18 3 or 24 4
15 + x
18 3 or 24 4
4+x
x + 7 (15) = 3 (15 + x ) 12 4 7 ⎡ ⎤ 12 ⋅ x + (15) = 12 ⋅ ⎡ 3 (15 + x ) ⎤ ⎢⎣ ⎥⎦ ⎢⎣ 4 ⎥⎦ 12 12 x + 7(15) = 9(15 + x ) 12 x + 105 = 135 + 9 x 3x = 30 x = 10
4 + 7 x = 3 (4 + x ) 12 4 7 ⎡ ⎤ 12 ⋅ 4 + x = 12 ⋅ ⎡ 3 (4 + x ) ⎤ ⎢⎣ 12 ⎥⎦ ⎢⎣ 4 ⎥⎦ 48 + 7 x = 9(4 + x ) 48 + 7 x = 36 + 9 x −2 x = −12 x=6
10 g of pure gold
6 oz of 14 karat gold
Copyright © Houghton Mifflin Company. All rights reserved.
46
49.
Chapter 1: Equations and Inequalities
Let t = the time it takes both electricians working together to wire the house.
50.
Let t = the time it takes them working together to print the report.
1 of the job every hour. 14 1 of the job every hour. The second electrician does 18
1 of the job every hour. 3 1 Printer B does of the job every hour. 4
1 t + 1 t =1 14 18 126 ⎡ 1 t + 1 t ⎤ = 126 ⋅1 ⎢⎣14 18 ⎥⎦ 9t + 7t = 126
1 t + 1 t =1 3 4 12 ⎡ 1 t + 1 t ⎤ = 12 ⋅1 ⎢⎣ 3 4 ⎥⎦ 4t + 3t = 12
The first electrician does
Printer A does
16t = 126
7t = 12
t = 7 7 hours 8
t = 1 5 hours 7
51.
Let t = the time it takes both painters working together to paint the kitchen. The painter can paint 1 of the kitchen every hour. 10 The apprentice can paint 1 of the kitchen every hour. 15 1 t + 1 t =1 10 15 ⎡ 1 t + 1 t ⎤ = 30 ⋅ 1 30 ⎢⎣ 10 15 ⎥⎦ 3t + 2t = 30 5t = 30 t = 6 hours
52.
Let t = the time it takes to deposit enough snow to open the beginning trail. The snow making machine does 1 of the job every hour. 16 The natural snow fall does 1 of the job every hour. 24 1 t + 1 t =1 16 24 1 1 ⎡ 48 t + t ⎤ = 48 ⋅ 1 ⎢⎣ 16 24 ⎥⎦ 3t + 2t = 48 5t = 48 t = 9 3 hours 5
53.
Let t = the time it takes the older machine to finish the job. The new machine does 1 of the job every hour. 12 1 The old machine does of the job every hour. 16 The new machine works for 4 hours: 4 1 = 1 . 12 3 The old machine completes the job. 1 + 1 t =1 3 16 1 t=2 16 3 t = 10 2 hours 3
54.
Let t = the time it takes the apprentice to finish the job. The mason does 1 of the job every hour. 12 The apprentice does 1 of the job every hour. 16
( )
(
)
The two people work together for 4 hours: 4 1 + 1 = 7 12 16 12 7 + 1 t =1 12 16 1 t= 5 16 12 t = 6 2 hours 3
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.2
55.
47
Let x = price of book 10.10 – x = price of bookmark. x = 10 + (10.10 − x)
56.
2 x = 20.10 x = 10.05
Let x = cost of yacht x = cost with 3 partners 3 x = cost with 4 partners 4
10.10 − 10.05 = 0.05 x x − = 4000 3 4 ⎡x x⎤ 12⎢ − ⎥ = 12(4000) ⎣3 4⎦ 4 x − 3 x = 48,000
The price of the book is $10.05. The price of the bookmark is $0.05.
x = 48,000 Cost = $48,000
....................................................... 57.
58.
Connecting Concepts 59.
117 x = 156(21 − x) 117 x = 3276 − 156 x 273x = 3276
400(0.5) = 5 x 200 = 5 x 40 = x
x = 12
8(100) + 40(5) = 160 x
A 40-lb force is needed to lift 400 lbs.
800 + 200 = 160 x 1000 = 160 x
12 ft from the 117-lb force
6.25 = x 6.25 ft 60.
180(5) + 4 x =1440(1) 900 + 4 x =1440 4 x = 540 x =135
61.
The second worker needs to apply a force of 135 pounds.
62.
1865t = 1100(2 − t ) 1865t = 2200 − 1100t 2965t = 2200 t ≈ 0.741989 d = 1865t
(t + 4 )( 1100 ) = 4.62( 1100 )t 1100t + 4400 = 5082t 4400 = 3982t 1.104972376 ≈ t
d ≈ 1383.81
The distance to the target is 1384 feet (to the nearest foot).
Copyright © Houghton Mifflin Company. All rights reserved.
d = 5082t d ≈ 5615 ft
48
Chapter 1: Equations and Inequalities
63.
F = 9 C + 32, Let C = F , then 5 F = 9 F + 32 5 5F = 9 F + 160 −4 F = 160
64.
F = −40o 1 1 1 1 x+ x+ x+5+ x+ 4 = x 6 12 7 2 1 1 1 ⎡1 ⎤ 84 ⎢ x + x + x + 5 + x + 4 ⎥ = 84 x 12 7 2 ⎣6 ⎦ 14 x + 7 x + 12 x + 420 + 42 x + 336 = 84 x 75 x + 756 = 84 x 756 = 9 x 84 = x
Diophantus was 84 years old when he died.
.......................................................
Prepare for Section 1.3
PS1. x 2 − x − 42 = ( x + 6)( x − 7)
PS2. 6 x 2 − x −15 = (2 x + 3)(3 x − 5)
PS3. 3+ −16 = 3 + 4i
PS4.
PS5.
−( −3) + ( −3)2 − 4(2)(1) 3 + 1 = =1 2(2) 4
−( −2) − ( −2)2 − 4( −3)(5) 2 − 64 = =1 2( −3) −6
PS6. (3 − i )2 − 6(3− i ) +10 = 9 − 6i + i 2 −18 + 6i +10 =0
Section 1.3 1.
x 2 − 2 x − 15 = 0 ( x + 3)( x − 5) = 0 x + 3 = 0 or x = −3
2. x−5 = 0 x=5
x 2 + 3x − 10 = 0 ( x − 2)( x + 5) = 0 or x + 5 = 0 x−2 = 0 x=2 x = −5
2 x2 − x = 1
3.
2
2x − x − 1 = 0 (2 x + 1)( x − 1) = 0 2x + 1 = 0 2x = − 1 x =−1 2
4.
2 x2 + 5x = 3 2x2 + 5x − 3 = 0 (2 x − 1)( x + 3) = 0 x+3= 0 2 x − 1 = 0 or x = −3 2x = 1 1 x= 2
5.
8 x 2 +189 x − 72 = 0 (8 x − 3)( x + 24) = 0
8x − 3 = 0 8x = 3 x=
3 8
or
6. x+2=0 x = −24
or
x −1 = 0 x =1
12 x 2 − 41x + 24 = 0 (4 x − 3)(3x −8) = 0 4 x −3 = 0 or 3 x −8 = 0 4x = 3 3x = 8 3 x= x=8 4 3
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Section 1.3
7.
49
3x 2 − 7 x = 0 x (3 x − 7 ) = 0 x = 0 or 3 x − 7 = 0 3x = 7 7 x= 3
8.
13.
11.
2 x 2 = 48
14.
x=8
x 2 = 81
12.
x=2
x 2 = 225 x = ± 225 x = ± 15
3x 2 = 144
15.
3x 2 + 12 = 0 3x 2 = −12 x 2 = −4
x 2 = 48
x = ± 24
x = ± 48
x = ±2 6
x = ±4 3
4 x 2 + 20 = 0
17.
x = ± −4 x = ± 2i
( x − 5)2 = 36
18.
x − 5 = ± 36 x − 5 = ±6 x = 5±6 x = 5 + 6 or x = 5−6 x = 11 x = −1
2
x = −5 x = ± −5 x = ±i 5 ( x − 3) 2 + 16 = 0
20.
x + 2 = ± −28 x + 2 = ± 2i 7 x = −2 ± 2i 7
x2 + 6x + 1 = 0
22.
x 2 + 6 x + 9 = −1 + 9 ( x + 3)2 = 8 x +3= ± 8
x 2 + 8 x − 10 = 0
23.
x 2 − 2 x + 1 = 15 + 1
( x + 4)2 = 26
( x − 1)2 = 16
x = −4 ± 26 x = −3 − 2 2
x 2 − 2 x − 15 = 0
x 2 + 8 x + 16 = 10 + 16 x + 4 = ± 26
x = −3 ± 2 2 or
x + 4 = ± 121 x + 4 = ± 11 x = −4 ± 11 x = −4 + 11 or x = −4 − 11 x=7 x = −15
( x + 2)2 = −28
x − 3 = ± −16 x − 3 = ± 4i x = 3 ± 4i
x = −3 + 2 2
( x + 4)2 = 121
( x + 2)2 + 28 = 0
( x − 3) 2 = −16
21.
( x − 8)( x − 2) = 0 or x − 2 = 0
3x = −8 8 x=− 3
4 x 2 = −20
19.
x −8 = 0
x = ± 81 x = ±9
x 2 = 24
16.
( x − 5)2 − 9 = 0 [( x − 5) − 3] [( x − 5) + 3] = 0
5 x = −8 8 x=− 5
[(3 x + 4) − 4][(3 x + 4 + 4] = 0 (3x)(3x + 8) = 0 3x = 0 or 3x + 8 = 0
x=0
9.
5x2 + 8x = 0 x(5 x + 8) = 0 x=0 or 5x + 8 = 0
(3 x + 4)2 − 16 = 0
10.
5 x 2 = −8 x
x = −4 + 26
or
x = −4 − 26
x − 1 = ± 16 x =1± 4 x=1+4 or x=5
Copyright © Houghton Mifflin Company. All rights reserved.
x=1–4 x = −3
50
24.
Chapter 1: Equations and Inequalities
x2 + 2 x − 8 = 0
25.
26.
x2 + 2 x + 1 = 8 + 1
x 2 + 4 x + 4 = −5 + 4
( x + 1)2 = 9
( x + 2)2 = −1
x 2 + 3x − 1 = 0 x 2 + 3x + 9 = 1 + 9 4 4
28.
2
2 x2 + 4 x − 1 = 0
or
x2 + 7 x − 2 = 0 x 2 + 7 x + 49 = 2 + 49 4 4
2 x 2 + 10 x − 3 = 0 2 x 2 + 10 x = 3 x 2 + 5x = 3 2 25 2 x + 5x + = 3 + 25 4 2 4 2
x = −1 ± 3 = −1 ± 2 x = −1 ± 6 = −1 ± 4 −2+ 6 , 2
x − 3 = ± −1 x = 3± i x = 3 + i or x = 3−i
⎛ 7⎞ 57 ⎜x+ ⎟ = 2⎠ 4 ⎝ 7 x + = ± 57 2 4 x = − 7 ± 57 2 2 −7 + 57 −7 − 57 x= , or x = 2 2 30.
2 x2 + 4 x = 1 x2 + 2 x = 1 2 1 2 x + 2x + 1 = + 1 2 2 ( x + 1) = 3 2 x +1 = ± 3 2
x=
( x − 3)2 = −1
2
⎛ 3 ⎞ 13 ⎜x+ ⎟ = 2⎠ 4 ⎝ 3 x + = ± 13 2 4 x = − 3 ± 13 2 2 −3 + 13 −3 − 13 x= , or x = 2 2 29.
x 2 − 6 x + 10 = 0 x 2 − 6 x + 9 = −10 + 9
x + 2 = ± −1 x + 2 = ±i x = −2 ± i x = −2 − i or x = −2 + i
x +1 = ± 9 x = −1 ± 3 x = −1 + 3 or x = −1 − 3 x=2 x = −4 27.
x2 + 4 x + 5 = 0
x=
3⋅2 2 2 6 =−2 ± 6 2 2 2
−2− 6 2
⎛ 5⎞ 31 ⎜x+ ⎟ = 2⎠ 4 ⎝ 5 x + = ± 31 2 4 x = − 5 ± 31 2 2 x=
−5 + 31 2
or
x=
Copyright © Houghton Mifflin Company. All rights reserved.
−5 − 31 2
Section 1.3
31.
51
3x 2 − 8 x + 1 = 0
32.
2
4 x 2 − 4 x + 15 = 0 4 x 2 − 4 x = −15 x 2 − x = − 15 4 1 15 2 x −x+ =− +1 4 4 4
3x − 8 x = −1 x2 − 8 x = − 1 3 3 16 1 2 8 x − x + = − + 16 3 9 3 9
2
2
⎛ 1⎞ 14 ⎜x− ⎟ =− 2 4 ⎝ ⎠ x − 1 = ± − 14 2 4 1 i x = ± 14 2 2
⎛ 4 ⎞ 13 ⎜x− ⎟ = 3⎠ 9 ⎝ 4 x − = ± 13 3 9 4 x = ± 13 3 3 x= 33.
4 + 13 3
or
x=
4 − 13 3
x2 − 2 x − 15 = 0, a = 1, b = −2, c = −15
x= 34.
−b ± b2 − 4ac 2a
x=
x=
−(−2) ± (−2)2 − 4(1)(−15) 2(1)
x=
37.
−1 ± 12 − 4(1)(−1) 2(1)
x=
−1 ± 1 + 4 −1 ± 5 = 2 2
x=
−1 + 5 2
or x =
−(−5) ± (−5)2 − 4(1)(−24) 2(1)
x2 + x + 1 = 0 x=
−1 ± 12 − 4(1)(1) 2(1)
−1 ± 1 − 4 −1 ± −3 = 2 2 1 3 1 3 x=− + i or x = − − i 2 2 2 2 x=
−1 − 5 2
2 x2 + 4 x + 1 = 0 x=
1 14 − i 2 2
x = 8 or x = −3
36.
x=
x=
5 ± 25 + 96 5 ± 121 5 ± 11 = = 2 2 2 5 + 11 16 5 − 11 −6 x= = = 8 or x = = = −3 2 2 2 2
2 ± 4 + 60 2 ± 64 2 ± 8 = = 2 2 2 2 + 8 10 2 − 8 −6 x= = =5 or x = = = −3 2 2 2 2 x = 5 or x = −3 x2 + x − 1 = 0
or
x 2 − 5 x − 24 = 0
x=
x=
35.
1 14 + i 2 2
38.
−4 ± 42 − 4(2)(1) 2(2)
2 x2 + 4 x − 1 = 0 x=
x = −4 ± 16 − 8 = −4 ± 8 4 4 4 2 2 2 2 − ± − ± x= = 4 2 x = −2 + 2 , or x = −2 − 2 2 2
−4 ± 42 − 4(2)( −1) 2(2)
x = −4 ± 16 + 8 = −4 ± 24 4 4 − 4 ± 2 6 − 2 ± 6 x= = 4 2 − 2 + 6 x= , or x = −2 − 6 2 2
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52
39.
Chapter 1: Equations and Inequalities
3x 2 − 5 x + 3 = 0 x=
40.
−( −5) ± ( −5)2 − 4(3)(3) 2(3)
x=
1 2 3 x + x −1 = 0 2 4 ⎛1 2 3 ⎞ 4 ⎜ x + x − 1⎟ = 4(0) 4 ⎝2 ⎠
42.
2 x 2 + 3x − 4 = 0 x=
2 2 1 x − 5x + = 0 3 2 1⎞ ⎛2 2 6 ⎜ x − 5 x + ⎟ = 6(0) 2⎠ ⎝3 4 x 2 − 30 x + 3 = 0
−3 ± 32 − 4(2)( −4) 2(2)
x=
x = −3 ± 9 + 32 4 3 41 − ± x= 4 x = −3 + 41 , or x = −3 − 41 4 4
43.
24 x 2 − 22 x − 35 = 0
x=
x = 22 ± 484 + 3360 48 22 3844 ± x= 48 = 22 ± 62 48 5 x = − , or x = 7 6 4
−( −30) ± ( −30)2 − 4(4)(3) 2(4)
30 ± 900 − 48 8 30 ± 852 x= 8 30 ± 2 213 x= 8 15 ± 213 x= 4 x=
44.
−(−22) ± (−22) 2 − 4(24)(−35) 2(24)
−( −5) ± ( −5)2 − 4(3)(4) 2(3)
x = 5 ± 25 − 48 = 5 ± −23 6 6 x = 5 ± i 23 6 x = 5 + 23 i, or x = 5 − 23 i 6 6 6 6
x = 5 ± 25 − 36 = 5 ± −11 6 6 x = 5 ± i 11 6 x = 5 + 11 i, or x = 5 − 11 i 6 6 6 6 41.
3x 2 − 5 x + 4 = 0
72 x 2 +13 x −15 = 0
x=
−13± (13)2 − 4(72)(−15) 2(72)
x = −13± 169 + 4320 144 13 4489 − ± x= 144 = −13± 67 144 5 x = − , or x = 3 9 8
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Section 1.3
45.
53
0.5 x 2 + 0.6 x − 0.8 = 0
x=
46.
−0.6 ± (0.6)2 − 4(0.5)(−0.8)
x=
2(0.5)
x = −0.6 ± 0.36 + 0.6 1 x = −0.6 ± 1.96 = −0.6 ±1.4 x = −2, or x = 4 5
47.
2 x2 − 5x − 7 = 0
x 2 + 3x + 3 = 0
48.
24 x 2 +10 x − 21= 0 2
2
b − 4ac = (10) − 4(24)(−21) =100 + 2016 = 2116 Two real solutions 56.
8x2 −5x +3 = 0 2
2(1.2)
x 2 + 3 x − 11 = 0
51.
b 2 − 4ac = (−2) 2 − 4(3)(10) = 4 − 120 = −116 No real solutions
x 2 − 20 x + 100 = 0
b 2 − 4ac = (12) 2 − 4(4)(9) = 144 − 144 = 0 One real solution
32 x 2 − 44 x +15 = 0 2
4 x 2 + 12 x + 9 = 0
52.
b 2 − 4ac = (−20) 2 − 4(1)(100) = 400 − 400 = 0 One real solution
54.
3x 2 − 2 x + 10 = 0
49.
b 2 − 4ac = (3) 2 − 4(1)(−11) = 9 + 44 = 53 Two real solutions
b 2 − 4ac = (3) 2 − 4(1)(3) = 9 − 12 = −3 No real solutions
53.
−0.4 ± (0.4)2 − 4(1.2)(−0.5)
x = −0.4 ± 0.16 + 2.4 2.4 − 0.4 ± 2.56 x= 2.4 = −0.4 ±1.6 2.4 5 x = − , or x = 1 6 2
b 2 − 4ac = (−5) 2 − 4( 2)(−7) = 25 + 56 = 81 Two real solutions
50.
1.2 x 2 + 0.4 x − 0.5 = 0
12 x 2 +15 x + 7 = 0
55.
2
b 2 − 4ac = (15) 2 − 4(12)(7) = 225 − 336 = −111 No real solutions
b − 4ac = (−44) − 4(32)(15) =1936 −1920 =16 Two real solutions 57.
2
b − 4ac = (−5) − 4(8)(3) = 25 − 96 = −71 No real solutions
31
31
31 2
31 2
2
⎛ 31 ⎞ a 2 + ⎜ ⎟ = 312 ⎝ 2⎠ ⎛ 31 ⎞ a 2 = 312 − ⎜ ⎟ ⎝ 2⎠
2
⎛ 31 ⎞ d = 312 − ⎜ ⎟ ⎝ 2⎠ d ≈ 26.8 in.
Copyright © Houghton Mifflin Company. All rights reserved.
2
54
Chapter 1: Equations and Inequalities
a 2 + b2 = c2
58. 2
2
(90) + (90) = c
59.
2
16, 200 = c 2 127.3 = c The distance is 127.3 ft.
a=4 b 3 a = 4b 3 a 2 + b2 = c2 2
⎛ 4b ⎞ 2 2 ⎜ ⎟ + b = 54 ⎝ 3 ⎠ 16b 2 + b 2 = 2916 9 25b 2 = 2916 9 b 2 =1049.76 b = 32.4 4(32.4) a= = 43.2 3 The TV is 32.4 in. high and 43.2 in. wide.
60.
250,000 = 40,000 + 20 x + 0.0001x 2
61.
19,000 = 38t 2 + 291t + 15,208 0 = 38t 2 + 291t − 3792 a = 38, b = 291, c = −3292
2
0 = 0.0001x + 20 x − 210,000 a = 0.0001, b = 20, c = −210,000 x=
−20 ± 202 − 4(0.0001)( −210,000) 2(0.0001)
t=
= −20 ± 484 = −20 + 22 = 10,000 books 0.0002 0.0002 (Reject negative x-value; x must be positive.)
62.
R = xp 16,500 = x(26 − 0.01x) 16,500 = 26 x − 0.01x
518,000 = −0.01x 2 + 168 x − 120,000 0 = −0.01x 2 + 168 x − 638,000 a = −0.01, b = 168, c = −638,000
2
−26 ± 262 − 4( −0.01)( −16,500) 2(−0.01) −26 ± 16 −26 ± 4 = = −0.02 −0.02 −26 + 4 −26 − 4 x= x= or −0.02 −0.02 −22 −30 = = −0.02 −0.02 = 1100 = 1500 1100 or 1500 items must be sold. x=
= −291 ± 585065 ≈ −291 ± 764.8954 76 76 t ≈ 6.23 or t ≈ −13.89 (not 0 ≤ t ≤ 14) 6.23 years after 1990 is 1996. 63.
0 = −0.01x 2 + 26 x − 16,500 a = −0.01, b = 26, c = −16,500
−291 ± 2912 − 4(38)( −3292) 2(38)
x=
−168 ± 1682 − 4( −0.01)( −638,000) 2( −0.01)
= −168 ± 2704 = −168 ± 52 −0.02 −0.02 168 52 − + or x= x = −168 − 52 −0.02 −0.02 116 220 − − = = 5800 = = 11000 −0.02 −0.02 5800 or 11,000 racquets must be sold.
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Section 1.3
64.
a.
55
Evaluate A = 0.72(1.28)h 2 with h = 7.
65.
Let w = width of region Then 132 − 3w = length. 2 Area = length(width) 576 = 132 − 3w ⋅ w 2 1152 = 132 w − 3w2 3w2 − 132 w + 1152 = 0 3( w2 − 44 w + 384) = 0 w2 − 44 w + 384 = 0 ( w − 32)( w − 12) = 0 or w − 32 = 0 w − 12 = 0 w = 32 w = 12 132 − 3w = 132 − 3(32) 132 − 3w = 132 − 3(12) 2 2 2 2 = 18 = 48 The region is either 32 feet wide and 18 feet long, or 12 feet wide and 48 feet long.
67.
Solve D = −45 x 2 + 190 x + 200 for x with D = 250.
A = 0.72(1.28)(7)2 = 0.72(1.28)(49) ≈ 45.2 square inches b.
Solve A = 0.72(1.28) h 2 for h with A = 92. 92 = 0.72(1.28)h 2 92 = 0.9216h 2 99.82639 ≈ h 2 h ≈ 99.82639 ≈ 10.0 inches
66.
Let x = side of cardboard. Then (x – 6) = length of box. Volume = (length)(width)(height) 126.75 = ( x − 6)( x − 6)(3) 126.75 = ( x 2 − 12 x + 36)(3) 126.75 = 3x 2 − 36 x + 108 0 = 3x 2 − 36 x − 18.75 a = 3, b = −36, c = −18.75
250 = −45 x 2 + 190 x + 200 0 = −45 x 2 + 190 x − 50 a = −45, b = 190, c = −50 x=
−(−36) ± (−36)2 − 4(3)(−18.75) 2(3) 36 ± 1521 36 ± 39 = = 6 6 36 39 75 + x = 36 − 39 = −3 x= = or 6 6 6 6 = −0.5 (no) = 12.5 Since length cannot be a negative number, each side of the cardboard is 12.5 inches.
= −190 ± 27100 ≈ −190 ± 164.2 −90 −90 − + 190 164.2 or x≈ x ≈ −190 − 164.2 −90 −90 − − 25.8 354.2 ≈ ≈ −90 −90 ≈ 0.3 mile ≈ 3.9 miles
x=
68.
Solve N = −5t 2 + 80t − 280 for t with N = 35. 2
−190 ± 1902 − 4( −45)( −50) 2( −45)
69.
Solve h = −16t 2 + 25.3t + 20 for t where h = 17.
35 = −5t + 80t − 280
17 = −16t 2 + 25.3t + 20
0 = −5t 2 + 80t − 315
0 = −16t 2 + 25.3t + 3
0 = −5(t 2 − 16t + 63)
t=
0 = −5(t − 7)(t − 9) t −7=0 t = 7 A.M.
or
t −9=0 t = 9 A.M.
−25.3 ± (25.3) 2 − 4( −16)(3) 2( −16)
−25.3 ± 832.09 −32 t = 1.7 or t = –0.11 He was in the air for 1.7 s. =
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56
70.
Chapter 1: Equations and Inequalities
Original candy bar lwh =V 5⋅ 2⋅0.5 = 5 cubic inches 80%(5) = 0.80(5) = 4 cubic inches Let w represent the width. Let w + 2.5 represent the length. lwh =V ( w + 2.5)( w)(0.5) = 4 ( w + 2.5) w = 8
71.
350 = −16t 2 + 220t 0 = −16t 2 + 220t − 350 a = −16, b = 220, c = −350 t=
w + 2.5w −8 = 0 −2.5 ± (2.5) 2 − 4(1)(−8) 2(1)
= −2.5 ± 38.25 2 w ≈ 4.5 in or w ≈ 1.8 in The dimensions are 1.8 in by 4.5 in by 0.5 in. 72.
When the ball hits the ground, the height is 0.
73.
Solve h = −16t 2 + 52t + 4.5 for t where h = 0. 0 = −16t 2 + 52t + 4.5 a = −16, b = 52, c = 4.5
Solve s = −16t 2 + 26.6t for t where s = 0. 2
0 = −16t + 26.6t 0 = t ( −16t + 26.6) t = 0 or −16t + 26.6 = 0 t ≈ 1.7 seconds
Solve s = 103.9t for t where s = 360 to find the time it takes the ball to reach the fence. 360 = 103.9t t ≈ 3.465 seconds Next, evaluate h = −16t 2 + 50t + 4.5 where t = 3.465 to determine if the ball is at least 10 feet in the air when it reaches the fence.
−52 ± 522 − 4( −16)(4.5) t= 2( −16) 52 2992 − ± ≈ −52 ± 54.699 t= −32 −32 t ≈ −52 − 54.699 −32 ≈ −106.699 −32 ≈ 3.3 seconds 74.
−220 ± 2202 − 4( −16)( −350) 2( −16)
= −220 ± 26000 ≈ −220 ± 161.245 −32 −32 − + 220 161.245 or t ≈ −220 − 161.245 t≈ −32 −32 ≈ −58.755 ≈ −381.245 −32 −32 ≈ 1.8 seconds ≈ 11.9 seconds
2
w=
Solve h = −16t 2 + 220t for t where h = 350.
h = −16(3.465) 2 + 50(3.465) + 4.5 h ≈ −14.3 No, the ball will not clear the fence.
75.
Solve h = 1 n ( n − 1) for n where h = 36. 2 1 36 = n ( n − 1) 2 72 = n ( n − 1) = n 2 − n 0 = n 2 − n − 72 = ( n − 9)( n + 8) or n+8=0 n−9=0 n = −8 (no) n = 9 people
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Section 1.3
76.
57
Solve D = 1.525 x 2 − 21.35 x + 72.225 for x where D = 100.
77.
Solve R = 37.5 x 2 − 225t + 342.5 for t where R = 1000. 1000 = 37.5 x 2 − 225t + 342.5
2
100 = 1.525 x − 21.35 x + 72.225
0 = 37.5 x 2 − 225t − 657.5 a = 37.5, b = −225, c = −657.5
2
0 = 1.525 x − 21.35 x − 27.775 a = 1.525, b = −21.35, c = −27.775 x=
t=
2
−( −21.35) ± ( −21.35) − 4(1.525)( −27.775) 2(1.525)
225 ± 149,250 225 ± 386.33 ≈ 75 75 + 225 386.33 − 225 386.33 t≈ or t ≈ 75 75 ≈ 8.2 ≈ −2.1 (not in the future) =
= 21.35 ± 625.25 ≈ 21.35 ± 25.005 3.05 3.05 x ≈ 21.35 + 25.005 or x ≈ 21.35 − 25.005 3.05 3.05 ≈ 15.198 ≈ −1.198 (not in the future) The year that the NARA will first exceed 100 petabytes is 15 years, after 2000, which is in 2015.
78.
Solve N = −0.015v 2 + 3v for v where N = 100.
Revenue for cellular phone software will first reach $1 billion 8.2 years from 2000, which is in 2008. 79.
a.
2
100 = −0.015v + 3v
If t = 0 represents the year 1995, then the year 2006 is represented by t = 11. Evaluate A = 0.05t 2 + 2.25t + 14 for t = 11.
0 = −0.015v 2 + 3v − 100 a = −0.015, b = 3, c = −100 v=
−225 ± ( −225)2 − 4(37.5)( −657.5) 2(37.5)
A = 0.05(11)2 + 2.25(11) + 14 = 0.05(121) + 2.25(11) + 14 = 6.05 + 24.75 + 14 = 44.8 million pounds
−3 ± 32 − 4( −0.015)( −100) 2( −0.015)
= −3 ± 9 − 6 −0.03 3 − ± 3 = −0.03
b.
v = −3 + 3 v = −3 − 3 −0.03 −0.03 ≈ 42 miles per hour ≈ 158 (not 0 ≤ v ≤ 90)
Solve A = 0.05t 2 + 2.25t + 14 for t where A = 50. 50 = 0.05t 2 + 2.25t + 14 0 = 0.05t 2 + 2.25t − 36 a = 0.05, b = 2.25, c = −36 t=
−2.25 ± 2.252 − 4(0.05)( −36) 2(0.05)
= −2.25 ± 5.0625 + 7.2 0.1 2.25 12.2625 − ± = ≈ −2.25 ± 3.50 0.1 0.1 − 2.25 + 3.50 t≈ t ≈ −2.25 − 3.50 or 0.1 0.1 ≈ −57.5 ≈ 12.5 (not 0 ≤ t ≤ 15) 12.5 years from 1995 will be in 2007.
....................................................... 80.
a.
b 2 − 4ac
Connecting Concepts b.
(−b)2 − 4(1)(−4) = b 2 +16 The discriminant is b 2 +16 , which is positive for any value of b.
b 2 − 4ac (−6) 2 − 4(1)(k ) = 36 − 4k 36 − 4k > 0 −4k > −36 k <9
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58
81.
Chapter 1: Equations and Inequalities
A line extending from A to MN is equal in length to b. Draw a line from A to Z (a point on MP) that is parallel to line OP and perpendicular to MP. Draw a line from A to N.
Since AN is also a radius of the circle, then its length is a . AZ is equal in length to OP, so OP = AZ = b. 2 We have a right triangle AZN, where angle AZN is the 90° angle. Also, ZN = ZP - NP, where ZP = a . Using the Pythagorean 2 Theorem, ( AZ ) 2 + ( NZ ) 2 = ( AN )2
()
b2 + ( ZP − NP )2 = a 2
(
b2 + a − NP 2
)
2
2
2 =a 4
2 2 b2 + a − aNP + ( NP ) 2 = a 4 4
( NP )2 − aNP + b2 = 0 or x 2 − ax + b2 = 0 , where x = NP Thus NP is a solution to x 2 − ab + b2 = 0 . Draw line AM. Since AM is also a radius of the circle, its length is a . 2
We have a right triangle AZM, where angle AZM is the 90° angle. Also, MZ = MP - ZP, where ZP = a . Using the Pythagorean 2 Theorem, ( AZ )2 + ( MZ )2 = ( AM )2
()
b2 + ( MP − ZP )2 = a 2
(
b2 + MP − a 2
)
2
2
2 =a 4
2 2 b2 + ( MP )2 − aMP + a = a 4 4
( MP )2 − aMP + b2 = 0 or x 2 − ax + b2 = 0 , where x = MP. Thus MP is a solution to x 2 − ab + b2 = 0 .
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Section 1.4
82.
59
4 b = − = −4 1 a c −21 ( −7)(3) = −21; = = −21 a 1
Yes, –7 and 3 are roots of x 2 + 4 x − 21 = 0.
84.
Yes,
(−12) 4 2+ 5 2− 5 4 b + = ; − =− = 9 3 3 3 3 a
86.
2+ 5 2− 5 and are roots of 9 x 2 − 12 x − 1 = 0. 3 3
−5 and 6 are roots of 2 x 2 − 7 x − 30 = 0. 2
b ( −2 ) =− =2 a 1 c ( 2) (1 + i )(1 − i ) = 1 − i 2 = 1 + 1 = 2; = =2 a 1
85.
⎛ 2 + 5 ⎞⎛ 2 − 5 ⎞ 4 − 5 1 c − 1 1 ⎜ ⎟⎜ ⎟= − ; = − ⎜ 3 ⎟⎜ 3 ⎟ 9 9 a 9 9 ⎝ ⎠⎝ ⎠ Yes,
( −7 ) 7 7 −5 b +6 = ; − = − = 2 2 2 2 a c ( −30) ⎛ −5 ⎞ = −15 ⎜ ⎟(6 ) = −15; = a 2 ⎝ 2 ⎠
83.
− 7 + 3 = −4; −
(1 + i ) + (1 − i ) = 2; −
Yes, 1 + i and 1 − i are roots of x 2 − 2 x + 2 = 0.
b ( −4) =− =4 a 1 c 12 = = 12 (2 + 3i )(2 − 3i ) = 4 − 9i 2 = 4 + 9 = 13; a 1
(2 + 3i ) + (2 − 3i ) = 4; −
No, 2 +3i and 2 −3i are not roots of x 2 − 4 x + 12 = 0.
....................................................... PS1. x 3 − 16 x
PS2. x 4 − 36 x 2
Prepare for Section 1.4 PS3. 82 / 3 = ( 3 8)2
x ( x 2 − 16)
x 2 ( x 2 − 36)
= 22
x ( x + 4)( x − 4)
x 2 ( x + 6)( x − 6)
=4
PS4. 163/ 2 = ( 16)3
PS5. (1 + x − 5)2
PS6. (2 − x + 3)2
= 43
12 + 2 x − 5 + ( x − 5)2
22 − 2(2) x + 3 + ( x + 3)2
= 64
1+ 2 x − 5 + x − 5
4−4 x+3+ x+3
x+2 x−5 −4
x−4 x+3+7
Section 1.4 1.
x3 − 25 x = 0
2.
x( x 2 − 25) = 0 x( x − 5)( x + 5) = 0 x = 0, x = 5, or x = −5
4.
x3 − 4 x2 − 2 x +8 = 0 x 2 ( x − 4) − 2( x − 4) = 0 2
( x − 4)( x − 2) = 0 x−4=0 x=4
or x 2 − 2 = 0
x3 − x = 0 x( x 2 −1) = 0 x( x −1)( x +1) = 0 x = 0, x = 1, or x = −1
5.
2 x5 −18 x3 = 0
x3 − 2 x 2 − x + 2 = 0
3.
x 2 ( x − 2) − ( x − 2) = 0 ( x − 2)( x 2 −1) = 0 ( x − 2)( x −1)( x +1) = 0 x = 2, x = 1, or x = −1 x 4 − 36 x 2 = 0
6.
2 x3 ( x 2 − 9) = 0
x 2 ( x 2 − 36) = 0
2 x3 ( x − 3)( x + 3) = 0
x 2 ( x − 6)( x + 6) = 0
x = 0, x = 3, or x = −3
x = 0, x = 6, or x = −6
2
x =2 x=± 2
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60
7.
Chapter 1: Equations and Inequalities
x 4 − 3x3 − 40 x 2 = 0
8.
x 4 + 3 x3 − 8 x − 24 = 0
x 2 ( x 2 − 3x − 40) = 0
x3 ( x + 3) −8( x + 3) = 0
x 2 ( x + 5)( x − 8) = 0
( x + 3)( x3 − 8) = 0 ( x + 3)( x − 2)( x 2 + 2 x + 4) = 0
x = 0, x = −5, or x = 8
x = −3, x = 2, or x =
−2 ± 22 − 4(1)(4) 2
−2 ± 4 − 16 2 −2 ± −12 x= 2 −2 ± 2i 3 x= 2 x = −1 ± i 3 x=
x = −1 + i 3 or x = −1 − i 3 Thus the solutions are − 3, 2, − 1 + i 3, − 1 − i 3. 9.
x 4 −16 x 2 = 0
x 4 −16 = 0
10.
x 2 ( x 2 −16) = 0
( x 2 − 4)( x 2 + 4) = 0
x 2 ( x − 4)( x + 4) = 0
( x − 2)( x + 2)( x 2 + 4) = 0
x = 0, x = 4, or x = −4
x = 2, x = −2, or x 2 = −4 x=± −4 x = 2i, x = −2i Thus the solutions are 2, −2, 2i, −2i.
11.
x3 −8 = 0
12.
( x − 2)( x 2 + 2 x + 4) = 0
x3 + 8 = 0 ( x + 2)( x 2 − 2 x + 4) = 0
x = 2, or x 2 + 2 x + 4 = 0 −2 ± 22 − 4(1)(4) 2 −2 ± −12 −2 ± 2i 3 = = −1 ± i 3 x= 2 2 x = −1 + i 3 or x = −1 − i 3
x = −2 or x =
x=
x=
−(−2) ± (−2)2 − 4(1)(4) 2(1) 2 ± 4 − 16 2 ± 2i 3 = =1± i 3 2 2
x = −2, x = 1 + i 3, or x = 1 − i 3
Thus the solutions are 2, − 1 + i 3, − 1 − i 3 13.
3 5 = x + 2 2x − 7
3(2 x − 7) = 5( x + 2) 6 x − 21 = 5 x + 10 6 x − 5 x = 10 + 21 x = 31
14.
4 7 = y+2 y−4 4( y − 4) = 7( y + 2) 4 y − 16 = 7 y + 14 4 y − 7 y = 14 + 16 − 3 y = 30 y = −10
15.
30 20 = 10 + x 10 − x
30(10 − x) = 20(10 + x) 300 − 30 x = 200 + 20 x − 30 x − 20 x = 200 − 300 − 50 x = −100 x=2
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Section 1.4
16.
61
6 4 = 8+ x 8− x
17.
6(8 − x) = 4(8 + x) 48 − 6 x = 32 + 4 x − 6 x − 4 x = 32 − 48 − 10 x = −16 8 x= 5
18.
12 3x = 2− x+4 x+4
12 ⎞ ⎛ 3x ⎞ ⎛ ( x + 4) ⋅ ⎜ ⎟ = ( x + 4) ⋅ ⎜ 2 − ⎟ x+4⎠ ⎝ x+4⎠ ⎝ 3x = 2( x + 4) − 12 3x = 2 x + 8 − 12 3x = 2 x − 4 3x − 2 x = −4 x = −4 No solution because each side is undefined when x = −4.
8 1 5 − = 2 m + 1 m − 2 2m + 1
1 ⎞ ⎛ 8 ⎛ 5 ⎞ − (2m + 1)( m − 2) ⎜ ⎟ = (2m + 1)(m − 2) ⎜ ⎟ ⎝ 2m + 1 m − 2 ⎠ ⎝ 2m + 1 ⎠ 8(m − 2) − 1(2m + 1) = 5( m − 2) 8m − 16 − 2m − 1 = 5m − 10 6m − 17 = 5m − 10 6m − 5m = −10 + 17 m=7 19.
2 + 9 = 3r r−3 r−3 ⎛ ⎞ ⎛ ⎞ ( r − 3) ⋅ ⎜ 2 + 9 ⎟ = ( r − 3) ⋅ ⎜ 3r ⎟ r − 3⎠ ⎝ ⎝ r − 3⎠ 2( r − 3) + 9 = 3r 2 r − 6 + 9 = 3r 2 r + 3 = 3r 2 r − 3r = −3 − r = −3 r=3 No solution because each side is undefined when r = 3.
20.
t +3= 4 t−4 t−4 ⎛ t ⎞ ⎛ ⎞ + 3⎟ = (t − 4 ) ⋅ ⎜ 4 ⎟ (t − 4) ⋅ ⎜ ⎝t−4 ⎠ ⎝t −4⎠ t + 3(t − 4) = 4 t + 3t − 12 = 4 4t − 12 = 4 4t = 4 + 12 4t = 16 t=4 No solution because each side is undefined when t = 4.
21.
5 − 3 = 4 x−3 x−2 x−3 ⎛ ⎞ ⎛ ⎞ ( x − 3)( x − 2) ⋅ ⎜ 5 − 3 ⎟ = ( x − 3)( x − 2) ⋅ ⎜ 4 ⎟ ⎝ x −3 x −2⎠ ⎝ x − 3⎠ 5( x − 2) − 3( x − 3) = 4( x − 2) 5x − 10 − 3x + 9 = 4 x − 8 2x − 1 = 4x − 8 2 x − 4 x = −8 + 1 −2 x = −7 x=7 2
22.
4 + 7 = 5 x −1 x + 7 x −1 ( x − 1)( x + 7 ) ⋅ ⎛⎜ 4 + 7 ⎞⎟ = ( x − 1)( x + 7 ) ⋅ ⎛⎜ 5 ⎞⎟ ⎝ x −1 x + 7 ⎠ ⎝ x −1⎠ 4( x + 7) + 7( x − 1) = 5( x + 7) 4 x + 28 + 7 x − 7 = 5 x + 35 11x + 21 = 5 x + 35 11x − 5 x = 35 − 21 6 x = 14 x=7 3
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62
23.
Chapter 1: Equations and Inequalities
x = x+4 x−3 x+2 x ( x + 2) = ( x + 4)( x − 3)
24.
x 2 + 2 x = x 2 + x − 12 2 x − x = −12 x = −12
25.
x 2 + x = x 2 + 2 x − 35 x − 2 x = −35 − x = −35 x = 35
x+3 = x−3 x+5 x−4 ( x + 3)( x − 4) = ( x − 3)( x + 5)
26.
x−4 −6=0
28.
x−4 =6 x − 4 = 36 x = 40
Check
10 − x = 4 10 − x = 16 −x = 6 x = −6 Check
40 − 4 − 6 = 0
16 = 4
6−6=0
4=4 The solution is –6.
x = 3+ 3− x
x = 5− x +5
30.
2
( x − 5) = ( 5 − x )2
x − 3= 3− x 2
( x − 3) = ( 3 − x )
2
2
x − 10 x + 25 = 5 − x
2
x 2 − 9 x + 20 = 0 ( x − 5)( x − 4) = 0 x = 5 or x = 4
x − 6x + 9 = 3 − x x 2 − 5x + 6 = 0 ( x − 3)( x − 2) = 0 x = 3 or x = 2
Check Check
10 − (−6) = 4
36 − 6 = 0 0=0 The solution is 40.
29.
x − 6 = x −1 x+4 x+2 ( x − 6)( x + 2) = ( x − 1)( x + 4) x 2 − 4 x − 12 = x 2 + 3x − 4 −4 x − 3x = −4 + 12 −7 x = 8 x = −8 7
x 2 − x − 12 = x 2 + 2 x − 15 − x − 2 x = −15 + 12 −3x = −3 x =1
27.
x = x+7 x − 5 x +1 x ( x + 1) = ( x + 7)( x − 5)
3 = 3+ 3− 3
3 = 3+ 0 3= 3 The solution is 3.
2 = 3+ 3− 2 2 = 3 +1 2 = 4 (No)
5 = 5−5 + 5
5= 0+5 5=5 The solution is 5.
Copyright © Houghton Mifflin Company. All rights reserved.
4 = 5− 4 + 5 4 = 1+ 5 4 = 6 (No)
Section 1.4
31.
63
3x − 5 − x + 2 = 1
32.
2
( 3x − 5) = (1 + x + 2)
2
( 5 x + 6)2 = (6 − 6 − x )2
3x − 5 = 1 + 2 x + 2 + x + 2
5 x + 6 = 36 − 12 6 − x + 6 − x
2x − 8 = 2 x + 2
6 x − 36 = −12 6 − x
( x − 4) 2 = ( x + 2)2
( x − 6)2 = (−2 6 − x )2
2
x − 8 x + 16 = x + 2
x 2 − 12 x + 36 = 4(6 − x)
x 2 − 9 x + 14 = 0 ( x − 7)( x − 2) = 0
x 2 − 12 x + 36 = 24 − 4 x x 2 − 8 x + 12 = 0 ( x − 6)( x − 2) = 0
x = 7, or x = 2 Check
x = 6, or x = 2
3(7) − 5 − 7 + 2 = 1
1=1
Check 6 − 6 + 30 + 6 = 6 0+6 = 6 6=6
3(2) − 5 − 2 + 2 = 1
6 − 2 + 10 + 6 = 6
16 − 9 = 1 4−3 =1
4 + 16 = 6 2+4=6 6=6 Both 6 and 2 check as solutions.
1 − 4 =1 1− 2 = 1 − 1 = 1 (No)
The solution is 7. 33.
6 − x + 5x + 6 = 6
2 x + 11 − 2 x − 5 = 2
34.
( 2 x + 11)2 = (2 + 2 x − 5)2 2 x + 11 = 4 + 4 2 x − 5 + 2 x − 5 12 = 4 2 x − 5 (3)2 = ( 2 x − 5)2 9 = 2x − 5 14 = 2 x 7=x
( x + 7 − 2) 2 = ( x − 9)2 x+7−4 x+7 +4= x−9 −4 x + 7 = −20 ( x + 7) 2 = (5)2 x + 7 = 25 x = 18 Check 18 + 7 − 2 = 18 − 9 25 − 2 = 9
Check 2(7) + 11 − 2(7) − 5 = 2 25 − 9 = 2 5− 3= 2 2=2 7 checks as the solution.
5− 2 = 3 3= 3 18 checks as a solution.
Copyright © Houghton Mifflin Company. All rights reserved.
64
35.
Chapter 1: Equations and Inequalities
x +7 + x−5 =6
x = 12 x − 35
36.
2
( x + 7) = (6 − x − 5)
2
2
x = ( 12 x − 35)2 x 2 = 12 x − 35
x + 7 = 36 − 12 x − 5 + x − 5
x 2 − 12 x + 35 = 0 ( x − 5)( x − 7) = 0 x = 5, or x = 7
12 x − 5 = 24 2
( x − 5) = (2) x −5= 4 x=9
2
Check 5 = 12(5) − 35
Check 9 + 7 + 9 − 5 = 6 16 + 4 = 6 4+2 =6 6=6 9 checks as a solution.
(2x)
2
38.
4 x 2 = 4 x + 15 4 x − 4 x − 15 = 0 (2 x + 3)(2 x − 5) = 0 3 5 x = − , or x = 2 2
7 = 49 7=7
3
7x − 3 = 3 2x + 7
3 11 = 3 11
The solution is 2.
⎡5⎤ ⎡5⎤ 2 ⎢ ⎥ = 4 ⎢ ⎥ + 15 ⎣2⎦ ⎣2⎦
−3 = −6 + 15
5 = 10 + 15
−3 = 9
5 = 25 5=5
−3 = 3
3
5 = 25 5=5
Check 3 7(2) − 3 = 3 2(2) + 7
⎡ 3⎤ ⎡ 3⎤ Check 2 ⎢− ⎥ = 4 ⎢− ⎥ + 15 ⎣ 2⎦ ⎣ 2⎦
39.
7 = 84 − 35
( 3 7 x − 3)3 = ( 3 2 x + 7)3 7x − 3 = 2x + 7 5 x = 10 x=2
= ( 4 x + 15)2
2
The solution is
5 = 60 − 35
5 and 7 check as solutions
2 x = 4 x + 15
37.
(No)
5 . 2
40.
3
2 x 2 + 5x − 3 = x 2 + 3 3
⎡ 3 2 x 2 + 5x − 3 ⎤ = ⎡ 3 x 2 + 3 ⎤ ⎢⎣ ⎥⎦ ⎣⎢ ⎦⎥
4
x 2 + 20 = 4 9 x 4
3
⎡⎣ 4 x 2 + 20 ⎤⎦ = ( 4 9 x )4 x 2 + 20 = 9 x
2 x 2 + 5x − 3 = x 2 + 3
x 2 − 9 x + 20 = 0 ( x − 4)( x − 5) = 0 x = 4, or x = 5
x 2 + 5x − 6 = 0 ( x − 1)( x + 6) = 0 x = 1, or x = −6 Check 3 2(1)2 + 5 − 3 = 3 (1) 2 + 3
Check
4 2
4 + 20 = 4 9(4) 4 36 = 4 36
34 =34 3
2(−6) 2 + 5(−6) − 3 = 3 (−6) 2 + 3 3 39 = 3 39
The solutions are 1 and –6.
7 = 12(7) − 35
4 2
5 + 20 = 4 9(5) 4 45 = 4 45
The solutions are 4 and 5.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.4
65
x1/ 3 = 2
41.
( x1/ 3 )3 = ( 2 )3
( x1/ 2 )2 = ( 5)2
x =8
x = 25
x2 / 5 = 9
43.
x1/ 2 = 5
42.
x 4 / 3 = 81
44.
( x 2 / 5 )5 / 2 = ( 9 )5 / 2
( x 4 / 3 )3 / 4 = (81)3 / 4
x = 243 x = −243, 243 x 3 / 2 = 27
45.
(x 47.
x = 27 x = −27, 27
)
3/ 2 2 / 3
= ( 27 ) x=9
x 3 / 4 = 125
46.
( x 3 / 4 )4 / 3 = (125)4 / 3
2/3
x = 625
2 x 2 / 3 − 16 = 59 2 x 2 / 3 = 75 x 2 / 3 = 25
48.
( x 2 / 3 )3 / 2 = ( 25)3 / 2
4 x 4 / 5 − 27 = 37 4 x 4 / 5 = 64 x 4 / 5 = 16
( x 4 / 5 )5 / 4 = (16 )5 / 4
x = 125 x = −125, 125
49.
x = 32 x = −32, 32
2 x 3 / 2 − 31 = 23 2 x 3 / 2 = 54 x 3 / 2 = 27
50.
( x 3 / 2 )2 / 3 = ( 27 )2 / 3
( x 3 / 5 )5 / 3 = (8)5 / 3
x=9
51.
x = 32
4 x 3 / 4 − 31 = 77 4 x 3 / 4 = 108 x 3 / 4 = 27
52.
x = 81
x = 243 x = −243, 243
x 4 − 9 x 2 + 14 = 0
54.
u 2 − 9u + 14 = 0
2
x =7 x=± 7
x 4 − 10 x 2 + 9 = 0
Let u = x 2 .
Let u = x 2 . (u − 7)(u − 2) = 0 or u=7
4 x 4 / 5 − 54 = 270 4 x 4 / 5 = 324 x 4 / 5 = 81
( x 4 / 5 )5 / 4 = (81)5 / 4
( x 3 / 4 )4 / 3 = ( 27 )4 / 3
53.
3 x 3 / 5 + 25 = 49 3 x 3 / 5 = 24 x3 / 5 = 8
u 2 − 10u + 9 = 0 (u − 9)(u − 1) = 0 u=2
u =9
2
2
x =2
x =9 x = ±3
x=± 2
The solutions are 7, − 7,
2, − 2.
or
u =1 2
x =1 x = ±1
The solutions are 3, − 3, 1, − 1.
Copyright © Houghton Mifflin Company. All rights reserved.
66
55.
Chapter 1: Equations and Inequalities
2 x 4 − 11x 2 + 12 = 0
56.
Let u = x 2 .
2u 2 − 11u + 12 = 0
6u 2 − 7 u + 2 = 0 (2u − 1)(3u − 2) = 0
(2u − 3)(u − 4) = 0 u= 3 2 2 3 x = 2
u= 1 2 2 1 x = 2
u=4
or
x2 = 4
x=± 3 =± 6 2 2 The solutions are
57.
6 x4 − 7 x2 + 2 = 0
Let u = x .
2
x=± 2 =± 6 3 3
x=± 1 =± 2 2 2
x = ±2 6 , − 6 , 2, − 2. 2 2
x 6 + x3 − 6 = 0
The solutions are
58.
Let u = x3.
2, − 2, 6, − 6. 2 2 3 3
6 x 6 + x 3 − 15 = 0
Let u = x 3 . 6u 2 + u − 15 = 0
u2 + u − 6 = 0 (u − 2)(u + 3) = 0 u=2
u=2 3 2 2 x = 3
or
(2u − 3)(3u + 5) = 0 or u= 3 2 x3 = 3 2 x = 3 3 = 3 3⋅ 4 2 2⋅4
u = −3
or
3
3
x =2
x = −3
x=32
x = 3 −3 = − 3 3
The solutions are 3 2 and − 3 3.
u=−5 3 3 5 x =− 3 x = − 3 5 = − 3 5⋅ 9 3⋅ 9 3
3 x = 12 2
3 x = − 45 3 3 3 45 12 The solutions are and − . 2 3
59.
x1 / 2 − 3 x1 / 4 + 2 = 0
60.
2 x1 / 2 − 5 x1 / 4 − 3 = 0
Let u = x1 / 4 .
Let u = x1 / 4 .
u 2 − 3u + 2 = 0 (u − 1)(u − 2) = 0 or u =1
2u 2 − 5u − 3 = 0 (2u + 1)(u − 3) = 0 1 or u=− 2 1 x1 / 4 = − 2
1/ 4
u=2 1/ 4
=1 x =2 x =1 x = 16 The solutions are 1 and 16. x
u =3 x1 / 4 = 3 x = 81
1 ⎞ ⎛ 1/ 4 since x1 / 4 ≥ 0 ⎟ =/ − ⎜x 2 ⎠ ⎝ The solution is 81.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.4
61.
67
3x 2 / 3 − 11x1 / 3 − 4 = 0
62.
Let u = x1 / 3.
Let u = x1 / 3. 6u 2 − 7u − 20 = 0
3u 2 − 11u − 4 = 0 (3u + 1)(u − 4) = 0 1 or u=− 3 1 x1 / 3 = − 3 1 x=− 27 The solutions are − 63.
(3u + 4)(2u − 5) = 0 4 3 4 1/ 3 x =− 3 64 x=− 27
u=4
u=−
x = 64 1 and 64. 27
9 x 4 = 30 x 2 − 25
64.
4 x 4 − 28 x 2 = −49 Let u = x 2 . 4u 2 − 28u + 49 = 0
(3u − 5)(3u − 5) = 0
(2u − 7)(2u − 7) = 0
u=5 3 x2 = 5 3
u=7 2 x2 = 7 2 x = ± 7 = ± 14 2 2
x = ± 5 = ± 15 3 3
The solutions are
15 and − 15 . 3 3
x2 / 5 − 1 = 0 x
2 / 5 5/ 2
)
14 and − 14 . 2 2
The solutions are
66.
2 x 2 / 5 − x1 / 5 = 6 Let u = x1 / 5.
=1 5/ 2
= ± (1) x = ±1 The solutions are 1 and − 1.
(x
5 2
x1 / 3 =
9u 2 − 30u + 25 = 0
2/5
u=
or
5 2 125 x= 8 64 125 The solutions are − and . 27 8
x1 / 3 = 4
Let u = x 2 .
65.
6 x 2 / 3 − 7 x1 / 3 − 20 = 0
2u 2 − u − 6 = 0 (2u + 3)(u − 2) = 0 3 2 3 1/ 5 x =− 2 u=−
⎡ 3⎤ x = ⎢− ⎥ ⎣ 2⎦ 243 x=− 32
u=2
or
5
The solutions are −
x1 / 5 = 2 x = 32
243 and 32. 32
Copyright © Houghton Mifflin Company. All rights reserved.
68
67.
Chapter 1: Equations and Inequalities
9 x − 52 x + 64 = 0
68.
Let u = x .
x = u.
Let 2
8u 2 − 38u + 9 = 0 (4u − 1)(2u − 9) = 0
9u − 52u + 64 = 0 (9u − 16)(u − 4) = 0 16 9 16 x= 9 x = 256 81 u=
1 4 1 x= 4 x= 1 16
9 2 9 x= 2 x = 81 4 1 81 The solutions are and . 16 4 u=
u=4
or
x =4 x = 16
The solutions are
69.
256 and 16. 81
Let x = the number of hours the assistant would take to build the fence working alone.
70.
of the job per hour. 1 worker 8 1 x
assistant
u=
Let x = the additional number of hours for the assistant to finish the job. In two hours, 1 of the job is done, 2 of the job is left to do.
x
8
or
3
The worker does 1 of the job per hour; the assistant does 1 5
1 x=2 14 3 2 x = ⋅ 14 3 28 x= = 9 1 hours 3 3
5
( 18 ) (5) + ( 1x ) (5) = 1 5 + 5 =1 8 x 8 x 5 + 5 = 1(8 x ) 8 x 5 x + 40 = 8 x 40 = 3x
(
8 x − 38 x + 9 = 0
)
40 = x 3 x = 13 13 hours
71.
Let x = number of rounds the golfer needs to play. 4(92) + 86 x 88 = 4+x 88(4 + x ) = 368 + 86 x 352 + 88 x = 368 + 86 x 2 x = 16 x = 8 rounds
72.
Let A = the age of the child. 1= A 2 A + 12 A + 12 = 2 A 12 = A The child is 12 years old.
73.
SMOG = w + 3
74.
SMOG = w + 3 4= w +3
6= w +3 3= w 9= w 9 words with three or more syllables.
1= w 1= w 1 word with three or more syllables.
Copyright © Houghton Mifflin Company. All rights reserved.
3
Section 1.4
69
75.
76.
L =π r
1 h= d 3 1 V= π 3 1 192 = π 3 2 192 = π 9 192(9) = r3 2π
r 2 + h2 2
15π = π r r + 4
2
15 = r r 2 + 16 225 = r 2 ( r 2 + 16) 4
2
0 = r + 16r − 225 Let u = r 2 .
u 2 + 16u − 225 = 0
⎡2 ⎤ r2 ⎢ r⎥ ⎣3 ⎦
r2
r ≈ 6.50 in. diameter d = 2 r ≈ 13.0 in.
78.
T = 2π
L 32
4 = 2π
L 32
L 32 4 L = 2 32 π 32(4) L= 2
π
d1 = 8 mm, d 2 = 12 mm 4 Vc = s3 Vs = π r 3 3 4 3 4 3 4 s = π (4) + π (6)3 = π (64 + 216) ≈ 1172.86 3 3 3 s ≈ 10.5 mm
r 2h
275 ≈ r 3 6.50 ≈ r
(u + 25)(u − 9) = 0 u = 9 or u = −25 (No) r2 = 9 r=3 The radius is 3 in. 77.
1 2 = (2 r ) = r 3 3
=
π2
L ≈ 12.969 The length is 13.0 ft (to the nearest tenth).
The side is approximately 10.5 mm. 79.
d =1.5
h
14 =1.5 h 2 (14) = h 3 28 = h 3 784 = h 9 The height is approximately 87 ft. Copyright © Houghton Mifflin Company. All rights reserved.
70
Chapter 1: Equations and Inequalities
....................................................... 80.
a.
s=
1 1 (a + b + c) = (5 + 6 + 7) = 9 2 2
r=
( s − a )( s − b)( s − c ) s
r=
(9 − 5)(9 − 6)(9 − 7) (4)(3)(2) = 9 9
Connecting Concepts 81.
a.
r = 24 ≈ 1.63 9 The radius is approximately 1.63 inches. b.
s= r=
1 1 (a + b + c) = (10 + 7 + 15) = 16 2 2 abc r= 4 s ( s − a )( s − b)( s − c ) s=
r=
(10)(7)(15) 4 16(16 − 10)(16 − 7)(16 − 15)
r=
10(7)(15) 10(7))(15) = ≈ 8.93 4 16(6)(9)(1) 4 ⋅ 4 ⋅ 6 ⋅ 3
The radius is approximately 8.93 in. b.
3 1 (a + a + a) = a 2 2 ( s − a )( s − b)( s − c ) s
⎡ 3 a − a⎤ ⎡ 3 a − a⎤ ⎡ 3 a − a⎤ ⎢ ⎦⎥ ⎣⎢ 2 ⎦⎥ ⎣⎢ 2 ⎦⎥ 2 = ⎣2 3a 2
a = side, a = b = c, r = 5 1 1 3a s = (a + b + c)= (a + a + a) = 2 2 2 abc a⋅a⋅a r= = 4 s ( s − a )( s − b)( s − c ) 4 s ( s − a)( s − a )( s − a ) 5=
a⋅ a⋅a 3 2 2= 2 2 2 = a = a 3a 12a 12 2 a 2= a=4 3 2 3
5=
a ⋅a ⋅a 4 3 a ⎡ 3 a − a⎤ ⎡ 3 a − a⎤ ⎡ 3 a − a⎤ 2 ⎣⎢ 2 ⎦⎥ ⎣⎢ 2 ⎦⎥ ⎢⎣ 2 ⎦⎥ 3 3 a3 = a = a2 4 4a 3 4 3 a ⎡ a ⎤ ⎡ a ⎤ ⎡ a ⎤ 4 3a 4 2 ⎣⎢ 2 ⎦⎥ ⎣⎢ 2 ⎦⎥ ⎢⎣ 2 ⎦⎥ 16
a =5 3 Each side is 5 3 in.
Each side is 4 3 in. 82.
T= s + s 4 1100 7100 T= + 7100 4 1100 T ≈ 27.5 seconds
83.
T= s 4
+ s 1100
4400T = 1100 s + 4 s Let
s = u. 0 = 4u 2 + 1100u − 4400T −1100 ± (1100)2 − 4(4)( −4400)T 8 −1100 ± 1,210,000 + 70,400T u= 8 −1100 + 1,210,000 + 70,400T s= 8 u=
⎛ −1100 + 1,210,000 + 70,400T ⎞ s=⎜ ⎟ 8 ⎝ ⎠ ⎛ −275 + 5 3025 + 176T ⎞ s=⎜ ⎟ 2 ⎝ ⎠
Copyright © Houghton Mifflin Company. All rights reserved.
2
2
Section 1.5
84.
71
T =3 ⎡ − 275 + 5 3025 + 176(3) ⎤ s=⎢ ⎥ 2 ⎥⎦ ⎢⎣
2
s ≈ 11.51762 ≈ 132.65 The distance is approximately 133 ft.
.......................................................
Prepare for Section 1.5 PS2. 3( −3)2 − 2( −3) + 5 = 38
PS1. {x | x > 5} PS3.
7+3 =2 7−2
PS4. 10 x 2 + 9 x − 9 = (3x + 5)(5 x − 3)
PS5.
x −3 , 2x −7≠0 2x −7
PS6. 2 x 2 − 11x + 15 = 0 (2 x − 5)( x − 3) = 0
It is undefined for x = 7 . 2
2x − 5 = 0 x=5 2
x −3=0 x=3
Section 1.5 1.
2 x + 3 < 11 2 x < 11 − 3 2x < 8 x<4
3x − 5 > 16
2.
−3( x + 2) ≤ 5 x + 7 −3x − 6 ≤ 5 x + 7 −8 x ≤ 13 x ≥ − 13 8
−4( x − 5) ≥ 2 x + 15 −4 x + 20 ≥ 2 x + 15 −6 x ≥ −5 x≤5 6
6.
⎧ 13 ⎫ ⎨x x ≥ − ⎬ 8⎭ ⎩
9.
4 x + 1 > −2 4 x > −3 3 x>− 4
⎧ ⎨x x ≤ ⎩
and and and
4 x + 1 ≤ 17 4 x ≤ 16 x≤4
⎧ ⎧ ⎫ 3⎫ 3 ⎨ x x > − ⎬ ∩ { x x ≤ 4} = ⎨ x − < x ≤ 4 ⎬ 4⎭ 4 ⎩ ⎩ ⎭
x + 4 > 3x + 16
4.
−2 x > 12 x < −6
{ x x > 7}
{x x < 4}
5.
3.
3x > 21 x>7
{ x x < −6}
7.
⎧ 5⎫ ⎨x x < − ⎬ 3⎭ ⎩
−4(3x − 5) > 2( x − 4) −12 x + 20 > 2 x − 8 −14 x > −28
8.
x<2
{ x x < 2}
5⎫ ⎬ 6⎭
10.
3( x + 7) ≤ 5(2 x − 8) 3x + 21 ≤ 10 x − 40 −7 x ≤ −61 x ≥ 61 7 ⎧ 61 ⎫ ⎨x x ≥ ⎬ 7⎭ ⎩
2 x + 5 > −16 2 x > −21 21 x>− 2
{
5x + 6 < 2 x + 1 3 x < −5 x<−5 3
and and
2x + 5 < 9 2x < 4
and
x<2
}
{
}
x x > − 21 ∩ { x x < 2} = x − 21 < x < 2 2 2
Copyright © Houghton Mifflin Company. All rights reserved.
72
11.
Chapter 1: Equations and Inequalities
10 ≥ 3 x − 1 ≥ 0 11 ≥ 3 x ≥ 1 11 1 x ≥ ≥ 3 3
12.
{x − 3 ≤ x ≤ 24}
⎧ 1 11 ⎫ ⎨x ≤ x ≤ ⎬ 3⎭ ⎩ 3
13.
x + 2 < −1 x < −3
or x + 3 ≥ 2 or x ≥ −1
14.
{ x x < −3}∪{ x x ≥−1} ={ x x <−3 15.
−4 x + 5 > 9 −4 x > 4 x < −1
or or or
4x + 1 < 5 4x < 4 x <1
or x ≥−1}
16.
2 x − 1 < −4 or 2 x − 1 > 4 2 x < −3 or 2x > 5 3 5 or x<− x> 2 2 3⎞ ⎛5 ⎞ ⎛ ⎜ − ∞, − ⎟ ∪ ⎜ , ∞ ⎟ 2⎠ ⎝2 ⎠ ⎝ 19.
4 − 5 x ≥ 24 4 − 5 x ≤ −24 or 4 − 5 x ≥ 24 − 5 x ≤ −28 − 5 x ≥ 20 28 x≥ x ≤ −4 5 28 ( −∞, − 4] ∪ ⎢⎡ , ∞ ⎞⎟ ⎣5 ⎠
x − 10 ≥ 2 or x − 10 ≥ 2
x≤8
x ≥ 12
(−∞, − 8] ∪ [12, ∞) 22.
2x − 5 ≥1 2x − 5 ≤ −1 or 2 x − 5 ≥ 1
−14 ≤ 3 x − 10 ≤ 14 − 4 ≤ 3 x ≤ 24 4 x ≤8 − ≤ 3 ⎡ 4 ⎤ ⎢− 3 , 8⎥ ⎦ ⎣ 23.
2 x < 16 x <8
x − 10 ≤ −2
x≥2
3x − 10 ≤ 14
or x > 3}
3x − 1 ≤ 5 3x ≤ 6 x≤2
(1, 8)
(−∞, − 8] ∪ [2, ∞ ) 21.
or or or
2x − 9 < 7 2< 1<
x + 3 ≤ −5 or x + 3 ≥ 5 x ≤ −8
2 x − 7 ≤ 15 2 x ≤ 22 x ≤ 11
−7 < 2 x − 9 < 7
20.
x+3 ≥5
or x + 2 ≤ 3 x ≤1 or
{x x ≤ 11}∪ {x x ≤ 2} = {x x ≤ 11} 18.
2 x −1 > 4
x +1 > 4 x>3
{ x x ≤1}∪{ x x > 3} ={ x x ≤1
{x x < −1}∪ {x x < 1} = {x x < 1} 17.
0 ≤ 2 x + 6 ≤ 54 −6 ≤ 2 x ≤ 48 −3 ≤ x ≤ 24
2x ≤ 4 x≤2
2x ≥ 6 x≥3
(−∞, 2] ∪ [3, ∞)
24.
3− 2x ≤ 5 −5 ≤ 3 − 2 x ≤ 5 − 8 ≤ − 2x ≤ 2 4 ≥ x ≥ −1
[−1, 4]
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.5
25.
73
26.
x −5 ≥ 0 (Note: The absolute value of any real number is greater than or equal to 0.) (−∞, ∞)
27.
(−∞, ∞)
28.
x−4 ≤0
x2 + 7 x > 0 x ( x + 7) > 0 The product x( x + 7) is positive. x = 0 is a critical value. x + 7 = 0 ⇒ x = −7 is a critical value.
2x + 7 ≤ 0 2x + 7 = 0 2 x = −7 7 x=− 2 ⎧ 7⎫ ⎨− ⎬ ⎩ 2⎭
(Note: No absolute value is less than 0.) x−4=0 x=4 {4}
29.
x −7 ≥ 0
30.
x ( x + 7)
x2 − 5x ≤ 0 x( x − 5) ≤ 0 The product x( x − 5) is negative or zero. x = 0 is a critical value. x − 5 = 0 ⇒ x = 5 is a critical value. x( x − 5) [0, 5]
(−∞, − 7) ∪ (0, ∞) 31.
33.
x 2 − 16 ≤ 0 ( x − 4)( x + 4) ≤ 0 The product ( x − 4)( x + 4) is negative or zero. x − 4 = 0 ⇒ x = 4 is a critical value. x + 4 = 0 ⇒ x = −4 is a critical value.
32.
x 2 − 49 > 0 ( x − 7)( x + 7) > 0 The product ( x − 7)( x + 7) is positive. x = 7 is a critical value. x + 7 = 0 ⇒ x = −7 is a critical value.
( x − 4)( x + 4)
( x − 7)( x + 7)
[−4, 4]
(−∞, − 7) ∪ (7, ∞)
x 2 + 7 x + 10 < 0 ( x + 5)( x + 2) < 0 The product ( x + 5)( x + 2) is negative. x + 5 = 0 ⇒ x = −5 is a critical value. x + 2 = 0 ⇒ x = −2 is a critical value.
34.
x2 + 5x + 6 < 0 ( x + 3)( x + 2) < 0 The product ( x + 3)( x + 2) is negative. x + 3 = 0 ⇒ x = −3 is a critical value. x + 2 = 0 ⇒ x = −2 is a critical value.
( x + 5)( x + 2)
( x + 3)( x + 2)
(−5, −2)
(−3, −2)
Copyright © Houghton Mifflin Company. All rights reserved.
74
35.
Chapter 1: Equations and Inequalities
x 2 − 3x ≥ 28
36.
x 2 + x − 30 < 0 ( x + 6)( x − 5) < 0 The product ( x + 6)( x − 5) is negative. x + 6 = 0 ⇒ x = −6 is a critical alue. x − 5 = 0 ⇒ x = 5 is a critical value.
x 2 − 3 x − 28 ≥ 0 ( x − 7)( x + 4) ≥ 0 The product ( x − 7)( x + 4) is positive or zero. x − 7 = 0 ⇒ x = 7 is a critical value. x + 4 = 0 ⇒ x = −4 is a critical value.
( x + 6)( x − 5)
( x − 7)( x + 4)
(−6, 5)
(−∞, − 4] ∪ [7, ∞) 37.
x+4 <0 x −1
38.
x+4 is negative. x −1 x + 4 = 0 ⇒ x = −4
x−2 >0 x+3 x−2 is positive. x+3 x − 2 = 0⇒ x = 2
The quotient
The quotient
x −1 = 0 ⇒ x = 1 The critical values are −4 and 1.
x + 3 = 0 ⇒ x = −3 The critical values are 2 and −3.
x+4 x −1
x−2 x+3
(−∞, − 3) ∪ (2, ∞)
(−4, 1) 39.
x 2 < − x + 30
x −5 ≥ 3 x +8 x −5 −3≥ 0 x +8 x − 5 − 3( x + 8) ≥0 x +8 x − 5 − 3x − 24 ≥ 0 x +8 −2 x − 29 ≥ 0 x +8 −2 x − 29 The quotient is positive or zero. x+8 29 − 2 x − 29 = 0 ⇒ x = − 8 x + 8 = 0 ⇒ x = −8 29 The critical values are − and − 8. 2
40.
x − 4 ≤1 x+6 x − 4 −1 ≤ 0 x+6 x − 4 − 1( x + 6) ≤0 x+6 x−4− x−6 ≤0 x+6 −10 ≤ 0 x+6 −10 is negative or zero. The quotient x+6 x + 6 = 0 ⇒ x = −6 The critical value is –6. −10 x+6
The denominator cannot equal zero ⇒ x ≠ −6.
−2 x − 29 x+8
(−6, ∞)
The denominator cannot equal zero ⇒ x ≠ −8. ⎞ ⎡ 29 ⎢− 2 , − 8 ⎟ ⎠ ⎣
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.5
41.
75
x ≥4 2x + 7 x −4≥0 2x + 7 x − 4(2 x + 7) ≥0 2x + 7 x − 8 x − 28 ≥ 0 2x + 7 −7 x − 28 ≥ 0 2x + 7 −7 x − 28 The quotient is positive or zero. 2x + 7 −7 x − 28 = 0 ⇒ x = −4
42.
25 16 5 3x − 5 = 0 ⇒ x = 3 25 5 The critical values are and . 16 3 16 x − 25 = 0 ⇒ x =
7 2x + 7 = 0 ⇒ x = − 2 7 The critical values are − 4 and − . 2
−7 x − 28 2x + 7 The denominator cannot equal zero ⇒ x ≠ −
16 x − 25 3x − 5
7. 2
The denominator cannot equal zero ⇒ x ≠
7⎞ ⎡ ⎢− 4, − 2 ⎟ ⎣ ⎠ 43.
x ≤ −5 3x − 5 x +5≤0 3x − 5 x + 5(3x − 5) ≤0 3x − 5 x + 15 x − 25 ≤ 0 3x − 5 16 x − 25 ≤ 0 3x − 5 16 x − 25 The quotient is negative or zero. 3x − 5
⎡ 25 5 ⎞ ⎢ 16 , 3 ⎟ ⎠ ⎣
( x + 1)( x − 4) <0 x−2 ( x + 1)( x − 4) is negative. The quotient x−2 x + 1 = 0 ⇒ x = −1 x−4 = 0⇒ x = 4 x−2 = 0⇒ x = 2 The critical values are –1, 4, and 2.
44.
x( x − 4) >0 x+5 The quotient
x ( x − 4) > 0 is positive. x+5
x=0 x−4 = 0⇒ x = 4 x + 5 = 0 ⇒ x = −5 The critical values are 0, 4, and –5.
( x + 1)( x − 4) x−2
x ( x − 4) x+5
(−∞, − 1) ∪ (2, 4)
(−5, 0) ∪ (4, ∞)
Copyright © Houghton Mifflin Company. All rights reserved.
5 . 3
76
45.
Chapter 1: Equations and Inequalities
x+2 ≤2 x−5 x+2 −2≤ 0 x−5 x + 2 − 2( x − 5) ≤0 x−5 x + 2 − 2 x + 10 ≤0 x −5 − x + 12 ≤0 x−5 − x + 12 The quotient is negative or zero. x −5 − x + 12 = 0 ⇒ x = 12
46.
x−5 = 0 ⇒ x = 5 The critical values are 12 and 5.
x−2 = 0⇒ x = 2 The critical values are 9 and 2.
− x + 12 x−5
47.
3x + 1 ≥4 x−2 3x + 1 −4≥ 0 x−2 3 x + 1 − 4( x − 2) ≥0 x−2 3x + 1 − 4 x + 8 ≥0 x−2 − x+9 ≥0 x−2 −x + 9 The quotient is positive or zero. x−2 −x + 9 = 0 ⇒ x = 9
−x + 9 x−2
The denominator cannot equal zero ⇒ x ≠ 5.
The denominator cannot equal zero ⇒ x ≠ 2.
(−∞, 5) ∪ [12, ∞)
(2, 9]
6 x 2 − 11x − 10 >0 x (3 x + 2)(2 x − 5) >0 x (3x + 2)(2 x − 5) is positive. The quotient x 2 3x + 2 = 0 ⇒ x = − 3 5 2x − 5 = 0 ⇒ x = 2 x=0 2 5 The critical values are − , , and 0. 3 2 (3x + 2)(2 x − 5) x
⎛ 2 ⎞ ⎛5 ⎞ ⎜ − , 0⎟ ∪ ⎜ , ∞⎟ ⎝ 3 ⎠ ⎝2 ⎠
48.
3x2 − 2 x − 8 ≥0 x −1 (3 x + 4)( x − 2) ≥0 x −1 (3 x + 4)( x − 2) is positive or zero. The quotient x −1 4 3x + 4 = 0 ⇒ x = − 3 x−2 =0⇒ x = 2 x −1 ⇒ x = 1 4 The critical values are − , 2, and 1. 3 (3 x + 4)( x − 2) x −1 The denominator cannot equal zero ⇒ x ≠ 1. ⎡ 4 ⎞ ⎢− 3 , 1⎟ ∪ [2, ∞) ⎣ ⎠
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.5
49.
77
x2 − 6 x + 9 ≤0 x−5 ( x − 3)( x − 3) ≤0 x−5 ( x − 3)( x − 3) The quotient is negative or zero. x −5 x−3 = 0 ⇒ x = 3 x−5 = 0 ⇒ x = 5 The critical values are 3 and 5.
50.
( x − 3)( x − 3) x−5
x 2 + 10 x + 25 ≥0 x +1 ( x + 5)( x + 5) ≥0 x +1 ( x + 5)( x + 5) The quotient is positive or zero. x +1 x + 5 = 0 ⇒ x = −5 x + 1 = 0 ⇒ x = −1 The critical values are −5 and −1. ( x + 5)( x + 5) x +1
The denominator cannot equal zero ⇒ x ≠ 5.
The denominator cannot equal zero ⇒ x ≠ −1.
(−∞, 5)
{−5} ∪ (−1, ∞)
51.
Plan A: 5 + 0.01x Plan B: 1 + 0.08x 5 + 0.01x < 1 + 0.08x 4 < 0.07x 57.1 < x Plan A is less expensive if you use more than 57 checks.
52.
Company A: 19 + 0.12m Company B: 12 + 0.21m 19 + 0.12m < 12 + 0.21m 7 < 0.09m 77.7 < m Company A is less expensive if you drive at least 78 miles.
53.
Let h = the height of the package. length + girth ≤ 130 length + 2(width) + 2(height) ≤ 130 34 + 2(22) + 2h ≤ 130 34 + 44 + 2h ≤ 130 78 + 2h ≤ 130 2h ≤ 52 h ≤ 26 The height must be more than 0 but less than or equal to 26 inches.
54.
−0.05 x + 1.73 < 1.25 −0.05 x < −0.48 x > 9.6 9.6 years after 2000 is in the year 2009.
55.
17.1895 x + 95.2065 > 600 17.1895 x > 504.7935 x > 29.366 29 months after September 2004 is in January 2007.
56.
Plan A: 15 + 1.49x 57. Plan B: 1.99x 1.99x < 15 + 1.49x 0.50x < 15 x < 30 Plan B is less expensive if fewer than 30 videos are rented.
F ≤ 104 9 68 ≤ C + 32 ≤ 104 5 9 36 ≤ C ≤ 72 5 5 5 ⎛9 ⎞ 5 (36) ≤ ⎜ C ⎟ ≤ (72) 9 9 ⎝5 ⎠ 9 ≤ 40° 20° ≤ C
58.
1.63 − μ < 2.33 1.79 −4.1707 < 1.63 − μ < 4.1707 −167.2 < − μ < −158.8 μ 167.2 > > 158.8 μ 158.8 < < 167.2 lb
59.
190 − μ < 2.575 2.45 −6.30875 < 1.90 − μ < 6.30875 −196.30875 < − μ < −183.69125 196.30875 > μ > 183.69125 183.7 < μ < 196.3 lb
63 < x + ( x + 2) + ( x + 4) < 81 63 < 3x + 6 < 81 < 75 57 < 3x < 25 19 < x x must be odd, thus x = 21 or x = 23. Therefore, the numbers are {21, 23, 25} or {23, 25, 27}.
−2.33 <
−2.575 <
60.
68 ≤
Copyright © Houghton Mifflin Company. All rights reserved.
78
61.
Chapter 1: Equations and Inequalities
Solve h − (2.47 f + 54.10) ≤ 3.72 for h where f = 32.24.
62.
h − (2.47 f + 54.10) ≤ 3.72
h − (3.32 r + 85.43) ≤ 4.57
h − [2.47(32.24) + 54.10] ≤ 3.72
h − [3.32(26.36) + 85.43] ≤ 4.57
h − (79.6328 + 54.10) ≤ 3.72
h − (87.5152 + 85.43) ≤ 4.57
h − 133.7328 ≤ 3.72
h − 172.9452 ≤ 4.57
or h − 133.7328 ≥ −3.72 h − 133.7328 ≤ 3.72 h ≤ 137.4528 h ≥ 130.0128 The height, to the nearest 0.1 cm, is from 130.0 cm to 137.5 cm. 63.
65.
or h − 172.9452 ≥ −4.57 h − 172.9452 ≤ 4.57 h ≤ 177.5152 h ≥ 168.3752 The potential stature, to the nearest 0.1 cm, is from 168.4 to 177.5 cm. 64.
R = 420 x − 2 x 2
To determine potential stature, solve h − (3.32r + 85.43) ≤ 4.57 for h where r = 26.36.
R = 312 x − 3 x 2
420 x − 2 x 2 > 0 2 x(210 − x) > 0 The product is positive. 2x = 0 ⇒ x = 0 210 − x = 0 ⇒ x = 210 Critical values are 0 and 210.
312 x − 3 x 2 ≥ 5925 3(− x 2 + 104 x − 1975) ≥ 0 3(− x + 25)( x − 79) ≥ 0 Critical values are 25 and 79.
2 x(210 − x)
3(− x + 25)( x − 79)
($0, $210)
[$25, $79]
−3x 2 + 312 x − 5925 ≥ 0
14.25 x + 350,000 < 50 x 14.25 x + 350,000 < 50 x −35.75 x < −350,000 x > 9790.2 At least 9791 books must be published.
66.
2 C = 0.00014 x + 12 x + 400,000 < 30
x 0.00014 x 2 + 12 x + 400,000 < 30 x 0.00014 x 2 − 18 x + 400,000 < 0
Solve 0.00014 x 2 − 18 x + 400,000 = 0 to find the critical values. −( −18) ± ( −18)2 − 4(0.00014)(400,000) 18 ± 324 − 224 18 ± 100 18 ± 10 = = = 2(0.00014) 0.00028 0.00028 0.00028 x = 18 + 10 x = 18 − 10 or 0.00028 0.00028 28 8 = = 0.00028 0.00028 = 100,000 ≈ 28,571.4 The critical values are 100,000 and ≈ 28,571.4. Since x is a non-negative integer, the intervals are (0, 26,571.4) , (28,571.4, 100,000) , and (100,000, ∞ ) . x=
Test 1:
0.00014(1)2 + 12(1) + 400,000 < 30 ⇒ 400,012.00014 < 30 , which is false. 1
Test 50,000:
0.00014(50,000)2 + 12(50,000) + 400,000 < 30 ⇒ 27 < 30 , which is true. 50,000
0.00014(150000)2 + 12(150000) + 400,000 < 30 ⇒ 35.6 < 30 , which is false 150000 The company should manufacture from 28,572 to 99,999 pairs of running shoes.
Test 150,000:
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.5
79
....................................................... 67.
69.
28 − 0.15 ≤ C ≤ 28 + 0.15 27.85 ≤ 2π r ≤ 28.15 27.85 28.15 ≤ r ≤ 2π 2π 4.432 ≤ r ≤ 4.480 The radius of the cylinder must be between 4.432 inches and 4.480 inches.
( x − 3) 2
Connecting Concepts 68.
735 ≤ πr h ≤ 765 735
πr
( x − 3) 2
( x − 4) 2 3
≤
h ≤
765
πr 2
( x − 1) 2
≥0 ( x − 4) 4 The quotient is positive or zero. x −1 = 0 ⇒ x = 1 x−4=0⇒ x = 4 Critical values are 1 and 4. ( x − 1) 2
2
( x − 4) 4
(−∞, 3) ∪ (3, 6) ∪ (6, ∞) 71.
2
735 765 ≤ h ≤ 4π 4π 58.5 ≤ h ≤ 60.9 The height of the beaker must be between 58.5 cm and 60.9 cm.
( x − 6)2 The quotient is positive. x−3 = 0 ⇒ x = 3 x−6 = 0⇒ x = 6 Critical values are 3 and 6.
( x − 6)
≤ 750 + 15
2
70.
>0
750 − 15 ≤ V
(−∞, 4) ∪ (4, ∞) 72.
≥0
( x + 3) The quotient is positive or zero. x−4 = 0⇒ x = 4 x + 3 = 0 ⇒ x = −3 Critical values are 4 and −3. ( x − 4) 2 ( x + 3)3
2x − 7 ≥0 ( x − 1) 2 ( x + 2)2 The quotient is positive or zero. 7 2x − 7 = 0 ⇒ x = 2 x −1 = 0 ⇒ x = 1 x + 2 = 0 ⇒ x = −2 7 Critical values are , 1, −2. 2 2x − 7
Denominator not 0 ⇒ x ≠ −3. (−3, ∞)
( x − 1)2 ( x + 2) 2 ⎡7 ⎞ ⎢ 2 , ∞⎟ ⎣ ⎠
73.
74.
1< x < 5 if x ≥ 0 if x < 0
1< x < 5
1 < −x < 5 − 1 > x > −5 (−5, − 1) ∪ (1, 5)
75.
2< x <3 if x ≥ 0 if x < 0
2< x<3
2 < −x < 3 − 2 > x > −3 (−3, − 2) ∪ (2, 3)
3≤ x < 7 if x ≥ 0 if x < 0
3≤ x<7
3 ≤ −x < 7 − 3 ≥ x > −7 (−7, − 3] ∪ [3, 7)
Copyright © Houghton Mifflin Company. All rights reserved.
80
76.
Chapter 1: Equations and Inequalities
77.
0< x ≤3 if x ≥ 0
0< x≤3
0 < x −α < δ , if x − a ≥ 0
δ >0 0<
x−a x 0 < −( x − a ) 0> x−a a> x
α<
if x < 0
0 < −x ≤ 3 0 > x ≥ −3 [−3, 0) ∪ (0, 3]
if x − a < 0
<δ < δ +α <δ > −δ > a −δ
( a − δ , a ) ∪ ( a, a + δ ) 78.
79.
0< x −5 < 2 if x − 5 ≥ 0 if x − 5 < 0
0< x−5 x 5< 0 < −( x − 5) 0> x−5 x 5>
<2 <7 <2 > −2 >3
(3, 5) ∪ (5, 7)
s = −16t 2 + v0t + s0 ,
s > 48, v0 = 64, s0 = 0
2
−16t + 64t > 48 2
−16t + 64t − 48 > 0 −16(t 2 − 4t + 3) > 0 −16(t − 1)(t − 3) > 0 The product is positive. The critical values are 1 and 3. (t − 1)(t − 3)
1 second < t < 3 seconds The ball is higher than 48 ft between 1 and 3 seconds. 80.
s = −16t 2 + v0t + s0 ,
s > 96, t > 0, v0 = 80, s0 = 32
2
−16t + 80t + 32 > 96 −16t 2 + 80t − 64 > 0 −16(t 2 − 5t + 4) > 0 −16(t − 1)(t − 4) > 0 The product is positive. The critical values are 1 and 4. (t − 1)(t − 4)
1 second < t < 4 seconds The ball is higher than 96 ft between 1 and 4 seconds.
.......................................................
Prepare for Section 1.6 PS2. 20 = k 1.52 45 = k
PS1. 1820 = k (28) 65 = k
PS4. k 4.5 ⋅ 32 82 (12.5) 4.5 2⋅ 32 = 28.125 8
PS3. k 3 52
(225) 32 = 27 5 PS5. The area becomes 4 times as large.
PS6. No. The volume becomes 9 times as large.
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Section 1.6
81
Section 1.6 1.
d = kt
2.
r = ks 2
3.
y=
k x
4.
p=
5.
m = knp
6.
t = krs3
7.
V = klwh
8.
u=
A = ks 2
10.
A = khr 2
11.
y = kx 64 = k ⋅ 48 64 = k 48 4 =k 3
14.
m = kn 92 = k ⋅ 23 92 = k 23 4=k
15.
9.
13.
17.
T = krs 2
18.
210 = k ⋅ 30 ⋅ 52 210 = k 30 ⋅ 52 7 =k 25 0.28 = k
21.
23.
u = kv w k 0.04 = ⋅ 8 0.04
s = k ⋅q 34 = k ⋅ 51 2 =k 3 p = 2 ⋅ 93 3 p = 62 semester hours
24.
p = kd 187.5 = k ⋅ 3 62.5 = k p = 62.5 ⋅ 7 p = 437.5 lb/ft 2
km1m2
r = kt 2
T = ktra 2
16.
C = kr 94.2 = k ⋅ 15 94.2 = k 15 6.28 = k
2
2 4 ⋅ 32 = k 2 4 ⋅ 36 1 =k 81
19.
V = klwh
3 t = kr s
20.
240 = k ⋅ 8 ⋅ 6 ⋅ 5
3 10 = k ⋅ 5 0.09 10 0.09 = k 53 2(0.3) =k 8 .06 = k 25 0.024 = k
240 = k 8⋅6⋅5 1= k
22.
d = k ⋅w 6 = k ⋅ 80 6 = k ⋅ 80 6 =k 80 3 =k 40 Therefore d =
25.
kv w2
12.
d2
144 = k ⋅ 108 144 = k 1082
0.04 0.04 = k 8 (0.04))0.2) =k 8 0.001 = k
V = kT 0.85 = k ⋅ 270 0.85 = k 270 0.17 = k 54 0.17 0.17 Thus V = T = ⋅ 324 = (0.17)6 = 1.02 liters 54 54
F=
k q
3 ⋅ 100 = 7.5 inches 40
j = k ⋅ d3 6 = k ⋅ (4)3 3 =k 32 3 p = ⋅ (5)3 32
p ≈ 11.7 fl oz
Copyright © Houghton Mifflin Company. All rights reserved.
26.
r = kv 2 140 = k ⋅ 602 140 = k 602 7 =k 180 7 Thus r = ⋅ 652 180 r ≈ 164.3 ft.
82
27.
Chapter 1: Equations and Inequalities
T =k l 1.8 = k 3 1.8 = k 3 1.03923 ≈ k a. T = 1.8 10 3
b.
= 1.8 30 3 = 0.6 30 ≈ 3.3 seconds
28.
A = kd 2
T =k l T = l k 2 = l 1.03923 4 = l 1.039232 3.7 ft ≈ l
29.
r=k t 30 = k 64 1920 = k r = 1920 48 r = 40 revolutions per minute
30.
f =k l 144 = k 20 2880 = k f = 2880 18 f = 160 vibrations per second
32.
l = k2 d 50 = k 2 10 5000 = k Thus I = 5000 d2 I = 5000 = 5000 225 152 I ≈ 22.2 footcandles
33.
a.
64 = k ⋅ 20 2 6 = k ⋅ 80 64 = k 400 4 =k 25 100 = 4 ⋅ d 2 25 625 = d 2 d = 25 ft
31.
l = k2 d 28 = k2 8 28 ⋅ 64 = k 1792 = k = 1792 l = 1792 16 42 l = 112 decibels
V = kr 2 h V1 = k (3r ) 2 h = 9( kr 2 h ) = 9V Thus the new volume is 9 times the original volume.
b.
V2 = kr 2 (3h ) = 3( kr 2 h ) = 3V Thus the new volume is 3 times the original volume.
c.
V3 = k (3r ) 2 (3h ) = k 9r 2 ⋅ 3 ⋅ h = 27( kr 2 h ) = 27V Thus the new volume is 27 times the original volume.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.6
34.
83
L = kwd 2
35.
200 = k ⋅ 2 ⋅ 62 200 = k 2 ⋅ 62 25 = k 9 Thus L = 25 ⋅ 4 ⋅ 4 2 9 1600 = 9 ≈ 178 lb 37.
V = knT P k (3n )T V1 = 1p 2
( )
6 ⋅ 102 = k 24 600 = 37.5 = k 16
⎛ ⎞ = 6 ⎜ knT ⎟ ⎝ p ⎠ = 6V Thus the new volume is 6 times larger than the original volume.
For Randy Johnson, ERA = kr i k (67) 2.32 = (260) 9.00 = k For Tom Glavine, 9(74) ERA = (224.2) = 2.97
38.
2 L = kbd l 2 800 = k ⋅ 4 ⋅ 8 12 800 ⋅ 12 = k 4 ⋅ 82 37.5 = k
37.5(3.5)(6) 2 16 ≈ 295 pounds
Thus L =
4 L = k ⋅ d2 h 4 6 = k ⋅ 22 10
36.
4
Thus L = 37.5 2⋅ 3 14 ≈ 15.5 tons 39.
2 F = kws r 2 k ⋅ 2800 = 1800 ⋅ 45 425 2800 ⋅ 425 = k 1800 ⋅ 452 14 ⋅ 425 = k 9 ⋅ 452 0.3264746 ≈ k
....................................................... 40.
S = kwd 3 when d = 10, w = 182 − 102 ≈ 15 ⇒ S ≈ k (15)(10)3 = 15,000k d = 12, w = 182 − 122 ≈ 13.4 ⇒ S ≈ k (13.4)(12)3 = 23,155k d = 14, w = 182 − 142 ≈ 11.3 ⇒ S ≈ k (11.3)(14)3 = 31,007k d = 16, w = 182 − 162 ≈ 8.2 ⇒ S ≈ k (8.2)(16)3 = 33,587k The strongest beam occurs when d = 16 inches.
41.
T = kd 3/ 2
365 = k ⋅ 933/ 2 365 = k 933 / 2 Thus
⋅ d 3/ 2 686 = 365 933/ 2
686 ⋅ 933 / 2 = d 3/ 2 365 ⎛ 686 ⋅ 933/ 2 ⎞ ⎜ ⎟ 365 ⎝ ⎠
(0.3264746) ⋅ 1800 ⋅ 552 450 ≈ 3950 pounds
Thus F =
2/3
=d 2/3
=d 93 ⎛⎜ 686 ⎞⎟ ⎝ 365 ⎠ 142 million miles ≈ d
Copyright © Houghton Mifflin Company. All rights reserved.
Connecting Concepts
84
Chapter 1: Equations and Inequalities
....................................................... 1.
No. The solution set of x = 3 is {3} but the solution
Assessing Concepts 2.
a<0
4.
Equal to 0
6.
c
set of x 2 = 9 is {–3, 3}. 3.
− a < −b
5.
False.
(
x + 3) = x + 6 x + 9 ≠ x + 9 2
7.
b
8.
a
9.
g
10.
f
....................................................... 1.
4.
7.
x − 2(5 x − 3) = −3( − x + 4) [1.1] x − 10 x + 6 = 3x − 12 −9 x + 6 = 3x − 12 −12 x = −18 x=3 2
2.
3x − 2 x − 1 = 3 [1.1] 4 8 2 8 ⎜⎛ 3x − 2 x − 1 ⎟⎞ = 8 ⎛⎜ 3 ⎞⎟ 8 ⎠ ⎝2⎠ ⎝ 4 2(3x ) − (2 x − 1) = 4(3) 6 x − 2 x + 1 = 12 4 x + 1 = 12 4 x = 11 x = 11 4
5.
x 2 − 5 x + 6 = 0 [1.3] ( x − 2)( x − 3) = 0 x − 2 = 0 or x − 3 = 0 x=2 x=3
8.
Chapter Review
3x − 5(2 x − 7) = −4(5 − 2 x ) [1.1]
3.
3x − 10 x + 35 = −20 + 8 x −7 x + 35 = −20 + 8 x −15 x = −55 x = 11 3
x + 1 =5 x+2 4 4( x + 2) ⎛⎜ x + 1 ⎞⎟ = 5(4)( x + 2) ⎝ x+2 4⎠ 4 x + x + 2 = 20( x + 2)
4x − 4x − 1 = 1 [1.1] 3 6 2 6 ⎜⎛ 4 x − 4 x − 1 ⎟⎞ = 6 ⎛⎜ 1 ⎞⎟ 6 ⎠ ⎝2⎠ ⎝ 3 2(4 x ) − (4 x − 1) = 3 8x − 4 x + 1 = 3 4x + 1 = 3 4x = 2 x=1 2 y −1 −1 = 2 y +1 y
6.
⎛ y −1 ⎞ ⎛ ⎞ − 1⎟ = y ( y + 1) ⎜ 2 ⎟ y ( y + 1) ⎜ ⎝ y +1 ⎠ ⎝ y⎠ y ( y − 1) − y ( y + 1) = 2( y + 1)
5 x + 2 = 20 x + 40
y 2 − y − y2 − y = 2 y + 2 −4 y = 2 y =−1 2
−15 x = 38 x = − 38 15
[1.5] [1.4] 6 x 2 + x − 12 = 0 [1.3] (3x − 4)(2 x + 3) = 0 3x − 4 = 0 or 2 x + 3 = 0 3x = 4 2 x = −3 x=4 x=−3 3 2
9.
3x 2 − x − 1 = 0 [1.3] x=
−( −1) ± ( −1)2 − 4(3)( −1) 2(3)
x = 1 ± 13 6 x = 1 + 13 6
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or
x = 1 − 13 6
Chapter Review
10.
85
x 2 − x + 1 = 0 [1.3] x=
11.
2 x ( x 2 − 4) = 0 2 x ( x − 2)( x + 2) = 0
x2 = 0 ⇒ x = 0
6 x 4 − 23x 2 + 20 = 0 [1.4]
x = 0, x = 2,
3x − 5 = 0 ⇒ x = 5 3 x = 0 or x = 5 3
14.
3x + 16 x − 12 = 0 [1.4]
Let u = x .
Let u = x 2 .
3u 2 + 16u − 12 = 0 (3u − 2)(u + 6) = 0 u=2 or 3 x=2 3 4 x= 9 Thus, x = 4 . 9
2
6u − 23u + 20 = 0 (3u − 4)(2u − 5) = 0 a=4 3 2 4 x = 3
15.
u=5 2 2 x =5 2
or
x=± 4 3
x=± 5 2
⎛ ⎞ x=± 2 ⎜ 3⎟ 3⎝ 3⎠
⎛ ⎞ x=± 5⎜ 2⎟ 2⎝ 2⎠
x=±2 3 3
x = ± 10 2
16.
x 2 − 15 = −2 x [1.4] 2
2
[1.4] 2
x 2 − 24 = 2 x
x 2 + 2 x − 15 = 0 ( x + 5)( x − 3) = 0 x = −5 or x = 3
x 2 − 2 x − 24 = 0 ( x − 6)( x + 4) = 0 x=6
( −5)2 − 15 = −2( −5) 10 = 10
Check
or
x =− 4 (6)2 − 24 = 2(6) 36 − 24 = 12
3 − 15 = −2(3) −6 = −6 The solutions are −5 and 3.
x = −6 No solution.
⎡⎣ x 2 − 24 ⎤⎦ = [ 2 x ]
x 2 − 15 = −2 x
2
u = −6
x 2 − 24 = 2 x 2
⎡⎣ x 2 − 15 ⎤⎦ = [ −2 x ]
Check
2 x 3 − 8 x = 0 [1.4]
12.
x 2 (3x − 5) = 0
−( −1) ± ( −1)2 − 4(1)(1) 2(1)
x = 1 ± −3 = 1 ± i 3 2 2 3 1 x= + i or x = 1 − 3 i 2 2 2 2
13.
3x 3 − 5 x 2 = 0 [1.4]
12 = 12 ( − 4)2 − 24 = 2( −4) 16 − 24 = −8 −8 = −8 The solutions are 6 and −4.
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or
x = −2
86
17.
Chapter 1: Equations and Inequalities
18.
3x + 4 + x − 3 = 5
2x + 2 − x + 2 = x − 6
[
3x + 4 = 5 − x − 3
[
3 x + 4 ] = [5 − x − 3 ] 2
2
( 2 x + 2) ( x + 2) = x + 5
x − 9 = −5 x − 3
2
⎡ ( 2 x + 2 ) ( x + 2 ) ⎤ = [ x + 5]2 ⎣ ⎦ ( 2 x + 2 ) ( x + 2 ) = x 2 + 10 x + 25
( x − 9 ) 2 = [ −5 x − 3 ]
2
x 2 − 18 x + 81 = 25( x − 3)
2 x 2 + 4 x + 2 x + 4 = x 2 + 10 x + 25
x 2 − 18 x + 81 = 25 x − 75
x 2 − 4 x − 21 = 0
x 2 − 43x + 156 = 0 ( x − 4)( x − 39) = 0 x=4 or x = 39
( x − 7)( x + 3) = 0
x=7 Check
3(4) + 4 + 4 − 3 = 5
2(7) + 2 − 7 + 2 = 7 − 6
2( −3) + 2 − −3 + 2 = −3 − 6
3(39) + 4 + 39 − 3 = 5
−4 − −1 = −9 2i − i = 3i i = 3i (No) The solution is 7. [1.4]
121 + 36 = 5 11 + 6 = 5 17 = 5 (No) The solution is 4. [1.4] 4 − 3x − 5 − x = 5 + x
[1.4]
4 − 3x − 5 − x ] = [ 5 + x ] 2
x = −3
or
16 − 9 = 1 4 −3=1 1=1
16 + 1 = 5 4 +1 = 5 5=5
[
2
−2 (4 − 3x )(5 − x ) = 5 x − 4 2
⎡⎣ −2 (4 − 3x )(5 − x ) ⎤⎦ = [5 x − 4]2
Check
4 − 3 ⎛⎜ 16 ⎞⎟ − 5 − 16 = 5 + 16 13 13 ⎝ 13 ⎠ 52 − ⎛ 48 ⎞ − 65 − 16 = 65 + 16 ⎜ ⎟ 13 ⎝ 13 ⎠ 13 13 13 13 4 − 13 2 − 13
2
4(4 − 3x )(5 − x ) = 25 x − 40 x + 16 4(20 − 19 x + 3x 2 ) = 25 x 2 − 40 x + 16 0 = 13x 2 + 36 x − 64 0 = (13x − 16)( x + 4) x = 16 13
or
2
−2 ( 2 x + 2 ) ( x + 2 ) = −2( x + 5)
2 x − 18 = −10 x − 3
19.
2
2 x + 2 − 2 ( 2 x + 2) ( x + 2) + x + 2 = x − 6
3x + 4 = 25 − 10 x − 3 + x − 3
Check
2x + 2 − x + 2 ] = [ x − 6]
49 = 81 13 13 7 = 9 (No) 13 13
4 − 3( −4) − 5 − ( −4) = 5 − 4 16 − 9 = 1
x = −4
4 −3=1 1=1
The solution is −4.
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Chapter Review
87
20.
3x + 9 − 2 x + 4 = x + 1
3(0) + 9 − 2(0) + 4 = 0 + 1
Check
2
⎡ 3x + 9 − 2 x + 4 ⎤ = ⎡ x + 1 ⎤ ⎣ ⎦ ⎣ ⎦
2
9− 4= 1 3− 2 =1 1=1
3x + 9 − 2 (3x + 9)(2 x + 4) + 2 x + 4 = x + 1 −2 (3x + 9)(2 x + 4) = −4 x − 12
3( −3) + 9) − 2( −3) + 4 = −3 + 1
2
⎡ (3x + 9)(2 x + 4) ⎤ = [ 2 x + 6]2 ⎣ ⎦
0 − − 2 = −2
2
0 − − 2 = −2
(3x + 9)(2 x + 4) = 4 x + 24 x + 36 2
2
− −2 = −2 (No)
6 x + 30 x + 36 = 4 x + 24 x + 36
The solution is 0.
2
2x + 6x = 0
[1.4]
2 x ( x + 3) = 0
x=0 21.
24.
or
x = −3
1 = 4(4 s 2 − 20s + 25)
1 = y2 + 6 y + 9
1 = 16s 2 − 80s + 100
0 = y2 + 6 y + 8 0 = ( y + 2)( y + 4)
0 = 16s 2 − 80s + 99 0 = (4 s − 11)(4 s − 9)
y = −2
s = 11 4
or
y = −4
25.
x + 5 = 4 [1.1]
or
28.
3x − 7 = 8 [1.1]
( x + 2)1/ 2 = 0 or ( x + 1)2 = 0 x+2=0 x +1= 0
3x − 7 = −8 3x = −1 x = −1 3
x 2 (3x − 4)1/ 4 + (3x − 4)5 / 4 = 0 [1.4]
x = −1
[1.5]
or
(3x − 4)1/ 4 = 0 or x 2 + 3x − 4 = 0 3x − 4 = 0 ( x + 4)( x − 1) = 0 x + 4 = 0 or x − 1 = 0 3x = 4 4 x= x = −4 x =1 3
( x + 2)1/ 2 ( x + 1)2 = 0
−3x ≥ −2 − 4 −3x ≥ −6 x≤2 (−∞, 2]
x − 3 = −2 x =1
(3x − 4)1/ 4 ( x 2 + 3x − 4) = 0
( x + 2)1/ 2 ( x 2 + 2 x + 1) = 0
−3x + 4 ≥ −2
or
3x − 7 = 8 3x = 15 x=5
2x = −6 x = −3
( x + 2)1/ 2 [1 + x 2 + 2 x ] = 0
x = −2
26.
2 x + 1 = −5
2x = 4 x=2
( x + 2)1/ 2 + x ( x + 2)3 / 2 = 0 [1.4]
x−3=2 x=5
2 x + 1 = 5 [1.1] 2 x + 1 = 5 or
x + 5 = −4 x = −9
x − 3 = 2 [1.1]
s=9 4
( x + 2)1/ 2 [1 + x ( x + 2) ] = 0
29.
23.
1 = 4 [1.4] (2 s − 5) 2
1 = ( y + 3)2
x + 5 = 4 or x = −1
27.
22.
1 = 1 [1.4] ( y + 3)2
30.
−2 x + 7 ≤ 5 x + 1 [1.5] −7 x ≤ −6 x≥6 7 ⎡6 , ∞ ⎢⎣ 7
)
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88
31.
Chapter 1: Equations and Inequalities
x 2 + 3x − 10 ≤ 0 [1.5] ( x + 5)( x − 2) ≤ 0 The product is negative or zero. x + 5 = 0 ⇒ x = −5 x−2=0⇒ x =2 Critical values are –5 and 2.
32.
( x + 5)( x − 2)
( x + 1)( x − 3) (−∞, − 1) ∪ (3, ∞)
[−5, 2] 33.
35.
37.
x 2 − 2 x − 3 > 0 [1.5] ( x + 1)( x − 3) > 0 The product is positive. x + 1 = 0 ⇒ x = −1 x−3= 0⇒ x =3 Critical values are –1 and 3.
61 ≤ 9 C + 32 ≤ 95 [1.5] 5 29 ≤ 9 C ≤ 63 5 145 ≤ C ≤ 35 9 ⎡ 145 , 35⎤ ⎣⎢ 9 ⎦⎥
34.
30 < 5 ( F − 32) < 65 [1.5] 9 54 < F − 32 < 117 86 < F (86, 149)
x 3 − 7 x 2 + 12 x ≤ 0 [1.5]
36.
< 149
x 3 + 4 x 2 − 21x > 0 [1.5]
x ( x 2 − 7 x + 12) ≤ 0 x ( x − 3)( x − 4) ≤ 0.
x ( x 2 + 4 x − 21) > 0 x ( x + 7)( x − 3) > 0.
The product is negative or zero.
The product is positive.
x=0 x−3=0⇒ x =3
x=0 x + 7 = 0 ⇒ x = −7
x−4=0⇒ x =4 The critical values are 0, 3, and 4.
x−3=0⇒ x =3 The critical values are 0, − 7, and 3.
x( x − 3)( x − 4)
x( x + 7)( x − 3)
(−∞, 0] ∪ [3, 4]
(−7, 0) ∪ (3, ∞)
x+3 > 0 [1.5] x−4 The quotient is positive. x + 3 = 0 ⇒ x = −3
38.
x−4 =0⇒ x = 4 The critical values are − 3 and 4. x+3 x−4
(−∞, − 3) ∪ (4, ∞)
x( x − 5) ≤ 0 [1.5] x+7 The quotient is negative or zero. x=0 x−5 = 0 ⇒ x = 5 x + 7 = 0 ⇒ x = −7 The critical values are 0, 5 and − 7. x( x − 5) x+7
Denominator ≠ 0 ⇒ x ≠ −7. (−∞, − 7) ∪ [0, 5]
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Chapter Review
39.
89
40.
2 x ≤ 10 [1.5] 3− x 2 x − 10 ≤ 0 3− x 2 x − 10(3 − x ) ≤0 3− x 2 x − 30 + 10 x ≤ 0 3− x 12 x − 30 ≤ 0 3− x The quotient is negative or zero. 12 x − 30 = 0 ⇒ x = 5 2 3− x = 0 ⇒ x = 3 The critical values are 5 and 3. 2 12 x − 30 3− x
x ≥ 1 [1.5] 5− x x −1 ≥ 0 5− x x − (5 − x ) ≥0 5− x x −5+ x ≥ 0 5− x 2x − 5 ≥ 0 5− x The quotient is positive or zero. 2x − 5 = 0 ⇒ x = 5 2 5− x = 0⇒ x = 5 The critical values are 5 and 5. 2 2x − 5 5− x Denominator ≠ 0 ⇒ x ≠ 5.
Denominator ≠ 0 ⇒ x ≠ 3.
( 41.
⎡5 , 5 ⎢⎣ 2
−∞, 5 ⎤⎥ ∪ (3, ∞) 2⎦
42.
3x − 4 < 2 [1.5] −2 < 3 x − 4 < 2 2 < 3x < 6 2< x <2 3 2, 2 3
45.
2 x − 3 ≥ 1 [1.5] 2 x − 3 ≥ 1 or 2 x − 3 ≤ −1 2x ≥ 4 x≥2
( ) 43.
)
2x ≤ 2 x ≤1
(−∞, 1] ∪ [2, ∞) 44.
0 < x − 2 <1 [1.5]
0 < x − a < b [1.5]
If x − 2 ≥ 0, then 2 < x < 3.
If x − a ≥ 0, then a < x < a + b.
If x − 2 < 0, then 0 < x − 2 < −1 2 > x > 1.
If x − a < 0, then 0 < x − a < −b a > x > a − b.
(1, 2) ∪ (2, 3)
[ a − b, a ) ∪ ( a , a + b ]
V = π r 2 h [1.2]
46.
V =h π r2
A [1.2] 1 + rt P(1 + rt ) = A P + Prt = A Prt = A − P P=
2 A = hb1 + hb2 2 A − hb2 = hb1
t = A− P Pr 48.
P = 2(l + w) [1.2] P = 2l + 2 w P − 2l = 2 w P − 2l = w 2
49.
e = mc 2 [1.2] e =m c2
A = h (b1 + b2 ) [1.2] 2 2 A = h (b1 + b2 )
47.
2 A − hb2 = b1 h 50.
F =G
m1m2 s2
Fs 2 = Gm1m2 Fs 2 = m 1 Gm2
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[1.2]
90
51.
54.
Chapter 1: Equations and Inequalities
Let x = the number [1.2] 1 1 1 x− x = 4+ x 2 4 5 1 ⎞ 1 ⎞ ⎛ ⎛1 20⎜ x − x ⎟ = 20⎜ 4 + x ⎟ 5 ⎠ 2 4 ⎝ ⎝ ⎠ 10 x − 5 x = 80 + 4 x 5 x = 80 + 4 x x = 80
52.
d = rt d = 8t 6(7 − t ) = 8t
P = 54 54 = 2l + 2 w 54 = 2(2 w − 9) + 2 w 54 = 4 w − 18 + 2 w 72 = 6 w 12 = w 2 w − 9 = 2(12) − 9 = 24 − 9 = 15 width = 12 ft, length = 15 ft [1.2]
Let x = cost last year Cost = last year + raise Let x = the number. 21 = x + 0.05 x 21 = 1.05 x 21 =x 1.05 20 = x
55.
The cost last year was $20.00 . [1.2]
57.
53.
4%
x
6%
5500 − x
42 − 6t = 8t 42 = 14t 3=t
d = 8(3) = 24 nautical miles [1.2] 56.
0.04 x + 0.06(5500 − x) = 295 0.04 x + 330 − 0.06 x = 295 − 0.02 x = −35 x = 1750 5500 − 1750 = 3750 $1750 in the 4% account $3750 in the 6% account [1.2]
Let x = monthly maintenance cost per owner 18 x = 24( x − 12) 18 x = 24 x − 288 − 6 x = −288 x = 48 18 x = 864 The total monthly maintenance cost is $864. [1.2]
58.
d = 6( 7 − t )
Let x = price of battery x + 20 = price of calculator x + x + 20 = 21 2 x + 20 = 21 2x = 1 x = 0.50 x + 20 = 20.50 Price of calculator is $20.50. [1.2] Price of battery is $0.50.
P = 40 A = 96 40 = 2l + 2 w 20 = l + w l = 20 − w 96 = lw 96 = (20 − w) w 96 = 20w − w2 w2 − 20 w + 96 = 0 ( w − 12)( w − 8) = 0 w = 12 or w = 8 l = 20 − 12 or l = 20 − 8 l =8 l = 12
Length = 8 in. and width = 12 in., or length = 12 in. and width = 8 in. [1.2]
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Chapter Review
91
59.
Part completed In 1 hour
Time Mason
x−9
Apprentic e
x
60.
1 x−9 1 x
1 ⎞ ⎛1 6⎜ + ⎟ =1 ⎝ x x−9⎠
Let x = number of adult tickets 4526 − x = number of student tickets 8 x + 2(4526 − x) = 33,196 8 x + 9052 − 2 x = 33,196 6 x = 24,144 x = 4024 4526 − x = 502 4024 adult tickets, 502 student tickets [1.2]
1 ⎞ ⎛1 6 x( x − 9)⎜ + ⎟ = 1x( x − 9) ⎝ x x−9⎠ 6( x − 9) + 6 x = x 2 − 9 x 6 x − 54 + 6 x = x 2 − 9 x 0 = x 2 − 21x + 54 0 = ( x − 18)( x − 3) x = 18 or x = 3 (Note : x = 3 ⇒ mason's time = − 6 hours. Thus x ≠ 3.) Apprentice takes 18 hours to build the wall. [1.4] 61.
62.
R = 72 x − 2 x 2 , R > 576 72 x − 2 x 2 > 576 0 > 2 x 2 − 72 x + 576 2 x 2 − 72 x + 576 < 0 x 2 − 36 x + 288 < 0 ( x − 24)( x − 12) < 0 The product is negative. x − 24 = 0 ⇒ x = 24 x − 12 = 0 ⇒ x = 12
1 d 4 V = 144 h=
Critical values are 24 and 12.
d = 2r ⇒ r =
(x − 24)(x − 12)
d 2
1 V = π r 2h 3
(12, 24)
2
1 ⎛d ⎞ ⎛1 ⎞ 144 = π ⎜ ⎟ ⎜ d ⎟ 3 ⎝2⎠ ⎝4 ⎠ 144 =
π d3 48
144(48) = d3 π d ≈ 13 ft 63.
The revenue is greater than $576 when the price is between $12 and $24. [1.5]
[1.2]
−1.96 < x − 50 < 1.96 5 −9.8 < 1.63 − x < 9.8 40.2 < − x < 59.8 x 41 < < 59, where x is an integer
64.
63.8 − μ < 1.645 0.45 −0.74025 < 63.8 − μ < 0.74025 −64.54025 < − μ < −63.05975 μ 64.5 > > 63.1 μ 63.1 < < 64.5 lb −1.645 <
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92
65.
Chapter 1: Equations and Inequalities
39 − μ < 1.96 0.53 −1.0388 < 39 − μ < 1.0388 −40.0388 < − μ < −37.9612 > 38.0 40.0 > μ μ < 40.0 lb 38.0 <
66.
−1.96 <
Let x = the score on the fifth test. 68 ≤ 82 + 72 + 64 + 95 + x ≤ 79 5 313 + x ≤ 79 68 ≤ 5 340 ≤ 313 + x ≤ 395 27 ≤ x ≤ 82 The student needs to earn a score in the interval [27, 82] to receive a C grade for the course. [1.5]
67.
Let C = the circumference, r = the radius, and d = the diameter. C = 2π r = π d 29.5 ≤ C ≤ 30.0 29.5 ≤ π d ≤ 30.0 29.5 ≤ d ≤ 30.0
π
68.
45 x 2 − 190 x + 100 = 0 9 x 2 − 38 x + 20 = 0 x=
π
The diameter of the basketball is from 9.39 to 9.55 inches. [1.5] A = km r2 9.8 =
70.
k (5.98 × 1026 ) (6,370,000)
9.8(6,370,000) 5.98 × 10
26
k (1.5)4 82
4(82 ) = k (1.5)4 4(82 )
=k
=k (1.5)4 k ≈ 50.5679
k ≈ 6.6497 × 10−13 A = 6.6497 ×210 r A=
4 L = kd2 h
4=
2
9.8(6,370,000)2 = k (5.98 × 1026 ) 2
−( −38) ± ( −38) 2 − 4(9)(20) 2(9)
= 38 ± 724 18 x ≈ 0.6 or x ≈ 3.6 More than 0.6 mi but less than 3.6 mi from the city center. [1.5]
9.39 ≤ d ≤ 9.55
69.
300 = −45 x 2 + 190 x + 200
−13
m
d L = 50.5679 h2
(6.6497 × 10−13 )(7.46 × 1024 ) (1,740,000)
L=
2
A ≈ 1.64 meters/sec2
[1.6]
4
50.5679(4)4
122 L ≈ 89.9 tons
[1.6]
....................................................... QR1.
QR2. EG = ED EG = 1 + x
x =1 1− x x x2 = 1 − x
EG = EF 2 + FG 2 = 12 + 22 = 5
2
x + x −1 = 0 −1 ± 12 − 4(1)( −1) −1 ± 1 + 4 −1 ± 5 x= = = 2(1) 2 2
QR3. Answers will vary.
Quantitative Reasoning
5 =1+ x x = 5 −1 AD = 2 + x = 2 + 5 − 1 = 1 + 5 = φ AB 2 2 2
QR4. Answers will vary.
QR5. Answers will vary.
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Chapter Test
93
....................................................... 1.
3.
Chapter Test
3(2 x − 5) + 1 = −2( x − 5) [1.1] 6 x − 15 + 1 = −2 x + 10 6 x − 14 = −2 x + 10 8 x = 24 x=3
2.
6 x 2 − 13x − 8 = (3x − 8)(2 x + 1) = 0 [1.3] 3x − 8 = 0 or 2 x + 1 = 0 8 x= x =−1 3 2
4.
x − 3 = 8 [1.1] x − 3 = 8 or x = 11
x − 3 = −8 x = −5
=−1 2
2 x2 − 8x + 1 = 0 ⇒ x2 − 4 x
[1.3]
x 2 − 4 x + 4 = − 1 + 4 ⇒ ( x − 2)2 = 7 2 2 x − 2 = ± 7 = ± 7 = ± 7 ⋅ 2 = ± 14 2 2 2 2⋅2 x = 2 ± 14 = 4 ± 14 = 4 ± 14 2 2 2 2
5.
3x 2 − 5 x − 1 = 0
[1.3]
6.
a = 3, b = −5, c = −1 x=
b2 − 4ac = (3)2 − 4(2)(1) = 9 − 8 = 1 The discriminant, 1, is a positive number. Therefore, there are two real solutions.
−( −5) ± ( −5)2 − 4(3)( −1) 2(3)
= 5 ± 25 + 12 = 5 ± 37 6 6
7.
ax − c = c( x − d ) ax − c = cx − cd ax − cx = c − cd x ( a − c) = c − cd
2 x 2 + 3x + 1 = 0 [1.3] a = 2, b = 3, c = 1
8.
x − 2 −1= 3− x
(
x − 2 − 1) = ( 3 − x ) 2
2
x − 2 − 2 x − 2 +1= 3− x 2x − 4 = 2 x − 2
x = c − cd , a ≠ c [1.2] a−c
x−2= x−2 ( x − 2)2 = ( x − 2 ) x2 − 4 x + 4 = x − 2 x 2 − 5x + 6 = 0 ( x − 3)( x − 2) = 0 x−3=0⇒ x =3 x−2=0⇒ x =2 Check
2 − 2 −1 = 3− 2 −1 = 1 (No)
3 − 2 −1 = 3 − 3 1−1 = 0 0=0 The solution is 3. [1.4]
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2
94
9.
Chapter 1: Equations and Inequalities
3x 2 / 3 + 10 x1/ 3 − 8 = 0 [1.4]
10.
1/ 3
Let u = x
3u 2 + 10u − 8 = 0 (3u − 2)(u + 4) = 0 u=2 3 1/ 3 2 x = 3
(x )
1/ 3 3
()
= 2 3 8 x= 27
3 −3= 5 [1.4] x+2 4 x+2 4( x + 2) ⎛⎜ 3 − 3 ⎞⎟ = 4( x + 2) ⎛⎜ 5 ⎞⎟ ⎝ x +2 4⎠ ⎝ x +2⎠ 4(3) − 3( x + 2) = 4(5) 12 − 3x − 6 = 20
u = −4
or
−3x = 14
x1/ 3 = −4 3
(x )
1/ 3 3
= ( −4 )
x = − 14 3
3
x = −64
2 x − 5 ≤ 11 or −3x + 2 > 14 2 x ≤ 16 −3x > 12 x≤8 x < −4 {x | x ≤ 8} ∪ {x | x < −4} = {x | x ≤ 8} [1.5]
11.
a.
12.
x 2 + x − 12 ≥ 0 x +1 ( x + 4)( x − 3) ≥0 x +1 The quotient is positive or zero. x + 4 = 0 ⇒ x = −4 x−3=0⇒ x =3 x + 1 = 0 ⇒ x = −1 Critical values are –4, 3, and −1. ( x + 4)( x − 3) x +1
b.
2x − 1 < 9 and −3x + 1 ≤ 7 2 x < 10 −3x ≤ 6 x<5 x ≥ −2 {x | x < 5} ∩ {x | x ≥ −2} = [ −2, 5) [1.5]
13.
x − 11 5 ≤ 9 32 32 − 9 ≤ x − 11 5 ≤ 9 32 32 32 7 10 ≤ x ≤ 11 7 8 16 The range is from 10 7 in. to 11 7 in. [1.5] 8 16
Denominator ≠ 0 ⇒ x ≠ −1. [−4, − 1) ∪ [3, ∞) [1.5] 14.
Let x = the rate of the current. Rate with current = 5 + x. Rate against current = 5 − x. d = rt 21 = (5 + x )t 9 = (5 − x )t 21 = t 9 =t 5+ x 5− x 21 = 9 5+ x 5− x 21(5 − x ) = 9(5 + x ) 105 − 21x = 45 + 9 x
15.
x
0.20
Remove x amount of 20%
x
1.00
Add x amount of 100%
6(0.20) − x(0.20) + x(1.00) = 6(0.50) 1.2 + 0.8 x = 3 0.8 x = 1.8 x = 2.25 liters [1.2]
60 = 30 x 2=x The current is 2 mph. [1.2]
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Cumulative Review
16.
95
Let x = number of hours the assistant needs to cover the parking lot. 6⎡ 1 + 1⎤ =1 ⎣⎢ 10 x ⎦⎥
17.
10 x (6) ⎡ 1 + 1 ⎤ = 10 x (1) ⎢⎣ 10 x ⎥⎦ 6 x + 60 = 10 x
10 + 0.18 x > 18 + 0.10 x 0.08 x > 8 x > 100 If you drive more than 100 miles, then company A is less expensive. [1.5]
−4 x = −60 x = 15 The assistant takes 15 hours to cover the parking lot. [1.4]
18.
0.5 = −0.0002348 x 2 + 0.0375 x
19.
2
0.0002348 x − 0.0375 x + 0.5 = 0
200(2 x 2 + 25) = 4500 x
−( −0.0375) ± ( −0.0375)2 − 4(0.0002348)(0.5) x= 2(0.0002348)
400 x 2 − 4500 x + 5000 = 0 4 x 2 − 45 x + 50 = 0 ( x − 10)(4 x − 5) = 0
= 0.0375 ± 0.00094 0.00047 0.0375 ± 0.0306 = 0.0004696 x ≈ 145.0 or x ≈ 14.7 More than 14.7 ft but less than 145.0 ft from a side line. [1.5] 20.
x 200 = 4500 2 x 2 + 25
x − 10 = 0 4 x − 5 = 0 x = 10 x = 1.25 More than 1.25 mi but less than 10 mi from the city center. [1.5]
v= k d 4= k 3000 k = 4 3000 = 40 30 v = 40 30 = 40 30 50 2500 v = 4 30 ≈ 4.4 miles/second [1.6] 5
.......................................................
Cumulative Review
1.
4 + 3( −5) = 4 − 15 = −11 [P.1]
3.
(3x − 5) 2 − ( x + 4)( x − 4) = (9 x 2 − 30 x + 25) − ( x 2 − 16) [P.3] 4.
2.
0.00017 = 1.7 × 10−4 [P.2] 8 x 2 + 19 x − 15 = (8 x − 5)( x + 3) [P.4]
= 9 x 2 − 30 x + 25 − x 2 + 16 = 8 x 2 − 30 x + 41 5.
7 x − 3 − 5 = 7 x − 3 − 5 x + 20 = 2 x + 17 [P.5] x−4 x−4 x−4
6.
a 2 / 3 ⋅ a1/ 4 = a 2 / 3 + 1/ 4 = a11/12 [P.2]
7.
(2 + 5i )(2 − 5i ) = 4 − 25i 2 = 4 + 25 = 29 [P.6]
8.
2(3x − 4) + 5 = 17 [1.1] 2(3x − 4) = 12 6 x = 20 x = 10 3
Copyright © Houghton Mifflin Company. All rights reserved.
96
Chapter 1: Equations and Inequalities
10.
2 x 2 − 4 x = 3 [1.3]
9.
2x − 4x − 3 = 0 x= 11.
−( −4) ± ( −4) − 4(2)( −3) 4 ± 2 10 2 ± 10 = = 2(2) 4 2 x = 3+ 9 − x
[1.4]
( x − 3)2 = ( 9 − x )
2
x2 − 6x + 9 = 9 − x x 2 − 5 x = x ( x − 5) = 0 x = 0 or x = 5 x 3 − 36 x = x ( x 2 − 36) = x ( x + 6)( x − 6) = 0 The solutions are 0, –6, 6. [1.4]
2 x − 6 = −4
2 x = 10 or x=5
2
x −3= 9− x
12.
2 x − 6 = 4 [1.1] 2x − 6 = 4
2
2x = 2 x =1
Check 0: 0 = 3+ 9 − 0
Check 5: 5 = 3+ 9 −5
0 = 3+ 9 0 = 3+ 3 0=6 No
5 = 3+ 4 5 = 3+ 2 5=5 The solution is 5.
13.
2 x 4 − 11x 2 + 15 = 0 Let u = x 2 . 2u 2 − 11u + 15 = (2u − 5)(u − 3) = 0 u−3=0 or 2u − 5 = 0 u=3 u = x2 = 5 2 x2 = 3 10 5 =± x=± x=± 3 2 2 The solutions are − 10 , 2
14.
3x − 1 > 2 or −3x + 5 ≥ 8 3x > 3 −3x ≥ 3 x >1 x ≤ −1 The solution is {x | x ≤ −1 or x > 1} . [1.5]
16.
x − 2 ≥ 4 ⇒ x − 2 − 4 ≥ 0 ⇒ x − 2 − 4(2 x − 3) ≥ 0 ⇒ x − 2 − 8 x + 12 ≥ 0 ⇒ −7 x + 10 ≥ 0 2x − 3 2x − 3 2x − 3 2x − 3 2x − 3 2x − 3 Solve −7 x + 10 = 0 and 2 x − 3 − 0 to find the critical values. −7 x + 10 = 0 2x − 3 = 0
x = 10 7
15.
3 . [1.4]
x−6 ≥2 ⇒ x−6≥2 or x − 6 ≤ −2 x≥8 x≤4 The solution is ( −∞, 4] ∪ [8, ∞) . [1.5]
x=3 2
The critical values are 10 and 3 . The intervals are ⎛⎜ −∞, 10 ⎞⎟ , ⎛⎜ 10 , 3 ⎞⎟ and ⎛⎜ 3 , ∞ ⎞⎟ 7 2 7 ⎠ ⎝ 7 2⎠ ⎝2 ⎠ ⎝ 0 2 2 2 − − 10 : ≥4⇒ ≥ 4 ⇒ ≥ 4 , which is false. Test 0, in the interval −∞, 7 2(0) − 3 3 −3
(
10 , − 3, 2
)
Test 1.45, in the interval ⎛⎜ 10 , 3 ⎞⎟ : 1.45 − 2 ≥ 4 ⇒ −0.55 ≥ 4 ⇒ 5.5 ≥ 4 , which is true. −0.1 ⎝ 7 2 ⎠ 2(1.45) − 3 Test 2, in the interval ⎛⎜ 3 , ∞ ⎞⎟ : 2 − 2 ≥ 4 ⇒ 0 ≥ 4 ⇒ 0 ≥ 4 , which is false. 1 ⎝2 ⎠ 2(2) − 3 3 The denominator cannot equal zero ⇒ x ≠ . 2 ⎧ ⎫ The solution is ⎨ x 10 ≤ x < 3 ⎬ . [1.5] 2⎭ ⎩ 7
Copyright © Houghton Mifflin Company. All rights reserved.
Cumulative Review
97
17.
18.
w
P = R−C = 200 x − 0.004 x 2 − (65 x + 320,000) = −0.004 x 2 + 135 x − 320,000 Profits must be greater than or equal to 600,000.
w +16 Perimeter = 2(Length) + 2(Width) 200 = 2( w + 16) + 2 w 200 = 2 w + 32 + 2 w 168 = 4 w 42 = w w = 42 w + 16 = 58 The width is 42 feet; the length is 58 feet. [1.2]
−0.004 x 2 + 135 x − 320,000 ≥ 600,000 −0.004 x 2 + 135 x − 920,000 ≥ 0 x=
−135 ± (135)2 − 4( −0.004)( −920,000) 2( −0.004)
= −135 ± 3505 −0.008 = 9475 or 24,275 9475 to 24,275 printers should be manufactured. [1.5]
19.
Let x = the score on the fourth test. 80 ≤ 86 + 72 + 94 + x < 90 and 0 ≤ x ≤ 100 4 252 x < 90 + 80 ≤ 4 320 ≤ 252 + x < 360 68 ≤ x < 108 [68, 108) ∩ [0, 100] = [68, 100] The fourth test score must be from 68 to 100. [1.5]
20.
600 p ≥ 100 100 − p 600 p ≥ 100(100 − p ) 600 p ≥ 10,000 − 100 p 700 p ≥ 10,000
p ≥ 14.3
and
600 p ≤ 180 100 − p 600 p ≤ 180(100 − p ) 600 p ≤ 18,000 − 180 p 780 p ≤ 18,000
p ≤ 23.1 They can expect to ticket from 14.3% to 23.1% of the speeders. [1.5]
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 2
Functions and Graphs Section 2.1 1.
3.
2.
a.
$31,500
b.
Increase from 2004 to 2005 33.5 − 32.9 = 0.6 Increase from 2005 to 2006 33.50 + 0.6 = 34.1 The per capita income for 2006 would be $34,100. Percent increase from 2004 to 2005 33.5 − 32.9 = 1.824% 32.9 Percent increase from 2005 to 2006 33.50(1.01824) = 34.111 The per capita income for 2006 would be $34,111.
c.
4.
a. b.
c.
When the cost of a game is $22, 50 million games can be sold. The projected numbers of sales decreases as the price of this game increases.
p
R = p⋅N
8 8 ⋅ 80 = 640 15 15 ⋅ 70 = 1050 22 22 ⋅ 60 = 1320 27 27 ⋅ 50 = 1350 31 31 ⋅ 40 = 1240 34 34 ⋅ 30 = 1020 36 36 ⋅ 20 = 720 37
d.
5.
d = ( −8 − 6)2 + (11 − 4)2
6.
37 ⋅ 10 = 370
The revenue increases to a certain point and then decreases as the price of the game increases.
d = ( −10 − ( −5))2 + (14 − 8)2
7.
d = ( −10 − ( −4))2 + (15 − ( −20))2
= ( −14)2 + (7)2
= ( −5)2 + (6)2
= ( −6)2 + (35)2
= 196 + 49
= 25 + 36
= 36 + 1225
= 245
= 61
= 1261
=7 5
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.1
8.
99
d = (36 − 40)2 + (20 − 32)2
9.
d = (0 − 5)2 + (0 − (−8))2
d = (5 − 0)2 + (13 − 0)2
10.
= (−4)2 + (−12)2
= (−5) 2 + (8)2
= 52 + 132
= 16 + 144
= 25 + 64
= 25 + 169
= 160
= 89
= 194
= 4 10 11.
d = ( 12 − 3)2 + ( 27 − 8)2
12.
d = (6 − 125)2 + (2 5 − 20)2
= (2 3 − 3)2 + (3 3 − 2 2) 2
= (6 − 5 5)2 + (2 5 − 2 5)2
= ( 3) 2 + (3 3 − 2 2)2
= (6 − 5 5)2 + 02
= 3 + (27 − 12 6 + 8)
= (6 − 5 5)2 = 6 − 5 5 = 5 5 − 6
= 3 + 27 − 12 6 + 8
Note: for another form of the solution,
= 38 − 12 6
d = (6 − 5 5)2 = 36 − 60 5 + 125 = 161 − 60 5
13.
d = ( − a − a ) 2 + ( −b − b ) 2
14.
d = (a − (a − b))2 + (a + b − b)2
= (−2a )2 + (−2b) 2
= ( a − a + b) 2 + ( a ) 2
= 4a 2 + 4b 2
= b2 + a 2
= 4(a 2 + b 2 )
= a 2 + b2
= 2 a 2 + b2
15.
d = (−2 x − x)2 + (3 x − 4 x)2 with x < 0
= (−3 x) 2 + ( − x)2
= 9 x2 + x2
= 9 x2 + x2
= 10 x 2
= 10 x 2 (Note: since x < 0, x 2 = − x)
(4 − x )2 + (6 − 0)2 = 10
(
d = (−2 x − x)2 + (3 x − 4 x)2 with x > 0
= (−3x)2 + ( − x)2
= − x 10
17.
16.
(4 − x )2 + (6 − 0)2
)
2
= 102
= x 10 (since x > 0, x 2 = x)
18.
(5 − 0)2 + ( y − ( −3)2 = 12
(
)
2
= 122
25 + y 2 + 6 y + 9 = 144
16 − 8 x + x 2 + 36 = 100
y 2 + 6 y − 110 = 0
x 2 − 8 x − 48 = 0 ( x − 12)( x + 4) = 0 x = 12 or x = −4 The points are (12, 0), ( − 4, 0).
(5) 2 + ( y + 3)2
−6 ± 62 − 4(1)( −110) 2(1) − ± + 440 6 36 y= 2 y = −6 ± 476 2 − ± 6 2 119 y= 2 y = −3 ± 119 y=
The points are (0, − 3 + 119), (0, − 3 − 119). Copyright © Houghton Mifflin Company. All rights reserved.
100
19.
22.
Chapter 2: Functions and Graphs
⎛ x + x2 y1 + y2 ⎞ M =⎜ 1 , 2 ⎟⎠ ⎝ 2
20.
⎛ x + x2 y1 + y2 ⎞ M =⎜ 1 , ⎟ 2 ⎠ ⎝ 2
= ⎛⎜ 1 + 5 , −1 + 5 ⎞⎟ 2 ⎠ ⎝ 2
= ⎛⎜ −5 + 6 , −2 + 10 ⎞⎟ 2 ⎠ ⎝ 2
= ⎛⎜ 6 , 4 ⎞⎟ ⎝2 2⎠ = (3, 2)
= ⎛⎜ 1 , 8 ⎞⎟ ⎝2 2⎠
21.
M = ⎛⎜ 6 + 6 , −3 + 11 ⎞⎟ 2 ⎠ ⎝ 2 = ⎛⎜ 12 , 8 ⎞⎟ ⎝ 2 2⎠ = (6, 4)
= ⎛⎜ 1 , 4 ⎞⎟ ⎝2 ⎠
⎛ 4 + ( −10) 7 + 7 ⎞ , M =⎜ ⎟ 2 2 ⎠ ⎝
23.
1.75 + ( −3.5) 2.25 + 5.57 ⎞ , M = ⎛⎜ ⎟ 2 2 ⎝ ⎠
24.
= ⎛⎜ − 1.75 , 7.82 ⎞⎟ 2 ⎠ ⎝ 2 = ( −0.875, 3.91)
= ⎛⎜ −6 , 14 ⎞⎟ ⎝ 2 2⎠ = ( −3, 7)
⎛ −8.2 + ( −2.4) , 10.1 + ( − 5.7) ⎞ ⎜ ⎟ 2 2 ⎝ ⎠ = ⎛⎜ − 10.6 , 4.4 ⎞⎟ 2 ⎠ ⎝ 2 = ( −5.3, 2.2)
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
⎛ 12 ⎞ Intercepts: ⎜ 0, ⎟, (6, 0 ) ⎝ 5⎠
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.1
40.
101
15 ⎞ ⎛ Intercepts: ⎜ 0,− ⎟, (5, 0) 4⎠ ⎝
41.
(0, 5 ), (0, − 5 ), (5, 0)
(0, 6 ), (0, − 6 ), (−6, 0)
42.
x = y2 − 6
x = − y2 + 5
43.
(0, 4), (0, −4), (−4, 0)
44.
(0, ± 2),( ± 2, 0)
45.
x2 + y 2 = 4
x = y3 − 2
x = | y | −4 46.
(0, 3 2 ), (−2, 0)
(0, 0)
47.
(0, ± 4), (±4, 0)
(0, ± 2), (±8, 0)
48.
| x − 4y | = 8 x2 = y 2
| x | + | y |= 4
49.
center (0, 0), radius 6
50.
center (0, 0), radius 7
51.
center (1, 3), radius 7
52.
center (2, 4), radius 5
53.
center (−2, −5), radius 5
54.
center (−3, −5), radius 11
55.
center (8, 0), radius 1
56.
center (0, 12), radius 1
57.
( x − 4)2 + ( y − 1) 2 = 22
60.
( x − 0 )2 + ⎛⎜ y − 2 ⎞⎟ = ( 11 ) 3⎠ ⎝
2
2
2
2
2
58.
( x − 5) 2 + ( y + 3) 2 = 42
59.
1⎞ ⎛ 1⎞ ⎛ ⎜x− ⎟ +⎜y − ⎟ = 2⎠ ⎝ 4⎠ ⎝
61.
( x − 0) 2 + ( y − 0) 2 = r 2
62.
( x − 0) 2 + ( y − 0) 2 = r 2
( 5 )2
(−3 − 0)2 + (4 − 0)2 = r 2
(5 − 0)2 + (12 − 0)2 = r 2
(−3)2 + 42 = r 2
52 + 122 = r 2
9 + 16 = r 2
25 + 144 = r 2 169 = 132 = r 2
25 = 52 = r 2 ( x − 0) 2 + ( y − 0)2 = 52
( x − 0)2 + ( y − 0)2 = 132
Copyright © Houghton Mifflin Company. All rights reserved.
102
63.
Chapter 2: Functions and Graphs
( x + 2)2 + ( y − 5)2 = r 2
64.
(1 + 2)2 + (7 − 5)2 = r 2
( x − 1)2 + ( y − 3)2 = r 2
32 + 22 = r 2
(4 − 1) 2 + ( −1 − 3)2 = r 2
9 + 4 = r2
( 13 ) = r 2 2 ( x + 2)2 + ( y − 5)2 = ( 13 )
32 + (−4)2 = r 2
2
13 =
9 + 16 = r 2 25 = 52 = r 2 ( x − 1)2 + ( y − 3)2 = 52 65.
x2 − 6x
+ y 2 = −5
2
66.
2
2
x 2 − 14 x
( x − 3)2 + ( y − 2)2 = 12 center (3, 2), radius 1
+ y2 + 8 y
2
= −56
68.
2
2
4 x2 + 4 x
x2 + x +
70.
+ y 2 = 63 4 2 63 + y = +1 4 4
1 4
2
2
2
2
+ y2 + 3 y = 15 2 4 2 2 1 3 9 15 x −x+ +y + y + = +1+9 4 2 4 4 4 4 x2 − x
⎛x − 1⎞ + ⎛ y + 3⎞ = ⎛ 5⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2⎠ ⎝ 2⎠ ⎝ ⎝2⎠ center ⎛⎜ 1 , − 3 ⎞⎟ , radius 5 2⎠ 2 ⎝2
( )
( x − 0)2 + ⎛⎜ y − 1 ⎞⎟ = 2 3⎠ ⎝ ⎛ 1⎞ center ⎜ 0, ⎟, radius 2 ⎝ 3⎠
⎛ 1 ⎞ center ⎜ − ,0 ⎟ , radius 4 ⎝ 2 ⎠
2
= 17
x 2 + ⎛⎜ y − 1 ⎞⎟ = 2 3⎠ ⎝
⎛ x + 1 ⎞ + ( y − 0) 2 = 42 ⎜ ⎟ 2⎠ ⎝
2
9 x2 + 9 y2 − 6 y
x2 + y2 − 2 y = 17 3 9 2 2 2 1 17 x +y − y + = +1 3 9 9 9
⎛ x + 1 ⎞ + y 2 = 16 ⎜ ⎟ 2⎠ ⎝
71.
= −25
2
( x − 5)2 + ( y + 1)2 = 12 center (5, −1), radius 1
+ 4 y 2 = 63
x2 + x
+ y2 + 2 y
x − 10 x + 25 + y + 2 y + 1 = −25 + 25 + 1
2
( x − 7) + ( y + 4) = 3 center (7, −4), radius 3 69.
x 2 − 10 x 2
x − 14 x + 49 + y + 8 y + 16 = −56 + 49 + 16 2
= −12
2
x − 6 x + 9 + y − 4 y + 4 = −12 + 9 + 4
2
( x − 3) + y = 2 center (3, 0), radius 2 67.
+ y2 − 4 y
2
x − 6 x + 9 + y = −5 + 9 2
x2 − 6x
2
72.
2
= − 25 4 2 2 9 25 25 x + 3x + + y − 5y + = − + 9 + 25 4 4 4 4 4 x 2 + 3x
+ y2 − 5 y
2
2
⎛ x + 3⎞ + ⎛ y − 5⎞ = ⎛ 3⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2⎠ ⎝ 2⎠ ⎝ ⎝2⎠ center ⎛⎜ − 3 , 5 ⎞⎟ , radius 3 2 ⎝ 2 2⎠
Copyright © Houghton Mifflin Company. All rights reserved.
2
Section 2.1
73.
103
d = ( −4 − 2)2 + (11 − 3)2
74.
= 36 + 64
d = (−3 − 7) 2 + (5 − (−2)) 2 = 100 + 49 = 149
Since the diameter is
= 100
149 , the radius is
149 . 2
⎛ 7 + ( − 3) (−2) + 5 ⎞ ⎛ 3 ⎞ Center is ⎜ , ⎟ = ⎜ 2, ⎟ 2 2 ⎝ ⎠ ⎝ 2⎠
= 10 Since the diameter is 10, the radius is 5. The center is the midpoint of the line segment from (2,3) to (-4,11). ⎛ 2 + (-4) 3 + 11 ⎞ , ⎟ = (−1,7) center ⎜ 2 ⎠ ⎝ 2
2 ⎛ 3⎞ 149 ⎞ ⎛ ( x − 2)2 + ⎜ y − ⎟ = ⎜⎜ ⎟⎟ 2⎠ ⎝ ⎝ 2 ⎠
2
( x + 1) 2 + ( y − 7) 2 = 5 2
75.
Since it is tangent to the x-axis, its radius is 11. 2
2
76.
2
Since it is tangent to the y-axis, its radius is 2. ( x + 2) 2 + ( y − 3) 2 = 2 2
( x − 7) + ( y − 11) = 11
.......................................................
Connecting Concepts
77.
78.
79.
80.
81.
82.
83.
84.
85.
87.
86.
⎛ x + 5 y +1⎞ , ⎜ ⎟ = (9, 3) 2 ⎠ ⎝ 2 therefore x + 5 = 9 2 x + 5 = 18 x = 13
88.
and
y +1 =3 2 y +1= 6 y=5
therefore x + 4 = −2 2 x + 4 = −4 x = −8
Thus (13, 5) is the other endpoint. 89.
⎛ x + ( −3) y + (−8) ⎞ , ⎜ ⎟ = (2, − 7) 2 2 ⎝ ⎠
⎛ x + 4 y + (−6) ⎞ , ⎜ ⎟ = (−2, 11) 2 ⎝ 2 ⎠ and
y + ( −6) = 11 2 y − 6 = 22 y = 28
Thus (−8, 28) is the other endpoint. 90.
⎛ x + 5 y + (−4) ⎞ , ⎜ ⎟ = (0, 0) 2 ⎝ 2 ⎠ y−4 =0 2 y−4=0 y=4
y −8 = −7 therefore x − 3 = 2 and 2 2 y − 8 = −14 x−3= 4 x=7 y = −6
therefore x + 5 = 0 2 x+5=0 x = −5
Thus (7, −6) is the other endpoint.
Thus (−5, 4) is the other endpoint.
Copyright © Houghton Mifflin Company. All rights reserved.
and
104
Chapter 2: Functions and Graphs
(3 − x )2 + (4 − y )2 = 5
91.
( −5 − x )2 + (12 − y )2 = 13
92.
(3 − x )2 + (4 − y )2 = 52
( −5 − x )2 + (12 − y )2 = 132
9 − 6 x + x 2 + 16 − 18 y + y 2 = 25
25 + 10 x + x 2 + 144 − 24 y + y 2 = 169
x2 − 6x + y2 − 8 y = 0 93.
x 2 + 10 x + y 2 − 24 y = 0
(4 − x ) 2 + (0 − y )2 + ( −4 − x )2 + (0 − y ) 2 = 10 (4 − x )2 + (0 − y ) 2 = 100 − 20 ( −4 − x )2 + (0 − y )2 + ( −4 − x )2 + ( − y )2 16 − 8 x + x 2 + y 2 = 100 − 20 ( −4 − x )2 + ( − y )2 + 16 + 8 x + x 2 + y 2 −16 x − 100 = −20 ( −4 − x )2 + ( − y )2 4 x + 25 = 5 ( −4 − x )2 + ( − y )2 16 x 2 + 200 x + 625 = 25 ⎡( −4 − x )2 + ( − y )2 ⎤ ⎣ ⎦ 16 x 2 + 200 x + 625 = 25 ⎡16 + 8 x + x 2 + y 2 ⎤ ⎣ ⎦ 16 x 2 + 200 x + 625 = 400 + 200 x + 25 x 2 + 25 y 2
Simplifying yields 9 x 2 + 25 y 2 = 225 . (0 − x )2 + (4 − y )2 − (0 − x )2 + ( −4 − y )2 = 6
94.
(
x 2 + (4 − y )2 − x 2 + (4 + y )2
)
2
= 62
x 2 + (4 − y )2 − 2 x 2 + (4 − y )2 x 2 + (4 + y 2 + x 2 + (4 + y )2 = 36 x 2 + 16 − 8 y + y 2 − 2 x 2 + (4 − y )2 x 2 + (4 + y 2 + x 2 + 16 + 8 y + y )2 = 36 2 x 2 + 2 y 2 − 4 = 2 x 2 + (4 − y )2 x 2 + (4 + y )2 ( x 2 + y 2 − 2)2 =
(
x 2 + (4 − y )2 x 2 + (4 + y )2
)
2
x 4 + x 2 y 2 − 2 x 2 + x 2 y 2 + y 4 − 2 y 2 − 2 x 2 − 2 y 2 + 4 = ( x 2 + (4 − y )2 )( x 2 + (4 + y ) 2 ) x 4 + x 2 y 2 − 4 x 2 − 4 y 2 + y 4 + 4 = ( x 2 + 16 − 8 y + y 2 )( x 2 + 16 + 8 y + y 2 ) x 4 + 2 x 2 y 2 − 4 x 2 − 4 y 2 + y 4 + 4 = x 4 + 16 x + 8 x 2 y + x 2 y 2 + 16 x 2 + 256 + 128 y + 16 y 2 −8 x 2 y − 128 y − 64 y 2 − 8 y 3 + x 2 y 2 + 16 y 2 + 8 y 3 + y 4 −36 x 2 + 28 y 2 = 252
95.
The center is (-3,3). The radius is 3. ( x + 3) 2 + ( y − 3) 2 = 32
96.
or
− 9 x 2 + 7 y 2 = 63.
⎛ 5 ⎞⎟ 5 5 ,− The center is ⎜ − . The radius is . ⎟ ⎜ 2 2 2 ⎠ ⎝ 2
2
⎞ ⎛ ⎞ ⎞ ⎛ ⎛ ⎜x+ 5 ⎟ +⎜ y+ 5 ⎟ =⎜ 5 ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝
....................................................... PS1. x 2 + 3x − 4 2
( −3) + 3( −3) − 4 = 9 − 9 − 4 = −4
2
Prepare for Section 2.2 PS2. D ={−3, −2, −1, 0, 2} R ={1, 2, 4, 5}
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.2
105
PS4. 2 x − 6 ≥ 0 2x ≥ 6 x≥3
PS3. d = (3 − ( −4))2 + ( −2 − 1)2 = 49 + 9 = 58
PS5.
PS6. a = 3x + 4, a = 6 x − 5
x2 − x − 6 = 0 ( x + 2)( x − 3) = 0 x+2=0 x = −2 –2, 3
3x + 4 = 6 x − 5 9 = 3x 3= x a = 3(3) + 4 = 13
x−3=0 x=3
Section 2.2 1.
Given f ( x ) = 3 x − 1, a. f (2) = 3(2) − 1 = 6 −1 =5 b. f ( −1) = 3( −1) − 1 = −3 − 1 = −4 c. f (0) = 3(0) − 1 = 0 −1 = −1 ⎛2⎞ ⎛2⎞ d. f ⎜ ⎟ = 3⎜ ⎟ − 1 ⎝3⎠ ⎝3⎠ = 2 −1 =1 e. f ( k ) = 3( k ) − 1 = 3k − 1 f. f ( k + 2) = 3( k + 2) −1 = 3k + 6 −1 = 3k + 5
2.
Given g ( x) = 2 x 2 + 3, a.
g (3) = 2(3)2 + 3 =18 + 3 = 21
3.
Given A( w) = w 2 + 5 , a.
= 5
b.
g (−1) = 2(−1)2 + 3 = 2+3 =5
b.
c.
g (0) = 2(0)2 + 3 = 0+3 =3
c.
2
d.
e.
⎛ ⎞ ⎛ ⎞ g ⎜ 1 ⎟ = 2⎜ 1 ⎟ +3 ⎝2⎠ ⎝2⎠ = 1 +3 2 =7 2
f.
A(2) = (2)2 + 5 = 9 =3 A(−2) = (−2)2 + 5 = 9 =3
d.
A(4) = 42 + 5 = 21
e.
A(r +1) = (r +1)2 + 5 = r 2 + 2r +1+ 5 = r 2 + 2r + 6
g (c) = 2(c)2 + 3 = 2c2 + 3
A(0) = (0)2 + 5
f.
g (c + 5) = 2(c + 5)2 + 3 2
= 2c + 20c + 50 + 3 = 2c 2 + 20c + 53
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A(−c) = (−c)2 + 5 = c2 + 5
106
4.
Chapter 2: Functions and Graphs
Given J (t ) = 3t 2 − t , a.
b.
5.
J (−4) = 3(−4)2 − (−4) = 48 + 4 = 52
Given f ( x ) = a.
J (0) = 3(0)2 − (0) = 0−0 =0
c.
d.
1 ⎛1⎞ ⎛1⎞ J ⎜ ⎟ = 3⎜ ⎟ − 3 ⎝ 3⎠ ⎝3⎠ 1 1 = − 3 3 =0
J (−c) = 3(−c)2 − (−c)
d.
J ( x +1) = 3( x +1)2 − ( x +1)
e.
2
= 3x + 6 x + 3− x −1 = 3x 2 + 5 x + 2 f.
f.
J ( x + h) = 3( x + h)2 − ( x + h)
1 1 = −2 2
⎛ ⎞ f ⎜− 3 ⎟= 1 ⎝ 5⎠ −3 5 1 = 3
= 3c 2 + c e.
1 1 = 2 2
f ( −2) =
b.
2
c.
f ( 2) =
1 , x
5
=1÷ 3 =1⋅ 5 = 5 5 3 3 1 1 f ( 2) + f ( −2) = + = 1 2 2 1 1 2 f ( c + 4) = = 2 2 c +4 c +4 f (2 + h) =
1 2+h
= 3x 2 + 6 xh + 3h2 − x − h 6.
Given T ( x) = 5, a. T (−3) = 5 b. T ( 0) = 5 c. d. e. f.
⎛2⎞ T⎜ ⎟ = 5 ⎝7⎠ T (3) + T (1) = 5 + 5 = 10 T ( x + h) = 5 T (3k + 5) = 5
7.
x , x
Given s ( x) =
8.
a.
s ( 4) =
4 4 = =1 4 4
a.
b.
s (5) =
5 5 = =1 5 5
b.
c.
s ( −2 ) =
−2 −2 = = −1 2 −2
d.
s (−3) =
−3 −3 = = −1 3 −3
e.
Since t > 0, t = t. s (t ) =
f.
c.
d.
t t = =1 t t
Since t < 0, t = −t.
e.
t t = = −1 t −t
f.
s (t ) =
x , x+4 0 0 r ( 0) = = =0 0+4 4 −1 −1 1 r ( −1) = = =− −1+ 4 3 3 −3 −3 = = −3 r ( −3) = 1 −3+ 4 ⎛1⎞ 1 ⎜ ⎟ 2 ⎛1⎞ 2 =⎝ ⎠ r⎜ ⎟ = 1 9⎞ ⎛ ⎝2⎠ +4 ⎜ ⎟ 2 ⎝2⎠ 1 9 1 2 1 = ÷ = ⋅ = 2 2 2 9 9 0.1 0.1 1 r (0.1) = = = 0.1 + 4 4.1 41 10.000 r (10,000) = 10,000 + 4 10,000 2500 = = 10,004 2501
Given r ( x) =
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.2
9.
107
a.
Since x = −4 < 2, use P ( x ) = 3 x + 1. P ( −4) = 3( −4) + 1 = −12 + 1 = −11
b.
Since x = 5 ≥ 2, use P ( x) = − x 2 + 11.
b.
P 5 = − 5 + 11 = −5 + 11 = 6 Since x = c < 2, use P ( x ) = 3 x + 1. P (c) = 3c + 1 Since k ≥ 1, then x = k + 1 ≥ 2,
c.
so use P ( x) = − x 2 + 11.
d.
c. d.
10.
Since t = −4 and 0 ≤ t ≤ 5, use Q (t ) = 4. Q ( 0) = 4 Since t = e and 6 < e < 7, then 5 < t ≤ 8, so use Q (t ) = −t + 9. Q (e ) = − e + 9 Since t = n and 1 < n < 2, then 0 ≤ t ≤ 5, so use Q(t ) = 4 Q ( 0) = 4
a.
( ) ( )2
Since t = m 2 + 7 and 1 < m ≤ 2,
P ( k + 1) = −( k + 1)2 + 11 = −( k 2 + 2k + 1) + 11
then 12 < m 2 ≤ 22
= − k 2 − 2k − 1 + 11
12 + 7 < m 2 + 7 ≤ 22 + 7
= − k 2 − 2k + 10
1 + 7 < m2 + 7 ≤ 4 + 7 8 < m 2 + 7 ≤ 11 thus 8 < t ≤ 11, so use Q(t ) = t − 7
Q( m 2 + 7) =
(m2 + 7)− 7
= m 2 = m = m since m > 0
11.
2x+3y = 7 3 y = −2 x + 7 y = − 2 x + 7 , y is a function of x. 3 3
12.
5x + y = 8 y = −5 x + 8, y is a function of x.
13.
− x + y2 = 2
14.
x2 − 2 y = 2
y2 = x + 2
−2 y = − x 2 + 2 y = 1 x 2 − 1, y is a function of x. 2
y = ± x + 2, y is a not function of x.
15.
y = 4 ± x , y is not a function of x since for each x > 0 there are two values of x.
16.
x2 + y2 = 9 y2 = 9 − x2 y = ± 9 − x 2 , y is a not function of x.
17.
y = 3 x , y is a function of x.
18.
y = x + 5, y is a function of x.
19.
y2 = x2
20.
y 3 = x3 3
y = x3 = x, y is a function of x.
y = ± x 2 , y is a not function of x.
21.
Function; each x is paired with exactly one y.
22.
Not a function; 5 is paired with 10 and 8.
23.
Function; each x is paired with exactly one y.
24.
Function; each x is paired with exactly one y.
25.
Function; each x is paired with exactly one y.
26.
Function; each x is paired with exactly one y.
27.
f ( x) = 3x − 4
Domain is the set of all real numbers.
28.
f ( x) = −2 x + 1
Domain is the set of all real numbers.
29.
f ( x) = x2 + 2
Domain is the set of all real numbers.
30.
f ( x ) = 3x 2 + 1
Domain is the set of all real numbers.
Copyright © Houghton Mifflin Company. All rights reserved.
108
Chapter 2: Functions and Graphs
{
4 x+2
31.
f ( x) =
33.
f ( x) = 7 + x
35.
f ( x) = 4 − x 2
37.
f ( x) =
}
Domain is x x ≠ −2 .
1 x+4
{
}
Domain is x x ≥ −7 .
{
}
Domain is x − 2 ≤ x ≤ 2 .
{
}
Domain is x x > −4 .
39.
{
6 x−5
32.
f ( x) =
34.
f ( x) = 4 − x
36.
f ( x ) = 12 − x 2
38.
f ( x) =
}
Domain is x x ≠ 5 .
1 5− x
{
}
Domain is x x ≤ 4 .
{
}
Domain is x − 2 3 ≤ x ≤ 2 3 .
{
}
Domain is x x < 5 .
40.
Domain: the set of all real numbers Domain: the set of all real numbers 41.
42.
Domain: the set of all real numbers 43.
44.
Domain: { x
− 6 ≤ x ≤ 6}
Domain: { x 0 ≤ x ≤ 4}
45.
46.
Domain: { x
47.
Domain: the set of all real numbers
− 3 ≤ x ≤ 3}
int ⎣⎡102 (2.3458) + 0.5⎦⎤ 10
2
Domain:
[ ] = int 235.08 = 235 = 2.35 100 100
48.
{x
}
0≤ x≤4
int [10(34.567) + 0.5] int [346.17] 346 = = = 34.6 10 10 10
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.2
109
49.
int ⎣⎡103 (34.05622) + 0.5⎦⎤
51.
int ⎡⎣104 (0.08951) + 0.5⎤⎦
53.
a.
103
104
b.
=
int [34,056.72] 34,056 = = 34.056 1000 1000
[ ] = int 895.6 = 895 = 0.0895 10,000 10,000
C (2.8) = 0.39 − 0.34int(1− 2.8) = 0.39 − 0.34int( −1.8) = 0.39 − 0.34( −2) = 0.39 + 0.68 = $1.07 c(w)
50.
int ⎡⎣100 (109.83) + 0.5⎤⎦
52.
int ⎡⎣103 (2.98245) + 0.5⎤⎦
54.
a.
Domain: [0, ∞)
b.
T (31,250) = 0.25(31,250 − 30,650) + 4220
100
[ ] = int 110.33 = 110 = 110 1 100
103
[ ] = int 2982.95 = 2982 = 2.982 1000 1000
= 0.25(600) + 4220 = 150 + 4220 = $4370
c.
T (78,900) = 0.28(78,900 − 74,200) + 15,107.50 = 0.28(4700) + 15,107.50 = 1316 + 15,107.50 = $16,423.50
55.
a. b. c. d.
Yes; every vertical line intersects the graph in one point. Yes; every vertical line intersects the graph in one point. No; some vertical lines intersect the graph at more than one point. Yes; every vertical line intersects the graph in at most one point.
56.
a. b. c. d.
Yes; every vertical line intersects the graph in at most one point. No; some vertical lines intersect the graph at more than one point. No; a vertical line intersects the graph at more than one point. Yes; every vertical line intersects the graph in one point.
57.
Decreasing on (−∞, 0] ; increasing on [0, ∞)
58.
Decreasing on (−∞, ∞)
59.
Increasing on (−∞, ∞)
60.
Increasing on (−∞, 2] ; decreasing on [2, ∞)
61.
Decreasing on (−∞, − 3] ; increasing on [ −3, 0] ; decreasing on [0, 3] ; increasing on [3, ∞)
62.
Increasing on (−∞, ∞)
64.
Constant on (−∞, ∞)
65.
Decreasing on (−∞, 0] ; constant on [0, 1]; increasing on [1, ∞)
66.
Constant on (−∞, 0] ; decreasing on [0, 3] ; constant on [3, ∞)
67.
g and F are one-to-one since every horizontal line intersects the graph at one point. f, V, and p are not one-to-one since some horizontal lines intersect the graph at more than one point.
68.
s is one-to-one since every horizontal line intersects the graph at one point. t, m, r and k are not one-to-one since some horizontal lines intersect the graph at more than one point.
63.
Constant on (−∞, 0] ; increasing on [0, ∞)
Copyright © Houghton Mifflin Company. All rights reserved.
110
69.
Chapter 2: Functions and Graphs
a.
2l + 2w = 50 2w = 50 − 2l w = 25 − l
b.
A = lw A = l (25 − l )
70.
a.
4 = 12 l d +l 4( d + l ) = 12l 4d + 4l = 12l 4d = 8l 1d =l 2 l (d ) = 1 d 2
b.
Domain: [0, ∞)
c.
l (8) = 1 (8) = 4 ft 2
A = 25l − l 2
71.
v(t ) = 80,000 − 6500t ,
0 ≤ t ≤ 10
73.
a.
C ( x) = 5(400) + 22.80 x = 2000 + 22.80 x
b.
R ( x) = 37.00 x
c.
P ( x) = 37.00 x − C ( x) = 37.00 − [2000 + 22.80 x] = 37.00 x − 2000 − 22.80 x = 14.20 x − 2000
72.
v(t ) = 44,000 − 4200t ,
74.
a.
V = lwh V = (30 − 2 x)(30 − 2 x)( x) V = (900 −120 x + 4 x 2 )( x) V = 900 x −120 x 2 + 4 x3
V = lwh ⇒ the domain of V is dependent on the domains of l, w, and h. Length, width and height must be positive values ⇒ 30 − 2 x > 0 and x > 0. −2 x > −30 x < 15 Thus, the domain of V is {x | 0 < x < 15}.
b.
Note x is a natural number. 75.
0≤t ≤8
15 = 15 − h 3 r 15 5= −h r 5r =15 − h h =15 − 5r h(r ) =15 − 5r
76.
r =2 h 4 r=2h 4 1 r= h 2 1 V = π r 2h 3
a.
b.
2
⎛ ⎞ ⎛ ⎞ V = 1 π ⎜ 1 h ⎟ h = 1 π ⎜ 1 h2 ⎟ h 3 ⎝2 ⎠ 3 ⎝4 ⎠ V = 1 π h3 12 77.
d = (3t ) 2 + (50) 2
78.
d = 9t 2 + 2500 meters, 0 ≤ t ≤ 60
t= t=
d r 1 + x2 3 − x + hours 2 8
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.2
79.
111
d = (45 − 8t )2 + (6t )2 miles where t is the number of hours after 12:00 noon
80.
A = xy
a.
A( x ) = x ⎛⎜ − 1 x + 4 ⎞⎟ ⎝ 2 ⎠ 2 1 A( x ) = − x + 4 x 2 b.
Circle C = 2π r x = 2π r r= x 2π
a.
Square C = 4s 20 − x = 4s s = 5− x 4
⎛ ⎞ Area = π r 2 = π ⎜ x ⎟ ⎝ 2π ⎠
2
82.
⎛ ⎞ Area = s 2 = ⎜ 5 − x ⎟ ⎝ 4⎠
2
= 25 − 5 x + x 2 16
2 2 Total Area = x + 25 − 5 x + x 4π 2 16 ⎛ 1 1⎞ 2 5 =⎜ + ⎟ x − x + 25 2 ⎝ 4π 16 ⎠
83.
x
0
4
8
12
16
20
Total Area
25
17.27
14.09
15.46
21.37
31.83
Domain: [0, 20].
a.
Left side triangle 2
2
c = 20 + (40 − x )
2
2
c = 30 + x
2
c = 900 + x 2
c = 400 + (40 − x )2
6
7
Area
3.5
6
8
6
3.5
Domain: (2, ∞)
b.
84.
Right side triangle 2
4
2 = x x−2
b.
c.
2
mPB = 0 − 2 = −2 x−2 x−2 0− y − y = m AB = x −0 x mPB = m AB −2 = − y x−2 x 2x = y x−2 Area = 1 bh = 1 xy 2 2 x 1 2 = x 2 x−2
a.
2
2
=x 4π
1
Domain: [0, 8].
c. 81.
x
Total length = 900 + x 2 + 400 + (40 − x )2
p
40
50
60
75
90
f(p)
4900
4300
3800
3200
2800
answers accurate to nearest 100 feet
b.
0
10
20
30
40
Total Length
74.72
67.68
64.34
64.79
70
Domain: [0, 40].
c. 85.
x
x
5
10
12.5
15
20
Y(x)
275
375
385
390
394
answers accurate to nearest apple
86.
x
100
200
500
750
1000
C(x)
57,121
59,927
65,692
69,348
72,507
answers accurate to nearest dollar
Copyright © Houghton Mifflin Company. All rights reserved.
112
87.
Chapter 2: Functions and Graphs
f (c) = c2 − c − 5 = 1
88.
g ( c ) = −2c 2 + 4c − 1 = −4 = −2 c 2 + 4 c + 3 = 0
c2 − c − 6 = 0 ( c − 3)( c + 2) = 0 c−3=0 c=3
or
c=
c+2=0 c = −2
−4 ± 42 − 4( −2)(3) 2( −2)
c = −4 ± 16 + 24 −4 c = −4 ± 40 −4 − ± 4 2 10 c= −4 c = 2 ± 10 2
89.
1 is not in the range of f(x), since x −1 only if x + 1 = x − 1 or 1 = −1. 1= x +1
91.
Set the graphing utility to “dot” mode.
90.
0 is not in the range of g(x), since 1 only if ( x − 3)(0) = 1 or 0 = 1. 0= x−3
WINDOW FORMAT Xmin=-4.7 Xmax=4.7 Xscl=1 Ymin=-5 Ymax=2 Yscl=1
92.
Set the graphing utility to “dot” mode. WINDOW FORMAT Xmin=-5 Xmax=5 Xscl=1 Ymin=-5 Ymax=3 Yscl=1
93.
WINDOW FORMAT Xmin=-4.7 Xmax=4.7 Xscl=1 Ymin=-5 Ymax=1 Yscl=1
94.
WINDOW FORMAT Xmin=-5 Xmax=5 Xscl=1 Ymin=-5 Ymax=5 Yscl=1
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Section 2.2
113
95.
96.
....................................................... 97.
f ( x) 32 = (9 − 3) − ( 4 − 2) = 6 − 2 = 4
Connecting Concepts f ( x) 74 = (−21 + 2) − (−12 + 2)
98.
= −19 − (−10) = −19 + 10 = −9
99.
f ( x) 02 = (16 − 12 − 2) − 0 = 2
101. a.
f ( x) 80 = 0 − 8 = − 8 = −2 2
100.
f (1, 7) = 3(1) + 5(7) − 2 = 3 + 35 − 2 = 36
b.
f (0, 3) = 3(0) + 5(3) − 2 = 13
c.
f ( −2, 4) = 3( −2) + 5(4) − 2 = 12
d.
f (4, 4) = 3(4) + 5(4) − 2 = 30
e.
f ( k , 2k ) = 3( k ) + 5(2k ) − 2 = 13k − 2
f.
f ( k + 2, k − 3) = 3( k + 2) + 5( k − 3) − 2 = 3k + 6 + 5k − 15 − 2 = 8k − 11
102. a.
g (3, − 4) = 2(3)2 − −4 + 3 = 18 − 4 + 3 = 17
b.
g ( −1, 2) = 2( −1)2 − 2 + 3 = 2 − 2 + 3 = 3
c.
g (0, − 5) = 2(0)2 − −5 + 3 = −2
d.
g ⎛⎜ 1 , − 1 ⎞⎟ = 2 1 2 ⎝2 4⎠
e.
g ( c, 3c ) = 2( c ) 2 − 3c + 3 = 2c 2 − 3c + 3
f.
()
2
− − 1 + 3 = 1 − 1 + 3 = 13 4 2 4 4 2
g ( c + 5, c − 2) = 2( c + 5) − c − 2 + 3
( 3c = 3c since c > 0) (Since c < 0, c − 2 < 0)
= 2c 2 + 20c + 50 − ( −( c − 2)) + 3 2
(Thus c − 2 = − c + 2)
2
= 2c + 20c + 50 + c − 2 + 3 = 2c + 21c + 51 103.
5 + 8 + 11 = 12 2 A(5, 8, 11) = 12(12 − 5)(12 − 8)(12 − 11)
s=
104.
C (18, 11) = 15(18) + 14(11) = 270 + 154 = $424
= 12(7)(4)(1) = 336 = 4 21
Copyright © Houghton Mifflin Company. All rights reserved.
114
105.
Chapter 2: Functions and Graphs
a 2 + 3a − 3 = a
106.
2
a + 2a − 3 = 0 ( a − 1)( a + 3) = 0 a =1
or
a =a a+5 a = a ( a + 5)
107.
108.
a = a 2 + 5a
a = −3
0 = a 2 + 4a 0 = a ( a + 4) a = 0 or a = −4
....................................................... PS1. d = 5 − ( −2) = 7
PS3.
Prepare for Section 2.3 PS2. The product of any number and its negative reciprocal is –1. −7 ⋅ 1 = − 1 7
−4 − 4 = −8 2 − ( −3) 5
PS4.
y − 3 = −2( x − 3) y − 3 = −2 x + 6 y = −2 x + 9
PS5. 3x − 5 y = 15 −5 y = −3x + 15
PS6.
y = 3x−3 5
y = 3x − 2(5 − x ) 0 = 3x − 2(5 − x ) 0 = 3x − 10 + 2 x 10 = 5 x 2=x
Section 2.3 1.
y −y 7−4 3 3 m= 2 1 = = =− x2 − x1 1 − 3 − 2 2
4.
m=
4−4 0 = =0 2 − (−3) 5
6.
m=
0−0 0 = =0 3−0 3
9.
7 1 − m= 2 2 = 7 − (−4) 3
11.
m=
13.
m=
2.
m=
1− 4 −3 3 = =− 5 − ( −2) 7 7 5.
7.
6 2 = 3⋅ 3 = 9 19 19 19 3
m=
3.
m=
2−0 1 =− 0−4 2
m = 4 − 0 = 4 undefined 0−0 0
−6 −2 − 4 =6 = − 4 − (−3) − 1
8.
m=
4 − (−1) 5 = − 3 − (−5) 2
2 − 4 −2 8 = =− 7 1 5 5 − 4 2 4
10.
m=
f (3 + h) − f (3) f (3 + h) − f (3) = 3+ h −3 h
12.
m=
f ( − 2 + h ) − f ( −2 + h ) 0 = =0 − 2 + h − (−2) h
f ( h ) − f ( 0) f ( h ) − f ( 0) = h−0 h
14.
m=
f ( a + h ) − f ( a ) f ( a + h) − f ( a ) = a+h−a h
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.3
15.
115
m=2 y-intercept (0, –4)
m = –1 y-intercept (0, 1)
16.
17.
m = −1
3
18.
m= 2 3
y-intercept (0, –2)
y-intercept (0, 4)
19.
m=0 y-intercept (0, 3)
20.
m=1 y-intercept (0, 0)
21.
m=2 y-intercept (0, 0)
22.
m = –3 y-intercept (0, 0)
23.
m = –2 y-intercept (0, 5)
24.
m=1 y-intercept (0, –4)
25.
m= −3
26.
m= −2
4
y-intercept (0, 4)
27.
Use y = mx + b with m = 1, b = 3. y = x+3
29.
Use y = mx + b with m = y=
31.
28.
3 1 ,b= . 4 2
30.
3 1 x + 4 2
Use y = mx + b with m = 0, b = 4. y=4
32.
Use y = mx + b with m = −2, b = 5. y = −2 x + 5 2 3 Use y = mx + b with m = − , b = . 3 4 2 3 y=− x + 3 4
Use y = mx + b with m = y=
33.
y − 2 = −4( x − (−3)) y − 2 = −4 x −12 y = −4 x −10
34.
y +1= −3( x + 5) y = −3x −15 −1 y = −3x −16
35.
3
y-intercept (0, –2)
1 , b = −1. 2
1 x −1 2
4 −1 −1 − 3 3 3 = =− 4 −4 3 y − 1 = − ( x − 3) 4 3 9 4 y = − x+ + 4 4 4 3 13 y = − x+ 4 4
m=
Copyright © Houghton Mifflin Company. All rights reserved.
36.
−8 − (−6) 2−5 −2 2 = = −3 3 2 y − (−6) = ( x − 5) 3 2 10 y+6 = x− 3 3 2 10 y = x− −6 3 3 2 28 y = x− 3 3
m=
116
37.
41.
45.
47.
Chapter 2: Functions and Graphs
m = −1 − 11 2−7 = −12 = 12 −5 5 12 y − 11 = ( x − 7) 5 12 y − 11 = x − 84 5 5 12 84 y= x− + 55 5 5 5 12 29 = x− 5 5 f ( x) = 1 − 4 x = 3 −4 x = 2 x=−1 2
38.
42.
−4 − 6 −3 − (−5) −10 = = −5 2 y − 6 = −5( x + 5) y − 6 = −5 x − 25 y = −5 x − 25 + 6 y = −5 x − 19
m=
f ( x) = 2 x + 2 = 4 3 2x = 2 3 x = 2 ⎛⎜ 3 ⎞⎟ ⎝2⎠ x=3
f ( x ) = 3x − 12 3x − 12 = 0 3x = 12
39.
f ( x ) = 2 x + 3 = −1
40.
f ( x ) = 4 − 3x = 7
2 x = −4 x = −2
43.
f ( x) = 3 − x = 5 2 x − =2 2 x = 2( −2)
−3 x = 3 x = −1
44.
f ( x ) = 4 x − 3 = −2 4x = 1 x=1 4
x = −4 46.
f ( x ) = −2 x − 4 −2 x − 4 = 0 −2 x = 4
x=4 The x-intercept of the graph of f ( x ) is (4,0).
x = −2 The x-intercept of the graph of f ( x ) (x) is (−2,0).
Xmin = − 4, Xmax = 6, Xscl=2, Ymin = −12.2, Ymax = 2, Yscl = 2
Xmin = − 4, Xmax = 6, Xscl=2, Ymin = −12.2, Ymax = 2, Yscl = 2
f ( x) = 1 x + 5 4
48.
f ( x) = − 1 x + 2 3
1 x+5 = 0 4 1 x = −5 4 x = −20 The x-intercept of the graph of f ( x ) is (−20,0).
− 1 x+2=0 3 − 1 x = −2 3 x =6 The x-intercept of the graph of f ( x ) is (6,0).
Xmin = −30, Xmax = 30, Xscl = 10, Ymin = −10, Ymax = 10, Yscl = 1
Xmin = −2, Xmax = 8, Xscl = 2, Ymin = −6, Ymax = 8, Yscl = 2
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Section 2.3
49.
117
Algebraic method: f1( x) = f 2 ( x) 4x+5 = x +6 3 x =1 x=1 3 Graphical method: Graph y = 4 x + 5 and y = x+6 They intersect at x =
50.
1 1 , y=6 . 3 3
Xmin = −6, Xmax = 6, Xscl = 2, Ymin = −7, Ymax =7.8, Yscl = 2
Xmin = −7.8, Xmax = 7.8, Xscl = 2, Ymin = −2 ,Ymax = 10, Yscl = 2 51.
Algebraic method:
f1 ( x ) = f 2 ( x )
f1( x) = f 2 ( x) −2 x −11= 3 x + 7 −5 x =18 x = − 18 5 Graphical method: Graph y = −2 x −11 and y = 3 x + 7. They intersect at x = −3.6, y = −3.8.
Algebraic method:
52.
2 x − 4 = − x + 12 3x = 16 x = 16 3 Graphical method: Graph y = 2 x − 4 and y = − x + 12
f1 ( x ) = f 2 ( x )
Algebraic method:
1 x+5= 2 x−7 2 3 1 ⎛ ⎞ 6 ⎜ x + 5 ⎟ = 6 ⎜⎛ 2 x − 7 ⎟⎞ ⎝2 ⎠ ⎝3 ⎠ 3x + 30 = 4 x − 42 72 = x Graphical method: Graph y = 1 x + 5 and 2 y = 2 x−7 3 They intersect at x = 72, y = 41.
1 2 They intersect at x = 5 , y = 6 . 3 3
Xmin = − 4, Xmax = 10, Xscl = 2, Ymin = −2, Ymax = 10, Yscl = 2 Xmin = − 20, Xmax = 120, Xscl = 20, Ymin = −20, Ymax = 100, Yscl = 20 53.
55.
m = 1505 −1482 = 2.875 28 − 20 The value of the slope indicates that the speed of sound in water increases 2.875 m for a one-degree increase in temperature.
a.
b.
m = 29 −13 ≈1.45 20 − 9 H (c) −13 =1.45(c − 9) H (c) =1.45c H (18) =1.45(18) ≈ 26 mpg
54.
56.
m = 4 −1 = 0.04 100 − 25 The value of the slope indicates that the file is being downloaded at 0.04 megabytes per second.
a.
b.
m = 799.1− 675.7 = 24.68 2005 − 2000 C (t ) − 675.7 = 24.68(t − 2000) C (t ) = 24.68t − 48,684.3 900 = 24.68t − 48,684.3 49584.3 = 24.68t 2009.1≈ t The debt will exceed $900 billion in 2009.
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118
57.
59.
Chapter 2: Functions and Graphs
m=
b.
60,000 = 2500t − 4,962,000 5,022,000 = 2500t 2008.8 = t The number of jobs will exceed 60,000 in 2008.
a.
b.
c. 61.
63,000 − 38,000 = 2500 2010 − 2000 N (t ) − 63,000 = 2500(t − 2010) N (t ) = 2500t − 4,962,000
a.
m = 240 −180 = 30 18 −16 B (d ) −180 = 30(d −16) B (d ) = 30d − 300 The value of the slope means that a 1-inch increase in the diameter of a log 32 ft long results in an increase of 30 board-feet of lumber that can be obtained from the log. B (19) = 30(19) − 300 = 270 board feet
Line A represents Michelle Line B represents Amanda Line C represents the distance between Michelle and Amanda.
58.
a.
b. c.
60.
a.
b. c.
62.
a. b.
c. 63.
a.
b.
65.
Find the slope of the line. 180 − 110 70 m= = ≈ 1.842 108 − 70 38 Use the point-slope formula to find the equation. y − y1 = m( x − x1)
64.
a.
T (180) = −10(180) + 2350 = 550o F After 3 hours, the temperature will be 550°F. m = 1640 − 800 = 42 60 − 40 E (T ) − 800 = 42(T − 40) E (T ) = 42T − 880 The value of the slope means that an additional 42 acre-feet of water evaporate for a one degree increase in temperature. E (75) = 42(75) −880 = 2270 acre-feet
m AB = 1− 9 = −4o F 8− 6 m AB = 1− 9 = −4o F 8−6 mDE = −4 − 5 = 9o F 5−6 The temperature changed most rapidly between points D and E. The temperature remained constant (zero slope) between points C and D.
Find the slope of the line. m = 11.2 − 76.5 = −65.3 ≈ −0.87 75 − 0 75 Use the point-slope formula to find the equation. y − y1 = m( x − x1) y − 76.5 = −0.87( x − 0) y − 76.5 = −0.87 x y = −0.87 x + 76.5
y − 110 = 1.842( x − 70) y − 110 = 1.842 x − 128.94 y = 1.842 x − 18.94 y = 1.842(90) − 18.94 y = 165.78 − 18.94 y = 146.84 ≈ 147
b.
P( x ) = 92.50 x − (52 x + 1782) P( x ) = 92.50 x − 52 x − 1782 P( x ) = 40.50 x − 1782
m = 2200 − 2150 = −10 15 − 20 T (t ) − 2200 = −10(t −15) T (t ) = −10t + 2350 The value of the slope means that the temperature is decreasing at a rate of 10 degrees per minute.
66.
y = −0.87( 25) + 76.5 y = −21.75 + 76.5 y = 54.75 ≈ 55 years
P( x ) = 124 x − (78.5 x + 5005) P( x ) = 124 x − 78.5 x − 5005 P( x ) = 45.5 x − 5005 45.5 x − 5005 = 0
40.50 x − 1782 = 0
45.5 x = 5005
40.50 x = 1782 x = 1782 40.50 x = 44, the break-even point
x = 5005 45.5 x = 110, the break-even point
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Section 2.3
67.
119
P( x ) = 259 x − (180 x + 10,270) P( x ) = 259 x − 180 x − 10, 270 P( x ) = 79 x − 10,270
68.
79 x − 10,270 = 0
6210 x − 1,602,180 = 0 6210 x = 1,602,180 1,602,180 x= 6210 x = 258, the break-even point
79 x = 10.270 x = 10.270 79 x = 130, the break-even point
69.
a. b. c. d.
C (0) = 8(0) + 275 = 0 + 275 = $275 C (1) = 8(1) + 275 = 8 + 275 = $283 C (10) = 8(10) + 275 = 80 + 275 = $355 The marginal cost is the slope of C ( x) = 8 x + 275, which is $8 per unit.
70.
71.
a. b. c.
C (t ) = 19,500.00 + 6.75t R (t ) = 55.00t P (t ) = R(t ) − C (t ) P (t ) = 55.00t − (19,500.00 + 6.75t ) P (t ) = 55.00t −19,500.00 − 6.75t P (t ) = 48.25t −19,500.00 48.25t =19,500.00 19,500.00 t= 48.25 t = 404.1451 days ≈ 405 days
72.
d.
P( x ) = 14,220 x − (8010 x + 1,602,180) P( x ) = 14,220 x − 8010 x − 1,602,180 P( x ) = 6210 x − 1,602,180
R (0) = 210(0) = $0 R (1) = 210(1) = $210 R (10) = 210(10) = $2100 The marginal revenue is the slope of R ( x) = 210 x, which is $210 per unit.
a. b. c. d.
m=
a.
b.
c.
117,500 − 98,000 19,500 = = 6.5 35,000 − 32,000 2000 P ( s ) − 98,000 = 6.5( s − 32,000) P ( s ) = 6.5s − 208,000 + 98,000 P ( s ) = 6.5s −110,000 P (50,000) = 6.5(50,000) −110,000 = 325,000 −110,000 = $215,000 Let 6.5s − 110,000 = 0. Then 6.5s = 110,000 s=
73.
75.
3 The graph of 3x + y = −24 has m = − . 4 3 y − 3 = − ( x − 1) 4 3 3 y = − x+ +3 4 4 3 15 y =− x+ 4 4
74.
The graph of y = − 1 x + 6 has m = − 1 . 2 2 1 y − 10 = − ( x − 4) 2 y = − 1 x + 2 + 10 2 y = − 1 x + 12 2
76.
110,000 ≈ 16,924 subscribers 6.5
The graph of x + y = 10 has m = −1. y + 1 = (−1)( x − 2) y = −x + 2 −1 y = −x +1
The graph of y = 5 x + 5 has m = 5 . 2 2 5 y + 2 = ( x − 10) 2 y = 5 x − 25 − 2 2 y = 5 x − 27 2
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120
Chapter 2: Functions and Graphs
77.
The graph of y = 3 x + 6 has m = 3 . 2 2 Thus we use a slope of − 2 . 3 2 y − 7 = − ( x + 9) 3 2 y = − x−6+7 3 2 y = − x +1 3
78.
The graph of y = − 2 x + 3 has m = − 2 . 3 3 Thus we use a slope of 3 . 2 3 y + 6 = ( x − 4) 2 y = 3 x−6−6 2 y = 3 x − 12 2
79.
The graph of y = 4 x − 9 has m = 4 . 7 7 4 y − 3 = ( x − 4) 7 y = 4 x − 16 + 3 7 7 4 5 y= x+ 7 7
80.
The graph of y = 9 x + 2 has m = 9 . 7 7 7 Thus we use a slope of − . 9 y + 3 = − 7 ( x − 1) 9 y = −7 x+ 7 −3 9 9 7 y = − x − 20 9 9
81.
The graph of y = 1 x + 9 has m = 1 . 2 2 Thus we use a slope of –2. y + 1 = −2( x + 3) y = −2 x − 6 − 1 y = −2 x − 7
82.
The graph of y = − 5 x + 7 has m = − 5 . 4 4 5 y + 5 = − ( x + 2) 4 y = −5 x − 5 −5 4 2 5 15 y=− x− 4 2
83.
The graph of x + y = 4 has m = −1. Thus we use a slope of 1. y − 2 = 1( x − 1) y = x −1+ 2 y = x +1
84.
The graph of 2 x − y = 7 has m = 2. 1 Thus we use a slope of − . 2 1 y − 4 = − ( x + 3) 2 1 3 8 y =− x− + 2 2 2 1 5 y =− x+ 2 2
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Section 2.3
85.
121
The equation of the line through (0,0) and P(3,4) has 4 slope . 3 The path of the rock is on the line through P(3,4) with 3 slope − , so y − 4 = − 3 ( x − 3). 4 4 y −4 = − 3 x+ 9 4 4 y = − 3 x+ 9 +4 4 4 y = − 3 x + 25 4 4 The point where the rock hits the wall at y = 10 is the point 3 25 of intersection of y = − x + and y = 10. 4 4
86.
3 25 − x+ = 10 4 4 −3 x + 25 = 40 −3 x = 15 x = −5 feet Therefore the rock hits the wall at (−5, 10). The x-coordinate is –5.
87.
a.
The path of the rock is on the line through P ( 15, 1) with slope − 15 so y −1= − 15( x − 15) y −1= − 15 x +15 y = − 15 x +15 +1 y = − 15 x +16 The point of impact with the wall at y = 14 is the point of
intersection of y = − 15 x + 16 and y = 14 intersect. − 15 x + 16 = 14 − 15 x = −2 2 x= ≈ 0.52 feet 15 ⎛ 2 ⎞ , 14 ⎟ . Therefore, the rock hits the wall at ⎜ 15 ⎝ ⎠ 2 The x-coordinate is or approximately 0.52. 15
h = 1 so Q (2 + h, [ 2 + h ] + 1) = Q (3, 32 + 1) = Q (3, 10) 2
m=
b.
The equation of the line through (0,0) and 1 . P ( 15 , 1) has slope 15
10 − 5 5 = =5 3− 2 1
h = 0.1 so Q (2 + h, [ 2 + h ] + 1) = Q (2.1, 2.12 + 1) = Q (2.1, 5.41) 2
m=
5.41 − 5 0.41 = = 4.1 2.1 − 2 0.1
c.
h = 0.01 so Q (2 + h, [ 2 + h ] + 1) = Q (2.01, 2.012 + 1) = Q (2.01, 5.0401)
d.
5.0401 − 5 0.0401 = = 4.01 2.01 − 2 0.01 As h approaches 0, the slope of PQ seems to be approaching 4.
e.
x1 = 2, y1 = 5, x2 = 2 + h, y2 = [2 + h]2 + 1
2
m=
2
[ 2 + h] + 1 − 5 = (4 + 4h + h2 ) + 1 − 5 = 4h + h2 = 4 + h y −y m= 2 1= x2 − x1 (2 + h) − 2 h h
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122
88.
Chapter 2: Functions and Graphs
a.
h = 1 so Q ( −1 + h, 3[ −1 + h ] ) = Q (0, 0) 2
m=
b.
0−3 −3 = = −3 0 − ( −1) 1
h = 0.1 so Q ( −1 + h, 3[ −1 + h ] ) = Q ( −0.9, 3( −0.9) 2 ) = Q ( −0.9, 2.43) 2
m=
2.43 − 3 −0.57 = = −5.7 0 . 9 ( 1 ) 0. 1 − − −
c.
h = 0.01 so Q ( −1 + h, 3[ −1 + h ] ) = Q ( −0.99, 3( −0.99)2 ) = Q ( −0.99, 2.9403)
d.
2.9403 − 3 −0.0597 = = −5.97 0.01 − 0.99 − (−1) As h approaches 0, the slope of PQ seems to be approaching −6.
e.
x1 = −1, y1 = 3, x2 = −1 + h, y2 = 3[ −1 + h ]
2
m=
2
y2 − y1 3[ −1 + h ] − 3 3(1 − 2h + h 2 ) − 3 3 − 6h + 3h 2 − 3 −6h + 3h 2 = = = = = −6 + 3h x2 − x1 ( −1 + h ) − ( −1) h h h 2
m=
89.
m=
( x + h ) 2 − x 2 x 2 + 2 xh + h 2 − x 2 2 xh + h 2 h (2 x + h ) = = = = 2x + h x+h−x h h h
90.
m=
4( x + h )2 − 4 x 2 4( x 2 + 2 xh + h 2 ) − 4 x 2 4 x 2 + 8 xh + 4h 2 − 4 x 2 8 xh + 4h 2 h(8 x + 4h ) = = = = = 8 x + 4h x+h−x h h h h
.......................................................
Connecting Concepts
y − y1 y2 − y1 ( x − x1 ) , the two-point form. for m in the point-slope form y − y1 = m( x − x1 ) to yield y − y1 = 2 x2 − x1 x2 − x1
91.
Substitute
92.
y − 0 = b − 0 ( x − a) 0−a y = b ( x − a) −a y = − bx + b a bx + y = b Then divide by b to produce x + y = 1. a a b
93.
y − 1 = 3 − 1 ( x − 5) 4−5 y − 1 = 2 ( x − 5) −1 y = −2( x − 5) y − 1 = −2 x + 10 y = −2 x + 10 + 1 y = −2 x + 11
94.
y − 7 = 6 − 7 ( x − 2) −1 − 2 y − 7 = −1 ( x − 2) −3 1 y − 7 = ( x − 2) 3 1 y−7= x− 2 3 3 y = 1 x − 2 + 21 3 3 3 1 19 y= x+ 3 3
95.
x y + = 1 with a = 3 and b = 5. a b x y + =1 3 5 ⎛x y⎞ 15⎜ + ⎟ = 15(1) ⎝3 5⎠ 5 x + 3 y = 15 Use
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Section 2.3
96.
98.
123
x y + = 1 with a = −2 and b = 7. a b x y + =1 −2 7 y⎞ ⎛ x 14⎜ + ⎟ = 14(1) ⎝−2 7⎠ − 7 x + 2 y = 14 Use
x y + = 1 Since (−3, 10) is on the line, − b 2b −3 + 10 = 1 − b 2b 2b ⎛⎜ 3 + 10 ⎞⎟ = 2b(1) ⎝ b 2b ⎠ 6 + 10 = 2b 16 = 2b 8=b x + y =1 −8 16 −2 x + y = 16
100. The slope of the radius from (0, 0) to (x, y) is 0.5, so
97.
99.
x y + = 1 with b = 3a. a b x + y =1 Since (5, 2) is on the line, a 3a 5 + 2 =1 a 3a ⎛ ⎞ 3a ⎜ 5 + 2 ⎟ = 3a (1) ⎝ a 3a ⎠ 15 + 2 = 3a 17 = 3a 17 = a 3 y x Thus + =1 ⎛ 17 ⎞ 3 ⎛ 17 ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 3⎠ ⎝ 3⎠ 3x + y = 1 17 17 3x + y = 17 Use
3(1 + h )3 − 3 3(1 + 3h + 3h 2 + h 3 ) − 3 = h 1+ h −1 2 3 = 3 + 9h + 9h + 3h − 3 h 2 3 = 9h + 9h + 3h h
=
h (9 + 9h + 3h 2 ) h
= 9 + 9h + 3h 2
y−0 y = = 0.5 thus y = 0.5 x. x−0 x
Substitute y = 0.5x into x 2 + y 2 = 25. x 2 + (0.5 x ) 2 = 25 x 2 + 0.25 x 2 = 25 1.25 x 2 = 25 x 2 = 20 x = ± 20 = ±2 5 If x = 2 5, then y = 0.5(2 5) = 5. If x = −2 5, then y = 0.5(−2 5) = − 5. The points are (2 5,
5) and ( − 2 5, − 5).
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124
Chapter 2: Functions and Graphs
101. The slope of the line through (3, 9) and (x, y) is
15 y − 9 15 , so = . 2 x−3 2
2( y − 9) = 15( x − 3)
Therefore
2 y − 18 = 15 x − 45 Substitute y = x 2 into this equation.
2 y − 15 x + 27 = 0 2x 2 − 15 x + 27 = 0 (2 x − 9)( x − 3) = 0
x = 9 or x = 3 2
If x =
2
81 9 ⎛ 9 81 ⎞ ⎛9⎞ , y = x2 = ⎜ ⎟ = ⇒ ⎜ , ⎟. 4 2 ⎝2 4 ⎠ ⎝2⎠
If x = 3, y = x 2 = (3)2 = 9 ⇒ (3, 9 ), but this is the point itself. ⎛ 9 81 ⎞ 15 ⎛ 9 81 ⎞ The point ⎜ , ⎟ is on the graph of y = x 2 , and the slope of the line containing (3, 9) and ⎜ , ⎟ is . ⎝2 4 ⎠ 2 ⎝2 4 ⎠ 3 y−2 3 102. The slope of the line through (3, 2) and (x, y) is , so = . 8 x−3 8 Therefore 8( y − 2) = 3( x − 3).
8 y − 16 = 3x − 9 8 y = 3x + 7
Substitute y = x + 1 into this equation.
8 x + 1 = 3x + 7
(8
x + 1 ) = ( 3x + 7 ) 2
2
64( x + 1) = 9 x 2 + 42 x + 49 64 x + 64 = 9 x 2 + 42 x + 49 0 = 9 x 2 − 22 x − 15 0 = (9 x + 5)( x − 3) x = − 5 or x = 3 9 5 If x = − , y = 9 If x = 3, y =
5 9 x +1 = − + = 9 9
4 2 ⎛ 5 2⎞ = ⇒ ⎜ − , ⎟. 9 3 ⎝ 9 3⎠
x + 1 = 3 + 1 = 4 = 2 ⇒ ( 3, 2 ) , but this is the point itself.
⎛ 5 2⎞ The point ⎜ − , ⎟ is on the graph of y = ⎝ 9 3⎠
⎛ 5 2⎞ 3 x + 1, and the slope of the line containing (3, 2) and ⎜ − , ⎟ is . ⎝ 9 3⎠ 8
....................................................... PS1. 3x 2 + 10 x − 8 = (3x − 2)( x + 4) PS3.
f ( −3) = 2( −3)2 − 5( −3) − 7 = 18 + 15 − 7 = 26
Prepare for Section 2.4 PS2. x 2 − 8 x = x 2 − 8 x + 16 = ( x − 4)2 PS4.
2 x2 − x − 1 = 0 (2 x + 1)( x − 1) = 0 2 x +1= 0 x=− 1 2
x −1= 0 x =1
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Section 2.4
125
PS5. x 2 + 3x − 2 = 0 x=
53 = −16t 2 + 64t + 5
PS6.
16t 2 − 64t + 48 = 0
−3 ± (3)2 − 4(1)( −2) 2(1)
t 2 − 4t + 3 = 0
= −3 ± 17 2
(t − 1)(t − 3) = 0 t =1, 3
Section 2.4 1.
d
2.
f
3.
b
4.
h
5.
g
6.
e
7.
c
8.
a
9.
f ( x) = ( x 2 + 4 x) +1
10.
= ( x 2 + 4 x + 4) +1− 4
= ( x 2 + 6 x + 9) −1− 9
= ( x + 2)2 − 3 standard form, vertex (−2, −3), axis of symmetry x = −2
11.
f ( x) = ( x 2 + 6 x) −1
f ( x ) = ( x 2 −8 x) + 5
= ( x + 3)2 −10 standard form, vertex (−3, −10), axis of symmetry x = −3
12.
= ( x 2 −8 x +16) + 5 −16
f ( x) = ( x 2 −10 x) + 3 = ( x 2 −10 x + 25) + 3 − 25
= ( x − 4)2 −11 standard form, vertex (4, −11), axis of symmetry x = 4
= ( x − 5)2 − 22 standard form, vertex (5, −22), axis of symmetry x = 5
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126
13.
Chapter 2: Functions and Graphs
f ( x) = ( x 2 + 3 x) +1
14.
⎛ ⎞ = ⎜ x 2 + 3 x + 9 ⎟ +1− 9 4⎠ 4 ⎝
⎛ ⎞ = ⎜ x 2 + 7 x + 49 ⎟ + 2 − 49 4 ⎠ 4 ⎝
2
2
⎛ ⎞ =⎜ x+ 3 ⎟ + 4 − 9 ⎝ 2⎠ 4 4 2
⎛ ⎞ =⎜ x+ 3 ⎟ ⎝ 2⎠ ⎛ 3 vertex ⎜ − , − ⎝ 2
15.
⎛ ⎞ = ⎜ x + 7 ⎟ + 8 − 49 ⎝ 2⎠ 4 4 2
⎛ ⎞ = ⎜ x + 7 ⎟ − 41 standard form, 4 ⎝ 2⎠ 7 ⎛ 7 41 ⎞ vertex ⎜ − , − ⎟ , axis of symmetry x = − 2 4 2 ⎠ ⎝
− 5 standard form, 4 5⎞ 3 ⎟ , axis of symmetry x = − 2 4⎠
f ( x) = − x 2 + 4 x + 2
16.
f ( x) = − x 2 − 2 x + 5
= −( x 2 − 4 x ) + 2
= −( x 2 + 2 x ) + 5
= −( x 2 − 4 x + 4) + 2 + 4
= −( x 2 + 2 x +1) + 5 +1
= −( x − 2)2 + 6 standard form, vertex (2, 6), axis of symmetry x = 2
17.
f ( x ) = ( x 2 + 7 x) + 2
f ( x) = −3 x 2 + 3 x + 7
= −( x +1)2 + 6 standard form, vertex (−1, 6), axis of symmetry x = −1
18.
2
f ( x) = −2 x 2 − 4 x + 5
= −3( x −1x) + 7
= −2( x 2 + 2 x) + 5
⎛ ⎞ = −3 ⎜ x 2 −1x + 1 ⎟ + 7 + 3 4⎠ 4 ⎝
= −2( x 2 + 2 x +1) + 5 + 2 = −2( x +1)2 + 7 standard form, vertex (−1, 7), axis of symmetry x = −1
2
⎛ ⎞ = −3 ⎜ x − 1 ⎟ + 28 + 3 4 4 ⎝ 2⎠ 2
⎛ ⎞ = −3 ⎜ x − 1 ⎟ + 31 4 ⎝ 2⎠
standard form,
1 ⎛ 1 31 ⎞ vertex ⎜ , ⎟ , axis of symmetry x = 2 ⎝2 4 ⎠
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Section 2.4
19.
21.
23.
25.
x=
127
10 −b = =5 2a 2(1)
20.
x=
6 −b = =3 2a 2(1)
y = f (5) = (5) 2 − 10(5) = 25 − 50 = −25 vertex (5, − 25)
y = f (3) = (3)2 − 6(3) = 9 − 18 = −9 vertex (3, − 9)
f ( x) = ( x − 5)2 − 25
f ( x) = ( x − 3) 2 − 9
x=
0 −b = =0 2a 2(1)
22.
x=
0 −b = =0 2a 2(1)
y = f (0) = (0) 2 − 10 = −10 vertex (0, − 10)
y = f (0) = (0) 2 − 4 = −4 vertex (0, − 4)
f ( x) = x 2 − 10
f ( x) = x 2 − 4
x=
−b −6 −6 = = =3 2a 2(−1) − 2
24.
x=
−b −4 −4 = = =2 2a 2(−1) − 2
y = f (3) = −(3) 2 + 6(3) + 1 − 9 + 18 + 1 = 10 vertex (3, 10)
y = f (2) = −(2) 2 + 4(2) + 1 = −4 + 8 + 1 =5 vertex (2, 5)
f ( x) = −( x − 3)2 + 10
f ( x) = −( x − 2) 2 + 5
x=
3 3 −b = = 2a 2(2) 4 2
⎛3⎞ ⎛3⎞ ⎛3⎞ y = f ⎜ ⎟ = 2⎜ ⎟ − 3⎜ ⎟ + 7 ⎝4⎠ ⎝4⎠ ⎝4⎠ ⎛ 9⎞ 9 = 2⎜ ⎟ − + 7 ⎝ 16 ⎠ 4 9 9 = − +7 8 4 9 18 56 = − + 8 8 8 47 = 8 ⎛ 3 47 ⎞ vertex ⎜ , ⎟ ⎝4 8 ⎠ ⎛ f(x) = 2⎜ x − ⎝
2
3⎞ 47 ⎟ + 4⎠ 8
26.
x=
10 10 5 −b = = = 2a 2(3) 6 3 2
⎛5⎞ ⎛5⎞ ⎛5⎞ y = f ⎜ ⎟ = 3⎜ ⎟ − 10⎜ ⎟ + 2 ⎝3⎠ ⎝3⎠ ⎝3⎠ ⎛ 25 ⎞ 50 = 3⎜ ⎟ − +2 ⎝ 9 ⎠ 3 25 50 = − +2 3 3 25 50 6 = − + 3 3 3 19 =− 3 ⎛ 5 19 ⎞ vertex ⎜ , − ⎟ 3⎠ ⎝3 2
5⎞ 19 ⎛ f(x) = 3⎜ x − ⎟ − 3 3 ⎝ ⎠
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128
27.
Chapter 2: Functions and Graphs
x=
−b −1 1 = = 2a 2(−4) 8
28.
2
6 6 3 −b = = =− 2a 2(−5) − 10 5 2
⎛1⎞ ⎛1⎞ ⎛1⎞ y = f ⎜ ⎟ = −4⎜ ⎟ + ⎜ ⎟ + 1 ⎝8⎠ ⎝8⎠ ⎝8⎠ 1 1 ⎛ ⎞ = −4⎜ ⎟ + + 1 ⎝ 64 ⎠ 8 1 1 = − + +1 16 8 1 2 16 =− + + 16 16 16 17 = 16 ⎛ 1 17 ⎞ vertex ⎜ , ⎟ ⎝ 8 16 ⎠
⎛ 3⎞ ⎛ 3⎞ ⎛ 3⎞ y = f ⎜ − ⎟ = −5⎜ − ⎟ − 6⎜ − ⎟ + 3 ⎝ 5⎠ ⎝ 5⎠ ⎝ 5⎠ 9 18 ⎛ ⎞ = −5⎜ ⎟ + +3 ⎝ 25 ⎠ 5 9 18 =− + +3 5 5 9 18 15 =− + + 5 5 5 24 = 5 ⎛ 3 24 ⎞ vertex ⎜ − , ⎟ ⎝ 5 5 ⎠
2
2
1⎞ 17 ⎛ f(x) = −4⎜ x − ⎟ + 8⎠ 16 ⎝ 29.
x=
3⎞ 24 ⎛ f(x) = −5⎜ x + ⎟ + 5⎠ 5 ⎝
f ( x) = x 2 − 2 x −1
30.
f ( x) = − x 2 − 6 x − 2
= ( x 2 − 2 x) −1
= −( x 2 + 6 x ) − 2
= ( x 2 − 2 x +1) −1−1
= −( x 2 + 6 x + 9) − 2 + 9
= ( x −1)2 − 2 vertex (1, −2) The y-value of the vertex is −2. The parabola opens up since a =1> 0. Thus the range is y y ≥ −2 .
{
f ( x) = 2 = x 2 − 2 x −1 0 = x2 − 2 x −3 0 = ( x − 3)( x +1) x − 3 = 0 or x + 1 = 0 x=3 x = −1
}
= −( x + 3)2 + 7 vertex (−3, 7) The y-value of the vertex is 7. The parabola opens down since a = −1 < 0. Thus the range is y y ≤ 7 .
{
}
f ( x) = 3 = − x 2 − 6 x − 2 x2 + 6 x + 5 = 0 ( x + 5)( x +1) = 0 x + 5 = 0 or x + 1 = 0 x = −5 x = −1
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.4
31.
129
f ( x) = −2 x 2 + 5 x −1
32.
⎛ ⎞ = −2 ⎜ x 2 − 5 x ⎟ −1 2 ⎠ ⎝ ⎛ ⎞ ⎛ ⎞ = −2 ⎜ x 2 − 5 x + 25 ⎟ −1+ 2 ⎜ 25 ⎟ 2 16 ⎠ ⎝ ⎝ 16 ⎠
f ( x) = 2 x 2 + 6 x − 5 = 2( x 2 + 3 x) − 5 ⎛ ⎞ ⎛ ⎞ = 2 ⎜ x2 + 3x + 9 ⎟ − 5− 2 ⎜ 9 ⎟ 4⎠ ⎝ ⎝4⎠ 2
2
⎛ ⎞ = 2 ⎜ x + 3 ⎟ −5− 9 2 ⎝ 2⎠
2
⎛ ⎞ = 2 ⎜ x + 3 ⎟ − 10 − 9 2 2 ⎝ 2⎠
⎛ ⎞ = −2 ⎜ x − 5 ⎟ − 8 + 25 ⎝ 4⎠ 8 8
2
⎛ ⎞ = −2 ⎜ x − 5 ⎟ + 17 8 ⎝ 4⎠
2
⎛ ⎞ = 2 ⎜ x + 3 ⎟ − 19 2 ⎝ 2⎠ ⎛ 3 19 ⎞ vertex ⎜ − , − ⎟ 2⎠ ⎝ 2
⎛ 5 17 ⎞ vertex ⎜ , ⎟ ⎝4 8 ⎠ 17 . 8 The parabola opens down since a = −2 < 0. ⎧ 17 ⎫ Thus the range is ⎨ y y ≤ ⎬. 8⎭ ⎩
The y-value of the vertex is
19 . 2 The parabola opens up since a =2 > 0. ⎧ 19 ⎫ Thus the range is ⎨ y y ≥ − ⎬. 2⎭ ⎩
The y-value of the vertex is −
f ( x) = 2 = −2 x 2 + 5 x −1 2 x2 −5x + 3 = 0 (2 x − 3)( x −1) = 0 2 x − 3 = 0 or x − 1 = 0 3 x= x =1 2
f ( x) =15 = 2 x 2 + 6 x − 5 0 = 2 x 2 + 6 x − 20 0 = 2( x 2 + 3 x −10) 0 = 2( x − 2)( x + 5) x − 2 = 0 or x + 5 = 0
x=2 33.
f ( x) = x 2 + 3x + 6
34.
x = −5
f ( x) = −2 x 2 − x +1
2
⎛ ⎞ = −2 ⎜ x 2 + 1 x ⎟ +1 2 ⎠ ⎝ ⎛ ⎞ ⎛ ⎞ = −2 ⎜ x 2 + 1 x + 1 ⎟ +1+ 2 ⎜ 1 ⎟ 2 16 ⎠ ⎝ ⎝ 16 ⎠
2
⎛ ⎞ = −2 ⎜ x + 1 ⎟ + 8 + 1 ⎝ 4⎠ 8 8
2
⎛ ⎞ = −2 ⎜ x + 1 ⎟ + 9 ⎝ 4⎠ 8
= ( x 2 + 3 x) + 6 ⎛ ⎞ = ⎜ x2 + 3x + 9 ⎟ + 6 − 9 4⎠ 4 ⎝ ⎛ ⎞ = ⎜ x + 3 ⎟ + 6− 9 4 ⎝ 2⎠
2
⎛ ⎞ = ⎜ x + 3 ⎟ + 24 − 9 4 4 ⎝ 2⎠
2
⎛ ⎞ = ⎜ x + 3 ⎟ + 15 4 ⎝ 2⎠ ⎛ 3 15 ⎞ vertex ⎜ − , ⎟ ⎝ 2 4⎠
⎛ 1 9⎞ vertex ⎜ − , ⎟ ⎝ 4 8⎠
15 . 4 The parabola opens up since a =1 > 0. ⎧ 15 ⎫ Thus the range is ⎨ y y ≥ ⎬. 4⎭ ⎩
The y-value of the vertex is
⎧ 15 ⎫ No, 3 ∉ ⎨ y y ≥ ⎬. 4⎭ ⎩
9 . 8 The parabola opens down since a = −2 < 0. ⎧ 9⎫ Thus the range is ⎨ y y ≤ ⎬. 8⎭ ⎩
The y-value of the vertex is
⎧ Yes, −2 ∈ ⎨ y y ≤ ⎩
9⎫ ⎬. 8⎭
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130
35.
Chapter 2: Functions and Graphs
f ( x) = x 2 + 8 x
36.
2
= −( x 2 + 6 x )
= ( x + 8 x +16) −16 2
= −( x 2 + 6 x + 9) + 9
= ( x + 4) −16 minimum value of –16 when x = −4
37.
f ( x) = − x 2 + 6 x + 2
= −( x + 3)2 + 9 maximum value of 9 when x = −3
38.
= −( x 2 −10 x) − 3
= −( x 2 − 6 x + 9) + 2 + 9
= −( x 2 −10 x + 25) − 3 + 25
f ( x) = 2 x 2 + 3 x +1
= −( x − 5)2 + 22 maximum value of 22 when x = 5
40.
2
2
⎛ ⎞ = 3 ⎜ x + 1 ⎟ − 12 − 1 ⎝ 6 ⎠ 12 12
2
⎛ ⎞ = 3 ⎜ x + 1 ⎟ − 13 6 ⎝ ⎠ 12
⎛ ⎞ = 2⎜ x+ 3 ⎟ + 8 − 9 ⎝ 4⎠ 8 8
2
⎛ ⎞ = 2⎜ x+ 3 ⎟ − 1 ⎝ 4⎠ 8 minimum value of −
1 3 when x = − 8 4
f ( x) = 5 x 2 −11
minimum value of −
42.
= 5( x 2 ) −11
13 1 when x = − 12 6
f ( x) = 3 x 2 − 41 = 3( x 2 ) − 41
= 5( x − 0)2 −11 minimum value of –11 when x = 0 43.
f ( x) = 3 x 2 + x −1 ⎛ ⎞ = 3 ⎜ x 2 + 1 x ⎟ −1 3 ⎠ ⎝ ⎛ 2 1 ⎞ ⎛ ⎞ = 3 ⎜ x + x + 1 ⎟ −1− 3 ⎜ 1 ⎟ 3 36 ⎠ ⎝ ⎝ 36 ⎠
⎛ ⎞ = 2 ⎜ x 2 + 3 x ⎟ +1 2 ⎠ ⎝ ⎛ 2 3 ⎞ ⎛ ⎞ = 2 ⎜ x + x + 9 ⎟ +1− 2 ⎜ 9 ⎟ 2 16 ⎠ ⎝ ⎝ 16 ⎠
41.
f ( x) = − x 2 +10 x − 3
= −( x 2 − 6 x ) + 2 = −( x − 3)2 +11 maximum value of 11 when x = 3
39.
f ( x) = − x 2 − 6 x
f ( x) = − 1 x 2 + 6 x +17 2 1 = − ( x 2 −12 x) +17 2 = − 1 ( x 2 −12 x + 36) +17 +18 2 1 = − ( x − 6)2 + 35 2 maximum value of 35 when x = 6
= 3( x − 0)2 − 41 minimum value of –41 when x = 0 44.
f ( x) = − 3 x 2 − 2 x + 7 4 5 ⎛ 2 8 ⎞ 3 = − ⎜ x + x ⎟+7 4⎝ 15 ⎠ ⎛ 2 8 ⎞ 3 = − ⎜ x + x + 16 ⎟ + 7 + 4 4⎝ 15 225 ⎠ 75 2
⎛ ⎞ = − 3 ⎜ x + 4 ⎟ + 529 4 ⎝ 15 ⎠ 75 529 4 4 maximum value of when x = − =7 75 75 15
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.4
45.
131
3 2 3 x + 27 = − ( x − 0) 2 + 27 64 64 The maximum height of the arch is 27 feet. h(10) = − 3 (10)2 + 27 64 = − 3 (100) + 27 64 = − 75 + 27 16 = − 75 + 432 16 16 357 = = 22 5 feet 16 16 3 2 h( x) = 8 = − x + 27 64 8 − 27 = − 3 x 2 64 −19 = − 3 x 2 64
h( x) = −
a. b.
c.
46.
l + w = 240 a. w = 240 − l b. A = l (240 − l ) A = 240l − l 2 c.
A = −l 2 + 240l A = −(l 2 − 240l ) A = −(l 2 − 240l +1202 ) +1202
A = −(l −120) 2 +1202 Thus l = 120 and w = 120 produce the greatest area.
64( −19) = −3x 2 64( −19) = x2 −3 64( −19) =x −3 8 19 = x 3 8 19 3 = x 3 8 57 = x 3 20.1 ≈ x h( x) = 8 when x ≈ 20.1 feet
47.
a.
b.
c.
3w + 2l = 600 3w = 600 − 2l w = 600 − 2l 3 A = w⋅l
48.
⎛ ⎞ A = ⎜ 600 − 2l ⎟ l ⎝ 3 ⎠ = 200l − 2 l 2 3 2 2 A = − (l − 300l ) 3 A = − 2 (l 2 − 300l +1502 ) +15,000 3 In standard form, A = − 2 (l −150)2 +15,000 3
The maximum area of 15,000 ft 2 is produced when 600 − 2(150) l = 150 ft and the width w = = 100 ft . 3
4 w + 2l =1200 2l =1200 − 4 w l = 1200 − 4 w 2 l = 600 − 2w A = w(600 − 2w) A = 600w − 2 w2 A = −2w2 + 600w A = −2( w2 − 300w) A = −2( w2 − 300w +1502 ) + 2⋅1502 A = −2( w −150)2 + 45,000 1200 − 4(150) = 300. 2 Thus the dimensions that yield the greatest enclosed area are w = 150 ft and l = 300 ft.
Thus when w = 150, the length l =
Copyright © Houghton Mifflin Company. All rights reserved.
132
49.
Chapter 2: Functions and Graphs
T (t ) = −0.7t 2 + 9.4t + 59.3
a.
50.
a.
9.4 ⎞ ⎛ t ⎟ + 59.3 = −0.7⎜ t 2 − 0.7 ⎠ ⎝ 94 ⎞ ⎛ = −0.7⎜ t 2 − t ⎟ + 59.3 7 ⎠ ⎝
⎛ ⎞ = −0.6 ⎜ t 2 − 32.1 t ⎟ − 350 0.6 ⎠ ⎝
2 2 ⎛ 94 ⎡ 47 ⎤ ⎡ 47 ⎤ ⎞ = −0.7⎜ t 2 − t + ⎢ ⎥ ⎟ + 59.3 + 0.7 ⎢ ⎥ ⎜ 7 ⎣7 ⎦ ⎣ 7 ⎦ ⎟⎠ ⎝
= −0.6(t − 26.75)2 + 79.3375
= −0.6(t 2 − 53.5t ) − 350 = −0.6 [t 2 − 53.5t + (26.75) 2 ] − 350 + 0.6(26.75) 2 2
≈ −0.6 ( t − 27 ) + 79 b.
The maximum number of larvae will survive at 27°C. The maximum number of larvae that will survive is 79.
c.
N(t) = 0 = −0.6t 2 + 32.1t − 350
2
⎛ 47 ⎞ ≈ −0.7⎜ t − ⎟ + 90.857 7 ⎠ ⎝ 2
5⎞ ⎛ ≈ −0.7⎜ t − 6 ⎟ + 91 7⎠ ⎝ The temperature is a maximum when 47 5 t= = 6 hours after 6:00 A.M. 7 7 5 Note (60 minutes) ≈ 43 minutes. 7 Thus the temperature is a maximum at 12:43 P.M.
t=
d.
51.
t = − b = − 82.86 = 0.14814 2a 2( −279.67)
52.
h( x) = −0.002 x 2 − 0.03 x + 8 h(39) = −0.002(39) − 0.03(39) + 8 = 3.788 > 3 Solve for x using quadratic formula. −0.002 x 2 − 0.03 x + 8 = 0
h(t ) = −9.8t 2 +100t h(t ) = −9.8(t − 5.1)2 + 254.9 The maximum height is 255 m.
54.
2
−32.1 ± 1030.41 − 840 −1.2 −32.1 ± 191.41 −32.1 ± 13.8 t= ≈ −1.2 −1.2 −32.1 + 13.8 −32.1 − 13.8 t= or t = −1.2 −1.2 = 15.25 ≈ 15 = 38.25 ≈ 38 Thus the x-intercepts to the nearest whole number for N(t) are (15, 0) and (38, 0). When the temperature is less than 15°C or greater than 38°C, none of the larvae survive.
h(t ) = −9.8(t 2 −10.2t )
E (0.14814) = −279.67(0.14814)2 + 82.86(0.14814) ≈ 6.1 The maximum energy is 6.1 joules.
53.
−32.1 ± (−32.1) 2 − 4(−0.6)( −350) 2(−0.6)
t=
The maximum temperature is approximately 91°F.
b.
N (t ) = −0.6t 2 + 32.1t − 350
h( x) = −0.0009 x 2 + 6 h(60.5) = −0.0009(60.5)2 + 6 ≈ 2.7 Since 2.7 is less than 5.4 and greater than 2.5, yes, the pitch is a strike.
x 2 +15 x − 4000 = 0 x= =
−15 ± (15)2 − 4(1)(−4000) 2(1) −15 ± 16, 225 2
, use positive value of x
x ≈ 56.2 Yes, the conditions are satisfied.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.4
55.
a.
133
E (v) = −0.018v 2 +1.476v + 3.4 ⎛ ⎞ = −0.018 ⎜ v 2 − 1.476 v ⎟ + 3.4 0.018 ⎠ ⎝ = −0.018 (v 2 − 82v) + 3.4
)
(
= −0.018 v 2 − 82v + 412 + 3.4 + 0.018(41) 2 2
b. 56.
= −0.018 (v − 41) + 33.658 The maximum fuel efficiency is obtained at a speed of 41 mph. The maximum fuel efficiency for this car, to the nearest mile per gallon, is 34 mpg.
h( x) = −0.0002348 x 2 + 0.0375 x ⎛ ⎞ = −0.0002348 ⎜ x 2 − 0.0375 x ⎟ 0.0002348 ⎠ ⎝ 2 2 ⎛ ⎡ ⎤ ⎞ ⎡ ⎤ = −0.0002348 ⎜ x 2 − 0.0375 x + ⎢ 1 ⋅ 0.0375 ⎥ ⎟ + 0.0002348 ⎢ 1 ⋅ 0.0375 ⎥ ⎜ 0.0002348 ⎣ 2 0.0002348 ⎦ ⎟⎠ ⎣ 2 0.0002348 ⎦ ⎝ 2 ⎛ ⎡ ⎤ ⎞ ≈ −0.0002348 ⎜ x 2 − 0.0375 x + ⎢ 1 ⋅ 0.0375 ⎥ ⎟ +1.5 ⎜ 0.0002348 ⎣ 2 0.0002348 ⎦ ⎟⎠ ⎝ The maximum height of the field, to the nearest tenth of a foot, is 1.5 feet.
57.
Let y = 0, then 0 = x 2 + 6 x 0 = x( x + 6) x = 0 or x + 6 = 0 x = −6 The x-intercepts are (0, 0) and (−6, 0).
58.
Let x = 0, then f ( x ) = 02 + 6(0) = 0 The y-intercept is (0, 0). 59.
Let y = 0, then 0 = −3 x 2 + 5 x − 6 x=
Let x = 0, then f ( x ) = −02 + 4(0) = 0 The y-intercept is (0, 0). 60.
63.
Let y = 0, then 0 = 2 x 2 + 3 x + 4
−5 ± 52 − 4(−3)(−6)
x=
2(−3) 2
61.
Let y = 0, then 0 = − x 2 + 4 x 0 = x ( − x + 4) x = 0 or −x + 4 = 0 x=4 The x-intercepts are (0, 0) and (4, 0).
−3 ± 32 − 4(2)(4) 2(2) 2
Since the discriminant 5 − 4( −3)(−6) = −47 is negative, there are no x-intercepts.
Since the discriminant 3 − 4(2)(4) = −23 is negative, there are no x-intercepts.
Let x = 0, then f ( x ) = −3(0)2 + 5(0) − 6 = −6 The y-intercept is (0, −6).
Let x = 0, then f ( x ) = 2(0) 2 + 3(0) + 4 = 4 The y-intercept is (0, 4).
−
b 296 =− = 740 2a 2(−0.2)
62.
−
b 810 =− = 675 2a 2(−0.6)
R (740) = 296(740) − 0.2(740) 2 = 109,520
R (675) = 810(675) − 0.6(675) 2 = 273,375
Thus, 740 units yield a maximum revenue of $109,520.
Thus, 675 units yield a maximum revenue of $273,375.
−
b 1.7 =− = 85 2a 2(−0.01)
64.
P (85) = −0.01(85) 2 + 1.7(85) − 48 = 24.25
Thus, 85 units yield a maximum profit of $24.25.
−
1.68 b = 11,760 =− 2a ⎛ 1 ⎞ ⎟⎟ 2⎜⎜ − ⎝ 14,000 ⎠
P (11,760) = −
(11,760) 2 + 1.68(11,760) − 4000 = 5878.40 14,000
Thus, 11,760 units yield a maximum profit of $5878.40. Copyright © Houghton Mifflin Company. All rights reserved.
134
65.
Chapter 2: Functions and Graphs
P ( x) = R ( x) − C ( x ) = x(102.50 − 0.1x ) − (52.50 x +1840)
P ( x) = R ( x) − C ( x) = x(210 − 0.25 x) − (78 x + 6399)
66.
= −0.1x 2 + 50 x −1840 The break-even points occur when R ( x) = C ( x) or P(x) = 0.
= −0.25 x 2 +132 x − 6399 − b = − 132 = 264 2a 2(−0.25)
Thus, 0 = −0.1x 2 + 50 x − 1840
P (264) = −0.25(264) 2 +132(264) − 6399 = $11,025, the maximum profit The break-even points occur when P(x) = 0.
−50 ± 502 − 4( −0.1)( −1840) x= 2( −0.1) −50 ± 1764 −0.2 −50 ± 42 = −0.2 x = 40 or x = 460
Thus, 0 = −0.25 x 2 + 132 x − 6399
=
−132 ± 1322 − 4(−0.25)(−6399) −132 ± 11025 = −0.5 2(−0.25) −132 ± 105 = ⇒ x = 54 or x = 474 −0.5
x=
The break-even points occur when x = 40 or x = 460.
The break-even points occur when x = 54 or x = 474. 67.
Let x = the number of people that take the tour. a. R ( x) = x(15.00 + 0.25(60 − x)) = x(15.00 +15 − 0.25 x) b.
= −0.25 x 2 + 30.00 x P ( x) = R ( x) − C ( x) = (−0.25 x 2 + 30.00 x ) − (180 + 2.50 x)
c.
= −0.25 x 2 + 27.50 x −180 b 27.50 − =− = 55 2a 2(−0.25)
P (55) = −0.25(55) 2 + 27.50(55) − 180
d. 68.
= $576.25 The maximum profit occurs when x = 55.
Let x = the number of parcels. a. b.
69.
R ( x) = xp = x(22 − 0.01x) = −0.01x 2 + 22 x P ( x) = R ( x) − C ( x) = (−0.01x 2 + 22 x ) − (2025 + 7 x)
c.
= −0.01x 2 +15 x − 2025 b 15 − =− = 750 2a 2(−0.01)
P (750) = −0.01(750) 2 + 15(750) − 2025
d. e.
= $3600 p(750) = 22 − 0.01(750) = $14.50 The break-even points occur when R ( x) = C ( x).
h(t ) = −16t 2 + 128t − b = − 128 = 4 seconds a. 2a 2( −16)
b.
h(4) = −16(4) 2 + 128(4) = 256 feet
c.
0 = −16t 2 +128t 0 = −16t (t − 8) −16t = 0 or t − 8 = 0 t =0 t =8 The projectile hits the ground at t = 8 seconds.
− 0.01x 2 + 22 x = 2025 + 7 x − 0.01x 2 + 15 x − 2025 = 0
− (15) ± 152 − 4(−0.01)(−2025) 2(−0.01) x = 150 or x = 1350 are the break-even points. Thus the minimum number of parcels the air freight company must ship to break even is 150. x=
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.4
70.
135
h(t ) = −16t 2 + 64t + 80 b 64 a. − =− =2 2a 2(−16)
b.
c.
h(2) = −16(2)2 + 64(2) + 80 = 144 feet b − = − 64 2a 2( −16) = 2 seconds 0 = −16t 2 + 64t + 80 0 = −16(t 2 − 4t − 5) 0 = −16(t − 5)(t +1) t − 5 = 0 or t +1 = 0 t =5 t = −1 No The projectile has height 0 feet at t = 5 seconds.
71.
y ( x) = −0.014 x 2 + 1.19 x + 5 − b = − 1.19 2a 2( −0.014) = 42.5
y (42.5) = −0.014(42.5)2 + 1.19(42.5) + 5 = 30.2875 ≈ 30 feet
72.
h(t ) = −204.8t 2 + 256t − b = − 256 2a 2( −204.8) = 0.625 h(0.625) = −204.5(0.625) 2 + 256(0.625) = 80.1171875 ≈ 80 inches
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136
Chapter 2: Functions and Graphs
73.
74.
y = a ( x − h) 2 + k y = a ( x − 0)2 + 6 y = ax 2 + 6 500 = a (2100)2 + 6
The perimeter is 48 = π r + h + 2r + h . Solve for h. 48 − π r − 2r = 2h 1 (48 − π r − 2r ) = h 2
494 = a (2100)2 494 = a 21002 0.000112018 ≈ a y = 0.000112018 x 2 + 6
Area = semicircle + rectangle 1 A = π r 2 + 2rh 2 1 2 ⎛1⎞ = π r + 2r ⎜ ⎟(48 − π r − 2r ) 2 ⎝2⎠ 1 2 = π r + r ( 48 − π r − 2r ) 2 1 2 = π r + 48r − π r 2 − 2r 2 2 ⎞ ⎛1 = ⎜ π − π − 2 ⎟ r 2 + 48r ⎝2 ⎠ ⎛ 1 ⎞ = ⎜ − π − 2 ⎟ r 2 + 48r ⎝ 2 ⎠
Graph the function A to find that its maximum occurs when r ≈ 6.72 feet.
Xmin = 0, Xmax = 14, Xscl = 1 Ymin = −50, Ymax = 200, Yscl = 50 1 (48 − π r − 2r ) 2 1 ≈ (48 − π(6.72) − 2(6.72)) 2 ≈ 6.72 feet
h=
Hence the optimal window has its semicircular radius equal to its height. Note: Using calculus it can be shown that the exact 48 value of r = h = . π+4
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Section 2.4
137
....................................................... 75.
f ( x) = x 2 − ( a + b) x + ab a. x-intercepts occur when y = 0.
Connecting Concepts
0 = x 2 − (a + b) x + ab
f ( x) = ax 2 + bx + c a. a < 0, b and c any real numbers b. a > 0, b and c any real numbers
0 = ( x − a)( x − b)
c.
76.
b 2 − 4ac > 0
x − a = 0 or x − b = 0
b.
77.
x=a x=b Thus the x-intercepts are (a, 0) and (b, 0). b ( a + b) a + b − = = which is the x-coordinate of 2a 2(1) 2 the midpoint of the segment joining (a, 0) and (b, 0).
Let f ( x) = ax 2 + bx + c. We know f ( 2) = a ( 2) 2 + b ( 2) + c = 1
78.
f (−3) = a (−3) 2 + b(−3) + c = 2
(1)
This implies c = 4 and from Equation (1) we have 4a + 2b + 4 = 1 or 4a + 2b = −2 ( 2) The x-value of the vertex is 2, and by the vertex formula we −b 2a
, which implies b = −4a.
9 ⎛⎜ b ⎞⎟ − 3b − 5 = 2 ⎝6⎠ 3 b − 3b − 5 = 2 2 − 3b = 7 2 b = 7 ⋅ 2 = − 14 3 −3 b a= 6 − 14 a= 3 6 a = − 14 = − 7 18 9 7 14 Hence f ( x) = − x 2 − x − 5. 9 3
Substituting –4a for b in Equation (2) gives us 4a + 2(−4a ) = −3 4a − 8a = −3 − 4a = −3 3 a= 4 3 Substituting for a in Equation (2) gives us 4 ⎛3⎞ 4⎜ ⎟ + 2b = −3 ⎝4⎠ 3 + 2b = −3 2b = −6 b = −3 Thus the desired quadratic function is 3 f ( x ) = x 2 − 3 x + 4. 4 79.
P = 32 = 2 x + 2w 16 = x + w a. w = 16 − x b. Area A = xw A = x(16 − x) A =16 x − x 2
81.
(1)
f (0) = 0 + 0 + c = −5, which implies c = −5. −b Now the vertex is (−3, 2), so − 3 = or 6a = b or 2a b a = . Thus, substituting in Equation (1) gives us 6 9( a ) − 3b − 5 = 2
f( 0 ) = a (0) 2 + b(0) + c = 4
have 2 =
Let f ( x) = ax 2 + bx + c. We know a < 0.
80.
A = 16 x − x 2 attains its maximum when x = − b = − 16 = 8. 2a 2( −1) Now x = 8 implies w = 16 − x w = 16 − 8 w=8 Thus the rectangle with perimeter 32 inches that has the largest area is the square with each side of length 8 inches.
The discriminant is b 2 − 4(1)(−1) = b 2 + 4, which is always positive. Thus the equation has two real zeros for all values of b.
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138
Chapter 2: Functions and Graphs
82.
The discriminant is b 2 − 4(−1)(1) = b 2 + 4, which is always positive. Thus the equation has two real zeros for all values of b.
83.
Increasing the constant c increases the height of each point on the graph by c units.
84.
Decreasing the coefficient a shrinks the graph of the parabola toward the x-axis.
85.
Let x = one number. Then 8 – x = the other number. P = x (8 − x ) = 8 x − x 2 , vertex at x = −b = −8 = 4. 2a −2 Thus, x = 4 and 8 − x = 4. The numbers are 4 and 4 .
86.
Let x = one number. Let x + 12 = the other number. P = x ( x + 12) = x 2 + 12x, vertex at x = −b = −12 = −6. 2a 2 Thus, x = −6 and x + 12 = −6 + 12 = 6. The numbers are –6 and 6.
87.
x1 = x, y1 = x3 , x2 = x + h, y2 = ( x + h)3 y −y ( x + h)3 − x3 x3 + 3hx 2 + 3h 2 x + h3 − x3 3hx 2 + 3h 2 x + h3 h(3 x 2 + 3hx + h 2 ) m= 2 1− = = = = 3x 2 + 3hx + h 2 x2 − x1 x+h−x h h h
88.
x1 = x, y1 = 4 x3 + x, x2 = x + h, y2 = 4( x + h)3 + ( x + h) y − y 4( x + h)3 + ( x + h) − (4 x3 + x) 4( x3 + 3hx 2 + 3h 2 x + h3 ) + x + h − 4 x3 − x = m= 2 1 = x2 − x1 x+h− x h 3 2 2 3 3 = 4 x +12hx +12h x + 4h + x + h − 4 x − x h 2 2 3 = 12hx +12h x + 4h + h h
=
h(12 x 2 +12hx + 4h 2 +1) h
=12 x 2 +12hx + 4h 2 +1
....................................................... PS1.
PS3.
Prepare for Section 2.5
f ( x) = x2 + 4 x − 6 − b = − 4 = −2 2a 2(1) x = −2
PS2.
f ( −2) = 2( −2)3 − 5( −2) = −16 + 10 = −6
PS4.
f (3) =
3(3)4 (3) 2 + 1
f ( −3) =
= 243 = 24.3 10
3( −3)4
( −3)2 + 1 f (3) = f ( −3)
− f (2) = −[2(2)3 − 5(2)] = −[16 − 10] = −6 f ( −2) = − f (2)
= 243 = 24.3 10
f ( −2) − g ( −2) = ( −2) 2 − [ −2 + 3] = 4 − 1 = 3 f ( −1) − g ( −1) = ( −1)2 − [ −1 + 3] = 1 − 2 = −1 f (0) − g (0) = (0) 2 − [0 + 3] = 0 − 3 = −3 f (1) − g (1) = (1)2 − [1 + 3] = 1 − 4 = −3 f (2) − g (2) = (2)2 − [2 + 3] = 4 − 5 = −1
PS5.
− a + a = 0, b + b = b 2 2 midpoint is (0, b)
PS6.
− a + a = 0, −b + b = 0 2 2 midpoint is (0, 0)
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Section 2.5
139
Section 2.5 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
a. No
b. Yes
14.
a. Yes
b. No
15.
a. No
b. No
16.
a. No
b. No
17.
a. Yes
b. Yes
18.
a. Yes
b. Yes
19.
a. Yes
b. Yes
20.
a. No
b. No
21.
a. Yes
b. Yes
22.
Not symmetric with respect to the origin since (− y ) = (− x) + 1 does not simplify to the original equation y = x + 1.
23.
No, since (− y ) = 3(− x) − 2 simplifies to (−y) = −3x – 2, which is not equivalent to the original equation y = 3 x − 2.
24.
Yes, since (− y ) = (− x)3 − (− x) simplifies to − y = − x 3 + x, which is equivalent to the original equation y = x 3 − x.
25.
Yes, since (− y ) = −( − x)3 implies − y = x 3 or y = − x 3 , which is the original equation.
26.
Yes, since (− y ) =
27.
Yes, since (− x) 2 + (− y ) 2 = 10 simplifies to the original equation.
28.
Yes, since (− x) 2 − (− y ) 2 = 4 simplifies to the original equation.
29.
Yes, since − y =
30.
Yes, since − y = − x simplifies to the original equation.
9 9 is equivalent to the original equation y = . − x ( ) x
−x simplifies to the original equation. −x
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140
Chapter 2: Functions and Graphs
31.
32.
symmetric with respect to the y-axis
33.
symmetric with respect to the x-axis
34.
symmetric with respect to the origin
35.
36.
symmetric with respect to the origin
symmetric with respect to the origin 37.
symmetric with respect to the origin
38.
39.
symmetric with respect to the line x = 2
symmetric with respect to the line x = 4
symmetric with respect to the line x = 2 40.
41.
42.
no symmetry
symmetric with respect to the x-axis, y-axis, and origin
symmetric with respect to the line x = 2 43.
Even since g (− x) = (− x) 2 − 7 = x 2 − 7 = g ( x).
44.
Even, since h(− x) = (− x) 2 + 1 = x 2 + 1 = h( x).
45.
Odd, since F (− x) = (− x)5 + (− x)3
46.
Neither, since G (− x) ≠ G ( x) and G (− x) ≠ −G ( x).
5
3
= −x − x = − F ( x). 47.
Even
48.
Even
49. Even
50.
Neither
51. Even
52.
Even
53.
Even
54. Neither
55.
Neither
56. Odd
57.
58.
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Section 2.5
59.
61.
a.
f ( x+ 2)
b.
f ( x) + 2
a.
f ( x + 3) ( −2 − 3, 5) = ( −5, 5) (0 − 3, − 2) = ( −3, − 2) (1− 3, 0) = ( −2, 0) f ( x ) +1 ( −2, 5 +1) = ( −2, 6) (0, − 2 +1) = (0, −1) (1, 0 +1) = (1, 1)
62.
a.
f (− x)
64.
b.
− f ( x)
a.
f (− x) ( − −1, 3) = (1, 3) ( −2, − 4) − f ( x) ( −1, − 3) (2, − −4) = (2, 4)
b.
63.
65.
141
b.
60.
a.
g ( x−1)
b.
g ( x) −1
a.
g ( x − 2) ( −3 + 2, −1) = ( −1, −1) (1+ 2, − 3) = (3, − 3) (4 + 2, 2) = (6, 2) g ( x) − 2 ( −3, −1− 2) = ( −3, − 3) (1, − 3 − 2) = (1, − 5) (4, 2 − 2) = (4, 0)
b.
66.
a.
− g ( x)
b.
g (− x)
a.
− g ( x) (4, − −5) = (4, 5) ( −3, − 2) g (− x) ( −4, − 5) ( − − 3, 2) = (3, 2)
b.
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142
Chapter 2: Functions and Graphs
67.
68.
69.
70.
1 y = − n( x ) + 1 2 1 y = − m( x) + 3 2 71.
a.
72.
b.
73.
b.
a.
74.
b.
75.
a.
a.
b.
76.
77.
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Section 2.5
143
78.
79.
81.
80.
82.
83. a.
b.
c.
84. a.
b.
c.
....................................................... 85.
a. b.
f ( x) =
2 2
( x + 1) + 1
f ( x) = −
+1
2 ( x − 2) 2 + 1
Connecting Concepts 86.
a.
f ( x ) = ( x − 2) 2 + ( x − 2) − 3 f ( x ) = ( x − 2) x − 3
b.
f ( x ) = − ⎣⎡( x − 3) 2 + ( x − 3) ⎦⎤ − 2 = − ⎡⎣ ( x − 3) x − 1 ⎤⎦ − 2 f ( x ) = (3 − x ) x − 1 − 2
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144
Chapter 2: Functions and Graphs
.......................................................
Prepare for Section 2.6 2 3 2 2 PS2. (3x − x + 2)(2 x − 3) = 6 x − 2 x + 4 x − 9 x + 3x − 6
PS1. (2 x 2 + 3x − 4) − ( x 2 + 3x − 5) = x 2 + 1
= 6 x 3 − 11x 2 + 7 x − 6 PS3.
f (3a ) = 2(3a )2 − 5(3a ) + 2
PS4.
f (2 + h ) = 2(2 + h )2 − 5(2 + h ) + 2 = 2h 2 + 8h + 8 − 5h − 10 + 2
= 18a 2 − 15a + 2
= 2h 2 + 3h PS6. 2 x − 8 = 0 x=4 Domain: x > 4 or [4, ∞)
PS5. Domain: all real numbers except x = 1
Section 2.6 1.
f ( x) + g ( x) = ( x 2 − 2 x −15) + ( x + 3)
2.
= x 2 − x −12 Domain all real numbers
= x 2 + x − 30 Domain all real numbers
f ( x) − g ( x) = ( x 2 − 2 x −15) − ( x + 3)
f ( x) − g ( x) = ( x 2 − 25) − ( x − 5)
= x 2 − 3 x −18 Domain all real numbers
= x 2 − x − 20 Domain all real numbers
f ( x) g ( x) = ( x 2 − 2 x −15)( x + 3)
f ( x) g ( x) = ( x 2 − 25)( x − 5)
= x3 + x 2 − 21x − 45 Domain all real numbers
= x3 − 5 x 2 − 25 x +125 Domain all real numbers
f ( x) / g ( x) = ( x 2 − 2 x −15) /( x + 3) = x − 5 Domain { x | x ≠ −3} 3.
f ( x ) + g ( x ) = (2 x + 8) + ( x + 4) = 3x +12 Domain all real numbers f ( x) − g ( x ) = (2 x + 8) − ( x + 4) = x + 4 Domain all real numbers f ( x) g ( x) = (2 x + 8)( x + 4)
f ( x) / g ( x) = ( x 2 − 25) /( x − 5) = x + 5 Domain { x | x ≠ 5} 4.
= 2 x 2 +16 x + 32 Domain all real numbers f ( x) / g ( x) = (2 x + 8) /( x + 4) = [ 2( x + 4)] /( x + 4)
f ( x ) + g ( x ) = ( x3 − 2 x 2 + 7 x ) + x = x3 − 2 x 2 + 8 x Domain all real numbers f ( x ) − g ( x ) = ( x3 − 2 x 2 + 7 x ) − x = x3 − 2 x 2 + 6 x Domain all real numbers
f ( x) + g ( x) = (5 x −15) + ( x − 3) = 6 x −18 Domain all real numbers f ( x) − g ( x) = (5 x −15) − ( x − 3) = 4 x −12 Domain all real numbers f ( x) g ( x) = (5 x −15)( x − 3) = 5 x 2 − 30 x + 45 Domain all real numbers f ( x) / g ( x) = (5 x −15) /( x − 3) = [5( x − 3)] /( x − 3)
= 2 Domain { x | x ≠ −4}
5.
f ( x) + g ( x) = ( x 2 − 25) + ( x − 5)
= 5 Domain { x | x ≠ 3}
6.
f ( x) + g ( x) = ( x 2 − 5 x − 8) + (− x) = x 2 − 6 x −8 Domain all real numbers f ( x) − g ( x) = ( x 2 − 5 x − 8) − (− x) = x 2 − 4 x −8 Domain all real numbers f ( x) g ( x) = ( x 2 − 5 x − 8)(− x)
f ( x ) g ( x ) = ( x3 − 2 x 2 + 7 x ) x = x 4 − 2 x3 + 7 x 2 Domain all real numbers f ( x ) / g ( x ) = ( x3 − 2 x 2 + 7 x ) / x = x 2 − 2 x + 7 Domain { x | x ≠ 0}
= − x3 + 5 x 2 + 8 x Domain all real numbers f ( x) / g ( x) = ( x 2 − 5 x −8 x) /(− x) = − x + 5 + 8 Domain { x | x ≠ 0} x
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Section 2.6
7.
145
f ( x ) + g ( x ) = (4 x − 7) + (2 x 2 + 3x − 5) = 2 x 2 + 7 x −12 Domain all real numbers f ( x ) − g ( x ) = (4 x − 7) − (2 x 2 + 3x − 5) = −2 x 2 + x − 2 Domain all real numbers f ( x ) g ( x ) = (4 x − 7)(2 x 2 + 3x − 5) = 6 x 3 −10 x 2 +12 x 2 − 20 x − 21x + 35 = 6 x 3 + 2 x 2 − 41x + 35 Domain all real numbers f ( x ) / g ( x ) = (4 x − 7) /(2 x 2 + 3x − 5) =
8.
⎧ ⎫ 4x −7 Domain ⎨ x | x ≠1, x ≠ − 5 ⎬ ⎩ ⎭ 2 2 x + 3x − 5 2
f ( x ) + g ( x ) = (6 x +10) + (3x 2 + x −10) = 3x 2 + 7 x Domain all real numbers f ( x ) − g ( x ) = (6 x +10) − (3x 2 + x −10) = −3x 2 + 5 x + 20 Domain all real numbers f ( x ) g ( x ) = (6 x +10)(3x 2 + x −10) =18 x 3 + 6 x 2 − 60 x + 30 x 2 +10 x −100 =18 x 3 + 36 x 2 − 50 x −100 Domain all real numbers f ( x ) / g ( x ) = (6 x + 10) /(3x 2 + x − 10) =
9.
{
f ( x) + g ( x) =
x−3 + x
Domain { x | x ≥ 3}
f ( x) − g ( x) =
x−3 − x
Domain { x | x ≥ 3}
f ( x) g ( x) = x x − 3
11.
}
6 x + 10 Domain x | x ≠ −2, x ≠ 5 3 3x 2 + x − 10
Domain { x | x ≥ 3}
f ( x) / g ( x) =
x−3 +x x
f ( x) + g ( x) =
4 − x 2 + 2 + x Domain {x | −2 ≤ x ≤ 2}
f ( x) − g ( x) =
4 − x 2 − 2 − x Domain {x | −2 ≤ x ≤ 2}
Domain { x | x ≥ 3}
10.
f ( x) + g ( x) =
x − 4 − x Domain {x | x ≥ 4}
f ( x) − g ( x) =
x − 4 + x Domain {x | x ≥ 4}
f ( x) g ( x) = − x x − 4
Domain { x | x ≥ 4}
x−4 x
Domain { x | x ≥ 4}
f ( x) / g ( x) = −
⎛ ⎞ f ( x) g ( x) = ⎜ 4 − x 2 ⎟(2 + x ) Domain {x | −2 ≤ x ≤ 2} ⎝ ⎠
12.
f ( x) / g ( x) =
4 − x2 2+ x
f ( x) + g ( x) =
x2 − 9 + x − 3
Domain {x | x ≤ −3 or x ≥ 3}
f ( x) − g ( x ) =
x2 − 9 − x + 3
Domain {x | x ≤ −3 or x ≥ 3}
Domain {x | −2 ≤ x ≤ 2}
⎛ ⎞ f ( x) g ( x) = ⎜ x 2 − 9 ⎟( x − 3) Domain {x | x ≤ −3 or x ≥ 3} ⎝ ⎠
f ( x) / g ( x) =
x2 − 9 x−3
Domain {x | x ≤ −3 or x ≥ 3}
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146
13.
Chapter 2: Functions and Graphs
( f + g )( x) = x 2 − x − 2
14.
( f + g )(5) = (5) 2 − (5) − 2 = 25 − 5 − 2 = 18
16.
2
⎛1⎞ ⎛1⎞ ⎛1⎞ ( f + g)⎜ ⎟ = ⎜ ⎟ − ⎜ ⎟ − 2 ⎝2⎠ ⎝2⎠ ⎝ 2⎠ 1 1 = − −2 4 2 9 =− 4
= 49 + 7 − 2 = 54
( f + g )( x) = x 2 − x − 2
17.
2
( f − g )( x) = x 2 − 5 x + 6
( f − g )( x) = x 2 − 5 x + 6
18.
( f − g )( −3) = (−3) 2 − 5(−3) + 6
( f − g )( x) = x 2 − 5 x + 6
( f + g )( x) = x 2 − x − 2
15.
( f + g )(−7) = (−7)2 − (−7) − 2
⎛2⎞ ⎛2⎞ ⎛ 2⎞ ( f + g)⎜ ⎟ = ⎜ ⎟ − ⎜ ⎟ − 2 ⎝3⎠ ⎝3⎠ ⎝ 3⎠ 4 2 = − −2 9 3 20 =− 9 19.
( f + g )( x) = x 2 − x − 2
( f − g )(24) = (24)2 − 5(24) + 6
= 9 + 15 + 6
= 576 − 120 + 6
= 30
= 462
20.
( f − g )( x) = x 2 − 5 x + 6 ( f − g )(0) = (0)2 − 5(0) + 6 =6
( f − g )(−1) = (−1)2 − 5(−1) + 6 = 1+ 5 + 6 = 12
21.
)
(
( fg )( x) = x 2 − 3 x + 2 ( 2 x − 4 )
22.
( fg )( x) = 2 x3 − 10 x 2 + 16 x − 8 ( fg )( −3) = 2(−3)3 − 10(−3)2 + 16(−3) − 8
= 2 x3 − 6 x 2 + 4 x − 4 x 2 + 12 x − 8
= −54 − 90 − 48 − 8
= 2 x3 − 10 x 2 + 16 x − 8
= −200
( fg )(7) = 2(7)3 − 10(7)2 + 16(7) − 8 = 686 − 490 + 112 − 8 = 300
23.
( fg )( x) = 2 x3 − 10 x 2 + 16 x − 8 3
24.
( fg )( −100) = 2(−100)3 − 10(−100)2 + 16( −100) − 8 = −2,000,000 − 100,000 − 1600 − 8
2
⎛2⎞ ⎛ 2⎞ ⎛2⎞ ⎛ 2⎞ ( fg ) ⎜ ⎟ = 2 ⎜ ⎟ − 10 ⎜ ⎟ + 16 ⎜ ⎟ − 8 ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ 16 40 32 = − + −8 125 25 5 −384 = = −3.072 125 25.
⎛ ⎜ ⎝ ⎛ ⎜ ⎝
f ⎞ x2 − 3x + 2 ⎟ ( x) = 2x − 4 g⎠ 1 1 f ⎞ ⎟ ( x) = x − 2 2 g⎠
⎛f ⎞ 1 1 ⎜ ⎟ (−4) = ( −4 ) − 2 2 g ⎝ ⎠ 1 = −2 − 2 1 5 = −2 or − 2 2
( fg )( x) = 2 x3 − 10 x 2 + 16 x − 8
= −2,101,608
26.
⎛ f ⎞ 1 1 ⎜ ⎟ ( x) = x − 2 2 ⎝g⎠ ⎛f ⎞ 1 1 ⎜ ⎟ (11) = (11) − 2 2 ⎝g⎠ 11 1 = − 2 2 10 = =5 2
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Section 2.6
27.
147
⎛f ⎞ 1 1 ⎜ ⎟ ( x) = x − 2 2 ⎝g⎠
28.
⎛ f ⎞⎛ 1 ⎞ 1 ⎛ 1 ⎞ 1 ⎜ ⎟⎜ ⎟ = ⎜ ⎟ − ⎝ g ⎠⎝ 2 ⎠ 2 ⎝ 2 ⎠ 2 1 1 = − 4 2 1 =− 4
29.
31.
⎛ f ⎞⎛ 1 ⎞ 1 ⎛ 1 ⎞ 1 ⎜ ⎟⎜ ⎟ = ⎜ ⎟ − ⎝ g ⎠⎝ 4 ⎠ 2 ⎝ 4 ⎠ 2 1 1 = − 8 2 3 =− 8
f ( x + h) − f ( x) [ 2( x + h) + 4] − (2 x + 4) = h h 2 x + 2(h) + 4 − 2 x − 4 = h 2 h = h =2 2 f ( x + h) − f ( x) [ ( x + h) − 6] − ( x − 6) = h h
=
x 2 + 2 x( h) + ( h) 2 − 6 − x 2 + 6 h
2 x ( h) + h 2 h = 2x + h
30.
32.
f ( x + h) − f ( x) [ 4( x + h) − 5] − (4 x − 5) = h h 4 x + 4(h) − 5 − 4 x + 5 = h 4(h) = h =4 2 2 ⎡ ⎤ f ( x + h) − f ( x) ⎢⎣( x + h) +11⎥⎦ − ( x +11) = h h
=
x 2 + 2 xh + (h)2 +11− x 2 −11 h
= 2 xh + h h = 2x + h
=
33.
⎛f ⎞ 1 1 ⎜ ⎟ ( x) = x − 2 2 ⎝g⎠
2
f ( x + h) − f ( x) 2( x + h)2 + 4( x + h) − 3− (2 x 2 + 4 x − 3) = h h 2 2 2 = 2 x + 4 xh + 2h + 4 x + 4h − 3 − 2 x − 4 x + 3 h 2 = 4 xh + 2h + 4h h = 4 x + 2h + 4
34.
f ( x + h) − f ( x) 2( x + h)2 − 5( x + h) − (2 x 2 − 5 x + 7) = h h 2 2 2 = 2 x + 4 xh + 2h − 5 x − 5h + 7 − 2 x + 5 x − 7 h 2
= 4 xh + 2h − 5h h = 4 x + 2h − 5
35.
f ( x + h) − f ( x) −4( x + h)2 + 6 − (−4 x 2 + 6) = h h 2 2 2 = −4 x − 8 xh − 4h + 6 + 4 x − 6 h
= −8 xh − 4h h = −8 x − 4h
2
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148
36.
Chapter 2: Functions and Graphs
f ( x) = −5 x 2 − 4 x f ( x + h) − f ( x) −5( x + h)2 − 4( x + h) − (−5 x 2 − 4 x) = h h =
−5 x 2 −10 x(h) − 5h 2 − 4 x 2 − 4h + 5 x 2 + 4 x h
−10 x(h) − 5h 2 − 4h h = −10 x − 5h − 4 =
37.
( g o f )( x) = g [ f ( x) ]
= g [ 3 x + 5] = 2 [ 3 x + 5]
= 6 x +10 − 7 = 6x +3
38.
( g o f )( x) = g [ f ( x) ]
= g [ 2 x − 7]
39.
= 6 x − 21+ 5 = 6 x −16 ( f o g )( x) = f [ g ( x) ]
= f [3 x + 2]
= 2 [3 x + 2] − 7 = 6x + 4−7 = 6x −3
( g o f )( x) = g ⎡⎢ x 2 + 4 x −1⎤⎥ ⎣ ⎦ 2 ⎡ ⎤ = ⎢ x + 4 x −1⎥ + 2 ⎣ ⎦
( g o f )( x) = g ⎡⎢ x 2 −11x ⎤⎥ ⎣ ⎦ 2 = 2 ⎡⎢ x −11x ⎤⎥ + 3 ⎣ ⎦
( g o f )( x) = g [ f ( x) ] = g ⎡⎢ x3 + 2 x ⎤⎥ ⎣ ⎦ 3 ⎡ = −5 ⎢ x + 2 x ⎤⎥ ⎣ ⎦ = −5 x3 −10 x
42.
= 3[ 2 x − 7 ] + 5
= 6 x − 21+ 2 = 6 x −19
= 2 x 2 − 22 x + 3 41.
= f [ 2 x − 7]
= 3[ 2 x − 7 ] + 2
= x 2 + 4 x +1
40.
( f o g )( x) = f [ g ( x)]
( g o f )( x) = g [ f ( x) ] = g ⎡⎢ − x − 7 ⎤⎥ ⎣ ⎦ = ⎡⎢ − x3 − 7 ⎤⎥ +1 ⎣ ⎦ 3
= − x3 − 6
( f o g )( x) = f [ x + 2] 2
= [ x + 2] + 4 [ x + 2]−1 = x 2 + 4 x + 4 + 4 x + 8 −1 = x 2 + 8 x +11 ( f o g )( x) = f [ 2 x + 3] 2
= [ 2 x + 3] −11[ 2 x + 3] = 4 x 2 +12 x + 9 − 22 x − 33 = 4 x 2 −10 x − 24 ( f o g )( x) = f [ g ( x)] = f [ −5 x ] 3
= [ −5 x ] + 2 [ − 5 x ] = −125 x3 −10 x
( f o g )( x) = f [ g ( x)] = f [ x +1]
3
= − [ x +1] − 7 = − x3 − 3 x 2 − 3 x −1− 7 = − x3 − 3 x 2 − 3 x − 8
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.6
43.
149
( g o f )( x) = g [ f ( x)] ⎡ ⎤ =g⎢ 2 ⎥ ⎣ x +1 ⎦ ⎡ ⎤ = 3 ⎢ 2 ⎥ −5 1 x + ⎣ ⎦ 5( x +1) 6 = − x +1 x +1 = 6 −5x −5 x +1 1 − = 5x x +1
44.
( g o f )( x) = g [ f ( x) ] = g ⎡⎣ x + 4 ⎤⎦ 1 x+4 = x+4 x+4 =
( f o g )( x) = f [ g ( x)]
= f [ 3 x − 5] = =
2
[3x − 5]+1 2 3x − 4
( f o g )( x) = f [ g ( x)] =f 1 x = 1 +4 x 1 = + 4x x 1 4x + = x = x 1+ 4 x x 2 = x + 4x x
45.
( g o f )( x) = g [ f ( x)] ⎡ ⎤ =g⎢ 1 ⎥ ⎣ x2 ⎦ =
⎡1 ⎤ ⎢ 2 ⎥ −1 ⎣x ⎦
2 = 1− x 2 x
( f o g )( x) = f [ g ( x)] = f ⎡⎣ x −1 ⎤⎦ 1 = ⎡ x −1 ⎤ ⎣ ⎦ 1 = x −1
2
2 = 1− x | x|
46.
( g o f )( x) = g [ f ( x)]
( f o g )( x) = f [ g ( x)]
⎡ ⎤ =g⎢ 6 ⎥ ⎣ x−2 ⎦ 3 = ⎡ 6 ⎤ 5⎢ ⎣ x − 2 ⎥⎦ = 3 ⎛ 30 ⎞ ⎜ ⎟ ⎝ x−2 ⎠ = 3⋅ x − 2 30 = x−2 10
⎡ ⎤ =f⎢3⎥ ⎣ 5x ⎦ 6 = ⎡3⎤ ⎢⎣ 5 x ⎥⎦ − 2 6 = ⎛ 3-10x ⎞ ⎜ ⎟ ⎝ 5x ⎠ = 30 x 3 −10 x
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150
47.
Chapter 2: Functions and Graphs
⎡ ( g o f )( x) = g ⎢ 3 ⎢⎣ 5 − x =− 2 ⎡ 3 ⎢ ⎣⎢ 5 − x =
⎤ ⎥ ⎥⎦
⎡ ⎤ ( f o g )( x) = f ⎢ − 2 ⎥ ⎣ x⎦ 3 = ⎡ 2⎤ 5− ⎢− ⎥ ⎣ x⎦ 3 = 5+ 2 x 3 = [5x + 2]
⎤ ⎥ ⎦⎥
−2 5 − x 3
x =
48.
3x 5x + 2
( f o g )( x) = f [ g ( x) ]
( g o f )( x) = g ⎡⎣ 2 x +1 ⎤⎦
= f ⎡⎢3 x 2 −1⎤⎥ ⎣ ⎦ 2 ⎤ ⎡ = 2 ⎢3 x −1⎥ +1 ⎣ ⎦
2
= 3 ⎣⎡ 2 x +1 ⎦⎤ −1 2
= 3 ( 2 x +1) −1 = 3(4 x 2 + 4 x +1) −1
= 6 x 2 − 2 +1
=12 x 2 +12 x + 3 −1
= 6 x 2 −1
=12 x 2 +12 x + 2 Use the results to work Exercises 49 to 64. 49.
( g o f )( x) = 4 x 2 + 2 x − 6
50.
( g o f )(4) = 4(4)2 + 2(4) − 6 = 64 + 8 − 6 = 66
51.
( f o g )(4) = 2(4)2 −10(4) + 3 = 32 − 40 + 3 = −5
( f o g )( x) = 2 x 2 −10 x + 3
52.
( f o g )(−3) = 2(−3) 2 −10(−3) + 3 =18 + 30 + 3 = 51
53.
57.
= 4−2−6 = −4
( g o h)( x) = 9 x 4 − 9 x 2 − 4
54.
(h o g )( x) = −3x 4 + 30 x3 − 75 x 2 + 4 ( h o g )(0) = −3(0) 4 + 30(0)3 − 75(0)2 + 4 =4
( f o f )( x) = 4 x + 9 ( f o f )(8) = 4(8) + 9 = 41
56.
( h o g )( x) = −3 x 4 + 30 x3 − 75 x 2 + 4 4
( g o f )( x) = 4 x 2 + 2 x − 6 ( g o f )(−1) = 4(−1)2 + 2(−1) − 6
( g o h)(0) = 9(0) 4 − 9(0)2 − 4 = −4
55.
( f o g )( x) = 2 x 2 −10 x + 3
3
58. 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ (h o g ) ⎜ 2 ⎟ = −3 ⎜ 2 ⎟ + 30 ⎜ 2 ⎟ − 75 ⎜ 2 ⎟ + 4 ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ 48 240 300 =− + − +4 625 125 25 = −48 +1200 − 7500 + 2500 625 3848 =− 625
( f o f )( x) = 4 x + 9 ( f o f )(−8) = 4(−8) + 9 = −23 ( g o h)( x) = 9 x 4 − 9 x 2 − 4 4
2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ( g o h) ⎜ − 1 ⎟ = 9 ⎜ − 1 ⎟ − 9 ⎜ − 1 ⎟ − 4 ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ 9 9 = − −4 81 9 = 1 −1− 4 9 = −4 8 or − 44 9 9
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.6
151
( g o f )( x) = 4 x 2 + 2 x − 6
59.
( g o f )( 3) = 4( 3)2 + 2( 3) − 6
61.
( f o g )( x) = 2 x 2 − 10 x + 3
60.
( f o g )( 2) = 2( 2)2 − 10( 2) + 3
= 12 + 2 3 − 6
= 4 − 10 2 + 3
= 6+2 3
= 7 − 10 2
( g o f )( x) = 4 x 2 + 2 x − 6
62.
( f o g )( x) = 2 x 2 −10 x + 3 ( f o g )(3k ) = 2(3k )2 −10(3k ) + 3
( g o f )(2c) = 4(2c)2 + 2(2c) − 6
=18k 2 − 30k + 3
= 16c 2 + 4c − 6 ( g o h)( x) = 9 x 4 − 9 x 2 − 4
63.
( g o h)( k + 1) = 9(k + 1) 4 − 9(k + 1)2 − 4 = 9(k 4 + 4k 3 + 6k 2 + 4k + 1) − 9k 2 − 18k − 9 − 4 = 9k 4 + 36k 3 + 54k 2 + 36k + 9 − 9k 2 − 18k − 13 = 9k 4 + 36k 3 + 45k 2 + 18k − 4
(h o g )( x) = −3x 4 + 30 x3 − 75 x 2 + 4
64.
(h o g )(k −1) = −3(k −1)4 + 30(k −1)3 − 75(k −1)2 + 4 = −3k 4 +12k 3 −18k 2 +12k − 3+ 30k 3 − 90k 2 + 90k − 30 − 75k 2 +150k − 75 + 4 = −3k 4 + 42k 3 −183k 2 + 252k −104 65.
a.
66.
r = 1.5t and A = πr 2
a.
2
so A(t ) = π [ r (t )]
= π (1.5t )2
w = 2 − 0.2t for 0 ≤ t ≤10 = −2 + 0.2t for 10 ≤ t ≤14 or w = 2 − 0.2t
2
A(2) = 2.25π (2) = 9π square feet ≈ 28.27 square feet b.
Note:
r = 1.5t h = 2r = 2(1.5t ) = 3t and 1 V = π r 2h so 3 1 V (t ) = π (1.5t )2 [3t ] 3
l = 3 − 0.5t for 0 ≤ t ≤ 6 = −3 + 0.5t for 6 ≤ t ≤14 or l = 3 − 0.5t
b.
l = lw = 3 − 0.5t 2-0.2t = (3 − 0.5t )(2 − 0.2t )
c.
A is increasing on [6, 8] and on [10, 14] ; and A is decreasing on [0, 6] and on [8, 10] .
d.
The highest point on the graph of A occurs when t = 0 seconds.
= 2.25π t 3 1 1 1 V = π r 2h = (π r 2 ) = hA 3 3 3 1 2 = ( 3t ) (2.25π t ) = 2.25π t 3 3
V (3) = 2.25π (3)3 = 60.75π cubic feet ≈ 190.85 cubic feet
Xmin = 0, Xmax =14, Xscl = 2, Ymin = −1, Ymax = 6, Yscl = 2
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152
67.
Chapter 2: Functions and Graphs
a.
2 2 2 Since d + 4 = s , 2
68.
2
d = s −16
The perimeter is an increasing function over 0 ≤ t ≤ 14. The graph of P = 2(3 + 0.5t ) + 2 2 − 0.2t is shown below.
d = s 2 −16 d = (48 − t )2 −16 ⋅ s = 48 − t = 2304 − 96t + t 2 −16 = t 2 − 96t + 2288 b.
s (35) = 48 − 35 =13 ft d (35) = 352 − 96(35) + 2288
Xmin = 0, Xmax = 14, Xscl = 5, Ymin = −1, Ymax = 22, Yscl =5
= 153 ≈12.37 ft
69.
(Y o F )( x) = Y ( F ( x)) converts x inches to yards. F takes x inches to feet, and then Y takes feet to yards.
71.
a.
70.
( I o F )( x) = I ( F ( x)) converts x yards to inches. F takes x yards to feet, and then I takes feet to inches.
On [0, 1] , a = 0 Δt = 1 − 0 = 1 C ( a + Δt ) = C (1) = 99.8 C ( a ) = C ( 0) = 0
C (1) − C (0) = 99.8 − 0 = 99.8 1 This is identical to the slope of the line through C (1) − C (0) (0, C(0)) and (1, C(1)) since m = = C (1) − C (0) 1− 0 On [0, 0.5] , a = 0 , Δt = 0.5 Average rate of change =
b.
C (0.5) − C (0) 78.1 − 0 = = 156.2 0.5 0.5 On [1, 2] , a = 1 , Δt = 2 − 1 = 1
Average rate of change = c.
C (2) − C (1) 50.1 − 99.8 = = −49.7 1 1 On [1, 1.5] , a = 1 , Δt = 1.5 − 1 = 0.5
Average rate of change = d.
C (1.5) − C (1) 84.4 − 99.8 −15.4 = = = −30.8 0.5 0.5 0.5 On [1, 1.25] , a = 1 , Δt = 1.25 − 1 = 0.25
Average rate of change = e.
Average rate of change = f.
C (1.25) − C (1) 95.7 − 99.8 −4.1 = = = −16.4 0.25 0.25 0.25
On [1, 1 + Δt ] , Con (1+ Δt ) = 25 (1+ Δt )3 −150 (1+ Δt ) 2 + 225 (1+ Δt ) 3
2
= 25(1+ 3( Δt ) + 3 ( Δt ) ) −150(1+ 2 ( Δt ) + ( Δt ) ) + 225(1+ Δt ) 2
3
2
= 25 + 75( Δt ) + 75 ( Δt ) ) + 25 ( Δt ) −150 − 300 ( Δt ) −150 ( Δt ) + 225 + 225 ( Δt ) 2
=100 − 75 ( Δt ) + 25 ( Δt ) Con (1) =100
3
3
Con (1+ Δt ) − Con (1) 100 − 75( Δt )2 + 25 ( Δt ) −100 = Δt Δt −75( Δt )2 + 25( Δt )3 = Δt = −75( Δt ) + 25( Δt )2
Average rate of change =
As Δt approaches 0, the average rate of change over [1, 1 + Δt ] seems to approach 0. Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.6
72.
a.
153
On [ 2, 3] , a = 2 Δt = 3 − 2 = 1 f (a + Δt ) = f (3) = 6 ⋅ 32 = 54 f (a ) = f (2) = 6 ⋅ 2 2 = 24
f (a + Δt ) − f (a ) f (3) − f (2) = = 54 − 24 = 30 feet per second Δt 1 This is identical to the slope of the line through (2, f (2)) f (3) − f (2) and (3, f (3)) since m = = f (3) − f (2). 3−2 On [ 2, 2.5] , a = 2 ,
Average velocity =
b.
Δt = 2 .5 − 2 = 0 .5 f (a + Δt ) = f (2.5) = 6( 2.5) 2 = 37.5
Average velocity = c.
On [ 2, 2.1] , a = 2
f (2.5) − f (2) 37.5 − 24 13.5 = = = 27 feet per second 0.5 0.5 0.5
Δt = 2 .1 − 2 = 0 .1 f (a + Δt ) = f (2.1) = 6( 2.1) 2 = 26.46
Average velocity = d.
On [ 2, 2.01] , a = 2
f (2.1) − f (2) 26.46 − 24 2.46 = = = 24.6 feet per second 0.1 0.1 0.1
Δt = 2.01 − 2 = 0.01 f (a + Δt ) = f (2.01) = 6( 2.01) 2 = 24.2406
Average velocity = e.
On [ 2, 2.001] , a = 2
f (2.01) − f (2) 24.2406 − 24 0.2406 = = = 24.06 feet per second 0.01 0.01 0.01
Δt = 2.001 − 2 = 0.001 f (a + Δt ) = f ( 2.001) = 6(2.001) 2 = 24.024006
Average velocity = f.
f (2.001) − f (2) 24.024006 − 24 0.24006 = = = 24.006 feet per second 0.001 0.001 0.001
On [ 2, 2 + Δt ] , Con
f (2 +Δt ) − f (2) 6(2 + Δt )2 − 24 6(4 + 4(Δt ) + (Δt )2 − 24 24 + 24(Δt ) + 6(Δt )2 − 24 = = = Δt Δt Δt ( Δt )
24Δt + 6(Δt ) 2 = 24 + 6( Δt ) Δt As Δt approaches 0, the average velocity seems to approach 24 feet per second. =
....................................................... 73.
( g o f )( x) = g [ f ( x) ]
= g [ 2 x + 3]
= 5(2 x + 3) +12 =10 x +15 +12 =10 x + 27 ( g o f )( x) = ( f o g )( x)
( f o g )( x) = f [ g ( x) ]
= f [5 x +12] = 2(5 x +12) + 3 =10 x + 24 + 3 =10 x + 27
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Connecting Concepts
154
74.
Chapter 2: Functions and Graphs
( g o f )( x) = g [ f ( x) ]
( f o g )( x) = f [ g ( x) ]
= g [ 4 x − 2]
= f [ 7 x − 4] = 4(7 x − 4) − 2 = 28 x −16 − 2 = 28 x −18
= 7(4 x − 2) − 4 = 28 x −14 − 4 = 28 x −18 ( g o f )( x) = ( f o g )( x) 75.
( g o f )( x) = g [ f ( x)]
( f o g )( x) = f [ g ( x)]
⎡ ⎤ = g ⎢ 6x ⎥ ⎣ x −1 ⎦
⎡ ⎤ = f ⎢ 5x ⎥ ⎣ x−2 ⎦ 6 ⎛⎜ 5 x ⎞⎟ = ⎝ x−2 ⎠ 5x −1
⎛ ⎞ 5⎜ 6x ⎟
= ⎝ x−1 ⎠ 6x −2
x−1 30 x = x−1 = 6 x−2 x+2 x−1
x−2
30 x x−1 4 x+ 2 x−1
= 30 x ⋅ x −1 x −1 2(2 x +1) = 15 x 2 x +1
30 x 30 x − x x 2 = = −2 5 x − x+ 2 x−2
4 x+2 x−2
= 30 x ⋅ x − 2 x − 2 2(2 x +1) = 15 x 2 x +1
( g o f )( x) = ( f o g )( x) 76.
( g o f )( x ) = g [ f ( x )]
( f o g )( x ) = f [ g ( x)]
⎡ ⎤ = g ⎢ 5x ⎥ ⎣ x+3 ⎦
⎡ ⎤ = f ⎢− 2 x ⎥ ⎣ x−4 ⎦ 5 ⎛⎜ − 2 x ⎞⎟ = ⎝ x−4 ⎠ 2x − +3
⎛ ⎞ 2 ⎜ 5x ⎟
= − ⎝ x+3 ⎠ 5x −4
x+3 10 x = − x+3 = − 5 x−4 x−12 x+3
= − 10 x ⋅ x + 3 x + 3 x −12 = − 10 x x −12 ( g o f )( x) = ( f o g )( x)
77.
( g o f )( x ) = g [ f ( x )]
= g [ 2 x + 3]
[ 2 x + 3]− 3 = 2
= 2x 2 =x
x−4 −10 x x−4 = = −2 x+3 x−12 x−4
10 x x+3 x−12 x+3
= −10 x ⋅ x − 4 x − 4 x −12 10 = x x −12
−10 x x−4 x−12 x−4
( f o g )( x) = f [ g ( x) ] ⎡ ⎤ = f ⎢ x −3 ⎥ ⎣ 2 ⎦ ⎡ ⎤ = 2 ⎢ x −3 ⎥ + 3 ⎣ 2 ⎦ = x − 3+ 3 =x
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.6
78.
155
( g o f )( x ) = g [ f ( x )]
= g [ 4 x − 5]
[ 4 x − 5] + 5 = 4
= 4x 4 =x
79.
( g o f )( x) = g [ f ( x) ] ⎡ ⎤ =g⎢ 4 ⎥ ⎣ x +1 ⎦ ⎡ ⎤ 4− ⎢ 4 ⎥ 1 + x ⎣ ⎦ = ⎡ 4 ⎤ ⎢⎣ x +1 ⎥⎦ 4x + 4− 4 = x +1 4 x +1 = 4 x ⋅ x +1 x +1 4 =x
80.
( g o f )( x) = g [ f ( x) ] ⎡ ⎤ =g⎢ 2 ⎥ ⎣1− x ⎦ ⎡ 2 ⎤ ⎢ ⎥−2 = ⎣1− x ⎦ ⎡ 2 ⎤ ⎢⎣1− x ⎥⎦ 2− 2+ 2x = 1− x 2 1− x = 2 x ⋅1− x 1− x 2 =x
81.
( g o f )( x) = g [ f ( x) ]
( f o g )( x) = f [ g ( x)] ⎡ ⎤ = f ⎢ x +5 ⎥ ⎣ 4 ⎦ ⎡ ⎤ = 4 ⎢ x +5 ⎥ −5 ⎣ 4 ⎦ = x + 5−5 =x ( f o g )( x) = f [ g ( x) ] ⎡ ⎤ = f ⎢ 4− x ⎥ ⎣ x ⎦ 4 = ⎡ 4− x ⎤ ⎢⎣ x ⎥⎦ +1 4 = 4− x + x x 4 = 4 x = 4⋅ x 4 =x ( f o g )( x ) = f [ g ( x )] ⎡ ⎤ = f ⎢ x−2 ⎥ ⎣ x ⎦ 2 = ⎡ x−2 ⎤ 1− ⎢ ⎣ x ⎥⎦ 2 = x− x+2 x 2 = ⋅ x 1 x− x+2 = 2⋅ x 1 2 =x
( f o g )( x) = f [ g ( x)]
= g ⎡⎢ x3 −1⎤⎥ ⎣ ⎦
= f ⎡ 3 x +1 ⎤ ⎣ ⎦
= 3 ⎡⎢ x3 −1⎤⎥ +1 ⎣ ⎦
= ⎡ 3 x +1 ⎤ −1 ⎣ ⎦ = x +1−1 =x
3
= x3 =x
3
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156
82.
Chapter 2: Functions and Graphs
( f o g )( x) = f [ g ( x)]
( g o f )( x ) = g [ f ( x )]
= f ⎡3 2 − x ⎤ ⎣ ⎦ = ⎡3 2 − x ⎤ + 2 ⎣ ⎦ = −(2 − x) + 2 = −2 + x + 2 =x
= g ⎡⎢ − x + 2 ⎤⎥ ⎣ ⎦ 3
= 3 2 − ⎡⎢ − x3 + 2 ⎤⎥ ⎣ ⎦ 3
= 2 + x3 − 2 3
= x3 =x
.......................................................
Prepare for Section 2.7
PS1. Slope: − 1 ; y-intercept: (0, 4) 3
PS2. 3x − 4 y =12 y = 3 x −3 4 3 Slope: ; y-intercept: (0, –3) 4
PS3. y = –0.45x + 2.3
PS4.
y + 4 = − 2 ( x − 3) 3 y = − 2 x−2 3
PS6.
f ( x1) − y1 + f ( x2 ) − y2 = (2)2 − 3− (−1) + (4)2 − 3−14
PS5.
f (2) = 3(2)2 + 4(2) −1=12 + 8 −1=19
= 4 − 3+1 + 16 − 3−14 = 2 +1 =3
Section 2.7 1.
The scatter diagram suggests no relationship between x and y.
2.
The scatter diagram suggest a nonlinear relationship between x and y.
3.
The scatter diagram suggests a linear relationship between x and y.
4.
The scatter diagram suggests a linear relationship between x and y.
5.
Figure A better approximates a graph that can be modeled by an equation than does Figure B. Thus Figure A has a coefficient of determination closer to 1.
6.
Figure A better approximates a graph that can be modeled by an equation than does Figure B. Thus Figure A has a coefficient of determination closer to 1.
7.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y = 2.00862069x + 0.5603448276 Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.7
8.
157
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y = −1.918918919x + 0.4594594595 9.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y = −0.7231182796x + 9.233870968 10.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y =0.6591216216x – 6.658108108 11.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y = 2.222641509x –7.364150943 12.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y = −2.301587302x + 4.813968254 13.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y = 1.095779221x2 – 2.69642857x + 1.136363636
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158
14.
Chapter 2: Functions and Graphs
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y = −0.5714285714x2 + 2.2x + 1.942857143 15.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y = −0.2987274717x2 – 3.20998141x + 3.416463667 16.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y = 1.414285714x2 + 1.954285714x – 2.705714286 17.
18.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
a.
y = 23.55706665x – 24.4271215
b.
y = 23.55706665(54) – 24.4271215 ≈ 1248 cm
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
a.
y = 3.410344828x + 65.09359606
b.
y = 3.410344828x + 65.09359606 ≈ 263 ft
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.7
19.
20.
21.
22.
23.
159
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
a.
y = 0.194224924x + 0.7978723404
b.
y = 0.194224924(32) + 0.7978723404 ≈ 4.3 m/s
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
a.
y = 6.357142857x + 90.57142857
b.
y = 6.357142857(7.5) + 90.57142857 = 138.25 or 138,000 bacteria
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
a.
y = 0.1628623408x – 0.6875682232
b.
y = 0.1628623408(158) – 0.6875682232 ≈ 25
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
a.
y = −0.6800298805x + 69.05129482
b.
5 feet 8 inches = 68 inches; y = −0.6800298805(68) + 69.05129482 ≈ 23
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
The value of r is close to 0. Therefore, no, there is not a strong linear relationship between the current and the torque.
Copyright © Houghton Mifflin Company. All rights reserved.
160
24.
Chapter 2: Functions and Graphs
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
The value of r is close to –1, so, yes, there is a strong correlation between a man’s age and his average remaining lifetime. 25.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
a.
26.
The value of r is close to –1, so, yes, there is a strong linear correlation.
b.
y = −0.9033088235x + 78.62573529
c.
y = −0.9033088235(25) + 78.62573529 ≈ 56 years
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y = 0.2711847067 x + 7.91528368 y = 0.2711847067(41) + 7.91528368 ≈19 meters per second
27.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y =113.3111246 x + 21.83605895 a. Positively b. 28.
y =113.3111246(9.5) + 21.83605895 ≈1098 calories
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
a.
y = −0.0074642857 x 2 + 1.148214286 x + 4.807142857
b.
y = −0.0074642857(65)2 + 1.148214286(65) + 4.807142857 ≈ 47.9 ft
Copyright © Houghton Mifflin Company. All rights reserved.
Section 2.7
29.
161
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
y = −0.6328671329x2 + 33.6160839x – 379.4405594 30.
31.
32.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
a.
y = −0.0004093949t2 + 0.2265681259t + 55.57907207
b.
1:00 P.M. is 7 hours, or 420 minutes, after 6:00 A.M. y = −0.0004093949(420)2 + 0.2265681259(420) + 55.57907207 ≈ 78.5°
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
a.
y = −0.0165034965x2 + 1.366713287x + 5.685314685
b.
y = −0.0165034965(50)2 + 1.366713287(50) + 5.685314685 ≈ 32.8 mpg
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
a.
y = 0.05208x2 – 3.56026x + 82.32999
b.
− b = − −3.56026 ≈ 34 kilometers per hour 2a 2(0.05208)
Copyright © Houghton Mifflin Company. All rights reserved.
162
Chapter 2: Functions and Graphs
....................................................... 33.
a.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here. 5-lb ball
y = 0.6130952381t2 – 0.0714285714t + 0.1071428571 10-lb ball
y = 0.6091269841t2 – 0.0011904762t – 0.3 15-lb ball
y = 0.5922619048t2 + 0.3571428571t – 1.520833333 b. 34.
All the regression equations are approximately the same. Therefore, there is one equation of motion.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.
a.
y = 454.1584409x – 40.78364910
b.
y = 454.1584409(1.5) – 40.78364910 ≈ 640 kilometers per second
Copyright © Houghton Mifflin Company. All rights reserved.
Connecting Concepts
Chapter Review
163
.......................................................
Exploring Concepts with Technology
Graphing Piecewise Functions with a Graphing Calculator 1.
Use Dot mode. Enter the function as Y1=X2*(X<2)-X*(X≥2) and graph this in the standard viewing window.
2.
Use Dot mode. Enter the function as Y1=(X2-X)*(X<2)+(-X+4)*(X≥2) and graph this in the standard viewing window.
3.
Use Dot mode. Enter the function as Y1=(-X2+1)*(X<0)+(X2-1)*(X≥0) and graph this in the standard viewing window.
4.
Use Dot mode. Enter the function as Y1=(X3-4X)*(X<1)+(X2-X+2)*(X≥1) and graph this in the standard viewing window.
.......................................................
Assessing Concepts
1.
Circle
2.
There are two values of y for one value of x.
3.
Let f ( x ) = x 2 , then f (3) = f ( −3) but 3 ≠ −3 .
4.
No. If f ( a ) = f (b) , then f is not a one-to-one function.
6.
All real numbers.
5.
f (0) is greater than 7.
7.
(–2, –5)
8.
9.
(–3, 5)
10.
(3, –10)
11.
(3, 1)
12.
(–4, –2)
f (1) is not defined.
....................................................... 1.
d = (7 − (−3))2 + (11− 2)2
[2.1]
Chapter Review 2.
= 102 + 92 = 100 + 81 = 181 3.
⎛ 2 + ( −3) 8 + 12 ⎞ ⎛ −1 20 ⎞ ⎛ 1 ⎞ , ⎜ ⎟ = ⎜ , ⎟ = ⎜ − , 10 ⎟ [2.1] 2 2 2 2 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
d = (−3 − 5)2 + (−8 − (−4))2
[2.1]
= (−8)2 + ( −4)2 = 64 +16 = 80 = 4 5 4.
⎛ −4 + 8 7 + (−11) ⎞ ⎛ 4 −4 ⎞ , ⎜ ⎟=⎜ , ⎟ = (2, − 2 ) [2.1] 2 ⎝ 2 ⎠ ⎝2 2 ⎠
Copyright © Houghton Mifflin Company. All rights reserved.
164
5.
Chapter 2: Functions and Graphs
center (3, −4), radius 9 [2.1]
x 2 +10 x + y 2 + 4 y = −20
6. 2
2
x +10 x + 25 + y + 4 y + 4 = −20 + 25 + 4 ( x + 5)2 + ( y + 2)2 = 9 center (−5, −2), radius 3 7.
( x − 2)2 + ( y + 3)2 = 52 [2.1]
8.
( x + 5)2 + ( y −1)2 = r 2 [2.1] (3 + 5) 2 + (1−1)2 = r 2 82 + 02 = r 2 82 = r 2 ( x + 5)2 + ( y −1)2 = 82
9.
a.
b.
f (1) = 3(1) 2 + 4(1) − 5 [2.2] = 3(1) + 4 − 5 = 3+ 4 − 5 =2 f (−3) = 3(−3)2 + 4(−3) − 5 = 3(9) −12 − 5 = 27 −12 − 5 =10
c.
f (t ) = 3t 2 + 4t − 5
d.
2
10.
a.
= 64 − 9 = 55
b.
= 39 c.
= 0 =0
2
= 3 x 2 + 6 xh + 3h 2 + 4 x + 4h − 5
d.
2
3 f (t ) = 3(3t + 4t − 5) = 9t +12t −15 f (3t ) = 3(3t )2 + 4(3t ) − 5 = 3(9t 2 ) +12t − 5 = 27t 2 +12t − 5
g (− x) = 64 − (− x) 2 = 64 − x 2
2
f.
g (8) = 64 − (8)2 = 64 − 64
= 3( x + 2 xh + h ) + 4 x + 4h − 5
e.
g (−5) = 64 − (−5)2 = 64 − 25
f ( x + h) = 3( x + h) + 4( x + h) − 5 2
g (3) = 64 − 32 [2.2]
e.
2 g (t ) = 2 64 − t 2
f.
g (2t ) = 64 − (2t )2 = 64 − 4t 2 = 4(16 − t 2 ) = 2 16 − t 2
Copyright © Houghton Mifflin Company. All rights reserved.
[2.1]
Chapter Review
11.
a.
165
( f o g )(3) = f [ g (3)] = f [3 − 8] = f [−5]
[2.6]
12.
a.
= (−5) + 4(−5) = 25 − 20
c.
=5 ( g o f )(−3) = g[ f (−3)] = g[(−3)2 + 4( −3)] = g[9 − 12] = g[−3] = [−3] − 8 = −11 ( f o g )( x) = f [ g ( x)] = f [ x − 8]
b.
= 2(6) 2 + 7 = 72 + 7 = 79 ( g o f )(−5) = g[ f (−5)]
c.
= g[2(−5)2 + 7] = g[57] =| 57 − 1 | = 56 ( f o g )( x ) = f [ g ( x )]
= f [| x − 1 |]
= 2( x − 1) 2 + 7
= x 2 − 16 x + 64 + 4 x − 32
= 2( x 2 − 2 x + 1) + 7
= x 2 − 12 x + 32 ( g o f )( x) = g[ f ( x)]
= 2x2 − 4x + 2 + 7
= g[ x 2 + 4 x)]
d.
= [ x2 + 4 x] − 8
= 2x2 − 4x + 9 ( g o f )( x ) = g[ f ( x)] = g[2 x 2 + 7]
= x2 + 4 x − 8
=| 2 x 2 + 7 − 1 | =| 2 x 2 + 6 | = 2 x2 + 6
13.
f ( x + h) − f ( x) 4( x + h)2 − 3( x + h) −1− (4 x 2 − 3 x −1) = h h =
[2.6]
4( x 2 + 2 xh + h 2 ) − 3 x − 3h −1− 4 x 2 + 3 x +1 h
2 2 2 = 4 x + 8 xh + 4h − 3x − 3h −1− 4 x + 3 x +1 h 2
= 8 xh + 4h − 3h h h(8 x + 4h − 3) = h = 8 x + 4h − 3
14.
g ( x + h ) − g ( x ) ( x + h )3 − ( x + h ) − ( x 3 − x ) = h h
[2.6]
3 2 2 3 3 = x + 3 x h + 3 xh + h − x − h − x + x h 2 2 3 = 3 x h + 3 xh + h − h h
=
[2.6]
= 2(| x − 1 |) 2 + 7
= ( x − 8)2 + 4( x − 8)
d.
= f [| ( − 5) − 1 |] = f [6]
2
b.
( f o g )( − 5) = f [ g ( − 5)]
h(3 x 2 + 3xh + h 2 −1) h
= 3 x 2 + 3 xh + h 2 −1
Copyright © Houghton Mifflin Company. All rights reserved.
166
Chapter 2: Functions and Graphs
15.
16.
f is increasing on [3, ∞) f is decreasing on (−∞, 3] [2.2]
f is increasing on [0, ∞) f is decreasing on (−∞, 0] [2.2]
17.
18.
f is increasing on [−2, 2] f is constant on (−∞, −2] ∪ [2, ∞) [2.2]
19.
f is constant on . . . , [−6, −5), [−5, −4), [−4, −3), [−3, −2), [−2, −1), [−1, 0), [0, 1), . . . [2.2] 20.
f is increasing on (−∞, ∞) [2.2]
f is increasing on (−∞, ∞) [2.2]
{
}
21.
Domain x x is a real number [2.2]
23.
Domain x − 5 ≤ x ≤ 5 [2.2]
25.
m=
{
}
−7 − 3 −10 = = −2 [2.3] 4 − (−1) 5 y − 3 = −2( x + 1) point - slope form y − 3 = −2 x − 2 y = −2 x + 1
{
}
22.
Domain x x ≤ 6 [2.2]
24.
Domain x x ≠ −3, x ≠ 5 [2.2]
26.
m=
{
11 − 0 11 [2.3] = 7−0 7 11 y − 0 = ( x − 0) 7 11 y= x 7
Copyright © Houghton Mifflin Company. All rights reserved.
}
Chapter Review
27.
167
[2.3] 3x − 4 y = 8 −4 y = −3 x + 8 y = 3 x−2 4 Slope of parallel line is y − 11 = y − 11 = y= y= y=
29.
28.
3 . 4
5 Slope of perpendicular line is . 2 5 y − (−7) = [ x − (−3)] 2 5 y + 7 = ( x + 3) 2 5 15 y+7 = x+ 2 2 5 15 y = x+ −7 2 2 5 15 14 y = x+ − 2 2 2 5 1 y = x+ 2 2
3 ( x − 2) 4 3 3 x− 4 2 3 3 x − + 11 4 2 3 3 22 x− + 4 2 2 3 19 x+ 4 2
f ( x) = ( x 2 + 6 x) +10
2 x = −5 y +10 [2.3] 5 y = −2 x +10 y = − 2 x+2 5
30.
[2.4]
f ( x ) = ( 2 x 2 + 4 x ) + 5 [2.4]
f ( x) = ( x 2 + 6 x + 9) +10 − 9
f ( x ) = 2( x 2 + 2 x ) + 5
f ( x) = ( x + 3)2 +1
f ( x ) = 2( x 2 + 2 x + 1) + 5 − 2 f ( x ) = 2( x + 1)2 + 3
31.
f ( x ) = − x 2 − 8 x + 3 [2.4]
32.
f ( x ) = −( x 2 + 8 x ) + 3
f ( x ) = (4 x 2 − 6 x) + 1 [2.4] 3 ⎞ ⎛ f ( x ) = 4⎜ x 2 − x ⎟ + 1 2 ⎠ ⎝ 3 9⎞ 9 ⎛ f ( x ) = 4⎜ x 2 − x + ⎟ + 1 − 2 16 4 ⎝ ⎠
f ( x ) = −( x 2 + 8 x + 16) + 3 + 16 f ( x ) = −( x + 4) 2 + 19
2
3⎞ 4 9 ⎛ f ( x ) = 4⎜ x − ⎟ + − 4⎠ 4 4 ⎝ 2
3⎞ 5 ⎛ f ( x ) = 4⎜ x − ⎟ − 4⎠ 4 ⎝
33.
f ( x) = −3 x 2 + 4 x − 5 [2.4]
34.
4 ⎞ ⎛ f ( x ) = −3⎜ x 2 − x ⎟ − 5 3 ⎠ ⎝ 4 4⎞ 4 ⎛ f ( x ) = −3⎜ x 2 − x + ⎟ − 5 + 3 9 3 ⎝ ⎠
f ( x) = x 2 − 6 x + 9 [2.4] f ( x) = ( x 2 − 6 x) + 9 f ( x ) = ( x 2 − 6 x + 9) + 9 − 9 f ( x ) = ( x − 3) 2 + 0
2
2⎞ 15 4 ⎛ f ( x ) = −3⎜ x − ⎟ − + 3⎠ 3 3 ⎝ 2
2⎞ 11 ⎛ f ( x ) = −3⎜ x − ⎟ − 3 3 ⎝ ⎠
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168
35.
Chapter 2: Functions and Graphs
−b −(−6) 6 = = = 1 [2.4] 2a 2(3) 6
36.
f (1) = 3(1)2 − 6(1) + 11
f (0) = 4(0) 2 − 10 = 0 − 101 = −10 Thus the vertex is (0, −10).
= 3(1) − 6 + 11 = 3 − 6 + 11 =8 Thus the vertex is (1, 8).
37.
−b −(60) −60 = = = 5 [2.4] 2a 2(−6) − 12
38.
f (5) = −6(5) 2 + 60(5) + 11 = −150 + 300 + 11 = 161 Thus the vertex is (5, 161). d=
mx1 + b − y1 1+ m
d= d= d= d=
2
−b −(−8) 8 = = = −4 [2.4] 2a 2(−1) − 2 f (−4) = 14 − 8( −4) − (−4) 2 = 14 + 32 − 16 = 30 Thus the vertex is (−4, 30).
= −6(25) + 300 + 11
39.
0 −b = = 0 [2.4] 2a 2(4)
, y = 2 x − 3, ( x1, y1) = (1, 3) [2.3]
40.
a. b.
2(1) + (−3) − 3 1 + 22 2−3−3 1+ 4 −4 5 4 5
=
4 5 5
41.
Revenue = 13x Profit = Revenue − Cost P =13x − (0.5 x +1050) P =13x − 0.5 x −1050 P =12.5 x −1050 c. Break even ⇒ Revenue = Cost 13 x = 0.5 x +1050 12.5 x =1050 x = 84 The company must ship 84 parcels. [2.5]
42.
[2.5] [2.5] 43.
The graph of y = x 2 − 7 is symmetric with respect to the y-axis. [2.5]
44.
The graph of x = y 2 + 3 is symmetric with respect to the x-axis. [2.5]
45.
The graph of y = x3 − 4 x is symmetric with respect to the origin. [2.5]
46.
The graph of y 2 = x 2 + 4 is symmetric with respect to the x-axis, y-axis, and the origin. [2.5]
47.
The graph of
48.
The graph of xy = 8 is symmetric with respect to the origin. [2.5]
x2 32
+
y2 42
= 1 is symmetric with respect to the x-axis, y-axis, and the origin. [2.5]
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Chapter Review
169
49.
The graph of y = x is symmetric with respect to the x-axis, y-axis, and the origin. [2.5]
50.
The graph of x + y = 4 is symmetric with respect to the origin. [2.5]
51.
52.
a. b.
Domain all real numbers Range y y ≤ 4
{
}
g is an even function [2.5]
53.
a.
Domain x − 4 ≤ x ≤ 4
54.
a. b.
{
{
{
}
}
Range y 0 ≤ y ≤ 4
Domain all real numbers Range y y ≥ 4
}
b.
g is an even function [2.5]
a.
Domain all real numbers Range y y is an even integer
b.
g is neither even nor odd [2.5]
g is an even function [2.5]
55.
56.
a. b.
57.
b.
Domain all real numbers Range all real numbers g is neither even nor odd [2.5]
a.
Domain all real numbers Range all real numbers g is an odd function [2.5]
F ( x ) = x 2 + 4 x − 7 [2.5]
58.
{
A( x) = x 2 − 6 x − 5 [2.5]
F ( x) = ( x 2 + 4 x) − 7
A( x) = ( x 2 − 6 x) − 5
F ( x ) = ( x 2 + 4 x + 4) − 7 − 4
A( x) = ( x 2 − 6 x + 9) − 5 − 9
F ( x) = ( x + 2) 2 − 11
A( x) = ( x − 3) 2 − 14
Copyright © Houghton Mifflin Company. All rights reserved.
}
170
59.
Chapter 2: Functions and Graphs
P ( x) = 3x 2 − 4 [2.5]
60.
G ( x) = 2 x 2 − 8 x + 3 [2.5] G ( x ) = 2( x 2 − 4 x ) + 3
P ( x) = 3( x − 0) 2 − 4
G ( x ) = 2( x 2 − 4 x + 4) + 3 − 8 G ( x ) = 2( x − 2) 2 − 5
61.
W ( x) = −4 x 2 − 6 x + 6 [2.5]
62.
3 ⎞ ⎛ W ( x) = −4⎜ x 2 + x ⎟ + 6 2 ⎠ ⎝ 3 9⎞ 9 ⎛ W ( x) = −4⎜ x 2 + x + ⎟ + 6 + 2 16 ⎠ 4 ⎝
T ( x) = −2 x 2 − 10 x [2.5]
T ( x) = −2( x 2 + 5 x) 25 ⎞ 25 ⎛ T ( x) = −2⎜ x 2 + 5 x + ⎟+ 4 ⎠ 2 ⎝ 2
5⎞ 25 ⎛ T ( x) = −2⎜ x + ⎟ + 2⎠ 2 ⎝
2
3⎞ 24 9 ⎛ W ( x) = −4⎜ x + ⎟ + + 4⎠ 4 4 ⎝ 2
3⎞ 33 ⎛ W ( x) = −4⎜ x + ⎟ + 4 4 ⎝ ⎠
63.
64.
[2.5]
[2.5] 65.
66.
[2.5]
[2.2] Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Review
171
67.
68.
[2.2]
[2.2] 69.
( f + g )( x) = x 2 − 9 + x + 3 [2.6]
70.
( f + g )( x) = x3 + 8 + x 2 − 2 x + 4 [2.6]
= x2 + x − 6 The domain is all real numbers.
= x3 + x 2 − 2 x +12 The domain is all real numbers.
( f − g )( x) = x 2 − 9 − ( x + 3)
( f − g )( x) = x3 + 8 − ( x 2 − 2 x + 4)
= x2 −9 − x −3
= x3 + 8 − x 2 + 2 x − 4
= x 2 − x −12 The domain is all real numbers.
= x3 − x 2 + 2 x + 4 The domain is all real numbers.
( fg )( x) = ( x 2 − 9)( x + 3)
( fg )( x) = ( x3 + 8)( x 2 − 2 x + 4)
= x3 + 3 x 2 − 9 x − 27 The domain is all real numbers.
= x5 − 2 x 4 + 4 x3 + 8 x 2 −16 x + 32 The domain is all real numbers.
⎛ f ⎞ x2 −9 ⎜⎜ ⎟⎟ ( x ) = x +3 ⎝g⎠ ( x − 3)( x + 3) = x +3 = x −3 The domain is { x x ≠ −3}.
⎛f ⎞ x3 + 8 ⎜⎜ ⎟⎟ ( x) = 2 x − 2x + 4 ⎝g⎠ =
( x + 2)( x 2 − 2 x + 4)
x2 − 2 x + 4 = x+2 The domain is restricted when x 2 − 2 x + 4 = 0 . x= x=
− ( −2) ± (−2) 2 − 4(1)(4) 2(1) 2 ± 4 − 16 2
2 ± − 12 which is not a real number 2 Therefore the domain is all real numbers. x=
71.
Let x = one of the numbers and 50 – x = the other number. Their product is given by y = x(50 − x ) = 50 x − x 2 = − x 2 + 50 x.
Now y takes on its maximum value when −b −50 −50 = = = 25. x= −2 2a 2(−1) Thus the two numbers are 25 and (50 – 25) = 25. That is, both numbers are 25. [2.4]
72.
Let x = the smaller number. Let x + 10 equal the larger number. The sum of their squares y is given by y = x 2 + ( x +10)2 = x 2 + x 2 + 20 x +100 = 2 x 2 + 20 x +100 Now y takes on its minimum value when −b −20 −20 x= = = = −5 2a 2(2) 4 Thus the numbers are –5 and (−5 + 10) = 5. [2.4]
Copyright © Houghton Mifflin Company. All rights reserved.
172
73.
Chapter 2: Functions and Graphs
s (t ) = 3t 2 [2.4]
74. 3(4)2 − 3(2) 2 4−2 3(16) − 3(4) = 2 − 48 12 = 2 36 = =18 ft/sec 2
a.
Average velocity =
b.
Average velocity =
c.
d.
e.
3(3)2 − 3(2)2 3− 2 3(9) − 3(4) = 1 = 27 −12 =15 ft/sec 1
3(2.5)2 − 3(2)2 2.5 − 2 3(6.25) − 3(4) = 0.5 18.75 −12 = 0.5 = 6.75 =13.5 ft/sec 0.5
s (t ) = 2t 2 + t [2.4] 2(5)2 + 5 −[2(3)2 + 3] 5−3 2(25) + 5 −[2(9) + 3] = 2 50 + 5 −[18 + 3] = 2 + − 50 5 18 − 3 = 2 = 34 =17 ft/sec 2
a.
Average velocity =
b.
Average velocity =
c.
Average velocity =
2(4) 2 + 4 −[2(3)2 + 3] 4−3 2(16) + 4 −[2(9) + 3] = 1 32 + 4 −[18 + 3] = 1 + − 32 4 18 − 3 = 1 = 15 =15 ft/sec 1
Average velocity =
3(2.01)2 − 3(2) 2 2.01− 2 3(4.0401) − 3(4) = 0.01 12.1203 −12 = 0.01 = 0.1203 =12.03 ft/sec 0.01 It appears that the average velocity of the ball approaches 12 ft/sec.
2(3.5)2 + 3.5 −[2(3)2 + 3] 3.5 − 3 2(12.25) + 3.5 −[2(9) + 3] = 0.5 24.5 + 3.5 −[18 + 3] = 0.5 = 24.5 + 3.5 −18 − 3 0.5 = 7 =14 ft/sec 0.5
Average velocity =
d.
e.
2(3.01)2 + 3.01−[2(3) 2 + 3] 3.01− 3 2(9.0601) + 3.01−[2(9) + 3] = 0.01 18.1202 + 3.01−[18 + 3] = 0.01 = 18.1202 + 3.01−18 − 3 0.01 = 0.1302 =13.02 ft/sec 2 It appears that the average velocity of the ball approaches 13 ft/sec. Average velocity = =
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here. [2.7]
75.
a. b.
y = 1.171428571x + 5.19047619 y = 1.171428571(12) + 5.19047619 ≈ 19 m/s
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Chapter Test
76.
173
a.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here. [2.7]
b.
h = 0.0047952048t 2 − 1.756843157t + 180.4065934 Empty ⇒ y = 0 ⇒ the graph intersects the x-axis. Graph the equation, and notice that it never intersects the x-axis.
Xmin = 0, Xmax = 400, Xscl = 100 Ymin = 0, Ymax = 200, Xscl = 50 c.
Thus, no, on the basis of this model, the can never empties. The regression line is a model of the data and is not based on physical principles.
.......................................................
Quantitative Reasoning
QR1.
The apparent attractor is 0.6. QR2.
The behavior is not chaotic. 0.59 and 0.74 are the approximate values of the two attractors. QR3. Four different values could be attractors: approximately 0.50, 0.58, 0.83, and 0.87. QR4. There are no attractors. QR5. The behavior of the function becomes more chaotic, and there are no attractors.
....................................................... 1.
⎛ x + x2 y1 + y2 ⎞ ⎛ −2 + 4 3 + ( −1) ⎞ ⎛ 2 2 ⎞ , , midpoint = ⎜ 1 =⎜ ⎟ = (1, 1) [2.1] ⎟=⎜ , 2 ⎠⎟ ⎝ 2 2 ⎠ ⎝2 2⎠ ⎝ 2 length = d = ( x1 − x2 ) 2 + ( y1 − y2 )2 = (−2 − 4) 2 + (3 − (−1))2 = ( −6) 2 + 42 = 36 + 16 = 52 = 2 13
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Test
174
2.
Chapter 2: Functions and Graphs
x = 0⇒ 0 = 2 y2 −4
x = 2 y 2 − 4 [2.1]
4 = 2 y2
y = 0 ⇒ x = 2(0)2 − 4 = −4 Thus the x-intercept is (−4, 0).
2 = y2 ± 2=y Thus the y-intercepts are (0, − 2) and (0,
3.
2). 4.
y = x + 2 + 1 [2.1]
x2 − 4 x + y 2 + 2 y − 4 = 0 2
2
( x − 4 x) + ( y + 2 y ) = 4 2
( x − 4 x + 4) + ( y 2 + 2 y +1) = 4 + 4 +1 ( x − 2)2 + ( y +1)2 = 9
center (2, −1), radius 3 5.
6.
x 2 −16 ≥ 0 ( x − 4)( x + 4) ≥ 0 The product is positive or zero. The critical values are 4 and −4.
{
a. b. c.
}
The domain is x x ≥ 4 or x ≤ −4 . [2.2] 7.
9.
a.
R = 12x
b.
P = revenue − cost P = 12 x − (6.75 x + 875) P = 11.25 x − 875
c.
break-even ⇒ P = 0 0 = 11.25 x − 875 875 = 11.25 x 78 ≈ x 78 parcels must be sent to break even. [2.4]
a.
8.
f ( x) = x 4 − x 2
[2.5]
f ( − x) = (− x)4 − (− x)2 = x 4 − x 2 = f ( x) f ( x) is an even function. b.
f ( x ) = x3 − x f ( − x ) = ( − x )3 − ( − x ) = − x 3 + x = − ( x3 − x ) = − f ( x ) f ( x ) is an odd function.
c.
increasing on (−∞, 2] never constant decreasing on [2, ∞ ) [2.2]
f ( x) = x −1 f ( − x) = − x −1 ≠ f ( x) not an even function f ( − x) = − x −1 ≠ − f ( x) not an odd function Neither
[2.5]
10.
[2.3] 3x − 2 y = 4 −2 y = − 3 x + 4 y = 3x−2 2 Slope of perpendicular line is − 2 . 3 y − y1 = m( x − x1 ) y + 2 = − 2 ( x − 4) 3 2 y+2=− x+8 3 3 2 8 y=− x+ −6 3 3 3 y =−2x+ 2 3 3
Copyright © Houghton Mifflin Company. All rights reserved.
[2.1]
Cumulative Review
11.
−
175
b −4 =− = 2 [2.4] 2a 2(1)
12.
= x2 + x − 3 ⎛ f ⎞ f ( x) ⎜ ⎟ ( x) = g ( x) ⎝g⎠ 2 x = −1 , x ≠ 2 x−2
f ( x) = x 2 + 1 [2.6]
14. 2
16.
f ( x) = x 2 − 2 x
g ( x) = 2 x + 5 [2.6]
( f o g )( x ) = f [ g ( x )] = f ( 2 x + 5)
2
f ( x + h ) − f ( x ) ( x + h ) + 1 − ( x + 1) = h h 2 2 2 = x + 2 xh + h + 1 − x − 1 h 2 h (2 x + h ) = 2 xh + h = h h = 2x + h
15.
= ( 2 x + 5) − 2 ( 2 x + 5) 2
= 4 x 2 + 20 x + 25 − 4 x − 10 = 4 x 2 + 16 x + 15
s (t ) = 5t 2 [2.6]
a.
Average velocity =
5(3)2 − 5(2)2 5(9) − 5(4) = = 45 − 20 = 25 ft/sec 3− 2 1
b.
Average velocity =
5(2.5)2 − 5(2)2 5(6.25) − 5(4) 31.25 − 20 = = = 22.5 ft/sec 2.5 − 2 0.5 0.5
c.
Average velocity =
5(2.01)2 − 5(2)2 5(4.0401) − 5(4) 20.2005 − 20 = = = 20.05 ft/sec 2.01 − 2 0.01 0.01
a.
b.
Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here. [2.7]
y = −7.98245614 x + 767.122807 y = −7.98245614(89) + 767.122807 ≈ 57 calories
....................................................... 1.
4.
[2.6]
= ( x 2 −1) + ( x − 2)
f (2) = 22 − 4(2) − 8 = 4−8−8 = −12 The minimum value of the function is –12.
13.
( f + g )( x ) = f ( x ) + g ( x )
Commutative Property of Addition [P.1]
2.
6 , 2 are not rational numbers
π
[P.1]
( −4 xy 2 )3 ( −2 x 2 y 4 ) = ( −64 x 3 y 6 )( −2 x 2 y 4 ) [P.2] = ( −64)( −2)( x 3+ 2 y 6+ 4 )
Cumulative Review 3.
3 + 4(2 x − 9) [P.1] 3 + 8 x − 36 8 x − 33
5.
24a 4 b3 = 4a 4 − 4 b3−5 = 4b −2 = 4 [P.2] 3 3 18a 4 b5 3b2
7.
x 2 + 6 x − 27 = ( x + 9)( x − 3) = x + 9 [P.5] ( x + 3)( x − 3) x + 3 x2 − 9
= 128 x 5 y10 6.
(2 x + 3)(3x − 7) = 6 x 2 − 5 x − 21 [P.3]
Copyright © Houghton Mifflin Company. All rights reserved.
176
Chapter 2: Functions and Graphs
8.
4( x − 1) 2(2 x − 1) −2 4 − 2 = − = 4x − 4 − 4x + 2 = [P.5] 2 x − 1 x − 1 (2 x − 1)( x − 1) (2 x − 1)( x − 1) (2 x − 1)( x − 1) (2 x − 1)( x − 1)
9.
6 − 2(2 x − 4) = 14 ⇒ 6 − 4 x + 8 = 14 ⇒ − 4 x = 0 ⇒ x = 0 [1.1]
10.
x2 − x − 1 = 0 ⇒
11.
(2 x − 1)( x + 3) = 4 ⇒ 2 x 2 + 5 x − 3 = 4 ⇒ 2 x 2 + 5 x − 7 = (2 x + 7)( x − 1) = 0 ⇒ x = − 7 or x = 1 [1.3] 2
12.
3x + 2 y = 15
x=
−( −1) ± ( −1)2 − 4(1)( −1) = 1 ± 1 + 4 = 1 ± 5 [1.3] 2(1) 2 2
[1.1]
3x = −2 y + 15 x = −2 y+5 3
13.
2
Let u = x . u2 − u − 2 = 0 (u − 2)(u + 1) = 0 u−2=0 or u + 1 = 0 u = −1 u=2 x2 = 2 x=± 2
15. 16.
3x − 1 < 5 x + 7 [1.5] −2 x < 8 x > −4
x 2 = −1 x = ±i
distance = [ −2 − 2]2 + [−4 − ( −3)]2 = ( −4)2 + ( −1)2 = 16 + 1 = 17 [2.1] G( x) = 2 x3 − 4 x − 7
[2.2]
17.
G ( −2) = 2( −2)3 − 4( −2) − 7 = 2( −8) + 8 − 7 = −15
18.
14.
x 4 − x 2 − 2 = 0 [1.4]
0
x
0.08
60
0.03
60 + x
19.
−1 − ( −3) −1 + 3 2 = = = − 1 [2.3] 2 −2 − 2 −2 − 2 − 4 1 The equation is y − ( −3) = − ( x − 2) ⇒ y = − 1 x − 2 2 2
The slope is m =
h ( x ) = −0.002 x 2 − 0.03x + 8 h(39) = −0.002(39) 2 − 0.03(39) + 8 = 3.788 ft Yes. [2.4]
0.08(60) + 0 x = 0.03(60 + x ) 4.8 = 1.8 + 0.03x 3 = 0.03x 100 = x 100 ounces of water [1.1] 20.
0.04°F/min [2.3]
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 3
Polynomial and Rational Functions Section 3.1 1.
2.
5 x 2 − 9 x + 10 − 10
x +3
3
2
x3 + 2 x2 − x + 1 +
x 2 + 4 x + 10 +
1 x −2
4.
x3 − 4 x 2 − 4 x − 3 −
7 x −1
x − 1 x 4 − 5x 3 + x − 4 x4 − x3 − 4x 3 −4 x 3 + 4 x 2 − 4 x2 + x −4 x 2 + 4x − 3x − 4 −3 x + 3 7 6.
25 x −3
x − 3 x3 + x2 − 2 x − 5 x 3 − 3x 2 4x2 − 2x 4 x 2 − 12 x 10x − 5 10 x − 30 25 7.
2
x + 4 6 x + 15x − 8 x + 2 6 x 3 + 24 x 2 − 9 x2 − 8x −9 x 2 − 36 x 28x + 2 28 x + 112 − 110
x − 2 x4 − 5 x 2 + 3x − 1 4 3 x − 2x 2x 3 − 5 x 2 2x 3 − 4 x 2 − x 2 + 3x − x2 + 2 x x −1 x−2 1 5.
x +4
3
x + 3 5 x + 6 x − 17 x + 20 5 x 3 + 15 x 2 − 9 x 2 − 17 x −9 x 2 − 27 x 10 x + 20 10 x + 30 − 10 3.
6 x 2 − 9 x + 28 − 110
x 2 + 5 x + 14 +
23 x −2
x − 2 x 3 + 3x 2 + 4 x − 5 x3 − 2 x 2 5x 2 + 4 x 5 x 2 − 10 x 14x − 5 14 x − 34 23
x 4 + 2 x3 + 2 x + 1 − 8
x−1
x −1 x5 + x 4 − 2 x3 + 2 x 2 − 3x − 7 x5 − x 4 2x 4 − 2x3 2 x 4 − 2 x3 0 + 2x 2 − 3 x 2 x2 − 2 x −x − 7 −x − 1 −8
8.
x 4 − 6 x 3 + 23x 2 − 89 x + 351 −
1396 x+4
x + 4 x 5 − 2 x 4 − x 3 + 3 x 2 − 5x + 8 x5 + 4 x 4 − 6x 4 − x 3 −6 x 4 − 24 x 3 23x 3 + 3x 2 23x 3 + 92 x 2 − 89 x 2 − 5 x −89 x 2 − 356 x 351x + 8 351x + 1404 − 1396
Copyright © Houghton Mifflin Company. All rights reserved.
178
Chapter 3: Polynomial and Rational Functions
9.
x 2 + 3x − 2 +
10.
− x+5 2 x 2 − x+1
− x +3 2 x 2 + x−3
x3 − x2 + 2 x − 1 +
12.
2 x 2 + 2 x − 3 2 x5 − x 3 + 5 x 2 − 9x + 6 5 4 2 x +2x − 3x 3 − 2 x 4 + 2x 3 + 5 x 2 −2 x 4 − 2 x 3 + 3x 2 4 x 3 − 2x 2 − 9 x 4 x3 + 4 x2 − 6x − 2 x 2 − 3x + 6 −2 x 2 − 2 x + 3 −x + 3
13.
2
−5 8 3
4 4
−7 24 17
6 6 12
−1
4
0 −4 −4
4
−2 4 2
14.
4
1 1
1
1
0 1 1
1
3 −2 1
16.
1 2
8 8
8x2 + 6
−8 155 147
6 25 31
−3
−4 −18 −22
6 6
0 24 24
5 96 101
−1 404 403
18.
0 1 1
0 1 1
0 1 1
5
6 6
1 735 736
0 66 66
17 −198 −181
−2 30 28
−3 140 137
−1 685 684
0 3420 3420
6 x 3 + 28 x 2 + 137 x + 684 + 3420 x −5
−1 1 0
20.
−1
1 1
0 −1 −1
0 1 1
0 −1 −1
x3 − x2 + x − 1 + 2 x +1
x 4 + x3 + x 2 + x + 1 21.
5
5 x 2 + 31x + 147 + 736 x−5
x 4 + 4 x 3 + 6 x 2 + 24 x + 101 + 403 x−4
19.
2 x+14 x 2 +1
6 x 2 − 22 x + 66 − 181 x+3
−10 16 6
0 4 4
5
5
4 x2 − 4 x + 2 + 1 x +1
17.
x 3 + 3x 2 − 3x − 10 +
x 2 +1 x 5 + 3 x 4 − 2 x 3 − 7 x 2 − x + 4 + x3 x3 4 3x − 3x 3 − 7x 2 + 3x 2 3x 4 3 − 3x −10x 2 − x −3 x 3 − 3x 2 −10 x + 2 x + 4 −10 x 2 −10 2 x +14
4 x 2 + 3x + 12 + 17 x−2
15.
3 x−12 x 2 −2 x+2
x 2 − 2 x + 2 3x 3 + x 2 − 5 x + 2 3 x 3 − 6x 2 + 6 x 7x 2 −11x + 2 7 x 2 −14 x +14 3x −12
2 x 2 − x + 1 2 x 4 + 5 x 3 − 6 x 2 + 4x + 3 2 x4 − x3 + x2 6x 3 − 7x 2 + 4x 6 x 3 − 3x 2 + 3x − 4x2 + x + 3 −4 x 2 + 2 x − 2 −x + 5 11.
3x + 7 +
−4
6
−3
4 0
0 6
3 0
22.
3 4
12
5
5
6
12
−9 −4
3 8
−6 0
12 x 2 − 4 x + 8
Copyright © Houghton Mifflin Company. All rights reserved.
1 1 2
Section 3.1
23.
2
179
1
0 2 2
1
1 4 5
0 10 10
1 20 21
0 42 42
1 84 85
0 170 170
4 340 344
24.
−3
1
0 −3 −3
1
0 9 9
0 −27 −27
0 81 81
1 −243 −242
−10 726 716
2
3
1 6 7
3
−5 30 25
1 14 15
26.
−2
4
10
−2
−6 16 10
28.
3
−1
0 −3 −3
−1
0 −20 −20
5 40 45
30.
3
−1 −220 −221
1
−20 −2210 −2230
32.
2
1
2 2 4
1
0 −27 −27
1 −81 −80
34.
−1
2 2
0 −63 −63
2 −189 −187
36.
−3
1
−5 8 3
1
0 −3 −3
−10 1200 1190
−1 54 53
3 15 18
−1 −18 −19
6
−1
8
1
1
4 57 61
3 −8 −5
0 1 1
−4
1 1
0 −183 −183
5 −40 −35
30 −280 −250
0 1 1
0 1 1
0 1 1
−1 1 0
0 −4 −4
0 16 16
20 −64 −64
0 176 176
−1 −704 −705
P( c ) = P( −4) = −705 -6 6 0
−3 1 −2
−25 9 −16
−6
38.
1
4 −6 −2
1
−1 2 1
0 48 48
40.
4
3 3
144 −144 0
A remainder of 0 implies that x + 3 is a factor of P( x).
−27 12 −15
−90 90 0
A remainder of 0 implies that x + 6 is a factor of P( x).
A remainder of 1 implies that x +1 is not a factor of P ( x). 41.
0 300 300
P (c) = P(1) = 0 0 −21 −21
1 −2 −1
−3
1
A remainder of 0 implies that x − 2 is a factor of P( x). 39.
−1 6 5
2
−1
P (c) = P(3) = −187 37.
−5 4 −1
P (c) = P(8) = −250
0 −9 −9
−10 3 −7
1
−5 80 75
0 20 20
6
P (c) = P (3) = −80 35.
3
2
P (c) = P(10) = −2230 33.
−1 −3 −4
P (c) = P(−3) = −183
−2 −20 −22
−2
0 3 3
2 x 4 + 5 x 3 + 20 x 2 + 75 x + 300 + 1190 x−4
P (c) = P (−2) = 45 31.
−1 −2 −3
0 2 2
P (c) = P (3) = 53
0 −8 −8
4
−3 8 5
4 2 2
P (c) = P( 2) = 25 29.
−1 −1 −2
0 1 1
− x 6 + x 5 − 2 x 4 + 2 x 3 − 3x 2 + 3 x − 4 − 1 x +1
x 5 − 3x 4 + 9 x 3 − 27 x 2 + 81x − 242 + 716 x+3
27.
−1 −1
x 7 + 2 x 6 + 5 x 5 + 10 x 4 + 21x 3 + 42 x 2 + 85 x + 170 + 344 x−2
25.
−1
4 12 16
−27 64 37
−36 148 112
A remainder of 112 implies that x − 4 is not a factor of P( x). 42.
3
1 1
0 3 3
−25 9 −16
0 −48 −48
144 −144 0
A remainder of 0 implies that x − 3 is a factor of P ( x).
Copyright © Houghton Mifflin Company. All rights reserved.
180
43.
Chapter 3: Polynomial and Rational Functions
5
1
−22 35 13
2 5 7
1
−50 65 15
−75 75 0
0 0 0
1 4
16
−8
9
14
4
16
4 −4
−1 8
2 16
4 8
2
3 3
49.
1
1 1
51.
−2
−8 6 −2
−10 −4 −14
28 −28 0
0 1 1
0 1 1
0 1 1
46.
3 3
53.
11
2 2
55.
2 3
57.
2
1 1
8 −6 2
10 −4 6
−18 22 4
−50 44 −6
3
−8
4
3
2 −6
−4 0
1 2 3
−20 20 0
66 −66 0
1 1
−1 4 3
−9 12 3
−11 12 1
−4 4 0
2
a.
8
1 1
−3 8 5
2 40 42
0 336 336
336 ways b.
6
10
-5 4
-2 -6
3 12
-6 0
−10 12 2
−2
−8 6 −2
1
52.
5
1
0 −2 −2
15
−
2
−1
2 5
12
4
5
−2 10
−4 0
5 −1 4
3 −4 −1
1
8 −8 0
−100 115 15
−34 30 −4
2
5
0 4 4
−2 25 23
0 5 5
1
6 −6 0
70 −60 10
−75 75 0 −153 150 −3
−5 1 −4
45 −45 0
−4 4 0
A remainder of 0 implies that x + 1 is a factor of P( x).
P (8) = 83 − 3(8)2 + 2(8) = 512 − 3(64) + 2(8) = 512 − 192 + 16 = 336 ways They are the same.
60.
2
2 2
−1 4 3
−7 6 −1
1 −2 −1
7 −2 5
−10 10 0
A remainder of 0 implies that x − 2 is a factor of P( x). P ( x) = ( x − 2)(2 x 4 + 3 x3 − x 2 − x + 5)
P ( x) = ( x − 4)( x + 3 x + 3 x + 1)
61.
9
P ( x) = ( x + 1)( x3 + 4 x 2 − x − 4)
A remainder of 0 implies that x − 4 is a factor of P( x). 3
−4
1
P ( x) = ( x − 2)( x + 3 x + 7)
4
9
4
58.
2
4 −4 0
10
1
56.
A remainder of 0 implies that x − 2 is a factor of P( x).
59.
3
50.
54.
−14 14 0
1 6 7
−1 2
4
2 −12 −10
−4 8 4
A remainder of 0 implies that x + 1 / 2 is a factor of P( x). 48.
−1 1 0
−23 15 −8
A remainder of 0 implies that x + 1 is a factor of P( x).
A remainder of 8 implies that x −1/ 4 is not a factor of P ( x ). 47.
−6 −9 −15
9 9
A remainder of 0 implies that x − 5 is a factor of P( x). 45.
−1
44.
62.
a.
7
1 1
−10 7 −3
35 −21 14
−50 98 48
24 336 360
0 2520 2520
2520 ways P (7) = 75 − 10(7) 4 + 35(7)3 − 50(7) 2 + 24(7) = 16,807 − 10(2401) + 35(343) − 50(49) + 24(7) = 16,807 − 24,010 + 12,005 − 2, 450 + 168 = 2520 ways They are the same.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.1
63.
181
8
a.
1.5
0.5 12 12.5
1.5
0 100 100
20
b.
a.
1 3
1 2
1 6
0
2
15
91
5 2
91 6
91
b.
1 1
7
a.
1
−6 12 6
−6 996 990
11 72 83
0 11880 11880
24
b.
6
1
0 70 70
0 490 490
11
b.
1 1
–3
1
10 –2 8 8 –3 5
1
1
9 2
325 6
650
−6 10632 10626
11 432 443
3 11 14
1
10 42 52
−8 312 304
9
b.
1
1 9 10
1
304 cubic inches –2
650
0 255024 255024
0 154 154
0 1694 1694
10 90 100
−8 900 892
1694 cubic inches 1 6 7
1
68.
54
−6 24 18
1 1
490 cubic inches a.
0
4
255,024 ways 3 7 10
1
67.
1 6
1 3
11,880 ways 66.
1 2
650 cans
12
a.
1 3
12
91cans 65.
0 610 610
610 cards
1 3
6
0.5 30 30.5
1.5
100 cards 64.
1.5
892 cubic inches 31 –16 15
30 –30 0
69.
1 1
0 1 1
1
–1 1 0
0 1 1
( x 3 − 1) ÷ ( x − 1) = x 2 + x + 1
15 –15 0
1
1 1
( x 3 + 10 x 2 + 31x + 30) ÷ ( x + 2) = x 2 + 8 x + 15
0 1 1
0 1 1
0 1 1
0 1 1
–1 1 0
( x 5 − 1) ÷ ( x − 1) = x 4 + x 3 + x 2 + x + 1
2
( x + 8 x + 15) ÷ ( x + 3) = ( x + 5) in.
1
1 1
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
–1 1 0
( x 7 − 1) ÷ ( x − 1) = x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 ( x 9 − 1) ÷ ( x − 1) = x8 + x 7 + x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 70.
2 1
–1 2 1
1
–14 2 –12
k –24 0
71.
–2
1 6 7
2
k = 24 72.
3 2
–25 21 –4
k –12 0
k = –12 3
14 k –6 –16 3 8 k–16 −6 − 2k + 32 = 0 −2k = −26 k = 13
–6 –2k+32 0
73.
–4
1
3 –4 1 –1 16 − 4k − 64 = 0 −4k = 48 k = −12
–8 4 –4
Copyright © Houghton Mifflin Company. All rights reserved.
k 16 k+16
16 –4k–64 0
182
Chapter 3: Polynomial and Rational Functions
.......................................................
Connecting Concepts
74.
P( c ) = P(1) = 1n − 1 = 0 Thus, by the Factor Theorem, (x – 1) is a factor of P(x) for any positive integer n.
75.
76.
18( −1)80 − 6(−1)50 + 4(−1) 20 − 2 = 18 − 6 + 4 − 2 = 14
77.
5(1) 48 + 6(1)10 − 5(1) + 7 = 5 + 6 − 5 + 7 = 13
i
1
−3
1
i −3 + i
1 −1 − 3i −3i
−3 3 0
A remainder of 0 implies that x − i is a factor of x 3 − 3 x 2 + x − 3.
78.
−2i
1
−2
1
−2i −2 − 2i
1 −4 − 4i − 3 +4i
−8 8 + 6i 6i
A remainder of 0 implies that x + 2i is a factor of x 4 − 2 x3 + x 2 − 8 x − 12.
....................................................... PS1. P( x ) = x 2 − 4 x + 6 − b = − −4 = − −4 = 2 2a 2(1) 2
Prepare for Section 3.2 PS2.
P( x ) = −2 x 2 − x + 1 − b = − −1 = − −1 = − 1 −4 2a 2( −2) 4
( )
P(2) = (2)2 − 4(2) + 6 = 4 −8+6 = 2 The minimum value is 2.
( ) ( ) ( ) ( ) 2
P − 1 = −2 − 1 − − 1 + 1 4 4 4 1 1 = −2 − − +1 16 4 8 1 2 =− + + =9 8 8 8 8 The maximum value is 9 . 8
PS3. P( x ) = x 2 + 2 x + 7 is a parabola that opens up. The x-value of the vertex is − b = − 2 = − 2 = −1 2a 2(1) 2 The graph decreases from the left until it reaches the vertex, and then it increases. P(x) increases on the interval [ −1, ∞).
PS4. P( x ) = −2 x 2 + 4 x + 5 is a parabola that opens downward. The x-value of the vertex is − b = − 4 = − 4 =1 2a 2( −2) −4 The graph increases from the left until it reaches the vertex, and then it decreases. P(x) decreases on the interval [1, ∞ ) .
PS5. x 4 − 5 x 2 + 4
PS6. P ( x) = 6 x 2 − x − 2
( x 2 − 4)( x 2 − 1) ( x + 2)( x − 2)( x + 1)( x − 1)
0 = 6 x2 − x − 2 0 = (3x − 2)(2 x + 1) 3x − 2 = 0 or 2 x + 1 = 0 x=2 x = −1 3
The x-intercepts are
2
( ) and ( − 12 , 0) . 2 ,0 3
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Section 3.2
183
Section 3.2 1.
Since an = 3 is positive and n = 4 is even, the graph of P goes up to its far left and up to its far right.
2.
Since an = −2 is negative and n = 3 is odd, the graph of P goes up to its far left and down to its far right.
3.
Since an = 5 is positive and n = 5 is odd, the graph of P goes down to its far left and up to its far right.
4.
Since an = −6 is negative and n = 4 is even, the graph of P goes down to its far left and down to its far right.
5.
P ( x ) = −4 x 2 − 3 x + 2 Since an = −4 is negative and n = 2 is even, the graph of P goes down to its far left and down to its far right.
6.
P ( x) = x 4 − 16 Since an = 1 is positive and n = 4 is even, the graph of P goes up to its far left and up to its far right.
7.
P( x ) = 1 x 3 + 5 x 2 − 1 2 2
Since an = 1 is positive and n = 3 is odd, the graph of P goes down to its far left and up to its far right. 2
8.
P( x ) = − 1 x 4 − 3 x 3 + 1 x − 3 4 4 2 2
Since an = − 1 is negative and n = 4 is even, the graph of P goes down to its far left and down to its far right. 2
9.
Up to the far left and down to the far right ⇒ a < 0.
10.
Down to the far left and down to the far right ⇒ a < 0.
11.
On a TI-83 calculator, the CALC feature is located above the TRACE key. There is a relative maximum of y ≈ 5.0 at x ≈ −2.1. There is a relative minimum of y ≈ −16.9 at x ≈ 1.4.
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184
Chapter 3: Polynomial and Rational Functions
12.
The step-by-step technique for a TI-83 calculator is illustrated in the solution to Exercise 15. The CALC feature is located above the TRACE key. There is a relative maximum of y ≈ 5.0 at x ≈ −3.1. There is a relative minimum of y ≈ −16.9 at x ≈ 0.4. 13.
The step-by-step technique for a TI-83 calculator is illustrated in the solution to Exercise 15. The CALC feature is located above the TRACE key. There is a relative maximum of y ≈ 31.0 at x ≈ −2.0. There is a relative minimum of y ≈ −77.0 at x ≈ 4. 14.
The step-by-step technique for a TI-83 calculator is illustrated in the solution to Exercise 15. The CALC feature is located above the TRACE key. There is a relative maximum of y ≈ 8.0 at x ≈ 1.0 There is a relative minimum of y ≈ −19.0 at x ≈ −2.0.
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Section 3.2
185
15.
The step-by-step technique for a TI-83 calculator is illustrated in the solution to Exercise 15. The CALC feature is located above the TRACE key. There is a relative maximum of y ≈ 2.0 at x ≈ 1.0. There is a relative minimum of y ≈ −14.0 at x ≈ −1.0, and another relative minimum of y ≈ −14.0 at x ≈ 3.0. 16.
The step-by-step technique for a TI-83 calculator is illustrated in the solution to Exercise 15. The CALC feature is located above the TRACE key. There is a relative maximum of y ≈ 9.0 at x ≈ 0.0. There is a relative minimum of y ≈ −16.0 at x ≈ −2.2, and another relative minimum of y ≈ −16.0 at x ≈ 2.2. 17.
19.
P( x ) = x 3 − 2 x 2 − 15 x
18.
0 = x ( x 2 − 2 x − 15)
0 = x ( x 2 − 6 x + 8)
0 = x ( x − 5)( x + 3) The zeros are 0, 5, –3.
0 = x ( x − 2)( x − 4) The zeros are 0, 2, 4.
P( x ) = x 4 − 13x 2 + 36
20.
0 = ( x 2 − 9)( x 2 − 4)
P( x ) = 4 x 4 − 37 x 2 + 9 0 = (4 x 2 − 1)( x 2 − 9) 0 = (2 x + 1)(2 x − 1)( x + 3)( x − 3)
0 = ( x + 3)( x − 3)( x + 2)( x − 2) The zeros are –3, 3, –2, 2.
21.
P( x ) = x 3 − 6 x 2 + 8 x
P( x ) = x5 − 5 x 3 + 4 x
The zeros are − 1 , 1 , − 3, 3. 2 2 22.
P( x ) = x 5 − 25 x 3 + 144 x
0 = x ( x 4 − 5 x 2 + 4)
0 = x ( x 4 − 25 x 2 + 144
0 = x ( x 2 − 4)( x 2 − 1)
0 = x ( x 2 − 16)( x 2 − 9)
0 = x ( x + 2)( x − 2)( x + 1)( x − 1) The zeros are 0, –2, 2, –1, 1.
0 = x ( x + 4)( x − 4)( x + 3)( x − 3) The zeros are 0, –4, 4, –3, 3.
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186
23.
Chapter 3: Polynomial and Rational Functions
P( x ) = 2 x 3 + 3x 2 − 23x − 42 3 2 3 −23 −42 6 27 12 2 9 4 −30 4
2
−23 44 21
3 8 11
2
P( x ) = 4 x 3 − x 2 − 6 x + 1 0 4 −1 −6 0 0 4 –1 −6
24.
−42 84 42
1
4
P(3) is negative; P(4) is positive. Therefore P(x) has a zero between 3 and 4. 25.
3
7 −6 1
3
3 −2 1
26.
7 −2 5
28.
4
7
−11
7
−15
4
6 13
19.5 8.5
12.75 19.75
29.625 14.625
P( x ) = x 4 − x 2 − x − 4 1.7 1 0 −1 1.7 2.89 1 1.7 1.89 1.8
1 1
0 1.8 1.8
−1 3.24 2.24
−4 3.7621 −0.2379
−1 4.032 3.032
−4 5.4576 1.4526
P( x ) = − x 4 + x 3 + 5x − 1 0.1 −1 1 0 −0.1 0.09 0.9 0.09 −1 0.2
−1 −1
1 −0.2 0.8
0 0.16 0.16
−1 0.5009 −0.4991
5 0.032 5.032
−1 1.0064 0.0064
P(0.1) is negative; P(0.2) is positive. Therefore P(x) has a zero between 0.1 and 0.2.
−16
−20
64
5
17.5 1.5
5.25 −14.75
−51.625 12.375
P( x ) = x 3 − x − 8 2.1 1 0 2.1 1 2.1 2.2
1 1
0 2.2 2.2
−1 4.41 3.41 −1 4.84 3.84
−8 7.161 −0.839 −8 8.448 0.448
P(2.1) is negative; P(2.2) is positive. Therefore P(x) has a zero between 2.1 and 2.2. 32.
5 0.009 5.009
5
P(3) is negative; P(3.5) is positive. Therefore P(x) has a zero between 3 and 3.5.
P(1.7) is negative; P(1.8) is positive. Therefore P(x) has a zero between 1.7 and 1.8. 31.
21 –38 –17
P( x ) = 5 x 3 − 16 x 2 − 20 x + 64 3 5 64 −16 −20 15 −3 −69 5 −1 −23 −5
3 12 =3.5
30.
−1 3.213 2.213
−2 –17 –19
P(1) is positive; P(2) is negative. Therefore P(x) has a zero between 1 and 2.
P(1) is negative; P(1.5) is positive. Therefore P(x) has a zero between 1 and 1.5. 29.
−21 4 –17
2 2
P( x ) = 4 x 4 + 7 x 3 − 11x 2 + 7 x − 15 1 4 7 −11 7 −15 4 11 0 7 4 11 0 7 −8
1 12 =1.5
1 –3 –2
P( x ) = 2 x 3 − 21x 2 − 2 x + 25 1 2 25 −21 −2 2 −19 −21 2 −19 −21 4 2
P(−3) is negative; P(−2) is positive. Therefore P(x) has a zero between −3 and −2. 27.
−6 3 –3
P(0) is positive; P(1) is negative. Therefore P(x) has a zero between 0 and 1.
P( x ) = 3x 3 + 7 x 2 + 3x + 7 3 7 3 7 −3 6 −27 −9 3 −2 9 −20 −2
−1 4 3
4
1 0 1
P( x ) = − x 3 − 2 x 2 + x − 3 −2.8 −1 −2 1 2.8 −2.24 −1 0.8 −1.24 −2.7
−1 −1
−2 2.7 0.7
1 −1.89 −0.89
−3 3.472 0.472 −3 2.403 −0.597
P(−2.8) is positive; P(−2.7) is negative. Therefore P(x) has a zero between −2.8 and −2.7.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.2
33.
187
P( x ) = ( x − 1)( x + 1)( x − 3) 0 = ( x − 1)( x + 1)( x − 3) x − 1 = 0 or x + 1 = 0 or x =1 x = −1
34.
0 = ( x + 2)( x − 6)2 or x − 6 = 0 x+2=0 x=6 x = −2
x−3=0 x=3
The exponent on (x + 2) is an odd integer. Therefore the graph of P(x) will cross the x-axis at (–2, 0).
The exponents on (x + 1), (x – 1), and (x – 3) are odd integers. Therefore the graph of P(x) will cross the x-axis at (−1, 0), (1, 0), and (3, 0).
35.
P ( x) = −( x − 3)2 ( x − 7)5 2
37.
The exponent on (x – 6) is an even integer. Therefore the graph of P(x) will intersect but not cross the x-axis at (6, 0). 36.
5
P ( x) = ( x + 2)3 ( x − 6)10
0 = −( x − 3) ( x − 7) x − 3 = 0 or x − 7 = 0 x=3 x=7
0 = ( x + 2)3 ( x − 6)10 x+2=0 or x − 6 = 0 x = −2 x=6
The exponent on (x – 7) is an odd integer. Therefore the graph of P(x) will cross the x-axis at (7, 0).
The exponent on (x + 2) is an odd integer. Therefore the graph of P(x) will cross the x-axis at (–2, 0).
The exponent on (x – 3) is an even integer. Therefore the graph of P(x) will intersect but not cross the x-axis at (3, 0).
The exponent on (x – 6) is an even integer. Therefore the graph of P(x) will intersect but not cross the x-axis at (6, 0).
P( x ) = (2 x − 3) 4 ( x − 1)15
38.
0 = (2 x − 3) 4 ( x − 1)15 2 x − 3 = 0 or x − 1 = 0
x=3 2
P( x ) = (5 x + 10)6 ( x − 2.7)5 0 = (5 x + 10)6 ( x − 2.7)5 or x − 2.7 = 0 5 x + 10 = 0
5 x = −10 x = −2
x =1
The exponent on ( x − 1) is an odd integer. Therefore the graph of P(x) will cross the x-axis at (1, 0).
The exponent on (5x + 10) is an even integer. Therefore the graph of P(x) will intersect but not cross the x-axis at (–2, 0).
( )
P( x ) = x 3 − 6 x 2 + 9 x
40.
0 = x ( x 2 − 6 x + 9)
x = 2.7
The exponent on (x – 2.7) is an odd integer. Therefore the graph of P(x) will cross the x-axis at (2.7, 0).
The exponent on (2x – 3) is an even integer. Therefore the graph of P(x) will intersect but not cross the x-axis at 3 , 0 . 2 39.
P ( x) = ( x + 2)( x − 6)2
P( x ) = x 4 + 3x 3 + 4 x 2 0 = x 2 ( x 2 + 3x + 4)
0 = x ( x − 3)2 x = 0 or x − 3 = 0 x=3
x2 = 0 x=0
The exponent on x is an odd integer. Therefore the graph of P(x) will cross the x-axis at (0, 0) The exponent on (x – 3) is an even integer. Therefore the graph of P(x) will intersect but not cross the x-axis at (3, 0).
or 0 = x 2 + 3x + 4 −3 ± 32 − 4(1)(4) −3 ± i 7 = 2(2) 4 Not a real number
x=
The exponent on x is an even integer. Therefore the graph of P(x) will intersect but not cross the x-axis at (0, 0).
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188
41.
Chapter 3: Polynomial and Rational Functions
Let P ( x ) = 0. 3
43.
42.
2
Let P ( x ) = 0.
x − x − 2x = 0
x 3 + 2 x 2 − 3x = 0
x ( x 2 − x − 2) = 0
x ( x 2 + 2 x − 3) = 0
x ( x − 2)( x + 1) = 0 x = 0, x = 2, x = –1 The graph crosses the x-axis at (0, 0), (2, 0), and (−1, 0).
x ( x + 3)( x − 1) = 0 x = 0, x = –3, x = 1 The graph crosses the x-axis at (0, 0), (−3, 0), and (1, 0).
Let x = 0. P(0) = 03 − 02 − 2(0) = 0 The y-intercept is (0, 0).
Let x = 0. P(0) = 03 + 2(0)2 − 3(0) = 0 The y-intercept is (0, 0).
x 3 has a positive coefficient and an odd exponent. Therefore, the graph goes down to the far left and up to the far right.
x 3 has a positive coefficient and an odd exponent. Therefore, the graph goes down to the far left and up to the far right.
Let P ( x ) = 0. 3
44.
Let P ( x ) = 0.
− x − 2 x + 5x − 6 = 0
− x 3 − 3x 2 + x + 3 = 0
x3 + 2 x 2 − 5x + 6 = 0
x 3 + 3x 2 − x − 3 = 0
2
2
1 1
2 2 4
−5 8 3
−6 6 0
1
1 1
3 1 4
−1 4 3
−3 3 0
( x − 2)( x 2 + 4 x + 3) = 0 ( x − 2)( x + 3)( x + 1) = 0 x = 2, x = –3, x = –1 The graph crosses the x-axis at (2, 0), (−3, 0), and (−1, 0).
( x − 1)( x 2 + 4 x + 3) = 0 ( x − 1)( x + 3)( x + 1) = 0 x = 1, x = –3, x = –1 The graph crosses the x-axis at (1, 0), (−3, 0), and (−1, 0).
Let x = 0. P(0) = −(0)3 − 2(0)2 + 5(0) + 6 = 6 The y-intercept is (0, 6).
Let x = 0. P(0) = −(0)3 − 3(0) 2 + (0) + 3 = 3 The y-intercept is (0, 3).
− x 3 has a negative coefficient and an odd exponent. Therefore, the graph goes up to the far left and down to the far right.
− x 3 has a negative coefficient and an odd exponent. Therefore, the graph goes up to the far left and down to the far right.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.2
45.
189
Let P ( x ) = 0. 4
3
46.
x − 4x + 2x + 4x − 3 = 0 1
1 1
3
1 1
Let P ( x ) = 0. x 4 − 6 x3 + 8x 2 = 0
2
−4 1 −3
2 −3 -1
4 −1 3
−3 3 0
−1 0 −1
3 −3 0
x 2 ( x 2 − 6 x + 8) = 0
−3 3 0
x 2 ( x − 2)( x − 4) = 0 x = 0, x = 2, x = 4 The graph intersects the x-axis but does not cross it at (0, 0). The graph crosses the x-axis at (0, 2) and (0, 4).
Let x = 0. P(0) = (0)4 − 6(0)3 + 8(0)2 = 0 The y-intercept is (0, 0). x 4 has a positive coefficient and an even exponent. Therefore, the graph goes up to the far left and up to the far right.
( x − 1)( x − 3)( x 2 − 1) = 0 ( x − 1)( x − 3)( x + 1)( x − 1) = 0 ( x − 1)2 ( x − 3)( x + 1) = 0 x = 1, x = 3, x = –1 The graph intersects the x-axis but does not cross it at (1, 0). The graph crosses the x-axis at (3, 0) and (−1, 0). Let x = 0. P(0) = (0)4 − 4(0)3 + 2(0)2 + 4(0) − 3 = −3 The y-intercept is (0, −3). x 4 has a positive coefficient and an even exponent. Therefore, the graph goes up to the far left and up to the far right.
47.
Let P ( x ) = 0. 3
48.
2
x + 6 x + 5 x − 12 = 0 ( x − 1)( x + 3)( x + 4) = 0 x = 1, x = –3, x = –4 The graph crosses the x-axis at (1, 0), (–3, 0), and (−4, 0). Let x = 0. P(0) = (0)3 + 6(0)2 + 5(0) − 12 = −12 The y-intercept is (0, –12). x 3 has a positive coefficient and an odd exponent. Therefore, the graph goes down to the far left and up to the far right.
Let P ( x ) = 0. − x3 + 4 x 2 + x − 4 = 0 − x 2 ( x − 4) + ( x − 4) = 0 ( − x 2 + 1)( x − 4) = 0 ( − x + 1)( x + 1)( x − 4) = 0 x = 1, x = –1, x = 4 The graph crosses the x-axis at (1, 0), (−1, 0), and (4, 0).
Let x = 0. P(0) = −(0)3 + 4(0) 2 + 0 − 4 = −4 The y-intercept is (0, –4). − x 3 has a negative coefficient and an odd exponent. Therefore, the graph goes up to the far left and down to the far right.
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190
49.
Chapter 3: Polynomial and Rational Functions
Let P ( x ) = 0. 1 −1
0
50. 7
−1 −1 −1 −1
6
Let P ( x ) = 0.
−6
x3 − 6 x 2 + 9 x = 0
6
x ( x − 3) 2 = 0 x = 0, x = 3 The graph crosses the x-axis at (0, 0), and (3, 0).
0
− x3 + 7 x − 6 = 0
Let x = 0. P(0) = (0)3 − 6(0)2 + 9(0) = 0 The y-intercept is (0, 0).
( x − 1)( − x 2 − x + 6) = 0 ( x − 1)( − x − 3)( x − 2) = 0 x = 1, x = –3, x = 2 The graph crosses the x-axis at (1, 0), (−3, 0), and (2, 0).
x 3 has a positive coefficient and an odd exponent. Therefore, the graph goes down to the far left and up to the far right.
Let x = 0. P(0) = −(0)3 + 7(0) − 6 = −6 The y-intercept is (0, –6). − x 3 has a negative coefficient and an odd exponent. Therefore, the graph goes up to the far left and down to the far right.
51.
Let P ( x ) = 0. 3
52.
2
Let P ( x ) = 0.
−x + 4x − 4x = 0
− x 4 + 2 x 3 + 3x 2 − 4 x − 4 = 0
− x ( x − 2)2 = 0 x = 0, x = 2 The graph crosses the x-axis at (0, 0), and (2, 0).
−( x − 2)2 ( x + 1)2 = 0 x = 2, x = –1 The graph intersects the x-axis but does not cross it at (2, 0) and (–1, 0). Let x = 0.
Let x = 0. P(0) = −(0)3 + 4(0)2 − 4(0) = 0 The y-intercept is (0, 0). − x 3 has a negative coefficient and an odd exponent. Therefore, the graph goes up to the far left and down to the far right.
P(0) = −(0)4 + 2(0)3 + 3(0)2 − 4(0) − 4 = −4 The y-intercept is (0, –4). − x 4 has a negative coefficient and an even exponent. Therefore, the graph goes down to the far left and down to the far right.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.2
53.
191
Let P ( x ) = 0. 4
54. 3
2
− x + 3x + x − 3x = 0
− x [ x 3 − 3x 2 − x + 3] = 0 − x ⎡⎣ x 2 ( x − 3) − 1( x − 3) ⎤⎦ = 0
− x ⎣⎡( x − 3)( x 2 − 1) ⎦⎤ = 0 − x ( x − 3)( x + 1)( x − 1) = 0 x = 0, x = 3, x = –1, x = 1 The graph crosses the x-axis at (0, 0), (3, 0), (–1, 0) and (1, 0). Let x = 0.
( )
P(0) = −(0)4 + 3(0)3 + 02 − 3(0) = 0 The y-intercept is (0, 0). − x 4 has a negative coefficient and an even exponent. Therefore, the graph goes down to the far left and down to the far right.
55.
Let P ( x ) = 0. 5
4
3
56. 2
Let P ( x ) = 0. 2 x 5 − 3x 4 − 4 x 3 + 3x 2 + 2 x = 0
x − x − 5x + x + 8 x + 4 = 0 3
Let P ( x ) = 0. 1 x4 + x3 − 2 x2 − x + 3 = 0 2 2 1 ( x − 1)2 ( x + 1)( x + 3) = 0 2 x = 1, x = –1, x = –3 The graph intersects the x-axis but does not cross it at (1, 0). The graph crosses the x-axis at (–1, 0) and (−3, 0). Let x = 0. P(0) = 1 (0)4 + (0)3 − 2(0)2 − 0 + 3 = 3 2 2 2 The y-intercept is 0, 3 . 2 4 1 x has a positive coefficient and an even exponent. 2 Therefore, the graph goes up to the far left and up to the far right.
2
x (2 x 4 − 3x 3 − 4 x 2 + 3x + 2) = 0
( x + 1) ( x − 2) = 0 x = –1, x = 2 The graph intersects the x-axis but does not cross it at (2, 0). The graph crosses the x-axis at (–1, 0). Let x = 0.
−1 2 2 4
−3 −4 −3 2 −2 5 −1 −2 −5 1 2 0 3
−5 1 2 4 −2 −2 −1 −1 0
2 2 2
2
x (2 x − 3x − 4 x + 3x + 2) = 0
P(0) = 05 − 04 − 5(0)3 + 02 + 8(0) + 4 = 4 The y-intercept is (0, 4).
x ( x + 1)( x − 2)(2 x 2 − x − 1) = 0 x ( x + 1)( x − 2)(2 x + 1)( x − 1) = 0
x 5 has a positive coefficient and an odd exponent. Therefore, the graph goes down to the far left and up to the far right.
x = 0, x = –1, x = 2, x = − 1 , x = 1 2
(
)
The graph crosses the x-axis at (–1, 0), − 1 , 0 (0, 0), 2
(1, 0), and (2, 0). Let x = 0. P(0) = 2(0)5 − 3(0) 4 − 4(0)3 + 3(0)2 + 2(0) = 0 The y-intercept is (0, 0). 2x 5 has a positive coefficient and an odd exponent. Therefore, the graph goes down to the far left and up to the far right.
57.
Shift the graph of P vertically upward 2 units.
58.
Shift the graph of P vertically downward 3 units.
59.
Shift the graph of P horizontally 1 unit to the right.
60.
Shift the graph of P horizontally 3 units to the left.
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192
Chapter 3: Polynomial and Rational Functions
61.
Shift the graph of P horizontally 2 units to the right and reflect this graph about the x-axis. Then shift the resulting graph vertically upward 3 units.
62.
Shift the graph of P horizontally 4 units to the left and vertically downward 5 units.
63.
Since both P( x ) and Q ( x ) have same value for an , they have the same far-left and far-right behavior.
64.
Since both P( x ) and Q ( x ) have same value for an , they have the same far-left and far-right behavior.
65.
a.
Volume = length × width × height V ( x ) = (15 − 2 x )(10 − 2 x ) x = [15(10 − 2 x ) − 2 x (10 − 2 x )] x = [150 − 30 x − 20 x + 4 x 2 ]x = [4 x 2 − 50 x + 150] x = 4 x 3 − 50 x 2 + 150 x
b.
x = 1.96 inches (to the nearest 0.01 inch) maximizes the volume of the box. 66.
V = ⎛⎜ 42 − 3x ⎞⎟ (18 − 2 x ) x = (21 − 1.5 x )(18 − 2 x ) x ⎝ 2 ⎠ Use a graphing utility to graph y = (21 − 1.5 x)(18 − 2 x) x Then use the maximum feature of the graphing utility to determine the x- and y-coordinates of the relative maximum.
67.
Use a graphing utility to graph y = (22 − 4 x)(16 − 2 x) x. Then use the maximum feature of the graphing utility to determine the x-and y-coordinates of the relative maximum.
Xmin = −1, Xmax = 15, Xscl = 2 Ymin = −200, Ymax = 1200, Yscl = 200
Xmin = −2, Xmax = 12, Xscl = 2 Ymin = −200, Ymax = 600, Yscl = 100
The value of x ≈ 3.571 inches will produce a box of maximum volume V ≈ 606.6 cubic inches.
The value of x ≈ 2.137 inches will produce a box of maximum volume V ≈ 337.1 cubic inches.
68.
a. b.
$264,000 5657 games
69.
$464,000
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Section 3.2
193
70.
a.
134 gazelles
b.
1013 gazelles
a.
20.69 milligrams
b.
1.968 hours × 60 minutes per hour ≈ 118 minutes after taking the medication
71.
91
72.
0
156
–11
73.
a.
D ( x) = (−0.0025)(4 x3 − 3 ⋅ 8 x 2 ) D(3) = (−0.0025)[4(3)3 − 3 ⋅ 8(3) 2 ] = (−0.0025)[4(27) − 3 ⋅ 8(9)] = (−0.0025)(108 − 216) D(3) = (−0.0025)( −108) = 0.27 foot = 0.27 foot × 12 inches per foot = 3.24 inches
b.
The beam achieves its maximum deflection, 4 feet from the end. The maximum deflection is 0.32 foot × 12 inches per foot = 3.84 inches. c.
The formula is valid on the interval (0, 4]. Therefore, D(5) cannot be determined by using the formula. However, 5 feet from one end of an 8-foot beam is 3 feet from the other end. Thus, the deflection at x = 5 is the same as the deflection where x = 3, which is 3.24 inches.
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194
74.
Chapter 3: Polynomial and Rational Functions
w2 + d 2 = 222 d 2 = 222 − w2 = 484 − w2 S = 1.15wd 2 = 1.15w(484 − w2 ) = 556.6w − 1.15w3
w ≈ 12.70 inches d 2 = 484 − w2 ≈ 484 − (12.70)2 ≈ 484 − 161.29 ≈ 322.71 d ≈ 322.71 ≈ 17.96 inches
....................................................... 75.
76.
77.
P(x – 3) shifts the graph horizontally three points to the right. (2 + 3, 0) = (5, 0)
80.
False. Let P(x) = x2 – 2x – 8, a = −3, and b = −5. Then P(a) = 7 and P(b) = 7. However, x = 4 is a zero of P and –3 < 4 < 5.
Xmin = −1, Xmax = 2, Xscl = 1 Ymin = −1, Ymax = 4, Yscl = 1
Xmin = −1, Xmax = 4, Xscl = 1 Ymin = −2, Ymax = 2, Yscl = 1
There is a real zero between 1 and 2.
There is a real zero between 3 and 4. 78.
Connecting Concepts
P(x + 1) –2 shifts the graph horizontally one points to the left and vertically two points down. (3 – 1, 5 – 2) = (2, 3)
79.
Shift the graph of y = x3 horizontally two units to the right and vertically upward 1 unit.
.......................................................
Prepare for Section 3.3 PS2.
PS1. P( x ) = 6 x 2 − 25 x + 14
0 = 6 x 2 − 25 x + 14 0 = (3x − 2)(2 x − 7) 3x − 2 = 0 x=2 3 PS3.
3
3 3
−21 27 6
2 2
or 2 x − 7 = 0 x=7 2 0 9 9
−2
3 −4 −1
4 2 6
−7 −12 −19
2 x 2 − x + 6 − 19 x+2
−3 18 15
−5 45 40
PS4. 1, 2, 3, 4, 6, 12
3x3 + 9 x 2 + 6 x + 15 + 40
x −3
PS5. ±1, ± 3, ± 9, ± 27
PS6.
P ( x ) = 4 x3 − 3 x 2 − 2 x + 5 P (− x) = 4(− x)3 − 3( − x)2 − 2(− x) + 5 P (− x) = −4 x3 − 3x 2 + 2 x + 5
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Section 3.3
195
Section 3.3 1.
P( x ) = ( x − 3)2 ( x + 5) The zeros are: –5 (multiplicity 1), 3 (multiplicity 2).
2.
P( x ) = ( x + 4)3 ( x − 1) 2 The zeros are: –4 (multiplicity 3), 1 (multiplicity 2).
3.
P( x ) = x 2 (3x + 5)2 The zeros are:
4.
P( x ) = x 3 (2 x + 1)(3x − 12)2 = x 3 (2 x + 1)[3( x − 4)]2
− 53 (multiplicity 2), 0 (multiplicity 2).
= 9 x 3 (2 x + 1)( x − 4)2 The zeros are:
− 12 (multiplicity 1), 0 (multiplicity 3), 4 (multiplicity 2). 5.
P ( x ) = ( x 2 − 4)( x + 3)2
6.
= ( x + 2)( x − 2)( x + 3) 2 The zeros are: −3 (multiplicity 2), –2 (multiplicity 1), 2 (multiplicity 1).
P( x ) = ( x + 4)3 ( x 2 − 9)2 = ( x + 4)3 [( x + 3)( x − 3)]2 = ( x + 4)3 ( x + 3)2 ( x − 3) 2 The zeros are: –4 (multiplicity 3), −3 (multiplicity 2), 3 (multiplicity 2).
7.
P( x ) = x 3 + 3x 2 − 6 x − 8 p = ± factors of 8 = ±1, ± 2, ± 4, ± 8 q = ± factors of 1 = ±1 p = possible rational zeros = ±1, ± 2, ± 4, ± 8 q
8.
P( x ) = x 3 − 19 x − 30 p = ± factors of 30 = ±1, ± 2, ± 3, ± 5, ± 6, ± 10, ± 15, ± 30 q = ± factors of 1 = ± 1 p = possible rational zeros = ± 1, ± 2, ± 3, ± 5, ± 6, ± 10, ± 15, ± 30 q
9.
P( x ) = 2 x 3 + x 2 − 25 x + 12 p = ± factors of 12 = ± 1, ± 2, ± 3, ± 4, ± 6, ± 12 q = ± factors of 2 = ± 1, ± 2 p = possible rational zeros = ± 1, ± 2, ± 3, ± 4, ± 6, ± 12, ± 1 , ± 3 2 2 q
10.
P( x ) = 3x 3 + 11x 2 − 6 x − 8 p = ± factors of 8 = ± 1, ± 2, ± 4, ± 8 q = ± factors of 3 = ± 1, ± 3, p = possible rational zeros = ± 1, ± 2, ± 4, ± 8, ± 1 , ± 2 , ± 4 , ± 8 3 3 3 3 q
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196
Chapter 3: Polynomial and Rational Functions
11.
P( x ) = 6 x 4 + 23x 3 + 19 x 2 − 8 x − 4 p = ± factors of 4 = ± 1, ± 2, ± 4 q = ± factors of 6 = ± 1, ± 2, ± 3, ± 6 p = possible rational zeros = ± 1, ± 2, ± 4, ± 1 , ± 1 , ± 1 , ± 2 , ± 4 2 3 6 3 3 q
12.
P( x ) = 2 x 3 + 9 x 2 − 2 x − 9 p = ± factors of 9 = ± 1, ± 3, ± 9 q = ± factors of 2 = ± 1, ± 2 p = possible rational zeros = ± 1, ± 3, ± 9, ± 1 , ± 3 , ± 9 2 2 2 q
13.
P( x ) = 4 x 4 − 12 x 3 − 3x 2 + 12 x − 7 p = ± factors of 7 = ± 1, ± 7 q = ± factors of 4 = ± 1, ± 2, ± 4 p = possible rational zeros = ± 1, ± 7, ± 1 , ± 7 , ± 1 , ± 7 2 2 4 4 q
14.
P( x ) = x 5 − x 4 − 7 x 3 + 7 x 2 − 12 x − 12 p = ± factors of 12 = ± 1, ± 2, ± 3, ± 4, ± 6, ± 12 q = ± factors of 1 = ± 1 p = possible rational zeros = ± 1, ± 2, ± 3, ± 4, ± 6, ± 12 q
15.
P( x ) = x 5 − 32 p = ± factors of 32 = ± 1, ± 2, ± 4, ± 8, ± 16, ± 32 q = ± factors of 1 = ± 1 p = possible rational zeros = ± 1, ± 2, ± 4, ± 8, ± 16, ± 32 q
16.
P( x ) = x 4 − 1 p = ± factors of 1 = ± 1 q = ± factors of 1 = ± 1 p = possible rational zeros = ± 1 q
Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.3
17.
197
3 −6 −6 1 4 1 4 −2 Don’t finish dividing. 1 is not an upper bound. 1
1
18.
1
−1
3 −6 −1 1 2 Don’t finish dividing. –1 is not a lower bound. 1
−2
1
−3
1
3 −6 4 −4 1 −1 −2 Don’t finish dividing. –4 is not a lower bound. −5
1
0 –19 −28 9 −3 1 −3 −10 Don’t finish dividing. –3 is not a lower bound.
−3
1
−4
1
−5
1
1
3
0 –19 −28 16 −4 1 −4 Don’t finish dividing. –4 is not a lower bound.
3 −6 −6 0 −3 1 0 −6 Don’t finish dividing. –3 is not a lower bound. 1
1
0 –19 −28 5 25 30 1 5 6 2 The smallest integer that is an upper bound is 5. 5
−6
3 −6 −6 −2 1 1 Don’t finish dividing. –2 is not a lower bound.
−4
1
0 –19 −28 4 16 1 4 –3 Don’t finish dividing. 3 is not an upper bound. 4
3 −6 −6 2 10 8 1 5 4 2 The smallest integer that is an upper bound is 2. 2
0 –19 −28 3 9 1 3 –10 Don’t finish dividing. 3 is not an upper bound. 3
0 –19 −28 25 −5 −30 1 6 −5 −40 The largest integer that is a lower bound is –5.
−6
3 −6 −6 10 −5 −20 1 4 −2 −26 The largest integer that is a lower bound is –5. 19.
1
1 10 −25 6 21 2 7 −4 Don’t finish dividing. 3 is not an upper bound. 3
2
1 10 −25 8 36 44 2 9 11 54 The smallest integer that is an upper bound is 4. 4
2
−3
2
1 10 −25 15 −6 2 −5 −10 Don’t finish dividing. –3 is not a lower bound. −4
2
1 10 −25 28 −8 −12 2 3 −7 −2 The largest integer that is a lower bound is –4.
20.
11 3 3 14 1 is not an upper bound.
−6 14 8
−9 8 −1
11 −6 −9 6 34 56 3 17 28 47 The smallest integer that is an upper bound is 2. 2
3
−4
3
11 −6 −9 4 −12 3 −1 −2 Don’t finish dividing. –4 is not a lower bound. −5
3
11 −6 −9 20 −15 −70 3 14 −4 −79 The largest integer that is a lower bound is –5.
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198
21.
Chapter 3: Polynomial and Rational Functions
23 19 −8 −4 6 29 48 40 6 29 48 40 36 The smallest integer that is an upper bound is 1. 1
6
−3
6
−4
6
22.
23 19 −8 −4 −18 6 5 Don’t finish dividing. –3 is not a lower bound.
3
–4
12 3 –12 7 –12 0 –4 0 3 Don’t finish dividing. 3 is not an upper bound. 4
–4
–4
−2
–4
1
1
12 4 –4 16 –1 is not a lower bound.
3 –16 –13
–12 13 1
0 0 1 1 1 1 1 1 is not an upper bound.
0 1 1
–9 2 9 10 –5 −15 –2 1 3 −24 The largest integer that is a lower bound is –5. 24.
−32 1 −31
−1
0 0 0 0 1 1 −1 −1 1 1 1 −1 −1 The largest integer that is a lower bound is –1. 27.
1
7 −1 −7 −12 −12 6 −2 1 −3 −1 Don’t finish dividing. –2 is not a lower bound.
−2
1
−3
1
7 −1 −7 −12 −12 12 24 −3 −15 −36 1 5 12 −4 −8 −48 The largest integer that is a lower bound is –3. 26.
1
1
1
7 −1 −7 −12 −12 4 12 20 108 384 1 3 5 27 96 372 The smallest integer that is an upper bound is 4.
0 0 0 0 −32 2 4 8 16 32 1 2 4 8 16 0 The smallest integer that is an upper bound is 2. 2
–2
7 −1 −7 −12 −12 3 6 1 2 −1 Don’t finish dividing. 3 is not an upper bound. 3
4
7 –1 6
0 1 1
–2
−5
12 3 –12 7 8 –40 74 –124 –4 20 –37 62 –117 The largest integer that is a lower bound is –2. 25.
–2
–9 2 9 8 –2 –1 Don’t finish dividing. –4 is not a lower bound.
12 3 –12 7 –16 –16 –52 –256 –4 –4 –13 –64 –249 The smallest integer that is an upper bound is 4. −1
–9 2 9 –2 –11 –9 –2 –11 –9 0 The smallest integer that is an upper bound is 1. −4
23 19 −8 −4 4 400 −24 −92 6 23 396 −1 −100 The largest integer that is a lower bound is –4. 23.
1
0 0 0 −1 1 1 1 1 1 1 1 1 0 The smallest integer that is an upper bound is 1. 1
−1
1
0 0 −1 1 −1 1 1 1 −1 0 The largest integer that is a lower bound is –1. 1
0 −1 −1
−32 −1 −33
P( x ) = x 3 + 3x 2 − 6 x − 8 has 1 change in sign ⇒ one positive zero. P( − x ) = − x 3 + 3x 2 + 6 x − 8 has 2 changes in sign ⇒ two or no negative zeros.
28.
P( x ) = x 3 − 19 x − 30 has 1 change in sign ⇒ one positive zero. P( − x ) = − x 3 + 19 x − 30 has 2 changes in sign ⇒ two or no negative zeros.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.3
29.
199
P( x ) = 2 x 3 + x 2 − 25 x + 12 has 2 changes in sign ⇒ two or no positive zeros. P( − x ) = −2 x 3 + x 2 + 25 x + 12 has 1 change in sign ⇒ one negative zero.
30.
P( x ) = 3x 3 + 11x 2 − 6 x − 8 has 1 change in sign ⇒ one positive zero. P( − x ) = −3x 3 + 11x 2 + 6 x − 8 has 2 changes in sign ⇒ two or no negative zeros.
31.
P( x ) = 6 x 4 + 23x 3 + 19 x 2 − 8 x − 4 has 1 change in sign ⇒ one positive zero. P( − x ) = 6 x 4 − 23x 3 + 19 x 2 + 8 x − 4 has 3 changes in sign ⇒ three or one negative zero.
32.
P( x ) = 2 x 3 + 9 x 2 − 2 x − 9 has 1 change in sign ⇒ one positive zero. P( − x ) = −2 x 3 + 9 x 2 + 2 x − 9 has 2 changes in sign ⇒ two or no negative zeros.
33.
P( x ) = 4 x 4 − 12 x 3 − 3x 2 + 12 x − 7 has 3 changes in sign ⇒ three or one positive zeros. P( − x ) = 4 x 4 + 12 x 3 − 3x 2 − 12 x − 7 has 1 change in sign ⇒ one negative zero.
34.
P( x ) = x 5 − x 4 − 7 x 3 + 7 x 2 − 12 x − 12 has 3 changes in sign ⇒ three or one positive zeros. P( − x ) = − x 5 − x 4 + 7 x 3 + 7 x 2 + 12 x − 12 has 2 changes in sign ⇒ two or no negative zeros.
35.
P( x ) = x 5 − 32 has 1 change in sign ⇒ one positive zero. P( − x ) = − x 5 − 32 has no changes in sign ⇒ no negative zeros.
36.
P( x ) = x 4 − 1 has 1 change in sign ⇒ one positive zero. P( − x ) = x 4 − 1 has 1 change in sign ⇒ one negative zero.
37.
P( x ) = 10 x 6 − 9 x 5 − 14 x 4 − 8 x 3 − 18 x 2 + x + 6 has 2 changes in sign ⇒ two or no positive zeros. P( − x ) = 10 x 6 + 9 x 5 − 14 x 4 + 8 x 3 − 18 x 2 − x + 6 has 4 change in sign ⇒ four, two or no negative zeros.
38.
P( x ) = 2 x 6 − 5 x 5 − 26 x 4 + 76 x 3 − 60 x 2 − 255 x + 700 has 4 changes in sign ⇒ four, two or no positive zeros. P( − x ) = 2 x 6 + 5 x 5 − 26 x 4 − 76 x 3 − 60 x 2 + 255 x + 700 has 2 changes in sign ⇒ two or no negative zeros.
39.
P( x ) = 12 x 7 − 112 x 6 + 421x 5 − 840 x 4 + 1038 x 3 − 938 x 2 + 629 x − 210 has 7 changes in sign ⇒ seven, five, three or one positive zeros. P( − x ) = −12 x 7 − 112 x 6 − 421x 5 − 840 x 4 − 1038 x 3 − 938 x 2 − 629 x − 210 has no changes in sign ⇒ no negative zeros.
40.
P( x ) = x 7 + 2 x 5 + 3x 3 + x has no changes in sign ⇒ no positive zeros. P( − x ) = − x 7 − 2 x 5 − 3x 3 − x has no changes in sign ⇒ no negative zeros.
41.
P ( x ) = x 3 + 3x 2 − 6 x − 8 one positive and two or no negative real zeros p = ±1, ± 2, ± 4, ± 8 q 2 1 3 −6 −8 2 10 8 1 5 4 0 x 2 + 5 x + 4 = ( x + 4)( x + 1) = 0 ⇒ x = −4, − 1 The zeros of P(x) are 2, −4, and –1.
42.
P( x ) = x 3 − 19 x − 30 one positive and two or no negative real zeros p = ±1, ± 2, ± 3, ± 5, ± 6, ± 10, ± 15, ± 30 q 5 1 –30 0 −19 5 25 30 1 0 5 6 x 2 + 5 x + 6 = ( x + 3)( x + 2) = 0 ⇒ x = −3, − 2 The zeros of P(x) are –3, –2, and 5.
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200
43.
Chapter 3: Polynomial and Rational Functions
P( x ) = 2 x 3 + x 2 − 25 x + 12 has two or no positive and one negative real zero. 3 2 p = ±1, ± 2, ± 3, ± 4, ± 6, ± 12, ± 1 , ± 3 q 2 2 2
−25 21 −4
1 6 7
12 −12 0
2 x 2 + 7 x − 4 = (2 x − 1)( x + 4) = 0 ⇒ x = 12 , − 4 . The zeros of P(x) are 3, 1 , −4. 2
44.
P( x ) = 3x 3 + 11x 2 − 6 x − 8 one positive and two or no negative real zeros p = ±1, ± 2, ± 4, ± 8, ± 1 , ± 2 , ± 4 , ± 8 q 3 3 3 3 1 3 11 −6 −8 3 14 8 3 14 8 0
45.
P( x ) = 6 x 4 + 23x 3 + 19 x 2 − 8 x − 4 one positive and three or one negative real zero 6 23 19 −2 −8 −4 6 4 −12 −22 6 11 0 −3 −2 −2
6
2
3x + 14 x + 8 = (3x + 2)( x + 4) = 0
6
⇒ x = − 23 , − 4
−3 2 −1
11 −12 −1
−2 2 0
6 x 2 − x − 1 = (3x + 1)(2 x − 1) = 0 ⇒ x = − 13 ,
The zeros of P(x) are 1, − 2 , − 4. 3
The zeros of P(x) are −2 (multiplicity 2), − 1 , 1 . 3
46.
P( x ) = 2 x 3 + 9 x 2 − 2 x − 9 one positive and two or no negative real zeros p = ±1, ± 3, ± 9, ± 1 , ± 3 , ± 9 q 2 2 2 1 2 9 −2 −9 2 11 9 2 11 9 0
47.
P( x ) = 2 x 4 − 9 x 3 − 2 x 2 + 27 x − 12 4 2 −9 −2 8 −4 2 −1 −6 1 2
2 x 2 + 11x + 9 = (2 x + 9)( x + 1) = 0 ⇒ x = − 92 , − 1 2
2
−1
−6
3
2
1 0
0 −6
−3 0
The zeros of P(x) are 4, 1 , 2
P ( x ) = 3x 3 − x 2 − 6 x + 2 two or no positive and one negative real zero p = ±1, ± 2, ± 1 , ± 2 q 3 3 1 3
49.
3
−1
−6
2
3
1 0
0 −6
−2 0
3x 2 − 6 = 0 ⇒ 3( x 2 − 2) = 0 ⇒ x = ± 2 The zeros of P(x) are
1, 3
2, − 2.
27 −24 3
2
−12 12 0
2 x 2 − 6 = 0 ⇒ 2( x 2 − 3) = 0 ⇒ x = ± 3
The zeros of P(x) are 1, − 9 , −1.
48.
1 2
3, − 3.
P( x ) = x 3 − 8 x 2 + 8 x + 24 two or no positive and one negative real zero p = ±1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24 q 6 1 8 24 −8 6 −12 −24 1 0 −2 −4 x2 − 2 x − 4 = 0 x=
−( −2) ± ( −2)2 − 4(1)( −4) 2(1)
= 2 ± 20 = 2 ± 2 5 = 1 ± 5 2 2
The zeros of P(x) are 6, 1 + 5, 1 − 5.
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Section 3.3
50.
201
P( x ) = x 3 − 7 x 2 − 7 x + 69 two or no positive and one negative real zero p = ±1, ± 3, ± 23, ± 69 q 1 69 −3 −7 −7 30 −3 −69 1 23 0 −10
P( x ) = 2 x 4 − 19 x 3 + 51x 2 − 31x + 5 four, two or no positive and no negative real zeros p = ±1, ± 5, ± 1 , ± 5 q 2 2 5 2 51 5 −19 −31 10 30 −45 −5 2 6 0 −9 −1
51.
1 2
x 2 − 10 x + 23 = 0 x=
−( −10) ± ( −10) 2 − 4(1)(23) 2(1)
2
−9
6
−1
2
1 −8
−4 2
1 0
2 x 2 − 8 x + 2 = 2( x 2 − 4 x + 1) = 0
= 10 ± 8 = 10 ± 2 2 = 5 ± 2 2 2
x=
The zeros of P(x) are −3, 5 + 2, 5 − 2.
−( −4) ± ( −4)2 − 4(1)(1) 4 ± 12 4 ± 2 3 = = =2± 3 2(1) 2 2
The zeros of P(x) are 5, 52.
1, 2
2 + 3, 2 − 3.
P( x ) = 4 x 4 − 35 x 3 + 71x 2 − 4 x − 6 three or one positive and one negative real zero p = ±1, ± 2, ± 3, ± 6, ± 1 , ± 3 , ± 1 , ± 3 q 2 2 4 4 3 4 71 −35 −4 −6 12 6 6 − 69 4 2 2 0 − 23
− 14
4
− 23
2
2
4
−1 − 24
6 8
−2 0
4 x 2 − 24 x + 8 = 4( x 2 − 6 x + 2) = 0 x=
−( −6) ± ( −6)2 − 4(1)(2) 6 ± 28 6 ± 2 7 = = = 3± 7 2(1) 2 2
The zeros of P(x) are 3, − 14 , 3 + 7, 3 − 7. 53.
P( x ) = 3x 6 − 10 x 5 − 29 x 4 + 34 x 3 + 50 x 2 − 24 x − 24 three or one positive and three or one negative real zeros 1 3 34 50 − 10 − 29 − 24 3 48 −7 − 36 −2 3 48 24 −7 − 36 −2 −1
−2
2
3
−7 −3 − 10
− 36 10 − 26
−2 26 24
48 − 24 24
3
− 10
− 26
24
24
3
−6 − 16
32 6
− 12 12
−24 9
3
2
− 24 24 0
24 − 24 0
− 23
3
−16
6
12
3
−2 −18
12 18
− 12 0
2
3x − 18 x + 18 = 3( x − 6 x + 6) = 0 ⇒ x − 6 x + 6 = 0 x=
−( −6) ± ( −6) 2 − 4(1)(6) 6 ± 12 6 ± 2 3 = = = 3± 3 2(1) 2 2
The zeros of P(x) are 1, −1, −2, − 23 , 3 + 3, 3 − 3. Copyright © Houghton Mifflin Company. All rights reserved.
202
54.
Chapter 3: Polynomial and Rational Functions
P ( x ) = 2 x 4 + 3x 3 − 4 x 2 − 3x + 2 two or no positive and two or no negative real zeros p = ±1, ± 2, ± 1 q 2 2 3 2 −2 −4 −3 2 4 −4 −2 2 1 0 −1 −2 −1
−1 −2 −3
2 2
−2 3 1
55.
P( x ) = x 3 − 3x − 2 one positive and two or no negative real zeros p = ±1, ± 2 q 2 1 0 −3 −2 2 4 2 1 2 1 0 x 2 + 2 x + 1 = ( x + 1)2 = 0 ⇒ x = −1 The zeros of P(x) are 2, −1 (multiplicity 2).
1 −1 0
2 x 2 − 3x + 1 = (2 x − 1)( x − 1) = 0 ⇒ x = 12 , 1 1, 2
The zeros of P(x) are −2, − 1, 56.
1.
P( x ) = 3x 4 − 4 x 3 − 11x 2 + 16 x − 4 three or one positive and one negative real zeros p = ±1, ± 2, ± 4, ± 1 , ± 2 , ± 4 q 3 3 3 1 3 16 −4 − 11 −4 3 4 −1 − 12 3 4 0 −1 − 12 2
3 3
−1 6 5
− 12 10 −2
57.
x2 − 2 x − 1 = 0
4 −4 0
−( −2) ± ( −2)2 − 4(1)( −1) 2(1) 2 8 ± 2 = = ± 2 2 = 1± 2 2 2
x=
2
3x + 5 x − 2 = (3x − 1)( x + 2) = 0 ⇒ x = 13 , − 2 The zeros of P(x) are 1, 2, 58.
1, 3
− 2.
P( x ) = x 3 − 2 x + 1 two or no positive and one negative real zeros p = ±1 q 1 1 0 1 −2 1 1 −1 1 1 0 −1 x2 + x − 1 = 0 −1 ± 12 − 4(1)( −1) −1 ± 5 = x= 2(1) 2 The zeros of P(x) are 1, −1 + 5 , −1 − 5 . 2 2
P( x ) = x 4 − 5 x 2 − 2 x = x ( x 3 − 5 x − 2) one positive and two or no negative real zeros p = ±1, ± 2 q 1 0 −2 −5 −2 4 2 −2 1 0 −2 −1
The zeros of P(x) are 0, − 2, 1 + 2, 1 − 2. 59.
P ( x ) = x 4 + x 3 − 3x 2 − 5 x − 2 one positive and three or one negative real zeros p = ±1, ± 2 q 1 1 −1 −3 −5 −2 0 3 2 −1 1 0 0 −3 −2 −1
1 1
0 −1 −1
−3 1 −2
−2 2 0
x 2 − x − 2 = ( x − 2)( x + 1) = 0 ⇒ x = 2, − 1 The zeros of P(x) are 2, −1 (multiplicity 3).
Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.3
60.
203
P( x ) = 6 x 4 − 17 x 3 − 11x 2 + 42 x
61.
= x (6 x 3 − 17 x 2 − 11x + 42) two or no positive and one negative real zeros p = ±1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21, ± 42, ± 1 , ± 3 , 2 2 q 7 21 1 2 7 14 1 7 ± , ± , ± , ± , ± , ± , ± , ± 2 2 3 3 3 3 6 6 2 6 42 −17 −11 12 −10 −42 6 0 −5 −21
1
1
The zeros of P(x) are 1(multiplicity 2), − 23 , 8.
2
1 −1
1 1
4 −1 3
6 −3 3
4 −3 1
3 −1 2
3 −2 1
1 −1 0
24 −24 0
2 x − 13x − 24 = (2 x + 3)( x − 8) = 0 ⇒ x = − 23 , 8
P( x ) = x 5 + 5 x 4 + 10 x 3 + 10 x 2 + 5 x + 1 no positive and five, three or one negative real zeros p = ±1 q 1 5 10 10 5 −1 −1 −4 −6 −4 1 4 6 4 1 −1
−11 −13 −24
2
The zeros of P(x) are 0, 2, 7 , − 3 .
62.
−15 2 −13
2 2
6 x 2 − 5 x − 21 = (3x − 7)(2 x + 3) = 0 ⇒ x = 73 , − 23 3
P( x ) = 2 x 4 − 17 x 3 + 4 x 2 + 35 x − 24 three or one positive and one negative real zeros p = ±1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24, ± 1 , ± 3 q 2 2 1 2 4 35 −17 −24 2 24 −15 −11 2 24 0 −15 −11
63.
1 −1 0
P( x ) = x 3 − 16 x = x ( x 2 − 16) one positive and one negative real zeros p = ±1, ± 2, ± 4, ± 8, ± 16 q x ( x 2 − 16) = x ( x + 4)( x − 4) ⇒ x = −4, 0, 4
The zeros of P(x) are −4, 0, and 4.
1 −1 0
2
x + 2 x + 1 = ( x + 1)2 ⇒ x = −1 The zeros of P(x) are −1 (multiplicity 5). 64.
P( x ) = x 3 − 4 x 2 − 3x = x ( x 2 − 4 x − 3) one positive and one negative real zeros p = ± 3, ± 1 q
65.
The original cube’s dimensions are n × n × n. The resulting solid measures n ⋅ n ⋅ ( n − 2). n ⋅ n ⋅ ( n − 2) = 567
n 2 ( n − 2) = 567 n 3 − 2n 2 − 567 = 0 9 1
x2 − 4 x − 3 = 0 −( −4) ± ( −4)2 − 4(1)( −3) 2(1) 4 28 4 ± = = ±2 7 =2± 7 2 2 The zeros of P(x) are 0, 2 + 7, 2 − 7. x=
66.
The new dimensions are n × (n − 1) × (n − 3). n ( n − 1)( n − 3) = 1560
1
0 63 63
−567 567 0
n = 9 inches.
67.
[( x )( x + 1)( x + 2)] − [(2)(1)( x )] = 112 x ( x 2 + 3x + 2) − 2 x = 112
n ( n 2 − 4n + 3) = 1560 3
−2 9 7
x 3 + 3x 2 + 2 x − 2 x = 112
2
n − 4n + 3n − 1560 = 0 13 1 3 −4 −1560 13 117 1560 1 9 120 0 The original cube was 13 inches on each edge.
4
x 3 + 3x 2 − 112 = 0 1 3 0 4 28 1 7 28
x = 4 inches
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−112 112 0
204
68.
Chapter 3: Polynomial and Rational Functions
x (2 x + 1)( x + 3) = 126
69.
a.
x (2 x 2 + 7 x + 3) = 126 2 x 3 + 7 x 2 + 3x − 126 = 0 3 2 7 3 −126 6 39 126 2 13 42 0 x = 3; 2 x + 1 = 2(3) + 1 = 7; x + 3 = 3 + 3 = 6 The box is 3 inches by 7 inches by 6 inches.
70.
b.
53 + 5(5) + 6 125 + 25 + 6 = 6 6 156 = = 26 pieces 6
P(5) =
n 3 + 5n + 6 = 64 6 n 3 + 5n + 6 = 384 n 3 + 5n − 372 = 0 7 1 0 5 −372 7 49 378 1 7 54 0 At least 7 cuts are needed to produce 64 pieces.
x 3 − (62 + 52 + 42 ) x − 2(6)(5)(4) = 0 ⇒ x 3 − (36 + 25 + 16) x − (2(6)(5)(4) = 0 ⇒ x 3 − 77 x − 240 = 0
x ≈ 10.04 71.
If 140 cannonballs are used, there are 7 rows in the pyramid. 72.
The minimum amount, to the nearest $1000, the company needs to spend on advertising is $293,000 73.
The company should decrease each dimension by 0.084 inch. 74.
Volume = cylinder + sphere = π r 2 h +
( 43 )π r 3 = π r 2 (6) + ( 43 )π r 3 = 9π
The radius is 1.098 feet. Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.3
75.
205
4 w + l = 81 ⇒ l = 81 − 4 w Volume =
w2 l = 4900 w2 (81 − 4 w) = 4900
When w = 12.9875, then l = 81 − 4 w = 81 − 4(12.9875) = 29.05 in. When w = 14, then l = 81 − 4 w = 81 − 4(14) = 25 in. Thus, the lengths can be 25 in. or 29.05 in.. 76.
0.65 hours × 60 minutes per hour = 39 minutes 3.63 hours = 3 hours +0.63 hours = 3 hours + 0.63(60) minutes = 3 hours 38 minutes After 39 minutes and after 3 hours 38 minutes. 77.
The giraffe is 16.9 feet tall. 78.
There are 9 cards in the group.
Copyright © Houghton Mifflin Company. All rights reserved.
206
79.
Chapter 3: Polynomial and Rational Functions
a.
T (5) = 0.23245(5)3 + 0.53797(5)2 + 7.88932(5) − 8.53299 = 0.23245(125) + 0.53797(25) + 7.88932(5) − 8.53299 = 29.05625 + 13.44925 + 39.4466 − 8.53299 = 73.41911 ≈ 73 seconds
b.
5 minutes = 5(60) = 300 minutes
Approximately 93,000 digits of π can be computed in 5 minutes.
....................................................... 80.
x= p
81.
x2 = p 2 x − p =0 P( x ) = x 2 − p p=± p q=±1 p =± p q Since,
Connecting Concepts
B = ⎛⎜ 28 + 1⎞⎟ ⎝ 2 ⎠ B = 15 |–3| = 3 < 15 1 = 1 <15 2 2 |5| = 5 < 15 The absolute value of each zero is less than B.
P( p) = p 2 − p ≠ 0
P (− p) = p 2 − p ≠ 0 There are no rational zeros. 82.
⎛ max of (| −5 |, | 2 |, | 8 |) ⎞ + 1⎟ B=⎜ |1 | ⎝ ⎠ B = ⎛⎜ 8 + 1⎞⎟ ⎝1 ⎠ B=9 |–1| = 1 < 9 |2| = 2 < 9 |4| = 4 < 9 The absolute value of each zero is less than B.
84.
⎛ max of (| −5 |, | −28 |, |15 |) ⎞ B=⎜ + 1⎟ |2| ⎝ ⎠
83.
⎛ max of (| −2 |, | 9 |, | 2 |, | −10 |) ⎞ B=⎜ + 1⎟ |1 | ⎝ ⎠ B = ⎛⎜ 10 + 1⎞⎟ ⎝1 ⎠ B = 11 |1 + 3i | = 10 < 11
|1 – 3i | = 10 < 11 |1| = 1 < 11 |–1| = 1 < 11 The absolute value of each zero is less than B.
⎛ max of (| −4 |, |14 |, | −4 |, |13 |) ⎞ + 1⎟ B=⎜ |1 | ⎝ ⎠ B = ⎛⎜ 14 + 1⎞⎟ ⎝ 1 ⎠ B = 15 |2 + 3i | = 13 < 15
|2 – 3i | = 13 10 < 15 | i | = 1 < 15 |–i | = 1 < 15 The absolute value of each zero is less than B.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.4
207
.......................................................
Prepare for Section 3.4 PS2. 2 − i 5
PS1. 3 + 2i PS3. ( x − 1)( x − 3)( x − 4)
PS4. ( x − (2 + i ))( x − (2 − i ))
( x − 2 − i )( x − 2 + i ) (( x − 2) − i )(( x − 2) + i )
2
( x − 1)( x − 7 x + 12) x ( x 2 − 7 x + 12) − 1( x 2 − 7 x + 12)
( x − 2)2 − i 2
x 3 − 7 x 2 + 12 x − x 2 + 7 x − 12
x 2 − 4 x + 4 − ( −1)
x 3 − 8 x 2 + 19 x − 12
x2 − 4 x + 4 + 1 x2 − 4 x + 5 PS6. x 2 − x + 5 = 0
PS5. x 2 + 9 = 0 x 2 = −9
x=
x = ± −9 x = ± 3i The solutions are 3i and –3i.
−( −1) ± ( −1) 2 − 4(1)(5) 2(1)
= 1 ± 1 − 20 = 1 ± −19 2 2 = 1 ± i 19 = 1 ± 19 i 2 2 2
The solutions are 1 + 19 i and 1 − 19 i 2 2 2 2
Section 3.4 1.
P( x ) = x 4 + x 3 − 2 x 2 + 4 x − 24 2 1 1 −2 2 6 1 3 4 −3
1 1
3 −3 0
4 0 4
2.
4 8 12 12 −12 0
x 2 + 4 = 0 ⇒ x 2 = −4 ⇒ x = ± −4 ⇒ x = ±2i The zeros are 2, − 3, 2i, − 2i. P( x ) = ( x − 2)( x + 3)( x − 2i )( x + 2i )
−24 24 0
P( x ) = x 3 − 3x 2 + 7 x − 5 1 1 −3 1 1 –2
7 –2 5
x2 − 2 x + 5 = 0 x=
−( −2) ± ( −2)2 − 4(1)(5) 2(1)
= 2 ± −16 = 2 ± 4i = 1 ± 2i 2 2 The zeros are 1, 1 + 2i, 1 − 2i. P( x ) = ( x − 1)( x − 1 + 2i )( x − 1 − 2i )
Copyright © Houghton Mifflin Company. All rights reserved.
−5 5 0
208
3.
Chapter 3: Polynomial and Rational Functions
P( x ) = 2 x 4 − x 3 − 4 x 2 + 10 x − 4 1 2
−2
4.
2
–1
–4
10
−4
2
1 0
0 –4
–2 8
4 0
0 −4 −4
–4 8 4
8 −8 0
2 2
−125 125 0
x 2 − 8 x + 25 = 0 x=
−( −8) ± ( −8)2 − 4(1)(25) 8 ± 64 − 100 = 2(1) 2
= 8 ± −36 = 8 ± 6i = 4 ± 3i 2 2 The zeros are 5, 4 + 3i, 4 − 3i. P( x ) = ( x − 5)( x − 4 − 3i )( x − 4 + 3i )
2
2 x − 4 x + 4 = 2( x 2 − 2 x + 2) = 0 x=
P( x ) = x 3 − 13x 2 + 65 x − 125 5 1 65 −13 5 −40 1 25 −8
−( −2) ± ( −2)2 − 4(1)(2) 2(1)
= 2 ± −4 = 2 ± 2i = 1 ± i 2 2 The zeros are 12 , − 2, 1 + i, 1 − i.
(
)
P( x ) = 2 x − 12 ( x + 2)( x − 1 − i )( x − 1 + i ) 5.
P( x ) = x 5 − 9 x 4 + 34 x 3 − 58 x 2 + 45 x − 13 1 1 34 45 −9 −58 1 26 −8 −32 1 26 13 −8 −32 1
−8 1 −7
1 1
1
26 −7 19
−7 1 −6
1 1
19 −6 13
−32 19 −13
6.
−13 13 0
13 −13 0
P( x ) = x 4 − 4 x 3 + 53x 2 − 196 x + 196 2 1 53 −4 −196 2 98 −4 1 49 −2 −98 2
−2 2 0
1 1
196 −196 0
−98 98 0
49 0 49
x 2 + 49 = 0 ⇒ x 2 = −49 ⇒ x = ± −49 = ±7i The zeros are 2 (multiplicity 2), 7i, −7i.
−13 13 0
P( x ) = ( x − 2)2 ( x − 7i )( x + 7i )
2
x − 6 x + 13 = 0 x=
−( −6) ± ( −6)2 − 4(1)(13) 6 ± 36 − 52 = 2(1) 2
= 6 ± −16 = 6 ± 4i = 3 ± 2i 2 2 The zeros are 1 (multiplicity 3), 3 + 2i , 3 − 2i. P( x ) = ( x − 1)3 ( x − 3 − 2i )( x − 3 + 2i ) 7.
P( x ) = 2 x 4 − x 3 − 15 x 2 + 23x + 15 2 −3 −1 −15 21 −6 2 6 −7
− 12
8.
23 −18 5
2
−7
6
5
2
−1 −8
4 10
−5 0
2 x 2 − 8 x + 10 = 2( x 2 − 4 x + 5) = 0 x=
15 −15 0
−( −4) ± ( −4)2 − 4(1)(5) 2(1)
P( x ) = 3x 4 − 17 x 3 − 39 x 2 + 337 x + 116 3 337 −4 −17 −39 116 −12 −308 3 77 29 −29
− 13
3
−29
77
29
3
−1 −30
10 87
−29 0
3x 2 − 30 x + 87 = 3( x 2 − 10 x + 29) = 0 x=
−( −10) ± ( −10) 2 − 4(1)(29) 2(1)
= 10 ± −16 = 10 ± 4i = 5 ± 2i 2 2
= 4 ± −4 = 4 ± 2i = 2 ± i 2 2 The zeros are −3, − 12 , 2 + i, 2 − i.
The zeros are −4, − 13 , 5 + 2i, 5 − 2i.
P( x ) = 2( x + 3) x + 12 ( x − 2 − i )( x − 2 + i )
P( x ) = 3( x + 4) x + 13 ( x − 5 − 2i )( x − 5 + 2i )
(
)
(
)
Copyright © Houghton Mifflin Company. All rights reserved.
116 −116 0
Section 3.4
9.
209
P( x ) = 2 x 4 − 14 x 3 + 33x 2 − 46 x + 40 4 2 33 −14 −46 8 36 −24 2 9 −6 −10 2
−6 4 −2
2 2
10.
40 −40 0
−10 10 0
9 −4 5
1 3
2 x2 − 2 x + 5 = 0
x=
= 2 ± −36 = 2 ± 6i = 1 ± 3 i 4 4 2 2 The zeros are 4, 2, 12 + 23 i, 12 − 23 i.
)(
)
28
−8
3
1 −12
−4 24
8 0
−( −4) ± ( −4)2 − 4(1)(8) 2(1)
(
12.
2
–9
18
–20
2
5 −4
–10 8
20 0
4 3
−( −2) ± ( −2) − 4(1)(4) 2(1)
1 3
2 + 2i, 2 − 2i.
) ( x − 2 − 2i)( x − 2 + 2i)
P( x ) = 3x 4 − 19 x 3 + 59 x 2 − 79 x + 36 3 59 –79 1 −19 –16 3 43 3 43 –36 −16
36 −36 0
3
−16
43
–36
3
4 −12
–16 27
36 0
2
3x 2 − 12 x + 27 = 3( x 2 − 4 x + 9) = 0
= 2 ± −12 = 2 ± 2i 3 = 1 ± i 3 2 2 The zeros are 25 , 1 + i 3, 1 − i 3.
(
1, 3
P( x ) = 3( x + 1) x −
2 x 2 − 4 x + 8 = 2( x 2 − 2 x + 4) = 0 x=
−13
The zeros are −1,
P( x ) = 2 x 3 − 9 x 2 + 18 x − 20 5 2
3
−8 8 0
= 4 ± −16 = 4 ± 4i = 2 ± 2i 2 2
P( x ) = ( x − 4)( x − 2) x − 12 − 23 i x − 21 + 23 i
11.
20 −28 −8
3x 2 − 12 x + 24 = 3( x 2 − 4 x + 8) = 0
−( −2) ± ( −2)2 − 4(2)(5) x= 2(2)
(
P( x ) = 3x 4 − 10 x 3 + 15 x 2 + 20 x − 8 3 15 −1 −10 13 −3 3 28 −13
x=
)
−( −4) ± ( −4)2 − 4(1)(9) 2(1)
= 4 ± −20 = 4 ± 2i 5 = 2 ± i 5 2 2 4 The zeros are 1, 3 , 2 + i 5, 2 − i 5.
P( x ) = 2 x − 25 ( x − 1 − i 3)( x − 1 + i 3)
(
)
P( x ) = 3( x − 1) x − 43 ( x − 2 − i 5)( x − 2 + i 5) 13.
P( x ) = 2 x 4 − x 3 − 2 x 2 + 13x − 6 2 −2 −1 −2 10 −4 2 8 −5 1 2
14.
−6 6 0
13 −16 –3
2
−5
8
–3
2
1 −4
–2 6
3 0
2 x 2 − 4 x + 6 = 2( x 2 − 2 x + 3) = 0 x=
2
−( −2) ± ( −2) − 4(1)(3) 2(1)
1 2
4
−4
13
–12
3
4
2 −2
–1 12
6 –6
−3 0
1 2
4
−2
12
–6
4
2 0
0 12
6 0
4 x 2 + 12 = 4( x 2 + 3) = 0 x 2 + 3 = 0 ⇒ x 2 = −3 ⇒ x = ± − 3 = ± i 3
= 2 ± −8 = 2 ± 2i 2 = 1 ± i 2 2 2 1 The zeros are −1, 2 , 1 + i 2, 1 − i 2.
(
P( x ) = 4 x 4 − 4 x 3 + 13x 2 − 12 x + 3
The zeros are
(
1 , (multiplicity 2), 2 1 2
P( x ) = 4 x − 2
)
i 3, − i 3.
( x − i 3)( x + i 3)
)
P( x ) = 2( x + 2) x − 12 ( x − 1 − i 2)( x − 1 + i 2)
Copyright © Houghton Mifflin Company. All rights reserved.
210
15.
Chapter 3: Polynomial and Rational Functions
P( x ) = 3x 5 + 2 x 4 + 10 x 3 + 6 x 2 − 25 x − 20 3 10 −1 2 6 −25 −3 1 −11 5 3 11 −1 −5 −20 −1
−1 −3 −4
3 3
4 3
−4 4 3 0 2 2 3x + 15 = 3( x + 5) = 0 3
2
11 4 15
−5 −15 −20
15 0 15
−20 20 0
16.
−20 20 0
−20 20 0
P( x ) = 2 x 6 − 11x 5 + 5 x 4 2 −2 −11 −4 2 −15 −1
2 2
2
2 2
1 2
2
2
x + 5 = 0 ⇒ x = −5 ⇒ x = ± − 5 = ± i 5
The zeros are −1, (multiplicity 2),
(
)
4, 3
2
i 5, − i 5.
+ 60 x 3 − 62 x 2 − 64 x + 40 5 60 −62 −64 30 −70 20 84 35 −10 −42 20
−15 −2 −17
35 17 52
−10 −52 −62
−42 62 20
−17 4 −13
52 −26 26
−62 52 −10
20 −20 0
−13 1 −12
26 −6 20
−10 10 0
40 −40 0
20 −20 0
2
2 x − 12 x + 20 = 2( x 2 − 6 x + 10) = 0
P( x ) = 3( x + 1)2 x − 43 ( x − i 5)( x + i 5)
x=
−( −6) ± ( −6)2 − 4(1)(10) 2(1)
= 6 ± −4 = 6 ± 2i = 3 ± i 2 2 The zeros are −2, − 1, 2, 12 , 3 + i, 3 − i.
(
)
P( x ) = 2( x + 2)( x + 1)( x − 2) x − 12 ( x − 3 − i )( x − 3 + i ) 1+i
17.
−5 2 + 2i −3 + 2i
2 2
1−i
−3 + 2i 2 – 2i −1
2 2
2x − 1 = 0 ⇒ x =
−i
i
1
5 + 3i
18.
−29 15 + 9i –14 + 9i
3 3
5 – 3i
1–i −1+i 0
3
3 3x + 1 = 0 ⇒ x = − 1
1 2
1
3
1
−i 3−i
–14 + 9i 15 – 9i 1
3
1 −1 – 3i −3i
3 −3 0
1 3 x + 3 = 0 ⇒ x = −3 The remaining zeros are i, −3. 2 + 7i
−6 2 + 7i –4 + 7i
1 1
2 – 7i
1 1
–5 + 3i 5 – 3i 0
The remaining zeros are 5 – 3i, − 1 .
−3i 3i 0
3–i
i
20.
92 –97 + 3i –5 + 3i
3
1. 2
The remaining zeros are 1 − i, 19.
−2 2 0
6 −5 – i 1–i
–4 + 7i 2 – 7i −2
71 –57 – 14i 14 – 14i 14 – 14i –4 + 14i 10
−146 126 + 70i –20 + 70i
530 −530 0
–20 + 70i 20 – 70i 0
−( −2) ± ( −2)2 − 4(1)(10) 2 ± −36 2 ± 6i = = = 1 ± 3i 2(1) 2 2 The remaining zeros are 2 − 7i, 1 + 3i, 1 − 3i. x 2 − 2 x + 10 = 0 ⇒ x =
Copyright © Houghton Mifflin Company. All rights reserved.
34 –34 0
Section 3.4
211
2 – 3i
21.
−4 2 – 3i –2 – 3i
1 1
−4 2 – 3i –2 – 3i
14 −13 1
13 −13 0
2 + 3i
1 1
–2 – 3i 2 + 3i 0
1 0 1
–2 – 3i 2 + 3i 0
x 2 + 1 = 0 ⇒ x 2 = −1 ⇒ x = ± i The remaining zeros are 2 + 3i, i, − i. 3i
22.
1
−6
1
3i –6 + 3i
–3i
1
–6 + 3i –3i –6
1 2
−6 1 −4
1 1
−64 54 + 39i –10 + 39i
22 –9 – 18i 13 – 18i
13 – 18i 18i 13
117 –117 – 30i –30i
–10 + 39i –39i –10
−90 90 0
–30i 30i 0
−10 10 0
13 −8 5
−( −4) ± ( −4)2 − 4(1)(5) 4 ± −4 4 ± 2i = = =2±i 2(1) 2 2 The remaining zeros are −3i , 2, 2 + i, 2 − i. x2 − 4 x + 5 = 0 ⇒ x =
1 + 3i
23.
−4 1 + 3i –3 + 3i
1 1
−30 25 + 15i –5 + 15i
19 –12 – 6i 7 – 6i
50 −50 0
1 – 3i
1 1
–3 + 3i 1 – 3i −2
2
−( −2) ± ( −2) − 4(1)(5) 2 ± −16 2 ± 4i = = = 1 ± 2i 2(1) 2 2 The remaining zeros are 1 − 3i, 1 + 2i , 1 − 2i. x2 − 2 x + 5 = 0 ⇒ x =
i
24.
−1
1
−4 –1 – i –5 – i
i 1
−i
3
–1 + i 1
–1 + i
1
−1
−4 1 – 5i –3 – 5i –5 – i
−i −1 3 2
1 1
−5 5 – 3i –3i –3 – 5i 5i −3
i −5
−5 6 1
−3 3 0 –3i 3i 0
−3 3 0
2
x + 2 x + 1 = ( x + 1)2 = 0 ⇒ x = −1 The remaining zeros are −i, 3, −1 (multiplicity 2) –2i
25.
2i
1
−3
1
–2i –3 – 2i
7 –4 + 6i 3 + 6i
−13 12 – 6i –1 – 6i
12 –12 + 2i 2i
–3 – 2i 2i −3
3 + 6i –6i 3
–1 – 6i 6i −1
2i –2i 0
1 1
1
1 1
−3 1 −2
3 −2 1
−4 4 0
−1 1 0
2
x − 2 x + 1 = ( x − 1)2 = 0 ⇒ x = 1 The remaining zeros are 2i, 1 (multiplicity 3).
Copyright © Houghton Mifflin Company. All rights reserved.
7 – 6i –2 + 6i 5
–5 + 15i 5 – 15i 0
212
26.
Chapter 3: Polynomial and Rational Functions
i
1
−8
1
i –8- i
18 –1 – 8i 17 – 8i
−8 8 + 17i 17i
17 −17 0
–i
1 1
–8 + i –i −8
2
−( −8) ± ( −8) − 4(1)(17) 8 ± −4 8 ± 2i = = =4±i 2(1) 2 2 The remaining zeros are −i, 4 + i , 4 − i. x 2 − 8 x + 17 = 0 ⇒ x =
27.
5 + 2i
1 1
5 – 2i
−17 5 + 2i –12 + 2i
112 –64 – 14i 48 – 14i
−333 268 + 26i –65 + 26i
–12 + 2i 5 – 2i −7
48 – 14i –35 + 14i 13
–65 + 26i 65 – 26i 0
1 1
x 2 − 7 x + 13 = 0 ⇒ x =
337 −337 0
−( −7) ± ( −7)2 − 4(1)(13) 7 ± −3 7 ± i 3 7 = = = ± 3i 2(1) 2 2 2 2
The remaining zeros are 5 − 2i, 7 + 3 i, 7 − 3 i. 2 2 2 2 28.
1 – 5i
2 2
1 + 5i
2 2
−1
−8 2 – 10i –6 – 10i –6 – 10i 2 + 10i −4 −4 −2 −6
2 2
2 x2 − 6x + 7 = 0 ⇒ x =
61 –56 + 20i 5 + 20i
−99 105 – 5i 6 – 5i
5 + 20i –4 – 20i 1
6 – 5i 1 + 5i 7
1 6 7
12 –19 – 35i –7 – 35i
182 −182 0
–7 – 35i 7 + 35i 0
7 −7 0
−( −6) ± ( −6)2 − 4(2)(7) 6 ± −20 6 ± 2i 5 3 = = = ± 5i 2(2) 4 4 2 2
The remaining zeros are 1 + 5i, − 1, 3 + 5 i, 3 − 5 i. 2 2 2 2 29.
1.5
2 2
−1 3 2
1 3 4
2 x 2 + 2 x + 4 = 2( x 2 + x + 2) = 0 ⇒ x =
−6 6 0 −1 ± (1) 2 − 4(1)(2) −1 ± −7 −1 ± i 7 = = =−1± 7i 2(1) 2 2 2 2
The solutions are 1.5, − 1 + 7 i, − 1 − 7 i. 2 2 2 2
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17 – 8i 8i 17
17i –17i 0
Section 3.4
213
30.
–0.75
4
2
2
3 –3 0
4
12 –12 0
16 0 16
4 x + 16 = 4( x + 4) = 0 ⇒ x 2 = −4 ⇒ x = ± −4 = ± 2i
The solutions are −0.75, 2i, − 2i. 31.
The solutions are −0.6, 0.75, 2.5 or − 2 , 3 , 5 . 3 4 2 32.
The solutions are −0.5, 1.3, 3.5 or − 1 , 4 , 7 . 2 3 2 33.
2
1 1
−4 2 −2
5 −4 1
−4 2 −2
4 −4 0
2
1 1
−2 2 0
1 0 1
x 2 + 1 = 0 ⇒ x 2 = −1 ⇒ x = ± −1 = ±i The solutions are 2 (multiplicity 2), i , − i.
Copyright © Houghton Mifflin Company. All rights reserved.
−2 2 0
214
Chapter 3: Polynomial and Rational Functions
34.
−2
1
4 −2 2
1 2
8 −4 4
16 −8 8
16 −16 0
−2
1 1
2 −2 0
4 0 4
8 −8 0
6 −2 4
12 −8 4
8 −8 0
2
x + 4 = 0 ⇒ x = −4 ⇒ x = ± −4 = ± 2i
The solutions are −2 (multiplicity 2), 2i, –2i. 35.
The solutions are –3 (multiplicity 2), 1 (multiplicity 2). 36.
3
1
3 3 6
1
−6 18 12
−28 36 8
x 2 + 4 x + 4 = ( x + 2)2 = 0 ⇒ x = −2 37.
P( x ) = ( x − 4)( x + 3)( x − 2)
−2
1 1
The solutions are 3, −2 (multiplicity 3). 38.
P( x ) = ( x + 1)( x − 1)( x + 5) P( x ) = ( x 2 − 1)( x + 5)
2
P( x ) = ( x − 4)( x + x − 6) 2
−24 24 0
2
P( x ) = x ( x + x − 6) − 4( x + x − 6)
P( x ) = x 2 ( x + 5) − 1( x + 5)
P( x ) = x 3 + x 2 − 6 x − 4 x 2 − 4 x + 24
P( x ) = x 3 + 5x 2 − x − 5
P( x ) = x 3 − 3x 2 − 10 x + 24
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Section 3.4
39.
215
P( x ) = ( x − 3)( x − 2i )( x + 2i )
40.
P( x ) = x ( x − i )( x + i )
P( x ) = ( x − 3)( x − [2i ] )
P( x ) = x ( x 2 − i 2 )
P( x ) = ( x − 3)( x 2 − 4i 2 )
P( x ) = x ( x 2 − [ −1])
P( x ) = ( x − 3)( x 2 − 4[ −1])
P( x ) = x ( x 2 + 1)
P( x ) = ( x − 3)( x 2 + 4)
P( x ) = x 3 + x
2
2
P( x ) = x ( x 2 + 4) − 3( x 2 + 4) P( x ) = x 3 + 4 x − 3x 2 − 12 P( x ) = x 3 − 3x 2 + 4 x − 12 41.
P( x ) = [ x − (3 + i )][ x − (3 − i )][ x − (2 + 5i )][ x − (2 − 5i )] P( x ) = ( x − 3 − i )( x − 3 + i )( x − 2 − 5i )( x − 2 + 5i ) P( x ) = [( x − 3) − i ][( x − 3) + i ][( x − 2) − 5i ][( x − 2) + 5i ] P( x ) = [( x − 3) 2 − i 2 ][( x − 2)2 − 25i 2 ] P( x ) = [( x 2 − 6 x + 9) − ( −1)][( x 2 − 4 x + 4) − (25[−1])] P( x ) = [( x 2 − 6 x + 9) + 1][( x 2 − 4 x + 4) + 25] P( x ) = ( x 2 − 6 x + 9 + 1)( x 2 − 4 x + 4 + 25) P( x ) = ( x 2 − 6 x + 10)( x 2 − 4 x + 29) P( x ) = x 2 ( x 2 − 4 x + 29) − 6 x ( x 2 − 4 x + 29) + 10( x 2 − 4 x + 29) P( x ) = x 4 − 4 x 3 + 29 x 2 − 6 x 3 + 24 x 2 − 174 x + 10 x 2 − 40 x + 290 P( x ) = x 4 − 10 x 3 + 63x 2 − 214 x + 290
42.
P( x ) = [ x − (2 + 3i )][ x − (2 − 3i )]( x + 5)( x − 2) P( x ) = [ x − 2 − 3i ][ x − 2 + 3i ]( x + 5)( x − 2) P( x ) = [( x − 2)2 − (3i )2 ]( x + 5)( x − 2) P( x ) = [( x 2 − 4 x + 4) − (9i 2 )]( x + 5)( x − 2) P( x ) = [( x 2 − 4 x + 4) − (9[ −1])]( x + 5)( x − 2) P( x ) = [( x 2 − 4 x + 4) + 9]( x + 5)( x − 2) P( x ) = ( x 2 − 4 x + 13)( x 2 + 3x − 10) P( x ) = x 2 ( x 2 + 3x − 10) − 4 x ( x 2 + 3x − 10) + 13( x 2 + 3x − 10) P( x ) = x 4 + 3x 3 − 10 x 2 − 4 x 3 − 12 x 2 + 40 x + 13x 2 + 39 x − 130 P( x ) = x 4 − x 3 − 9 x 2 + 79 x − 130
43.
P( x ) = [ x − (6 + 5i )][ x − (6 − 5i )]( x − 2)( x − 3)( x − 5) P( x ) = [ x − 6 − 5i ][ x − 6 + 5i ]( x − 2)( x 2 − 8 x + 15) P( x ) = [( x − 6)2 − (5i )2 ][ x ( x 2 − 8 x + 15) − 2( x 2 − 8 x + 15)] P( x ) = [( x 2 − 12 x + 36) − (25i 2 )]( x 3 − 8 x 2 + 15 x − 2 x 2 + 16 x − 30) P( x ) = [( x 2 − 12 x + 36) − (25[ −1])]( x 3 − 10 x 2 + 31x − 30) P( x ) = ( x 2 − 12 x + 36 + 25)( x 3 − 10 x 2 + 31x − 30) P( x ) = ( x 2 − 12 x + 61)( x 3 − 10 x 2 + 31x − 30) P( x ) = x 2 ( x 3 − 10 x 2 + 31x − 30) − 12 x ( x 3 − 10 x 2 + 31x − 30) + 61( x 3 − 10 x 2 + 31x − 30) P( x ) = x 5 − 10 x 4 + 31x 3 − 30 x 2 − 12 x 4 + 120 x 3 − 372 x 2 + 360 x + 61x 3 − 610 x 2 + 1891x − 1830 P( x ) = x 5 − 22 x 4 + 212 x 3 − 1012 x 2 + 2251x − 1830
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216
44.
Chapter 3: Polynomial and Rational Functions
Note: 2 x − 1 = 0 if and only if x =
1 2
. Therefore, if x =
1 2
45.
(that is , 3 is a zero), then 4 x − 3 = 0.
(that is, 1 is a zero), then 2 x − 1 = 0.
(
2
P( x ) = x −
1 2
Note: 4 x − 3 = 0 if and only if x = 43 . Therefore, if x =
(
) [ x − (4 − i )][ x − (4 + i)]
4
)
P( x ) = x − 43 [ x − (2 + 7i )][ x − (2 − 7i )]
P( x ) = (2 x − 1)[ x − 4 + i ][ x − 4 − i ]
P( x ) = (4 x − 3)[ x − 2 − 7i ][ x − 2 + 7i ]
P( x ) = (2 x − 1)[( x − 4)2 − (i )2 ]
P( x ) = (4 x − 3)[( x − 2)2 − (7i )2 ]
P( x ) = (2 x − 1)[ x 2 − 8 x + 16 − ( −1)]
P( x ) = (4 x − 3)[( x 2 − 4 x + 4) − 49i 2 ]
P( x ) = (2 x − 1)[ x 2 − 8 x + 16 + 1]
P( x ) = (4 x − 3)[( x 2 − 4 x + 4) − 49( −1)]
P( x ) = (2 x − 1)( x 2 − 8 x + 17)
P( x ) = (4 x − 3)( x 2 − 4 x + 4 + 49)
P( x ) = 2 x ( x 2 − 8 x + 17) − 1( x 2 − 8 x + 17)
P( x ) = (4 x − 3)( x 2 − 4 x + 53)
P( x ) = 2 x 3 − 16 x 2 + 34 x − x 2 + 8 x − 17
P( x ) = 4 x ( x 2 − 4 x + 53) − 3( x 2 − 4 x + 53)
P( x ) = 2 x 3 − 17 x 2 + 42 x − 17
P( x ) = 4 x 3 − 16 x 2 + 212 x − 3x 2 + 12 x − 159 P( x ) = 4 x 3 − 19 x 2 + 224 x − 159
46.
(
P( x ) = x − 14
) ( x + 15 ) ( x − i )( x + i )
47.
P( x ) = (4 x − 1)(5 x + 1)( x − i )( x + i ) P( x ) = (20 x 2 − x − 1)( x 2 − i 2 ) P( x ) = (20 x 2 − x − 1)( x 2 − [ −1]) P( x ) = (20 x 2 − x − 1)( x 2 + 1) P( x ) = 20 x 2 ( x 2 + 1) − x ( x 2 + 1) − 1( x 2 + 1) P( x ) = 20 x 4 + 20 x 2 − x 3 − x − x 2 − 1 P( x ) = 20 x 4 − x 3 + 19 x 2 − x − 1
If 2 – 5i is a zero, then 2 + 5i is also a zero. P( x ) = [ x − (2 − 5i )][ x − (2 + 5i )]( x + 4) P( x ) = [ x − 2 + 5i ][ x − 2 − 5i ]( x + 4) P( x ) = [( x − 2)2 − (5i )2 ]( x + 4) P( x ) = [ x 2 − 4 x + 4 − 25i 2 ]( x + 4) P( x ) = [ x 2 − 4 x + 4 − 25( −1)]( x + 4) P( x ) = [ x 2 − 4 x + 4 + 25]( x + 4) P( x ) = ( x 2 − 4 x + 29)( x + 4) P( x ) = x 2 ( x + 4) − 4 x ( x + 4) + 29( x + 4) P( x ) = x 3 + 4 x 2 − 4 x 2 − 16 x + 29 x + 116 P( x ) = x 3 + 13x + 116
48.
If 1 – 3i is a zero, then 1 + 3i is also a zero. P( x ) = [ x − (3 + 2i )][ x − (3 − 2i )]( x − 7) P( x ) = ( x 2 − 6 x + 13)( x − 7) P( x ) = x 3 − 6 x 2 + 13x − 7 x 2 + 42 x − 91 P( x ) = x 3 − 13x 2 + 55 x − 91
49.
If 4 + 3iand 5 - i are zeros, then 4 – 3i and 5 + i are also zeros. P( x ) = [ x − (4 + 3i )][ x − (4 − 3i )][ x − (5 − i )][ x − (5 + i )] P( x ) = ( x 2 − 8 x + 25)( x 2 − 10 x + 26) P( x ) = x 4 − 10 x 3 + 26 x 2 − 8 x 3 + 80 x 2 − 208 x + 25 x 2 − 250 x + 650 P( x ) = x 4 − 18 x 3 + 131x 2 − 458 x + 650
Copyright © Houghton Mifflin Company. All rights reserved.
3 4
Section 3.4
50.
217
If i and 3 – 5i are zeros, then –i and 3 + 5i are also zeros. P( x ) = ( x − i )( x + i )[ x − (3 − 5i )][ x − (3 + 5i )] 2
51.
P( x ) = ( x 2 + x − 2)( x − 3)[ x − 1 − 4i ][ x − 1 + 4i ]
2
P( x ) = ( x − i )[ x − 3 + 5i ][ x − 3 − 5i ] 2
2
P( x ) = ( x + 2)( x − 1)( x − 3)[ x − (1 + 4i )][ x − (1 − 4i )] P( x ) = [ x 3 − 2 x 2 − 5 x + 6][( x − 1)2 − (4i )2 ]
2
P( x ) = [ x − ( −1)][( x − 3) − (5i ) ]
P( x ) = ( x 3 − 2 x 2 − 5 x + 6)[ x 2 − 2 x + 1 − 16i 2 ]
P( x ) = ( x 2 + 1)[ x 2 − 6 x + 9 − 25i 2 ]
P( x ) = ( x 3 − 2 x 2 − 5 x + 6)[ x 2 − 2 x + 1 + 16]
2
2
2
2
2
2
P( x ) = ( x + 1)[ x − 6 x + 9 − 25( −1)]
P( x ) = ( x 3 − 2 x 2 − 5 x + 6)[ x 2 − 2 x + 17]
P( x ) = ( x + 1)[ x − 6 x + 9 + 25]
P( x ) = x 5 − 4 x 4 + 16 x 3 − 18 x 2 − 97 x + 102
P( x ) = ( x + 1)( x − 6 x + 34) P( x ) = x 2 ( x 2 − 6 x + 34) + 1( x 2 − 6 x + 34) P( x ) = x 4 − 6 x 3 + 34 x 2 + x 2 − 6 x + 34 P( x ) = x 4 − 6 x 3 + 35 x 2 − 6 x + 34
52.
P( x ) = ( x + 5)( x − 3)2 [ x − (2 + i )][ x − (2 − i )] P( x ) = ( x + 5)( x 2 − 6 x + 9)[ x − 2 − i ][ x − 2 + i ] P( x ) = [ x 3 − x 2 − 21x + 45][( x − 2)2 − (i )2 ] P( x ) = ( x 3 − x 2 − 21x + 45)[ x 2 − 4 x + 4 − i 2 ] P( x ) = ( x 3 − x 2 − 21x + 45)[ x 2 − 4 x + 4 + 1] P( x ) = ( x 3 − x 2 − 21x + 45)[ x 2 − 4 x + 5] P( x ) = x 5 − 5 x 4 − 12 x 3 + 124 x 2 − 285 x + 225
....................................................... 53.
P( x ) = a ( x + 1)( x − 2)( x − 3) P(1) = a (1 + 1)(1 − 2)(1 − 3) 12 = a (2)( −1)( −2) 12 = 4a 3= a P( x ) = 3( x + 1)( x − 2)( x − 3)
54.
P( x ) = a ( x − 3i )( x + 3i )( x − 2) P( x ) = a ( x 2 − 9i 2 )( x − 2) P( x ) = a[ x 2 − 9(i 2 )]( x − 2) P( x ) = a[ x 2 − 9( −1)]( x − 2) P( x ) = a ( x 2 + 9)( x − 2)
2
P( x ) = (3x + 3)( x − 5 x + 6) P( x ) = 3x ( x 2 − 5 x + 6) + 3( x 2 − 5 x + 6) P( x ) = 3x 3 − 15 x 2 + 18 x + 3x 2 − 15 x + 18 P( x ) = 3x 3 − 12 x 2 + 3x + 18
Connecting Concepts
P (3) = a (32 + 9)(3 − 2) P (3) = a (9 + 9)(1) P (3) = 18a 27 = 18a 3 2
=a
P( x ) = 23 ( x 2 + 9)( x − 2) P( x ) = 23 ( x 3 − 2 x 2 + 9 x − 18) x − 27 P( x ) = 23 x 3 − 3x 2 + 27 2
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218
55.
Chapter 3: Polynomial and Rational Functions
P( x ) = a ( x − 3)( x + 5)[ x − (2 + i )][ x − (2 − i )] P( x ) = a ( x 2 + 2 x − 15)[ x − 20 − i ][ x − 2 + i ] P( x ) = a ( x 2 + 2 x − 15)[( x − 2)2 − i 2 ] P( x ) = a ( x 2 + 2 x − 15)[ x 2 − 4 x + 4 − ( −1)] P( x ) = a ( x 2 + 2 x − 15)[ x 2 − 4 x + 4 + 1] P( x ) = a ( x 2 + 2 x − 15)( x 2 − 4 x + 5) P(1) = a (12 + 2[1] − 15)(12 − 4[1] + 5) 48 = a (1 + 2 − 15)(1 − 4 + 5) 48 = a ( −12)(2) 48 = −24a −2 = a P( x ) = −2( x 2 + 2 x − 15)( x 2 − 4 x + 5) P( x ) = −2[ x 2 ( x 2 − 4 x + 5) + 2 x ( x 2 − 4 x + 5) − 15( x 2 − 4 x + 5) P( x ) = −2[ x 4 − 4 x 3 + 5 x 2 + 2 x 3 − 8 x 2 + 10 x − 15 x 2 + 60 x − 75) P( x ) = −2( x 4 − 2 x 3 − 18 x 2 + 70 x − 75) P( x ) = −2 x 4 + 4 x 3 + 36 x 2 − 140 x + 150
56.
P( x ) = a (2 x − 1)[ x − (1 − i )][ x − (1 + i )] P( x ) = a (2 x − 1)[ x − 1 + i ][ x − 1 − i ] P( x ) = a (2 x − 1)[( x − 1)2 − i 2 ] P( x ) = a (2 x − 1)[( x − 1)2 − ( −1)] P( x ) = a (2 x − 1)( x 2 − 2 x + 1 + 1) P( x ) = a (2 x − 1)( x 2 − 2 x + 2) P(4) = a[2(4) − 1][(4)2 − 2(4) + 2] 140 = a[8 − 1][16 − 8 + 2] 140 = a (7)(10) 140 = 70a 2=a P( x ) = 2(2 x − 1)( x 2 − 2 x + 2) P( x ) = (4 x − 2)( x 2 − 2 x + 2) P( x ) = 4 x ( x 2 − 2 x + 2) − 2( x 2 − 2 x + 2) P( x ) = 4 x 3 − 8 x 2 + 8 x − 2 x 2 + 4 x − 4 P( x ) = 4 x 3 − 10 x 2 + 12 x − 4
57.
P( x ) = x 3 − x 2 − ix 2 − 9 x + 9 + 9i = x 3 + ( −1 − i ) x 2 − 9 x + (9 + 9i ) 1+i 1 –1 – i 9 + 9i 1–i 1 0 −9 −9 1+i 0 –9 – 9i 1–i –2i 1 0 0 1 1–i –9 – 2i −9 Zero remainder implies 1 + i is a zero. Non-zero remainder implies 1 – i is not a zero. The Conjugate Pair Theorem does not apply because some of the coefficients of the polynomial function are not real numbers.
58.
a.
Answers will vary.
b.
Answers will vary.
c.
No such polynomial function exists because the nonreal complex zeros must occur in pairs.
d.
Answers will vary.
....................................................... PS1.
x 2 − 9 = ( x + 3)( x − 3) = x − 3 , x ≠ −3 x − 2 x − 15 ( x + 3)( x − 5) x − 5 2
Prepare for Section 3.5 PS2.
−1 + 4 3 = = 3 =−3 2 ( −1)2 − 2( −1) − 5 1 + 2 − 5 −2
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Section 3.5
PS3.
219
2( −3)2 + 4( −3) − 5 2(9) − 12 − 5 = 3 −3 + 6 = 18 − 12 − 5 = 1 3 3
PS4.
2 x 3 + x 2 − 15 x = 0 x (2 x 2 + x − 15) = 0 x (2 x − 5)( x + 3) = 0
x=0
or 2 x − 5 = 0
or x + 3 = 0
x=5 2 PS5. The degree of the numerator, x 3 + 3x 2 − 5 , is 3;
PS6.
2
the degree of the denominator, x − 4 , is 2.
x = −3
x + 4 + 72x − 11 x − 2x x 2 − 2 x x 3 + 2 x 2 − x − 11 x3 − 2 x 2 4 x2 − x 4 x2 − 8x 7 x − 11
Section 3.5 1.
3.
5.
7.
F ( x) = 1 ⇒ x ≠ 0 x The domain is all real numbers except 0. 2 F ( x ) = x 2 − 3 ⇒ no restrictions on denominator. x +1 The domain is all real numbers.
2x − 1 2x − 1 = ⇒ x ≠ 3, 6 2 2 x 2 − 15 x + 18 (2 x − 3)( x − 6) 3 The domain is all real numbers except and 6. 2 F ( x) =
2 x2 2 x2 = ⇒ x ≠ 0, 6, −2 2 x − 4 x − 12 x x ( x − 6)( x + 2) The domain is all real numbers except 0, 6 and –2. F ( x) =
3
2.
4.
6.
8.
x 2 + 3x = 0 x( x + 3) = 0 x = 0 or x + 3 = 0 x = −3 Vertical asymptotes: x = 0, x = −3
10.
11.
6 x 2 − 5x − 4 = 0 (3x − 4)(2 x + 1) = 0 3x − 4 = 0 or 2 x + 1 = 0
12.
4 3
3 x3 + 4 ⇒ x ≠ 5, −5 F ( x ) = x2 + 4 = x − 25 ( x − 5)( x + 5) The domain is all real numbers except 5 and –5.
3x − 2 3x − 2 = ⇒ x ≠ 3, 6 4 4 x 2 − 27 x + 18 (4 x − 3)( x − 6) 3 The domain is all real numbers except and 6. 4 F ( x) =
2 3x 2 F ( x ) = 32 x = ⇒ x ≠ 5, − 5 x − 5 ( x − 5)( x + 5)
The domain is all real numbers except
9.
x=
2 ⇒ x≠3 x−3 The domain is all real numbers except 3. F ( x) =
x2 − 4 = 0 ( x + 2)( x − 2) = 0 or x − 2 = 0 x+2=0 x = −2 x=2 Vertical asymptotes: x = 2, x = −2 x3 − 8 = 0 ( x − 2)( x 2 + 2 x + 4) = 0 Vertical asymptotes: x = 2
x = − 12
Vertical asymptotes: x = − 1 , x = 4 2
3
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5 and − 5 .
220
13.
Chapter 3: Polynomial and Rational Functions
4 x 3 − 25 x 2 + 6 x = 0 x (4 x 2 − 25 x + 6) = 0 x (4 x − 1)( x − 6) = 0
x = 0 or 4 x − 1 = 0 x=1 4
x 4 − 81 = 0
14.
( x 2 + 9)( x 2 − 9) = 0 ( x 2 + 9)( x − 3)( x + 3) = 0 x − 3 = 0 or x + 3 = 0 x=3 x = −3 Vertical asymptotes: x = 3, x = −3
x−6=0 x=6
or
Vertical asymptotes: x = 0, x = 1 , x = 6 4 15.
Horizontal asymptote: y =
4 =4 1
16.
17.
Horizontal asymptote: y =
15,000 = 30 500
18.
19.
Horizontal asymptote: y = 4 = 12 1 3
21.
Vertical asymptote: x = −4 Horizontal asymptote: y = 0
22.
Vertical asymptote: x = 2 Horizontal asymptote: y = 0
23.
Vertical asymptote: x = 3 Horizontal asymptote: y = 0
24.
Vertical asymptote: x = −2 Horizontal asymptote: y = 0
25.
Vertical asymptote: x = 0 Horizontal asymptote: y = 0
26.
Vertical asymptote: x = 0 Horizontal asymptote: y = 0
27.
Vertical asymptote: x = −4 Horizontal asymptote: y = 1
28.
Vertical asymptote: x = 2 Horizontal asymptote: y = 1
29.
Vertical asymptote: x = 2 Horizontal asymptote: y = −1
20.
Horizontal asymptote: y = 0 ⎛ ( x + 5)2 − 25 ⎞ Horizontal asymptote: y = 6000 ⎜⎜ 2 ⎟⎟ ⎝ ( x + 5) ⎠ y = 6000 = 6000 1 2 (2 x − 3)(3x + 4) = 6 x 2 − x − 12 (1 − 2 x )(3 − 5 x ) 10 x − 11x + 3 Horizontal asymptote: y = 6 = 3 10 5
F ( x) =
Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.5
221
30.
Vertical asymptote: x = 1 Horizontal asymptote: y = −1
31.
Vertical asymptotes: x = 3, x = −3 Horizontal asymptote: y = 0
32.
Vertical asymptotes: x = 2, x = −2 Horizontal asymptote: y = 0
33.
Vertical asymptotes: x = −3, x = 1 Horizontal asymptote: y = 0
34.
Vertical asymptotes: x = 4, x = −2 Horizontal asymptote: y = 0
35.
Vertical asymptote: x = −2 Horizontal asymptote: y = 1
36.
Vertical asymptote: x = −1, x = 1 Horizontal asymptote: y = 2
37.
Vertical asymptote: none Horizontal asymptote: y = 0
38.
Vertical asymptote: x = 3 Horizontal asymptote: y = 1
39.
Vertical asymptotes: x = 3, x = −3 Horizontal asymptote: y = 2
40.
Vertical asymptote: none Horizontal asymptote: y = 3
41.
Vertical asymptotes:
42.
Vertical asymptotes: x = 5, x = 1 Horizontal asymptote: y = 2
43.
−4
x = −1 + 2 , x = −1 − 2 Horizontal asymptote: y = 1
5 −1 28 −12 3 27 −7 27 F ( x) = 3 x − 7 + x+4 Slant asymptote: y = 3 x − 7 3
Copyright © Houghton Mifflin Company. All rights reserved.
222
Chapter 3: Polynomial and Rational Functions
x +1+
44.
x −1
45.
2
x − 3x + 5
x 2 − 3x + 5 x 3 − 2 x 2 + 3x + 4
x3 − 1
x3
1 1 − = x− x x2 x2 x2 Slant asymptote: y = x 2
=
x 3 − 3x 2 + 5 x x2 − 2 x + 4 x 2 − 3x + 5 x −1 x −1 F ( x) = x + 1 + x 2 − 3x + 5 Slant asymptote: y = x + 1 46.
4000 + 20 x + 0.0001x 2 4000 = + 20 + 0.0001x x x Slant Asymptote: y = 0.0001x + 20 F ( x) =
2 − x − 2 + −3x 3− 3x + 5 x −1
48.
47.
49.
x 3 − 1 − x 4 − 2 x 3 − 3x 2 + 4 x − 1 − x4
+x − 2 x 3 − 3x 2 + 3x − 1 −2 x 3 +2 2 − 3x + 3x − 3
5
−4
15 8 −20 −25 −4 −5 −17 − 17 F ( x ) = −4 x − 5 + x −5 Slant asymptote: y = −4 x − 5 x2 − 4 4 = x− x x Slant asymptote: y = x Vertical asymptote: x = 0 F ( x) =
2 F ( x ) = − x − 2 + −3x 3− 3x + 5 x −1 Slant asymptote: y = − x − 2
50.
x 2 + 10 1 5 = x+ 2x 2 x 1 Slant asymptote: y = x 2 Vertical asymptote: x = 0 F ( x) =
51.
−3
1 1
−3 −3 −6
−4 18 14
2
x − 3x − 4 14 = x−6+ x+3 x+3 Slant asymptote: y = x − 6 Vertical asymptote: x = −3 F ( x) =
Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.5
52.
223
2x + 5 = 0 2 x = −5 5 x=− 2
45 1 13 4 x− + 2 4 2x + 5
53.
45 4 = 1 x − 13 + 2 4 2x +5
2 2
2 x + 5 x2 − 4x − 5
5 8 13
3 52 55
2 x2 + 5x + 3 55 = 2 x + 13 + x−4 x−4 Slant asymptote: y = 2 x + 13 Vertical asymptote: x = 4 F ( x) =
2 x 5 13 − x−5 2 13 65 − x− 2 4 45 4
x2 +
2 F ( x) = x − 4 x − 5 2x+5
4
Slant asymptote: y = 1 x − 13 2 4 5 Vertical asymptote: x = − 2
54.
−3
0 −9 36 −12 4 27 −12 27 F ( x ) = 4 x − 12 + x+3 Slant asymptote: y = 4 x − 12 Vertical asymptote: x = −3 4
55.
−2
1 1
−1 −2 −3
0 6 6
x2 − x 6 = x −3+ x+2 x+2 Slant asymptote: y = x – 3 Vertical asymptote: x = −2 F ( x) =
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224
Chapter 3: Polynomial and Rational Functions
x 56.
1
1 1
1 1 2
0 2 2
57.
x2 + x 2 F ( x) = = x+2+ x −1 x −1 Slant asymptote: y = x + 2 Vertical asymptote: x = 1
2
x − 4 x3 +1 x3 − 4 x 4x + 1
58.
x3 3x
2
−
1 3x
2
=
1 1 x− 3 3x 2
1 x 3 Vertical asymptote: x = 0
Slant asymptote: y =
3
F ( x ) = x2 + 1 = x + 42x + 1 x −4 x −4 Slant asymptote: y = x Vertical asymptotes: x = 2, x = −2
59.
60.
61.
62.
63.
64.
65.
66.
67.
F ( x) =
a.
0.43(1000) + 76,000 430 + 76,000 76,430 = = = $76.43 1000 1000 1000 0.43(10,000) + 76,000 4,300 + 76,000 80,300 C (10,000) = = = = $8.03 10,000 10,000 10,000 0.43(100,000) + 76,000 43,000 + 76,000 119,000 = = C (100,00) = = $1.19 100,000 100,000 100,000
b.
y = 0.43. As the number of golf balls that are produced increases, the average cost per golf ball approaches $0.43.
C (1000) =
Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.5
68.
a.
225
C (1000) = C (10,000) = C (100,000) =
0.001(1000)2 + 54(1000) + 175,000 230,000 = = $230 1000 1000 0.001(10,000)2 + 54(10,000) + 175,000 815,000 = = $81.50 10,000 10,000 0.001(100,000) 2 + 54(100,000) + 175,000 15,575,000 = = $155.75 100,000 100,000
b.
The minimum cost per CD player is approximately $80.46. Producing approximately 13,229 CD players will minimize the average cost per CD player. 69.
a. b.
2000( 40) 80,000 = = $1,333.33 100 − 40 60 2000(80) 160,000 C (80) = = = $8,000 100 − 80 20 C ( 40) =
c.
70.
a.
0.0006(1000)2 + 9(1000) + 401,000 410,060 = = $410.60 1000 1000 0.0006(10,000)2 + 9(10,000) + 401,000 551,000 C (10,000) = = = $55.10 10,000 10,000 0.0006(100,000)2 + 9(100,000) + 401,000 7,301,000 = $73.01 C (100,000) = = 100,000 100,000 C (1000) =
b.
The minimum cost per telephone is approximately $40.02. Producing approximately 25,852 telephones will minimize the average cost per telephone. 71.
a.
R(0) = R(7) = R(15) =
b.
22.8(0) 2 + 177(0) + 5900 33.8(0) 2 + 266(0) + 15,200 22.8(7)2 + 177(7) + 5900 33.8(7)2 + 266(7) + 15,200
= 5900 ≈ 0.38816 ≈ 38.8% 15,200 = 8256.2 ≈ 0.44108 ≈ 44.1% 18,718.2
22.8(15) 2 + 177(15) + 5900 33.8(15) 2 + 266(15) + 15,200
=
13,685 ≈ 0.51073 ≈ 51.1% 26,795
22.8 ≈ 67.5% 33.8
Copyright © Houghton Mifflin Company. All rights reserved.
226
72.
Chapter 3: Polynomial and Rational Functions
a.
≈ 39.7% b.
≈ 53.5% 73.
a.
S (2) = S (4) = S (10) =
150(2) 1.5(2)2 + 80 150(4) 1.5(4)2 + 80 150(10)
= 300 ≈ 3.488 thousand ≈ 3500 86 = 600 ≈ 5.769 thousand ≈ 5800 104
1.5(10)2 + 80
= 1500 ≈ 6.522 thousand ≈ 6500 230
b.
In the seventh month
74.
75.
c.
(n < m) The sales will approach zero.
a.
= 402.5 ≈ 201 milligrams 2 0.04(5)2 + 1 0.5(10) + 400 405 M (10) = = = 81 milligrams 5 0.04(10)2 + 1
b.
( n < m ) ⇒ as t → ∞, M → 0 milligrams.
M (5) =
0.5(5) + 400
a.
r ≈ 3.8 centimeters b.
No. The degree of the numerator is not exactly one more than the degree of the denominator.
c.
As the radius r increases without bound, the surface area approaches twice the area of a circle with radius r. Since V = π r 2 h, then h = V 2 . h → 0 as r → ∞ so as the radius increases without bound, the surface area πr of the can approaches the area of the top and bottom of the can, two circles with radius r.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 3.5
76.
a.
b.
227
10(2) 20 5 = = ohms 10 + 2 12 3 10(20) 200 20 ohms RT (20) = = = 10 + 20 30 3 10 R2 RT = → 10 = 10 ohms R2 + 10 1 RT (2) =
....................................................... 77.
Horizontal asymptote: y=2 2=
Connecting Concepts 78.
2
3x + 2
+
−20 x − 20
2 x + 3x + 4 x2 + 4x + 7
x2 + 4
x 2 + 4 3 x 3 + 2 x 2 − 8 x − 12
2 x 2 + 8 x + 14 = 2 x 2 + 3 x + 4 5 x + 10 = 0 x = −2 The graph of F intersects its horizontal asymptote at (−2,2).
3x 3
− 12 x 2
2 x − 20 x − 12 + 8 − 20 x − 20 Slant asymptote: y = 3 x + 2
3x + 2 =
3 x 3 + 2 x 2 − 8 x − 12 x2 + 4
3 x 3 + 2 x 2 + 12 x + 8 = 3 x 3 + 2 x 2 − 8 x − 12 20 x = −20 x = −1 y = 3(−1) + 2 = −1 The graph of F intersects its slant asymptote at (−1,−1). 79.
Horizontal asymptote: y = 1 1=
80.
Consider the rational function f ( x) =
4 3 2 x + x + x + 3x + 2 x
x3 + x 2 + 4 x + 1 x3 + 1
3
x 4 x 3 x 2 + 3x + 2 + + x3 x3 x 2 ( x + 2)( x + 1) = x +1+ x3
f ( x) =
x3 + 1 = x3 + x 2 + 4 x + 1 x2 + 4 x = 0 x( x + 4) = 0 x = 0, or x = −4 The graph of F intersects its horizontal asymptote at (0, 1) and (−4,1).
x +1 =
x 4 + x3 + x 2 + 3x + 2 x3
x 4 + x3 = x 4 + x3 + x 2 + 3x + 2 0 = x 2 + 3x + 2 0 = ( x + 2)( x + 1) x = −2,−1 Thus, the graph of f intersects the slant asymptote y = x + 1 when x = −2 and − 1, that is, at the points (−2, −1) and (–1, 0).
Copyright © Houghton Mifflin Company. All rights reserved.
228
Chapter 3: Polynomial and Rational Functions
.......................................................
Exploring Concepts with Technology
Finding Zeros of a Polynomial Using Mathematica 1.
In(2):= NSolve[x^4-3x^3+x-5==0] Out[2]= {{x -> -1.14039}, {x -> 0.536692 – 1.06842 I}, {x -> 0.536692 + 1.06842 I}, {x -> 3.067}}
2.
In(3):= NSolve[3x^3-4x^2+x-3==0] Out[3]= {{x -> -0.102814 – 0.799511 I}, {x -> -0.1022814 + 0.799511 I}, {x -> 1.53896}}
3.
In(4):= NSolve[4x^5-3x^3+2x^2-x+2==0] Out[4]= {{x -> -1.25095},, (x -> – 0.156173 – 0.685216 I}, {x -> -0.156173 + 0.685216 I}, {x -> 0.781647 – 0.445283 I}, {x -> 0.781647 + 0.445283 I}}
4.
In(5):= NSolve[-3x^4-6x^3+2x-8==0] Out[5]= {{x -> -1.60199 – 0.623504 I}, {x -> -1.600199 + 0.623504 I}. {x -> 0.601988 – 0.743846 I}, {x -> 0.601988 + 0.7344846 I}}
.......................................................
Assessing Concepts
1.
True
2.
True
3.
False, Descartes’ Rule of Signs indicates that
4.
R( x ) =
x 3 − x 2 + x − 1 has three or one positive zeros. 5.
P( x ) = x − 1 + i is one example.
6.
c
7.
d
8.
f
9.
a
10.
b
11.
e
12.
f
3x 2 is one example. ( x − 2)( x − 5)
....................................................... 1.
3
−11 12 1
4 4
3.
−2
3 3
0 −6 −6
5 3 8 −5 12 7
−2 24 22
4 x2 + x + 8 +
Chapter Review 2.
22 x −3
1
3x 2 − 6 x + 7 + [3.1]
−13 x+2
4.
1 2
−18 5 −13
0 5 5
5
[3.1]
1 −14 −13
5
2 −13 −11
2
7
16
−10
2
1 8
4 20
10 0
Copyright © Houghton Mifflin Company. All rights reserved.
5 x 2 + 5 x − 13 +
[3.1]
2 x 2 + 8 x + 20 [3.1]
−11 x −1
Chapter Review
5.
5
229
−10 15 5
3 3
7.
4
1 1
9.
−2
6 6
11.
1
3
1 13.
1
1 1
−5 24 19
2 4 6 0 −12 −12
0 0 0
55 −55 0
1 76 77
−12 24 12 −26 15 −11
2 3 5 −1 1 0
−36 25 −11
−2 0 −2
15.
6.
8.
33 −33 0 1 −2 −1
1 1
P (4) = 77 [3.1] 8 −24 −16
−7
3x 2 + 5 x − 11 [3.1]
1 32 33
6 −14 −8
−65 56 −9
0
−10 −4 −14
8 14 22
−4
−1
4 −4 P (−2) = 33 [3.1]
10.
3
−4
[3.1]
4
−8 15 7
5 5
12.
1 −1 0
9 −7 2
2
14.
1 2
−8 0 −8
8 −8 0
2
−6 69 63
2 21 23
3
−8
3
2
1 4
2 −6
−3 0
16.
x3 + 2 x 2 − 8 x − 9 [3.1]
P (−1) = 22 [3.1]
0 189 189 −31 32 1
2
[3.1]
−63 63 0
−9 567 558 4 −4 0
P (3) = 558 [3.1]
[3.1]
[3.1]
17.
[3.2] [3.2]
[3.2] 18.
19.
20.
[3.2]
[3.2]
[3.2] 21.
p = ±1, ± 2, ± 3, ± 6 [3.3] q
22.
p = ±1, ± 2, ± 3, ± 5, ± 6, ± 10, ± 15, ± 30, ± 1 , ± 3 , ± 5 , ± 15 [3.3] q 2 2 2 2
23.
p = ±1, ± 2, ± 3, ± 4, ± 6, ± 12, ± 1 , ± 2 , ± 4 , ± 1 , ± 2 , ± 3 , ± 4 , ± 6 , ± 12 , ± 1 , ± 2 , ± 4 [3.3] q 3 3 3 5 5 5 5 5 5 15 15 15
24.
p = ±1, ± 2, ± 4, ± 8, ± 16, ± 32, ± 64 [3.3] q
25.
26.
1 1 1 2 p = ±1, ± 2, ± , ± , ± , ± [3.3] q 6 3 2 3
27.
P ( x ) = x 3 + 3x 2 + x + 3 P ( − x ) = − x 3 + 3x 2 − x + 3 no positive and three or one negative real zeros [3.3]
29.
P( x ) = x 4 − x − 1 P( − x ) = x 4 + x − 1 one positive and one negative real zeros [3.3]
28.
P ( x ) = x 4 − 6 x 3 − 5 x 2 + 74 x − 120 P ( − x ) = x 4 + 6 x 3 − 5 x 2 − 74 x − 120 three or one positive and one negative real zeros [3.3]
p = ±1 [3.3] q
Copyright © Houghton Mifflin Company. All rights reserved.
230
30.
31.
Chapter 3: Polynomial and Rational Functions
P( x ) = x5 − 4 x 4 + 2 x 3 − x 2 + x − 8 P( − x ) = − x5 − 4 x 4 − 2 x 3 − x 2 − x − 8 five, three or one positive and no negative real zeros [3.3]
1
1
6 1 7
1
−10 10 0
3 7 10
x 2 + 7 x + 10 = 0 ( x + 5)( x + 2) = 0 x = −5 or x = −2
The zeros of x3 + 6 x 2 + 3 x − 10 are 1, − 5, and − 2. [3.3] 32.
2
−10 2 −8
1 1
31 −16 15
−30 30 0
x 2 − 8 x + 15 = 0 ( x − 5)( x − 3) = 0 x = 5 or x = 3
The zeros of x3 − 10 x 2 + 31x − 30 are 2, 5, and 3. [3.3] 33.
−2
6
35 −12 23
6
72 −46 26
60 −52 8
−2
16 −16 0
6
23 −12 11
6
26 −22 4
8 −8 0
The zeros of 6 x 4 + 35 x3 + 72 x 2 + 60 x + 16 are − 2 (multiplicity 2), − 4 /3, and − 1/2. [3.3] 34.
−1
2
2
7
5
7
3
2
−1 6
−3 2
−1 6
−3 0
−3
2
6
2
6
2
−6 0
0 2
−6 0
2x 2 + 2 = 0 2 x 2 = −2 x 2 = −1 x = ± −1 x = ±i
The zeros of 2 x 4 + 7 x3 + 5 x 2 + 7 x + 3 are − 1 / 2, − 3, i, and − i. [3.4] 35.
1
−4 1 −3
1 1
6 −3 3
−4 3 −1
1 −1 0
1
1 1
−3 1 −2
3 −2 1
−1 1 0
x2 − 2 x + 1 = 0 ( x − 1)( x − 1) = 0 x = 1 or x = 1
The zero of x 4 − 4 x3 + 6 x 2 − 4 x + 1 is 1 (multiplicity 4). [3.3] 36.
−1
2
x=
2
−7
22
13
2 x 2 − 8 x + 26 = 0
4 26
−13 0
2( x 2 − 4 x + 13) = 0
2
−1 −8
−(−4) ± (−4)2 − 4(1)(13) 4 ± 16 − 52 4 ± −36 4 ± 6i = = = = 2 ± 3i 2(1) 2 2 2
The zeros of 2 x3 − 7 x 2 + 22 x + 13 are − 1 / 2, 2 + 3 i, and 2 − 3 i. [3.4] 37.
1 – 2i
1 + 2i
1
−4
1
1 – 2i –3 – 2i
6 –7 + 4i –1 + 4i
–4 7 + 6i 3 + 6i
–3 – 2i 1 + 2i 2
–1 + 4i –2 – 4i 3
3 + 6i –3 – 6i 0
1 1
15 15 0
x 2 − 2 x − 3 = ( x − 3)( x + 1) = 0 x = 3, x = −1 The remaining zeros are 1 + 2i, 3, and –1. [3.4] Copyright © Houghton Mifflin Company. All rights reserved.
6x 2 + 11x + 4 = 0 (3x + 4)(2 x + 1) = 0 x = − 4 or x = − 1 3 2
Chapter Review
38.
231
2+i
1 1
2–i
1 1
−1 2+i 1+i
−17 1 + 3i –16 + 3i
1+i 2–i 3
–16 + 3i 6 – 3i −10
55 –35 – 10i 20 – 10i
−50 50 0
20 – 10i –20 + 10i 0
x 2 + 3x − 10 = ( x + 5)( x − 2) = 0 x = −5, x = 2 The remaining zeros are 2 – i, –5, and 2. [3.4] 39.
( x − 4)( x + 3)(2 x −1) = ( x 2 − x −12)(2 x −1) = 2 x3 − x 2 − 2 x 2 + x − 24 x +12 = 2 x3 − 3 x 2 − 23 x +12
40.
[3.4]
( x − 2)( x + 3)( x − i )( x + i) = ( x 2 + x − 6)( x 2 +1) = x 4 + x 2 + x3 + x − 6 x 2 − 6 = x 4 + x3 − 5 x 2 + x − 6
41.
[3.4]
( x −1)( x − 2)( x − 5i )( x + 5i ) = ( x 2 − 3 x + 2)( x 2 + 25) = x 4 + 25 x 2 − 3 x3 − 75 x + 2 x 2 + 50 = x 4 − 3 x3 + 27 x 2 − 75 x + 50
42.
[3.4]
( x + 2)( x + 2)[ x − (1+ 3i )][ x − (1− 3i )] = ( x 2 + 4 x + 4)( x 2 − 2 x +10) = x 4 − 2 x3 +10 x 2 + 4 x3 − 8 x 2 + 40 x + 4 x 2 − 8 x + 40 = x 4 + 2 x3 + 6 x 2 + 32 x + 40
43.
x2 ⇒ no restrictions on denominator. x +7 The domain is all real numbers. [3.4] F ( x) =
2
[3.4] 44.
45.
x + 2 = 0 ⇒ vertical asymptote : x = −2 3 = 3 ⇒ horizontal asymptote : y = 3 [3.5] 1
47.
48. x + 1 = 0 ⇒ vertical asymptote : x = −1 5 11 −1 2 f ( x) = 2 x + 3+ 8 −2 −3 x +1 ⇒ slant asymptote: y = 2 x + 3 [3.5] 2 3 8
46.
2 2 F ( x ) = 3x2 + 2 x − 5 = 3x + 2 x − 5 ⇒ x ≠ 1 , 4 6 6 x − 25 x + 4 (6 x − 1)( x − 4) 1 The domain is all real numbers except and 4. [3.4] 6
x2 + 2x − 3 = 0 ( x + 3)( x − 1) = 0 ⇒ vertical asymptotes : x = −3, x = 1 2 = 2 ⇒ horizontal asymptote: y = 2 [3.5] 1 2x2 + x + 7 = 0 − 1 ± 12 − 4(1)(7) − 1 ± 1 − 28 − 1 ± − 27 = = 2(1) 2 2 x is not a real number ⇒ vertical asymptote : none x=
6 = 3 ⇒ horizontal asymptote : y = 3 [3.5] 2
49.
50.
51.
52.
[3.5] [3.5]
[3.5]
Copyright © Houghton Mifflin Company. All rights reserved.
[3.5]
232
Chapter 3: Polynomial and Rational Functions
53.
54.
55.
[3.5]
[3.5] 57.
58.
56.
[3.5]
b.
5.75(5000) + 34,200 62,950 = = $12.59 5000 5000 5.75(50,000) + 34,200 321,700 = ≈ $6.43 C (50,000) = 50,000 50,000 y = 5.75. As the number of skateboards that are produced increases, the average cost per skateboard approaches $5.75. [3.5]
a.
F (1) =
60 = 60 = 15o F 12 + 2(1) + 1 4
b.
F (4) =
60 = 60 = 2.4o F 42 + 2(4) + 1 25
c.
F (t ) → 0o F as t → ∞. [3.5]
a.
C (5000) =
59.
a.
As the radius of the blood vessel gets smaller, the resistance gets larger.
b.
As the radius of the blood vessel gets larger, the resistance gets smaller. [3.5]
....................................................... QR1. S (6) = 3.6(6)3 − 36.8(6)2 + 145.2(6) + 8 ≈ 412 3
Quantitative Reasoning QR2.
2
S (8) = 3.6(8) − 36.8(8) + 145.2(8) + 8 ≈ 737.6 There will be about 412,000 hybrid vehicles sold in 2010, and 737,600 hybrid vehicles sold in 2012.
2.5 ⎛⎜ ⎝
96,000 ⎞ ⎛ 96,000 ⎞ + 3600 ⎟ = 2.5 ⎜ ⎟ 18 ⎠ ⎝ 18 + x ⎠ 240,000 240,000 = + 3600 18 18 + x 240,000(18 + x ) = 18(240,000) + 3600(18)(18 + x ) 240,000 x = 1,116,400 + 64,800 x 175,200 x = 1,166,400 x ≈ 6.7 additional mpg
QR3. M (1) = 0.2416(1)3 + 2.3106(1)2 − 1.2373(1) + 25.00 = 26.2879
M (2) = 0.2416(2)3 + 2.3106(2)2 − 1.2373(2) + 25.00 = 33.4846 M (3) = 0.2416(3)3 + 2.3106(3)2 − 1.2373(3) + 25.00 = 47.8777 M (4) = 0.2416(4)3 + 2.3106(4)2 − 1.2373(4) + 25.00 = 70.7548 M (5) = 0.2416(5)3 + 2.3106(5)2 − 1.2373(5) + 25.00 = 103.4035 M (6) = 0.2416(6)3 + 2.3106(6)2 − 1.2373(6) + 25.00 = 147.1114 M (7) = 0.2416(7)3 + 2.3106(7)2 − 1.2373(7) + 25.00 = 203.1661 M (8) = 0.2416(8)3 + 2.3106(8)2 − 1.2373(8) + 25.00 = 272.8552 M (1) + M (2) + M (3) + M (4) + M (5) + M (6) + M (7) + M (8) = 904.9412 ≈ $900 QR4.
[3.5]
3 ⎛⎜ ⎝
96,000 ⎞ ⎛ 96,000 ⎞ + 3600 + 900 ⎟ = 3⎜ ⎟ 18 ⎠ ⎝ 18 + x ⎠ 288,000 288,000 = + 4500 18 18 + x 288,000(18 + x ) = 18(288,000) + 4500(18)(18 + x ) 288,000 x = 1,458,000 + 81,000 x 207,000 x = 1,458,000 x ≈ 7.0 additional mpg
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Test
233
....................................................... 1.
4.
−2
−1 −12 −13
3
5 4 2 −6 3 6 −1 − 13 [3.1] 3x 2 − x + 6 + x+2
−2
Chapter Test
−3
7 6 13 −3 P (−2) = 43 [3.1]
2.
−5 48 43
2 −26 −24
3.
5. P ( x) = −3 x3 + 2 x 2 − 5 x + 2 [3.2] Since An = −3 is negative and n = 3 is odd, the graph of P goes up to the far left and down to the far right.
3x3 + 7 x 2 − 6 x = 0 [3.2]
6.
x(3 x 2 + 7 x − 6) = 0 x(3x − 2)( x + 3) = 0 x = 0, 3 x − 2 = 0, or x + 3 = 0 2 3
P ( x) = ( x 2 − 4) 2 (2 x − 3)( x + 1)3
8.
2
3
P ( x) = ( x − 2) ( x + 2) (2 x − 3)( x + 1) The zeros of P are 2 (multiplicity 2), –2 (multiplicity 2),
−3 −1 1 4 2 2 2 1 1 3 Because P(1) and P(2) have different signs, P must have a real zero between 1 and 2. [3.2]
4
2
5 8 2 13 upper bound: 4 −5
−23 52 29
−38 116 78
5 −23 25 −10 2 2 −5 lower bound: −5 [3.3] 11.
1 2
24 312 336 −38 −10 −48
2
2
−3
−11
6
2
1 −2
−1 −12
−6 0
2
p = ±1, ± 3 q = ±1, ± 2, ± 3, ± 6 1 3 1 1 [3.3] p = ±1, ± 3, ± , ± , ± , ± 2 2 3 6 q
3 (multiplicity 1), and –1 (multiplicity 3). [3.3] 2
9.
[3.1]
P ( x) = 2 x3 − 3 x 2 − x + 1 [3.2] 1 2 1 −3 −1 2 −1 −2 2 −1 −2 −1
2
x = −3
The zeros of 3x 3 + 7 x 2 − 6 x = 0 are 0, 2 , and − 3. [3.2] 3
2
1
x 4 − 4 x3 + 7 x 2 − 6 x + 2.
x=
7.
7 −6 2 −4 1 −3 4 −2 1 −3 4 −2 0 A remainder of 0 implies that x −1 is a factor of 1
10.
P ( x) = x 4 − 3x3 + 2 x 2 − 5 x + 1 P ( − x ) = x 4 + 3 x3 + 2 x 2 + 5 x + 1 four, two, or no positive and no negative real zeros [3.3]
24 240 264
2 x 2 − 2 x − 12 = 0 2( x + 2)( x − 3) = 0 x = −2 or x = 3
12.
2 + 3i
6 6
2 − 3i
6 6
The zeros of 2 x3 − 3x 2 − 11x + 6 are 1/2, − 2, and 3. [3.3]
−5 12 + 18i 7 + 18i 7 + 18i 12 − 18i 19
12 −40 + 57i −28 + 57i −28 + 57i 38 − 57i 10
207 −227 + 30i −20 + 30i −20 + 30i 20 − 30i 0
6 x 2 + 19 x + 10 = 0 (3 x + 2)(2 x + 5) = 0 3x + 2 = 0 2x + 5 = 0 x = −2 / 3
x = −5 / 2 4
The zeros of 6x − 5 x3 +12 x 2 + 207 x +130 are 2 + 3i, 2 − 3i, − 2 / 3, and − 5/ 2. [3.4]
Copyright © Houghton Mifflin Company. All rights reserved.
130 −130 0
234
13.
Chapter 3: Polynomial and Rational Functions
P ( x) = x( x 4 − 6 x3 + 14 x 2 − 14 x + 5) 1 1 5 −6 14 −14 1 −5 9 −5 1 9 0 −5 −5
1
1 1
−5 1 −4
14.
P ( x) = [ x − (1+ i)][ x − (1− i )]( x − 3)( x) [3.4] = ( x 2 − 2 x + 2)( x − 3)( x) = ( x3 − 5 x 2 + 8 x − 6)( x) = x 4 − 5 x3 + 8 x 2 − 6 x
−5 5 0
9 −4 5
x2 − 4 x + 5 = 0 x=
−( −4) ± ( −4)2 − 4(1)(5) 2(1)
x = 4 ± 16 − 20 = 4 ± −4 2 2 4 2 i ± x= =2±i 2 The zeros of x 5 − 6x 4 + 14 x 3 − 14 x 2 + 5 x are 0, 1 (multiplicity 2), 2 + i , and 2 − i. [3.4]
15.
f ( x) =
3x2 − 2 x + 1 x2 − 5x + 6
x2 − 5x + 6 = 0
( x − 3)( x − 2) = 0 x=3 x=2 vertical asymptotes: x = 3, x = 2 [3.5]
17.
16.
f ( x) =
2x2 − 1
horizontal asymptote: y = 18.
[3.5] 19.
3x2 − 2 x + 1
a.
b.
[3.5]
w(t ) = 70t + 120 t + 40 70(1) + 120 70 + 120 190 = = ≈ 5 words per minute w(1) = 1 + 40 41 41 70(10) + 120 700 + 120 820 w(10) = = = ≈ 16 words per minute 10 + 40 50 50 70(20) + 120 1400 + 120 1520 w(20) = = = ≈ 25 words per minute 20 + 40 60 60 As t → ∞, w(t ) → 70 = 70 words per minute [3.5] 1
Copyright © Houghton Mifflin Company. All rights reserved.
3 [3.5] 2
Cumulative Review
20.
235
length = 25 – 2(2x) = 25 – 4x width = 18 – 2x height = x volume = length × width × height = (25 − 4 x)(18 − 2 x)( x) = (450 − 122 x + 8 x 2 )( x) = 450 x − 122 x 2 + 8 x3 = 8 x3 − 122 x 2 + 450 x
The value of x (to the nearest 0.001 inch) that will produce a box with the maximum volume is 2.42 inches. The maximum volume (to the nearest 0.1 cubic inch) is 487.9 cubic inches. [3.3]
....................................................... 1.
3.
3 + 4i = 3 + 4i ⋅ 1 + 2i = 3 + 10i + 8i 2 [P.6] 1 − 2i 1 − 2i 1 + 2i 12 − 4i 2 3 + 10i + 8( −1) 3 + 10i − 8 = = 1 − 4( −1) 1+ 4 = −5 + 10i = −1 + 2i 5
Cumulative Review 2.
x=
2x + 5
) = (2 + 2
x −1
)
Check 2: 2(2) + 5 − (2) − 1 = 2 4 + 5 − 2 −1 = 2 9− 1=2 3 −1 = 2 2=2 Yes Check 10: 2(10) + 5 − (10) − 1 = 2 20 + 5 − 10 − 1 = 2 25 − 9 = 2 5−3= 2 2=2 Yes The solutions are x = 2, x = 10. [1.4]
2
2x + 5 = 4 + 4 x − 1 + x − 1 2x + 5 = 3 + 4 x − 1 + x x + 2 = 4 x −1
(
( x + 2) 2 = 4 x − 1
)
2
x 2 + 4 x + 4 = 16( x − 1) x 2 + 4 x + 4 = 16 x − 16 x 2 − 12 x + 20 = 0 ( x − 2)( x − 10) = 0 x = 2, x = 10 4.
x − 3 ≤ 11 −11 ≤ x − 3 ≤ 11 −8 ≤ x ≤ 14 {x | −8 ≤ x ≤ 14}
[1.5]
−( −1) ± ( −1)2 − 4(1)( −1) 1 ± 5 = 2(1) 2
x = 1 − 5 , x = 1 + 5 [1.3] 2 2
2x + 5 − x − 1 = 2 2x + 5 = 2 + x − 1
(
x2 − x − 1 = 0 a = 1, b = −1, c = −1
5.
d = (2 − 7)2 + [5 − ( −11)]2 [2.1] = (2 − 7)2 + (5 + 11)2
6.
Shift the graph of y = x 2 two units to the right and four units up. [2.5]
= ( −5) 2 + (16)2 = 25 + 256 = 281
Copyright © Houghton Mifflin Company. All rights reserved.
236
Chapter 3: Polynomial and Rational Functions
P( x ) = x 2 − 2 x − 3
7.
P( x + h ) − P ( x ) [( x + h )2 − 2( x + h ) − 3] − ( x 2 − 2 x − 3) x 2 + 2 xh + h 2 − 2 x − 2h − 3 − x 2 + 2 x + 3 = = h h h 2 xh h h 2 2 + − = = 2x + h − 2 [2.6] h
8.
f ( x) = 2 x 2 + 5 x − 3 g ( x) = 4 x − 7 ( f o g )( x) = f [ g ( x)] = f (4 x − 7)
[2.6]
9.
( f − g )( x ) = f ( x ) − g ( x )
[2.6]
= x 3 − 2 x + 7 − ( x 2 − 3x − 4) = x 3 − 2 x + 7 − x 2 + 3x + 4 = x 2 − x 2 + x + 11
= 2(4 x − 7)2 + 5(4 x − 7) − 3 = 2(16 x 2 − 56 x + 49) + 5(4 x − 7) − 3 = 32 x 2 − 112 x + 98 + 20 x − 35 − 3 = 32 x 2 − 92 x + 60
10.
12.
−2
−2 −4 16 −28 4 14 −32 3 2 59 4 x − 8 x + 14 x − 32 + [3.1] x+2 4
0 −8 −8
−5 64 59
11.
3
2
2 P(3) = 141 [3.1]
0 6 6
−3 18 15
The leading term has a negative coefficient. The graph of P(x) goes down to the far right. [3.2]
13.
The relative maximum (to the nearest 0.0001) is 0.3997. [3.2] 14.
P( x ) = 3x 4 − 4 x 3 − 11x 2 + 16 x − 4 [3.3] p = ± factors of 4 = ±1, ± 2, ± 4 q = ± factors of 3 = ±1, ± 3 p = ±1, ± 2, ± 4, ± 1 , ± 2 , ± 4 3 3 3 q
15.
P( x ) = x 3 + x 2 + 2 x + 4 has no changes of sign. There are no positive real zeros. P( − x ) = − x 3 + x 2 − 2 x + 4 has three changes of sign. There are three or one negative real zeros. [3.3]
Copyright © Houghton Mifflin Company. All rights reserved.
4 45 49
−6 147 141
Cumulative Review
16.
237
P( x ) = x 3 + x + 10 no positive and one negative real zeros p = ±1, ± 2, ± 5, ± 10 q 1 0 1 10 −2 4 −2 −10 1 5 0 −2 x=
17.
If 3 + i is a zero of P(x), then 3 – i is also a zero. P ( x) = [ x − (3 + i )][ x − (3 − i )]( x + 2) = [ x − 3 − i ][ x − 3 + i ]( x + 2) = [( x − 3)2 − i 2 ]( x + 2) = [ x 2 − 6 x + 9 − (−1)]( x + 2) = [ x 2 − 6 x + 9 + 1]( x + 2)
−( −2) ± ( −2)2 − 4(1)(5) 2(1)
= ( x 2 − 6 x + 10)( x + 2) = x 2 ( x + 2) − 6 x( x + 2) + 10( x + 2)
= 2 ± −16 = 2 ± 4i = 1 ± 2i 2 2 The zeros are –2, 1 – 2i, 1 + 2i. [3.4]
18.
P( x ) = x 3 − 2 x 2 + 9 x − 18 three or one positive and no negative real zeros p = ±1, ± 2, ± 3, ± 6, ± 9, ± 18 q 2 1 9 −2 −18 2 0 18 1 0 9 0 x2 + 9 = 0
= x3 + 2 x 2 − 6 x 2 − 12 x + 10 x + 20 = x3 − 4 x 2 − 2 x + 20
19.
F ( x) =
4 x2 x + x−6 2
Vertical asymptotes: x2 + x − 6 = 0 ( x + 3)( x − 2) = 0 x = –3, x = 2 Horizontal asymptote: y = 4 ⇒ y = 4 [3.5] 1
x 2 = −9 x = ± −9 x = ±3i P( x ) = ( x − 2)( x + 3i )( x − 3i ) [3.4]
20.
3 2 F ( x ) = x +24 x + 1 x +4 x+4 +1 x2 + 4 x3 + 4 x2
x3
+ 4x 2
4x − 4x + 1 + 16 4x2 − 4 x − 15 The slant asymptote is y = x + 4. [3.5]
Copyright © Houghton Mifflin Company. All rights reserved.
[3.4]
Chapter 4
Exponential and Logarithmic Functions Section 4.1 1.
If f (3) = 7 , then f −1 (7) = 3.
2.
If g ( −3) = 5 , then g −1 (5) = −3.
3.
If h −1 ( −3) = −4 , then h(−4) = −3.
4.
If f −1 (7) = 0 , then f (0) = 7.
5.
If 3 is in the domain of f −1 , then f [ f −1 (3)] = 3.
6.
a.
If f is a one-to-one function and f (0) = 5 , then f −1 (5) = 0.
b.
If f is a one-to-one function and f (1) = 2 , then f −1 (2) = 1.
7.
The domain of the inverse function f −1 is the range of f .
8.
The range of the inverse function f −1 is the domain of f .
9.
10.
11.
12.
Yes, the inverse is a function.
Yes, the inverse is a function. 13.
14.
Yes, the inverse is a function. 17.
15.
No, the inverse relation is not a function.
f ( x ) = 4 x; g ( x ) = x 4 x x =4 =x f [ g ( x )] = f 4 4 g [ f ( x )] = g (4 x ) = 4 x = x 4 Yes, f and g are inverses of each other.
() ()
Yes, the inverse is a function. 16.
No, the inverse relation is not a function. 18.
Yes, the inverse is a function.
No, the inverse relation is not a function.
f ( x ) = 3 x; g ( x ) = 1 3x 1 f [ g ( x )] = f =3 1 = 1 ≠ x 3x 3x x No, f and g are not inverses of each other.
( ) ( )
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.1
19.
239
f ( x ) = 4 x − 1; g ( x ) = 1 x + 1 4
20.
4
(4 4) = 4( 1 x + 1 ) −1 = x +1−1 4 4
2
2
2
(2 2) = 2(1 x − 3) + 3 = x − 3 + 3 2 2
g [ f ( x )] = g 1 x − 3 4
4
4
=x Yes, f and g are inverses of each other. f ( x ) = − 1 x − 1 ; g ( x ) = −2 x + 1
=x Yes, f and g are inverses of each other. 22.
2
2
3
3
f [ g ( x )] = f 1 x − 2 3 3
= − 1 ( −2 x + 1) − 1 = x − 1 − 1 2
f ( x ) = 3x + 2; g ( x ) = 1 x − 2
( ) = 3( 1 x − 2 ) + 2 = x − 2 + 2 3 3
f [ g ( x )] = f ( −2 x + 1) 2
2
=x
= 1 (4 x − 1) + 1 = x − 1 + 1
2
2
= 1 (2 x + 3) − 3 = x + 3 − 3
=x g [ f ( x )] = g (4 x − 1)
21.
2
f [ g ( x )] = f (2 x + 3)
f [ g ( x )] = f 1 x + 1
4
f ( x) = 1 x − 3 ; g ( x) = 2 x + 3
2
= x −1 ≠x No, f and g are not inverses of each other.
=x g [ f ( x )] = g (3x + 2) = 1 (3x + 2) − 2 = x + 2 − 2 3
3
3
3
=x Yes, f and g are inverses of each other.
23.
5 ; g ( x) = 5 + 3 x−3 x 5 +3 f [ g ( x )] = f x 5 = = 5 = 5⋅ x 5 + 3−3 5 5 x x =x f ( x) =
24.
( ) ( )
( )
g [ f ( x )] = g
( x 5− 3)
5 + 3= x − 3+ 3 5 x−3 =x Yes, f and g are inverses of each other.
( )
=
25.
f ( x ) = x 3 + 2; g ( x ) = 3 x − 2
f [ g ( x )] = f ( 3 x − 2 ) 3
= (3 x − 2 ) + 2 = x − 2 + 2 =x
g [ f ( x )] = g ( x 3 + 2 ) 3 3
f ( x) = 2 x ; g ( x) = x x −1 x−2 x f [ g ( x )] = f x−2 2x 2x 2 x x−2 = x−2 = x−2 = x − 1 x − ( x − 2) 2 x−2 x−2 x−2 = 2x ⋅ x − 2 x−2 2 =x g [ f ( x )] = g 2 x x −1 2x 2x 2x 1 1 x x x − − = = = −1 2 x − 2 2 x − 2( x − 1) 2 x −1 x −1 x −1 = 2x ⋅ x − 1 x −1 2 =x Yes, f and g are inverses of each other.
26.
f ( x ) = ( x + 5)3 ; g ( x ) = 3 x − 5
f [ g ( x )] = f ( 3 x − 5)
3
= ( 3 x − 5 + 5) = x 3 =x
(
g [ f ( x )] = g ( x + 5)3 3 3
= x +2−2 = x =x Yes, f and g are inverses of each other.
3
)
= 3 ( x + 5)3 − 5 = x + 5 − 5 =x Yes, f and g are inverses of each other.
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Chapter 4: Exponential and Logarithmic Functions
240 27.
The inverse of {( −3, 1), ( −2, 2), (1, 5), (4, − 7)} is {(1, − 3), (2, − 2), (5, 1), ( −7, 4)}.
28.
The inverse of {( −5, 4), ( −2, 3), (0, 1), (3, 2), (7, 11)} is {(4, − 5), (3, − 2), (1, 0), (2, 3), (11, 7)}.
29.
The inverse of {(0, 1), (1, 2), (2, 4), (3, 8), (4, 16)} is {(1, 0), (2, 1), (4, 2), (8, 3), (16, 4)}.
30.
The inverse of {(1, 0), (10, 1), (100, 2), (1000, 3), (10000, 4) is {(0, 1), (1, 10), (2, 100), (3, 1000), (4, 10,000). f ( x) = 2 x + 4
31.
32.
f ( x) = 4 x − 8
x = 2y + 4
1x+2= y 4 f −1 ( x ) = 1 x + 2 4
f ( x ) = −3 x − 8 x = −3 y − 8 x + 8 = −3 y
1x+7 = y 3 3 −1 f ( x) = 1 x + 7 3 3
f ( x ) = −2 x + 5 x = −2 y + 5 x − 5 = −2 y
35.
−1 x − 8 = y
f ( x) = − x + 3 x = −y + 3 y = −x + 3
36.
−1 x+ 5 = y 2
3
f −1 ( x ) = − x + 3
2
f −1 ( x ) = − 1 x + 5
( x) = − 1 x − 8
f
x + 7 = 3y
x +8 = 4y
1 x−2= y 2 f −1 ( x ) = 1 x − 2 2
3 3 −1
x = 3y − 7
x = 4y −8
x − 4 = 2y
34.
f ( x ) = 3x − 7
33.
2
3
2
37.
f ( x) = 2 x , x ≠ 1 x −1 2y x= y −1 x ( y − 1) = xy − x = 2 y xy − 2 y = y ( x − 2) = x y= x x−2 −1 f ( x) = x , x ≠ 2 x−2
38.
x , x≠2 x−2 y x= y−2 x ( y − 2) = xy − 2 x = y xy − y = y ( x − 1) = 2 x y = 2x x −1 −1 f ( x) = 2 x , x ≠ 1 x −1
39.
f ( x ) = x − 1 , x ≠ −1 x +1 y −1 x= y +1 x ( y + 1) = xy + x = y − 1 xy − y = − x − 1 y − xy = y (1 − x ) = x + 1 y = x +1 1− x f −1 ( x ) = x + 1 , x ≠ 1 1− x
40.
f ( x ) = 2 x − 1 , x ≠ −3 x+3 2y −1 x= y+3 xy + 3x = 2 y − 1 xy − 2 y = −3x − 1 y = 3x + 1 2− x −1 3 f ( x) = x + 1 , x ≠ 2 2− x
41.
f ( x ) = x 2 + 1, x ≥ 0
42.
f ( x ) = x 2 − 4, x ≥ 0
43.
f ( x) =
x + 4 = y2
x − 1 = y2 x −1 = y
x+4 = y
−1
−1
f ( x ) = x − 1, x ≥ 1 Note: Do not use ± with the radical because the domain of f , and thus the
f ( x ) = x + 4, x ≥ −4 Note: Do not use ± with the radical because the domain of f , and thus the
range of f −1 , is nonnegative.
range of f −1 , is nonnegative.
44.
f ( x ) = x − 2, x ≥ 2 x=
x = y2 − 4
x = y2 + 1
y−2
f ( x) = 4 − x , x ≤ 4 x = 4− y
2
2
x = y−2
x =4− y
2
y = − x2 + 4
x +2= y f −1 ( x ) = x 2 + 2, x ≥ 0 Note: The range of f , is nonnegative, therefore the
f −1 ( x ) = − x 2 + 4, x ≥ 0 Note: The range of f , is non-negative, therefore the
domain of f −1 is also nonnegative.
domain of f −1 is also non-negative.
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Section 4.1
241
f ( x ) = x 2 + 4 x , x ≥ −2
45.
f ( x ) = x 2 − 6 x, x ≤ 3
46.
x = y2 + 4 y
x = y2 − 6 y
x + 4 = y2 + 4 y + 4
x + 9 = y2 − 6 y + 9
x + 4 = ( y + 2)2
x + 9 = ( y − 3)2
x+4 = y+2
− x+9 = y−3
y = x+4 −2 f
−1
− x +9 +3= y
f −1 ( x ) = − x + 9 + 3, x ≥ −9
( x ) = x + 4 − 2, x ≥ −4
Note: The range of f , is non-negative, therefore the domain of f
−1
range of f −1 must also be non-positive.
is also non-negative.
f ( x ) = x 2 + 4 x − 1, x ≤ −2
47.
Note: Because the range of f , is non-positive, the
f ( x ) = x 2 − 6 x + 1, x ≥ 3
48.
x = y2 + 4 y − 1
x = y2 − 6 y + 1
x + 1 = y2 + 4 y
x − 1 = y2 − 6 y
x + 1 + 4 = y2 + 4 y + 4
x − 1 + 9 = y2 − 6 y + 9
x + 5 = ( y + 2)2
x + 8 = ( y − 3)2
− x+5 = y+2
x+8 = y −3
− x+5−2= y
x +8 +3= y
−1
49.
f ( x ) = − x + 5 − 2, x ≥ −5 Note: Because the range of f , is non-positive, the
f −1 ( x ) = x + 8 + 3, x ≥ −8 Note: The range of f , is non-negative, therefore the
range of f −1 must also be non-positive.
domain of f −1 is also non-negative.
V ( x) = x3 x= y 3
50.
3
x=y
V −1 ( x ) = 3 x V −1 ( x ) finds the length of a side of a cube given the volume. 51.
f ( x ) = 5 ( x − 32) 9 x = 5 ( y − 32) 9 9 x = y − 32 5
f ( x ) = 12 x x = 12 y x =y 12 −1 f ( x) = x 12 f −1 ( x ) converts x inches into feet.
52.
a. b.
9 x + 32 = y 5 f −1 ( x ) = 9 x + 32 5 f −1 ( x ) is used to convert x degrees Celsius to an equivalent Fahrenheit temperature.
S (96) = 3 (96) + 18 = $162 2 S ( x ) = 3 x + 18 2 x = 3 y + 18 2 x − 18 = 3 y 2 2 x − 12 = y 3 S −1 ( x ) = 2 x − 12 3 −1 2 S (399) = (399) − 12 = $254 3
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
242 s ( x ) = 2 x + 24 x = 2 y + 24 x − 24 = 2 y
53.
54.
1 x − 12 = y 2 s −1 ( x ) = 1 x − 12 2
K ( x ) = 1.3x − 4.7 x = 1.3 y − 4.7 x + 4.7 = 1.3 y x + 4.7 = y 1.3 K −1 ( x ) = x + 4.7 1.3
55.
E ( s ) = 0.05s + 2500 s = 0.05 y + 2500 s − 2500 = 0.05 y 1 s − 2500 = y 0.05 0.05 20s − 50,000 = y E −1 ( s ) = 20s − 50,000 The executive can use the inverse function to determine the value of the software that must be sold in order to achieve a given monthly income.
56.
No. It is not a one-to-one function. For a given cost, there is more than one weight that can be associated with that cost.
57.
a. b.
p(10) ≈ 0.12 = 12% ; p (30) ≈ 0.71 = 71% The graph of p, for 1 ≤ n ≤ 60 , is an increasing function. Thus p has an inverse that is a function.
c.
p −1 (0.223) represents the number of people required to be in the group for a 22.3% probability that at least two of the people will share a birthday.
a.
D f (13) = 2(13) − 1 = 25 O f (24) = 2(24) − 1 = 47 (space) f (36) = 2(36) − 1 = 71 Y f (34) = 2(34) − 1 = 67 O f (24) = 47 U f (30) = 2(30) − 1 = 59 R f (27) = 2(27) − 1 = 53 (space) f (36) = 71 H f (17) = 2(17) − 1 = 33 O f (24) = 47 M f (22) = 2(22) − 1 = 43 E f (14) = 2(14) − 1 = 27 W f (32) = 2(32) − 1 = 63 O f (24) = 47 R f (27) = 53 K f (20) = 2(20) − 1 = 39 The code is 25 47 71 67 47 59 53 71 33 47 43 27 63 47 53 39.
59.
c. 60.
58.
b.
a. b.
Answers will vary. No. L is not a one-to-one function.
f −1 (49) = 49 + 1 = 25 2 −1 33 f (33) = + 1 = 17 2 −1 47 + 1 = 24 f (47) = 2 f −1 (45) = 45 + 1 = 23 2 −1 27 + 1 = 14 f (27) = 2 f −1 (71) = 71 + 1 = 36 2
P H O N E (space)
f −1 (33) = 17 H f −1 (47) = 24 O f −1 (43) = 43 + 1 = 22 M 2 f −1 (27) = 14 E The message is PHONE HOME.
Answers will vary.
g ( x) = 2 x + 3 x = 2y + 3 x−3 = y 2 −1 g ( x) = x − 3 2
g −1 (59) = 59 − 3 = 28 2 −1 31 g (31) = − 3 = 14 2 −1 39 g (39) = − 3 = 18 2 g −1 (73) = 73 − 3 = 35 2
S E I Z
g −1 (31) = 14 E
The message is SEIZE THE DAY.
g −1 (75) = 75 − 3 = 36 (space) 2
g −1 (61) = 61 − 3 = 29 T 2 −1 37 g (37) = − 3 = 17 H 2 g −1 (31) = 14 E g −1 (75) = 36 (space) g −1 (29) = 29 − 3 = 13 D 2 g −1 (23) = 23 − 3 = 10 A 2 g −1 (71) = 71 − 3 = 34 Y 2
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Section 4.1
61.
63.
243
f (2) = 7, f (5) = 12, and f (4) = c. Because f is an increasing linear function, and 4 is between 2 and 5, then f (4) is between f (2) and f (5) . Thus, c is between 7 and 12. f is a linear function, therefore f −1 is a linear function.
62.
f (1) = 13, f (4) = 9, and f (3) = c. Because f is a decreasing linear function, and 3 is between 1 and 4, then f (3) is between f (1) and f (4) Thus, c is between 9 and 13.
64.
f is a linear function, therefore f −1 is a linear function.
f (2) = 3 ⇒ f −1 (3) = 2
f (5) = −1 ⇒ f −1 ( −1) = 5
f (5) = 9 ⇒ f −1 (9) = 5
f (9) = −3 ⇒ f −1 ( −3) = 9
Since 6 is between 3 and 9, f −1 (6) is between 2 and 5. 65.
Since −2 is between –1 and –3, f −1 ( −2) is between 5 and 9.
g is a linear function, therefore g −1 is a linear function.
66.
g is a linear function, therefore g −1 is a linear function.
g −1 (3) = 4 ⇒ g (4) = 3
g −1 ( −2) = 5 ⇒ g (5) = −2
g −1 (7) = 8 ⇒ g (8) = 7 Since 5 is between 4 and 8, g (5) is between 3 and 7.
g −1 (0) = −3 ⇒ g ( −3) = 0 Since 0 is between 5 and –3, then g (0) is between –2 and 0.
....................................................... 67.
68.
Connecting Concepts
f ( x ) = mx + b, m ≠ 0 y = mx + b x = my + b x − b = my x−b = y m b 1 −1 f ( x) = x − m m b⎞ 1 ⎛ and the y-intercept is ⎜ 0, − ⎟ . The slope is m⎠ m ⎝
f ( x ) = ax 2 + bx + c, a > 0, x > − b 2a
{
}
2⎫ ⎧ Domain of f is x x > − b , Range of f is ⎨ y y ≥ 4ac − b ⎬ . 4a ⎭ 2a ⎩
{
}
2⎫ ⎧ Domain of f −1 is ⎨ x x ≥ 4ac − b ⎬ , Range of f −1 is y y > − b . 2a 4 a ⎩ ⎭
y = ax 2 + bx + c x = ay 2 + by + c x − c = a y2 + b y a b2 + x − c = ⎛ y 2 + b y + b2 ⎞ Complete the square. ⎜ ⎟ a a 4a 2 4a 2 ⎠ ⎝
(
)
(
b2 + 4ax − 4ac = y + b 2a 4a 2
(
)
2
2 a + b + 4ax2 − 4ac = y + b 2a 4a
)
{
}
Choose the positive root, since the Range of f −1 is y y > − b . 2a
2 Thus f −1 ( x ) = − b + b + 4ax2 − 4ac 2a 4a
2 2 f −1 ( x ) = −b + b + 4ax − 4ac , a ≠ 0, x ≥ 4ac − b 2a 4a
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
244 69.
The reflection of f across the line given by y=x yields f. Thus f is its own inverse.
70.
The reflection of f across the line given by y=x yields f. Thus f is its own inverse.
71.
There is at most one point where each horizontal line intersects the graph of the function. The function is a one-to-one function.
72.
There is at most one point where each horizontal line intersects the graph of the function. The function is a one-to-one function.
73.
A horizontal line intersects the graph of the function at more than one point. Thus, the function is not a one-to-one function.
74.
A horizontal line intersects the graph of the function at more than one point. Thus, the function is not a one-to-one function.
....................................................... PS2. 3−4 = 1 = 1 = 1 34 3⋅3⋅3⋅3 81
PS1. 23 = 2⋅ 2⋅2 = 8
PS3.
PS5.
Prepare for Section 4.2
1
4+ 1 22 + 2−2 = 4 = 16 +1 = 17 2 2 8 8
PS4. 32 − 3−2 9 − 9 81−1 80 40 = = = = 2 2 18 18 9
f ( x) =10 x
PS6.
f (−1) =10−1 = 1 10 f (0) =100 =1
() f ( −1) = ( ) = 2 f (0) = ( ) =1 f (1) = ( ) = f (2) = ( ) = f ( x) =
1 2
1 2
f (1) =101 =10 f (2) =102 =100
1 2
1 2
x
1 2 0
−1
1
2
1 2
1 4
Section 4.2 1.
f (0) = 30 = 1
2.
f (4) = 34 = 81
4.
g (0) = 40 = 1 g ( −1) = 4 −1 = 1 4
7.
( ) =4 j (4) = ( 1 ) = 1 2 16 j ( −2) = 1 2
4
−2
5.
f (3) = 53 = 125 f ( −2) = 5−2 = 1 25
3.
g (3) = 103 = 1000
( ) = 94 h( −3) = ( 3 ) = 8 2 27 h(2) = 3 2
2
6.
−3
8.
g ( −2) = 10−2 = 1 100
( ) = 25 h(3) = ( 2 ) = 8 5 125 h( −1) = 2 5
3
( ) =4 j (5) = ( 1 ) = 1 4 1024 j ( −1) = 1 4
−1
5
Copyright © Houghton Mifflin Company. All rights reserved.
−1
Section 4.2
245
9.
f (3.2) = 23.2 ≈ 9.19
10.
f ( −1.5) = 3−1.5 ≈ 0.19
11.
g (2.2) = e2.2 ≈ 9.03
12.
g ( −1.3) = e −1.3 ≈ 0.27
13.
h( 2) = 5
2
14.
π h(π ) = 5π ≈ 0.11 h(π ) = 5 ≈ 0.11
15.
f ( x ) = 5 x is a basic exponential graph.
19.
f ( x ) = 10 x
22.
f ( x) = 5 2
≈ 9.74
g ( x ) = 1 + 5− x is the graph of f ( x ) reflected across the y-axis and moved up 1 unit. h( x ) = 5x + 3 is the graph of f ( x ) moved to the left 3 units. k ( x ) = 5x + 3 is the graph of f ( x ) moved up 3 units. a. k ( x ) 16.
b. g ( x )
c. h( x )
d. f ( x )
( ) is an exponential function with a base between 0 and 1. g ( x ) = ( 1 ) is the graph of f ( x ) reflected across the y-axis. 4 h( x ) = ( 1 ) is the graph of f ( x ) moved 2 units to the right. 4 k ( x ) = 3 ( 1 ) is the graph of f ( x ) stretched vertically by a factor of 3. 4 x
f ( x) = 1 4
−x
x −2
x
a. k ( x )
b. f ( x )
c. g ( x )
d. h( x )
17.
f ( x ) = 3x
18.
f ( x) = 4 x
20.
f ( x) = 6x
21.
f ( x) = 3 2
23.
f ( x) = 1 3
()
x
()
x
24.
()
f ( x) = 2 3
()
x
x
25.
Shift the graph of f vertically upward 2 units.
26.
Shift the graph of f vertically downward 3 units.
27.
Shift the graph of f horizontally to the right 2 units.
28.
Shift the graph of f horizontally to the left 5 units.
29.
Reflect the graph of f across the y-axis.
30.
Reflect the graph of f across the x-axis.
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
246 Stretch the graph of f vertically away from the x-axis by a factor of 2.
32.
33.
Reflect the graph of f across the y-axis, and then shift this graph vertically upward 2 units.
34.
Shift the graph of f horizontally 3 units to the right, and then shift this graph vertically upward 1 unit.
35.
Shift the graph of f horizontally 4 units to the right, and then reflect this graph across the x-axis.
36.
Reflect the graph of f across the y-axis, and then reflect this graph across the x-axis.
37.
Reflect the graph of f across the y-axis, and then shift this graph vertically upward 3 units.
38.
Shift the graph of f horizontally 2 units to the left, stretch this graph away from the x-axis by a factor of 3, and then shift this graph vertically downward 1 unit.
31.
39.
x −x f ( x) = 3 + 3 2
40.
x −x f ( x) = e + e 2
43.
f ( x) =
10 , with x ≥ 0 1 + 0.4e −0.5 x
Horizontal asymptote: y = 10 47.
a. b.
2
41.
f ( x ) = − e( x − 4)
44.
46.
f ( x) =
x −x f ( x) = e − e 2
No horizontal asymptote.
Horizontal asymptote: y = 0
No horizontal asymptote 45.
2
Horizontal asymptote: y = 0
No horizontal asymptote 42.
f ( x ) = 4 ⋅ 3− x
Shrink the graph of f vertically towards the x-axis by a factor of 1 .
f ( x ) = 0.5e − x
Horizontal asymptote: y = 0 10 , with x ≥ 0 1 + 1.5e −0.5 x
Horizontal asymptote: y = 10
f ( x ) = 1.353(1.9025) x ⇒ f (9) = 1.353(1.9025)9 ≈ 442 million connections
10 years after January 1, 1998 is in 2008.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.2
48.
a.
247
A(t ) = 200e −0.014t ⇒ f (45) = 200e −0.014(45) ≈ 107 mg
b.
It will take 99 minutes. 49.
50.
51.
a.
d ( p ) = 25 + 880e −0.18 p ⇒ d (8) = 25 + 880e −0.18(8) ≈ 233 items per month
b.
d ( p ) = 25 + 880e −0.18 p ⇒ d (18) = 25 + 880e −0.18(18) ≈ 59 items per month As p → ∞, d ( p ) → 25. The demand will approach 25 items per month.
a.
I (t ) = 24,000 − 22,000e −0.005t ⇒ I (10) = 24,000 − 22,000e −0.005(10) ≈ $3072.95
b.
I (t ) = 24,000 − 22,000e −0.005t ⇒ I (100) = 24,000 − 22,000e −0.005(100) ≈ $10,656.33 As t → ∞, I (t ) → 24,000. The monthly income will approach $24,000.
a.
P (t ) = 100 ⋅ 22t ⇒ P (3) = 100 ⋅ 22(3) = 100 ⋅ 26 = 100 ⋅ 64 = 6400 bacteria P (t ) = 100 ⋅ 22t ⇒ P (6) = 100 ⋅ 22(6) = 100 ⋅ 212 = 100 ⋅ 4096 = 409,600 bacteria
b.
11.6 hours 52.
a. b.
I ( x ) = 100e −1.5 x ⇒ I (1) = 100e −1.5(1) ≈ 22.3%
5 millimeters 53.
a. b.
P( x ) = (0.9) x ⇒ P(3.5) = (0.9)3.5 ≈ 69.2%
For a transparency of 45%, the UV index is 7.6. 54.
a. b. c.
3600 ⇒ P (0) = 3600 = 3600 = 3600 = 3600 = 3600 = 450 bass 8 1 + 7e −0.05t 1 + 7e −0.05(0) 1 + 7e0 1 + 7(1) 1 + 7 3600 P(t ) = 3600 ⇒ P(12) = ≈ 744 bass 1 + 7e −0.05t 1 + 7e −0.05(12) As t → ∞, P (t ) → 3600. The bass population will increase, approaching 3600. P(t ) =
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
248 55.
a.
n +1
B(n ) = 3
2
n +1
B(n ) = 3
2
− 3 ⇒ B(5) = 35+1 − 3 = 36 − 3 = 729 − 3 = 726 = 363 beneficiaries 2 2 2 2 − 3 ⇒ B(10) = 310+1 − 3 = 311 − 3 = 177147 − 3 = 177144 = 88,572 beneficiaries 2 2 2 2
b.
13 rounds 56.
a. b.
I ( x ) = 100e −0.95 x ⇒ I (2) = 100e −0.95(2) = 100e −1.9 ≈ 15.0%
about 0.73 foot 57.
a. b.
T (t ) = 65 + 115e −0.042t ⇒ T (10) = 65 + 115e −0.042(10) = 65 + 115e −0.42 ≈ 141o F
28.3 minutes 58.
a. b.
T (t ) = 75 + 95e −0.12t ⇒ T (2) = 75 + 95e −0.12(2) = 75 + 95e −0.24 ≈ 149.7o F
8.3 minutes
59.
c.
As t → ∞, T (t ) → 75. Therefore, room temperature is 75o F.
a. b.
f ( n ) = (27.5)2( n −1) /12 ⇒ f (40) = (27.5)2(40−1) /12 = (27.5)239 /12 = (27.5)23.25 ≈ 261.63 vibrations per second No. The function f ( n ) is not a linear function. Therefore, the graph of f ( n ) does not increase at a constant rate.
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Section 4.2
249
....................................................... 60.
x −x cosh( x ) = e + e is an even function. That is, prove 2 cosh( − x ) = cosh(x ).
Connecting Concepts 61.
x −x cosh( x ) = e + e 2
Proof:
x −x sinh( x ) = e − e is an odd function. That is, prove 2 sinh( − x ) = −sinh(x ). x −x sinh( x ) = e − e 2
Proof:
−x x cosh( − x ) = e + e 2
−x x sinh( − x ) = e − e 2
cosh( − x ) =
−x x sinh( − x ) = − e + e 2
( e x − e− x )
2 cosh( − x ) = F ( x )
sinh( − x ) =
( e x − e− x )
2 sinh( − x ) = − F ( x ) 62.
63.
64.
65.
domain: (−∞, ∞)
domain: (−∞, ∞)
66.
67.
domain: 68.
domain:
Let f ( x ) = 2 x and g ( x ) = x 2 + 4 . Then h( x ) = 2
70.
( −∞, 0]
( x2 +4)
= f [ x 2 + 4] = f [ g ( x ) ] = ( f o g ) ( x ) .
69.
[ 0, ∞ )
Let f ( x ) = e x and g ( x ) = 2 x − 5 . Then h( x ) = e
( 2 x −5 )
= f [2 x − 5] = f [ g ( x )] = ( f o g ) ( x ) .
x −x By definition the average of two numbers is their sum divided by 2. The expression e + e shows that f is the average of 2
g ( x ) = e x and h( x ) = e − x .
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Chapter 4: Exponential and Logarithmic Functions
250
....................................................... PS2. 3− x = 1 = 1 3x 27 1 = 1 3x 33 x=3
PS1. 2 x = 16 x
2 =2 x=4
PS4.
Prepare for Section 4.3
4
f ( x) = 2 x x+3 2y x= y+3 xy + 3x = 2 y 3x = 2 y − xy = y (2 − x ) 3x = y 2− x −1 f ( x ) = 3x 2−x
PS5.
PS3. x 4 = 625 x 4 = 54 x=5
g ( x) = x − 2
PS6. The domain is the set of all positive real numbers.
x−2≥0 x≥2 The domain is {x | x ≥ 2} .
Section 4.3 1.
1 = log10 ⇒ 101 = 10
2.
4 = log10,000 ⇒ 104 = 10,000
3.
2 = log8 64 ⇒ 82 = 64
4.
3 = log 4 64 ⇒ 43 = 64
5.
0 = log7 x ⇒ 70 = x
6.
−4 = log3 1 ⇒ 3−4 = 1 81 81
7.
ln x = 4 ⇒ e4 = x
8.
ln e2 = 2 ⇒ e2 = e2
9.
ln1 = 0 ⇒ e0 = 1
10.
ln x = −3 ⇒ e −3 = x
11.
2 = log(3x + 1) ⇒ 102 = 3x + 1
12.
1 = ln ⎛ x + 1 ⎞ ⇒ e1/ 3 = x + 1 ⎜ 2 ⎟ 3 x2 ⎝ x ⎠
13.
32 = 9 ⇒ log3 9 = 2
14.
53 = 125 ⇒ log5 125 = 3
15.
4 −2 = 1 ⇒ log 4 1 = −2 16 16
16.
100 = 1 ⇒ log1 = 0
17.
b x = y ⇒ logb y = x
18.
2 x = y ⇒ log 2 y = x
19.
y = e x ⇒ ln y = x
20.
51 = 5 ⇒ log5 5 = 1
21.
100 = 102 ⇒ log100 = 2
22.
2 −4 = 1 ⇒ log 2 1 = −4 16 16
23.
e2 = x + 5 ⇒ 2 = ln( x + 5)
24.
3x = 47 ⇒ log3 47 = x
25.
log 4 16 = 2 because 42 = 16
27.
−3
3
= ⎛⎜ 2 ⎞⎟ = 8 27 ⎝ 3⎠
26.
log3/ 2 8 = −3 because ⎛⎜ 3 ⎞⎟ 27 ⎝2⎠
log3 1 = −5 because 3−5 = ⎛⎜ 1 ⎞⎟ = 1 243 243 ⎝ 3⎠
28.
logb 1 = 0 because b0 = 1
29.
ln e3 = 3 because e3 = e3
30.
logb b = 1 because b1 = b
31.
log 1 = −2 because 10−2 = 12 = 1 100 100 10
32.
log1,000,000 = 6 because 106 = 1,000,000
5
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Section 4.3
251
()
−4
33.
log0.5 16 = log1/ 2 16 = −4 because 1 2
34.
log0.3 100 = log3/10 100 = −2 because ⎛⎜ 3 ⎞⎟ 9 9 ⎝ 10 ⎠
35.
4log1000 = 12 ⇒ log10004 = 12 because 1012 = (103 ) = (1000 ) 4
37.
= 24 = 16 −2
2
= ⎛⎜ 10 ⎞⎟ = 100 9 ⎝ 3⎠ 36.
log5 1252 = 6 because 56 = 125
38.
3log11 161,051 = 15 ⇒ log11161,0513 = 15
4
2 log7 2401 = 8 ⇒ log7 24012 = 8
because 1115 = (115 ) = (161,051)
2 because 78 = ( 74 ) = ( 2401)
3
2
39.
log3 5 9 = 2 ⇒ log3 91/5 = 2 5 5 because 32/5 = ( 32 )
1/5
41.
y = log 4 x x=4
46.
= (169 )
42.
44.
y
47.
y
45.
2/7
2/7 = ( 343)
y = log12 x x = 12 y
y
y = log1/ 2 x x = (1/ 2)
y = log5 / 2 x x = (5 / 2)
y = log 6 x x=6
1/ 3
= ( 36 )
2log7 7 343 = 6 ⇒ log7 3432/7 = 6 7 7 because 76/7 = ( 73 )
5/ 3
y
y = log8 x x=8
49.
1/3
5log13 3 169 = 10 ⇒ log131695/3 = 10 3 3 5/3
log6 3 36 = 2 ⇒ log6 361/3 = 2 3 3 because 62/3 = ( 62 )
1/ 5
= (9)
because 1310/3 = (132 )
43.
40.
3
48.
y
y = log1/ 4 x x = (1/ 4) y
50.
y = log 7 / 3 x x = (7 / 3) y
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Chapter 4: Exponential and Logarithmic Functions
252 51.
f ( x ) = log5 ( x − 3)
52.
x −3> 0 x>3 The domain is (3, ∞ ).
54.
H ( x ) = log1/ 4 ( x 2 + 1)
55.
P( x ) = ln( x 2 − 4)
x > −1 True for all real numbers. The domain is ( −∞, ∞ ).
⎛ 2 ⎞ h ( x ) = ln ⎜ x ⎟ ⎝ x −4⎠
58.
x3 − x > 0
60.
( x 2 + 7 x + 10) > 0 ( x + 5)( x + 2) > 0 Critical values are –5 and –2. Product is positive. (−∞, −5) ∪ ( −2, ∞)
62.
4x − 8 = 0 4x = 8 x=2 The domain is ( −∞, 2) ∪ (2, ∞ )
64.
x−4=0 x=4 The domain is ( −∞, 4) ∪ (4, ∞ )
x ( x − 1) > 0 x ( x + 1)( x − 1) > 0 Critical values are 0, −1 and 1. Product is positive. −1 < x < 0 or x > 1 (−1,0) ∪ (1, ∞)
(
)
The domain is 11 , ∞ . 2 3x − 7 > 0 3x > 7 x>7 3
x4 − x2 > 0 x 2 ( x + 1)( x − 1) > 0 Critical values are 0, −1 and 1 Product is positive. x < −1 or x > 1 ( −∞, −1) ∪ (1, ∞)
2
63.
J ( x ) = ln ⎜⎛ x − 3 ⎟⎞ ⎝ x ⎠ x−3 >0 x The critical values are 3 and 0. The quotient is positive. x < 0 or x > 3 The domain is ( −∞, 0) ∪ (3, ∞).
x 2 ( x 2 − 1) > 0
x2 > 0 x−4 The critical values are 0 and 4. The quotient is positive. x>4 The domain is (4, ∞ ).
2 x − 11 > 0 2 x > 11 x > 11 2
56.
( x + 2)( x − 2) > 0 The critical values are –2 and 2. The product is positive. The domain is ( −∞, − 2) ∪ (2, ∞ ).
2
61.
k ( x ) = log 2 / 3 (11 − x ) 11 − x > 0 − x > −11 x < 11 The domain is ( −∞, 11).
x −4>0
x +1> 0
59.
53.
2
2
57.
k ( x ) = log 4 (5 − x ) 5− x > 0 − x > −5 x<5 The domain is ( −∞, 5).
( )
The domain is 7 , ∞ . 3 65.
Shift 3 units to the right.
66.
Shift 3 units to the left.
67.
Shift 2 units up.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.3
253
68.
Shift 4 units down.
71.
Shift 4 units to the right and 1 unit up
73.
The graph of f ( x ) = log5 ( x − 2) is the graph of y = log5 x shifted 2 units to the right.
69.
Shift 3 units up.
70.
72.
Shift 2 units up.
Shift 3 units to the right and 1 unit down
The graph of g ( x ) = 2 + log5 x is the graph of y = log5 x shifted 2 units up. The graph of h( x ) = log5 ( − x ) is the graph of y = log5 x reflected across the y-axis. The graph of k ( x ) = − log5 ( x + 3) is the graph of y = log5 x reflected across the x-axis and shifted left 3 units. a. k ( x ) 74.
The graph of The graph of The graph of The graph of a. k ( x )
b. f ( x )
c. g ( x )
d. h( x )
f ( x ) = ln x + 3 is the graph of y = ln x shifted 3 units up. g ( x ) = ln( x − 3) is the graph of y = ln x shifted 3 units to the right. h( x ) = ln(3 − x ) is the graph of y = ln x shifted 3 units to the left and reflected across the y-axis. k ( x ) = − log5 ( x + 3) is the graph of y = log5 x reflected across the x-axis and shifted left 3 units. b. h( x )
c. g ( x )
d. f ( x )
75.
76.
77.
78.
79.
80.
81.
82.
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
254 83.
85.
84.
a. b.
r (t ) = 0.69607 + 0.60781ln t ⇒ r (9) = 0.69607 + 0.60781ln 9 ≈ 2.0%
45 months 86.
a.
S (t ) = 5 + 29ln(t + 1) S (0) = 5 + 29ln(0 + 1) = 5 words per minute S (3) = 5 + 29ln(3 + 1) ≈ 45.2 words per minute
b.
6.9 months 87.
N ( x ) = 2750 + 180ln ⎛⎜ x + 1⎞⎟ ⎝ 1000 ⎠
a.
b. 88.
20,000 ⎞ + 1⎟ = 2750 + 180ln ( 21) ≈ 3298 units N (20,000) = 2750 + 180ln ⎛⎜ ⎝ 1000 ⎠ 40,000 ⎞ ⎛ + 1⎟ = 2750 + 180ln ( 41) ≈ 3418 units N (40,000) = 2750 + 180ln ⎜ ⎝ 1000 ⎠ 60,000 + 1⎞⎟ = 2750 + 180ln ( 61) ≈ 3490 units N (60,000) = 2750 + 180ln ⎛⎜ ⎝ 1000 ⎠ N (0) = 2750 + 180ln ⎛⎜ 0 + 1⎞⎟ = 2750 + 180ln (1) = 2750 + 180(0) = 2750 + 0 = 2750 units ⎝ 1000 ⎠
BSA = 0.0003207 ⋅ H 0.3 ⋅ W (0.7285−0.0188log W ) BSA = 0.0003207 ⋅ (162.56)0.3 ⋅ (49,895.2)(0.7285−0.0188log 49,895.2) ≈ 1.50 square meters
89.
BSA = 0.0003207 ⋅ H 0.3 ⋅ W (0.7285−0.0188log W ) BSA = 0.0003207 ⋅ (185.42)0.3 ⋅ (81,646.6) (0.7285−0.0188log81,646.6) ≈ 2.05 square meters
90.
M ( x ) = −2.51log x + 1, 0 < x ≤ 1 a. b. c. d.
( ) ( )
( ) ( )
M 1 = −2.51log 1 + 1 = 3.51 10 10 M 1 = −2.51log 1 + 1 ≈ 7.53 400 400 The star with an apparent magnitude of 12 is brighter than a star with an apparent magnitude of 15. M(x) is a decreasing function.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.3
255
91.
N = int( x log b) + 1 a. N = int(10log 2) + 1 = 3 + 1 = 4 digits b. N = int(200log 3) + 1 = 95 + 1 = 96 digits c. N = int(4005log 7) + 1 = 3384 + 1 = 3385 digits d. N = int(2,0996,001log 2) + 1 = 6,320,429 + 1 = 6,320,430 digits
92.
a.
99 = 387,420,489 9(9
9
)
= 9387,420,489 N = int(387,420,489log 9) + 1 = 369,693,100 digits
369,693,100 digits ×
b.
1 page × 1 ream ≈ 739.4 reams 1000 digits 500 pages
.......................................................
Connecting Concepts
93.
f (x) and g (x) are inverse functions
94.
f (x) and g (x) are inverse functions
95.
The domain of the inverse is the range of the function. Range of f: {y | –1 < y < 1}. The domain of the function is the range of the inverse. Range of g: all real numbers.
96.
Domain: all real numbers; range {y | 0 < y < 1}.
....................................................... PS1. log 3 + log 2 ≈ 0.77815
PS2. ln8 − ln 3 ≈ 0.98083
log 6 ≈ 0.77815
ln 8 ≈ 0.98083 3
PS4.
2 ln 5 ≈ 3.21888 ln ( 5
2
) ≈ 3.21888
()
PS5.
ln 5 ≈ 1.60944 log5 ≈ 1.60944 log e
Prepare for Section 4.4 3log 4 ≈ 1.80618
PS3.
log(43 ) ≈ 1.80618
PS6.
log8 ≈ 0.90309 ln8 ≈ 0.90309 ln10
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
256
Section 4.4 1.
logb ( xyz ) = logb x + logb y + logb z
3 ln z = ln z 3 − ln x − ln y xy
2.
= 3ln z − 1 ln x − 1 ln y 2 2 3.
5.
ln x4 = ln x − ln z 4 z = ln x − 4 ln z
log 2
y
4.
log5
xy 2 z4
= log5 x + log5 y 2 − log5 z 4 = log5 x + 2 log5 y − 4 log5 z
x = log x − log y 3 2 2
(
)
logb x 3 y = logb x + logb y1/ 3
6.
3
= logb x + 1 logb y 3
= log 2 x1/ 2 − log 2 y 3 = 1 log 2 x − 3log 2 y 2 1/ 2
1/ 2 1/ 2 = log7 x z2 = log7 x1/ 2 + log7 z1/ 2 − log7 y 2 = 1 log7 x + 1 log7 z − 2log7 y 2 2 y
xz = log ( xz ) 7 y2 y2
7.
log7
8.
ln 3 x 2 y = ln ( x 2 y1/ 2 )
9.
ln ( e2 z ) = ln e2 + ln z = 2ln e + ln z = 2 + ln z
10.
ln ( x1/ 2 y 2 / 3 ) = ln x1/ 2 + ln y 2 / 3 = 1 ln x + 2 ln y 2 3
11.
⎛ 3 ⎞ log 4 ⎜ z 3 ⎟ = log 4 z1/ 3 − log 4 42 − log 4 y 3 = 1 log 4 z − 2 log4 4 − 3log4 y = 1 log4 z − 2 − 3log4 y 3 3 ⎝ 16 y ⎠
12.
4⎞ ⎛ log5 ⎜ xz ⎟ = log5 x1/ 2 + log5 z 4 − log5 53 = 1 log5 x + 4log5 z − 3log5 5 = 1 log5 x + 4log5 z − 3 2 2 ⎝ 125 ⎠
13.
log x z = log ( xz1/ 2 )
14.
⎛3 2 ⎞ ln ⎜ x2 ⎟ = ln x 2 / 3 − ln z 2 = 2 ln x − 2ln z 3 ⎝ z ⎠
15.
ln
16.
⎛ 2 ⎞ ln ⎜ x −3z ⎟ = ln x 2 + ln z1/ 2 − ln y −3 = 2ln x + 1 ln z + 3ln y 2 y ⎝ ⎠
17.
log( x + 5) + 2log x = log( x + 5) + log x 2 = log[ x 2 ( x + 5)]
1/ 3
1/ 2
= ln ( x 2 / 3 y1/ 6 ) = ln x 2 / 3 + ln y1/ 6 = 2 ln x + 1 ln y 3 6
= log x1/ 2 z1/ 4 = 1 log x + 1 log z 2 4
( 3 z e ) = ln ( ze1/ 2 )1/ 3 = ln z1/ 3e1/ 6 = ln z1/ 3 + ln e1/ 6 = 1 ln z + 1 ln e = 1 ln z + 1 3
6
3
6
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Section 4.4
257
18.
3 4 3log 2 t − 1 log 2 u + 4 log 2 v = log 2 t 3 − log 2 u1/ 3 + log 2 v 4 = log2 t 3 + log 2 3 u + log2 v 4 = log 2 t3 v 3 u
19.
ln( x 2 − y 2 ) − ln( x − y ) = ln
20.
1 log ( x + 5) − 3log y = log ( x + 5)1/ 2 − log y 3 = log x + 5 − log y 3 = log 8 8 8 8 8 8 2 8
21.
3log x + 1 log y + log( x + 1) = log x 3 + log y1/ 3 + log( x + 1) = log x 3 + log 3 y + log( x + 1) = log ⎡ x 3 ⋅ 3 y ( x + 1) ⎤ ⎣ ⎦ 3
22.
ln( xz ) − ln( x y ) + 2 ln
23.
⎛ xy 2 ⎞ log ( xy 2 ) − log z = log ⎜ ⎟ ⎝ z ⎠
24.
⎛ y1/ 2 z ⎞ ln ( y1/ 2 z ) − ln z1/ 2 = ln ⎜ 1/ 2 ⎟ = ln ( y1/ 2 z1/ 2 ) or ln yz ⎝ z ⎠
25.
⎛ x2 y4 ⎞ 2 ( log6 x + log6 y 2 ) − log6 ( x + 2) = log6 x 2 + log6 y 4 − log6 ( x + 2) = log6 ⎜ ⎟ ⎝ x+2⎠
26.
2 1 log x − log y + 2log ( x + 2) = log x1/ 2 − log y + log ( x + 2)2 = log ⎡ x ( x + 2) ⎤ ⎥ 3 3 3 3 3 3 3⎢ 2 y ⎣ ⎦
27.
⎡ ( x + 4) 4 ⎤ 2ln( x + 4) − ln x − ln ( x 2 − 3) = ln( x + 4)4 − ln x − ln ( x 2 − 3) = ln ⎢ ⎥ 2 ⎣ x ( x − 3) ⎦
28.
⎛ 3xy ⎞ ⎛ 3y ⎞ log(3x ) − (2 log x − log y ) = log(3 x ) − log x 2 + log y = log ⎜ 2 ⎟ = log ⎜ ⎟ ⎝ x ⎠ ⎝ x ⎠
29.
⎡ (2 x + 5) w ⎤ ln(2 x + 5) − ln y − 2ln z + 1 ln w = ln(2 x + 5) − ln y − ln z 2 + ln w1/ 2 = ln ⎢ ⎥ 2 yz 2 ⎣ ⎦
30.
x ( y + 3)( y + 2) ⎤ logb x + logb ( y + 3) + logb ( y + 2) − logb ( y 2 + 5 y + 6) = logb ⎡⎢ ⎥ = logb x ⎣ ( y + 2)( y + 3) ⎦
31.
⎡ ( x + 3)( x − 3) y 3 ⎤ ⎡ ( x + 3) y 3 ⎤ ln ( x 2 − 9 ) − 2ln( x − 3) + 3ln y = ln( x + 3)( x − 3) − ln( x − 3)2 + ln y 3 = ln ⎢ ln = ⎥ ⎢ ⎥ ⎣ x−3 ⎦ ( x − 3)2 ⎣ ⎦
32.
( x + 3)( x + 4) ⎤ x+3 logb ( x 2 + 7 x + 12 ) − 2logb ( x + 4) = logb ( x + 3)( x + 4) − logb ( x + 4)2 = logb ⎡⎢ ⎥ = logb x + 4 2 + ( x 4) ⎣ ⎦
33.
log7 20 =
x2 − y2 ( x + y )( x − y ) = ln = ln( x + y ) x− y x− y
y ⎛ = ln( xz ) − ln( x y ) + ln ⎜ z ⎝
x+5 y3
2 ⎛ y⎞ y2 ⎞ y2 y 2 −1/ 2 y 3/ 2 = ln ⎜ xz ⋅ 2 ⎟ = ln = ln = ln ⎟ ⎜ (x y ) z ⎟ z⎠ z z z y ⎝ ⎠
( )
log 20 ≈ 1.5395 log 7
34.
log5 37 =
log 37 ≈ 2.2436 log 5
35.
log11 8 =
Copyright © Houghton Mifflin Company. All rights reserved.
log8 ≈ 0.8672 log11
Chapter 4: Exponential and Logarithmic Functions
258 37.
log 1 3 ≈ −0.6131 log6 1 = 3 log 6
log 17 ≈ 0.6447 log 9
40.
log4 7 =
log5.5 ≈ 3.1035 log 3
43.
logπ e =
46.
g ( x ) = log8 (5 − x ) =
49.
h( x ) = log3 ( x − 3)2
log 22 ≈ 0.7901 log 50
36.
log50 22 =
39.
log9 17 =
42.
log
45.
f ( x ) = log 4 x =
48.
t ( x ) = log9 (5 − x ) =
3
5.5 =
log x log 4
log(5 − x ) log 9
log 7 ≈ 0.7018 log 4
log e ≈ 0.8735 log π log(5 − x ) log8
38.
log 7 8 ≈ −0.1215 log3 7 = 8 log 3
41.
log
44.
logπ 15 =
47.
g ( x ) = log8 ( x − 3) =
50.
J ( x ) = log12 ( − x ) =
log( x − 3)2 log 3 2log( x − 3) = log 3
2
17 =
log17 ≈ 8.1749 log 2 log 15 ≈ 1.1828 log π log( x − 3) log8
log( − x ) log12
=
log x − 2
51.
F ( x ) = − log5 x − 2 = −
53.
False. log10 + log10 = 1 + 1 = 2 but log(10 + 10) = log 20 ≠ 2 .
55.
True.
58.
False. log 100 = log10 = 1 10 log100 2 but = =2 log10 1
log5
log x − 8 log 2
52.
n( x ) = log 2 x − 8 =
54.
False. log(10 ⋅ 10) = log(102 ) = 2 but log10 ⋅ log10 = 1 ⋅ 1 = 1
56.
False. log10 ⋅ log10 = 1 ⋅ 1 = 1 but log10 + log10 = 1 + 1 = 2
57.
False. log100 − log10 = 2 − 1 = 1 but log(100 − 10) = log 90 ≠ 1
59.
False.
log100 2 = =2 log10 1 but log100 − log10 = 2 − 1
60.
True.
=1 Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.4
259
61.
False. (log10)2 = 12 = 1 but 2log10 = 2(1) = 2
63.
log3 5 ⋅ log5 7 ⋅ log7 9 =
64.
log5 20 ⋅ log20 60 ⋅ log60 100 ⋅ log100 125 =
=
log 20 log 60 log100 log125 ⋅ ⋅ ⋅ log5 log 20 log 60 log100 log 20 log5
⋅
log 60 log 20
⋅
log100 log 60
⋅
ln 500501 = 501ln 500 ≈ 3113.52
66.
n (log S − log S ) f 0
Sn = S0 ⋅ 10 N
68.
1 (log 500,000− log1,000,000)
S1 = 250,000 ⋅ 10 4
= 870,551
= 198,818
2 (log 500,000 − log1,000,000) = 1,000,000 ⋅ 10 5
2 (log100,000 − log 250,000)
S2 = 250,000 ⋅ 10 4
= 757,858
= 158,114
3 (log 500,000− log1,000,000) = 1,000,000 ⋅ 10 5
3 (log100,000 − log 250,000)
S3 = 250,000 ⋅ 10 4
= 659,754 S4
= 125,743
4 (log 500,000 − log1,000,000) = 1,000,000 ⋅ 10 5
4 (log100,000 − log 250,000)
S4 = 250,000 ⋅ 10 4
= 574,349 S5
n (log S − log S ) f 0
Sn = S0 ⋅ 10 N
1 (log100,000 − log 250,000)
S1 = 1,000,000 ⋅ 10 5
S3
⎛ ⎞ ln ⎜ 1300 ⎟ = ln 50−300 = −300ln 50 ≈ −1174 ⎝ 50 ⎠ ⎛ ⎞ ln ⎜ 1233 ⎟ = ln151−233 = −233ln151 ≈ −1169 ⎝ 151 ⎠ 1 is smaller. 50300
ln 500501 is larger.
S2
log125 log125 log53 = = log5 log5 log100
3log5 3 log5 =3 = log5 log5
ln 506500 = 500ln 506 ≈ 3113.27
67.
True.
log5 log 7 log 9 log5 log 7 log 9 log 9 log 32 2log 3 2 log 3 ⋅ ⋅ = ⋅ ⋅ = = = = =2 log 3 log5 log 7 log 3 log5 log 7 log 3 log 3 log 3 log 3
=
65.
62.
= 100,000 The scales are 1:198,818; 1:158,114; 1:125,743; 1:100,000.
5 (log 500,000− log1,000,000) = 1,000,000 ⋅ 10 5
= 500,000 The scales are 1:870,551; 1:757,858; 1:659,754; 1:574,349; 1:500,000. 69.
pH = − log[H + ]
70.
pH = − log[1.26 × 10−3 ] pH = 2.9 2.9 < 7; vinegar is an acid.
pH = − log[3.97 × 10−11 ] pH = 10.4 10.4 > 7; milk of magnesia is a base 71.
pH = − log[H + ]
72.
9.5 = − log[H + ] 10−9.5 = 10log[H
pH = − log[H + ] 5.6 = − log[H + ]
−9.5 = log[H + ] +
pH = − log[H + ]
log[H + ] = −5.6 10log[H ] = 10−5.6 +
]
[H + ] = 3.16 × 10−10 mole per liter
[H + ] = 2.51 × 10−6 mole per liter
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
260 73.
dB( I ) = 10log ⎛⎜ I ⎞⎟ ⎝ I0 ⎠ a.
⎛ 1.58 × 108 ⋅ I 0 ⎞ dB(1.58 × 108 ⋅ I 0 ) = 10log ⎜ ⎟⎟ ⎜ I0 ⎝ ⎠
⎛ 10,800 ⋅ I 0 ⎞ dB(10,800 ⋅ I 0 ) = 10log ⎜ ⎟ I0 ⎝ ⎠ = 10log(10,800)
b.
= 10log(1.58 × 108 )
c.
≈ 40.3 decibels
≈ 82.0 decibels ⎛ 3.16 × 1011 ⋅ I 0 ⎞ dB(3.16 × 1011 ⋅ I 0 ) = 10log ⎜ ⎟⎟ ⎜ I0 ⎝ ⎠
⎛ 1.58 × 1015 ⋅ I 0 ⎞ dB(1.58 × 1015 ⋅ I 0 ) = 10log ⎜ ⎟⎟ ⎜ I0 ⎝ ⎠
d.
= 10log(3.16 × 1011 ) ≈ 115.0 decibels
74.
= 10log(1.58 × 1015 ) ≈ 152.0 decibels
⎛ I pain ⎞ 125 = 10log ⎜ ⎟ ⎝ I0 ⎠ ⎛ I pain ⎞ 12.5 = log ⎜ ⎟ ⎝ I0 ⎠ I pain 1012.5 = I0
dB( I ) = 10log ⎛⎜ I ⎞⎟ ⎝ I0 ⎠ ⎛I ⎞ 175 = 10log ⎜ Bronco ⎟ I 0 ⎝ ⎠ ⎛I ⎞ 17.5 = log ⎜ Bronco ⎟ I 0 ⎝ ⎠ I 1017.5 = Bronco I0
I Bronco 1017.5 ⋅ I 0 = 12.5 I pain 10 ⋅ I 0 17.5
= 1012.5 10 = 1017.5−12.5 = 105 = 100,000 times more intense
1012.5 ⋅ I 0 = I pain
1017.5 ⋅ I 0 = I Bronco 75.
⎛I ⎞ 110 = 10log ⎜ 110 ⎟ ⎝ I0 ⎠
⎛ ⎞ dB( I ) = 10log ⎜ I ⎟ ⎝ I0 ⎠
⎛I ⎞ 11 = log ⎜ 110 ⎟ ⎝ I0 ⎠ I 1011 = 110 I0
⎛I ⎞ 120 = 10log ⎜ 120 ⎟ ⎝ I0 ⎠ ⎛I ⎞ 12 = log ⎜ 120 ⎟ ⎝ I0 ⎠ I 1012 = 120 I0
I120 1012 ⋅ I 0 = I110 1011 ⋅ I 0 12
= 1011 = 1012 −11 = 101 10 = 10 times more intense
1011 ⋅ I 0 = I110
1012 ⋅ I 0 = I120 10
10
76.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ dB(2 I ) − dB ( I ) = 10log ⎜ 2 I ⎟ − 10log ⎜ I ⎟ = log ⎜ 2 I ⎟ I I 0 0 ⎝ ⎠ ⎝ ⎠ ⎝ I0 ⎠
⎛ ⎞ − log ⎜ I ⎟ ⎝ I0 ⎠
⎡ (2)10 ( I )10 ( I )10 ⎡ (2 I )10 ( I 0 )10 ⎤ 0 ⎢ = log ⎢ ⋅ = ⋅ log ⎥ 10 10 10 ⎢ ( I ) ⎦⎥ ( I )10 ⎣⎢ ( I 0 ) ⎣ ( I0 ) 77.
10
⎛ 2I ⎞ ⎜ ⎟ I = log ⎝ 0 ⎠
10
⎛ I ⎞ ⎜ ⎟ ⎝ I0 ⎠
⎡⎛ ⎞10 ⎛ I ⎞10 ⎤ = log ⎢⎜ 2 I ⎟ ⋅ ⎜ 0 ⎟ ⎥ ⎢⎝ I 0 ⎠ ⎝ I ⎠ ⎥ ⎣ ⎦
⎤ ⎥ = log 210 = 10log 2 ≈ 3.0103 decibels ⎥ ⎦
⎛ 100,000 I 0 ⎞ 5 M = log ⎜ ⎟ = log100,000 = log10 = 5 I0 ⎝ ⎠
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Section 4.4
78.
261
⎛ ⎞ M = log ⎜ I ⎟ ⎝ I0 ⎠
79.
⎛ 398,107,000 I 0 ⎞ M = log ⎜ ⎟ I0 ⎝ ⎠ = log 398,107,000 ≈ 8.6
81.
⎛ ⎞ log ⎜ I ⎟ = M ⎝ I0 ⎠
80.
⎛ ⎞ log ⎜ I ⎟ = 6.5 ⎝ I0 ⎠ I = 106.5 I0
⎛ ⎞ l og ⎜ I ⎟ = M ⎝ I0 ⎠ ⎛ ⎞ log ⎜ I ⎟ = 9.5 ⎝ I0 ⎠ I = 109.5 I0
I = 106.5 I 0
I = 109.5 I 0
I ≈ 3,162,277.7 I 0
I = 3,162, 277,660 I 0
⎛ ⎞ M = log ⎜ I ⎟ ⎝ I0 ⎠ ⎛I ⎞ ⎛I ⎞ I M 5 = log ⎜ 5 ⎟ ⇒ 5 = log ⎜ 5 ⎟ ⇒ 105 = 5 ⇒ 105 I 0 = I 5 I0 ⎝ I0 ⎠ ⎝ I0 ⎠ ⎛I ⎞ ⎛I ⎞ I M 3 = log ⎜ 3 ⎟ ⇒ 3 = log ⎜ 3 ⎟ ⇒ 103 = 3 ⇒ 103 I 0 = I 3 I I I 0 ⎝ 0⎠ ⎝ 0⎠ I 5 105 I 0 105 = = = 105−3 = 102 = 100 to 1 I 3 103 I 0 103
• short cut: begin with this line
82.
109.5 = 109.5−8.3 = 101.2 ≈ 15.8 times more intense 108.3
83.
108.9 = 108.9 −7.1 = 101.8 to 1 ≈ 63 to 1 1 107.1
84.
108.2 = 108.2 −6.9 = 101.3 to 1 ≈ 20 to 1 1 106.9
85.
M = log A + 3log8t − 2.92 = log18 + 3log[8(31)] − 2.92 ≈ 5.5
86.
M = log A + 3log8t − 2.92 = log 26 + 3log[8(17)] − 2.92 ≈ 4.9
87.
Let r = logb M and s = logb N .
88.
Let x = logb M .
Then M = b r and N = b s . Consider the quotient of M and N M N M N M logb N M logb N
89.
a. c.
Then M = b x . M = bx
r = bs b
(M ) p = (bx )
= br − s
p
M p = b xp logb M p = xp
=r−s
logb M p = p logb M
= logb M − logb N
b. M ≈4 M ≈6 When t = 40, M = log A + 3log8t − 2.92 = log50 + 3log[8(40)] − 2.92 ≈ 6.3 When t = 30, M = log A + 3log8t − 2.92 = log1 + 3log[8(30)] − 2.92 ≈ 4.2 The results from parts a. and b. are close to the magnitudes of 6.3 and 4.2 produced by the amplitude time difference formula.
....................................................... PS1. 36 = 729 ⇒ log3 729 = 6
PS2. log5 625 = 4 ⇒ 54 = 625
Prepare for Section 4.5 PS3. a x + 2 = b ⇒ log a b = x + 2
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Chapter 4: Exponential and Logarithmic Functions
262 PS4.
4a = 7bx + 2cx 7bx + 2cx = 4a x (7b + 2c) = 4a x = 4a 7b + 2c
PS5.
165 = 300 1 + 12 x 165(1 + 12 x ) = 300
PS6.
165 + 1980 x = 300 1980 x = 135 x = 135 = 3 1980 44
A = 100 + x 100 − x A(100 − x ) = 100 + x 100 A − Ax = 100 + x 100 A − 100 = Ax + x 100( A − 1) = x ( A + 1) 100( A − 1) x= A +1
Section 4.5 1.
2 x = 64 x
2 =2 x=6
5.
2.
6
x
13.
3.
5
3 =3 x=5
25 x + 3 = 1 8
6.
25 x + 3 = 2 −3 5 x + 3 = −3 5 x = −6 x =−6 5 9.
3x = 243
34 x −7 = 1 9
7.
34 x −7 = 3−2 4 x − 7 = −2 4x = 5 x=5 4
5 x = 70
10.
6 x = 50
49 x = 1 343
9x = 1 243
4.
7 2 x = 7 −3 2 x = −3
32 x = 3−5 2 x = −5
x=−3 2
x=−5 2
x
⎛2⎞ = 8 ⎜ ⎟ 125 ⎝ 5⎠ x
⎛2⎞ =⎛2⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 5⎠ ⎝ 5⎠ x=3
8.
x
3
⎛2⎞ = ⎛2⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 5⎠ ⎝ 5⎠ x = −2
3− x = 120
11.
x
⎛ 2 ⎞ = 25 ⎜ ⎟ 4 ⎝ 5⎠ −2
7 − x = 63
12.
log(5 x ) = log 70
log(6 x ) = log 50
log(3− x ) = log120
log(7 − x ) = log 63
x log5 = log 70 log 70 x= log5
x log 6 = log 50 log 50 x= log 6
− x log 3 = log120
− x log 7 = log 63
102 x + 3 = 315 log102 x + 3 = log 315 (2 x + 3) log10 = log 315 2 x + 3 = log 315
x=
14.
106− x = 550 (6 − x ) log10 = log550 6 − x = log550 x = 6 − log550
log120 −x = log 3 log120 x=− log 3
15.
log 63 log 7 log 63 x=− log 7
−x =
e x = 10 ln e x = ln10
log 315 − 3 2
Copyright © Houghton Mifflin Company. All rights reserved.
x = ln10
Section 4.5
16.
263
e x +1 = 20
21− x = 3x +1
17.
log 21− x = log 3x +1
ln e x +1 = ln 20 x + 1 = ln 20 x = ln 20 − 1
(1 − x ) log 2 = ( x + 1) log 3 log 2 − x log 2 = x log 3 + log 3 log 2 − x log 2 − x log 3 = log 3 − x log 2 − x log 3 = log 3 − log 2 − x (log 2 + log 3) = log 3 − log 2 (log 3 − log 2) (log 2 + log 3) log 2 − log 3 log 2 − log 3 or x= log 2 + log 3 log 6
x=−
3x − 2 = 42 x +1
18.
log 3x − 2 = log 42 x +1 ( x − 2) log 3 = (2 x + 1) log 4 x log 3 − 2log 3 = 2 x log 4 + log 4 x log 3 − 2log 3 − 2 x log 4 = log 4 x (log 3 − 2 x log 4 = log 4 + 2log 3 log 4 + 2log 3 x= log 3 − 2log 4
20.
53 x = 3x + 4
log 22 x −3 = log5− x −1 (2 x − 3) log 2 = ( − x − 1) log 5 2 x log 2 − 3log 2 = − x log 5 − log 5 2 x log 2 + x log 5 − 3log 2 = − log5 2 x log 2 + x log 5 = 3log 2 − log 5
x (2 log 2 + log 5) = 3log 2 − log 5 x=
21.
log( x 2 + 19) = 2 x 2 + 19 = 102 x 2 + 19 = 100 x 2 = 81 x = ± 81 x = 9, − 9
3log 2 − log 5 2 log 2 + log 5
log(4 x − 18) = 1 4 x − 18 = 101 4 x − 18 = 10 4 x = 28 x=7
log53 x = log 3x + 4 3x log5 = ( x + 4)log 3 3x log5 = x log 3 + 4log 3 3x log5 − x log 3 = 4log 3 x (3log5 − log 3) = 4log 3 4log 3 x= 3log5 − log 3 22.
22 x −3 = 5− x −1
19.
23.
ln( x 2 − 12) = ln x x 2 − 12 = x x 2 − x − 12 = 0 ( x − 4)( x + 3) = 0 x = 4 or x = −3 (No; not in domain.) x=4
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
264 24.
25.
log(2 x 2 + 3x) = log(10 x + 30)
log 2 + log 2 ( x − 4) = 2 log 2 x ( x − 4) = 2
2 x 2 + 3 x = 10 x + 30
log 2 ( x 2 − 4 x ) = 2
2 x 2 − 7 x − 30 = 0 (2 x + 5)( x − 6) = 0
22 = x 2 − 4 x
x = − 5 or x = 6
0 = x2 − 4 x − 4
−5, 6
x=
2
2
4 ± 16 − 4(1)( −4) 2
x = 4±4 2 2 x =2±2 2
2
2 − 2 2 is not a solution because the logarithm of a negative
number is not defined. The solution is x = 2 + 2 2 . 26.
log3 x + log 3( x + 6) = 3
27.
log(5 x − 1) = 2 + log( x − 2) log(5 x − 1) − log( x − 2) = 2 (5 x − 1) log =2 ( x − 2) (5 x − 1) 102 = ( x − 2) 100( x − 2) = 5 x − 1 100 x − 200 = 5 x − 1 95 x = 199 x = 199 95
29.
ln(1 − x) + ln(3 − x) = ln 8 ln[(1 − x)(3 − x)] = ln 8 (1 − x)(3 − x) = 8
log3 x ( x + 6) = 3 33 = x ( x + 6) 27 = x 2 + 6 x 0 = x 2 + 6 x − 27 0 = ( x + 9)( x − 3) x=3 x = −9 log3 (−9) is not defined. The solution is x = 3.
28.
1 + log(3x − 1) = log(2 x + 1) 1 = log(2 x + 1) − log(3x − 1) (2 x + 1) (3x − 1) 2 + 1 x 10 = 3x − 1 10(3x − 1) = (2 x + 1) 1 = log
3 − 4 x + x2 = 8 x2 − 4 x − 5 = 0 ( x + 1)( x − 5) = 0 x = −1 or x = 5 (No; not in domain.) The solution is x = −1.
30 x − 10 = (2 x + 1) 28 x = 11
x = 11 28
30.
log(4 − x) = log( x + 8) + log(2 x + 13) log(4 − x) = log[( x + 8)(2 x + 13)] 4 − x = ( x + 8)(2 x + 13) 4 − x = 2 x 2 + 29 x + 104
31.
log x 3 − 17 = 1 2 1 log ( x 3 − 17 ) = 1 2 2 101 = x 3 − 17
2
0 = 2 x + 30 x + 100
27 = x 3
2
0 = 2( x + 15 x + 50) 0 = 2( x + 5)( x + 10) x = −5 or x = −10 (No; not in domain.) The solution is x = −5.
3
27 = x 3 3= x The solution is x = 3. 3
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.5
265
log x 3 = (log x )2
32.
3log x = (log x )
33.
log(log x ) = 1
34.
ln(ln x) = 2 e2 = ln x
1
2
10 = log x 10
2
(log x ) − 3log x = 0 log x (log x − 3) = 0
10
2
=x
ee = x
log x = 0 or log x − 3 = 0 x =1
35.
log x = 3 x = 1000
ln(e3 x ) = 6 3 x ln e = 6 3x(1) = 6
36.
ln x = 1 ln ⎛⎜ 2 x + 5 ⎞⎟ + 1 ln 2 2 ⎝ 2⎠ 2 ln x = 1 ln 2 ⎛⎜ 2 x + 5 ⎞⎟ 2 2⎠ ⎝ 1 ln x = ln(4x + 5) 2
3x = 6 x=2
ln x = ln(4 x + 5)1/ 2 x = 4x + 5 2
x = 4x + 5 0 = x2 − 4 x − 5 0 = ( x − 5)( x + 1) x = 5, − 1
Check:
ln5 = 1 ln ⎛⎜ 10 + 5 ⎞⎟ + 1 ln2 2 ⎝ 2⎠ 2 1.609 = 1.2628 + 0.3465 1.609. = 1.609
ln( − 1) = 1 ln ⎛⎜ −2 + 5 ⎞⎟ + 1 2 ⎝ 2⎠ 2 x = −1 is not a solution because ln(−1) is not defined.
The solution is x = 5. 37.
log7 (5 x ) − log7 3 = log7 (2 x + 1)
( )
38.
39.
eln( x −1) = 4 ln eln( x −1) = ln 4 ln( x − 1)ln e = ln 4 ln( x − 1)(1) = ln 4 ( x − 1) = 4 x=5
log 4 x + log 4 ( x − 2) = log4 15 log 4 x ( x − 2) = log 4 15
log7 5 x = log7 (2 x + 1) 3 5x = 2 x + 1 3 5x = 6 x + 3 −3 = x x = –3 is not a solution because log7 ( −15) is undefined. No solution.
x 2 − 2 x = 15 x 2 − 2 x − 15 = 0 ( x − 5)( x + 3) = 0 x = 5, − 3 x = –3 is not a solution because log 4 ( −3) is undefined. The solution is x = 5. 40.
10log(2 x + 7) = 8 log10log(2 x + 7) = log8 log(2 x + 7) = log8 2x + 7 = 8 2x = 1 x=1 2
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
266 41.
10 x − 10− x = 20 2
42.
10 x + 10− x = 8 2
10 x (10 x − 10− x ) = 40 (10 x )
10 x (10 x + 10− x ) = (16)10 x
102 x − 1 = 40 (10 x )
102 x + 1 = 16(10 x )
102 x − 40 (10 ) − 1 = 0
102 x − 16 (10 x ) + 1 = 0
Let u = 10 x.
Let u = 10 x.
x
u 2 − 40u − 1 = 0 u=
u=
40 ± 402 − 4 (1)( −1)
2 u = 16 ± 256 − 4 2 u = 16 ± 6 7 2 u =8±3 7
2 = 40 ± 1600 + 4 2 ± 40 1604 = 2 ± 40 2 401 = 2 = 20 ± 401
10 x = 20 + 401
( x = log ( 20 +
16 ± 162 − 4 (1)(1)
10 x = 8 ± 3 7
( ) x = log (8 ± 3 7 )
x log 10 = log 8 ± 3 7
) 401 )
log 10 x = log 20 + 401
43.
10 x + 10− x = 5 10 x − 10− x
44.
10 x + 10− x = 5 (10 x − 10− x )
10 x (10 x + 10− x ) = 5 (10 x − 10− x )10 x 102 x + 1 = 5 (102 x − 1) 4 (102 x ) = 6 2 (102 x ) = 3
(10 x )2 = 3 2
x
10 =
3 2
x log 10 = log 3 2 x = log 3 2
10 x − 10− x = 1 10 x + 10− x 2 10 x (10 x − 10− x ) = 1 (10 x + 10− x )(10 x ) 2 2x 10 − 1 = 1 (102 x + 1) 2 2x 10 − 1 = 1 (102 x ) + 1 2 2 2x 2x ) 3 1 ( 10 − 10 = 2 2 1 (102 x ) = 3 2 2 102 x = 3 2 x log 10 = log 3 2 x = log 3 log 3 x= 2
x = 1 log ⎛⎜ 3 ⎞⎟ 2 ⎝2⎠
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.5
45.
267
e x + e − x = 15 2 e x ( e x + e − x ) = (30)e x
46.
e x ( e x − e − x ) = (30)( e x )
e2 x + 1 = e x (30) e
2x
e2 x − 1 = 30e x
x
− 30e + 1 = 0
e2 x − 30e x − 1 = 0
x
Let u = e .
Let u = e x .
2
u − 30u + 1 = 0
u 2 − 30u − 1 = 0
u = 30 ± 900 − 4 2 ± 30 896 u= 2 u = 30 ± 8 14 2 u = 15 ± 4 14
30 ± 900 − 4( −1) 2 ± 30 904 u= = 30 ± 2 226 2 2 u = 15 ± 226 u=
e x = 15 ± 226
x
e = 15 ± 4 14
x ln e = ln (15 ± 226)
x ln e = ln (15 ± 4 14)
x = ln (15 + 226)
x = ln (15 ± 4 14) 47.
1 =4 e x − e− x 1 = 4( e x − e − x ) x
x
e x − e − x = 15 2
x
1( e ) = 4( e )( e − e
48. −x
e x + e− x = 3 e x − e− x e x ( e x + e − x ) = 3( e x − e − x )e x
)
e2 x + 1 = 3e2 x − 3
e x = 4( e2 x − 1)
4 = 2e 2 x
e x = 4e2 x − 4
2 = e2 x
0 = 4e2 x − e x − 4
ln 2 = 2 x ln e ln 2 = x 2
x
Let u = e . 0 = 4u 2 − u − 4 1 ± 1 − 4(4)( −4) 8 1 65 ± u= 8 + 1 65 x e = 8 ⎛ ⎞ x ln e = ln ⎜ 1 + 65 ⎟ 8 ⎝ ⎠ u=
x = ln (1 + 65) − ln 8 49.
2− x +3 = x + 1
50.
3x − 2 = −2 x − 1
Graph f = 2− x +3 − ( x + 1). Its x-intercept is the solution. x ≈ 1.61
Graph f = 3x − 2 + 2 x + 1. Its x-intercept is the solution. x ≈ −0.53
Xmin = −4, Xmax = 4, Xscl = 1,
Xmin = −8, Xmax = 8, Xscl = 2,
Ymin = −4, Ymax = 4, Yscl = 1
Ymin = −8, Ymax = 8, Yscl = 2
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
268 51.
53.
e3− 2 x − 2 x = 1
52.
Graph f = e3− 2 x − 2 x − 1. Its x-intercept is the solution. x ≈ 0.96
Graph f = 2e x + 2 + 3x − 2. Its x-intercept is the solution. x ≈ −1.05
Xmin = −4, Xmax = 4, Xscl = 1,
Xmin = −4, Xmax = 4, Xscl = 1,
Ymin = −4, Ymax = 4, Yscl = 1
Ymin = −4, Ymax = 4, Yscl = 1
3 log 2 ( x − 1) = − x + 3
54.
2 log(2 − 3x ) − 2 x + 1. log 3 Its x-intercept is the solution. x ≈ 0.38
Xmin = −4, Xmax = 4, Xscl = 1,
Xmin = −4, Xmax = 4, Xscl = 1,
Ymin = −4, Ymax = 4, Yscl = 1
Ymin = −4, Ymax = 4, Yscl = 1
Graph f =
ln(2 x + 4) + 1 x = −3 2
56.
Graph f = ln (2 x + 4) + 1 x + 3. 2 Its x-intercept is the solution. x ≈ −1.93
2 x +1 = x 2 − 1
2ln(3 − x) + 3x = 4 Graph f = 2ln (3-x) + 3 x − 4. Its x-intercepts are the solutions. x ≈ 0.81, 2.91
Xmin = −4, Xmax = 4, Xscl = 1, Ymin = −4, Ymax = 4, Yscl = 1
Xmin = −4, Xmax = 4, Xscl = 1, Ymin = −4, Ymax = 4, Yscl = 1
57.
2 log3 (2 − 3 x) = 2 x − 1
3 log(x − 1) + x − 3. log 2 Its x-intercept is the solution. x ≈ 2.20 Graph f =
55.
2e x + 2 + 3x = 2
58.
ln( x ) = − x 2 + 4
Graph f = 2 x +1 − x 2 + 1. Its x-intercept is the solution. x ≈ −1.34
Graph f = ln x + x 2 − 4 . Its x-intercept is the solution. x ≈ 1.84
Xmin = −4, Xmax = 4, Xscl = 1,
Xmin = −4, Xmax = 4, Xscl = 1,
Ymin = −4, Ymax = 4, Yscl = 1
Ymin = −4, Ymax = 4, Yscl = 1
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.5
59.
a.
269
P (0) = 8500(1.1)0 = 8500(1) = 8500
60.
a.
R (1) = 145e −0.092 ≈ 132 beats per minute
P (2) = 8500(1.1)2 = 10, 285 15,000 = 8500(1.1)t
b.
b.
80 = 145e0.092t ln80 = ln145 − 0.092t ln e ln 80 − ln145 =t −0.092 6=t 6 minutes
a.
⎛1⎞ 1/ 2 A ⎜ ⎟ = 80 ( 0.727 ) ≈ 68 ⎝2⎠ A = 68 mg
ln15,000 = 8500(1.1)t ln 51,000 = ln 8500 + t ln (1.1) ln15,000 − ln 8500 =t ln (1.1) 6≈t The population will reach 15,000 in 6 years.
61.
a.
T (10 ) = 36 + 43e
−0.058(10 )
= 36 + 43e −0.58
62.
T ≈ 60o F 45 = 36 + 43e−0.058t
b.
R ( 0 ) = 145e0 = 145 beats per minute
t
50 = 80 ( 0.727 )
b.
ln ( 45 − 36 ) = ln 43 − 0.058t ln e ln ( 45 − 36 ) − ln 43 −0.058
ln 50 = ln 80 + t ln 0.727 ln 50 − ln 80 =t ln 0.727 t ≈ 1.47 hours ≈ 88 minutes
=t
t ≈ 27 minutes 63.
114 = 198 − (198 − 0.9)e −0.23 x
64.
−84 = −197.1e −0.23 x
−73 = −93.4e −0.21x
84 = e −0.23 x 197.1 ln 84 = −0.23x 197.1 84 197.1 x= ≈ 3.7 years −0.23
73 = e −0.21x 93.4 ln 73 = −0.21x 93.4 73 93.4 x= ≈ 1.2 years −0.21
( )
65.
21 = 94 − (94 − 0.6)e −0.21x
( )
5 + 29ln(t + 1) = 65 Graph f = 29ln( x + 1) − 60. Its x-intercept is the solution. x ≈ 6.9 months
66.
Xmin = −4, Xmax = 10, Xscl = 1, Ymin = −4, Ymax = 4, Yscl = 1
0.37ln x + 0.05 = 2.9 Graph f = 0.37ln x − 2.85. Its x-intercept is the solution. x ≈ 2200 thousand people or 2,200,000 people
Xmin = −800, Xmax = 3200, Xscl = 800, Ymin = −1, Ymax = 1, Yscl = 1
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
270 67.
Consider the first function for time less than 10 seconds. 275 = −2.25 x 2 + 56.26 x − 0.28
68.
a.
0 = −2.25 x 2 + 56.26 x − 275.28 −56.26 ± (56.26)2 − 4( −2.25)( −275.28) 2( −2.25) x = 6.67 or 18.33 18.33 s > 10 s, so it is not a solution. The solution is 6.67 s. x=
363.4 − 88.4ln x = 50 Graph f = 313.4 − 88.4ln x. Its x-intercept is the solution. x ≈ 34.65 2(34.65) = 69.3 m
Consider the second function for time greater than 10 seconds. Xmin = −40, Xmax = 80, Xscl = 10, Ymin = −10, Ymax = 10, Yscl = 1
275 = 8320(0.73) x 275 = (0.73) x 8320 ln 275 = ln 0.73x 8320 ln 275 = x ln 0.73 8320 ln 275 8320 ≈ 10.83 x= ln 0.73 The solutions are 6.67 s and 10.83 s.
( ) ( )
b.
( )
568.2 − 161.5ln x = 125 Graph f = 443.2 − 161.5ln x. Its x-intercept is the solution. x ≈ 15.55 2(15.55) = 31.1 m
Xmin = −20, Xmax = 40, Xscl = 10, Ymin = −10, Ymax = 10, Yscl = 1
69.
a.
b. c. d. 71.
70.
48 hours P = 100 As the number of hours of training increases, the test scores approach 100%.
a.
b. c. d.
b. c. d. 72.
In 27 years or 2026 B = 1000 As the number of years increases, the bison population approaches but never exceeds 1000.
a.
45 weeks P = 100 The more experience a person has, the closer the person’s score is to 100%.
a.
b. c. d.
21 hours Y = 50,000 The number of yeast approaches but never exceeds 50,000.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.5
73.
a.
c. 74.
271
b.
When T = 100, r ≈ 0.019, or 1.9% t=−
9 ⎛ 24 + v ⎞ ln ⎜ ⎟ 24 ⎝ 24 − v ⎠
1.5 = −
9 ⎛ 24 + v ⎞ ln ⎜ ⎟ 24 ⎝ 24 − v ⎠
a.
24 (1.5 )
75.
a.
e
4
( 24 ) − e
4
− v − e v = 24 − 24e
(
50 = e0.64t − 1 100 e0.64t + 1 0.64 t 0.5 = e0.64t − 1 e +1
0.5( e0.64t + 1) = e0.64t − 1 0.5e01.64t + 0.5 = e0.64t − 1 0.5e0.64t − e0.64t = −1.5
v = 24 + v
4
−0.5e0.64t = −1.5 4
e0.64t = 3
)
v −1 − e4 = 24 v=
⎛ 0.64t ⎞ v = 100 ⎜ e0.64t − 1 ⎟ +1⎠ ⎝e ⎛ 0.64t ⎞ 50 = 100 ⎜ e0.64t − 1 ⎟ +1⎠ ⎝e
⎛ 24 + v ⎞ = ln ⎜ ⎟ 9 ⎝ 24 − v ⎠ ⎛ 24 + v ⎞ 4 = ln ⎜ ⎟ ⎝ 24 − v ⎠ 24 + v e4 = 24 − v − = +v 4 v 24 24 ) e (
0.64t = ln 3 t = ln 3 0.64 t ≈ 1.72 In approximately 1.72 seconds, the velocity will be 50 feet per second.
24 − 24e4
−1 − e4 v ≈ 23.1367 The velocity after 1.5 seconds is approximately 23.14 feet per second.
76.
When r = 3%, or 0.03, T ≈78 years
b.
The vertical asymptote occurs when the denominator of 24 + v is zero, or when v = 24. 24 − v
b.
The horizontal asymptote is the value of ⎡ e0.64t − 1 ⎤ 100 ⎢ ⎥ as t → ∞. Therefore, the ⎢⎣ e0.64t + 1 ⎥⎦ horizontal asymptote is v = 100 feet per second.
c.
The velocity of the object cannot reach or exceed 24 feet per second.
c.
The object cannot fall faster than 100 feet per second.
a.
77.
Graph V = 400,000 − 150,000(1.005) x and V = 100,000. They intersect when x ≈ 138.97. After 138 withdrawals, the account has $101,456.39. After 139 withdrawals, the account has $99,963.67. The designer can make at most 138 withdrawals and still have $100,000.
b.
When s = 100, t ≈ 2.6 seconds.
Xmin = 0,Xmax = 200,Xscl = 25 Ymin = −50000,Ymax = 350000,Yscl = 50000
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
272 78.
a.
h( x ) = 10( e x / 20 + e − x / 20 ) The lowest height of the cable is in the middle, where x = 0. h(0) = 10( e0 / 20 + e −0 / 20 ) = 10( e0 + e0 ) = 10(1 + 1) = 10(2) = 20 feet
b.
h(10) = 10( e10 / 20 + e −10 / 20 ) = 10( e1/ 2 + e −1/ 2 ) ≈ 22.6 feet 24 = 10( e x / 20 + e − x / 20 ) ⇒ 2.4 = e x / 20 + e − x / 20 ⇒ 2.4e x / 20 = ( e x / 20 + e − x / 20 )e x / 20
c.
2.4e
x / 20
= e2 x / 20 + e0 = ( e x / 20 ) 2 + 1 ⇒ 2.4e x / 20 = (e x / 20 )2 1. Let u = e x / 20 . Then 2.4u = u 2 + 1
0 = u 2 − 2.4u + 1 ⇒ u =
−( −2.4) ± ( −2.4)2 − 4(1)(1) 2.4 ± 1.76 2.4 ± 1.3266 = ≈ 2 2(1) 2
e x / 20 = 2.4 + 1.76 2 2.4 + 1.76 x / 20 = ln 2 x = 20ln 2.4 + 1.76 ≈ 12.4 feet 2
or
e x / 20 = 2.4 − 1.76 2 2.4 − 1.76 x / 20 = ln 2 x = 20ln 2.4 − 1.76 ≈ −12.4 no negative height 2
.......................................................
Connecting Concepts
79.
The second step because log 0.5 < 0. Thus the inequality sign must be reversed.
80.
The third step. log2 (8 + 8) does not equal log 2 8 + log2 8.
81.
log( x + y ) = log x + log y log( x + y ) = log xy Therefore x + y = xy x − xy = − y x(1 − y ) = − y −y x= 1− y y x= y −1
82.
No. The domain of g ( x ) includes negative numbers; the domain of f ( x ) does not. Thus, for any negative value of x, f ( x ) ≠ g ( x ) .
83.
Since e0.336 ≈ 1.4,
84.
2.2 = e− k ln 2.2 = − k ln e
F ( x) = (1.4) x ≈ (e0.336 ) x = e0.336 x = G ( x)
− ln 2.2 = k −0.788 ≈ k
....................................................... 12(2)
PS1.
A = 1000 ⎛⎜ 1 + 0.1 ⎞⎟ 12 ⎠ ⎝
PS3.
0.5 = e14k ln 0.5 = ln e14k ln 0.5 = 14k ln 0.5 = k 14 −0.0495 ≈ k
= 1220.39
Prepare for Section 4.6 PS2.
PS4.
A = 600 ⎛⎜ 1 + 0.04 ⎞⎟ 4 ⎠ ⎝
4(8)
= 824.96
0.85 = 0.5t / 5730 ln 0.85 = ln 0.5t / 5730 t ln 0.5 5730 5730ln 0.85 = t ln 0.5 1340 ≈ t ln 0.85 =
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.6
PS5.
273
70 5 + 9e −12 k 6(5 + 9e −12 k ) = 70 6=
30 + 54e
−12 k
−3
2
4,000,000 = 3n +1 − 3
= 70
3,999,997 = 3n +1
−12 k
= 40 = 20 e 27 −12 k = ln 20 ln e 27 −12k = ln 20 27 1 k = − ln 20 12 27 k ≈ 0.025
54e
n +1
2,000,000 = 3
PS6.
ln 3,999,997 = ln 3n +1 ln 3,999,997 = ( n + 1)ln 3 ln 3,999,997 = n +1 ln 3 ln 3,999,997 −1 = n ln 3 12.8 ≈ n
−12 k
Section 4.6 1.
3.
a.
t = 0 hours, N ( 0 ) = 2200 ( 2 ) = 2200 bacteria
b.
t = 3 hours, N ( 3) = 2200 ( 2 ) = 17,600 bacteria
a.
0
2.
3
N (t ) = N 0 ekt where N 0 = 24600 N (5) = 22,600e
4.
a.
t = 3 years, f ( 3) = 12,400 (1.14 ) ≈ 18,400
b.
t = 4.25 years, f ( 4.25) = 12,400 (1.14 )
a.
k (5)
58,100 = 53,700e4 k 58,100 = e4 k 53,700 58,100 ⎞ 4k ln ⎛⎜ ⎟ = ln ( e ) ⎝ 53,700 ⎠ 58,100 ⎞ ln ⎛⎜ ⎟ = 4k ⎝ 53,700 ⎠ 1 ⎡ln ⎛ 58,100 ⎞ ⎤ = k 4 ⎢⎣ ⎜⎝ 53,700 ⎟⎠ ⎥⎦ 0.01969 ≈ k
0.01368 ≈ k N (t ) = 22,600e0.01368t t = 15 N (15) = 22,600e0.01368(15) = 22,600e0.2052 ≈ 27,700
b.
4.25
≈ 21,600
N (t ) = N 0 ekt where N 0 = 53,700 N (4) = 53,700ek (4)
24,200 = 22,600e5k 24,200 = e 5k 22,600 24,200 ⎞ 5k ln ⎛⎜ ⎟ = ln ( e ) ⎝ 22,600 ⎠ 24,200 ⎞ ln ⎛⎜ ⎟ = 5k ⎝ 22,600 ⎠ 1 ⎡ln 24, 200 ⎤ = k 5 ⎣⎢ 22,600 ⎦⎥
b.
3
N (t ) = 53,700e0.01969t t = 12 N (12) = 53,700e0.01969(12) = 53,700e0.23628 ≈ 68,000
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
274 N (t ) = N 0 ekt where N 0 = 362,300
5.
N (t ) = N 0 ekt where N 0 = 276,400
6.
N (4) = 362,300ek (4) 379,700 = 362,300e 379,700 = e4k 362,300 379,700 ⎞ 4k ln ⎛⎜ ⎟ = ln ( e ) ⎝ 362,300 ⎠ 379,700 ⎞ ln ⎛⎜ ⎟ = 4k ⎝ 362,300 ⎠ 1 ⎡ln 379,700 ⎤ = k 4 ⎢⎣ 362,300 ⎥⎦ 0.011727 ≈ k
291,800 = 276,400e4 k 291,800 = e4k 276,400 291,800 ⎞ 4k ln ⎛⎜ ⎟ = ln ( e ) ⎝ 276,400 ⎠ 291,800 ⎞ ln ⎛⎜ ⎟ = 4k ⎝ 276,400 ⎠ 1 ⎡ln 291,800 ⎤ = k 4 ⎢⎣ 276,400 ⎥⎦ 0.013555 ≈ k
N (t ) = 362,300e0.011727 t t=9
N (t ) = 276,400e0.013555 t t =8
N (9) = 362,300e0.011727(9)
N (9) = 276,400e0.013555(8)
= 362,300e ≈ 402,600
7.
N (4) = 276,400ek (4)
4k
0.105543
= 276,400e0.10844 ≈ 308,100
a.
c.
Since A = 4 micrograms are present when t = 0, find the time t at which half remains—that is when A = 2.
b.
A(5) = 4e −0.23 ≈ 3.18 micrograms
d.
1 = 4e −0.046t 1 = e −0.046t 4 1 = −0.046t ln 4 ln 1 4 =t −0.046t 30.14 ≈ t The amount of sodium-24 will be 1 microgram after 30.14 hours.
2 = 4e −0.046t
() ()
1 = e −0.046t 2 ln 1 = −0.046t 2 ln 1 2
() ( ) =t
−0.046 15.07 ≈ t The half-life of sodium-24 is about 15.07 hours.
8.
N (t ) = N 0 ekt N (138) = N 0 e138k 138k
0.5N 0 = N 0 e
138k
0.5 = e
138k
ln 0.5 = ln e ln 0.5 = 138k ln e ln 0.5 = 138k ln 0.5 = k 138 −0.005023 ≈ k
9.
N ( t ) = N 0 ( 0.5)
t / 5730
N ( t ) = 0.45 N 0 0.45 N 0 = N 0 ( 0.5) ln ( 0.45) =
t / 5730
t ln 0.5 5730
10.
N ( t ) = N 0 ( 0.5)
t /138
N (730) = N 0 (0.5)730 / 138 ≈ 0.0256 N 0 After 2 years (730 days), only 2.56% of the polonium sample will remain.
5730 ln 0.45 = t ln 0.5 6601 ≈ t The bone is about 6601 years old.
N ( t ) = N 0(0.5)t /138 ≈ N 0 e −0.005023t Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.6
275
N ( t ) = N 0 ( 0.5)
11.
t / 5730
N ( t ) = N 0 ( 0.5)
12.
N ( t ) = 0.75 N 0
N ( t ) = 0.65 N 0 t / 5730
t / 5730
0.75 N 0 = N 0 ( 0.5) ln 0.75 = t ln 0.5 5730 ln 0.75 =t 5730 ln 0.5 2378 ≈ t The Rhind papyrus is about 2378 years old.
13.
a.
0.65 N 0 = N 0 ( 0.5) ln 0.65 = t ln 0.5 5730 ln 0.65 =t 5730 ln 0.5 3561 ≈ t The bone is about 3600 years old.
P = 8000, r = 0.05, t = 4, n = 1
14.
a.
4
⎛ 0.045 ⎞ B = 22,000⎜1 + ⎟ ≈ $24,024.55 1 ⎠ ⎝
7
b.
⎛ 0.05 ⎞ t = 7, B = 8000 ⎜1 + ⎟ ≈ $11, 256.80 1 ⎠ ⎝
a.
P = 38,000, r = 0.065, t = 4, n = 1
10
16.
b.
⎛ 0.045 ⎞ t = 10, B = 22,000⎜1 + ⎟ 1 ⎠ ⎝
a.
P = 12,500, r = 0.08, t = 10, n = 1
4
10
⎛ 0.065 ⎞ B = 38,000⎜1 + ⎟ ≈ $48,885.72 1 ⎠ ⎝
17.
4(365)
b.
⎛ 0.065 ⎞ n = 365, B = 38,000⎜1 + ⎟ 365 ⎠ ⎝
c.
⎛ 0.065 ⎞ n = 8760, B = 38,000⎜1 + ⎟ 8760 ⎠ ⎝
P = 15,000, r = 0.1, t = 5 B = 15,000e
5(0.1)
19.
ln 2 r ln 2 t= 0.0784 t ≈ 8.8 years
21.
B = Pert
t=
⎛ 0.08 ⎞ B = 12,500⎜1 + ⎟ 1 ⎠ ⎝
≈ $49,282.20
4(8760 )
≈ $49,283.30 18.
Let B = 3P
ln 3 r ln 3 t= 0.076 t ≈ 14 years
t=
r = 0.076
⎛ 0.08 ⎞ n = 365, B = 12,500⎜1 + ⎟ 365 ⎠ ⎝
c.
0.08 ⎞ ⎛ n = 8760, B = 12,500⎜1 + ⎟ ⎝ 8760 ⎠
22.
ln 3 r ln 3 t= 0.055 t ≈ 20 years
24.
ln 3 0.055 t = 20 years
t=
t=
t=
≈ $27,816.82 87,600
P = 32,000, r = 0.08, t = 3
ln 2 2 ln 2 t= 0.0588 t ≈ 11.8 years
3 = e rt ln 3 = rt ln e ln 3 t= r 23.
3650
b.
20.
3P = Pert
≈ $34,165.33
≈ $26,986.56
B = 32,000e3(0.08 ) ≈ $40,679.97
≈ $24,730.82
r = 0.0784
P = 22,000, r = 0.045, n = 1, t = 2 2
⎛ 0.05 ⎞ B = 8000⎜1 + ⎟ ≈ $9724.05 1 ⎠ ⎝
15.
t / 5730
r = 0.0588
r = 0.055
r = 0.055
Copyright © Houghton Mifflin Company. All rights reserved.
≈ $27,819.16
Chapter 4: Exponential and Logarithmic Functions
276 25.
27.
29.
31.
a. b.
1900 0.16
c.
P(0) =
a. b.
157,500 0.04
c.
P (0) =
a. b.
2400 0.12
c.
P(0) =
a=
26.
1900 = 200 1 + 8.5e −0.16(0) 28.
157,500 = 45,000 1 + 2.5e −0.04(0)
30.
2400 = 300 1 + 7e −0.12(0)
c − P0 5500 − 400 = = 12.75 P0 400 c 1 + ae −bt 5500 P (2) = 1 + 12.75e − b(2) 5500 780 = 1 + 12.75e −2b P(t ) =
32.
a. b.
32,550 0.08
c.
P(0) =
a. b.
51 0.03
c.
P(0) =
a. b.
320 0.12
c.
P(0) =
a=
32,550 = 18,600 1 + 0.75e −0.08(0)
51 = 25 1 + 1.04e −0.03(0)
320 = 20 1 + 15e −0.12(0)
c − P0 9500 − 6200 = = 0.53226 P0 6200 c 1 + ae − bt 9500 P (8) = 1 + 0.53226e −b(8) 9500 7100 = 1 + 0.53226e −8b P (t ) =
780(1 + 12.75e −2b ) = 5500
7100(1 + 0.53226e −8b ) = 9500
780 + 9945e −2b = 5500
7100 + 3779e −8b = 9500
9945e −2b = 4720
3779e −8b = 2400
e −2b = 4720 9945 −2 b ln e = ln 4720 9945 4720 −2b = ln 9945 1 b = − ln 4720 2 9945 b ≈ 0.37263 5500 P(t ) = 1 + 12.75e −0.37263 t
e −8b = 2400 3779 −8b ln e = ln 2400 3779 −8b = ln 2400 3779 1 b = − ln 2400 8 3779 b ≈ 0.05675 9500 P(t ) = 1 + 0.53226e −0.05675 t
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.6
33.
a=
277
c − P0 100 − 18 = = 4.55556 P0 18
34.
c 1 + ae − bt 100 P (3) = 1 + 4.55556e − b(3) 100 30 = 1 + 4.55556e −3b 30(1 + 4.55556e −3b ) = 100 P (t ) =
c − P0 P0 c − 3200 a= 3200 3200a + 3200 = c a=
c 1 + ae − bt + 3200 P(22) = 3200a−0.056(22) 1 + ae 3200 5565 = 3200a −+1.232 1 + ae 5565(1 + ae −1.232 ) = 3200a + 3200 P(t ) =
30 + 136.67e −3b = 100 136.67e −3b = 70 e −3b = 70 136.67 ln e −3b = ln 70 136.67 −3b = ln 70 136.67 b = − 1 ln 70 3 136.67 b ≈ 0.22302 100 P(t ) = 1 + 4.55556e −0.22302 t
35.
a.
b.
37.
a.
625,000 1 + 3.1e −0.045 t 625,000 R (1) = ≈ $158,000 1 + 3.1e −0.045(1) 625,000 R (2) = ≈ $163,000 1 + 3.1e −0.045(2) 625,000 R(t ) = , as t → ∞, R(t ) → $625,000 1 + 3.1e −0.045 t R(t ) =
a=
c − P0 1600 − 312 = = 4.12821 312 P0
5565 + 5565ae −1.232 = 3200a + 3200 5565ae −1.232 − 3200a = −2365 a (5565e −1.232 − 3200) = −2365 −2365 5565e −1.232 − 3200 a = 1.5 c = 3200a + 3200 = 3200(1.5) + 3200 = 8000 8000 P(t ) = 1 + 1.5e −0.056 t a=
36.
38.
A(t ) =
b.
A(t ) =
a.
a=
c 1 + ae − bt 1600 P (6) = 1 + 4.12821e −b(6) 1600 416 = 1 + 4.12821e −6b 416(1 + 4.12821e −6b ) = 1600
b.
c − P0 3400 − 240 = = 13.167 240 P0 P(t ) =
416 + 1717.34e −6b = 1600
310 + 4081.6677e − b = 3400
−6 b
= 1184 −6 b = 1184 e 1717.34 ln e −6b = ln 1184 1717.34 −6b = ln 1184 1717.34 1 b = − ln 1184 6 1717.34 b ≈ 0.06198 1600 P(t ) = 1 + 4.12821e −0.06198 t 1600 ≈ 497 wolves P(10) = 1 + 4.12821e −0.06198(10)
1650 , as t → ∞, A(t ) → 1650 cars 1 + 2.4e −0.055 t
c 1 + ae −bt 3400 P(1) = 1 + 13.16667e − b(1) 3400 310 = 1 + 13.16667e − b 310(1 + 13.16667e − b ) = 3400
P(t ) =
1717.34e
1650 1 + 2.4e −0.055 t 1650 A(1) = ≈ 504 cars 1 + 2.4e −0.055(1) 1650 A(2) = ≈ 524 cars 1 + 2.4e −0.055(2)
a.
b.
4081.6677e − b = 3090 e − b = 3090 4081.6677 ln e − b = ln 3090 4081.6677 −b = ln 3090 4081.6677 b = − ln 3090 4081.6677 b ≈ 0.27833 3400 P(t ) = 1 + 13.16667e −0.27833 t 3400 ≈ 1182 groundhogs P(7) = 1 + 13.16667e −0.27833(7)
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Chapter 4: Exponential and Logarithmic Functions
278 39.
a.
a=
c − P0 8500 − 1500 = = 4.66667 1500 P0
40.
a.
c 1 + ae − bt 8500 P (2) = 1 + 4.66667e − b(2) 8500 1900 = 1 + 4.66667e −2b
c − P0 5500 − 800 = = 5.875 800 P0 c 1 + ae − bt 5500 P (1) = 1 + 5.875e − b(1) 5500 900 = 1 + 5.875e − b
P(t ) =
P (t ) =
1900(1 + 4.66667e −2b ) = 8500
900(1 + 5.875e −b ) = 5500
1900 + 8866.673e −2b = 8500
900 + 5287.5e − b = 5500
8866.673e
−2 b
5287.5e − b = 4600
= 6600
6600 8866.673 ln e −2b = ln 6600 8866.673 −2b = ln 6600 8866.673 b = − 1 ln 6600 2 8866.673 b ≈ 0.14761 8500 P(t ) = 1 + 4.66667e −0.14761t 8500 4000 = 1 + 4.66667e −0.14761t
e − b = 4600 5287.5 ln e − b = ln 4600 5287.5 −b = ln 4600 5287.5 b = − ln 4600 5287.5 b ≈ 0.13929 5500 P(t ) = 1 + 5.875e −0.13929 t
e −2b =
b.
a=
2000 =
b.
5500 1 + 5.875e −0.13929 t
4000(1 + 4.66667e −0.14761t ) = 8500
2000(1 + 5.875e −0.13929 t ) = 5500
1 + 4.66667e −0.14761t = 2.125
1 + 5.875e −0.13929 t = 2.75
4.66667e −0.14761t = 1.125 e −0.14761t = 1.125 4.66667 −0.14761 t = ln 1.125 ln e 4.66667 −0.14761t = ln 1.125 4.66667 1 t=− ln 1.125 0.14761 4.66667 t ≈ 9.6 The population will exceed 4000 in 2007 + 9 = 2016.
5.875e −0.13929 t = 1.75 e −0.13929 t = 1.75 5.875 −0.13929 t = ln 1.75 ln e 5.875 −0.13929t = ln 1.75 5.875 1 t=− ln 1.75 0.13929 5.875 t ≈ 8.7 The population will exceed 2000 in 2003 + 8 = 2011.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.6
41.
a.
279
A = 34o F, T0 = 75o F, Tt = 65o F, t = 5. Find k .
42.
a.
N ( 2 ) = 100 (1.04 − 0.992 ) N ( 2 ) = 6 wpm
b.
N ( 40 ) = 100 (1.04 − 0.9940 ) N ( 40 ) ≈ 37 wpm
65 = 34 + ( 75 − 34 ) e −5k 31 = 41e −5k 31 = e −5k 41 ⎛ 31 ln ⎜ ⎞⎟ = −5k ⎝ 41 ⎠
c.
0.6 = 1.04 − 0.99t
k = − 1 ln ⎛⎜ 31 ⎞⎟ 5 ⎝ 41 ⎠ k ≈ 0.056
b.
0.44 = 0.99t ln 0.44 = ln 0.99t ln 0.44 = t ln 0.99 82 ≈ t After 82 hours of practice, a student can expect to type 60 wpm.
A = 34o F, k = 0.056, T0 = 75o F, t = 30 Tt = 34 + ( 75 − 34 ) e
−30( 0.056 )
Tt = 34 + (41)e −1.68 Tt ≈ 42o F
c.
60 = 100 (1.04 − 0.99t )
Tt = 36o F, k = 0.056, Tt = 75o F, A = 34o F 36 = 34 + (75 − 34 )e−0.056t 2 = 41e −0.056t t ≈ 54 minutes
43.
a.
44.
10% of 80,000 is 8000.
a.
0.4 = 1 − e −0.03 t
0.1 = 1 − e −0.0005 t −0.9 = − e −0.0005 t
−0.6 = − e −0.03 t
0.9 = e −0.0005 t
0.6 = e −0.03 t
ln 0.9 = −0.0005t ln e
ln ( 0.6 ) = ln e −0.03 t
ln 0.9 = −0.0005t
ln ( 0.6 ) = −0.03t
ln 0.9 = t −0.0005 211 h ≈ t
b.
40% of 1,200,000 is 480,000. 480,000 = 1,200,000 (1 − e −0.03 )
8000 = 80,000 (1 − e −0.0005 t )
ln ( 0.6 ) −0.03
t ≈ 17 days
50% of 80,000 is 40,000. 40,000 = 80,000 (1 − e −0.0005t ) 0.5 = 1 − e −0.0005 t −0.5 = − e −0.0005 t 0.5 = e −0.0005 t ln ( 0.5) = ln ( e −0.0005 t )
=t
b.
960,000 = 1,200,000 (1 − e −0.03t ) 0.8 = 1 − e −0.03t 0.2 = e −0.03t ln 0.2 = t −0.03 t ≈ 54 days
ln ( 0.5) = −0.0005t ln ( 0.5)
=t −0.0005 1386 h ≈ t
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Chapter 4: Exponential and Logarithmic Functions
280 45.
V (t ) = V0 (1 − r )t
46.
a.
I (0) = 6(1 − e 0 ) = 0 amps
b.
I (0.5) = 6(1 − e −2.5(0.5) ) ≈ 4.28 amps
t
0.5V0 = V0 (1 − 0.20 ) t
0.5 = (1 − 0.20 ) 0.5 = 0.8t
I (t ) − 1 = − e −2.5t 6 I (t ) 1− = e −2.5t 6 I (t ) ln 1 − = −2.5t 6
ln 0.5 = ln 0.8t ln 0.5 = t ln 0.8
(
ln 0.5 =t ln 0.8 3.1 years ≈ t 47.
c. d. 49.
48.
20 = 64(1 − e−t / 2 )
c.
d.
(
I (t )
)
a.
b.
50 = 64(1 − e−t / 2 )
0.625 = 1 − e−t / 2
0.78125 = 1 − e−t / 2
e−t / 2 = 0.375 −t / 2 = ln 0.375 t ≈ 0.98 seconds The horizontal asymptote is v = 32. As time increases, the object’s velocity approaches but never exceeds 32 ft/sec.
e−t / 2 = 0.21875 −t / 2 = ln 0.21875 t ≈ 3.0 seconds The horizontal asymptote is v = 64. As time increases, the object’s velocity approaches but never exceeds 64 ft/sec.
a.
b.
)
t = − 2 ln 1 − 6 5
a.
b.
I (t ) = 6(1 − e −2.56t )
c.
c. d. 50.
The graphs of s = 32t + 32(e −t − 1) and s = 50 intersect when t ≈ 2.5 seconds. The slope m of the secant line containing (1, s(1)) s (2) − s (1) ≈ 24.56 ft/sec and (2, s(2)) is m = 2 −1 The average speed of the object was 24.56 feet per second between t = 1 and t = 2.
51.
a.
b. c.
d.
The graphs of s = 64t + 128(e −t / 2 − 1) and s = 50 intersect when t ≈ 2.1 seconds. The slope m of the secant line containing (1, s(1)) s (2) − s (1) ≈ 33.5 ft/sec and (2, s(2)) is m = 2 −1 The average speed of the object was 33.5 feet per second between t = 1 and t = 2.
52.
Xmin = 0,Xmax = 80,Xscl = 10,
Xmin = 0,Xmax = 24,Xscl = 3,
Ymin = −10,Ymax = 110,Yscl = 15
Ymin = −50,Ymax = 350,Yscl = 50
When P = 75%, t ≈ 45 hours.
When N = 140 circuits, t ≈ 11 weeks. Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.6
281
....................................................... 53.
Connecting Concepts 54.
a.
A(1) = 0.51/ 2 ≈ 0.71 gram
b.
A(4) = 0.54 / 2 + 0.5(4−3) / 2 = 0.52 + 0.51/ 2 ≈ 0.96 gram
Use the TRACE feature and the graph of ⎧0.5x / 2 0≤ x<3 ⎪⎪ x/2 ( x − 3) / 2 + 0.5 y = ⎨0.5 3≤ x < 6 ⎪ x/2 ( x − 3) / 2 (x − 6)/2 + 0.5 + 0.5 x≥6 ⎪⎩0.5 to determine the value of x when y = 0.25 the first time.
A(9) = 0.59 / 2 + 0.5(9−3) / 2 + 0.5(9−6) / 2
c.
= 0.54.5 + 0.53 + 0.51.5 ≈ 0.52 gram
Xmin = 0, Xmax = 16, Xscl = 2, Ymin = 0, Ymax = 2, Yscl = 0.25 After 11.1 hours
55.
N (t ) = 22,755e0.0287 t
56.
= 22,755 ( e0.0287 )
⎛7⎞ V (3) = 350,000⎜ ⎟ ⎝8⎠
3/ 2
a.
⎛7⎞ V (5) = 350,000⎜ ⎟ ⎝8⎠
5/ 2
b. c.
0.10(350,000) = 350,000 (.875)
t
t
≈ 22,755 (1.0291) The annual growth rate is 2.91%.
≈ 286,471 gallons ≈ 250,662 gallons t/2
0.10 = 0.875t / 2 ln 0.10 = t ln 0.875 2 t = 2 ln 0.10 ≈ 34 ln 0.875 t ≈ 34 hours
....................................................... PS1. Decreasing PS3. P(0) =
PS2. Decreasing
108 = 108 = 36 1 + 2e −0.1(0) 1 + 2 10 =
PS5.
Prepare for Section 4.7
20 1 + 2.2e −0.05 t
1.05(0) = 840 PS4. N (0) = 840e
PS6. P(t ) =
55 1 + 3e −0.08t
10(1 + 2.2e −0.05 t ) = 20 10 + 22e −0.05 t = 20 e −0.05 t = 10
22
ln e
−0.05 t
= ln 10
22
−0.05t = ln 10
There is a horizontal asymptote at P = 55.
22
t = −20ln 10
22
t ≈ 15.8
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Chapter 4: Exponential and Logarithmic Functions
282
Section 4.7 1.
2.
5
3.
6
0 0
decreasing exponential function; decreasing logarithmic function
increasing exponential function 8
4.
5.
0
decreasing exponential function; decreasing logarithmic function 6.
30 0
decreasing logarithmic function
increasing logarithmic function
increasing logarithmic function
7.
y ≈ 0.99628(1.20052) x ; r ≈ 0.85705 8.
y ≈ 1.48874(2.50469) x ; r ≈ 0.99999 9.
y ≈ 1.81505(0.51979) x ; r ≈ −0.99978
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.7
283
10.
y ≈ 4.23016(0.83937) x ; r ≈ −0.99999 11.
y ≈ 4.89060 − 1.35073ln x; r ≈ −0.99921 12.
y ≈ 18.02743 − 0.94970ln x; r ≈ −0.99997 13.
y ≈ 14.05858 + 1.76393ln x; r ≈ 0.99983 14.
y ≈ 60.08692 + 3.36076ln x; r ≈ 0.99932 15.
y≈
235.58598 1 + 1.90188e −0.05101x
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Chapter 4: Exponential and Logarithmic Functions
284 16.
y≈
710.56899 1 + 3.06263e −0.02968 x
y≈
2098.68307 1 + 1.19794e −0.06004 x
y≈
5398.79784 1 + 2.40005e −0.16010 x
17.
18.
19.
a. b. c.
t
Linear: p ≈ 0.22129t + 3.99190; r ≈ 0.99288 . Exponential: p ≈ 4.05326 (1.04460 ) ; r ≈ 0.99412 The exponential model’s r is closer to 1. 16
p = 4.05326 (1.04460 )
≈ $8.15
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Section 4.7
285
20.
a. b. c.
Linear: p ≈ 0.04461t + 0.14475 . Logarithmic: p ≈ −11.98408 + 3.59664 ln t Linear: r ≈ 0.96031. Logarithmic: r ≈ 0.97250. The logarithmic model provides a slightly better fit. p ≈ −11.98408 + 3.59664ln(109) ≈ 4.89 lbs per capita per day
a.
Exponential: T ≈ 0.06273(1.07078) F
b.
T ≈ 0.06273(1.07078)65 ≈ 5.3 hours
21.
22.
P ≈ 10.147(0.89104) x P ≈ 10.147(0.89104) 24 ≈ 0.6 newtons/cm 2
23.
a.
T ≈ 0.7881(1.07259) F
b.
T ≈ 0.07881(1.07259)65 ≈ 7.5 hours 7.5 – 5.3 = 2.2 hours.
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Chapter 4: Exponential and Logarithmic Functions
286 24.
Exponential: y ≈ 48.5557(0.99872) x Logarithmic: y ≈ 61.735786 − 4.104476ln x a. Exponential: y ≈ 48.5557(0.99872) 48 ≈ 45.7 Logarithmic: y ≈ 61.735786 − 4.104476ln 48 ≈ 45.8 The decreasing logarithmic best models the data. b. 25.
y ≈ 61.735786 − 4.104476ln108 ≈ 42.52 s
An increasing logarithmic model provides a better fit because of the concave-downward nature of the graph.
26.
c.
1,534,427 1 + 2.233148e −0.043679 t 1,534,427 P (60) ≈ ≈ 1,320,000 people 1 + 2.233148e −0.043679 (60) c ≈ 1,534,000
a.
p ≈ 7.862(1.026) y
b.
p ≈ 7.862(1.026)60 ≈ 36 cm
a. b.
T ≈ 28.502 − 5.372ln x T ≈ 28.502 − 5.372 ln(50) = 7.5 ml/L
a. b.
P(t ) ≈
27.
28.
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Section 4.7
287
29.
a.
Linear: pH ≈ 0.01353q + 7.02852; r ≈ 0.956627. Logarithmic: pH ≈ 6.10251 + 0.43369ln q; r ≈ 0.999998. The logarithmic model provides the better fit.
b.
2.09749 8.2 ≈ 6.10251 + 0.43369 ln q ⇒ 2.09749 ≈ 0.43369 ln q ⇒ 2.09749 ≈ ln q ⇒ q ≈ e 0.43369 ≈ 126.0 0.43369
a.
y ≈ 3.05401(1.0179) x
b.
8 ≈ 3.05401(1.0179) x ⇒
a.
p ≈ 3200(0.91894)t ; 200 ≈ 3200(0.91894)t ⇒ 1 ≈ (0.91894)t ⇒ ln 1 ≈ ln(0.91894)t 16 16 ⇒ ln1 − ln16 ≈ t ln 0.91894 ⇒ t ≈ − ln16 ln 0.91894 ⇒ t ≈ 32.8 years after 1980 ⇒ in 2012 No. The model fits the data perfectly because there are only two data points.
30.
8 8 ≈ (1.0179) x ⇒ ln ≈ ln(1.0179) x 3.05401 3.05401 ⇒ ln8 − ln 3.05401 ≈ x ln1.0179 ⇒ x ≈ ln8 − ln 3.05401 ln1.0179 ⇒ x ≈ 54.3 years after 1960 ⇒ in 2014
31.
b.
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Chapter 4: Exponential and Logarithmic Functions
288 32.
a. b. c.
Linear: y ≈ 0.83545 x + 73.48, r ≈ 0.87361 ; Logarithmic: y ≈ 72.35354 + 3.78794ln x, r ≈ 0.95898 The logarithmic regression provides the better fit. y ≈ y ≈ 72.35354 + 3.78794ln12 ≈ 81.8 inches (6 feet 9.8 inches)
a. b. c.
Exponential: S ≈ 7062.46390(0.956776)t , r ≈ −0.89618 ; Logarithmic: S ≈ 6995.50673 − 841.74326ln t , r ≈ −0.96240 The logarithmic regression provides the better fit. S (11) ≈ 6995.50673 − 841.74326ln11 ≈ 4977 sites
a.
T − 70o ≈ 96.16777(0.93787)t
b.
80 − 70 ≈ 96.16777(0.93787)t
33.
34.
10 10 ≈ 0.93787t ⇒ ln ≈ ln 0.93787t 96.16777 96.16777 ⇒ ln10 − ln 96.16777 ≈ t ln 0.93787 ⇒ t ≈ ln10 − ln 96.16777 ≈ 35 min ln 0.93787
10 ≈ 96.16777(0.93787)t ⇒
35.
a. b.
11.26828 1 + 2.74965e −0.02924t As t → ∞, P (t ) → 11 billion people t →∞, P(t ) →11 billion people P(t ) ≈
Copyright © Houghton Mifflin Company. All rights reserved.
Section 4.7
36.
289
The graph of the logarithmic regression equation passes through both of the data points.
.......................................................
Connecting Concepts
37.
A and B have different exponential regression functions. 38.
a.
The x-coordinate of the first ordered pair is 0, and 0 is not in the domain of y = ln x.
b.
Use a horizontal translation. For instance, add 1 to each of the x-coordinates. Find the logarithmic regression function for this new data. Remember that each x-value in the regression represents x – 1 in the original data.
a. b.
Exponential: y ≈ 1.81120(1.61740) x ; r ≈ 0.96793. Power: y ≈ 2.09385( x1.40246 ); r ≈ 0.99999. The power regression provides the better fit.
a.
The power regression function, t ≈ 1.11088(l 0.50113 ), provides the better fit.
b.
12 ≈ 1.11088(l 0.50113 ) ⇒
39.
40.
12 ≈ (l 0.50113 ) ⇒ ⎛ 12 ⎞ ⎜ ⎟ 1.11088 ⎝ 1.11088 ⎠
(1/ 0.50113)
≈ (l 0.50113 )(1/ 0.50113) ≈ 115.4 feet
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Chapter 4: Exponential and Logarithmic Functions
290
.......................................................
Exploring Concepts with Technology
Using a Semilog Graph to Model Exponential Decay 1.
a.
b. c.
2.
Xmin = −1, Xmax = 31, Xscl = 4, Ymin = −1, Ymax = 7, Yscl = 1 m = 4.3 − 3.3 ≈ −0.0909 4 − 15 ln A − 4.3 = −0.0909 ( t − 4 )
a.
b. c.
ln A = −0.0909t + 4.664 d.
e
ln A
= e −0.0909 t + 4.664
Xmin = −1, Xmax =7.1, Xscl = 1, Ymin = −1, Ymax =3, Yscl = 0.5 m = 2.785 − 2.754 = 0.031 ≈ 0.0155 3 −1 2 ln B − 2.574 = 0.0155(t − 1) ln B = 0.0155t + 2.7385 eln B = e0.0155 t + 2.7385
d.
A = e −0.0909 t e4.664 A=e
B = e0.0155 t e2.7385
−0.0909 t 4.664
B = 15.46e0.0155 t
e
e.
e.
Xmin = −5, Xmax = 35, Xscl = 5, Ymin = −10, Ymax = 110, Yscl = 15 f.
f.
A(t ) = 106e −0.0909t At t = 0 there is A = 106 mg present We must find t where A = 1 (106 ) = 53 mg. 2
Xmin = −1, Xmax = 7, Xscl = 1, Ymin = 14, Ymax = 18, Yscl = 0.5 If B = 17.5, use a graphing calculator to determine that the point of intersection of y1 = 15.46e0.0155 t and y 2 = 17.5 is t ≈ 8 years. 1986 + 8 = 1994 Or algebraically, solve for t: 17.5 = 15.46e0.0155 t 17.5 = e0.0155 t 15.46 1.13195 = e0.0155 t ln (1.13195) = 0.0155 t ln e ln(1.13195) =t 0.0155 8≈t 1986 + 8 = 1994
53 = 106e−0.0909 t 53 = e−0.0909 t 106 1 = e−0.0909 t 2
ln 0.5 = −0.0909t ln 0.5 t= ≈ 7.6 days −0.0909
.......................................................
Assessing Concepts
1.
False; f ( x ) = x 2 does not have an inverse function.
2.
True
3.
True
4.
False; h( x) is not an increasing function for 0 < b < 1 .
5.
False; j ( x) is not an increasing function for 0 < b < 1 .
6.
c
7.
b
9.
a
11.
d
12.
g
8.
f
Copyright © Houghton Mifflin Company. All rights reserved.
10.
e
Chapter Review
291
....................................................... 1.
3.
⎛ x +5⎞ ⎛ x +5⎞ F [G ( x)] = F ⎜ ⎟ = 2⎜ ⎟ − 5 = x + 5 − 5 = x [4.1] 2 ⎝ ⎠ ⎝ 2 ⎠ 2x − 5 + 5 2x G[ F ( x)] = G (2 x − 5) = = =x 2 2 Yes, F and G are inverses.
Chapter Review 2.
( )
h[k ( x)] = h x 2 = x 2 = x k[h( x)] = k
( x) = ( x)
2
[4.1]
=x
Yes, h and k are inverses.
3
⎛ 3 ⎞ x −1 + 3 3 + 3( x − 1) 3 + 3 x − 3 3x = = = = x [4.1] l[m( x)] = l ⎜ ⎟= 3 3 3 3 ⎝ x −1⎠ x −1
3 3x 3x ⎛ x + 3⎞ = = =x m[l ( x)] = m ⎜ ⎟ = x +3 + − x x x 3 3 ⎝ ⎠ −1 x
Yes, l and m are inverses. 2x
4.
⎛ 2 x ⎞ x −5 − 5 2 x − 5( x − 5) 2 x − 5 x + 25 −3x + 25 p[q ( x)] = p ⎜ = = = ≠ x [4.1] ⎟= 2(2 x) 4x 4x ⎝ x − 5 ⎠ 2 2x
( x −5 )
No, p and q are not inverses. y = 3 x − 4 [4.1] x = 3y − 4 x + 4 = 3y x+4 =y 3
5.
f
9.
−1
( x) = 1 x + 4 3 3
log5 25 = x [4.3]
g
10.
x
13.
y = −2 x + 3 [4.1] x = −2 y + 3 x − 3 = −2 y x−3 =y −2
6.
−1
2 x =−1 y−2 2 1 x+2=− y 2
−2( x + 2) = y
( x) = − 1 x + 3 2 2
log3 81 = x [4.3]
y = − 1 x − 2 [4.1] 8.
7.
h −1( x) = −2 x − 4
11.
x
ln e3 = x [4.3]
5 = 25
3 = 81
e =e
5 x = 52 x=2
3x = 34 x=4
x=3
32 x + 7 = 27 [4.5] 32 x + 7 = 33 2x + 7 = 3 2 x = −4 x = −2
14.
5 x − 4 = 625 [4.5] 5 x − 4 = 54 x−4=4 x =8
x
15.
12.
3
2x = 1 8
y = 1 [4.1] x 1 x= y xy = 1 y=1 x −1 1 k ( x) = x
ln eπ = x x
[4.3]
π
e =e
x =π
[4.5]
2 x = 2 −3 x = −3
16.
27 ( 3x ) = 3−1 [4.5] 27 ( 3x ) = 1 3 x 3 = 1 81 3x = 3−4 x = −4
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
292 log x 2 = 6
17.
[4.5]
18.
106 = x 2 1,000,000 = x
2
1 log x = 5 2 log x = 10
[4.5]
19.
1010 = x
± 1,000,000 = x ±1000 = x
10log 2 x = 14 [4.5] 2 x = 14 x=7
22.
23.
24.
25.
26.
28.
29.
31.
33.
x 2 = 64 x = ±8
30.
32.
log 4 64 = 3
[4.3]
34.
3
4 = 64
37.
53 = 125 [4.3] log5 125 = 3
41.
log b
43.
45.
2
eln x = 64 [4.5]
x = ±1010
21.
27.
20.
log1/ 2 8 = −3 [4.3] ⎛1⎞ ⎜ ⎟ ⎝2⎠
−3
35.
(
=8
2
4 = 4 [4.3]
42.
logb
ln xy 3 = ln x + 3 ln y [4.4]
44.
ln
2 log x + 1 log ( x + 1) = log ( x 2 3 x + 1 ) [4.4] 3
46.
5log x − 2 log( x + 5) = log
ln1 = 0 [4.3] e0 = 1
2) = 4
100 = 1 [4.3] log10 1 = 0
x2 y3 = 2 log b x + 3 log b y − log b z [4.4] z
36.
4
39.
38.
210 = 1024 [4.3] log2 1024 = 10
log
40.
81/ 2 = 2 2 [4.3] log8 2 2 = 1 2
x = 1 log x − 2log y + log z [4.4] ( b b b ) y z 2 = 1 logb x − 2logb y − logb z 2 2
xy z4
= 1 ( ln x + ln y ) − 4 ln z [4.4] 2 1 = ln x + 1 ln y − 4 ln z 2 2
Copyright © Houghton Mifflin Company. All rights reserved.
x5 [4.4] ( x + 5)2
Chapter Review
293
47.
1 ln 2 xy − 3ln z = ln 2 xy [4.4] 2 z3
49.
log 5 101 =
51.
log 4 0.85 =
53.
4 x = 30
ln x − (ln y − ln z ) = ln
log 101 ≈ 2.86754 [4.4] log 5
50.
log 3 40 =
log 40 ≈ 3.35776 [4.4] log 3
log 0.85 ≈ −0.117233 [4.4] log 4
52.
log 8 0.3 =
log 0.3 ≈ −0.578989 [4.4] log 8
54.
[4.5]
x
log 4 = log 30 x log 4 = log 30 log 30 x= log 4
56.
5 x +1 = 41 ( x + 1) log5 = log 41
ln ( 3x ) − ln ( x − 1) = ln 4
55.
[4.5]
log 41 log5 log 41 x= −1 log5
57.
ln ( 3x ⋅ 2 ) = 1
3x = 4 x − 4 4=x
ln x + 2 e ( ) = 6 [4.5]
58.
log 2 x +1) 10 ( = 31 [4.5]
( x + 2) = 6
ln ( 6 x ) = 1
2 x + 1 = 31 2 x = 30
x+2=6 x=4
1
e = 6x e=x 6 4 x + 4− x = 2 4 x − 4− x
x = 15
5 x + 5− x = 8 2
60.
5 x ( 5 x + 5− x ) = 16 ( 5x )
4 x ( 4 x + 4− x ) = 2 ( 4 x − 4− x ) 4 x
52 x + 1 = 16 ( 5x )
42 x + 1 = 2 ( 42 x − 1)
52 x − 16 ( 5 x ) + 1 = 0
42 x + 1 = 2 ( 42 x − 1)
Let 5 x = u
42 x − 2 ⋅ 42 x + 3 = 0
u 2 − 16u + 1 = 0
42 x = 3 2 x ln 4 = ln 3 x = ln 3 2 ln 4
[4.5]
ln 3x = ln 4 x −1 3x = 4 x −1 3x = 4 ( x − 1)
x +1=
ln ( 3x ) + ln 2 = ln1 [4.5]
59.
x xz = ln [4.4] y z y
48.
u=
16 ±
162 − 4 (1)(1)
2 16 252 ± u= 2 16 6 7 ± u= 2 u =8±3 7
[4.5]
5x = 8 ± 3 7 x= 61.
log ( log x ) = 3
[4.5]
62.
ln ( ln x ) = 2
3
10
=x
1000
=x
10
63.
log x − 5 = 3
2
10 = log x (10 3 )
[4.5]
ln (8 ± 3 7 ) [4.5] ln 5
e = ln x e
( e2 )
=x
3
10 = x − 5 106 = x − 5 106 + 5 = x x = 1,000,005
Copyright © Houghton Mifflin Company. All rights reserved.
[4.5]
Chapter 4: Exponential and Logarithmic Functions
294 64.
log x + log ( x − 15) = 1
65.
log x ( x − 15) = 1
3 =x 81 = x
1 = log5 x 2
[4.5]
5 = x2
15 ± 152 − 4 (1)( −10 )
± 5=x
[4.5]
2
x = 15 ± 265 2 + 15 265 x= 2
[4.5]
log5 x 3 = log5 16 x [4.5]
68.
25 = 16log4 x [4.5] 25 = 4
2
25 = 4log4 x
x = 16 x=4
⎛ I ⎞ m = log ⎜ ⎟ [4.4] ⎝ I0 ⎠ ⎛ 51,782,000 I 0 ⎞ = log ⎜ ⎟ I0 ⎝ ⎠ = log 51,782,000 ≈ 7.7
69.
2 log 4 x
3
x = 16 x
70.
70 = log5 x 2
4
0 = x 2 − 15 x − 10
67.
log7 (log5 x 2 ) = 0
4 = log3 x
10 = x 2 − 15 x
x=
66.
log 4 (log3 x ) = 1
2
25 = x 2 ±5 = x 5= x
M = log A + 3log8t − 2.92 [4.4] = log18 + 3log8(21) − 2.92 = log18 + 3log168 − 2.92 ≈ 5.0
71.
⎛I ⎞ log ⎜ 1 ⎟ = 7.2 ⎝ I0 ⎠ I1 = 107.2 I0
⎛I ⎞ log ⎜ 2 ⎟ = 3.7 ⎝ I0 ⎠ I2 = 103.7 I0
and
I1 = 107.2 I 0
[4.4]
I 2 = 103.7 I 0
I1 107.2 I 0 103.5 3162 = = ≈ I 2 103.7 I 0 1 1 3162 to 1
72.
I1 = 600 = 10 x [4.4] I2
73.
pH = − log ⎡⎢ H3O + ⎤⎥ ⎣ ⎦
[4.4]
5.4 = − log ⎡⎢ H3O + ⎤⎥ [4.4] ⎣ ⎦
74.
= − log ⎡⎢ 6.28 × 10−5 ⎤⎥ ⎣ ⎦ ≈ 4.2
log 600 = log10 x log 600 = x 2.8 ≈ x
−5.4 = log ⎡⎢ H3O + ⎤⎥ ⎣ ⎦ 10−5.4 = H3O + 0.00000398 ≈ H3O + H3O + ≈ 3.98 × 10−6
75.
P = 16,000, r = 0.08, t = 3 [4.6] a. b.
0.08 ⎞ ⎛ B = 16,000⎜1 + ⎟ 12 ⎠ ⎝
36
76.
≈ $20,323.79
B = 16,000e 0.08(3)
P = 19,000, r = 0.06, t = 5 [4.6] 1825
a.
0.06 ⎞ ⎛ B = 19,000⎜1 + ⎟ 365 ⎠ ⎝
b.
B = 19,000e 0.3 ≈ $25,647.32
B = 16,000e 0.24 ≈ $20,339.99
Copyright © Houghton Mifflin Company. All rights reserved.
≈ $25,646.69
Chapter Review
77.
295
S (n ) = P (1 − r )n , P = 12,400, r = 0.29, t = 3 [4.6]
78.
N ( t ) = N 0 e −0.12t
a.
S (n ) = 12,400(1 − 0.29 )3 ≈ $4438.10
N (10 ) = N 0 e N (10 ) N0
[4.6]
−0.12(10 )
= e −1.2 = .301
N (10 ) N0
= 30.1% healed
100% − 30.1% = 69.9% healed b.
N (t ) N0
= 0.5
0.5 = e −0.12t ln 0.5 = −0.12t ln 0.5 = t −0.12 t ≈ 6 days c.
N (t ) N0
= 0.1
0.1 = e −0.12t ln 0.1 = −0.12t ln 0.1 = t −0.12 t ≈ 19 days 79.
N (2) = 5
N ( 0) = 1 k 0 1 = N 0e ( )
80.
5 = e2 k ln 5 = 2k k = ln 5 ≈ 0.8047 2
1 = N0
N (0) = N 0 = 2 and N (3) = N 0 e 3k = 2e 3k = 11 e3k = 11 2 3k 11 e = 2 3k = ln ⎛⎜ 11 ⎞⎟ ⎝2⎠
Thus N ( t ) = e0.8047 t [4.6]
k = 1 ln ⎛⎜ 11 ⎞⎟ 3 ⎝2⎠ ≈ 0.5682
Thus N (t ) = 2e 0.5682t [4.6] 81.
4 = N (1) = N 0e k and thus
4 = e k . Now, we also N0 5
⎛ 4 ⎞ 1024 ⎟ = . have N (5) = 5 = N 0e5k = N 0 ⎜⎜ ⎟ N 04 ⎝ N0 ⎠ N0 = 4
1024 ≈ 3.783 5
82.
1 = N (0 ) = N 0 and 2 = N (− 1) = N 0e− k .
Since N 0 = 1 , we have 2 = 1 ⋅ e− k . ln 2 = − k k ≈ −0.6931 Thus N ( t ) = e−0.6931 t . [4.6]
Thus 4 = 3.783ek . ⎛ 4 ⎞ ln⎜ ⎟=k ⎝ 3.783 ⎠ k ≈ 0.0558
Thus N 0 = 3.783e0.0558 t . [4.6] Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
296 83.
a.
N (1) = 25, 200ek (1) = 26,800 [4.6] k 26,800 e = 25, 200 ⎛ 26,800 ⎞ k ln e = ln ⎜ ⎟ ⎝ 25, 200 ⎠ k ≈ 0.061557893
84.
P(t) = 0.5 t / 5730 = 0.96 [4.6]
t log 0.5 = log 0.96 5730 log 0.96 t = 5730 log 0.5
N (t ) = 25, 200e0.061557893 t
b.
)
(
log 0.5 t / 5730 = log 0.96
⎛ log 0.96 ⎞ t = 5730 ⎜ ⎟ ≈ 340 years ⎝ log 0.5 ⎠
N (7) = 25, 200e0.061557893(7) = 25, 200e0.430905251 ≈ 38,800
85.
Answers will vary.
86.
a.
b. c. 87.
a.
exponential: R ≈ 179.949 ( 0.968094t ) , r ≈ −0.99277 logarithmic: R ≈ 171.19665 − 35.71341ln t , r ≈ −0.98574 The exponential equation provides a better fit to the data.
R ≈ 179.949 ( 0.968094108 ) ≈ 5.4 per 1000 live births [4.7] P(t ) =
mP0
P0 + ( m − P0 )e − kt
P(3) = 360 =
88.
128 = 128 = 128 = 128 = 21 1 3 6 1 + 5e −0.27(0) 1 + 5e0 1 + 5
a.
P(0) =
b.
As t → ∞, e−0.27t → 0. P(t ) → 128 = 128 = 128 [4.6] 1 + 5(0) 1
1400(210)
210 + (1400 − 210)e − k (3) 294000 360 = 210 + 1190e −3k
360 ( 210 + 1190e −3k ) = 294000
210 + 1190e −3k = 294000 360 −3k 29400 1190e = − 210 36 e −3k = 29400 / 36 − 210 1190 −3k ⎛ 29400 / 36 − 210 ⎞ ln e = ln ⎜ ⎟ 1190 ⎝ ⎠ −3k = ln ⎛⎜ 29400 / 36 − 210 ⎞⎟ 1190 ⎝ ⎠
b.
k = − 1 ln ⎛⎜ 29400 / 36 − 210 ⎞⎟ 3 ⎝ 1190 ⎠ k ≈ 0.2245763649 294000 1400 P(t ) = = 210 + 1190e −0.22458t 1 + 17 e −0.22458t 3 294000 P(13) = 210 + 1190e −0.22458(13) 294000 = 210 + 1190e −2.919492744 ≈ 1070 coyotes [4.6]
Copyright © Houghton Mifflin Company. All rights reserved.
Quantitative Reasoning
297
....................................................... QR1.
Exponential: S ≈ 5.94860(1.72192)t Logistic: S ≈
244.56468 1 + 39.38651e −0.57829 t
QR2. Exponential: S ≈ 5.94860(1.72192)10 ≈ 1363.1 million 244.56468 Logistic: S ≈ ≈ 218.1 million 1 + 39.38651e −0.57829 (10) QR3.
S ≈ 58.73554 + 16.75801ln t QR4. S ≈ 58.73554 + 16.75801ln(6) ≈ 88.8 million QR5.
S≈
85.24460 1 + 14.0040e −0.70591 t
QR6. Answers will vary.
Copyright © Houghton Mifflin Company. All rights reserved.
Quantitative Reasoning
Chapter 4: Exponential and Logarithmic Functions
298
....................................................... y = 2 x − 3 [4.1]
1.
Chapter Test 2.
x = 2y − 3 x + 3 = 2y 1x+ 3 = y 2 2 f −1( x) = 1 x + 3 2 2
x [4.1] 4x − 8 y x= 4y − 8 x(4 y − 8) = y f ( x) = y =
4 xy − 8 x = y 4 xy − y = 8 x y (4 x − 1) = 8 x 8x y= 4x − 1 8x f −1( x) = 4x − 1 4x − 1 ≠ 0 ⇒ 4x ≠ 1 ⇒ x ≠ 1
4 − 1 Domain of f : all real numbers except 1 . 4 −1
Range of f
Range of f 3.
a.
logb (5 x − 3) = c [4.3]
b.
bc = 5 x − 3
4.
logb
z2 y
3
x
5.
: all real numbers except 2.
3x / 2 = y log3 y =
= logb z 2 − logb y 3 − logb x1/ 2 [4.4]
= domain of f ⇒ 4 x − 8 ≠ 0 ⇒ x ≠ 2 .
−1
x 2
log10 ( 2 x + 3) − 3log10 ( x − 2 ) = log10 ( 2 x + 3) − log10 ( x − 2 ) = log10 2 x + 33 [4.4] ( x − 2)
1 = 2logb z − 3logb y − logb x 2
6.
10.
log12 [4.4] log 4 ≈ 1.7925
log 4 12 =
45− x = 7 x 5− x
7.
8.
[4.5] x
ln 4 = ln 7 (5 − x ) ln 4 = x ln 7 5ln 4 − x ln 4 = x ln 7 5ln 4 = x ln 7 + x ln 4 5ln 4 = x (ln 7 + ln 4) 5ln 4 = x ln 28
11.
12. log( x + 99) − log(3 x − 2) = 2 [4.5] x + 99 log =2 3x − 2 x + 99 = 102 3x − 2 x + 99 = 100(3 x − 2) x + 99 = 300 x − 200 −299 x = −299 x =1
9.
[4.5] 5 x = 22 x log5 = log 22 log 22 x= log5 x ≈ 1.9206
ln(2 − x) + ln(5 − x) = ln(37 − x) ln(2 − x)(5 − x) = ln(37 − x) (2 − x)(5 − x) = (37 − x) 10 − 7 x + x 2 = 37 − x x 2 − 6 x − 27 = 0 ( x − 9)( x + 3) = 0 x = 9 (not in domain) or x = −3 x = −3 [4.5]
Copyright © Houghton Mifflin Company. All rights reserved.
3
Chapter Test
13.
a.
299
(
) 12(5) = 20,000 (1 + 0.078 ) 12
nt A = P 1+ r n
= 20,000(1.0065) = $29,502.36 b.
r⎞ ⎛ A = P ⎜1 + ⎟ ⎝ n⎠
14.
[4.6]
(
nt
(
2 = 1+
(
)
0.04 12t 12 0.04 12t
2P = P 1 +
60
[4.6]
12
)
)
0.04 12t 12 0.04 ln 2 = 12t ln 1 + 12
ln 2 = ln 1 +
A = Pert
= 20,000e0.078(5) = 20,000e0.39 = $29,539.62
12t =
(
(
ln 2
ln 1 +
0.04 12
)
)
1 ln 2 t= ⋅ 12 ln 1 + 0.04
(
t ≈ 17.36 years 15.
a.
⎛ I ⎞ M = log ⎜ ⎟ [4.4] ⎝ I0 ⎠ ⎛ 42,304,000 I 0 ⎞ = log ⎜ ⎟ I0 ⎝ ⎠ = log 42,304,000 ≈ 7.6
16.
a.
12
)
N (3) = 34600ek (3) = 39800
34600e3k = 39800 39800 e3k = 34600 ⎛ 398 ⎞ 3k ln e = ln ⎜ ⎟ ⎝ 346 ⎠ ⎛ 398 ⎞ 3k = ln ⎜ ⎟ ⎝ 346 ⎠ ⎛ 398 ⎞ k = 1 ln ⎜ 3 ⎝ 346 ⎟⎠ k ≈ 0.0466710767 N (t ) = 34600e0.0466710767 t [4.6]
b.
⎛I ⎞ log ⎜ 1 ⎟ = 6.3 ⎝ I0 ⎠ I1 = 106.3 I0
and
⎛I ⎞ log ⎜ 2 ⎟ = 4.5 ⎝ I0 ⎠ I2 = 104.5 I0
I1 = 106.3 I 0
b.
N (10) = 34600e0.0466710767(10) = 34600e0.466710767 ≈ 55,000
I 2 = 104.5 I 0
I1 106.3 I 0 101.8 63 = = ≈ I 2 104.5 I 0 1 1 Therefore the ratio is 63 to 1.
17.
P (t ) = 0.5 t / 5730 = 0.92 [4.6] log 0.5 t / 5730 = log 0.92 t log 0.5 = log 0.92 5730 t log 0.92 = 5730 log 0.5 ⎛ log 0.92 ⎞ t = 5730 ⎜ ⎟ ⎝ log 0.5 ⎠ t ≈ 690 years
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
300 18.
a.
y = 1.671991998(2.471878247) x
b.
y = 1.671991998(2.471878247)7.8 ≈ 1945
[4.7]
19.
20.
a.
Logarithmic: d ≈ 67.35501 + 2.54015ln t ; Logistic: d ≈
b.
Logarithmic: d ≈ 67.35501 + 2.54015ln(12) ≈ 73.67 m ; 72.03783 ≈ 72.03 m [4.7] Logistic: d ≈ 1 + 0.15279e −0.67752 (12)
a.
a=
c − P0 1100 − 160 = = 5.875 P0 160
72.03783 1 + 0.15279e −0.67752 t
b.
P(t ) =
1100 ≈ 457 raccoons 1 + 5.875e −0.20429(7)
c 1 + ae − bt 1100 P(1) = 1 + 5.875e − b(1) 1100 190 = 1 + 5.875e − b P (t ) =
190(1 + 5.875e − b ) = 1100 190 + 1116.25e − b = 1100 1116.25e − b = 910 e − b = 910 1116.25 −b ln e = ln 910 1116.25 −b = ln 910 1116.25 b = − ln 910 1116.25 b ≈ 0.20429 1100 P(t ) = [4.6] 1 + 5.875e −0.20429 t
....................................................... 1.
x − 4 ≤ 2 ⇒ −2 ≤ x − 4 ≤ 2 ⇒ 2 ≤ x ≤ 6 . The solution is [2, 6]. [1.5]
Copyright © Houghton Mifflin Company. All rights reserved.
Cumulative Review
Cumulative Review
2.
3.
301
The critical values are: − x + 6 = 0 or 2 x − 6 = 0 x=6 x=3 The intervals are: ( −∞, 3),(3, 6), and (6, ∞). The quotient − x + 6 is positive or zero. 2x − 6
x ≥ 1 [1.5] 2x − 6 x −1≥ 0 2x − 6 x − 2x − 6 ≥ 0 2x − 6 2x − 6 x − 2x + 6 ≥ 0 2x − 6 −x + 6 ≥ 0 2x − 6
2
The denominator ≠ 0 ⇒ x ≠ 3. The solution is {x | 3 < x ≤ 6}.
Find the y-value of the vertex of [2.4]
4.
d = (11 − 5)2 + (7 − 2)2 [2.1]
−x + 6 2x − 6
h(t ) = −16t 2 + 44t + 8 . − b = − 44 = 1.375 2a 2( −16)
2
= 6 + 5 = 36 + 25 = 61 ≈ 7.8
h(1.375) = −16(1.375)2 + 44(1.375) + 8 = 38.25 feet f ( x) = 2 x + 1
5.
[2.6]
6.
2
(g o
g ( x) = x − 5 f )( x ) = g[ f ( x )] = g (2 x + 1)
f ( x ) = 3x − 5 x = 3y − 5
7.
[4.1]
L = kwd 2 1500 = k (4)(8) 1500 = 256k 1500 = 375 = k 256 64 375 L= wd 2 64 L = 375 (6)(10)2 64 L ≈ 3500 pounds
x + 5 = 3y 1x+5= y 3 3 f −1 ( x ) = 1 x + 5 3 3
= (2 x + 1)2 − 5 = 4 x2 + 4 x + 1 − 5 = 4 x2 + 4 x − 4
8.
[1.6] 2
P( x ) = x 4 − 3x 3 + x 2 − x − 6 has three changes of sign. There are three or one positive real zeros. P( − x ) = x 4 + 3x 3 + x 2 + x − 6 has one change of sign. There is one negative real zero. [3.3]
9.
P( x ) = x 4 − 5 x 3 + x 2 + 15 x − 12 has three or one positive and one negative real zeros. [3.3] p = ±1, ± 2, ± 3, ± 4, ± 6, ± 12 are the possible rational zeros. q 1 1 1 15 −5 −12 1 12 −4 −3 1 12 0 −4 −3 2
2
x − 3 = 0 ⇒ x = 3 ⇒ x = ± 3 The zeros are 1, 4, − 3, 10.
4
1 1
−4 4 0
−3 0 −3
3.
P( x ) = ( x − 2)[ x − (1 − i )][ x − (1 + i )] = ( x − 2)[ x − 1 + i ][ x − 1 − i ] = ( x − 2)[( x − 1) + i ][( x − 1) − i ] 2
2
2
12 −12 0
2
[3.4]
2
= ( x − 2)[( x − 1) − i ] = ( x − 2)[ x − 2 x + 1 − ( −1)] = ( x − 2)[ x − 2 x + 1 + 1] = ( x − 2)( x − 2 x + 2) = x ( x 2 − 2 x + 2) − 2( x 2 − 2 x + 2) = x 3 − 2 x 2 + 2 x − 2 x 2 + 4 x − 4 = x 3 − 4 x 2 + 6 x − 4 11.
12.
r ( x ) = 3x − 5 [3.5] x−4
Vertical asymptote: x − 4 = 0 ⇒ x = 4
Horizontal asymptote: (n = m) ⇒ y = 3 ⇒ y = 3 1
4 . The denominator of R(x) will not equal zero for any real value of x. Thus, there are no restrictions on the domain. x +1 The domain is all real numbers. R(x) is positive for all values of x. The smallest denominator value is 1, thus the highest R(x) value is 4. The range is { y | 0 < y ≤ 4}. [3.5] R( x ) =
2
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 4: Exponential and Logarithmic Functions
302 13.
f ( x ) = 0.4 x is a decreasing function since 0.4 < 1. [4.2]
14.
log 4 x = y ⇒ 4 y = x [4.3]
16.
11,650,600 I 0 M = log I = log = log11,650,600 ≈ 7.1 [4.4] I0 I0
17.
2e x = 15 ⇒ e x = 7.5 ⇒ ln e x = ln 7.5 ⇒ x = ln 7.5 ⇒ x ≈ 2.0149 [4.5]
18.
0.5N 0 = N 0 e5730k ⇒ 0.5 = e5730k ⇒ ln 0.5 = ln e5730k ⇒ ln 0.5 = 5730k ⇒ k = ln 0.5 ≈ −0.000121 [4.6] 5730
15.
53 = 125 ⇒ log5 125 = 3 [4.3]
N (t ) = N 0 e −0.000121t ⇒ 0.94 N 0 = N 0 e −0.000121t ⇒ 0.94 = e −0.000121t ⇒ ln 0.94 = ln e −0.000121t
⇒ ln 0.94 = −0.000121t ⇒ t = e x − e − x = 12 2
19.
ln 0.94 ≈ 510 years old −0.000121
[4.5]
e x ( e x − e − x ) = (24)e x e2 x − 1 = e x (24) e2 x − 24e x − 1 = 0
Let u = e x . u 2 − 24u − 1 = 0 24 ± 576 − 4( −1) 2 24 580 ± u= 2 24 2 145 ± u= 2 u = 12 ± 145 u=
e x = 12 ± 145 x ln e = ln (12 + 145) cannot take ln of a negative x = ln (12 + 145) x ≈ 3.1798 20.
a.
a=
c − P0 450 − 160 290 = = = 1.8125 [4.7] P0 160 160
450 450 ⇒ 205 = ⇒ 205(1 + 1.8125e3k ) = 450 ⇒ 205 + 371.5625e3k = 450 1 + 1.8125ek (3) 1 + 1.8125e3k ⇒ 371.5625e3k = 245 ⇒ e3k = 245 ⇒ ln e3k = ln 245 371.5625 371.5625 ln 245 ln 371.5625 − ⇒ 3k = ln 245 − ln 371.5625 ⇒ k = ⇒ k ≈ −0.13882 3 450 P(t ) ≈ 1 + 1.8125e −0.13882t 450 ≈ 310 wolves P(10) ≈ 1 + 1.8125e −0.13882(10) P(3) =
b.
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 5
Trigonometric Functions Section 5.1 1.
90o − 15o = 75o
2.
180o − 15o = 165o
90o − 87o = 3o 180o − 87o = 93o
90° 89°60′ = ′ −70°15 −70°15′ 19°45′
5.
90° 89°59′60′′ = −56°33′15′′ −56°33′15′′ 33°26′45′′
180° 179°59′60′′ = −56°33′15′′ −56°33′15′′ 123°26′45′′
6.
90° 89°59′60′′ = −19°42′05′′ −19°42′05′′ 70°17′55′′
180° 179°59′60′′ = −19°42′05′′ −19°42′05′′ 160°17′55′′
7.
π −1 2 π −1
8.
π − 0.5 2 π − 0.5
9.
π−π=π 2 4 4 π − π = 3π 4 4
10.
π−π=π 2 3 6 π − π = 2π 3 3
11.
π − 2π = π 2 5 10 π − 2π = 3π 5 5
12.
π−π=π 2 6 3 π − π = 5π 6 6
13.
610 o = 250 o + 360 o α is a quadrant III angle coterminal with an angle of measure 250°.
14.
765 o = 45 o + 2 ⋅ 360 o α is a quadrant I angle coterminal with an angle of measure 45°.
15.
− 975 o = 105 o − 3 ⋅ 360 o α is a quadrant II angle coterminal with an angle of measure 105°.
16.
− 872 o = 208 o − 3 ⋅ 360 o α is a quadrant III angle coterminal with an angle of measure 208°.
17.
2456 o = 296 o + 6 ⋅ 360 o α is a quadrant IV angle coterminal with an angle of measure 296°.
18.
− 3789 o = 171o − 11 ⋅ 360 o α is a quadrant II angle coterminal with an angle of measure 171°.
19.
On a TI-83 graphing calculator, the degree symbol, ° , and the DMS function are located in the ANGLE menu.
20.
On a TI-83 graphing calculator, the degree symbol, ° , and the DMS function are located in the ANGLE menu.
24.56° = 24°33′36′′
180° 179°60′ = ′ −70°15 −70°15′ 109°45′
4.
90o 89°60′ = ′ −22°43 −22°43′ 67°17′
3.
180° 179°60′ = ′ −22°43′ −22°43 157°17′
110.24° = 110°14′24′′
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304
21.
Chapter 5: Trigonometric Functions
On a TI-83 graphing calculator, the degree symbol, ° , and the DMS function are located in the ANGLE menu.
22.
18.96° = 18°57′36′′
64.158° = 64°9′28.8′′ 23.
On a TI-83 graphing calculator, the degree symbol, ° , and the DMS function are located in the ANGLE menu.
24.
A TI-83 calculator needs to be in degree mode to convert a DMS measure to its equivalent degree measure. On a TI-83 both the degree symbol, ° , and the minute symbol, ' , are located in the ANGLE menu. The second symbol, '' , is entered by pressing ALPHA followed by [ '' ] which is located on the plus sign, [+], key.
26.
25°25′12′′ = 25.42° 27.
A TI-83 calculator needs to be in degree mode to convert a DMS measure to its equivalent degree measure. On a TI-83 both the degree symbol, ° , and the minute symbol, ' , are located in the ANGLE menu. The second symbol, '' , is entered by pressing ALPHA followed by [ '' ] which is located on the plus sign, [+], key.
63°29′42′′ = 63.495°
A TI-83 calculator needs to be in degree mode to convert a DMS measure to its equivalent degree measure. On a TI-83 both the degree symbol, ° , and the minute symbol, ' , are located in the ANGLE menu. The second symbol, '' , is entered by pressing ALPHA followed by [ '' ] which is located on the plus sign, [+], key.
183°33′36′′ = 183.56°
On a TI-83 graphing calculator, the degree symbol, ° , and the DMS function are located in the ANGLE menu.
224.282° = 224°16′55.2′′
3.402° = 3°24′7.2′′ 25.
On a TI-83 graphing calculator, the degree symbol, ° , and the DMS function are located in the ANGLE menu.
28.
A TI-83 calculator needs to be in degree mode to convert a DMS measure to its equivalent degree measure. On a TI-83 both the degree symbol, ° , and the minute symbol, ' , are located in the ANGLE menu. The second symbol, '' , is entered by pressing ALPHA followed by [ '' ] which is located on the plus sign, [+], key.
141°6′9′′ = 141.1025°
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.1
29.
305
A TI-83 calculator needs to be in degree mode to convert a DMS measure to its equivalent degree measure. On a TI-83 both the degree symbol, ° , and the minute symbol, ' , are located in the ANGLE menu. The second symbol, '' , is entered by pressing ALPHA followed by [ '' ] which is located on the plus sign, [+], key.
30.
A TI-83 calculator needs to be in degree mode to convert a DMS measure to its equivalent degree measure. On a TI-83 both the degree symbol, ° , and the minute symbol, ' , are located in the ANGLE menu. The second symbol, '' , is entered by pressing ALPHA followed by [ '' ] which is located on the plus sign, [+], key.
19°12′18′′ = 19.205°
211°46′48′′ = 211.78° 31.
30o = 30o ⎛⎜ π o ⎞⎟ = π ⎝ 180 ⎠ 6
32.
−45o = −45o ⎛⎜ π o ⎞⎟ = − π 4 ⎝ 180 ⎠
33.
90o = 90o ⎛⎜ π o ⎞⎟ = π ⎝ 180 ⎠ 2
34.
15o = 15o ⎛⎜ π o ⎞⎟ = π ⎝ 180 ⎠ 12
35.
165o = 165o ⎛⎜ π ⎞⎟ = 11π ⎝ 180o ⎠ 12
36.
315o = 315o ⎛⎜ π o ⎞⎟ = 7π ⎝ 180 ⎠ 4
37.
420o = 420o ⎛⎜ π ⎞⎟ = 7π ⎝ 180o ⎠ 3
38.
630o = 630o ⎛⎜ π ⎞⎟ = 7π ⎝ 180o ⎠ 2
39.
585o = 585o ⎛⎜ π o ⎞⎟ = 13π 4 ⎝ 180 ⎠
40.
135o = 135o ⎛⎜ π o ⎞⎟ = 3π ⎝ 180 ⎠ 4
41.
−9o = −9o ⎛⎜ π o ⎞⎟ = − π 20 ⎝ 180 ⎠
42.
−110o = −110o ⎛⎜ π o ⎞⎟ = − 11π 18 ⎝ 180 ⎠
43.
7π = 7π ⎛ 180o ⎞ = 420o ⎜ ⎟ 3 3 ⎝ π ⎠
44.
π = π ⎛ 180o ⎞ = 45o ⎜ ⎟ 4 4⎝ π ⎠
45.
π = π ⎛ 180o ⎞ = 36o ⎜ ⎟ 5 5⎝ π ⎠
46.
o⎞ ⎛ − 2π = − 2π ⎜ 180 ⎟ = −120o 3 3 ⎝ π ⎠
47.
π = π ⎛ 180o ⎞ = 30o ⎜ ⎟ 6 6⎝ π ⎠
48.
π = π ⎛ 180o ⎞ = 20o ⎜ ⎟ 9 9⎝ π ⎠
49.
3π = 3π ⎛ 180o ⎞ = 67.5o ⎜ ⎟ 8 8 ⎝ π ⎠
50.
11π = 11π ⎛ 180o ⎞ = 110o ⎜ ⎟ 18 18 ⎝ π ⎠
51.
11π = 11π ⎛ 180o ⎞ = 660o ⎜ ⎟ 3 3 ⎝ π ⎠
52.
6π = 6π ⎛ 180o ⎞ = 216o ⎜ ⎟ 5 5 ⎝ π ⎠
53.
o⎞ ⎛ − 5π = − 5π ⎜ 180 ⎟ = −75o 12 12 ⎝ π ⎠
54.
o⎞ ⎛ − 4π = − 4π ⎜ 180 ⎟ = −144o 5 5 ⎝ π ⎠
55.
o⎞ ⎛ 1.5 = 1.5 ⎜ 180 ⎟ ≈ 85.94o ⎝ π ⎠
56.
o⎞ ⎛ −2.3 = −2.3 ⎜ 180 ⎟ ≈ −131.78o ⎝ π ⎠
57.
133o = 133o ⎛⎜ π o ⎞⎟ ≈ 2.32 ⎝ 180 ⎠
58.
427o = 427o ⎛⎜ π o ⎞⎟ ≈ 7.45 ⎝ 180 ⎠
59.
o⎞ ⎛ 8.25 = 8.25 ⎜ 180 ⎟ ≈ 472.69o ⎝ π ⎠
60.
−90o = −90o ⎛⎜ π o ⎞⎟ ≈ −1.57 ⎝ 180 ⎠
61.
θ=s r = 8 =4 2 o = 4 ⎛⎜ 180 ⎞⎟ ≈ 229.18o ⎝ π ⎠
62.
63.
θ=s r = 12.4 ≈ 2.38 5.2 o = 2.38 ⎛⎜ 180 ⎞⎟ ≈ 136.63o ⎝ π ⎠
θ=s r = 4 ≈ 0.57 7 o = 0.57 ⎛⎜ 180 ⎞⎟ ≈ 32.74o ⎝ π ⎠
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306
64.
67.
Chapter 5: Trigonometric Functions
θ=s r = 84.3 ≈ 2.35 35.8 o = 84.3 ⎛⎜ 180 ⎞⎟ ≈ 134.92o 35.8 ⎝ π ⎠
65.
s = rθ
68.
73.
76.
79.
θ = 3 ( 2π ) 8 = 3π 4
s = rθ = 3 ⎛⎜ 7π ⎞⎟ ⎝ 2 ⎠ = 32.99 ft
s = rθ
69.
θ = 3 ( 2π )
72.
θ2 =
2 = 3π
⎛ ⎞ = 5 ⋅ (144o ) ⎜ π o ⎟ ⎝ 180 ⎠ ≈ 12.57 m 71.
θ2 =
r1 θ1 r2
ω=θ
t 2π = 60 = π radian/sec 30
ω=θ
t ( 2π ) 200 = 60 = 20π radians/sec 3
v = ωr
74.
s = rθ = 6 ⎛⎜ 5π ⎞⎟ ⎝ 6 ⎠ = 5π in.
b.
s = rθ 24 = 6θ θ = 4 radians
= 2π radians or 360o
75.
ω=θ
78.
ω=v
t ( 2π ) 50 = 60 = 5π radians/sec 3
= 2π 86,400 ≈ 7.27 × 10−5 radian/sec
77.
ω=θ t
=
(
2π 33 13
r
)
= 55 ⋅ 5280 ⋅ 12 3600 14 ≈ 69.14 radians/sec
60 10π radians/sec = 9 ≈ 3.49 radians per second 80.
a.
ω=θ t
v = ωr
r1θ1 = r2θ 2
81.
= 500 ⋅ 2π ⋅ 60 ⋅ 18 12 ⋅ 5280 ≈ 54 mph
83.
r2 ⋅ θ2 r1
= 1.2 ( 240 o ) ⎛⎜ π o ⎞⎟ 0.8 ⎝ 180 ⎠
= 14 ( 150 o ) ⎛⎜ π o ⎞⎟ 28 ⎝ 180 ⎠ 5π = radians or 75o 12
= 450 ⋅ 2π ⋅ 60 ⋅ 15 12 ⋅ 5280 ≈ 40 mph
82.
66.
= (8 ) π 4 ≈ 6.28 in.
⎛ ⎞ = 25 ⋅ ( 42o ) ⎜ π o ⎟ ⎝ 180 ⎠ ≈ 18.33 cm 70.
s = rθ
(3.5)(150 ⋅ 2π ) = (1.75)θ 2 600π = θ 2 300(2π ) = θ 2 The rear gear is making 300 revolutions. The rear gear and tire are making same number of revolutions Tire is 12 inches = 1 ft s = 1 ft(300)(2π ) = 1885 ft
Speed for outer ring: v=s t 2π (32) 60 2 = ⋅ 3.75 5280 ≈ 36.5567 mph
Speed for inner ring: v=s t 2π (38) 60 2 = ⋅ 3.75 5280 ≈ 43.4111 mph
The outer swing has a greater speed of 43.4111 − 36.5567 ≈ 6.9 mph.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.1
84.
307
For the lead horse: d1 = π (202) ≈ 634.6 ft
85.
s = rθ ⎛ 1o ⎞ ⎛ π ⎟⎜ = 93,000,000(31′) ⎜ ⎜ 60 ′ ⎟ ⎜⎝ 180 o ⎝ ⎠ ≈ 840,000 mi
d t1 = 1 = 634.6 ≈ 26.008 sec r 24.4
For the second horse: d 2 = π (206.5) ≈ 648.7 ft
⎞ ⎟ ⎟ ⎠
d2 = 648.7 ≈ 24.9 ft/s 26.008 t The second horse must go 24.9 ft/s. r=
86.
a. b.
θ = 36 (2π) = 6π radians 60 s = rθ
87.
5
a.
ω=θ
88.
t
2π 1.61 hours ≈ 3.9 radians per hour v= s t 2π r = 1.61 hours 2π (625 km + 6370 km) = 1.61 hours 2π (6995 km) = 1.61 hours ≈ 27,300 km per hour
b.
When the rear tire makes one revolution, the bicycle travels s = rθ = 2π r = 2π (30 inches) = 60π inches.
90.
The angular velocity of point A is ω = θ = 2π = 2 ⎛⎜ π ⎞⎟ . t t ⎝t⎠ When the bicycle travels 60π inches, point B on the front tire travels though an angle of s 60π inches θ= = = 3π . r 20 inches The angular velocity of point B is ω = θ = 3π = 3 ⎛⎜ π ⎞⎟ . t t ⎝t⎠ Thus, point B has the greater angular velocity. b.
91.
a.
7.5° = 7.5°
Point A and point B travel a linear distance of 60π inches in the same amount of time. Therefore, both points have the same linear velocity.
Solving the formula θ = s for r, r s we have r =
θ
= 520 miles π /24 ≈ 3970 miles
Solving the formula θ = s for s yields s = rθ . r Solving the formula θ = s for r yields r = s . r θ s = rθ s = rθ t t r v= θ t
v = rθ t v = r ⎛⎜ θ ⎞⎟ ⎝t⎠ v = rω
v = rθ t s v= t
v = rθ t r rω = θ t
ω=θ t
Thus, all of the formulas are valid.
( )( )
1′ = 1′ ⋅ 1° ⋅ π = π radians 60′ 180° 10,800
1 nautical mile = s = rθ = (3960 statute miles) ⎛⎜ π ⎞⎟ ≈ 1.15 statute miles ⎝ 10,800 ⎠ b.
(180π ° ) = 24π
=
s = 6.25 ⎛⎜ 6π ⎞⎟ ⎝ 5 ⎠ s ≈ 23.6 ft
89.
a.
Earth’s circumference = 2π r ≈ 2π (3960 statute miles) ⎛⎜ 1 nautical mile ⎞⎟ ≈ 7920π nautical miles ⎝ 1.15 statute miles ⎠ 1.15 7920 π The question, then, is what percent of is 2217. 1.15 2217 ≈ 0.10 = 10% 7920π /1.15
Copyright © Houghton Mifflin Company. All rights reserved.
308
92.
Chapter 5: Trigonometric Functions
12 o ⋅
π ≈ 0.067π 180 o s = rθ s = 485 ( 0.067π ) s ≈ 102 ft
....................................................... 93.
A = 1 r 2θ 2 = 1 ( 52 ) π 2 3 ≈ 13 in 2
94.
96.
A = 1 r 2θ 2 2 1 = ( 30 ) 62 o ⎛⎜ π o ⎞⎟ 2 ⎝ 180 ⎠
97.
()
= 487 ft 2
A = 1 r 2θ 2 2 = 1 ( 2.8) 2 ≈ 31 ft 2
Connecting Concepts 95.
( 5π2 )
⎛ o ⎞ 25o 47′ = 25o + 47′ ⎜ 1 ⎟ ⎝ 60′ ⎠ ° = 25 47 60 Convert to radians. ° 25 47 ⋅ π = 1547π radians 60 180° 10,800
98.
s = rθ
A = 1 r 2θ 2 2 = 1 (120 ) 0.65 2 = 4680 cm 2
⎛ o ⎞ 40o 45′ = 40 o + 45′ ⎜ 1 ⎟ ⎝ 60′ ⎠ = 40.75o convert to radians. 40.75o ⋅ π o = 4075π = 163π radians 18,000 720 180 s = rθ s = 3960 ⎛⎜ 163π ⎞⎟ ⎝ 720 ⎠ ≈ 2820
⎛ ⎞ s = 3960 ⎜ 1547π ⎟ ⎝ 10,800 ⎠ ≈ 1780
To the nearest 10 miles, New York City is 2820 miles north of the equator.
To the nearest 10 miles, Miami is 1780 miles north of the equator.
....................................................... PS1.
1 ⋅ 3= 3 3 3 3
⎛ ⎞ ⎛ ⎞ PS4. ⎛⎜ a ⎞⎟ ÷ ⎜ 3 a ⎟ = ⎛⎜ a ⎞⎟ ⋅ ⎜ 2 ⎟ ⎝2⎠ ⎝ 2 ⎠ ⎝2⎠ ⎝a 3⎠ = 1 ⋅ 3 3 3 = 3 3
2 ⋅ 2=2 2= 2 2 2 2
PS2.
PS5.
2=x 2 5 5 2 =x 2 x ≈ 3.54
Prepare for Section 5.2 ⎛2⎞ ⎛a⎞ PS3. a ÷ ⎜ ⎟ = a ⋅ ⎜ ⎟ = 2 ⎝2⎠ ⎝a⎠ PS6.
3= x 3 18 18 3 = x 3 x=6 3 x ≈ 10.39
Section 5.2 1.
r = 52 + 122 r = 25 + 144 + 169 r = 13
y 12 = r 13 x 5 cosθ = = r 13 y 12 tanθ = = x 5 sinθ =
r 13 = y 12 r 13 secθ = = x 5 5 x cotθ = = y 12 cscθ =
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.2
2.
309
r = 32 + 7 2
sinθ =
r = 9 + 49 + 58
3.
3 3 58 x = = 58 r 58 y 7 tanθ = = x 3
x = 7 2 − 42
cscθ =
r 7 = y 4
secθ =
r 7 7 33 = = x 33 33
cot θ =
x 33 = y 4
cscθ =
r 9 = =3 y 3
secθ =
9 3 2 r = = x 6 2 4
cot θ =
x 6 2 = =2 2 3 y
5.
cscθ =
29 r = y 5
r = 2 2 + 52
sin θ =
r = 4 + 25 = 29
6.
x = 82 − 52
x = 2+
( 3)
2
x=
( 10 ) − ( 5 ) 2
x = 10 − 5 = 5
2
r 29 = x 2 x 2 cot θ = = y 5
cosθ =
secθ =
y 5 = r 8 x 39 cosθ = = r 8
cscθ =
r 8 = y 5
secθ =
8 8 39 r = = 39 x 39
sin θ =
x = 4+3 = 7
8.
y 5 5 29 = = r 29 29
x 2 2 29 = = r 29 29 y 5 tan θ = = x 2
x = 64 − 25 = 39
7.
y 4 4 33 = = x 33 33
y 3 1 = = r 9 3 x 6 2 2 2 = cosθ = = 9 3 r 3 1 2 y = = tan θ = = 4 x 6 2 2 2
x=6 2
r 58 = x 3 x 3 cotθ = = y 7
y 4 = r 7 x 33 cosθ = = r 7
sin θ =
x = 81 − 9 = 72
r 58 = y 7
secθ =
tan θ =
x = 92 − 32
cscθ =
cosθ =
sin θ =
x = 49 − 16 = 33
4.
7 7 58 y = = 58 r 58
tan θ =
y 5 5 39 = = x 39 39
cot θ =
x 39 = y 5
sin θ =
y = r
3 21 = 7 7
cscθ =
r = y
cosθ =
x 2 2 7 = = r 7 7
secθ =
tan θ =
y 3 = 2 x
sin θ =
y 5 2 = = r 2 10
cscθ =
10 r = = 2 y 5
cosθ =
x 5 2 = = r 2 10
secθ =
10 r = = 2 x 5
tan θ =
y 5 = =1 5 x
cot θ =
x = y
7 21 = 3 3
r 7 = 2 x x 2 2 3 cot θ = = = 3 y 3
5 =1 5
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310
9.
Chapter 5: Trigonometric Functions
opposite side = 62 − 32
opp 3 3 3 = = hyp 6 2 adj 3 1 cosθ = = = hyp 6 2
cscθ =
= 36 − 9 = 27 = 3 3
opp 3 3 = = 3 adj 3
cot θ =
tan θ =
10.
sin θ =
= 0.36 = 0.6
hypotenuse = 52 + 62 = 25 + 36
opposite side =
cscθ =
adj 5 5 61 = = hyp 61 61 opp 6 tan θ = = adj 5
secθ =
2
2 − 12
opp 1 2 = = hyp 2 2
cscθ =
adj 1 2 = = hyp 2 2 opp 1 tan θ = = =1 adj 1
secθ =
sin θ =
= 2 −1 = 1 = 1
13.
tan θ =
y 3 = x 4
For exercises 16 to 18, since tan θ = 16.
sin θ =
y 4 = r 5
hyp 2 = = 2 opp 1
hyp 2 = = 2 adj 1 adj 1 = =1 cot θ = opp 1
cosθ =
For exercises 13 to 15, since sin θ =
hyp 61 = opp 6
hyp 61 = adj 5 adj 5 cot θ = = opp 6
cosθ =
= 61
12.
cscθ =
opp 6 6 61 = = hyp 61 61
sin θ =
adj 3 1 3 = = = opp 3 3 3 3
hyp 1 5 = = opp 0.6 3 hyp 1 5 secθ = = = adj 0.8 4 adj 0.8 4 cot θ = = = opp 0.6 3
opp 0.6 = = 0.6 = 3 hyp 1 5 adj 0.8 = = 0.8 = 4 cos θ = hyp 1 5 opp 0.6 3 = = tan θ = adj 0.8 4
opposite side = 12 − 0.82 = 1 − 0.64
11.
hyp 6 2 2 3 = = = opp 3 3 3 3 hyp 6 secθ = = =2 adj 3
sin θ =
y 3 = , y = 3, r = 5, and x = 52 − 32 = 4 . r 5 sec θ = r = 5 14. x 4 y 4 = , y = 4, x = 3, and r = 32 + 42 = 5 . x 3 cot θ = x = 3 17. y 4
15.
cos θ = x = 4 r 5
18.
sec θ = r = 5 x 3
For exercises 19 to 21, since sec β = r = 13 , r = 13, x = 12, and y = 132 − 122 = 25 = 5 . x 12 cot β = x = 12 cos β = x = 12 19. 20. 21. r 13 y 5
csc β = r = 13 y 5
For exercises 22 to 24, since cos θ x = 2 , x = 2, r = 3, and y = 32 − 22 = 9 − 4 = 5 . r 3
y 5 = r 3
22.
sin θ =
25.
sin 45o + cos 45o = 2 + 2 = 2 2 2
23.
secθ =
r 3 = x 2
24.
26.
tan θ =
y 5 = x 2
csc 45o − sec 45o = 2 − 2 = 0
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.2
311
csc 60o sec 30o + cot 45o = 2 3 ⋅ 2 3 + 1 3 3 4 = +1 3 7 = 3
27.
sin 30o cos 60o − tan 45o = 1 ⋅ 1 − 1 2 2 = 1 −1 4 =−3 4
29.
sin 30o cos 60o + tan 45o = 1 ⋅ 1 + 1 = 1 + 1 = 5 2 2 4 4
30.
sec 30o cos 30o − tan 60o cot 60o = 2 3 ⋅ 3 − 3 ⋅ 3 = 1 − 1 = 0 3 2 3
31.
sin
33.
sin π + tan π = 2 + 3 = 3 2 + 2 3 4 6 2 3 6
35.
sec π cos π − tan π = 2 ⋅ 1 − 3 = 1 − 3 = 3 − 3 3 3 6 2 3 3 3
36.
cos π tan π + 2 tan π = 2 ⋅ 3 + 2 ⋅ 3 = 6 + 2 3 = 6 + 12 3 4 6 3 2 3 6 6
37.
2 csc π − sec π cos π = 2 ⋅ 2 − 2 ⋅ 3 = 2 2 − 3 4 3 6 2
38.
3 tan π + sec π sin π = 3 ⋅ 1 + 2 3 ⋅ 3 = 3 + 1 = 4 4 6 3 3 2
39.
tan 32o ≈ 0.6249
40.
sec 88o ≈ 28.6537
41.
cos 63o 20′ ≈ 0.4488
42.
cot 55o50′ ≈ 0.6787
43.
cos 34.7o ≈ 0.8221
44.
tan 81.3o ≈ 6.5350
45.
sec 5.9o ≈ 1.0053
46.
sin π ≈ 0.5878 5
47.
tan π ≈ 0.4816 7
48.
sec 3π ≈ 2.6131 8
49.
csc 1.2 ≈ 1.0729
50.
sin 0.45 ≈ 0.4350
π 3
+ cos
π 6
=
3 3 3 + = 2⋅ = 3 2 2 2
51.
28.
π
− sec
π
= 2−2 = 0
32.
csc
34.
sin π cos π − tan π = 3 ⋅ 2 − 1 = 6 − 1 = 6 − 4 3 4 4 2 2 4 4
6
3
52.
sin 52o =
h 12 o
h = 12 sin 52 h ≈ 9.5 ft
tan 52o =
d 31
d = 31 tan 52o d ≈ 40 m
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312
Chapter 5: Trigonometric Functions
53.
54.
tan19o = 32.0 x 32.0 x= = 92.9 in. tan19o
d = rt d t= r
cos 57.3o =
d , 12.5
d = 12.5 cos 57.3o
12.5 cos 57.3o 11 t ≈ 0.61 h ≈ 37 min Time of closest approach; 3:00 P.M.+37 min is 3:37 P.M. t=
55.
56.
cos 38o =
d + 0.33 6
d = 6 cos 38o + 0.33 d ≈ 5.1 ft
tan 68.9o =
h 116
h = 116 tan 68.9o h ≈ 301 m
57.
58.
240 mi ⎛ 1 hr ⎞ ⎜ ⎟ 4 min hr ⎝ 60 min ⎠ d = 16 mi h sin 6o = d d=
h = 16 sin 6o h ≈ 1.7 mi
sin 5o = d= t=
80 d 80 sin 5o
ft = 917.9 ft
d r
mi mi ft 1h = 9 ⋅ 5280 ⋅ h h mi 60 min 0 ( 5280 ) ft ft = = 792 60 min min 917.9 ft t= 792ft/min t ≈ 1.2 min
r =9
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.2
313
59.
60.
tan 27.8° = x 131 x = 131tan 27.8° = 69.1 ft height = 69.1 + 5.5 = 74.6 ft
tan 43o = d1 =
2500 d1
tan 27 o =
2500
d2 =
2500
tan 27 o 2500 2500 + d = d1 + d 2 = tan 43o tan 27 o d ≈ 2680.9 + 4906.5 d ≈ 7600 ft
61.
sin 0.056o = d=
670,900 d 670,900 o
62.
tan 43
sin 46.5o =
r 149,000,000
r = 149,000,000sin 46.5o km ≈ 108,000,000 km
km
sin 0.056 ≈ 686,000,000 km
63.
o
2500 d2
A = 1 bh 2 = 1 (2a sin θ )(a cos θ ) 2 = a 2 sin θ cosθ
64.
65.
From exercise 63, find the area of one triangle A = a 2 sin θ cosθ = (4) 2 sin 60o cos 60o = 16 3 ⋅ 1 2 2 =4 3 Find the area of the hexagon, A = 6(4 3) = 24 3 in.2
66.
tan 36.4 =
d 350 + x
tan 51.9o = x=
tan 36.4° =
d 350 +
d tan 51.9°
350 tan 36.4° d= 1 − tan 36.4° tan 51.9° d ≈ 612 ft
d x d tan 51.9o
tan 22o =
h 240
h = 240 tan 22o d = 80 + h d = 80 + 240 tan 22o d ≈ 180 ft
Copyright © Houghton Mifflin Company. All rights reserved.
314
Chapter 5: Trigonometric Functions
67.
68.
tan 42o = x=
h=
h x
tan 37.8o = h
tan 42o
tan 37.8o =
h 100 + x h 100 +
x=
h tan 42o
H tan 36.7o
o 1 − tan 37.8o
h ≈ 5.60 × 10 ft
−55.5(tan 36.7)(tan 32.1) tan 32.1 − tan 36.7 H ≈ 220 ft
H=
h ≈ 560 ft
a.
70.
h tan 53.6° = 412 h = 412 tan 53.6° h ≈ 559 feet
b.
H − 55.5 tan 32.1o
H H = − 55.5 tan 36.7o tan 32.1o H tan 32.1 = H tan 36.7 − 55.5(tan 36.7)(tan 32.1) H ( tan 32.1 − tan 36.7 ) = −55.5(tan 36.7)(tan 32.1)
100 tan 37.8o tan 42.0 2
69.
x=
x 235 x = 235tan 46.3°
tan 46.3° =
x ≈ 246 feet
( AC )2 = 4122 + 5592 AC = 4122 + 5592 AC = 482, 225 AC ≈ 694.4 feet
h 246 h = 246 tan 65.5°
tan 65.5° =
h ≈ 540 feet
x 694.4 x = 694.4 tan15.5°
tan15.5° =
x ≈ 193 feet
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Section 5.2
315
....................................................... 71.
Connecting Concepts 72.
Consider the right triangle formed by A, B and the midpoint of AC. h a h = a sin θ
sin θ =
r = 6 2 − 32
1 bh 2 1 A = b ( a sin θ ) 2 1 A = ab sin θ 2 A=
r = 27 r =3 3 r ≈ 5.2 m
73.
74.
if θ = 45o sin θ = d1 =
d1 = d 2
d = 82 + 8.52
3 d1
d ≈ 11.7 ft 3
sin 45o
d = 2d1 =
6 sin 45o
d ≈ 8.5 ft
.......................................................
Prepare for Section 5.3
PS1. − 4 3
PS2.
5 ⋅ 5 =5 5= 5 10 2 2 5 5
PS3.
PS4. 2π − 9π = 10π − 9π = π 5 5 5 5
PS5.
3π − 1π = 2π = π 2 2 2
PS6.
120 − 180 = −60 = 60
Copyright © Houghton Mifflin Company. All rights reserved.
(−3) 2 + (−5) 2 = 9 + 25 = 34
316
Chapter 5: Trigonometric Functions
Section 5.3 1.
x = 2 , y = 3 , r = 2 2 + 32 = 13 cscθ =
13 3
sin θ =
2 2 13 x = = 13 r 13 y 3 tan θ = = x 2
secθ =
13 2
cosθ =
cosθ =
x = −2 , y = 3 , r = sin θ =
cot θ =
(− 2)2 + (3)2
3 13 3 y = = 13 r 13
2 13 x −2 = =− 13 r 13 y 3 3 tan θ = = =− x −2 2
= 13
x 8 89 −8 = =− r 89 89 y −5 5 tan θ = = = x −8 8
y 0 = =0 r 5 x −5 = −1 cosθ = = 5 r 0 y tan θ = = =0 x −5
sin θ =
cscθ =
58 7
3 3 58 x = = 58 r 58 y 7 tan θ = = x 3
secθ =
58 3
x = −3 , y = 5 , r =
cot θ =
(− 3)2 + 52
3 7
= 34
5 5 34 y = = 34 r 34
34 5 34 secθ = − 3 3 cot θ = − 5 cscθ =
−3 3 34 x = =− 34 r 34 5 5 y =− tan θ = = x −3 3
13 2
cosθ =
2 3
6.
x = −6 , y = −9 , r =
cscθ = −
89 5
sin θ =
secθ = −
89 8
cosθ =
cot θ =
(− 5)2 + (0)2
7 58 7 y = = 58 r 58
sin θ =
= 89
y −5 5 89 = =− r 89 89
x = −5 , y = 0 , r =
13 3
cot θ = −
cosθ =
7.
4.
secθ = −
(− 8)2 + (− 5)2
x = −8 , y = −5 , c = sin θ =
2 3
cscθ =
cosθ =
5.
x = 3 , y = 7 , r = 32 + 7 2 = 58
3 13 3 y = = 13 r 13
sin θ =
3.
2.
8.
cscθ is undefined secθ = −1 cot θ is undefined
= 117 = 3 13
y −9 3 3 13 = =− =− r 3 13 13 13
x 2 2 13 −6 = =− =− r 3 13 13 13 y −9 3 tan θ = = = x −6 2
8 5
=5
(− 6)2 + (− 9)2
x = 0 , y = 2 , r = 02 + 22 = 2 y 2 sin θ = = = 1 r 2 x 0 cosθ = = = 0 r 2 y 2 tan θ = = ⇒ undefined x 0
cscθ = −
13 3
secθ = −
13 2
cot θ =
2 3
cscθ = 1 secθ is undefined cot θ = 0
9.
sin180o = 0
10.
cos 270o = 0
11.
tan180o = 0
12.
sec 90o = undefined
13.
csc 90o = 1
14.
cot 90o = 0
15.
cos π = 0 2
16.
sin 3π = −1 2
17.
tan π = undefined 2
18.
cot π = undefined
19.
sin π = 1 2
20.
cos π = −1
21.
sin θ > 0 in quadrants I and II. cos θ > 0 in quadrants I and IV. quadrant I
22.
23.
cosθ > 0 in quadrants I and IV. tan θ < 0 in quadrants II and IV. quadrant IV
tan θ < 0 in quadrants II and IV. sin θ < 0 in quadrants III and IV. quadrant IV
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.3
317
24.
sin θ < 0 in quadrants III and IV. cosθ > 0 in quadrants I and IV. quadrant IV
27.
sin θ = −
28.
2 cot θ = −1 = x , x = −1, y = 1 in quadrant II, r = ( −1) + 12 = 2 , cos θ = −1 = − 2 2 y 2
29.
csc θ = 2 = r , r = 2, y = 1, x = ± y
30.
sec θ = 2 3 = r , r = 2 3, x = 3, y = ± x 3
31.
θ is in quadrant IV, sinθ = − 1 =
32.
θ is in quadrant III, tan θ = 1 =
33.
cos θ = 1 , θ is in quadrant I or IV. 2
25.
sin θ < 0 in quadrants III and IV. cos θ < 0 in quadrants II and III. quadrant III
y −y 3 1 y = = , y = −1, r = 2, x = ± 2 2 − (− 1)2 = ± 3 , x = − 3 in quadrant III , tan θ = = x − 3 3 2 r
2
(
2 2 ) − 12 = ±1 , x = −1 in quadrant II , cot θ = −1 = −1 1
( 2 3 ) 2 − 32
= ± 3, y = − 3 in quadrant IV , sin θ = − 3 = − 1 2 2 3
y , y = −1, r = 2, x = 2 2 − 12 = 3 , tan θ = −1 = − 3 3 r 3
y , y = −1, x = −1, r = x
(− 1)2 + (− 1)2 34.
θ is in quadrant I, x = 1, y = 3 , r = 2
35.
r 2 2 3 = = y 3 3
cos θ = − 1 ,θ is in quadrant II or III. 2
sin θ =
37.
= 2 , cosθ =
−1 2
=−
2 2
tan θ = 1 , θ is in quadrant I or III. sin θ = 2 , θ is in quadrant III or IV. 2 θ is in quadrant III, x = 2, y = 2, r = 2 sec θ = r = 2 = 2 x 2
tan θ = 3 , θ is in quadrant I or III.
cscθ =
tan θ < 0 in quadrants II and IV. cosθ < 0 in quadrants II and III. quadrant II
26.
36.
3 , θ is in quadrant I or II. 2
sec θ = 2 3 , θ is in quadrant I or IV. 3 1 sin θ = − , θ is in quadrant II or IV. 2
θ is in quadrant II, x = −1, y = 3 , r = 2
θ is in quadrant IV, x = 3 , y = −1, r = 2
3 x −1 cot θ = = =− y 3 3
cot θ =
θ = 160o Since 90o < θ < 180o ,
θ + θ ′ = 180o θ ′ = 20o
38.
3 x = =− 3 y −1
θ = 255o Since 180o < θ < 270o , θ ′ + 180o = θ θ ′ = 75o
Copyright © Houghton Mifflin Company. All rights reserved.
318
39.
Chapter 5: Trigonometric Functions
θ = 351o
40.
Since 270o < θ < 360o ,
41.
θ = 48o
θ = θ ′ = 360o
Since 0o < θ < 48o , θ′ =θ
θ ′ = 9o
θ ′ = 48o
θ=
11π 5
42.
θ is coterminal withα =
α′ =α =θ ′ π θ′ =
π 2
3π < θ < −2π , 2 θ ′ + θ = 2π
Since −
10π , θ > 2π = 5
Since 0 < α <
θ = −6
,
11π 10π π − = . 5 5 5
θ ′ = 2π − 6 ≈ 0.28
5
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.3
43.
θ=
319
8 3
Since
44.
π
<θ <π,
2 θ +θ′ = π
8 θ′ = π − 3
θ=
18π 7
45.
14π , 7 θ is coterminal with
θ > 2π =
18π 14π 4π α= − = . 7 7 7 π 3π Since < α < , 2 2 α +α′ = π
α′ = π −
θ = 1406o = 326o + 3 ⋅ 360o θ is coterminal with α = 326o. Since 270o < α < 360o ,
α + α ′ = 360o
4π 7
3π 7 3π θ′ = 7
=
Copyright © Houghton Mifflin Company. All rights reserved.
α ′ = 34o θ ′ = 34o
320
46.
Chapter 5: Trigonometric Functions
θ = 840o = 120o + 2 ⋅ 360o
47.
θ > 360o θ is coterminal with α = 120°. o
o
Since 90 < α < 180 ,
α + α ′ = 180 α ′ = 60
o
o
θ = −475o = 245o − 2 ⋅ 360o θ is coterminal with α = 245° Since 180° < α < 270°, α ′ + 180° = α α ′ = 245° − 180° α ′ = 65° θ ′ = 65°
48.
θ = −650o = 70o − 2 ⋅ 360o θ is coterminal with α = 70o. Since 0o < α < 90o ,
α ′ = α = 70o θ ′ = α ′ = 70o
θ ′ = 60o
49.
θ = 225° is in quadrant III. 225° − 180° = 45° so θ ′ = 45°. Thus, sin 225° = − sin 45° = −
51.
θ = 405° is in quadrant I. 405° − 360° = 45° soθ ′ = 45°. Thus, tan 405° = tan 45° = 1.
50.
θ = 300° is in quadrant IV. 360° − 300° = 60° so θ ′ = 60°. 1 Thus, cos 300° = cos 60° = . 2
2 . 2
52.
θ = 150° is in quadrant II. 180° − 150° = 30° so θ ′ = 30°. 1 1 = Thus, sec150° = cos150° − cos30° −2 1 2 3 = = =− . 3 − 3/2 3
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.3
53.
321
4 3
θ = π is in quadrant III.
54.
θ=
17π 16π π = + is coterminal 4 4 4
with
π 4
so cos
in quadrant I and θ ′ =
π 4
7π is in quadrant III. 6
7π π π − π = so θ ′ = . 6 6 6
4 π π π − π = so θ ′ = . 3 3 3 4π 1 1 = = Thus, csc 3 sin 4π / 3 − sin π / 3 1 2 2 3 . = =− =− 3 3 − 3/2
55.
θ=
Thus, cot
56.
,
17π 2 π . = cos = 4 4 2
7π π cos π / 6 3/2 = cot = = = 3. 6 6 sin π / 6 1/ 2
−π is in quadrant IV. 3 π ⎛ π⎞ π 0 − ⎜ − ⎟ = so θ ′ = . 3 ⎝ 3⎠ 3
θ=
⎛ −π Thus, tan ⎜ ⎝ 3
π − sin π / 3 ⎞ ⎟ = − tan = 3 cos π / 3 ⎠ =
57.
θ = 765o = 720o + 45o is coterminal
58.
with 45o in quadrant I and θ ′ = 45o , so sec 765o = sec 45o =
1 cos 45
o
=
1 2 /2
= 2.
− 3/2 = − 3. 1/2
θ = −510o = −720o + 210o is coterminal with 210° in quadrant III and θ ′ = 30°, so csc ( −510° ) =
1 sin ( −510° )
1 1 = = −2. − sin 30° −1/ 2
Copyright © Houghton Mifflin Company. All rights reserved.
322
59.
Chapter 5: Trigonometric Functions
θ = 540o = 360o + 180o is coterminal with 180o , so cot 540o = cot 180o =
60.
cos180o sin 180
o
=
θ = 570o = 360o + 210o is coterminal with 210o in quadrant III and θ ′ = 30o ,
−1 , 0
so cos 570o = − cos 30o = −
which is undefined.
3 . 2
61.
sin 127o ≈ 0.798636
62.
sin(−257o ) ≈ 0.974370
63.
cos( −116o ) ≈ −0.438371
64.
cot 398 o ≈ 1.27994
65.
sec 578 o ≈ −1.26902
66.
sec 740 o ≈ 1.06418
67.
⎛ π⎞ sin ⎜ − ⎟ ≈ −0.587785 ⎝ 5⎠
68.
cos
3π ≈ 0.222521 7
69.
csc
70.
tan (−4.12) ≈ −1.48584
71.
sec (−4.45) ≈ −3.85522
72.
csc 0.34 ≈ 2.99862
73.
⎛ ⎞ sin 210 o − cos 330o tan 330o = − 1 − 3 ⎜ − 3 ⎟ = − 1 + 1 = 0 2 2 ⎝ 3 ⎠ 2 2
74.
⎛ ⎞ tan 225 o + sin 240o cos60o = 1 + ⎜ − 3 ⎟ 1 = 1 − 3 = 4 − 3 4 4 ⎝ 2 ⎠2
75.
o 1 3 ⎛ 1 ⎞ ⎛ 3 ⎞⎟ sin 30 + cos 30 = ⎜ ⎟ + ⎜ = + =1 ⎜ ⎟ 2 2 4 4 ⎝ ⎠ ⎝ ⎠
76.
cos π sin
77.
sin
3π π π 1 1 3 tan − cos = ( −1)(1) − = −1 − = − 2 4 3 2 2 2
78.
cos
7π 4π 7π 2 + cos = tan 4 3 6 2
2
2
o
2
o
⎛ 7π 11π 2 ⎞⎟ ⎛⎜ 3 ⎞⎟ 2 3 3 2 +2 3 − tan = −1⎜ − − − = + = ⎜ 2 ⎟ ⎜ 3 ⎟ 4 6 2 3 6 ⎝ ⎠ ⎝ ⎠
⎛
( 3 ) + ⎜⎜ − ⎝
2
3⎞ 6 3 − = ⎟⎟ = 2 ⎠ 2 2
6− 3 2
2
79.
sin 2
5π 5π ⎛ 2⎞ ⎛ 2⎞ 1 1 + cos 2 = ⎜⎜ − ⎟⎟ + ⎜⎜ − ⎟ = + =1 4 4 ⎝ 2 ⎠ ⎝ 2 ⎟⎠ 2 2
80.
tan 2
7π 7π 2 − sec2 = ( −1) − 4 4
( 2)
2
= 1 − 2 = −1
Copyright © Houghton Mifflin Company. All rights reserved.
9π ≈ −1.70130 5
Section 5.3
323
....................................................... 81.
Connecting Concepts
sin θ = 1 , θ is in quadrant I or quadrant II 2
82.
θ = 120o , 300o
θ = 30o , 150o 83.
cos θ = − 3 , θ is in quadrant II or quadrant III 2
84.
cscθ = − 2 θ is in quadrant III or IV
86.
θ = 225o , 315o
88.
91.
94.
,
3
92.
3
y
2
+1 = =
tan θ = −1 θ is in quadrant II or IV θ = 3π , 7π 4
4
90.
sec θ = −2 3 3 θ is in quadrant II or III θ = 5π , 7π 6 6
93.
1 + tan 2 θ = sec 2 θ
6
cos θ = − 1 2 θ is in quadrant II or III
θ=
cot 2 θ + 1 = csc2 θ
x2
tan θ = − 3 3 θ is in quadrant II or IV
6
π 2π ,
87.
θ = 5π , 11π
3
sin θ = 3 2 θ is in quadrant I or II
θ=
89.
π 5π 3
cot θ = − 1 θ is in quadrant II or IV
θ = 135o , 315o
cos θ = 1 2 θ is in quadrant I or IV
θ=
tan θ = 1, θ is in quadrant I or quadrant III
θ = 45o , 225o
θ = 150o , 120o 85.
tan θ = − 3,θ is in quadrant II or quadrant IV
y2
1+
2π 4π , 3 3
x2
x2
= r2 x = sec 2 θ
96.
x2 + y 2 y2 r2 y2
cos(90o − θ ) = sin θ y y = r r
sin(90o − θ ) = cos θ x=x r r
....................................................... x2 + y2 = 1 (0) 2 + (1) 2 = 1 Yes
PS4. C = 2π r = 2π (1) = 2π
x2 + y2 2
95.
= csc2 θ
PS1.
=
PS2.
Prepare for Section 5.4
x2 + y2 = 1 2
2
⎛ 1 ⎞ + ⎛ 3 ⎞ =1 ⎜ ⎟ ⎜ ⎟ ⎝2⎠ ⎝ 2 ⎠ 1 + 3 =1 4 4 Yes
PS5. even
PS3.
x2 + y 2 = 1 2
2
⎛ 2⎞ ⎛ 3⎞ ⎟ =1 ⎜ ⎟ +⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ 2 + 3 ≠1 4 4 No
PS6. neither
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324
Chapter 5: Trigonometric Functions
Section 5.4 1.
t=
π
6 y = sin t = sin
π 6
1 = 2
2.
x = cos t
= cos =
π
x = cos t
π
= cos
4
2 = 2
3 2
π 4
2 = 2
The point on the unit circle corresponding to t = π is
⎛ 3 , 1 ⎞. ⎜ 2 2⎟ ⎝ ⎠
⎛ 2 , 2 ⎞. ⎜ 2 2 ⎟ ⎝ ⎠
7π 6 y = sin t
t=
4
4.
x = cos t 7π = cos 6 3 =− 2
4π 3 y = sin t
t=
x = cos t 4π = cos 3 1 =− 2
4π = sin 3 3 =− 2
The point on the unit circle corresponding to t = 7π is
The point on the unit circle corresponding to t = 4π is
⎛ − 3 , − 1 ⎞. ⎜ 2 2 ⎟⎠ ⎝
⎛ − 1 , − 3 ⎞. ⎜ 2 2 ⎟⎠ ⎝
6
5π 3 y = sin t
t=
5π = sin 3 3 =− 2
3
6. x = cos t 5π = cos 3 1 = 2
t=−
π
6 y = sin t
x = cos t
⎛ π⎞ = cos ⎜ − ⎟ ⎝ 6⎠
⎛ π⎞ = sin ⎜ − ⎟ ⎝ 6⎠ 1 =− 2
=
3 2
The point on the unit circle corresponding to t = 5π is
The point on the unit circle corresponding to t = − π is
⎛ 1 , − 3 ⎞. ⎜2 2 ⎟⎠ ⎝
⎛ 3 , − 1 ⎞. ⎜ 2 2 ⎟⎠ ⎝
3
7.
4 y = sin t
The point on the unit circle corresponding to t = π is
7π = sin 6 1 =− 2
5.
π
= sin
6
6
3.
t=
11π 6 y = sin t
t=
= sin =−
11π 6
1 2
6
8.
x = cos t 11π = cos 6 3 = 2
t =0 y = sin t = sin 0 =0
x = cos t = cos 0 =1
The point on the unit circle corresponding to t = 0 is (1,0 ) .
The point on the unit circle corresponding to t = 11π is 6
⎛ 3 , − 1 ⎞. ⎜ 2 2 ⎟⎠ ⎝
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.4
9.
325
t =π y = sin t
10.
x = cos t = sin π = cos π =0 = −1 The point on the unit circle corresponding to t = π is (−1,0).
7π 4 y = sin t
t=−
x = cos t
⎛ 7π ⎞ = sin ⎜ − ⎟ ⎝ 4 ⎠ 7π = − sin 4 2 = 2
⎛ 7π ⎞ = cos ⎜ − ⎟ ⎝ 4 ⎠ 7π = cos 4 2 = 2
The point on the unit circle corresponding to t = − 7π is 4
⎛ 2 , 2 ⎞. ⎜ 2 2 ⎟ ⎝ ⎠
11.
2π 3 y = sin t
t=−
12.
x = cos t
⎛ 2π ⎞ = sin ⎜ − ⎟ ⎝ 3 ⎠ 2π = − sin 3 3 =− 2
⎛ 2π ⎞ = cos ⎜ − ⎟ ⎝ 3 ⎠ 2π = cos 3 1 =− 2
t = −π y = sin t
x = cos t
= sin ( −π )
= cos ( −π )
= − sin π
= cos π = −1
=0
The point on the unit circle corresponding to t = −π is ( − 1,0).
The point on the unit circle corresponding to t = − 2π is 3
⎛ − 1 , − 3 ⎞. ⎜ 2 2 ⎟⎠ ⎝
13.
tan 11π = − tan π = − 3 6 6 3
14.
cot 2π = − cot π = − 3 3 3 3
15.
cos ⎛⎜ − 2π ⎞⎟ = − cos π = − 1 3 2 ⎝ 3 ⎠
16.
sec ⎛⎜ − 5π ⎞⎟ = − sec π = − 2 3 6 3 ⎝ 6 ⎠
17.
csc ⎛⎜ − π ⎞⎟ = − csc π = − 2 3 3 3 ⎝ 3⎠
18.
tan (12π ) = tan 0 = 0
19.
sin 3π = − sin π = −1 2 2
20.
cos 7π = cos π = 1 3 3 2
21.
sec − 7π = − sec π = − 2 3 6 6 3
22.
sin ⎛⎜ − 5π ⎞⎟ = sin π = 3 3 2 ⎝ 3 ⎠
23.
sin1.22 ≈ 0.9391
24.
cos 4.22 ≈ −0.4727
25.
csc ( −1.05 ) ≈ −1.1528
26.
sin ( −0.55 ) ≈ −0.5227
27.
tan 11π ≈ −0.2679 12
28.
cos 2π ≈ 0.3090 5
29.
cos − π ≈ 0.8090 5
30.
csc8.2 ≈ 1.0630
31.
sec1.55 ≈ 48.0889
32.
cot 2.11 ≈ −0.5983
33.
a.
sin 2 ≈ 0.9
b.
cos 2 ≈ –0.4
a.
sin 4.1 ≈ –0.8
b.
cos 4.1 ≈ –0.6
34.
37.
a.
sin 3 ≈ 0.1
b.
cos 3 ≈ –1.0
sin t = 0.4 when t = 0.4 or 2.7
35.
38.
( )
a.
sin 5.4 ≈ –0.8
b.
cos 5.4 ≈ 0.6
cos t = 0.8 when t = 0.6 or 5.6
36.
39.
( )
sin t = –0.3 when t = 3.4 or 6.0
Copyright © Houghton Mifflin Company. All rights reserved.
326
40.
Chapter 5: Trigonometric Functions
cos t = –0.7 when t = 2.3 or 3.9
41.
f (− x) = −4sin( − x) = 4sin x
42.
f ( − x ) = −2cos ( − x ) = −2cos x
= − f ( x)
= f ( x)
The function defined by f ( x ) = −4sin x is an odd function. 43.
G ( − x ) = sin ( − x ) + cos ( − x )
44.
F ( − x ) = tan ( − x ) + sin ( − x )
45.
= − tan x − sin x
= − sin x + cos x
= −F ( x)
The function defined by G ( x ) = sin x + cos x is neither an even
The function defined by F ( x ) = tan x + sin x is an odd function.
nor an odd function.
46.
The function defined by f ( x ) = −2cos x is an even function.
( ) C ( − x ) = cos − x −x = − cos x x = −C ( x )
47.
v ( − x ) = 2sin ( − x ) cos ( − x )
The function defined by sin ( x ) S ( x) = is an even function. x 48.
w ( − x ) = − x tan ( − x )
= −2sin x cos x
= x tan x
= −v ( x )
= w( x)
The function defined by w ( x ) = x tan x is an even function.
The function defined by v ( x ) = 2sin x cos x is an odd function.
cos x The function defined by C ( x ) = x is an odd function.
( ) S ( − x ) = sin − x −x = − sin x = sin x x −x = S ( x)
49.
2π
50.
2π
51.
π
52.
π
53.
2π
54.
2π
55.
56.
cos t = x cos ( −t ) = x cos ( −t ) = cos t
57.
y x −y tan ( t − π ) = −x tan ( t − π ) = tan t tan t =
cos t = x cos(π + t ) = − x cos t = − cos(π + t )
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Section 5.4
327
58.
59.
sin ( −t ) = − y
sin ( t − π ) = − y
sin t = y
64.
67.
sin t ⋅ cos t cos t = sin t
tan t cos t =
1 sec t = cos t tan t sin t cos t = 1 ⋅ cos t cos t sin t = 1 = csc t sin t
65.
1 − sec2 t = 1 −
=
1 − sin 2 t = cos2 t cot 2 t cos2 t sin 2 t
68.
2
63.
1
66.
2
cos t
1 csc t = sin t cot t cos t sin t = 1 ⋅ sin t sin t cos t = 1 = sec t cos t 1 − csc2 t = − ( csc2 t − 1) = − cot 2 t
cos 2 t − 1 cos 2 t − sin 2 t cos 2 t
= − tan 2 t
csc2 t − cot t = 1 + cot 2 t − cot t cot t cot t 2 = 1 + cot t − cot t cot t cot t = tan t + cot t − cot t = tan t
71.
= cos2 t ⋅ sin 2 t cos t = sin 2 t
cot t sin t =
=
2 2 tan t − sec t = tan t − 1 + tan t tan t tan t
sec ( −t ) =
cos t ⋅ sin t sin t = cos t
62.
1 x
1 x sec ( −t ) = sec t
sin ( t − π ) = − sin t
2 = tan t − 1 − tan t tan t tan t = tan t − cot t − tan t = − cot t
70.
sec t =
sin t = y
sin ( −t ) = − sin t
61.
60.
69.
1 − cos2 t = sin 2 t tan 2 t sin 2 t cos2 t 2
= sin 2 t ⋅ cos2 t sin t = cos2 t
1 1 + = 1 + cos t + 1 − cos t 1 − cos t 1 + cos t (1 − cos t ) (1 + cos t ) 2 = 1 − cos2 t = 22 = 2csc2 t sin t
Copyright © Houghton Mifflin Company. All rights reserved.
328
72.
Chapter 5: Trigonometric Functions
1 1 1 + sin t + 1 − sin t + = 1 − sin t 1 + sin t (1 − sin t )(1 + sin t )
= =
73.
2 1 − sin 2 t 2 cos 2 t
sin t cos t + tan t + cot t cos t sin t = sin t tan t cos t ⎛ sin 2 t + cos 2 t ⎞ cos t ⎟ =⎜ ⎜ sin t ⋅ cos t ⎟ sin t ⎝ ⎠
= 2sec2 t
= =
sin 2 t + cos 2 t sin 2 t 1 sin 2 t
= csc2 t
74.
1 − sin t csc t − sin t sin t = 1 csc t sin t =
75.
sin 2 t (1 + cot 2 t ) = sin 2 t ( csc2 t ) = sin 2 t ⋅
1 sin 2 t
=1
1 − sin 2 t 1
= cos 2 t
76.
cos2 t (1 + tan 2 t ) = cos2 t sec2 t = cos2 t ⋅
77.
1 cos2 t
sin 2 t + cos 2 t = 1 sin 2 t = 1 − cos 2 t sin t = ± 1 − cos 2 t
=1
Because 0 < t <
π 2
,sin t is positive.
Thus, sin t = 1 − cos 2 t .
78.
1 + tan 2 t = sec2 t
79.
tan 2 t = sec2 t − 1
csc t = ± 1 + cot 2 t
tan t = ± sec2 t − 1 Because
csc2 t = 1 + cot 2 t
Because
3π < t < 2π , tan t is negative. 2
π 2
< t < π , csc t is positive.
Thus, csc t = 1 + cot 2 t .
Thus, tan t = − sec 2 t − 1.
80.
sec2 t = 1 + tan 2 t
81.
sec t = ± 1 + tan 2 t 3π Because π < t < , sec t is negative. 2 2
Thus, sec t = − 1 + tan t .
( )
d (t ) = 1970cos π t 64
d (24) = 1970cos ⎛⎜ π ⋅ 24 ⎞⎟ ⎝ 64 ⎠ = 1970cos ⎛⎜ 3π ⎞⎟ ⎝ 8 ⎠ ≈ 750 miles
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.4
82.
329
T (t ) = −41cos ⎛⎜ π t ⎞⎟ + 36 ⎝6 ⎠
83.
2 2 cos t − 1 = cos t − 1 = − sin t cos t cos t cos t
March 5: T (2) = −41cos ⎛⎜ π ⋅ 2 ⎞⎟ + 36 ⎝6 ⎠ = −41cos ⎛⎜ π ⎞⎟ + 36 ⎝3⎠ = −41(0.5) + 36 = 15.5° F July 20: T (6.5) = −41cos ⎛⎜ π ⋅ 6.5 ⎞⎟ + 36 ⎝6 ⎠ = −41cos ⎛⎜ 13π ⎞⎟ + 36 ⎝ 12 ⎠ ≈ −41( −0.9659258263) + 36 ≈ 75.6° F
84.
1 2 2 2 + t t 1 tan 1 sec 1 t = 1 ⋅ cos t = cos = = = = 1 ⋅ 1 = csc t sec t tan t + sin t tan t tan t tan t cos2 t sin t sin t cos t sin t cos t cos t
85.
1 2 2 2 + t t 1 cot 1 csc 1 t = 1 ⋅ sin t = sin = = = = 1 ⋅ 1 = csc t sec t cot t + cos t sin 2 t cos t sin t cos t sin t cos t cot t cot t cot t sin t
86.
2 2 sin t − 1 = sin t − 1 = − cos t sin t sin t sin t
87.
(1 − sin t )2 = 1 − 2sin t + sin 2 t
88.
(1 − cos t )2 = 1 − 2cos t + cos2 t
89.
(sin t − cos t )2 = sin 2 t − 2sin t cos t + cos2 t = 1 − 2sin t cos t
90.
(sin t + cos t )2 = sin 2 t + 2sin t cos t + cos2 t = 1 + 2sin t cos t
91.
(1 − sin t )(1 + sin t ) = 1 − sin 2 t
92.
(1 − cos t )(1 + cos t ) = 1 − cos2 t
= cos2 t
93.
= sin 2 t
sin t + 1 + cos t = ( sin t ) ( sin t ) + (1 + cos t ) (1 + cos t ) 1 + cos t sin t sin t (1 + cos t )
94.
2 2 = sin t + 1 + 2 cos t + cos t sin t (1 + cos t ) ( ) = 2 + 2 cos t = 2 1 + cos t sin t (1 + cos t ) sin t (1 + cos t ) = 2 = 2 csc t sin t
1 − sin t − 1 1 = 1 − sin t − sin t + 1 cos t tan t + sec t cos t cos t cos t t − 1 sin = − cos t cos t sin t + 1 (1 − sin t ) (1 + sin t ) − ( cos t ) ( cos t ) = cos t (1 + sin t ) 2 2 1−1 = 1 − sin t − cos t = cos t (1 + sin t ) cos t (1 + sin t ) =0
95.
cos 2 t − sin 2 t = ( cos t − sin t )( cos t + sin t )
96.
sec2 t − csc2 t = ( sec t − csc t )( sec t + csc t )
97.
tan 2 t − tan t − 6 = ( tan t + 2 ) ( tan t − 3)
98.
cos2 t + 3cos t − 4 = ( cos t − 1) ( cos t + 4 )
Copyright © Houghton Mifflin Company. All rights reserved.
330
99.
Chapter 5: Trigonometric Functions
2sin 2 t − sin t − 1 = ( 2sin t + 1) ( sin t − 1)
100.
4 cos2 t + 4 cos t + 1 = ( 2 cos t + 1) ( 2 cos t + 1) 2 = ( 2 cos t + 1)
....................................................... 101.
csc t = 2,
sin t =
0
Connecting Concepts 102.
cos 2 t + sin 2 t = 1
1 2 = csc t 2
sin t = ± 1 − cos 2 t
cos 2 t + sin 2 t = 1
Because
cos t = ± 1 − sin 2 t cos t is positive in quadrant I. 2
cos t = 1 − ⎛⎜ 2 ⎞⎟ = ⎝ 2 ⎠ cos t =
103.
(2)
1 2
sin t = −
sin t = 1 , π < t < π 2 2 sin t tan t = cos t sin t tan t = ± 1 − sin 2 t
104.
2
cot t = 3 , π < t < 3π 3 2 cos t cot t = sin t cos t = cot t sin t
1 sin 2 t
1
1 2
sin t = ±
()
Becauseπ < t < 3π ,sin t is negative.
2 1− 1 2
1 + cot 2 t 2
1 tan t = − 2 3 2
1
cos t = − cot t =−
3 tan t = − 3
3 3
1 + cot 2 t 1
1 + ⎛⎜ 3 ⎞⎟ ⎝ 3 ⎠ 1 cos t = − 2
sin 2 t + cos2 t = 1 sin 2 t sin 2 t
106.
= csc2 t
107.
3 2
1 + cot 2 t = csc2 t =
Because π < t < π , tan t is negative.
105.
3π < t < 2π , sin t is negative. 2
2 sin t = − 1 − 1
2 2
tan t = −
3π < t < 2π 2
cos t = 1 , 2
2
=
− 3 3 ⋅ 3 2
sin 2 t + cos2 t = 1 cos2 t cos2 t = sec 2 t
( cos t − 1) ( cos t + 1) = cos2 t − 1
108.
( sec t − 1) ( sec t + 1) = sec2 t − 1
= − (1 − cos2 t )
= 1 + tan 2 t − 1
= − sin 2 t
= tan 2 t
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Section 5.5
331
....................................................... PS1. sin 3π ≈ 0.7 4
PS2. cos 5π ≈ −0.7 4
PS4. Compress each point on the graph of y = f (x) toward the y-axis by a factor of 1 . 2
PS5.
Prepare for Section 5.5 PS3. Reflect the graph of y = f (x) across the x-axis to produce y = – f (x).
2π = 2π ⋅ 3 = 6π 1 1 1 3
PS6.
2π = 2π ⋅ 5 = 5π 2 1 2 5
Section 5.5 1.
4.
y = − 1 sin x 2 a = − 1 = 1 , p = 2π 2 2
3.
5.
y = 1 sin 2π x 2 1 a = , p = 2π = 1 2 2π
6.
y = 2sin π x 3 a = 2, p = 2π = 6 π /3
y = −2sin x 2
8.
y = − 1 sin x 2 2 1 1 a = − = , p = 2π = 4π 2 2 1/ 2
9.
y = 1 cos x 2 1 a = , p = 2π 2
y = −3cos x
11.
y = cos x 4
12.
y = cos3x
y = 1 cos 2π x 2 1 a = , p = 2π = 1 2 2π
15.
y = 4.7sin 0.8π t a = 4.7, p = 2π = 2.5 0.8π
18.
y = 2sin x a = 2, p = 2π
2.
y = sin 2π 3
a = 1, p = 2π = 3π 2/3 7.
a = −2 = 2, p = 2π = 4π 1/ 2 10.
a = −3 = 3, p = 2π
a = 1, p = 2π = π 2
a = 1, p = 2π 3
a = 1, p = 2π = 8π 1/ 4
y = 2cos π x 3 a = 2, p = 2π = 6 π /3
14.
16.
y = 3 cos 4π 4 3 a = , p = 2π = π 4 4 2
17.
19.
y = 1 sin x, a = 1 , p = 2π 2 2
13.
20.
y = sin 2 x
y = −3cos 2 x 3
a = −3 = 3, p = 2π = 3π 2/3 y = 2.3cos0.005π t a = 2.3, p = 2π = 400 0.005π
y = 3 cos x, a = 3 , p = 2π 2 2
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332
Chapter 5: Trigonometric Functions
21.
y = 3cos x, a = 3, p = 2π
22.
y = − 3 sin x, a = − 3 = 3 , p = 2π 2 2 2
23.
y = − 7 cos x, a = − 7 = 7 , p = 2π 2 2 2
24.
y = 3sin x, a = 3, p = 2π
25.
y = −4sin x, a = −4 = 4, p = 2π
26.
y = −5cos x, a = −5 = 5, p = 2π
27.
y = cos3x, a = 1, p = 2π 3
28.
y = sin 4 x, a = 1, p = π 2
29.
y = sin 3x , a = 1, p = 4π 2 3
30.
y = cos π x, a = 1, p = 2
31.
y = cos π x, a = 1, p = 4 2
32.
y = sin 3π x, a = 1, p = 8 4 3
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Section 5.5
333
33.
y = sin 2π x, a = 1, p = 1
34.
y = cos3π x, a = 1, p = 2 3
35.
y = 4 cos x , a = 4, p = 2π = 4π 2 1/ 2
36.
y = 2 cos 3x , a = 2, p = 2π = 8π 4 3/ 4 3
37.
y = −2 cos x , a = −2 = 2, p = 2π = 6π 3 1/ 3
38.
y = − 4 cos3x, a = − 4 = 4 , p = 2π 3 3 3 3
39.
y = 2sin π x, a = 2, p = 2π = 2
40.
y = 1 sin π x , a = 1 , p = 2π = 6 π /3 2 3 2
41.
y = 3 cos π x , a = 3 , p = 2π = 4 π /2 2 2 2
42.
y = cos π x , a = 1, p = 2π = 6 π /3 3
43.
y = 4sin 2π x, a = 4 = 4, p = 2π = 3 3 2π / 3
44.
y = 3cos 3π x, a = 3, p = 2π = 4 2 3π / 2 3
45.
y = 2 cos 2 x, a = 2, p = 2π = π 2
46.
y = 1 sin 2.5 x, a = 1 , p = 2π = 4π 2 2 2.5 5
π
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334
Chapter 5: Trigonometric Functions
47.
y = −2sin1.5 x, a = −2 = 2, p = 2π = 4π 1.5 3
49.
y = 2sin
52.
55.
58.
61.
63.
51.
y = −2cos3 x
x 3
54.
2 y = − 3sin x 3
y=
1 x y = − cos 2 2
53.
y = − 2sin
y = − 3cos π x
56.
y = − 2cos
2π 1 = 6π , b = , a = 2 b 3 1 y = 2cos x 3
59.
2π = 2, b = π , a = 2 b y = −2cos π x a.
Frequency =
f ( x ) = 2sin
π 2
57.
x
2x 2 = 3π , b = , a = 2 b 3 2 y = 2sin x 3
62.
Amplitude, a = 4, period, b = 2π = π 2 V = 4sin π t , 0 ≤ t ≤ 8 ms
64.
4 cycles 1 = cycle/ms 8 ms 2
2x 2π = 3π , a = 2, p = 3 2/3
y = − 3 cos5 x, a = − 3 = 3 , p = 2π 4 4 4 5
1 sin 3 x 2
50.
b.
65.
x 2
48.
60.
2π 4π 3 3 = ,b= ,a= b 3 2 2 3 3 y = sin x 2 2
2π = 1, b = 2π , a = 3 b y = −3sin 2π x a.
b.
66.
2π = π , b = 2, a = 1 b y = cos 2 x
Amplitude, 3/ 5 = 1.5 , period, b = 2π = 0.8π 2.5 2.5 V = 1.5sin 0.8π t , 0 ≤ t ≤ 5 ms Frequency =
f ( x ) = −3cos
2 cycles = 0.4 cycle/ms 5 ms
3x 2π 8π = , a = −3 = 3, b = 4 3/ 4 3
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.5
67.
335
y1 = 2cos x , a = 2, p = 2π = 4π 2 1/ 2 y2 = 2cos x, a = 2, p = 2π
68.
y1 = sin 3π x, p = 1, p = 2π = 2 3π 3 y2 = sin π x , p = 1, p = 2π = 6 3 π /3
69.
y = 1 x sin x 2
70.
y = 1 x + sin x 2
71.
y = − x cos x
72.
y = − x + cos x
73.
y = esin x
74.
y = ecos x
The maximum value of ecos x is e. The minimum value of ecos x is 1 ≈ 0.3679. e
The maximum value of esin x is e. The minimum value of esin x is 1 ≈ 0.3679. e
The function defined by y = ecos x is periodic with a period of 2π .
The function defined by y = esin x is periodic with a period of 2π .
....................................................... 75.
a=2 p=
2π = 3π b 2 b= 3
2 y = 2sin x 3
76.
a=4 p=
2π =2 b b =π
Connecting Concepts 77.
a = 2.5 p=
y = 4sin π x
2π = 3.2 b 5π b= 8
y = 2.5sin
Copyright © Houghton Mifflin Company. All rights reserved.
5π x 8
336
78.
Chapter 5: Trigonometric Functions
a=3
79.
a=3
80.
2π π = b 2 b=4
p = 2π = 2.5 b b = 4π 5
y = 3cos 4 x
y = 3cos 4π x 5
p=
a = 4.2 p=
2π =1 b b = 2π
y = 4.2cos 2π x
....................................................... PS1. tan π ≈ 1.7 3
PS2. cot π ≈ 0.6 3
PS4. Shift the graph of y = f (x) 2 units to the right and up 3 units.
PS5. π = π ⋅ 2 = 2π 1 1 1 2
Prepare for Section 5.6 PS3. Stretch each point on the graph of y = f (x) away from the x-axis by a factor of 2 to produce y = 2 f (x). PS6.
π −3/ 4
= π ⋅ − 4 = π ⋅ 4 = 4π 1 3 1 3 3
Section 5.6 1.
y = tan x is undefined for π + kπ , k an integer. 2
2.
y = cot x is undefined for kπ , k an integer.
3.
y = sec x is undefined for π + kπ , k an integer. 2
4.
y = csc x is undefined for kπ , k an integer.
5.
p = 2π
6.
p =π
7.
p =π
8.
p = 2π
9.
p = π = 2π 1/ 2
10.
p=π 2
11.
p = 2π 3
12.
p = 2π = 4π 1/ 2
13.
p=π 3
14.
p = π = 3π 2/3 2
15.
p = 2π = 8π 1/ 4
16.
p = 2π = π 2
17.
p = π =1
18.
p= π =3 π /3
19.
p= π =5 π /5
20.
p= π =2 π /2
21.
p=
2π = 8.5 π / 4.25
22.
p = 2π = 5 π / 2.5
23.
y = 3tan x, p = π
24.
y = 1 tan x, p = π 3
25.
y = 3 cot x, p = π 2
26.
y = 4cot x, p = π
27.
y ( x ) = 2sec x, p = 2π
28.
y = 3 sec x, p = 2π 4
π
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Section 5.6
337
29.
y = 1 csc x, p = 2π 2
30.
y = 2csc x, p = 2π
31.
y = 2 tan x , p = π = 2π 2 1/ 2
32.
y = −3tan 3x, p = π 3
33.
y = −3cot x , p = π = 2π 2 1/ 2
34.
y = 1 cot 2 x, p = π 2 2
35.
y = −2 csc x , p = 2π = 6π 3 1/ 3
36.
y = 3 csc 3x , p = 2π 2 3
37.
y = 1 sec 2 x, p = 2π = π 2 2
38.
y = −3sec 2 x , p = 2π = 3π 3 2/3
39.
y = −2sec π x , p = 2π = 2
40.
y = 3csc π x , p = 2π = 4 π /2 2
41.
y = 3tan 2π x, p = π = 1 2π 2
42.
y = − 1 cot π x , p = π = 2 π /2 2 2
43.
y = 2 csc 3x, p = 2π 3
44.
y = sec x , p = 2π = 4π 2 1/ 2
π
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338
Chapter 5: Trigonometric Functions
45.
y = 3sec π x, p = 2π = 2
46.
y = csc π x , p = 2π = 4 π /2 2
47.
y = 2 cot 2 x, p = π 2
48.
y = 1 tan x , p = π = 2π 2 2 1/ 2
49.
y = 3tan π x, p = π = 1
50.
y = cot π x , p = π = 2 π /2 2
51.
π = 2π , b = 3
π
π
b
3
2
52.
y = cot 3 x 2
54.
57.
2π = 2, b = π b y = csc π x
y = tan x
π = 2π , b = 1 b
2
53.
y = tan 1 x 2
55.
58.
2π = 8π , b = 3 b 3 4 y = sec 3 x 4
y = sec x = sec x
56.
59.
2π = 3π , b = 2 3 b y = csc 2 x 3 2π = 1, b = 2π b y = sec 2π x
y = csc x
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.6
60.
339
y = cot x
61.
a. tan x = h 1.4 h = 1.4 tan x
62.
b. cos x = 1.4 d d = 1.4 = 1.4sec x cos x
a.
sin x = 3.5 d d = 3.5 = 3.5csc x sin x
b.
d = 3.5csc x d (1) = 3.5csc(1) ≈ 4.16 mi d (1.2) = 3.5csc(1.2) ≈ 3.76 mi
c.
d. The graph of d is above the graph of h, however, the distance between the graphs approaches
0 as x approaches π . 2
....................................................... 63.
66.
69.
π
=
π
,b = 3
b 3 y = tan 3 x
2π 5π 4 = ,b = b 2 5 4 y = csc x 5
2π 4 = 1.5, b = π b 3 4π y = csc x 3
64.
π
67.
π
=
π
,b = 2
b 2 y = cot 2 x
b
= 2, b =
y = cot
70.
π 2
π 2
Connecting Concepts 65.
68.
x
2π 3π 8 = ,b = b 4 3 8 y = sec x 3
π
= 0.5, b = 2π b y = tan 2π x
2π 2 = 3, b = π b 3 2π y = sec x 3
....................................................... PS1.
y = 2sin 2 x amplitude = 2, period = π
PS4. maximum at 2
PS2.
y = 2 cos x 3 3
amplitude = 2 , period = 6π 3 PS5. minimum at –3
Prepare for Section 5.7 PS3.
y = −4sin 2π x amplitude = 4, period = 1
PS6. f (x) = cos x is symmetric to y–axis.
Copyright © Houghton Mifflin Company. All rights reserved.
340
Chapter 5: Trigonometric Functions
Section 5.7 1.
a = 2, p = 2π , phase shift = π 2
2.
a = −3 = 3, p = 2π , phase shift = −π
3.
a = 1, p = 2π = π , phase shift = π / 4 = π 2 2 8
4.
a = 3 , p = 2π = 4π , phase shift = −π / 3 = − 2π 4 1/ 2 1/ 2 3
5.
a = −4 = 4, p = 2π = 3π , phase shift = −π / 6 = − π 2/3 2/3 4
6.
a = 3 = 3 , p = 2π = 8π , phase shift = 3π / 4 = 3π 2 2 1/ 4 1/ 4
7.
a = 5 , p = 2π , phase shift = 2π 4 3 3
8.
a = 6, p = 2π = 6π , phase shift = π / 6 = π 1/ 3 1/ 3 2
9.
p = π , phase shift = π / 4 = π 2 2 8
10.
p = π = 2π , phase shift = − π = 2π 1/ 2 1/ 2
11.
p = 2π = 6π , phase shift = −π = −3π 1/ 3 1/ 3
12.
p = 2π , phase shift = π / 6 = π 3 3 18
13.
p = 2π = π , phase shift = π / 8 = π 2 2 16
14.
p = 2π = 8π , phase shift = π / 2 = 2π 1/ 4 1/ 4
15.
p = π = 4π , phase shift = −3π = −12π 1/ 4 1/ 4
16.
p = π , phase shift = π / 4 = π 2 2 8
17.
y = sin ⎛⎜ x − π ⎞⎟ 2⎠ ⎝
18.
y = sin ⎛⎜ x + π ⎞⎟ 6⎠ ⎝ 0 ≤ x + π ≤ 2π 6 − π ≤ x ≤ 11π 6 6
0 ≤ x − π ≤ 2π 2 π ≤ x ≤ 5π 2 2
period = 2π , phase shift = − π 6
period = 2π , phase shift π 2
19.
y = cos ⎛⎜ x + π ⎞⎟ ⎝2 3⎠ 0 ≤ x + π ≤ 2π 2 3 π − ≤ x ≤ 5π 3 2 3 2 π − ≤ x ≤ 10π 3 3
20.
y = cos ⎛⎜ 2 x − π ⎞⎟ 3⎠ ⎝
0 ≤ 2 x − π ≤ 2π 3 π ≤ 2 x ≤ 7π 3 3 π ≤ x ≤ 7π 6 6
period = 4π , phase shift = − 2π 3
period = π , phase shift = π 6
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.7
21.
y = tan ⎛⎜ x + π ⎞⎟ 4⎠ ⎝
341
22.
period = π , phase shift = − π 4
y = 3 cot ⎛⎜ 3x + π ⎞⎟ 2 4⎠ ⎝
23.
−π < x −π < π 2 2 π < x < 3π 2 2 period = π , phase shift = π
−π < x + π < π 2 4 2 3 π π − <x< 4 4
24.
y = tan ( x − π )
y = 2 cot ⎛⎜ x − π ⎞⎟ ⎝2 8⎠ π x 0< − <π 2 8 π < x < 9π 8 2 8 π < x < 9π 4 4
period = 2π , phase shift = π 4
25.
y = sec ⎛⎜ x + π ⎞⎟ 4⎠ ⎝
26.
0 ≤ 2 x + π ≤ 2π −π ≤ 2 x ≤ π
0 ≤ x + π ≤ 2π 4 π 7 − ≤x≤ π 4 4
0 < 3x + π < π 4 π 3 − < 3x < π 4 4
−π ≤ x ≤ π 2 2 period = π , phase shift = − π 2
period = 2π , phase shift = − π 4
−π <x<π 12 4
y = csc ( 2 x + π )
period = π , phase shift = − π 3 12
27.
y = csc ⎛⎜ x − π ⎞⎟ ⎝3 2⎠ x 0 ≤ − π ≤ 2π 3 2 π ≤ x ≤ 5π 2 3 2 3π ≤ x ≤ 15π 2 2
period = 6π , phase shift = 3π 2
28.
y = sec ⎛⎜ 2 x + π ⎞⎟ 6⎠ ⎝ 0 ≤ 2 x + π ≤ 2π 6 11 π − ≤ 2x ≤ π 6 6 11 π π − ≤x≤ 12 12 period = π , phase shift = − π 12
29.
y = −2sin ⎛⎜ x − 2π ⎞⎟ ⎝3 3 ⎠ 2 x 0 ≤ − π ≤ 2π 3 3 2π ≤ x ≤ 8π 3 3 3 2π ≤ x ≤ 8π period = 6π , phase shift = 2π
Copyright © Houghton Mifflin Company. All rights reserved.
342
30.
Chapter 5: Trigonometric Functions
y = − 3 sin ⎛⎜ 2 x + π ⎞⎟ 2 ⎝ 4⎠
31.
0 ≤ 2 x + π ≤ 2π 4 π 7 − ≤ 2x ≤ π 4 4 π π 7 − ≤x≤ 8 8
y = −3cos ⎛⎜ 3x + π ⎞⎟ 4⎠ ⎝
32.
y = −4 cos ⎛⎜ 3x + 2π ⎞⎟ ⎝ 2 ⎠ 3 x + 2π ≤ 2π 0≤ 2 −2π ≤ 3x ≤ 0 2 4 − π ≤ x≤0 3 period = 4 π , phase shift = − 4 π 3 3
0 ≤ 3x + π ≤ 2π 4 7 π − ≤ 3x ≤ π 4 4 − π ≤ x ≤ 7π 12 12 2 π period = , phase shift = − π 3 12
period = π , phase shift = − π 8
33.
y = sin x + 1, p = 2π
34.
y = − sin x + 1, p = 2π
35.
y = − cos x − 2, p = 2π
36.
y = 2sin x + 3, p = 2π
37.
y = sin 2 x − 2, p = π
38.
y = − cos x + 2, p = 4π 2
39.
y = 4cos (π x − 2 ) + 1, p = 2
40.
41.
y = − sin(π x + 1) − 2, p = 2 phase shift = − 1
phase shift = 2
π
42.
y = −3cos ( 2π x − 3) + 1, p = 1 phase shift = 3 2π
y = 2sin ⎜⎛ π x + 1⎟⎞ − 2, p = 4 ⎝ 2 ⎠ 2 phase shift = −
π
π
43.
y = sin ⎛⎜ x − π ⎞⎟ − 1 , p = 2π 2⎠ 2 ⎝ phase shift = π 2
44.
y = −2 cos ⎛⎜ x + π ⎞⎟ + 3, p = 2π 3⎠ ⎝ phase shift = − π 3
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.7
343
45.
y = tan x − 4, p = 2π 2
46.
y = cot 2 x + 3, p = π 2
47.
y = sec 2 x − 2, p = π
48.
y = csc x + 4, p = 6π 3
49.
y = csc x − 1, p = 4π 2
50.
y = sec ⎛⎜ x − π ⎞⎟ + 1, p = 2π 2⎠ ⎝ phase shift = π 2
y = x + cos x 2
51.
y = x − sin x
53.
y = x + sin 2 x
54.
y = 2 x − sin x 3
55.
56.
y = − sin x + cos x
57.
sine curve, a = 1, 2π = π , b = 2 b c phase shift = − = π , c = − π 3 b 6 y = sin ⎛⎜ 2 x − π ⎞⎟ 3⎠ ⎝
58.
59.
52.
cosecant curve, 2π = 4π , b = 1 2 b c phase shift = − = 2π , c = −π b x ⎛ ⎞ y = csc ⎜ − π ⎟ ⎝2 ⎠
60.
y = sin x + cos x
cosine curve, a = 1, 2π = 2π , b = 1 b phase shift = − c = π , c = −π b translation one unit upward
y = cos ( x − π ) + 1
tangent curve, π = π , b = 1 b c phase shift = − = π , c = − π b 2 2 π ⎛ ⎞ y = tan ⎜ x − ⎟ 2⎠ ⎝
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344
Chapter 5: Trigonometric Functions
61.
secant curve, 2π = 2π , b = 1 b c = π , c = −π = − phase shift b 2 2 y = sec ⎛⎜ x − π ⎞⎟ 2⎠ ⎝
63.
a.
62.
phase shift: − c = − −1.25π = 1.25(6) = 7.5 months π /6 b 2 π 2 π period: = = 2(6) = 12 months b π /6
64.
sine curve, 2π = 2π , b = 1 b phase shift = − c = − π , c = π b 2 2 y = sin ⎛⎜ x + π ⎞⎟ or y = cos x 2⎠ ⎝
a.
phase shift: − c = b
(− 127 π ) ⎛π ⎞ ⎜ ⎟ ⎝6⎠
= ⎛⎜ − 7π ⎞⎟ ⋅ ⎛⎜ 6 ⎞⎟ = 3.5 months ⎝ 12 ⎠ ⎝ π ⎠
period: 2π = 2π = 2(6) = 12 months b π /6 b.
c. 65.
First graph y1 = 4.1cos ⎛⎜ π t ⎞⎟ . Because the phase shift ⎝6 ⎠ is 7.5 months, shift the graph of y1 7.5 units to the
First graph y1 = 2.7 cos ⎛⎜ π t ⎞⎟ . Because the phase ⎝6 ⎠ shift is 3.5 months, shift the graph of y1 3.5 units
right to produce the graph of y2 . Now shift the graph
to the right to produce the graph of y2 . Now shift
of y2 upward 7 units to product the graph of S.
the graph of y2 upward 4 units to product the graph of S.
7.5 months after January 1 is the middle of August.
y = 2.3sin 2π t + 1.25t + 315 t = 16 between 1990 and 2006 y = 2.3sin 2π (16 ) + 1.25 (16 ) + 315 y ≈ 335 ppm difference ≈ 335 − 315 ≈ 20 ppm
67.
b.
66.
c.
3.5 months after January 1 is the middle of April.
a.
(cos t , sin t )
b.
(cos t + 5, 0)
c.
As A rotates counterclockwise from (1, 0) to (–1, 0), point B moves horizontally from (6, 0) to (4, 0). As A continues to rotate counterclockwise from (–1, 0) back to (1, 0), point B moves horizontally from (4, 0) back to (6, 0).
5 rpm a = 7 p = 0.2 min
p = 2π b 2 0.2 = π b b = 10π s = 7cos10π t + 5
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.7
68.
345
current: a=5
69.
1 2π = ⇒ b = 120π 60 b
= π radians/sec 5
no phase shift
in t seconds, θ increased by π t radians. 5 π s tan t = 5 400
i = 5cos(120π t ) voltage: a = 180
400 tan π t = s for 0 ≤ t < 2.5 or 7.5 < t ≤ 10 5
1 2π = ⇒ b = 120π 60 b
phase shift = = − =−
Change 6 rpm to radians/second. 6 rpm = 6 ⋅ 2π radians ⋅ 1 minute 1 minute 60 sec
c b c 3π = −0.005 ⇒ c = 0.6π or 120π 5
3π ⎞ ⎛ V = 180cos ⎜120π t + ⎟ 5 ⎠ ⎝ 70.
p = 12, 2π = 12, b = π b 6 π π c − = 3 c = −3 ⋅ = − 6 2 b π⎞ ⎛π Curve 1. y = 0.6sin ⎜ x − ⎟ + 12 2⎠ ⎝6
71.
π⎞ ⎛π Curve 2. y = 1.3sin ⎜ x − ⎟ + 12 2⎠ ⎝6
amplitude A = 3 k =9 24 = 2 cycles 12 = 1 cycle 2π π = 12 ⇒ B = B 6
72.
phase shift = −
π
y = 3cos t + 9 6 At 6:00 P.M., t = 12.
π⎞ ⎛π Curve 3. y = 2sin ⎜ x − ⎟ + 12 6 2⎠ ⎝ π⎞ ⎛π Curve 4. y = 2.7sin ⎜ x − ⎟ + 12 2⎠ ⎝6 Domain: 0 ≤ x ≤ 12
y = 3cos
y = sin x − cos x 2
π
⎛ 7π 5π ⎞ + f (7) = 30cos ⎜ ⎟ + 90 4 ⎠ ⎝ 12 11π = 30cos + 90 6 ⎛ 3⎞ = 30 ⎜⎜ ⎟⎟ + 90 = 15 3 + 90 ⎝ 2 ⎠ f (7) ≈ 116°
y = 3+9
y = 2sin 2 x − cos x
⎞ ⎟ + 90 ⎠
At 1:00 P.M., t = 7.
⋅ 12 + 9 6 y = 3cos 2π + 9
74.
c c = −15 = − b π /12 5π c= 4
5π ⎛π f (t ) = 30cos ⎜ t + 4 ⎝ 12
y = 12 ft
73.
amplitude = 30 p = 24, 2π = 24, b = π b 12
75.
y = 2 cos x + sin x 2
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346
Chapter 5: Trigonometric Functions
76.
y = − 1 cos 2 x + sin x 2 2
77.
y = x sin x 2
78.
79.
y = x sin x 2
80.
y = x cos x 2 2
81.
y = x cos x
82.
....................................................... 83.
sine function, a = 2, p = π
84.
2π = π , b = 2 b
cotangent function, p = π 2
85.
87.
tangent function, p = 2π
π = 2π ⇒ b = 1
g [ h ( x )] = ( cos x ) + 2 2
2
b
phase shift = − c = − π , c = π 4 6 b ⎛ ⎞ 2 π y = 3cos ⎜ x + ⎟ 6⎠ ⎝3
phase shift = − c = π ⇒ c = − π 4 b 2 π x ⎛ ⎞ y = tan ⎜ − ⎟ ⎝2 4⎠
88.
h [ g ( x )] = ( sin x ) + 2 ( sin x ) + 1 2
= cos2 x + 2
π =π , b=2 b
cosine function, a = 3, p = 3π 2π = 3π , b = 2 3 b
phase shift = − c = π , c = − 2π 3 b 3 ⎛ ⎞ 2 π y = 2sin ⎜ 2 x − ⎟ 3 ⎠ ⎝
86.
Connecting Concepts
= sin 2 x + 2sin x + 1
2
phase shift = − c = − π , c = π 4 2 b y = cot ⎛⎜ 2 x + π ⎞⎟ 2⎠ ⎝
89.
sin x → 1 as x → 0 x
90.
y = 2 + sec x , p = 2π = 4π 2 1/ 2
91.
y = x sin x
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.8
92.
347
y = x cos x
....................................................... PS1.
PS4.
3 2π
Prepare for Section 5.8
PS2. 5 2
18 = 9 = 3 2
PS3. 4cos 2π (0) = 4cos 0 = 4
PS5. 4cos ⎛⎜ 16 ⋅ 2π ⎞⎟ = 4cos(4 ⋅ 2π ) ⎝ 4 ⎠ = 4cos8π = 4
PS6.
y = 4cos π x
3.
y = 3cos
Section 5.8 1.
y = 2sin 2t
2.
amplitude = 2 2π p= =π 2 1 1 frequency = = p π
4.
y = 4sin 3t
amplitude =
amplitude = p=
2π =4 π /2
frequency =
1 1 = p 4
6.
amplitude = 4 2π =2 p= frequency =
1 1 = p 2
y = 5cos 2π t amplitude = 5 2π p= =1 2π 1 1 frequency = = = 1 p 1
y = 2sin
πt 3
amplitude = 2 p = 2π = 6 π /3 frequency 1 = 1 p 6
π
8.
3 4
y = 4cos π t
2t 3
amplitude = 3 2π p= = 3π 2/3 1 1 frequency = = p 3π
2π = 6π 1/ 3 1 1 frequency = = p 6π 5.
3 πt y = sin 4 2
2 3
p=
amplitude = 4 2π p= 3 1 1 3 = frequency = = p 2π / 3 2π 7.
2 t y = cos 3 3
9.
a=4 1 = 1 =2 frequency 1.5 3 2π = 2 ⇒ b = 3π 3 b p=
y = 4cos3π t
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348
10.
Chapter 5: Trigonometric Functions
frequency = 0.8 1 1 5 p= = = frequency 0.8 4 2π 5 8 = ⇒b= π b 4 5 amplitude = 4
11.
amplitude = 2 p =π 2π =π ⇒b = 2 b
14.
amplitude = 3
amplitude = 4 p=
15.
π
2 2π π = ⇒b=4 b 2
10 πt 3
amplitude = 1 p=2 2π = 2⇒b =π b y = sin π t
y = 4sin 4t
p =1
frequency = 1
2π = 1 ⇒ b = 2π b
p=
y = 3sin 2π t
2π = 1 ⇒ b = 2π b
amplitude = 4 frequency = 4 1 p= =1 frequency 4 2π = 1 ⇒ b = 8π b 4
y = 2sin 2π t
y = 4sin 8π t
17.
amplitude = 2
p = 0.6 a =1 2π 10 = 0.6 ⇒ b = π b 3 y = cos
3 4π y = cos t 2 3
y = 2sin 2t
16.
12.
3 amplitude = 2 2π 4π = 1.5 ⇒ b = b 3
8 y = 4cos π t 5
13.
p = 1.5
1 = 1 =1 frequency 1
18.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 5.8
19.
349
amplitude = 1 2 frequency = 2 ⇒ p = π π 2 2π = π ⇒ b = 4 b 2
20.
π
amplitude = 5
23.
amplitude =
1 2
24.
1 2π y = cos t 2 3
y = 5cos t 4 amplitude = 4 p=
y = 2.5cos π t
p=3 2π 2 =3⇒b = π b 3
π
26.
π
amplitude = 2 feet, a = −2
period = p =
1 2π
k 1 = m 2π
a.
b. 31.
28.
8 1 1 1 = ⋅ = 32 2π 2 4π
amplitude = 2 p =π 2π =π ⇒b = 2 b
amplitude = 1.5 feet, a = −1.5 frequency = f =
1 = 4π f
⎛ 1 y = a cos 2π ft = −2cos 2π ⎜ ⎝ 4π 1 = −2cos t 2 29.
period = p = ⎞ ⎟t ⎠
frequency = f = 392π = 196 cycles/s 2π 1 period = p = s 196
1 2π
k 1 = m 2π
amplitude = a = 78 − 4 = 37 2 2 π period = p = = π 45 22.5
π t + 41 ( 22.5 )
3 1 1 1 = ⋅ = 27 2π 3 6π
1 1 = = 6π f 1/ 6π
⎛ 1 y = −1.5cos 2π ft = −1.5cos 2π ⎜ ⎝ 6π 1 = −1.5cos t 3 30.
amplitude = 9.3 ft, a = 9.3 period = p = 2π = 0.16 12.5 h = 9.3sin 0.16π t
32.
amplitude = a = 6.5 − 4.5 = 1 2 2 π π period = p = = 60 30
The amplitude needs to increase.
h = −37cos
2π t 5
y = 2cos 2t
y = 4cos 4t
frequency = f =
amplitude = 5 p=5 2π 2π =5⇒b = 5 b y = 5cos
2 2π π = ⇒b=4 b 2
27.
amplitude = 2.5 frequency = 0.5 ⇒ p = 2 2π = 2 ⇒ b = π b
y = 3cos 2t
1 frequency = ⇒ p = 8 8 2π π =8⇒b= b 4
25.
21.
2π = π ⇒ b = 2 b
y = 1 cos 4t 2
22.
amplitude = 3 frequency 1 ⇒ p = π
d = cos π t + 5.5 30
Copyright © Houghton Mifflin Company. All rights reserved.
⎞ ⎟t ⎠
350
33.
Chapter 5: Trigonometric Functions
a.
The pseudoperiod is 2π = π . There are 10 ≈ 3 π
2
34.
a.
complete oscillations of length π in 0 ≤ t ≤ 10. b.
a.
f ( t ) < 0.01 for all t > 59.8 ( nearest tenth ) .
The pseudoperiod is 2π = 1. There are 10 complete 2π
b.
37.
a.
f ( t ) < 0.01 for all t > 10.5 ( nearest tenth ) .
Xmi n = 10, Xm ax = 20, Xscl = 1, Ymin = −0.01, Ymax = 0.01, Yscl = 0.005
36.
a.
oscillations of length 1 in 0 ≤ t ≤ 10. b.
2π
1
complete oscillations of length 2π in 0 ≤ t ≤ 10.
Xmin = 56, Xm ax = 65, Xscl = 1, Ymin = −0.01, Ymax = 0.01, Yscl = 0.005
35.
The pseudoperiod is 2π = 2π . There are 10 ≈ 1
The pseudoperiod is 2π = 2. There are 10 = 5 π
2
complete oscillations of length 2 in 0 ≤ t ≤ 10.
f ( t ) < 0.01 for all t > 71.0 ( nearest tenth ) .
b.
f ( t ) < 0.01 for all t > 17.2 ( nearest tenth ) .
Xmin = 70, Xm ax = 73, Xscl = 1
Xmi n = 15, Xm ax = 20, Xscl = 1,
Ymin = −0.01, Ymax = 0.01, Yscl = 0.005
Ymin = −0.01, Ymax = −.01, Yscl = 0.005
The pseudoperiod is 2π = 1. There are 10 complete 2π
38.
a.
oscillations of length 1 in 0 ≤ t ≤ 10.
( )
The pseudoperiod is 2π = 2 . There are 10 = 15 3π 3 2/3 complete oscillations of length 2 in 0 ≤ t ≤ 10. 3
b.
f ( t ) < 0.01 for all t > 9.1 ( nearest tenth ) .
b.
Xmin = 8, Xm ax = 10, Xscl = 1, Ymin = −0.01, Ymax = 0.01, Yscl = 0.005
39.
a.
The pseudoperiod is 2π = 1. There are 10 complete 2π
Xmin = 22.5, Xma x = 24, Xscl = 1, Ymin = −0.01, Ymax = 0.01, Yscl = 0.005
40.
a.
oscillations of length 1 in 0 ≤ t ≤ 10.
b.
f ( t ) < 0.01 for all t > 6.1( nearest tenth ) .
X min = 5, X max = 7, Xscl = 1, Ymin = −0.01, Ymax = .01, Yscl = 0.005
f ( t ) < 0.01 for all t > 23.0 ( nearest tenth ) .
The pseudoperiod is 2π = 1. There are 10 complete 2π
oscillations of length 1 in 0 ≤ t ≤ 10.
b.
f ( t ) < 0.01 for all t > 4.6 ( nearest tenth ) .
X min = 4, X max = 6, Xscl = 1, Ymin = −0.01, Ymax = 0.01, Yscl = 0.005
Copyright © Houghton Mifflin Company. All rights reserved.
Exploring Concepts with Technology
351
....................................................... 41.
43.
45.
m 9m , p2 = 2π = 3 p1 k k Increasing the main mass to 9m will triple the period. p1 = 2π
yes
Connecting Concepts 42.
The frequency will double.
44.
no
Xmin = 0, Xma x = 15, Xscl = 1,
Xmin = 0, Xma x = 10, Xscl = 1,
Ymin = −1, Ymax = 1.5, Yscl = 0.25
Ymin = −3, Ymax = 5, Yscl = 1
yes
46.
yes
Xmin = 0, Xmax = 10, Xscl = 1,
Xmin = 0, Xmax = 15, Xscl = 1,
Ymin = −3, Ymax = 9, Yscl = 1
Ymin = −1, Ymax = 1.5, Yscl = 0.25
.......................................................
Exploring Concepts with Technology
Sinusoidal Families 1.
All three sine graphs have, a period of 2π , x-intercepts at nπ , and no phase shift, but their amplitudes are 2, 4, and 6 respectively. 2.
All three sine graphs have x-intercepts at n, an amplitude of 1, and no phase shift, but their periods are 2, 1, and 0.5 respectively, and y = sin 2π x and y = 4π x have additional x-intercepts at 0.5n and 0.25n respectively. 3.
All three sine graphs have a period of 2π and an amplitude of 1, but their phase shifts are −π / 4, − π / 6, and − π /12, respectively. Copyright © Houghton Mifflin Company. All rights reserved.
352
4.
Chapter 5: Trigonometric Functions
Yes, the calculator has displayed all three graphs. All three sine graphs have an amplitude of 1, a period of 2π , and a phase shift of −(2n − 1)π .
.......................................................
Assessing Concepts
1.
True
2.
False; sec2 θ − tan 2 θ = 1 is an identity.
3.
False; 1 rad ≈ 57.3°.
4.
True
5.
π
6.
(0, 1)
7.
The period is
2π = 8 . 3π / 4 3
8.
Shift the graph of y1 to the left π units. 2
9.
All real numbers except multiples of π .
10.
The vertical asymptotes are x = π and x = 3π . 2 2
4
....................................................... 1.
complement: 90° − 65° = 25° [5.1] supplement:180° − 65° = 115°
2.
Chapter Review
θ = 980° = 260° + 2 ⋅ 360° [5.3] θ is coterminal with α = 260° and θ ' = α '. Since 180° < α < 270°, 180° + α ' = α 180° + a′ = 260° α ' = 80° θ = 80°
3.
2 = 2 ⎜⎛ 180° ⎟⎞ [5.1] ⎝ π ⎠ = 114.59°
6.
θ=
s 12 [5.1] = r 40 = 0.3
For exercises 8 to 11, cscθ = 8.
cos θ = x = 5 [5.2] r 3
10.
sin θ =
y 2 [5.2] = r 3
4.
315° = 315° ⎛⎜ π ⎞⎟ [5.1] ⎝ 180° ⎠ 7 π = 4 7.
w=
5.
s = rθ = 3 ( 75° ) ⎛⎜ π ⎞⎟ ⎝ 180° ⎠ [5.1] = 3.93 m
V 50 63360 = ⋅ [5.1] r 16 3600 ≈ 55 rad/sec
3 r = , r = 3, y = 2, and x = 32 − 22 = 5. 2 y 9.
cot θ = x = 5 [5.2] y 2
11.
sec θ = r = 3 = 3 5 [5.2] 5 x 5
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Chapter Review
353
12.
2
x = 1, y = −3, r = 12 + ( −3) = 10
13.
15.
3 3 10 =− 10 10
cosθ =
cscθ = −
10 3
secθ = 10
sec 150° =
b.
⎛ 3π ⎞ tan ⎜ − ⎟ = 1 ⎝ 4 ⎠
c.
cot ( −225° ) = −1
d.
cos
a.
b.
16.
a.
b.
17.
a.
b.
(
a.
cos 123° ≈ −0.5446
b.
cot 4.22 ≈ 0.5365
c.
sec612° ≈ −3.2361
d.
tan
2π ≈ 3.0777 [5.3] 5
)
2 3 x = , x = − 3, r = 2, y = − 22 − − 3 = −1 [5.3] 2 r
y 1 =− r 2
tan φ =
y −1 3 = = x − 3 3
3 y = , y = 3, x = −3, r = 3 x
secφ =
( −3 ) 2 + (
3
)
2
= 2 3 [5.3]
r 2 3 2 3 = =− 3 x −3
csc φ =
sin φ = −
14.
2π 1 [5.3] =− 3 2
sin φ =
tan φ = −
tan θ =
2 2 3 =− 3 − 3
a.
cos φ = −
−3 = −3 1 1 cot θ = − 3
1 10 = 10 10
sin θ = −
r 2 3 = =2 y 3
(
)
2 2 , y = − 2, r = 2, x = − 22 − − 2 = 2 [5.3] 2
cos φ =
x 2 = 2 r
cot φ =
x 2 = = −1 y − 2
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354
18.
Chapter 5: Trigonometric Functions
a.
W (π ) = ( −1,0 ) [5.4]
b.
3⎞ ⎛ π ⎞ ⎛1 W ⎜ − ⎟ = ⎜⎜ , − ⎟ 2 ⎟⎠ ⎝ 3⎠ ⎝2
c.
⎛ 5π W⎜ ⎝ 4
19.
= sin x tan x = f ( x)
The function defined by f ( x) = sin( x) tan( x) is an even function. 21.
cos (π + t ) = − x
tan ( −t ) = −
cos t = x
1+
sin 2 φ cos 2 φ
y x
y x tan ( −t ) = − tan t
cos (π + t ) = − cos t
22.
[5.4]
f ( − x ) = sin ( − x ) tan ( − x ) = ( − sin x )( − tan x )
2 2⎞ ⎞ ⎛ ,− ⎟ ⎟ = ⎜⎜ − 2 ⎟⎠ ⎠ ⎝ 2
20.
f ( x ) = sin ( x ) tan ( x )
tan t =
sin φ
2
= 1 + tan φ [5.4]
23.
+1 tan φ + 1 cos φ = cot φ + 1 cos φ + 1
[5.4]
sin φ
= sec2 φ
sin φ + cos φ cos φ = cos φ +sin φ sin φ
=
sin φ ( sin φ + cos φ )
cos φ ( cos φ + sin φ )
= tan φ 24.
cos 2 φ + sin 2 φ 1 [5.4] = csc φ csc φ = sin φ
25.
sin 2 φ (tan 2 φ + 1) = sin 2 φ sec2 φ [5.4] =
sin 2 φ cos2 φ
= tan 2 φ
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Chapter Review
26.
1+
1 2
tan φ
355
= =
tan 2 φ + 1 2
tan φ
27.
[5.4]
cos2 φ 2
1 − sin φ
sec2 φ
−1 =
1 − sin 2 φ
1 − sin 2 φ = 1−1
− 1 [5.4]
=0
tan 2 φ
1 cos 2 φ = sin 2 φ cos 2 φ
=
1 sin 2 φ
= csc2 φ 28.
y = 3cos ( 2 x − π ) [5.5]
29.
2π 2π = =π 2 b −π π c = phase shift = − = − 2 2 b
no amplitude; period =
a = 3 = 3; period =
30.
32.
y = 2 tan 3x [5.6]
31.
3π ⎞ ⎛ y = −4sec ⎜ 4 x − ⎟ [5.6] 2 ⎠ ⎝ 2π 2π π = = no amplitude; period = 4 2 b −3π / 2 3π c = phase shift = − = − 4 8 b
=
π 3
33.
2π ⎞ ⎛ y = cos ⎜ 2 x − ⎟ + 2 [5.5] 3 ⎠ ⎝ 2π 2π a = 1 = 1; period = = =π 2 b −2π / 3 π c = phase shift = − = − 2 3 b
π⎞ ⎛ y = 2csc ⎜ x − ⎟ − 3 [5.6] 4⎠ ⎝ 2π 2π = = 2π no amplitude; period = 1 b −π / 4 π c = phase shift = − = − 1 4 b
34.
y = 2 cos π x, p = 2π = 2
35.
y = − sin 2 x , p = 2π = 3π 3 2/3
36.
37.
y = cos ⎛⎜ x − π ⎞⎟ , p = 2π 2⎠ ⎝
38.
y = 1 sin ⎜⎛ 2 x + π ⎟⎞ , p = 2π = π 2 ⎝ 4⎠ 2
39.
phase shift = π 2
b
phase shift = 0
π⎞ ⎛ y = −2sin ⎜ 3 x + ⎟ [5.5] 3⎠ ⎝ 2π 2π a = −2 = 2; period = = 3 b c π /3 π phase shift = − = − =− b 3 9
π
π
phase shift − π 8
y = 2sin 3x , p = 2π = 4π 2 3/ 2 3
y = 3cos3 ( x − π ) , p = 2π 3 phase shift = π
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356
40.
Chapter 5: Trigonometric Functions
y = − tan x , p = π = 2π 2 1/ 2
41.
y = 2 cot 2 x, p = π 2
42.
y = tan ⎛⎜ x − π ⎞⎟ , p = π 2⎠ ⎝
phase shift = π 2
43.
y = − cot ⎛⎜ 2 x + π ⎞⎟ , p = π 4⎠ 2 ⎝
44.
phase shift = − π 8
y = −2 csc ⎜⎛ 2 x − π ⎟⎞ , p = 2π = π 3⎠ 2 ⎝
45.
y = 3 sec ⎛⎜ x + π ⎞⎟ , p = 2π 4⎠ ⎝
phase shift = − π 4
phase shift = π 6
46.
y = 3sin 2 x − 3
47.
y = 2cos3x + 3
48.
y = − cos ⎛⎜ 3x + π ⎞⎟ + 2 2⎠ ⎝
49.
y = 3sin ⎛⎜ 4 x − 2π ⎞⎟ − 3 3 ⎠ ⎝
50.
y = 2 − sin 2 x
51.
y = sin x − 3 cos x
52.
53.
h 1.14 h = 1.14sin 4.5° ≈ 0.089 mi
sin 4.5° =
[5.2]
h 8.55 h = 8.55 tan 55.3 ≈ 12.3 feet [5.2] tan 55.3° =
Copyright © Houghton Mifflin Company. All rights reserved.
Quantitative Reasoning
54.
357
Speed for inner ring: v= s t 2π (14.5) = 24 ≈ 3.79609 ft/s
Speed for outer ring: v=s t 2π (21) = 24 ≈ 5.497787 ft/s
55.
The outer swing has a greater speed of 5.497787 − 3.79609 ≈ 1.7 ft/s. [5.1]
56.
y = 2.5sin 50t
80 + x 80 x = + h h h x (2) cot 37° = h x Substitute for in equation (1). h 80 cot18° = + cot 37° h 80 Solve for h. = cot18° − cot 37° h 1 h = 80 cot18° − cot 37° 80 h= ≈ 46 ft cot18° − cot 37° [5.2] (1) cot18° =
57.
[5.8]
amplitude = 0.5 [5.8] 1 f = 2π p =π
amplitude = 2.5 π 2π 2π = = p= 50 25 b 1 25 frequency = = p π
k 1 = m 2π
58.
f ( t ) < 0.01 for all t > 7.2 [5.8]
20 1 = 5 π
⎛1⎞ y = −0.5cos 2π ft = −0.5cos 2π ⎜ ⎟ t ⎝π ⎠ y = −0.5cos 2t
Xmin = 5, Xmax = 10, Xcsl = 1 Ymin = −.01, Ymax = .01, Yscl = .005
....................................................... QR1. a.
c.
e.
QR2.
2π = m ⇒ 3π n
2π n = 3π m ⇒ period = 6π 2π (3) = 3π (2)
π /2 = m ⇒ 2π / 3 n
π n = 2π m
⇒ period = 2π 3 π (4) = 2π (3) 2 3 5/ 2 = m ⇒ 5 n = 3 m ⇒ period = 7.5 3/ 2 n 2 2 5 (3) = 3 (5) 2 2
3 =m ⇒ 2.5 n
2
3n = 2.5m ⇒ period = 15 s 3(5) = 2.5(6)
Quantitative Reasoning b.
d.
f.
2/3 = m ⇒ n 4
2 n = 4m 3
⇒ period = 4
2 (6) = 4(1) 3 3π / 2 = m ⇒ 3π n = 8π m ⇒ period = 24π 8π / 3 n 2 3 3π (16) = 8π (9) 2 3 4π / 5 = m ⇒ 4π n = 4π m ⇒ period = 4π n 4π 5 4π (5) = 4π (1) 5
QR3. 1.25 = m ⇒ 2.25 n
1.25n = 2.25m ⇒ period = 11.25 s 1.25(9) = 2.25(5)
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358
QR4.
Chapter 5: Trigonometric Functions
6n = 4.5m = 27 w ⇒ period = 54 s 6(9) = 4.5(12) = 27(2)
....................................................... 1.
4.
150° = 150° ⎛⎜ π ⎞⎟ ⎝ 180° ⎠ 5 π = 6
[5.1]
rev [5.1] sec rev ⎛ 2π rad ⎞ w=6 ⎜ ⎟ sec ⎝ rev ⎠ w = 12π rad/sec
2.
5.
w=6
π−
11 π π= 12 12
Chapter Test s = rθ
3.
[5.1]
[5.1]
⎛ π ⎞ s = 10 ( 75° ) ⎜ ⎟ ⎝ 180° ⎠ s ≈ 13.1 cm
v = rw [5.1] = 8 ⋅ 10 = 80 cm/sec
6.
r = 7 2 + 32 r = 58
secθ =
7.
9.
csc 67° ≈ 1.0864
[5.2]
11π [5.4] 6 x = cos t y = sin t
t=
=
3 2
=−
8.
10.
tan
π 6
cos
sec2 t − 1 sec2 t
1 2
π 3
π 2
[5.2]
1 1 ⋅ − 1 [5.3] 3 2 1 = −1 2 3 =
=
3 −1 6
=
3 −6 6
1 −1 2t cos = 1 cos 2 t 1−cos 2 t cos 2 t
=
⎛ 3 1⎞ W ( x, y ) = W ⎜⎜ , − ⎟⎟ 2⎠ ⎝ 2
− sin
58 7
[5.4]
1 cos 2 t 2
= 1 − cos t = sin 2 t
11.
period =
π b
=
π 3
[5.6]
12.
a = −3 = 3; period = 2π = 2π = π b 2
phase shift = −
π 4
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[5.7]
Chapter Test
13.
359
period = π = 3 [5.7] π /3
14.
y = 3cos 1 x, p = 4π 2
phase shift = − c = − π / 6 = − 1 2 π /3 b
15.
y = −2sec 1 x , 2
p = 4π
16.
Shift the graph [of y = 2sin(2 x)] [5.7] π units to the right and down 1 unit. 4
17.
y = 2 − sin x 2
19.
18.
y = sin x − cos 2 x
20.
p=5 a = 13 y = 13 cos
5=
2π 2π ,b = b 5
2π 2π t or y = 13 sin t 5 5
tan 42.2° = h x h tan 42.2° = h cot 422.2°
x=
tan 37.4o = =
h 5.24 x h 5.24 + h cot 42.2°
Solve for h. h 5.24 + h cot 42.2° tan 37.4°(5.24 + h cot 42.2°) = h 5.24 tan 37.4° + h tan 37.4° cot 42.2° = h h − h tan 37.4° cot 42.2° = 5.24 tan 37.4° h(1 − tan 37.4° cot 42.2°) = 5.24 tan 37.4° 5.24 tan 37.4° h= 1 − tan 37.4° cot 42.2° h ≈ 25.5 meters tan 37.4o =
The height of the tree is approximately 25.5 meters. [5.2]
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[5.8]
360
Chapter 5: Trigonometric Functions
.......................................................
Cumulative Review 3 ÷1 = 3⋅2= 3 2 2 2 1
1.
x 2 − y 2 = ( x + y )( x − y ) [P.4]
2.
3.
A = 1 bh [P.4] 2 = 1 (4)(6) 2 = 12 in 2
4.
−x = − x = − f ( x) [2.5] (− x) 2 + 1 x 2 + 1 Odd function
5.
x [4.1] 2x − 3 y x= 2y − 3 x(2 y − 3) = 2 xy − 3 x = y 2 xy − y = y (2 x − 1) = 3 x y = 3x 2x − 1 f −1 ( x) = 3 x 2x − 1
6.
Domain: (−∞, 4) ∪ (4, ∞) [2.2/3.5]
7.
Range: [0, 2] [2.2]
8.
Shift the graph of y = f (x) horizontally 3 units to the right. [2.5]
9.
Reflect the graph of y = f (x) across the y-axis. [2.5]
10.
300o = 300o ⎛⎜ π o ⎞⎟ = 5π [5.1] ⎝ 180 ⎠ 3
11.
5π = 5π ⎛ 180o ⎞ = 225o [5.1] ⎜ ⎟ 4 4 ⎝ π ⎠
12.
f ⎛⎜ π ⎞⎟ = sin ⎛⎜ π + π ⎞⎟ = sin ⎛⎜ π ⎞⎟ = 1 [5.3] ⎝2⎠ ⎝3⎠ ⎝3 6⎠
13.
f ⎛⎜ π ⎞⎟ = sin ⎛⎜ π ⎞⎟ + sin ⎛⎜ π ⎞⎟ = 3 + 1 = 3 + 1 2 ⎝3⎠ ⎝3⎠ ⎝6⎠ 2 2
14.
⎞ ⎛ ⎞ ⎛ cos2 45o + sin 2 60o + = ⎜ 2 ⎟ + ⎜ 3 ⎟ = 2 + 3 = 5 [5.2] 4 4 4 ⎝ 2 ⎠ ⎝ 2 ⎠
16.
θ = 210o [5.3] Since 180o < θ < 270o , θ ′ + 180o = θ θ ′ = 30o
f ( x) =
15.
negative [5.3]
17.
θ = 2π 3
[5.3]
18.
[5.2]
[P.5]
f (− x) =
2
Domain: ( −∞, ∞ ) [5.4]
19.
Range: [–1, 1]
Since π < θ < π , 2 θ +θ′ = π
θ′ = π
3
20.
tan θ =
opp 3 = adj 4
hypotenuse = 32 + 42 = 9 + 16
sin θ =
opp 3 = hyp 5
[5.2]
= 25 =5
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2
[5.4]
Chapter 6
Trigonometric Identities and Equations Section 6.1 1.
tan x csc x cos x =
sin x 1 ⋅ ⋅ cos x = 1 cos x sin x
2.
tan x sec x sin x = tan x ⋅ 1 ⋅ sin x cos x = tan x ⋅ sin x cos x = tan x ⋅ tan x = tan 2 x
3.
4sin 2 x − 1 (2sin x − 1)(2sin x + 1) = = 2sin x − 1 2sin x + 1 2sin x + 1
4.
5.
(sin x − cos x)(sin x + cos x) = sin 2 x − cos 2 x
6.
2
2
= 1 − cos x − cos x
sin 2 x − 2 sin x + 1 (sin x − 1) 2 = = sin x − 1 sin x − 1 sin x − 1
(tan x)(1 − cot x) = tan x − tan x cot x = tan x − 1
= 1 − 2cos 2 x 7.
9.
11.
1 1 cos x sin x − = − sin x cos x sin x cos x sin x cos x cos x − sin x = sin x cos x cos x(1 + sin x) cos x = 1 − sin x (1 − sin x)(1 + sin x) cos x(1 + sin x) = 1 − sin 2 x cos x(1 + sin x) = cos 2 x (1 + sin x) 1 sin x = = + cos x cos x cos x = sec x + tan x
1 − tan 4 x sec2 x
= =
(1 + tan 2 x)(1 − tan 2 x)
8.
10.
12.
sec2 x 2
1 3 cos x 3sin x + = + sin x cos x sin x cos x sin x cos x cos x + 3sin x = sin x cos x sin x(1 + cos x) sin x = 1 − cos x (1 − cos x)(1 + cos x) sin x(1 + cos x) = 1 − cos 2 x sin x(1 + cos x) = sin 2 x 1 + cos x = sin x 1 cos x = + sin x sin x = csc x + cot x sin 4 x − cos 4 x = (sin 2 x + cos 2 x)(sin 2 x − cos2 x) = 1(sin 2 x − cos 2 x)
2
sec x(1 − tan x)
= sin 2 x − cos 2 x
2
sec x
= 1 − tan 2 x 13.
1 + tan 3 x (1 + tan x)(1 − tan x + tan 2 x) = 1 + tan x 1 + tan x = 1 − tan x + tan 2 x
14.
(
)
sin x cos x tan x − sin x cos x cos x − sin x = cot x cot x sin x − sin x = cot x =0
Copyright © Houghton Mifflin Company. All rights reserved.
362
15.
Chapter 6: Trigonometric Identities and Equations
sin x − 2 + 1
sin x = sin x − 1 sin x
=
sin x − 2 + 1
sin x ⋅ sin x 1 sin x sin x − sin x 2
16.
sin x − 2sin x + 1
sin 2 x − 1 (sin x − 1)(sin x − 1) = (sin x − 1)(sin x + 1) sin x − 1 = sin x + 1
17.
(sin x + cos x)2 = sin 2 x + 2 sin x cos x + cos 2 x
18.
= sin 2 x + cos 2 x + 2 sin x cos x = 1 + 2 sin x cos x 19.
cos x(1 − sin x) cos x = 1 + sin x (1 + sin x)(1 − sin x) cos x(1 − sin x) = 1 − sin 2 x cos x(1 − sin x) = cos 2 x 1 − sin x = cos x 1 sin x = − cos x cos x = sec x − tan x cos x
21.
sin x sin x sin x(1 + cos x) − sin x(1 − cos x) − = 1 − cos x 1 + cos x (1 − cos x)(1 + cos x ) sin x + sin x cos x − sin x + sin x cos x = 1 − cos 2 x 2sin x cos x = sin 2 x 2cos x = sin x = 2cot x
(tan x + 1)2 = tan 2 x + 2 tan x + 1 = 1 + tan 2 x + 2 tan x = sec 2 x + 2 tan x
20.
sin x(1 − cos x) sin x = 1 + cos x (1 + cos x)(1 − cos x) sin x(1 − cos x) = 1 − cos 2 x sin x(1 − cos x) = sin 2 x 1 − cos x = sin x 1 cos x = − sin x sin x = csc x − cot x cos x
sin x
cot x + tan x sin x + cos x = 1 sec x
22.
cos x cos x sin x + sin x cos x = sin x cos x ⋅ 1 sin x cos x cos x 2 2
cos x + sin x sin x 1 = sin x = csc x =
23.
cos x tan x + 2cos x − tan x − 2 cos x(tan x + 2) − (tan x + 2) = tan x + 2 tan x + 2 (tan x + 2)(cos x − 1) = tan x + 2 = cos x − 1
24.
2 sin x cot x + sin x − 4 cot x − 2 sin x(2 cot x + 1) − 2(2 cot x + 1) = 2 cot x + 1 2 cot x + 1 (2 cot x + 1)(sin x − 2) = 2 cot x + 1 = sin x − 2
sin x
+ cot x + tan x sin x cos x = 1 csc x sin x cos x sin x + sin x cos x = sin x cos x ⋅ 1 sin x cos x sin x 2 2
cos x + sin x cos x 1 = cos x = sec x =
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.1
363
25.
1 − sin x 1 sin x = − = sec x − tan x cos x cos x cos x
26.
cos x − 1 cos x 1 = − = cot x − csc x sin x sin x sin x
27.
sin 2 x − cos 2 x = sin 2 x − (1 − sin 2 x)
28.
sin 2 x − cos 2 x = 1 − cos 2 x − cos 2 x
= sin 2 x − 1 + sin 2 x
= 1 − 2cos 2 x
= 2 sin 2 x − 1 29.
1 sin 2 x
+
1 cos 2 x
= =
cos2 x + sin 2 x
30.
tan 2 x
sin 2 x cos 2 x 1 2
1
−
1 cot 2 x
= =
2
sin x cos x
= csc2 x sec2 x
cot 2 x − tan 2 x tan 2 x cot 2 x (csc2 x − 1) − (sec2 x − 1) 1
= csc2 x − 1 − sec2 x + 1 = csc 2 x − sec 2 x
31.
1 − cos x cos x
sec x − cos x = =
32.
=
sin 2 x cos x = sin x tan x 1 1 +1 +1 sin x sin x sin x = ⋅ 1 1 −1 − 1 sin x sin x sin x
34.
1 + sin x 1 − sin x (1 + sin x) (1 + sin x) = ⋅ 1 − sin x 1 + sin x
= =
=
1 + 2 sin x + sin 2 x
=
1 − sin 2 x 1 + 2 sin x + sin 2 x
=
cos 2 x 1 cos 2 x
+
2 sin x cos 2 x
+
1 1 1 1 + + sin x cos x sin x cos x sin x cos x = ⋅ 1 1 1 1 sin x cos x − − sin x cos x sin x cos x
cos x + sin x cos x − sin x cos x + sin x cos x − sin x = ⋅ cos x − sin x cos x − sin x
=
=
sin x cos x + cos x sin x
sin 2 x + cos 2 x sin x cos x 1 = sin x cos x = csc x sec x
1 − cos 2 x cos x
=
33.
tan x + cot x =
cos 2 x − sin 2 x 2
cos x − 2 sin x cos x + sin 2 x cos 2 x − sin 2 x 1 − 2 sin x cos x
sin 2 x cos 2 x
2
= sec x + 2 tan x sec x + tan 2 x 35.
sin 4 x − cos 4 x = (sin 2 x + cos 2 x)(sin 2 x − cos2 x ) = 1(sin 2 x − cos 2 x) = sin 2 x − (1 − sin 2 x) = sin 2 x − 1 + sin 2 x = 2 sin 2 x − 1
36.
sin 6 x + cos6 x = (sin 2 x + cos 2 x)(sin 4 x − sin 2 x cos 2 x + cos 4 x) = sin 4 x − sin 2 x cos 2 x + cos 4 x
Copyright © Houghton Mifflin Company. All rights reserved.
364
37.
39.
Chapter 6: Trigonometric Identities and Equations
1 1 1 + cos x = ⋅ 1 − cos x 1 − cos x 1 + cos x 1 + cos x = 1 − cos 2 x 1 + cos x = sin 2 x
38.
sin x cos x sin x − cos x − = 1 − sin x 1 − sin x 1 − sin x
40.
cos 2 x 1 − sin 2 x = 1 − sin x 1 − sin x (1 − sin x)(1 + sin x) = 1 − sin x = 1 + sin x
tan x cot x tan x − cot x − = 1 + tan x 1 + tan x 1 + tan x
sin x cos x − = sin x sin x 1 sin x − sin x sin x
=
41.
43.
45.
tan x cot x − = tan x tan x 1 tan x + tan x tan x 2
1 − cot x csc x − 1
1 − cot x cot x + 1 (1 − cot x)(1 + cot x) = cot x + 1 = 1 − cot x
=
(1 − cos x) − (1 + cos x) 1 1 − = 1 + cos x 1 − cos x (1 + cos x)(1 − cos x) 1 − cos x − 1 − cos x = 1 − cos 2 x −2 cos x = sin 2 x = −2 cot x csc x
1 1 + csc x + csc x sin x sin x sin x = ⋅ 1 1 sin x − sin x − sin x sin x sin x 1+1 = 1 − sin 2 x 2 = cos 2 x cos x 1+ 1 cot x + 1 + csc x = sin x + sin x 1 cos x 1 + csc x cot x 1+ sin x sin x = cos x + sin x + 1 sin x + 1 cos x cos 2 x + (sin x + 1) 2 = cos x(sin x + 1) 2 2 = cos x + sin x + 2sin x + 1 cos x(sin x + 1) = 1 + 2sin x + 1 cos x(sin x + 1) 2(1 + sin x) = cos x(sin x + 1) = 2sec x
42.
44.
1 1 (1 + sin x) − (1 − sin x) − = 1 − sin x 1 + sin x (1 − sin x)(1 + sin x) 1 + sin x − 1 + sin x = 1 − sin 2 x 2sin x = cos 2 x = 2 tan x sec x 2 ⎛ ⎞ ⎟ 2 cot x = tan x ⎜ tan x cot x + tan x tan x ⎜⎜ 1 + tan x ⎟⎟ ⎝ tan x ⎠ 2 = 1 + tan 2 x = 2 sec2 x = 2 cos2 x
46.
1 − 1 cos 2 x sin 2 x 2 2 = sin 2x − cos2 x sin x cos x sin 2 x − cos 2 x = sin x cos 2x sin2 x cos x sin x cos x sin x cos x sin x − cos x = cos x sin x sin x cos x = tan x − cot x sin x cos x
sec2 x − csc 2 x =
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.1
47.
365
1 + sin x 1 + sin x 1 + sin x = ⋅ 1 − sin x 1 − sin x 1 + sin x
48.
(1 + sin x)2
=
1 − sin 2 x (1 + sin x)2
=
cos x ⋅ sin x sin x cot x 2 cos x = cos x / sin x = 2 sin x cos x +
cos x + cot x sin x = cot x
cos 2 x 1 + sin x , cos x > 0 = cos x
49.
sin 3 x + cos3 x (sin x + cos x)(sin 2 x − sin x cos x + cos 2 x) 50. = sin x + cos x sin x + cos x = sin 2 x − sin x cos x + cos2 x = 1 − sin x cos x
1 − sin x 1 + sin x (1 − sin x)2 − (1 − sin x)2 − = 1 + sin x 1 − sin x (1 + sin x)(1 − sin x) =
1 − 2 sin x + sin 2 x − 1 − 2 sin x − sin 2 x
=
1 − sin 2 x −4 sin x
cos 2 x = −4 tan x sec x
51.
sec x − 1 sec x + 1 (sec x − 1)2 − (sec x + 1) 2 − = sec x + 1 sec x − 1 (sec x − 1)(sec x + 1) = =
sec2 x − 2sec x + 1 − sec 2 x − 2sec x − 1 sec2 x − 1 −4sec x
1 cos x 1 + cos x − cos x(1 − cos x) − = 1 − cos x 1 + cos x (1 − cos x)(1 + cos x) = =
tan 2 x 2
−4 cos x ⋅ cos x sin 2 x = −4csc x cot x =
52.
=
1 + cos x − cos x + cos 2 x 1 − cos 2 x 1 + cos 2 x sin 2 x 2 2
sin x
1 + sin x cos x (1 + sin x )(1 − sin x ) − cos x (cos x ) 1 − sin 2 x − cos2 x cos2 x − cos2 x − = = = =0 cos x 1 − sin x cos x (1 − sin x ) cos x (1 − sin x ) cos x (1 − sin x )
54.
( sin x + cos x + 1) 2 = sin 2 x + sin x cos x + sin x + cos x sin x + cos2 x + cos x + sin x + cos x + 1 = 1 + 2sin x cos x + 2sin x + 2 cos x + 1 = 2 ( sin x cos x + cos x + sin x + 1) = 2 ( sin x + 1) ( cos x + 1) sin x 1 sec x + tan x cos x + cos x cos x = ⋅ 1 − sin x cos x sec x − tan x cos x
cos x
1 + sin x = 1 − sin x 1 + sin x 1 + sin x = ⋅ 1 − sin x 1 + sin x = =
(1 + sin x)2 1 − sin 2 x (1 + sin x)2 cos 2 x
Copyright © Houghton Mifflin Company. All rights reserved.
1 + 1 − sin 2 x
sin 2 x sin 2 x
= 2csc2 x − 1
53.
55.
−
=
sin 2 x
366
56.
Chapter 6: Trigonometric Identities and Equations
sin 3 x − cos3 x (sin x − cos x )(sin 2 x + sin x cos x + cos2 x ) = sin x + cos x sin x + cos x sin x − cos x (sin x − cos x )(1 + sin x cos x ) = ⋅ sin x − cos x sin x + cos x = = =
=
=
(sin 2 x − 2sin x cos x + cos2 x )(1 + sin x cos x ) sin 2 x − cos2 x (1 − 2sin x cos x )(1 + sin x cos x ) sin 2 x − cos2 x 1 − sin x cos x − 2sin 2 x cos2 x sin 2 x − cos2 x 1 sin 2 x
2 x cos2 x sin x sin 2 x sin 2 x − cos2 x sin 2 x sin 2 x 2
− sin x 2cos x − 2sin
csc2 x − cot x − 2 cos x 1 − cot 2 x
57.
Identity
58.
Identity
59.
Identity
60.
Identity
61.
Identity
62.
Identity
63.
Not an identity
64.
Not an identity
65.
Not an identity. If x = π / 4, the left side is 2 and the right side is 1.
66.
Not an identity. If x = π / 6, the left side is
67.
Not an identity. If x = 0°, the left side is
68.
Not an identity. If x = π , the left side is 1 and the right side is −1.
69.
Not an identity. If x = 0, the left side is –1 and the right side is 1.
70.
Not an identity. If x = π , the left side is 1 and the right side is −1.
3 and the right side is 2 3 / 3.
3 / 2 and the right side is (2 + 3) / 2.
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Section 6.1
367
....................................................... 71.
1 − sin x + cos x 1 − sin x + cos x 1 + sin x + cos x = ⋅ 1 + sin x + cos x 1 + sin x + cos x 1 + sin x − cos x
= = =
Connecting Concepts 72.
1 − tan x + sec x (1 − tan x + sec x)(1 + tan x + sec x) = 1 + tan x − sec x (1 + tan x − sec x)(1 + tan x + sec x)
1 − sin 2 x + 2sin x cos x − cos 2 x
=
1 + 2sin x + sin 2 x − cos 2 x 2
2
1 − (sin x + cos x) + 2sin x cos x 2
=
2
2sin 4 x + 2sin 2 x cos2 x − 3sin 2 x − 3cos 2 x 2sin 2 x
= = =
1 + 2 tan x + tan 2 x − sec2 x 1 + 2sec x − (sec2 x − 1) + sec2 x
1 + 2 tan x + tan 2 x − (tan 2 x + 1) 2 + 2sec x = 2 tan x 1 + sec x = tan x
1 + 2sin x + sin x − (1 − sin x) 1 − 1 + 2sin x cos x
1 + 2sin x + sin 2 x − 1 + sin 2 x 2sin x cos x 2sin x cos x = = 2 x(1 + sin x) 2sin 2sin x + 2sin x cos x = 1 + sin x 73.
1 + 2sec x − tan 2 x + sec2 x
2sin 2 x(sin 2 x + cos 2 x) − 3(sin 2 x + cos 2 x) 2sin 2 x (2sin 2 x − 3)(sin 2 x + cos2 x) 2sin 2 x 2sin 2 x − 3 2sin 2 x 2sin 2 x
3 − 2sin 2 x 2sin 2 x 3 = 1 − csc2 x 2 =
74.
4 tan x sec2 x − 4 tan x − sec2 x + 1 4 tan 3 x − tan 2 x
= = =
4 tan x (sec2 x − 1) − (sec2 x − 1) 4 tan 3 x − tan 2 x (4 tan x − 1)(sec2 x − 1) tan 2 x(4 tan x − 1) tan 2 x
tan 2 x =1 75.
sin x cos x cos x sin x − cos x sin x tan x + sin x − 2 sin x = sin x − cos x sin x(tan x − 1) = sin x − cos x sin x(tan x − 1) cos x = sin x cos x − cos x cos x tan x(tan x − 1) = tan x − 1 = tan x
sin x(tan x + 1) − 2 tan x cos x = sin x − cos x
sin x tan x + sin x − 2
Copyright © Houghton Mifflin Company. All rights reserved.
368
76.
Chapter 6: Trigonometric Identities and Equations
sin 2 x cos x + cos3 x − sin 3 x cos x − sin x cos3 x 2
1 − sin x
= =
cos x(sin 2 x + cos 2 x) − sin x cos x(sin 2 x + cos 2 x) 1 − sin 2 x cos x − sin x cos x
1 − sin 2 x cos x(1 − sin x) = (1 − sin x)(1 + sin x) cos x = 1 + sin x
77.
sin 4 x + cos 4 x = sin 4 x + 2 sin 2 x cos 2 x + cos4 x − 2 sin 2 x cos 2 x = (sin 2 x + cos 2 x)2 − 2 sin 2 x cos 2 x = 1 − 2 sin 2 x cos 2 x
78.
tan 4 x + sec 4 x = tan 4 x − 2 tan 2 x sec 2 x + sec 4 x + 2 tan 2 x sec 2 x = (tan 2 x − sec 2 x)2 + 2 tan 2 x sec 2 x = 1 + 2 tan 2 x sec2 x
.......................................................
Prepare for Section 6.2 PS2. sin ⎜⎛ π + π ⎟⎞ = sin ⎜⎛ 5π ⎟⎞ = 1 ⎝2 3⎠ ⎝ 6 ⎠ 2
PS1. cos ⎛⎜ π − π ⎞⎟ = cos ⎛⎜ π ⎞⎟ = 1 ⎝2 6⎠ ⎝3⎠ 2 cos ⎛⎜ π ⎞⎟ cos ⎛⎜ π ⎞⎟ + sin ⎛⎜ π ⎞⎟ sin ⎛⎜ π ⎞⎟ = 0 ⋅ 3 + 1 ⋅ 1 = 1 2 2 2 ⎝2⎠ ⎝2⎠ ⎝6⎠ ⎝6⎠ 1 Both functional values equal . 2
sin ⎜⎛ π ⎟⎞ cos ⎜⎛ π ⎟⎞ + cos ⎜⎛ π ⎟⎞ sin ⎜⎛ π ⎟⎞ = 1 ⋅ 1 + 0 ⋅ 3 = 1 2 2 2 ⎝2⎠ ⎝2⎠ ⎝3⎠ ⎝3⎠ 1 Both functional values equal to . 2
PS3. sin(90o − 30o ) = sin(60o ) = 3 = cos(30o ) 2 o o o sin(90 − 45 ) = sin(45 ) = 2 = cos(45o ) 2 sin(90o − 120o ) = sin( −30o ) = − 1 = cos(120o ) 2 For each of the given values of θ , the functional values are equal.
PS4. tan ⎛⎜ π − π ⎞⎟ = tan ⎛⎜ π ⎞⎟ = cot ⎛⎜ π ⎞⎟ ⎝2 6⎠ ⎝3⎠ ⎝6⎠ ⎛ ⎞ ⎛ ⎞ π π π tan ⎜ − ⎟ = tan ⎜ ⎟ = cot ⎛⎜ π ⎞⎟ ⎝2 4⎠ ⎝4⎠ ⎝4⎠ ⎛ ⎞ ⎛ ⎞ ⎛ 4π ⎞ π π π 4 5 tan ⎜ − ⎟ = tan ⎜ − ⎟ = cot ⎜ ⎟ 3 ⎠ ⎝2 ⎝ 6 ⎠ ⎝ 3 ⎠ For each of the given values of θ , the functional values are equal.
PS5. tan ⎛⎜ π − π ⎞⎟ = tan ⎛⎜ π ⎞⎟ = 3 ⎝3 6⎠ ⎝6⎠ 3 ⎛ ⎞ ⎛ ⎞ π π 3 3− 3 2 3 tan ⎜ ⎟ − tan ⎜ ⎟ 3− 3 ⎝3⎠ ⎝6⎠ = 3 = 3 3 = 3 = 3 ⎛ ⎞ ⎛ ⎞ + 1 1 2 3 π π 3 1 + tan ⎜ ⎟ tan ⎜ ⎟ 1 + 3 ⋅ ⎝3⎠ ⎝6⎠ 3
PS6. For k is any integer, the value of (2k + 1) will result in odd integers. Thus sin[(2k + 1)π ] will be 0.
Both functional values equal
3. 3
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.2
369
Section 6.2 1.
sin(45° + 30°) = sin 45° cos30° + cos 45° sin 30°
2.
sin(330° + 45°) = sin 330° cos 45° + cos330° sin 45°
2 3 2 1 = ⋅ + ⋅ 2 2 2 2 6 2 6+ 2 = + = 4 4 4 3.
cos(45° − 30°) = cos 45° cos30° + sin 45° sin 30°
1 2 3 2 =− ⋅ + ⋅ 2 2 2 2 2 6 − 2+ 6 =− + = 4 4 4 4.
cos(120° − 45°) = cos120° cos 45° + sin120° sin 45°
2 3 2 1 ⋅ + ⋅ 2 2 2 2 6 2 6+ 2 = + = 4 4 4
1 2 3 2 =− ⋅ + ⋅ 2 2 2 2 2 6 − 2+ 6 =− + = 4 4 4
=
5.
7.
tan 45° − tan 30° 1 + tan 45° tan 30° 3 3− 3 1− 3− 3 3 = = 3 = = 2− 3 ⎛ 3 ⎞ 3+ 3 3+ 3 1 + 1⎜⎜ ⎟⎟ 3 ⎝ 3 ⎠
tan(45° − 30°) =
5π π 5π π ⎛ 5π π ⎞ − ⎟ = sin cos − cos sin sin ⎜ 4 6 4 6 4 6 ⎝ ⎠ 2 3 ⎛ 2⎞ 1 =− ⋅ − ⎜− ⎟⋅ 2 2 ⎜⎝ 2 ⎟⎠ 2 =−
9.
6.
=
8.
6 2 − 6+ 2 + = 4 4 4
3π π 3π π ⎛ 3π π ⎞ + ⎟ = cos cos − sin sin cos ⎜ 4 6 4 6 ⎝ 4 6⎠
10.
11.
6
4
4π π 4π π ⎛ 4π π ⎞ + ⎟ = sin sin ⎜ cos + cos sin 3 4 3 4 3 4 ⎝ ⎠ =−
3 2 ⎛ 1⎞ 2 ⋅ + ⎜− ⎟⋅ 2 2 ⎝ 2⎠ 2
=−
6 2 6+ 2 − =− 4 4 4
π π π π ⎛π π ⎞ cos ⎜ − ⎟ = cos cos + sin sin 4 3 4 3 ⎝4 3⎠ =
12.
3 +1 3 +3 = 3 = 3 1 − 3 ⋅ 1 3− 3 3 3
( 3 − 1)2 4 − 2 3 = = 2− 3 3 −1 2
2 1 2 3 ⋅ + ⋅ 2 2 2 2 2 6 2+ 6 = + = 4 4 4
2 3 2 1 ⋅ − ⋅ 2 2 2 2 6 2 6+ 2 =− − =− 4 4 4 =−
π π ⎛ π π ⎞ tan 6 + tan 4 tan ⎜ + ⎟ = ⎝ 6 4 ⎠ 1 − tan π tan π
tan 240° − tan 45° 1 + tan 240° tan 45° 3 −1 3 −1 3 −1 = = = 1 + ( 3)(1) 1 + 3 3 +1
tan(240° − 45°) =
11π π tan − tan ⎛ 11π π ⎞ 6 4 tan ⎜ − ⎟= 4 ⎠ 1 + tan 11π tan π ⎝ 6 6 4 =
=
3 +3 3+ 3 9+6 3 +3 ⋅ = 9−3 3− 3 3+ 3
=
=
12 + 6 3 = 2+ 3 6
=
− 3 −1 3
1 + ⎛⎜ − 3 ⎞⎟ (1) ⎝ 3 ⎠
=
− 3 −1 3
1− 3 3
=
( −3 − 3)(3 + 3) (3 − 3)(3 + 3)
−9 − 6 3 − 3 −12 − 6 3 = 9−3 6 = −2 − 3
Copyright © Houghton Mifflin Company. All rights reserved.
− 3 −3 3− 3
370
Chapter 6: Trigonometric Identities and Equations
13.
cos 212° cos122° + sin 212° sin122° = cos(212° − 122°) = cos90° = 0
14.
sin167° cos107° − cos167° sin107° = sin(167° − 107°) = sin 60° =
15.
sin
17.
5π π 5π π π 1 ⎛ 5π π ⎞ cos − cos sin = sin ⎜ − ⎟ = sin = 12 4 12 4 6 2 ⎝ 12 4 ⎠
7π π − tan 12 4 = tan ⎛ 7π − π ⎞ = tan π = 3 ⎜ ⎟ 7π π 3 ⎝ 12 4 ⎠ 1 + tan tan 12 4 tan
3 2
16.
18.
cos
π 12
tan
π
cos
π 4
+ tan
− sin
π 12
sin
π
π 1 ⎛π π⎞ = cos ⎜ + ⎟ = cos = 4 3 2 ⎝ 12 4 ⎠
π
3 = tan ⎛ π + π ⎞ = tan π = undefined ⎜ ⎟ π 2 ⎝6 3⎠ 1 − tan tan 6 3 6
π
19.
sin 42o = cos(90o − 42o ) = cos 48o
20.
cos80o = sin(90o − 80o ) = sin10o
21.
tan15o = cot(90o − 15o ) = cot 75o
22.
cot 2o = tan(90o − 2o ) = tan 88o
23.
sec 25o = csc(90o − 25o ) = csc65o
24.
csc84o = sec(90o − 84o ) = sec6o
25.
sin 7 x cos 2 x − cos 7 x sin 2 x = sin(7 x − 2 x) = sin 5 x
26.
sin x cos3x + cos x sin 3 x = sin( x + 3 x) = sin 4 x
27.
cos x cos 2 x + sin x sin 2 x = cos( x − 2 x) = cos(− x) = cos x
28.
cos 4 x cos 2 x − sin 4 x sin 2 x = cos(4 x + 2 x) = cos6 x
29.
sin 7 x cos3 x − cos 7 x sin 3 x = sin(7 x − 3x) = sin 4 x
30.
cos x cos5 x − sin x sin 5 x = cos( x + 5 x) = cos 6 x
31.
cos 4 x cos(−2 x) − sin 4 x sin( −2 x) = cos 4 x cos 2 x + sin 4 x sin 2 x = cos(4 x − 2 x) = cos 2 x
32.
sin(− x)cos3x − cos(− x)sin 3 x = − sin x cos3x − cos x sin 3x = −(sin x cos3x + cos x sin 3 x) = − sin( x + 3 x) = − sin 4 x
33.
x 2x x 2x ⎛ x 2x ⎞ = sin ⎜ + sin cos + cos sin ⎟ = sin x 3 3 3 3 ⎝3 3 ⎠
34.
cos
3x x 3x x x ⎛ 3x x ⎞ cos + sin sin = cos ⎜ − ⎟ = cos 4 4 4 4 2 ⎝ 4 4⎠
35.
tan 3x + tan 4 x = tan(3x + 4 x ) = tan 7 x 1 − tan 3 x tan 4 x
36.
tan 2 x − tan 3 x = tan(2 x − 3x ) = tan(− x) = − tan x 1 + tan 2 x tan 3 x
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.2
37.
4 4 3 tan α = − , sin α = , cos α = − , 3 5 5 15 15 8 tan β = , sin β = − , cos β = − 8 17 17 a. sin(α − β ) = sin α cos β − cosα sin β
b.
c.
39.
371
⎛ 4 ⎞⎛ 8 ⎞ ⎛ 3 ⎞⎛ 15 ⎞ = ⎜ ⎟⎜ − ⎟ − ⎜ − ⎟⎜ − ⎟ ⎝ 5 ⎠⎝ 17 ⎠ ⎝ 5 ⎠⎝ 17 ⎠ 32 45 77 =− − =− 85 85 85 cos(α + β ) = cos α cos β − sin α sin β
c.
24 24 7 , sin α = , cos α = , 7 25 25 8 15 8 sin β = − , cos β = − , tan β = 17 17 15 a. sin(α + β ) = sin α cos β + cosα sin β tan α =
b.
⎛ 3 ⎞ ⎛ 8 ⎞ ⎛ 4 ⎞ ⎛ 15 ⎞ = ⎜− ⎟⎜− ⎟ − ⎜ ⎟⎜− ⎟ ⎝ 5 ⎠ ⎝ 17 ⎠ ⎝ 5 ⎠ ⎝ 17 ⎠ 24 60 84 = + = 85 85 85 tan α − tan β tan(α − β ) = 1 + tan α tan β
c.
⎛ 24 ⎞⎛ 15 ⎞ ⎛ 7 ⎞⎛ 8 ⎞ = ⎜ ⎟⎜ − ⎟ + ⎜ ⎟⎜ − ⎟ ⎝ 25 ⎠⎝ 17 ⎠ ⎝ 25 ⎠⎝ 17 ⎠ 360 56 416 =− − =− 425 425 425 cos(α + β ) = cos α cos β − sin α sin β
⎛ 7 ⎞ ⎛ 15 ⎞ ⎛ 24 ⎞ ⎛ 8 ⎞ = ⎜ ⎟⎜− ⎟ − ⎜ ⎟⎜− ⎟ ⎝ 25 ⎠ ⎝ 17 ⎠ ⎝ 25 ⎠ ⎝ 17 ⎠ 105 192 87 =− + = 425 425 425 tan α − tan β tan(α − β ) = 1 + tan α tan β
=
− 4 − 15 24 8 = 3 8 ⋅ 1 − 60 24 1 + − 4 15 24 3 8
=
24 − 8 24 − 8 105 7 15 = 7 15 ⋅ 192 8 24 105 1+ 1+ 105 7 15
=
−32 − 45 77 = 24 − 60 36
=
360 − 56 304 = 105 + 192 297
− 4 − 15 3
( )( )
3 4 3 sin α = , cos α = , tan α = , 5 5 4 5 12 12 cos β = − , sin β = , tan β = − 13 13 5 a. sin(α − β ) = sin α cos β − cos α sin β
b.
38.
3 ⎛ 5 ⎞ 4 ⎛ 12 ⎞ = ⎜− ⎟ − ⎜ ⎟ 5 ⎝ 13 ⎠ 5 ⎝ 13 ⎠ 15 48 =− − 65 65 63 =− 65 cos(α + β ) = cos α cos β − sin α sin β 4 ⎛ 5 ⎞ 3 ⎛ 12 ⎞ = ⎜− ⎟ − ⎜ ⎟ 5 ⎝ 13 ⎠ 5 ⎝ 13 ⎠ 20 36 =− − 65 65 56 =− 65 tan α − tan β tan(α − β ) = 1 + tan α tan β
( ) ( )( )
3 − − 12 5 = 4 3 1+ − 12 4 5 3 + 12 20 = 4 5 ⋅ 36 20 1− 20
=
15 + 48 63 =− 20 − 36 16
40.
( )( )
24 7 24 , cos α = − , tan α = − , 25 25 7 4 3 3 cos β = − , sin β = − , tan β = 5 5 4 a. cos( β − α ) = cos β cos α + sin β sin α sin α =
b.
= − 4 ⎛⎜ − 7 ⎞⎟ + ⎛⎜ − 3 ⎞⎟ 24 5 ⎝ 25 ⎠ ⎝ 5 ⎠ 25 28 = − 72 = − 44 125 125 125 sin(α + β ) = sin α cos β + cos α sin β 24 ⎛ 4 ⎞ ⎛ 7 ⎞⎛ 3 ⎞ ⎜ − ⎟ + ⎜ − ⎟⎜ − ⎟ 25 ⎝ 5 ⎠ ⎝ 25 ⎠⎝ 5 ⎠ 96 21 =− + 125 125 75 3 =− =− 125 5 tan α + tan β tan(α + β ) = 1 − tan α tan β =
c.
=
=
− 24 + 3 7
4
( 7 )( 34 )
1 − − 24
− 24 + 3 28 7 4⋅ 1 + 72 28 28
−96 + 21 = 28 + 72 75 3 =− =− 100 4
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372
41.
Chapter 6: Trigonometric Identities and Equations
4 3 4 sin α = − , cos α = − , tan α = , 5 5 3 12 5 5 cos β = − , sin β = , tan β = − 13 13 12 a. sin(α − β ) = sin α cos β − cos α sin β
b.
c.
42.
12 ⎞ ⎛ 3 ⎞ 5 = ⎛⎜ − 4 ⎞⎛ ⎟⎜ − ⎟ − ⎜ − ⎟ ⎝ 5 ⎠⎝ 13 ⎠ ⎝ 5 ⎠ 13 = 48 + 15 = 63 65 65 65 cos(α + β ) = cos α cos β − sin α sin β
7 24 7 , cos α = , tan α = − , 25 25 24 8 15 15 cos β = , sin β = − , tan β = − 17 17 8 a. sin(α + β ) = sin α cos β + cos α sin β sin α = −
b.
12 ⎞ ⎛ 4 ⎞ 5 = ⎛⎜ − 3 ⎞⎛ ⎟⎜ − ⎟ − ⎜ − ⎟ ⎝ 5 ⎠⎝ 13 ⎠ ⎝ 5 ⎠ 13 = 36 + 20 = 56 65 65 65 tan α + tan β tan(α + β ) = 1 − tan α tan β
⎛ 7 ⎞ ⎛ 8 ⎞ 24 ⎛ 15 ⎞ = ⎜− ⎟⎜ ⎟ + ⎜− ⎟ ⎝ 25 ⎠ ⎝ 17 ⎠ 25 ⎝ 17 ⎠ 56 360 416 =− − =− 425 425 425 cos(α − β ) = cos α cos β + sin α sin β 24 ⎛ 8 ⎞ ⎛ 7 ⎞ ⎛ 15 ⎞ ⎜ ⎟+⎜− ⎟⎜− ⎟ 25 ⎝ 17 ⎠ ⎝ 25 ⎠ ⎝ 17 ⎠ 192 105 297 = + = 425 425 425 tan α + tan β tan(α + β ) = 1 − tan α tan β =
c.
( ) ( )( )
4+ − 5 12 = 3 4 −5 1− 3 12 4− 5 4− 5 = 3 12 = 3 12 ⋅ 36 1 + 29 1 + 20 36 36 36
= 48 − 15 = 33 36 + 20 56
43.
15 8 8 , sin α = , tan α = , 17 17 15 3 4 3 sin β = − , cos β = − , tan β = 5 5 4 a. sin(α + β ) = sin α cos β + cos α sin β cos α =
8 ⎛ 4 ⎞ 15 ⎛ 3 ⎞ ⎜− ⎟ + ⎜− ⎟ 17 ⎝ 5 ⎠ 17 ⎝ 5 ⎠ 32 45 77 =− − =− 85 85 85 cos(α − β ) = cos α cos β + sin α sin β
=
44.
c.
= 15 ⎛⎜ − 4 ⎞⎟ + 8 ⎛⎜ − 3 ⎞⎟ 17 ⎝ 5 ⎠ 17 ⎝ 5 ⎠ = − 60 − 24 = − 84 85 85 85 tan α − tan β tan(α − β ) = 1 + tan α tan β 8 −3 8 −3 8 −3 60 = 15 4 = 15 4 = 15 4 ⋅ 1 + 24 1 + 24 60 1+ 8 3 60 60 15 4 32 − 45 13 = =− 60 + 24 84
()
24
416 −56 − 360 =− 192 − 105 87
7 24 24 , sin α = , tan α = − , 25 25 7 12 5 12 sin β = − , cos β = , tan β = − 13 13 5 a. sin(α + β ) = sin α cos β + cos α sin β cos α = −
=
b.
( ) ( )( )
− 7 + − 15
8 1 − − 7 − 15 24 8 7 15 − − − 7 − 15 192 = 24 8 = 24 8 ⋅ 1 − 105 1 − 105 192 192 192
=
24 ⎛ 5 ⎞ ⎛ 7 ⎞ ⎛ 12 ⎞ ⎜ ⎟ + ⎜− ⎟⎜− ⎟ 25 ⎝ 12 ⎠ ⎝ 25 ⎠ ⎝ 13 ⎠ 120 84 204 = + = 325 325 325 cos(α + β ) = cos α cos β − sin α sin β =
b.
c.
⎛ 7 ⎞ 5 24 ⎛ 12 ⎞ = ⎜− ⎟ − ⎜− ⎟ ⎝ 25 ⎠ 13 25 ⎝ 13 ⎠ 35 288 253 =− + = 325 325 325 tanα − tan β tan(α − β ) = 1+ tanα tan β =
( 5 ) = − 247 + 125 = − 247 + 125 ⋅ 35 1+ 288 35 1+ ( − 24 )( −12 ) 1+ 288 35 35 7 5 − 24 − −12 7
= −120 + 84 = − 36 35 + 288 323
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Section 6.2
45.
373
3 4 4 cos α = − , sin α = − , tan α = , 5 5 3 5 12 5 sin β = , cos β = , tan β = 13 13 12 a. sin(α − β ) = sin α cos β − cos α sin β
b.
c.
46.
5⎞ = ⎛⎜ − 4 ⎞⎟ 12 − ⎛⎜ − 3 ⎞⎛ ⎟⎜ ⎟ ⎝ 5 ⎠ 13 ⎝ 5 ⎠⎝ 13 ⎠ = − 48 + 15 = − 33 65 65 65 cos(α + β ) = cos α cos β − sin α sin β
8 15 15 , sin α = − , tan α = − , 17 17 8 24 7 24 sin β = − , cos β = − , tan β = 25 25 7 a. sin(α − β ) = sin α cos β − cos α sin β cos α =
b.
= ⎛⎜ − 3 ⎞⎟ 12 − ⎛⎜ − 4 ⎞⎟ 5 ⎝ 5 ⎠ 13 ⎝ 5 ⎠ 13 = − 36 + 20 = − 16 65 65 65 tan α + tan β tan(α + β ) = 1 − tan α tan β
⎛ 15 ⎞⎛ 7 ⎞ 8 ⎛ 24 ⎞ = ⎜ − ⎟⎜ − ⎟ − ⎜ − ⎟ ⎝ 17 ⎠⎝ 25 ⎠ 17 ⎝ 25 ⎠ 105 192 297 = + = 425 425 425 cos(α + β ) = cosα cos β − sin α sin β 8 ⎛ 7 ⎞ ⎛ 15 ⎞⎛ 24 ⎞ ⎜ − ⎟ − ⎜ − ⎟⎜ − ⎟ 17 ⎝ 25 ⎠ ⎝ 17 ⎠⎝ 25 ⎠ 56 360 416 =− − =− 425 425 425 tan α + tan β tan(α + β ) = 1 − tan α tan β =
c.
4+ 5 4+ 5 = 3 12 = 3 12 1 − 20 1− 4 5 36 3 12 4+ 5 = 3 12 ⋅ 36 = 48 + 15 = 63 1 − 20 36 36 − 20 16 36
( )
47.
3 4 3 sin α = , cos α = , tan α = , 5 5 4 5 5 12 tan β = , sin β = − , cos β = − 12 13 13 a. sin(α + β ) = sin α cos β + cos α sin β
b.
c.
49.
=
=
36 − 20 16 = 48 + 15 63
48.
15 15 8 , sin α = , cos α = , 8 17 17 7 7 24 tan β = − , sin β = − , cos β = 24 25 25 a. sin(α − β ) = sin α cos β − cos α sin β tan α =
15 ⎛ 24 ⎞ 8 ⎛ 7 ⎞ ⎜ ⎟ − ⎜− ⎟ 17 ⎝ 25 ⎠ 17 ⎝ 25 ⎠ 360 56 416 = + = 425 425 425 cos(α − β ) = cos α cos β + sin α sin β =
b.
c.
8 ⎛ 24 ⎞ 15 ⎛ 7 ⎞ = ⎜ ⎟ + ⎜− ⎟ 7 ⎝ 25 ⎠ 17 ⎝ 25 ⎠ 192 105 87 = − = 425 425 425 tan α + tan β tan(α + β ) = 1 − tan α tan β
( ) ( )
15 + − 7 15 − 7 192 24 = 8 = 8 24 ⋅ 105 192 15 7 + 1 − 1− 192 8 24 360 − 56 304 = = 192 + 105 297
( )( )
π π ⎛π ⎞ cos ⎜ − θ ⎟ = cos cosθ + sin sin θ 2 2 ⎝2 ⎠ = 0 ⋅ cosθ + 1 ⋅ sin θ = sin θ
( )( )
−105 + 192 56 + 360 87 = 416
⎛ 4 ⎞ ⎛ 12 ⎞ ⎛ 3 ⎞ ⎛ 5 ⎞ = ⎜ ⎟⎜− ⎟ + ⎜ ⎟⎜− ⎟ ⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠ 48 15 63 =− − =− 65 65 65 tan α − tan β tan(α − β ) = 1 + tan α tan β 3− 5 3− 5 48 4 12 = 4 12 ⋅ 1 + 15 48 1+ 3 5 48 4 12
− 15 + 24 56 8 7 7 ⋅ = 8 1 + 360 56 1 − − 15 24 56 8 7
=
⎛ 3 ⎞ ⎛ 12 ⎞ ⎛ 4 ⎞ ⎛ 5 ⎞ = ⎜ ⎟⎜− ⎟ + ⎜ ⎟⎜− ⎟ ⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠ 36 20 56 =− − =− 65 65 65 cos(α − β ) = cos α cos β + sin α sin β
=
− 15 + 24
50.
cos(θ + π ) = cosθ cos π − sin θ sin π = cosθ (−1) − sin θ (0) = − cosθ
Copyright © Houghton Mifflin Company. All rights reserved.
374
51.
Chapter 6: Trigonometric Identities and Equations
sin ⎛⎜ θ + π ⎞⎟ = sin θ cos π + cos θ sin π 2⎠ 2 2 ⎝ = sin θ (0) + cos θ (1)
52.
sin(θ + π ) = sin θ cos π + cosθ sin π = sin θ (−1) + cosθ (0) = − sin θ
54.
tan 2θ = tan(θ + θ )
= cos θ
53.
π π ⎞ tan θ + tan 4 ⎛ tan ⎜θ + ⎟ = 4 ⎠ 1 − tan θ tan π ⎝
tan θ + tan θ 1 − tan θ tan θ 2 tan θ = 1 − tan 2 θ
4
=
tan θ + 1 = 1 − tan θ
55.
3π 3π ⎛ 3π ⎞ cos ⎜ − θ ⎟ = cos cosθ + sin sin θ 2 2 2 ⎝ ⎠ = 0(cosθ ) + (−1)sin θ
56.
3π 3π ⎛ 3π ⎞ sin ⎜ + θ ⎟ = sin cosθ + cos sin θ 2 2 2 ⎝ ⎠ = ( −1) cosθ + (0)sin θ
= − sin θ 57.
= − cosθ
⎛π ⎞ cos (π / 2 − θ ) cot ⎜ − θ ⎟ = ⎝2 ⎠ sin (π / 2 − θ ) =
58.
cos(π + θ ) sin(π + θ ) cos π cosθ − sin π sin θ = sin π cosθ + cos π sin θ (−1)cosθ − (0)sin θ = (0)cosθ + ( −1)sin θ − cosθ = − sin θ = cot θ
cot(π + θ ) =
( cos π2 ) cosθ + (sin π2 ) sinθ (sin π2 ) cosθ − ( cos π2 ) sinθ
(0) cosθ + (1)sin θ (1)cosθ − (0)sin θ sin θ = cosθ = tan θ =
59.
csc(π − θ ) =
1 sin(π − θ )
60.
1 sin π cosθ − cos π sin θ 1 = (0)cosθ − (−1)sin θ 1 = sin θ = cscθ =
61.
sin 6 x cos 2 x − cos 6 x sin 2 x = sin(6 x − 2 x) = sin 4 x = sin(2 x + 2 x) = sin 2 x cos 2 x + cos 2 x sin 2 x = 2sin 2 x cos 2 x
1 sec ⎛⎜ π − θ ⎞⎟ = ⎝2 ⎠ cos π − θ 2
( )
1 cos π cos θ + sin π sin θ 2 2 1 = (0) cos θ + (1)sin θ = 1 sin θ = csc θ =
62.
cos5 x cos3 x + sin 5 x sin 3 x = cos(5 x − 3 x) = cos 2 x = cos( x + x) = cos x cos x − sin x sin x
63.
cos(α + β ) + cos(α − β ) = cosα cos β − sin α sin β + cosα cos β + sin α sin β = 2cosα cos β
64.
cos(α − β ) − cos(α + β ) = cosα cos β + sin α sin β − cosα cos β + sin α sin β = 2sin α sin β
65.
sin(α + β ) + sin(α − β ) = sin α cos β + cosα sin β + sin α cos β − cosα sin β = 2sin α cos β
Copyright © Houghton Mifflin Company. All rights reserved.
= cos 2 x − sin 2 x
Section 6.2
66.
67.
69.
71.
375
sin(α − β ) − sin(α + β ) = sin α cos β − cosα sin β − sin α cos β − cosα sin β = −2cos α sin β cos(α − β ) cosα cos β + sin α sin β = sin(α + β ) sin α cos β + cosα sin β cos α cos β sin α sin β + sin α cos β sin α cos β = sin α cos β cosα sin β + sin α cos β sin α cos β cot α + tan β = 1 + cot α tan β sin( x + h) − sin x sin x cos h + cos x sin h − sin x = h h sin x(cos h − 1) cos x sin h = + h h cos h − 1 sin h = sin x + cos x h h
⎛π ⎞ ⎡π ⎤ sin ⎜ + α − β ⎟ = sin ⎢ + (α − β ) ⎥ ⎝2 ⎠ ⎣2 ⎦
π
70.
72.
π
sin(α + β ) sin α cos β + cosα sin β = sin(α − β ) sin α cos β − cosα sin β sin α cos β cosα sin β + sin α cos β sin α cos β = sin α cos β cosα sin β − sin α cos β sin α cos β 1 + cot α tan β = 1 − cot α tan β cos( x + h) − cos x cos x cos h − sin x sin h − cos x = h h cos x(cos h − 1) sin x sin h = − h h cos h − 1 sin h = cos x − sin x h h
⎛π ⎞ ⎡π ⎤ cos ⎜ + α + β ⎟ = cos ⎢ + (α + β ) ⎥ ⎝2 ⎠ ⎣2 ⎦
π
= cos cos(α + β ) − sin sin(α + β ) 2 2 = (0)cos(α + β ) − (1)sin(α + β )
= cos(α − β )
= − sin(α + β )
= cosα cos β + sin α sin β
= −(sin α cos β + cosα sin β )
sin 3 x = sin(2 x + x) = sin 2 x cos x + cos 2 x sin x = sin( x + x)cos x + cos( x + x)sin x = (sin x cos x + cos x sin x)cos x + (cos x cos x − sin x sin x)sin x = 2sin x cos 2 x + sin x cos 2 x − sin 3 x = 3sin x cos 2 x − sin 3 x = 3sin x(1 − sin 2 x) − sin 3 x = 3sin x − 3sin 3 x − sin 3 x = 3sin x − 4sin 3 x
74.
π
cos(α − β ) + cos sin(α − β ) 2 2 = (1) cos(α − β ) + (0)sin(α − β )
= sin
73.
68.
cos3 x = cos(2 x + x) = cos 2 x cos x − sin 2 x sin x = cos( x + x)cos x − sin( x + x)sin x = (cos x cos x − sin x sin x) cos x − (sin x cos x + cos x sin x)sin x = cos3 x − cos x sin 2 x − 2cos x sin 2 x = cos3 x − 3cos x sin 2 x = cos3 x − 3cos x(1 − cos 2 x) = cos3 x − 3cos x + 3cos3 x = 4cos3 x − 3cos x
Copyright © Houghton Mifflin Company. All rights reserved.
376
Chapter 6: Trigonometric Identities and Equations
75.
cos(θ + 3π ) = cosθ cos3π − sin θ sin 3π = (cosθ )(−1) − (sin θ )(0) = − cosθ
77.
tan(θ + π ) =
79.
sin(θ + 2kπ ) = sin θ cos(2kπ ) + cosθ sin 2kπ = (sin θ )(1) + (cosθ )(0) = sin θ
80.
We consider two cases, (1) k an odd and (2) k an even integer.
81.
83.
tan θ + tan π 1 − tan θ tan π tan θ + 0 = 1 − (tan θ )(0) = tan θ
(1)
sin(θ − kπ ) = sin θ cos( kπ ) − cos θ sin( kπ ) = (sin θ )(−1) − (cos θ )(0) = − sin θ , provided k is odd
(2)
sin(θ − kπ ) = sin θ cos( kπ ) − cos θ sin( kπ ) = (sin θ )(1) − (cosθ )(0) = sin θ , provided k is even
(
)
y = sin π − x and y = cos x both have the following graph. 2
y = sin 7 x cos 2 x − cos 7 x sin 2 x and y = sin 5 x both have the following graph.
76.
sin(θ + 2π ) = sin θ cos 2π + cosθ sin 2π = (sin θ )(1) + (cosθ )(0) = sin θ
78.
cos[θ + (2k + 1)π ] = cosθ cos[(2k + 1)π ] − sin θ sin[(2k + 1)π ] = (cosθ )(−1) − (sin θ )(0) = − cosθ
82.
y = cos( x + π ) and y = − cos x both have the following graph.
84.
y = sin 3 x and y = 3sin x − 4sin 3 x both have the following graph.
....................................................... 85.
sin( x − y ) ⋅ sin( x + y ) = (sin x cos y − cos x sin y )(sin x cos y + cos x sin y ) = sin 2 x cos 2 y − cos 2 x sin 2 y
86.
sin( x + y + z ) = sin[ x + ( y + z )] = sin x cos( y + z ) + cos x sin( y + z ) = sin x(cos y cos z − sin y sin z ) + cos x (sin y cos z + cos y sin z ) = sin x cos y cos z − sin x sin y sin z + cos x sin y cos z + cos x cos y sin z
Copyright © Houghton Mifflin Company. All rights reserved.
Connecting Concepts
Section 6.3
377
87.
cos( x + y + z ) = cos[ x + ( y + z )] = cos x cos( y + z ) − sin x sin( y + z ) = cos x[cos y cos z − sin y sin z ] − sin x[sin y cos z + cos y sin z ] = cos x cos y cos z − cos x sin y sin z − sin x sin y cos z − sin x cos y sin z
88.
sin( x + y ) sin x cos y + cos x sin y = sin x sin y sin x sin y sin x cos y cos x sin y = + sin x sin y sin x sin y = cot y + cot x
90.
ER =
89.
cos( x − y ) cos x cos y + sin x sin y = cos x sin y cos x sin y cos x cos y sin x sin y = + cos x sin y cos x sin y = cot y + tan x
2sin10° + sin 20° 2sin(10° + 20°) 2sin10° + sin 20° = 2sin 30° ER ≈ 0.69
....................................................... PS1. sin 2α = sin(α + α )
PS2. cos 2α = cos(α + α ) = cos α cos α − sin α sin α
= sin α cos α + cos α sin α = 2sin α cos α
= cos 2 α − sin 2 α 3 3 2 = 2 = 3 = 3 1+ cos(60o ) 1+ 1 3 2 2 ⎛ o⎞ sin(90o ) tan ⎜ 90 ⎟ = tan(45o ) =1 = 1 =1 ⎝ 2 ⎠ 1+ cos(90o ) 1+ 0 3 3 o⎞ ⎛ sin(120o ) tan ⎜ 120 ⎟ = tan(60o ) = 3 2 2 = = = 3 ⎝ 2 ⎠ 1+ cos(120o ) 1− 1 1 2 2 For each of the given values of α , the functional values are equal.
⎛ o⎞ PS4. tan ⎜ 60 ⎟ = tan(30o ) = 3 3 ⎝ 2 ⎠
PS3. tan 2α = tan(α + α )
= tan α + tan α 1 − tan α tan α = 2 tan α 1 − tan 2 α
PS5. Let α = 45o ; then the left side of the equation is 1, and
the right side of the equation is
Prepare for Section 6.3
2.
sin(60o )
PS6. Let α = 60o ; then the left side of the equation is 3 , and the right side of the equation is 1 . 2 4
Section 6.3 1.
4.
2sin 2α cos 2α = sin 2(2α ) = sin 4α
2.
2cos 2 2 β − 1 = cos 2(2 β ) = cos 4 β
5.
2sin 3θ cos3θ = sin 2(3θ ) = sin 6θ
3.
cos 2 3α − sin 2 3α = cos 2(3α ) = cos 6α
6.
1 − 2sin 2 5 β = cos 2(5β ) = cos10 β cos 2 6α − sin 2 6α = cos 2(6α ) = cos12α
Copyright © Houghton Mifflin Company. All rights reserved.
378
7.
9.
Chapter 6: Trigonometric Identities and Equations
2 tan 3α 1 − tan 2 3α
2 tan 4θ
8.
= tan 2(3α )
1 − tan 2 4θ
= tan 6α
= tan 2(4θ ) = tan 8θ
2
4 3 sin α 3/ 5 3 ⎛4⎞ cos α = − , sin α = 1 − ⎜ ⎟ = , tan α = = =− 5 5 cos α −4 / 5 4 ⎝5⎠ cos 2α = cos2 α − sin 2 α
sin 2α = 2sin α cos α
2
⎛ 3⎞⎛ 4 ⎞ = 2⎜ ⎟⎜ − ⎟ ⎝ 5 ⎠⎝ 5 ⎠ 24 =− 25
⎛ 4⎞ ⎛ 3⎞ = ⎜− ⎟ −⎜ ⎟ ⎝ 5⎠ ⎝ 5⎠ 16 9 = − 25 25 7 = 25
tan 2α =
2
=
=
2 tan α 1 − tan 2 α
( 4 ) = − 64 2 1− 9 1 − (− 3 ) 16 4 2 −3
−6
4 ⋅ 16 = −24 1 − 9 16 16 − 9 16
=−
10.
cos α =
2
24 7 sin α −7 / 25 7 ⎛ 24 ⎞ , sin α = − 1 − ⎜ ⎟ = − , tan α = = =− 25 25 cos α 24 / 25 24 ⎝ 25 ⎠ cos 2α = cos2 α − sin 2 α
sin 2α = 2sin α cos α
2
⎛ 7 ⎞⎛ 24 ⎞ = 2 ⎜ − ⎟⎜ ⎟ ⎝ 25 ⎠⎝ 25 ⎠ 336 =− 625
tan 2α =
⎛ 24 ⎞ ⎛ 7 ⎞ = ⎜ ⎟ − ⎜− ⎟ ⎝ 25 ⎠ ⎝ 25 ⎠ 576 49 = − 625 625 527 = 625
2
=
=
2 tan α 1 − tan 2 α
( 24 ) 2 1 − (− 7 ) 24 2 − 7
−7
12 ⋅ 576 1 − 49 576 576
=−
11.
sin α =
24 7
336 336 =− 576 − 49 527
2
8 15 sin α 8 /17 8 ⎛ 8⎞ , cos α = − 1 − ⎜ ⎟ = − , tan α = = =− 17 17 cos α −15/17 15 ⎝ 17 ⎠
sin 2α = 2sin α cos α ⎛ 8 ⎞ ⎛ 15 ⎞ = 2⎜ ⎟⎜ − ⎟ ⎝ 17 ⎠ ⎝ 17 ⎠ 240 =− 289
cos 2α = cos2 α − sin 2 α 2
⎛ 15 ⎞ ⎛ 8⎞ = ⎜− ⎟ − ⎜ ⎟ ⎝ 17 ⎠ ⎝ 17 ⎠ 225 64 = − 289 289 161 = 289
tan 2α = 2
=
=
2 tan α 1 − tan 2 α
( 15 ) = − 1615 2 1 − 64 1 − (− 8 ) 225 15 2 −8
− 16
15 ⋅ 225 = −240 1 − 64 225 225 − 64 225
=−
240 161
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.3
12.
379
sin α = −
2
9 40 sin α −9 / 41 9 ⎛ 9⎞ , cos α = − 1 − ⎜ − ⎟ = − , tan α = = = 41 41 cos α −40 / 41 40 ⎝ 41 ⎠ cos 2α = cos2 α − sin 2 α
sin 2α = 2sin α cos α
2
⎛ 9 ⎞ ⎛ 40 ⎞ = 2⎜ − ⎟⎜ − ⎟ ⎝ 41 ⎠ ⎝ 41 ⎠ 720 = 1681
13.
tan α = −
tan 2α =
⎛ 40 ⎞ ⎛ 9⎞ = ⎜− ⎟ − ⎜− ⎟ ⎝ 41 ⎠ ⎝ 41 ⎠ 1600 81 = − 1681 1681 1519 = 1681
2
2 tan α 1 − tan 2 α
=
( 40 ) = 209 2 1 − 81 1− ( 9 ) 1600 40
=
9 1600 720 20 ⋅ = 1 − 81 1600 1600 − 81 1600
=
720 1519
2 9
24 24 7 , r = 242 + 72 = 576 + 49 = 625 = 25, sin α = − , cos α = 7 25 25
cos 2α = cos2 α − sin 2 α
sin 2α = 2sin α cos α
2
⎛ 24 ⎞ ⎛ 7 ⎞ = 2⎜ − ⎟⎜ ⎟ ⎝ 25 ⎠ ⎝ 25 ⎠ 336 =− 625
⎛ 7 ⎞ ⎛ 24 ⎞ = ⎜ ⎟ −⎜− ⎟ ⎝ 25 ⎠ ⎝ 25 ⎠ 49 576 = − 625 625 527 =− 625
tan 2α = 2
=
=
2 tan α 1 − tan 2 α
( 7) 2 1 − ( − 24 ) 7 2 − 24
− 48
7 ⋅ 49 1 − 576 49 49
336 −336 = = 49 − 576 527 14.
4 4 3 tan α = , r = 42 + 32 = 16 + 9 = 25 = 5, sin α = , cos α = 3 5 5 cos 2α = cos2 α − sin 2 α
sin 2α = 2sin α cos α
2
⎛ 4 ⎞⎛ 3⎞ = 2⎜ ⎟⎜ ⎟ ⎝ 5 ⎠⎝ 5⎠ 24 = 25
⎛ 3⎞ ⎛4⎞ =⎜ ⎟ −⎜ ⎟ ⎝ 5⎠ ⎝5⎠ 9 16 = − 25 25 7 =− 25
2
tan 2α = 2 tan α 1 − tan 2 α 8 2 4 3 = = 3 2 1 − 16 1− 4 9
() (3)
=
8 3 ⋅ 9 = 24 1 − 16 9 9 − 16 9
= − 24 7 15.
2
15 225 289 − 225 64 8 15/17 15 ⎛ 15 ⎞ , cos α = 1 − ⎜ ⎟ = 1 − = = = , tan α = = 17 289 289 289 17 8 /17 8 ⎝ 17 ⎠ 2 tan α tan 2α = sin 2α = 2sin α cos α cos 2α = cos2 α − sin 2 α 1 − tan 2 α 2 2 ⎛ 15 ⎞ ⎛ 8 ⎞ 8 15 ⎛ ⎞ ⎛ ⎞ = 2⎜ ⎟⎜ ⎟ 15 =⎜ ⎟ −⎜ ⎟ 2 15 ⎝ 17 ⎠ ⎝ 17 ⎠ 8 ⎝ 17 ⎠ ⎝ 17 ⎠ = = 4 2 1 − 225 240 64 225 1 − 15 64 = = − 8 289 289 289 15 64 240 161 = 4 ⋅ = =− 225 64 64 − 225 1− 289 sin α =
( ) ( ) 64
240 =− 161
Copyright © Houghton Mifflin Company. All rights reserved.
380
16.
Chapter 6: Trigonometric Identities and Equations 2 −3 3 3 9 4 ⎛ 3⎞ sin α = − , cos α = − 1 − ⎜ − ⎟ = − 1 − = − , tan α = 5 = 5 25 5 ⎝ 5⎠ −4 4 5
2
sin 2α = 2sin α cos α
2
cos 2α = cos α − sin α 2
⎛ 3⎞⎛ 4 ⎞ = 2⎜ − ⎟⎜ − ⎟ ⎝ 5⎠⎝ 5 ⎠ 24 = 25
⎛ 4⎞ ⎛ 3⎞ = ⎜− ⎟ − ⎜− ⎟ 5 ⎝ ⎠ ⎝ 5⎠ 16 9 = − 25 25 7 = 25
2
tan 2α = 2 tan α 1 − tan 2 α 3 2 3 4 = = 2 2 1− 9 1− 3 16
() (4)
=
3 2
⋅ 16 = 24 1 − 9 16 16 − 9
= 24 7 17.
cos α =
2
40 1600 9 −9 / 41 9 ⎛ 40 ⎞ , sin α = − 1 − ⎜ ⎟ = − 1 − = − , tan α = =− 41 1681 41 40 / 41 40 ⎝ 41 ⎠ cos 2α = cos2 α − sin 2 α
sin 2α = 2sin α cos α
2
⎛ 9 ⎞ ⎛ 40 ⎞ = 2⎜ − ⎟⎜ ⎟ ⎝ 41 ⎠ ⎝ 41 ⎠ 720 =− 1681
⎛ 40 ⎞ ⎛ 9⎞ = ⎜ ⎟ − ⎜− ⎟ ⎝ 41 ⎠ ⎝ 41 ⎠ 1600 81 = − 1681 1681 1519 = 1681
2
tan 2α = 2 tan α 1 − tan 2 α 2 − 9 − 9 40 20 = = 2 1 − 81 9 1− − 1600 40
( ) ( )
=
− 9
20
1 − 81
1600
=
18.
16
−720 = − 720 1600 − 81 1519
2
4 16 3 −3/ 5 3 ⎛4⎞ cos α = , sin α = − 1 − ⎜ ⎟ = − 1 − = − , tan α = =− 5 5 25 5 4 / 5 4 ⎝ ⎠ cos 2α = cos2 α − sin 2 α
sin 2α = 2sin α cos α
2
⎛ 3⎞⎛ 4 ⎞ = 2⎜ − ⎟⎜ ⎟ ⎝ 5⎠⎝ 5 ⎠ 24 =− 25
⎛4⎞ ⎛ 3⎞ = ⎜ ⎟ −⎜− ⎟ 5 ⎝ ⎠ ⎝ 5⎠ 16 9 = − 25 25 7 = 25
2
tan 2α = 2 tan α 1 − tan 2 α 2 −3 −3 4 2 = = 2 1− 9 3 1− − 16 4
( ) ( )
=
−3
2 ⋅ 16 = −24 1 − 9 16 16 − 9 16
= − 24 7 19.
⋅ 1600 1600
(
)
6cos2 x = 6 1 + cos 2 x = 3 (1 + cos 2 x ) 2
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.3
20.
381
(
)(
)
2
2
sin 4 x cos4 x = 1 − cos 2 x 1 + cos 2 x 2 2 1 = (1 − cos 2 x + cos2 2 x )(1 + cos 2 x + cos2 2 x ) 16 = 1 (1 − 2cos2 2 x + cos4 2 x ) 16 2⎞ ⎛ = 1 ⎜ 1 − 2 1 + cos 4 x + 1 + cos 4 x ⎟ 16 ⎝ 2 2 ⎠
(
) (
)
(
)
= 1 ⎛⎜ 1 − 1 − cos 4 x + 1 + 1 cos 4 x + 1 1 + cos8 x ⎞⎟ 16 ⎝ 4 2 4 2 ⎠ 1 3 1 1 = − cos 4 x + cos8 x 16 8 2 8 = 1 3 − cos 4 x + 1 cos8 x 32 4 4
( (
21.
(
)
)
2
cos4 x = 1 + cos 2 x 2 1 = (1 + 2cos 2 x + cos2 2 x ) 4 = 1 1 + 2cos 2 x + 1 + cos 4 x 4 2 1 = ( 3 + 4 cos 2 x + cos 4 x ) 8
(
23.
(
)(
22.
)
)
2
150° 2 1 − cos150° =+ 2
sin 75° = sin
)
(
)(
24.
)
26.
)
2
sin 2 x cos4 x = 1 − cos 2 x 1 + cos 2 x 2 2 2 1 = (1 − cos 2 x ) (1 + cos 2 x ) 8 = 1 1 − 1 + cos 4 x (1 + cos 2 x ) 8 2 1 1 1 = − cos 4 x (1 + cos 2 x ) 8 2 2 = 1 (1 + cos 2 x − cos 4 x − cos 2 x cos 4 x ) 16
( (
sin 4 x cos2 x = 1 − cos 2 x 1 + cos 2 x 2 2 = 1 (1 − cos2 2 x ) (1 − cos 2 x ) 8 = 1 1 − 1 + cos 4 x (1 − cos 2 x ) 8 2 1 1 1 = − cos 4 x (1 − cos 2 x ) 8 2 2 = 1 (1 − cos 2 x − cos 4 x + cos 4 x cos 2 x ) 16
( (
25.
)
(
)
)
)
3
sin 6 x = 1 − cos 2 x 2 = 1 (1 − 3cos 2 x + 3cos2 2 x − cos3 2 x ) 8 = 1 ⎛⎜ 1 − 3cos 2 x + 3 + 3 cos 4 x − cos 2 x 1 + cos 4 x ⎞⎟ 8⎝ 2 2 2 ⎠ 1 = ( 5 − 7cos 2 x + 3cos 4 x − cos 2 x cos 4 x ) 16
(
210° 2 1 + cos 210° =− 2
cos105° = cos
27.
135° 2 1 − cos135° = sin135° 1 − (− 2 / 2) = 2/2
tan 67.5° = tan
=
1 − (− 3 / 2) 2
=−
1 + (− 3 / 2) 2
=
2+ 3 4
=−
2− 3 4
=
=
2+ 3 2
=−
2− 3 2
=
Copyright © Houghton Mifflin Company. All rights reserved.
2+ 2 2 ⋅ 2 2
2 2+2 2 = 2 +1
)
382
28.
31.
34.
Chapter 6: Trigonometric Identities and Equations
330° 2 1 + cos330° =− 2
cos165° = cos
=−
1 + ( 2 / 2) 2
=
1 − ( − 2 / 2) 2
=−
2+ 3 2
=−
2+ 2 4
=
2+ 2 4
=−
2+ 2 2
=
2+ 2 2
45° 2 1 − cos 45° =+ 2
sin
α 2
135° 2 1 + cos135° =+ 2
cos 67.5° = cos
32.
=
1− 2 / 2 2
=
=
2− 2 4
=
=
2− 2 2
=
5π 5π / 4 = cos 8 2 1 + cos5π / 4 =− 2
sin α =
35.
cos
33.
sin
1 + (− 2 / 2)
1− 2 / 2 2
2− 2 4
=
2− 2 4
2− 2 2
=
2− 2 2
36.
sin
=
2− 3 4
=
2+ 2 4
=
2− 3 2
=
2+ 2 2
2− 2 4
=−
2− 2 2 2
5 25 12 ⎛5⎞ =− . , cosα = − 1 − ⎜ ⎟ = − 1 − 13 169 13 ⎝ 13 ⎠
=
=
1 + cos α 2
1 − (−12 /13) 2
=
1 − 12 /13 2
=
=
13 + 12 26
=
13 − 12 26
=
=
25 5 = 26 26
=
1 1 = 26 26
=
26 26
5 26 26
α 2
(
=
=−
cos
3π 3π / 4 = sin 8 2 1 − cos3π / 4 =+ 2
1− 3 / 2 2
=
1 − cosα 2
7π 7π / 4 = sin 8 2 1 − cos 7π / 4 =+ 2 =
2/2
5π 5π / 6 = cos 12 2 1 + cos5π / 6 =+ 2
1− 2 / 2 2
=
=
225° 2 1 − cos 225° =+ 2
sin112.5° = sin
1+ 3 /2 2
=−
37.
30.
=−
sin 22.5° = sin
cos
315° 2 1 + cos315° =− 2
cos157.5° = cos
29.
tan
α 2
=
1 − cos α sin α 1 + 12
13 5 13
13 + 12 5 =5
Copyright © Houghton Mifflin Company. All rights reserved.
1− − 2 / 2 2
)
Section 6.3
38.
2
7 49 24 ⎛ 7 ⎞ , cos α = − 1 − ⎜ − ⎟ = − 1 − =− 25 625 25 ⎝ 25 ⎠
sin α = − sin
39.
383
α 2
=
1 − cos α 2
=−
1 + cosα 2
=
1 − (−24 / 25) 2
=−
1 − 24 / 25 2
=
25 + 24 50
=−
25 − 24 50
=
49 50
=−
1 50
=
7 2 10
=−
cos α = −
cos
cos α = sin
41.
α 2
tan
α 2
= =
1 − cos α sin α 1 + 24
25
−7
25
25 + 24 = −7 = −7
2 10
2
cos α = − 1 + cos α 2 2 1 8 /17 − =− 2 = − 17 − 8 34 9 =− 34 3 = − 34 34
tan
α 2
= =
1 − cos α sin α 1 + 18 17
− 15 17
17 + 8 = −15 5 =− 3
2
12 144 5 ⎛ 12 ⎞ , sin α = 1 − ⎜ ⎟ = 1 − = 13 13 169 13 ⎝ ⎠
=
1 − cos α 2
=
1 − 12 /13 2
=
13 − 12 = 26
=
26 26
tan α =
2
8 64 15 ⎛ 8⎞ , sin α = − 1 − ⎜ − ⎟ = − 1 − =− 17 17 289 17 ⎝ ⎠
sin α = 1 − cos α 2 2 1 − (−8 /17) = 2 + 17 8 = 34 25 = 34 5 = 34 34 40.
α
cos
1 26
α 2
=
1 + cos α 2
=
1 + 12 /13 2
=
13 + 12 = 26
=
tan α = 1 − cos α 2 sin α =
25 26
1 − 12
13 5 13
= 13 − 12 5 1 = 5
5 26 26
4 4 3 , r = 32 + 42 = 25 = 5, sin α = , cos = 3 5 5
sin α = 1 − cos α 2 2
cos α = 1 + cos α 2 2
tan
α 2
=
= 1 − 3/ 5 2
= 1 + 3/ 5 2
=
= 5−3 10
= 5+3 10
=
= 1 5
= 4 5
= 5 5
=2 5 5
1 − cos α sin α 1− 3 4 5
5
5−3 4 1 = 2
Copyright © Houghton Mifflin Company. All rights reserved.
384
42.
Chapter 6: Trigonometric Identities and Equations
tan α = −
sin
α 2
8 8 15 , r = 82 + 152 = 64 + 225 = 17, sin α = , cos = − 15 17 17
=
1 − cosα 2
=
1 − (−15 /17) 2
cos
α 2
17 + 15 32 = = 34 34
=
43.
cos α = sin
α 2
44.
α 2
= =
2
=
1 17
=
17 17
= =
1 − cosα sin α
( 17 ) = 17 + 15
1 − − 15 8 17
8
=4
α
α
1 − cosα sin α
=−
1 + cos α 2
1 − 24 / 25 2
=−
1 + 24 / 25 2
=
25 − 24 1 = 50 50
=−
25 + 24 49 =− 50 50
=−
cos
2
2 10
=−
tan
2
=
1 − 24
25 = 25 − 24 7 −7 − 25
7 2 10
1 7
2
9 81 40 ⎛ 9⎞ , cosα = 1 − ⎜ − ⎟ = 1 − = 41 41 1681 41 ⎝ ⎠ 1 − cos α 2
cos
α 2
=−
1 + cos α 2
tan
1 − 40 / 41 = 2
1 + 40 / 41 =− 2
41 − 40 1 = = 82 82
41 + 40 81 =− =− 82 82
=
47.
1 + (−15 /17) 2
α
2
1 − cosα 2
=
=
tan
24 576 7 ⎛ 24 ⎞ , sin α = − 1 − ⎜ ⎟ = − 1 − =− 25 625 25 ⎝ 25 ⎠
sin α = − sin
45.
4 17 17
=
=
1 + cosα 2
17 − 15 2 = = 34 34
16 4 = 17 17
=
=
82 82
=−
1 (2sin 3x cos3 x) 2 1 = sin 2(3x) 2 1 = sin 6 x 2
sin 3 x cos3 x =
sin 2 x + cos 2 x = sin 2 x + cos 2 x − sin 2 x = cos 2 x
α 2
= =
1 − 40
41 = 41 − 40
−9
41
=−
9 82 82
46.
1 − cosα sin α
1 9
cos 8 x = cos 2(4 x) = cos2 4 x − sin 2 4 x
48.
cos 2 x = cos 2 x − sin 2 x sin 2 x sin 2 x 2 2 = cos2 x − sin 2 x sin x sin x = cot 2 x − 1
Copyright © Houghton Mifflin Company. All rights reserved.
−9
Section 6.3
49.
385
1 + cos 2 x 1 + 2 cos 2 x − 1 = sin 2 x 2 sin x cos x
50.
2 cos 2 x 2 sin x cos x = cot x =
51.
53.
55.
sin 2 x 2
1 − sin x
=
2 sin x cos x
52.
2
cos x = 2 tan x
cos 2 x = cos 2 x − sin 2 x cos 2 x cos 2 x 2 2 = cos 2 x − sin 2 x cos x cos x = 1 − tan 2 x sin 2 x − tan x = 2sin x cos x − =
54.
sin x cos x
56.
1 1 = 1 − cos 2 x 1 − 1 + 2sin 2 x 1 = 2sin 2 x = 1 csc 2 x 2 cos 2 x − sin 2 x cos 2 x = 2 sin x cos x sin 2 x = cot 2 x 2sin x cos x
sin 2 x cos x − sin x cos 2 x = tan 2 x 2
sin 2 x − cot x = 2 sin x cos x −
2sin x cos 2 x − sin x cos x
=
2 sin 2 x cos x − cos x sin x
=
cos 4 x − sin 4 x = (cos 2 x + sin 2 x)(cos 2 x − sin 2 x)
58.
sin 4 x = 2 sin 2 x cos 2 x = 2(2 sin x cos x)(cos 2 x − sin 2 x)
= cos 2 x − sin 2 x = cos 2 x 59.
cos x sin x
cos x(2 sin 2 x − 1) sin x = cot x(− cos 2 x) = − cot x cos 2 x
sin x (2cos 2 x − 1) cos x = tan x cos 2 x =
57.
=
2
= 4 sin x cos3 x − 4 sin 3 x cos x
cos 2 x − 2sin 2 x cos 2 x − sin 2 x + 2sin 4 x = cos2 x (1 − 2sin 2 x ) − sin 2 x(1 − 2sin 2 x) = (1 − 2sin 2 x)(cos 2 x − sin 2 x) = cos 2 x cos 2 x = cos 2 2 x
60.
2cos 4 x − cos2 x − 2sin 2 x cos 2 x + sin 2 x = cos 2 x(2cos 2 x − 1) − sin 2 x(2cos 2 x − 1) = (2cos 2 x − 1)(cos 2 x − sin 2 x) = cos 2 x ⋅ cos 2 x = cos 2 2 x
61.
cos 4 x = cos 2(2 x)
62.
sin 4 x = sin 2(2 x)
= 2cos 2 x − 1
= 2 sin 2 x cos 2 x
= 2(2cos 2 x − 1)2 − 1
= 2(2 sin x cos x)(1 − 2 sin 2 x)
= 2(4cos 4 x − 4cos 2 x + 1) − 1
= 4 sin x cos x − 8 sin 3 x cos x
2
= 8cos 4 x − 8cos 2 x + 1
Copyright © Houghton Mifflin Company. All rights reserved.
386
63.
Chapter 6: Trigonometric Identities and Equations
cos3 x − cos x = cos(2 x + x) − cos x = cos 2 x cos x − sin 2 x sin x − cos x = (2cos 2 x − 1) cos x − 2sin x cos x ⋅ sin x − cos x = 2cos3 x − cos x − 2sin 2 x cos x − cos x = 2cos3 x − 2cos x − 2sin 2 x cos x = 2cos3 x − 2cos x − 2(1 − cos 2 x) cos x = 2cos3 x − 2cos x − 2cos x + 2cos3 x = 4cos3 x − 4 cos x
64.
sin 3 x + sin x = sin(2 x + x) + sin x = sin 2 x cos x + cos 2 x sin x + sin x = (2sin x cos x) cos x + (1 − 2sin 2 x)sin x + sin x = 2sin x cos 2 x + sin x − 2sin 3 x + sin x = 2sin x(1 − sin 2 x) + 2sin x − 2sin 3 x = 2sin x − 2sin 3 x + 2sin x − 2sin 3 x = 4sin x − 4sin 3 x
65.
sin 3 x + cos3 x = (sin x + cos x)(sin 2 x − sin x cos x + cos 2 x) 2sin x cos x ⎞ ⎛ = (sin x + cos x) ⎜ sin 2 x + cos 2 x − ⎟ 2 ⎝ ⎠ 1 ⎛ ⎞ = (sin x + cos x) ⎜ 1 − sin 2 x ⎟ ⎝ 2 ⎠
66.
cos3 x − sin 3 x = (cos x − sin x)(cos 2 x + sin x cos x + sin 2 x) 2 sin x cos x ⎞ ⎛ = (cos x − sin x) ⎜ cos 2 x + sin 2 x + ⎟ 2 ⎝ ⎠ ⎛ 1 ⎞ = (cos x − sin x) ⎜1 + sin 2 x ⎟ ⎝ 2 ⎠
67.
sin 2
x ⎡ 1 − cos x ⎤ = ⎢± ⎥ 2 ⎣⎢ 2 ⎦⎥
2
68.
cos 2
x ⎡ 1 + cos x ⎤ = ⎢± ⎥ 2 ⎣⎢ 2 ⎦⎥
1 − cos x 2 1 − cos x sec x = ⋅ 2 sec x sec x − 1 = 2 sec x
1 + cos x 2 1 + cos x sec x = ⋅ 2 sec x sec x + 1 = 2 sec x
=
69.
71.
tan
x 1 − cos x = 2 sin x 1 cos x = − sin x sin x = csc x − cot x
x x ⎛ x⎞ 2sin cos = sin 2 ⎜ ⎟ 2 2 ⎝2⎠ = sin x
2
=
70.
72.
tan
x sin x = 2 1 + cos x
cos 2
=
sin x cos x 1 + cos x cos x cos x
=
tan x sec x + 1
x x ⎛ x⎞ − sin 2 = cos 2 ⎜ ⎟ 2 2 ⎝2⎠ = cos x
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.3
73.
387 2
x x⎞ x x x ⎛ 2 x + 2 sin cos + sin 2 ⎜ cos + sin ⎟ = cos 2 2⎠ 2 2 2 2 ⎝ x x ⎛ x⎞ = cos 2 + sin 2 + sin 2 ⎜ ⎟ 2 2 ⎝2⎠ = 1 + sin x
74.
tan 2
x ⎛ 1 − cos x ⎞ =⎜ ⎟ 2 ⎝ sin x ⎠ = =
2
(1 − cos x)2 sin 2 x (1 − cos x)2 1 − cos 2 x
(1 − cos x)2 (1 − cos x)(1 + cos x) 1 − cos x = 1 + cos x
=
1
= cos x
− cos x
cos x 1 + cos x cos x cos x
=
75.
sin 2
2
⎛ 1 − cos x ⎞ x sec x = ⎜⎜ ± ⎟⎟ sec x 2 2 ⎝ ⎠
76.
1 − cos x ⋅ sec x 2 1 = (sec x − 1) 2
2
⎛ 1 + cos x ⎞ x sec x = ⎜⎜ ± ⎟⎟ sec x 2 2 ⎝ ⎠ 1 + cos x ⋅ sec x 2 1 = (sec x + 1) 2
=
=
2
77.
cos 2
sec x − 1 sec x + 1
⎛ ⎞ cos 2 x − cos x = ⎜ ± 1 + cos x ⎟ − cos x 2 2 ⎝ ⎠ = 1 + cos x − cos x 2 = 1 + cos x − 2cos x 2 − x 1 cos = 2 2 x = sin 2
2
78.
79.
sin 2 x − cos 2 x = − ⎛⎜ cos 2 x − sin 2 x ⎞⎟ 2 2 2 2⎠ ⎝ ⎛ ⎞ x = − cos 2 ⎜ ⎟ ⎝2⎠ = − cos x
80.
81.
sin 2 x − cos x = 2sin x cos x − cos x = (cos x)(2sin x − 1)
82.
⎛ ⎞ sin 2 x + cos x = ⎜ ± 1 − cos x ⎟ + cos x 2 2 ⎝ ⎠ = 1 − cos x + cos x 2 = 1 − cos x + 2cos x 2 + x 1 cos = 2 2 x = cos 2
cos 2 x − sin 2 x = cos x 2 2 = 2sin x cos x 2sin x = 1 csc x sin 2 x 2 cos 2 x = 1 − 2sin 2 x sin 2 x sin 2 x 2 = 12 − 2sin2 x sin x sin x = csc 2 x − 2
Copyright © Houghton Mifflin Company. All rights reserved.
388
83.
85.
Chapter 6: Trigonometric Identities and Equations
tan 2 x = 2 tan 2x 1 − tan x 2 tan x tan x = 1 − tan 2 x tan x tan x 2 = cot x − tan x
84.
2 2 2cos 2 x = 2 ( cos x − sin x ) sin 2 x 2sin x cos x 2 2 = cos x − sin x sin x cos x sin x cos x = cot x − tan x
sin 2 x + 1 − cos 2 x 1 − cos 2 x + 1 − cos 2 x = sin x(1 + cos x) sin x(1 + cos x)
86.
1 2x 1 = csc 2 2 2sin 2 x
2
2(1 − cos 2 x) = sin x(1 + cos x) 2(1 − cos x)(1 + cos x) = sin x(1 + cos x) 2(1 − cos x) = sin x x = 2 tan 2
1 sin 2 x 1 = 2sin x cos x = 1 csc x sec x 2
87.
csc 2 x =
89.
cos
x ⎛ x⎞ = cos 2 ⎜ ⎟ 5 ⎝ 10 ⎠ = 1 − 2 sin 2
x 10
=
1
(
1− cos x 2
2 ± =
1
2
( 1−cos2 x )
=
)
2
1 1 − cos x
1 ⋅ 1 + cos x 1 − cos x 1 + cos x x = 1 + cos2x = 1 + cos 1 − cos x sin 2 x = 12 + cos2 x sin x sin x = csc2 x + cot x csc x =
1 cos 2 x 1 = 2cos 2 x − 1 2 1 = ⋅ sec 2 x 2 2cos x − 1 sec x 2 = sec x2 2 − sec x
88.
sec 2 x =
90.
sec2
x 1 = 2 cos 2 x
2
=
= =
1 ⎛ ± 1+ cos x ⎞ ⎜ ⎟ 2 ⎝ ⎠ 1
2
1+ cos x 2
2 1 + cos x
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.3
91.
a.
389
1 − cos π 4 M= =1÷ π / 4) 2 ( sin 2 1
=1÷ =1÷ =
b.
M sin sin
1− 2 2 2
α 2
α
2
=1 =
1 M
α
⎛ 1 ⎞ = sin −1 ⎜ ⎟ ⎝M ⎠ −1 ⎛ 1 ⎞ α = 2sin ⎜ ⎟ ⎝M ⎠ 2
2− 2 4 2
2− 2 ≈ 2.61
c.
1 ⎛ 1 ⎞ decreases. So if sin −1 ⎜ ⎟ ⎝M ⎠ M decreases, then α decreases. As M increases
....................................................... 92.
94.
96.
y = sin 2 x + cos 2 x and y = cos 2 x both have the following graph.
x x y = sin cos and y = sin x do not have the same graph. 2 2
sin 3 x + cos3 x (sin x + cos x)(sin 2 x − sin x cos x + cos 2 x) = sin x + cos x sin x + cos x 2sin x cos x 2 2 = sin x + cos x − 2 = 1 − 1 sin 2 x 2
Connecting Concepts 93.
y=
sin 2 x 1 − sin 2 x
and y = 2 tan x both have the following graph.
2
95.
97.
x x⎞ ⎛ y = ⎜ cos + sin ⎟ and y = 1 + sin x both have the 2 2 ⎝ ⎠ following graph.
cos 4 x = cos 2 x ⋅ cos 2 x cos 2 x + 1 cos 2 x + 1 = ⋅ 2 2
(
)
= 1 cos 2 2 x + 2cos 2 x + 1 4
⎛ cos 4 x + 1 ⎞ = 1⎜ + 2cos 2 x + 1⎟ 4⎝ 2 ⎠ = 1 cos 4 x + 1 + 1 cos 2 x + 1 8 8 2 1 = cos 4 x + 1 cos 2 x + 3 8 2 8
Copyright © Houghton Mifflin Company. All rights reserved.
4
390
98.
Chapter 6: Trigonometric Identities and Equations
− sin x sin x − sin 2 x sin x − 2sin x cos x sin x − 2sin x cos x sin x(1 − 2cos x ) x = = = = = − tan cos x + cos 2 x cos x + 2cos2 x − 1 2cos2 x + cos x − 1 (2cos x − 1)(cos x + 1) cos x + 1 2
....................................................... PS1.
1 [sin(α + β ) + sin(α − β )] 2 = 1 [sin α cos β + cos α sin β + sin α cos β − cos α sin β ] 2 = sin α cos β
Prepare for Section 6.4 PS2.
PS3. sin π − sin π = 0 − 1 = − 1 PS4. 6 2 2 ⎛π +π ⎞ ⎛π +π ⎞ ⎜ ⎟ ⎜ 6 ⎟ = 2cos⎛ 7π ⎞cos⎛ 5π ⎞ 2cos⎜ 2 6 ⎟sin ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 1 ⎠ ⎛ ⎛ 7π ⎞ ⎞ ⎛ ⎛ 5π ⎞ ⎞ ⎜ 1+ cos⎜ ⎟ ⎟ ⎜ 1− cos⎜ ⎟ ⎟ ⎝ 6 ⎠ ⎟+⎜ ⎝ 6 ⎠⎟ 2⎜⎜ − ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠
1 [ cos(α + β ) + cos(α − β )] 2 = 1 [ cos α cos β − sin α sin β + cos α cos β − sin α sin β ] 2 = cos α cos β
(
)
()
()
2 sin x + π = 2 ⎡sin x cos π + cos x sin π ⎤ ⎢⎣ 4 4 4 ⎥⎦ ⎡ ⎛ ⎞ ⎛ ⎞⎤ = 2 ⎢ ( sin x ) ⎜ 2 ⎟ + ( cos x ) ⎜ 2 ⎟ ⎥ 2 2 ⎝ ⎠ ⎝ ⎠⎦ ⎣ = sin x + cos x
1− 3 1+ 3 2 2 = −2 2 2 = − 1− 3 = − 1 4 4 1 =− 2 Both functional values equal − 1 . 2 PS5. Answers will vary.
PS6.
(−1)2 + ( 3 ) = 1 + 3 = 4 = 2 2
Section 6.4 1.
1 [sin( x + 2 x) + sin( x − 2 x)] 2 = sin 3 x + sin(− x)
2sin x cos 2 x = 2 ⋅
1 [cos(4 x − 2 x) − cos(4 x + 2 x)] 2 = cos 2 x − cos 6 x
2.
2sin 4 x sin 2 x = 2 ⋅
= sin 3 x − sin x
1 [sin(6 x + 2 x) − sin(6 x − 2 x)] 2 1 = [sin 8 x − sin 4 x ] 2
1 [cos(3x + 5 x) + cos(3x − 5 x)] 2 1 = [ cos8 x + cos(−2 x)] 2 1 = (cos8 x + cos 2 x) 2
3.
cos 6 x sin 2 x =
4.
cos3 x cos5 x =
5.
2sin 5 x cos3 x = sin(5 x + 3 x) + sin(5 x − 3 x) = sin 8 x + sin 2 x
6.
2sin 2 x cos 6 x = sin(2 x + 6 x) + sin(2 x − 6 x) = sin 8 x + sin(−4 x) = sin 8 x − sin 4 x
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.4
391
7.
sin x cos5 x = 1 [ cos( x − 5 x) − cos( x + 5 x) ] 2 = 1 [ cos(−4 x) − cos6 x ] 2 = 1 (cos 4 x − cos 6 x) 2
9.
cos 75° cos15° =
11.
1 [cos(75° + 15°) + cos(75° − 15°)] 2 1 = (cos 90° + cos 60°) 2 1⎛ 1⎞ = ⎜0 + ⎟ 2⎝ 2⎠ 1 = 4
1 [sin(3x + x) − sin(3x − x)] 2 1 = (sin 4 x − sin 2 x) 2
8.
cos 3x sin x =
10.
sin105° cos15° =
1 [sin(105° + 15°) + sin(105° − 15°)] 2 1 = (sin120° + sin 90°) 2 1⎛ 3 ⎞ = ⎜⎜ + 1⎟⎟ 2⎝ 2 ⎠ 3+2 4
=
cos157.5° sin 22.5° = 1 [sin(157.5° + 22.5°) − sin(157.5° − 22.5°)] 2 = 1 (sin180° − sin135°) 2 ⎛ ⎞ 1 = ⎜0 − 2 ⎟ 2⎝ 2 ⎠ =− 2 4
12.
sin195° cos15° = 1 [sin(195° + 15°) + sin(195° − 15°) ] 2 = 1 (sin 210° + sin180°) 2 = 1 ⎛⎜ − 1 + 0 ⎞⎟ 2⎝ 2 ⎠ 1 =− 4
14.
sin 11π sin 7π = 1 ⎡⎢ cos 11π − 7π − cos 11π + 7π ⎤⎥ 12 12 2 ⎣ 12 12 12 12 ⎦
(
) = 1 ( cos π − cos 3π ) 2 3 2 1 1 = ( − 0) 2 2
(
)
13.
15.
=1
⎡ ⎤ sin 13π cos π = 1 ⎢sin ⎛⎜ 13π + π ⎞⎟ + sin ⎛⎜ 13π − π ⎞⎟ ⎥ ⎝ 12 12 ⎠ ⎦ 12 12 2 ⎣ ⎝ 12 12 ⎠ = 1 ⎛⎜ sin 7π + sin π ⎞⎟ 2⎝ 6 ⎠ ⎛ ⎞ 1 1 = ⎜ − + 0⎟ 2⎝ 2 ⎠ 1 =− 4
(
) ( = 1 ⎡⎢sin 2π + sin ( − π ) ⎤⎥ 2⎣ 3 2 ⎦ 2 π π 1 = ( sin − sin ) 2 3 2 = 1 ⎛⎜ 3 − 1⎞⎟ 2⎝ 2 ⎠
4
= 3 −2 4
16.
cos
17π 7π 1 ⎡ ⎛ 17π 7π ⎞ ⎛ 17π 7π ⎞ ⎤ sin = sin ⎜ + − ⎟ − sin ⎜ ⎟⎥ 12 12 2 ⎢⎣ ⎝ 12 12 ⎠ ⎝ 12 12 ⎠ ⎦ 1⎛ 5π ⎞ = ⎜ sin 2π − sin ⎟ 2⎝ 6 ⎠
17.
)
sin π cos 7π = 1 ⎡⎢sin π + 7π + sin π − 7π ⎤⎥ 12 12 2 ⎣ 12 12 12 12 ⎦
4θ + 2θ 4θ − 2θ cos 2 2 = 2sin 3θ cosθ
sin 4θ + sin 2θ = 2sin
1⎛ 1⎞ = ⎜0 − ⎟ 2⎝ 2⎠ 1 =− 4
Copyright © Houghton Mifflin Company. All rights reserved.
392
Chapter 6: Trigonometric Identities and Equations
5θ + 3θ 5θ − 3θ sin 2 2 = −2sin 4θ sin θ
18.
cos 5θ − cos 3θ = −2sin
20.
sin 7θ − sin 3θ = 2cos
22.
cos 3θ + cos5θ = 2cos
7θ + 3θ 7θ − 3θ sin 2 2 = 2cos5θ sin 2θ 3θ + 5θ 3θ − 5θ cos 2 2 = 2 cos 4θ cos(−θ )
3θ + θ 3θ − θ cos 2 2 = 2cos 2θ cosθ
19.
cos 3θ + cosθ = 2cos
21.
cos 6θ − cos 2θ = −2 sin
23.
cosθ + cos 7θ = 2 cos
6θ + 2θ 6θ − 2θ sin 2 2 = −2 sin 4θ sin 2θ
θ − 7θ cos 2 2 = 2 cos 4θ cos(−3θ )
= 2 cos 4θ cosθ
24.
= 2 cos 4θ cos3θ
3θ + 7θ 3θ − 7θ cos 2 2 = 2sin 5θ cos(−2θ )
sin 3θ + sin 7θ = 2sin
25.
5θ + 9θ 5θ − 9θ cos 2 2 = 2sin 7θ cos(−2θ )
sin 5θ + sin 9θ = 2sin
= 2sin 5θ cos 2θ
= 2sin 7θ cos 2θ
5θ + θ 5θ − θ sin 2 2 = −2sin 3θ sin 2θ
26.
cos 5θ − cosθ = −2sin
28.
sin 2θ + sin 6θ = 2sin
θ + 7θ
2θ + 6θ 2θ − 6θ cos 2 2 = 2sin 4θ cos(−2θ )
2θ + θ 2θ − θ sin 2 2 3 1 = −2sin θ sin θ 2 2
27.
cos 2θ − cosθ = −2sin
29.
cos θ − cosθ = −2sin 2
θ +θ
2
2
4
32.
3θ + θ 3θ − θ 3θ θ + sin = 2sin 4 2 cos 4 2 4 2 2 2 5 1 = 2sin θ cos θ 8 8
cosθ + cos
θ 2
= 2cos
θ +θ
θ −θ
2
2
2 cos
31.
sin
θ 2
− sin
θ 3
θ +θ
θ −θ
2
2
= 2cos 2 = 2cos
3 sin 2
5 1 θ sin θ 12 12
2
cos (α + β ) + cos (α − β ) = cos α cos β − sin α sin β + cosα cos β + sin α sin β = 2cos α cos β
34.
)
4
3 1 = 2cos θ cos θ 4 4 33.
4
= 2sin 3 θ sin 1 θ 4
sin
(
2
= −2sin 3 θ sin − 1 θ
= 2sin 4θ cos 2θ
30.
θ −θ
sin 2
cos (α − β ) − cos (α + β ) = cosα cos β + sin α sin β − cosα cos β + sin α sin β = 2sin α sin β
Copyright © Houghton Mifflin Company. All rights reserved.
3
Section 6.4
35.
393
1 [sin(3x + x) − sin(3x − x)] 2 = sin 4 x − sin 2 x = 2sin 2 x cos 2 x − sin 2 x = sin 2 x(2cos 2 x − 1)
2cos3x sin x = 2 ⋅
1 [sin(5 x + 3x) + sin(5 x − 3x)] 2 1 = (sin8 x + sin 2 x) 2 1 = (2sin 4 x cos 4 x + 2sin x cos x) 2 = sin 4 x cos 4 x + sin x cos x
36.
sin 5 x cos3 x =
38.
sin 3 x cos x =
= 2sin x cos x ⎡⎢ 2(1 − 2sin 2 x) − 1⎤⎥ ⎣ ⎦ = 4sin x cos x − 8sin 3 x cos x − 2sin x cos x = 2sin x cos x − 8cos x sin 3 x 37.
1 [cos(5 x + 7 x) + cos(5 x − 7 x)] 2 = cos12 x + cos(−2 x) = cos12 x + cos 2 x
2 cos 5 x cos 7 x = 2 ⋅
= cos 2 6 x − sin 2 6 x + 2 cos 2 x − 1
1 [sin(3x + x) + sin(3x − x)] 2 1 = (sin 4 x + sin 2 x) 2 1 = (2sin 2 x cos 2 x + sin 2 x) 2 1 = [ (sin 2 x(2cos 2 x + 1)] 2 1 = ⋅ 2sin x cos x ⎡⎢ 2(1 − 2sin 2 x) + 1⎤⎥ ⎣ ⎦ 2 = sin x cos x(2 − 4sin 2 x + 1) = sin x cos x(3 − 4sin 2 x)
39.
= 2(1 − 2sin 2 x)sin x
5 x + 3x 5 x − 3x sin 2 2 = 2sin 4 x sin x = −2(2sin 2 x cos 2 x sin x)
= 2sin x − 4sin 3 x
= −4 ⎡⎢ 2sin x cos x(2cos 2 x − 1)sin x ⎤⎥ ⎣ ⎦
3x + x 3x − x sin 2 2 = 2 cos 2 x sin x
sin 3 x − sin x = 2 cos
40.
cos5 x − cos3 x = −2sin
= −8sin 2 x(2cos3 x − cos x) 41.
2x + 4 2 x − x4 cos 2 2 = 2cos3 x cos(− x) = 2cos3 x cos x
sin 2 x + sin 4 x = 2cos
= 2cos x sin(2 x + x) = 2cos x(sin 2 x cos x + cos 2 x sin x) = 2cos x[(2sin x cos x)cos x + (2cos 2 x − 1)sin x] = 2cos x sin x(4cos 2 x − 1) 42.
3x + x 3x − x cos 2 2 = 2 cos 2 x cos x
cos 3 x + cos x = 2 cos
2
= 2(2 cos x − 1) cos x = 4 cos3 x − 2 cos x
43.
2cos 3 x + x sin 3 x − x sin 3 x − sin x 2 2 = cos3 x − cos x −2sin 3 x + x sin 3 x − x 2
cos 2 x =− sin 2 x = − cot 2 x
Copyright © Houghton Mifflin Company. All rights reserved.
2
394
44.
Chapter 6: Trigonometric Identities and Equations 5 x +3 x 5 x −3 x cos5 x − cos3 x −2sin 2 sin 2 = sin 5 x + sin 3x 2sin 5 x +3 x cos 5 x −3 x 2
45.
2
sin 5 x + sin 3 x 4sin x cos3 x − 4sin 3 x cos x
sin 4 x sin x =− sin 4 x cos x = − tan x
46.
=
2sin 5 x +3 x cos 5 x −3 x 2
4x+2 x 4 x−2 x cos 4 x − cos 2 x −2sin 2 sin 2 = sin 2 x − sin 4 x 2cos 2 x + 4 x sin 2 x − 4 x 2
2
− sin 3 x sin x = cos3 x sin(− x) − sin 3 x sin x = − cos3 x sin x = tan 3x 1 [sin( x + y + x − y) + sin( x + y − x + y)] 2 1 = [sin 2 x + sin 2 y ] 2 1 = [ 2sin x cos x + 2sin y cos y ] 2 = sin x cos x + sin y cos y
47.
sin( x + y ) cos( x − y ) =
48.
sin( x + y ) sin( x − y ) =
1 [cos( x + y − x + y) − cos( x + y + x − y )] 2 1 = [ cos 2 y − cos 2 x ] 2 1⎡ = ⎢1 − 2sin 2 y − 1 + 2sin 2 x ⎤⎥ ⎦ 2⎣ = sin 2 x − sin 2 y
49.
a = −1, b = −1, k =
( −1)2 + ( −1)2
α is a third quadrant angle. sin β =
−1
1
= 2 2 β = 45° α = −180° + 45° = −135° y = 2 sin( x − 135°)
= 2,
50.
a = 3, b = −1, k =
(
−3
)
2
α is a fourth quadrant angle. 1 1 = 2 2 β = 30°
sin β = −
α = −30° y = 2sin( x − 30°)
Copyright © Houghton Mifflin Company. All rights reserved.
2
4sin x cos x(cos 2 x − sin 2 x) sin 4 x cos x = 2sin x cos x cos 2 x 2sin 2 x cos 2 x cos x = sin 2 x cos 2 x = 2cos x
2
+ ( −1) = 2,
Section 6.4
395 2
51.
2 1 3 3⎞ ⎛1⎞ ⎛ , k = ⎜ ⎟ + ⎜⎜ − a= , b=− ⎟ = 1, 2 2 ⎝ 2 ⎠ ⎝ 2 ⎟⎠ α is a fourth quadrant angle.
2
52.
⎛ 3 ⎞ ⎛ 1 ⎞2 3 1 , b = − , k = ⎜⎜ ⎟⎟ + ⎜ − ⎟ = 1, 2 2 ⎝ 2 ⎠ ⎝ 2⎠ α is a fourth quadrant angle.
a=
−1/ 2 1 = 1 2 β = 30°
3 3 2 sin β = = 1 2 β = 60°
sin β =
−
α = −30°
α = −60°
y = sin( x − 30°)
y = sin( x − 60°) 2
53.
2
1 1 2 ⎛1⎞ ⎛1⎞ a = , b =− , k = ⎜ ⎟ +⎜ ⎟ = , 2 2 2 ⎝2⎠ ⎝2⎠ α is a fourth quadrant angle.
2
54.
−1/ 2 2 = 2 2/2 β = 45° α = −45°
2 ⎛ 3 1 3⎞ ⎛ 1⎞ , b = − , k = ⎜⎜ − ⎟⎟ + ⎜ − ⎟ = 1, 2 2 ⎝ 2 ⎠ ⎝ 2⎠ α is a third quadrant angle.
a=−
−1/ 2 1 = 1 2 β = 30°
sin β =
sin β =
α = 30° − 180° = −150° y = sin( x − 150°)
2 y= sin( x − 45°) 2
55.
2
2
2
a = −3, b = 3, k = (−3) + 3 = 3 2,
56.
α is a second quadrant angle. 3 1 2 = − 2 3 2 2 β = 45°
sin β =
β = 45° α = 45°
y = 3 2 sin( x + 135°)
y = sin( x + 45°)
a = π , b = −π , k = π 2 + (−π )2 = π 2,
58.
a = −0.4, b = 0.4, k = (−0.4)2 + 0.42 = 0.4 2,
α is a second quadrant angle.
α is a fourth quadrant angle. −π 1 2 sin β = = = 2 2 π 2 β = 45°
59.
2/2 2 = 1 2
sin β =
α = 180° − 45° = 135°
57.
0.4 1 2 = = 2 0.4 2 2 β = 45°
sin β =
α = −45°
α = 180° − 45° = 135°
y = π 2 sin( x − 45°)
y = 0.4 2 sin( x + 135°)
a = −1, b = 1, k = (−1) 2 + 12 = 2,
60.
2
a = − 3, b = −1, k = (− 3)2 + ( −1) = 2,
α is a second quadrant angle.
α is a third quadrant angle.
sin β = 1 = 1 = 2 2 2 2
sin β = − 1 = 1 2 2
β =π
4
α = π − π = 3π 4
4 ⎛ y = 2 sin ⎜ x + 3π ⎞⎟ ⎝ 4 ⎠
2
⎛ 2⎞ ⎛ 2⎞ 2 2 , b= , k = ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ = 1, 2 2 ⎝ 2 ⎠ ⎝ 2 ⎠ α is a first quadrant angle. a=
β =π
6
α = π − π = − 5π 6
6 ⎛ ⎞ 5 π y = 2sin ⎜ x − ⎟ 6 ⎠ ⎝
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396
Chapter 6: Trigonometric Identities and Equations 2
61.
⎛ 3 ⎞ ⎛ 1 ⎞2 3 1 , b = , k = ⎜⎜ ⎟⎟ + ⎜ ⎟ = 1, 2 2 ⎝ 2 ⎠ ⎝2⎠ α is a first quadrant angle.
62.
a=
α is a first quadrant angle.
β =π
β=
6
y = sin ⎛⎜ x + π ⎞⎟ 6⎠ ⎝
π⎞ ⎛ y = 2sin ⎜ x + ⎟ 3⎠ ⎝
a = −10, b = 10 3, k = (−10)2 + (10 3)2 = 20,
sin β =
β=
64.
β =π
π
3
3
π
α = −π
=
3
y = 6sin ⎛⎜ x − π ⎞⎟ 3⎠ ⎝
a = −5, b = 5, k = (−5)2 + 52 = 5 2,
66.
α is a second quadrant angle.
β=
a = 3, b = −3, k = 32 + (−3)2 = 3 2,
α is a fourth quadrant angle.
5 2 = 2 5 2
sin β =
π 4
a = 3, b = −3 3, k = (3)2 + (−3 3)2 = 6, sin β = −3 3 = 3 6 2
2π 3 3 2π ⎞ ⎛ y = 20sin ⎜ x + ⎟ 3 ⎠ ⎝
sin β =
3
α is a fourth quadrant angle.
10 3 3 = 20 2
α =π −
β=
3π α =π − = 4 4 3π ⎞ ⎛ y = 5 2 sin ⎜ x + ⎟ 4 ⎠ ⎝
π
y = − sin x − 3 cos x
y = sin x + 3 cos x
π⎞ ⎛ y = 2sin ⎜ x + ⎟ 3⎠ ⎝
−3 2 = 2 3 2
π 4
α =−
π 4
π⎞ ⎛ y = 3 2 sin ⎜ x − ⎟ 4⎠ ⎝ 68.
2π ⎞ ⎛ y = 2sin ⎜ x − ⎟ 3 ⎠ ⎝
70.
3
π
α=
α is a second quadrant angle.
67.
π
α =π
6
65.
3 3 = 2 2
sin β =
sin β = 1/ 2 = 1 1 2
63.
a = 1, b = 3, k = ( 3)2 + (1)2 = 2,
y = − 3 sin x + cos x
69.
π⎞ ⎛ y = 2 2 sin ⎜ x + ⎟ 4⎠ ⎝
5π ⎞ ⎛ y = 2sin ⎜ x + ⎟ 6 ⎠ ⎝
71.
y = − 3 sin x − cos x 5π ⎞ ⎛ y = 2sin ⎜ x − ⎟ 6 ⎠ ⎝
y = 2sin x + 2cos x
72.
y = − sin x + cos x 3π ⎞ ⎛ y = 2 sin ⎜ x + ⎟ 4 ⎠ ⎝
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.4
73.
397
74.
y = −5sin x + 5 3 cos x 2π ⎞ ⎛ y = 10sin ⎜ x + ⎟ 3 ⎠ ⎝
75.
3π ⎞ ⎛ y = 2sin ⎜ x + ⎟ 4 ⎠ ⎝
76.
y = 6 3 sin x − 6cos x
π⎞ ⎛ y = 12sin ⎜ x − ⎟ 6⎠ ⎝
77.
a. b.
c. 78.
a. b.
c.
) (
p (t ) = 2sin 2π ⋅ 1336t + 2π ⋅ 770t sin 2π ⋅ 1336t − 2π ⋅ 770t 2 2 = 2sin(2106π t )sin(556π t ) 1336 + 770 = 2106 = 1053 cycles per second 2 2 p (t ) = sin(2π ⋅ 1336t ) + sin(2π ⋅ 852t )
(
y = 5 2 sin x + 5 2 cos x
π⎞ ⎛ y = 10sin ⎜ x + ⎟ 4⎠ ⎝
p (t ) = sin(2π ⋅ 1336t ) + sin(2π ⋅ 770t )
(
y = − 2 sin x + 2 cos x
) (
p (t ) = 2sin 2π ⋅ 1336t + 2π ⋅ 852t sin 2π ⋅ 1336t − 2π ⋅ 852t 2 2 = 2sin(2188π t )sin(484π t ) 1336 + 852 = 2188 = 1094 cycles per second 2 2
)
)
79.
Identity
80.
Identity
81.
Identity
82.
Identity
83.
Identity
84.
Identity
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398
Chapter 6: Trigonometric Identities and Equations
....................................................... 85.
Connecting Concepts
Let x = α + β and y = α − β . x + y = α + β + α − β and x − y = α + β − (α − β ) x + y = 2α x − y = 2β x+ y x− y α= 2β = 2 2 cos(α − β ) + cos(α + β ) = 2cosα cos β x+ y x− y ⎡x + y x − y⎤ ⎡x+ y x− y⎤ − cos ⎢ ⎥ + cos ⎢ 2 + 2 ⎥ = 2cos 2 cos 2 2 2 ⎣ ⎦ ⎣ ⎦ x+ y x− y cos y + cos x = 2cos cos 2 2
86.
cos( x − y ) = cos x cos y + sin x sin y cos( x + y ) = cos x cos y − sin x sin y (cos x cos y + sin x sin y ) − (cos x cos y − sin x sin y ) = cos( x − y ) − cos( x + y ) cos x cos y + sin x sin y − cos x cos y + sin x sin y = cos( x − y ) − cos( x + y ) 2sin x sin y = cos( x − y ) − cos( x + y ) sin x sin y = 1 [cos( x − y ) − cos( x + y )] 2
87.
x + y = 180° y = 180° − x sin x + sin y = sin x + sin(180° − x) = sin x + sin 180° cos x − cos180° sin x = sin x + 0(cos x) − ( −1) sin x = 2 sin x
89.
sin 2 x + sin 4 x + sin 6 x = 2sin
90.
sin 4 x − sin 2 x + sin 6 x = 2sin
91.
10 x +8 x 10 x −8 x cos10 x + cos8 x 2cos 2 cos 2 = sin10 x − sin 8 x 2cos 10 x +8 x sin 10 x −8 x
88.
x + y = 360° y = 360° − x cos x + cos y = cos x + cos(360° − x) = cos x + cos360° cos x + sin 360° sin x = cos x + (1)cos x + (0)sin x = 2cos x
2x + 4x 2x − 4x cos + 2sin 3x cos3 x 2 2 = 2sin 3x cos x + 2sin 3 x cos3 x = 2sin 3 x(cos x + cos3 x) = 2sin 3 x ⎛⎜ 2cos x + 3x cos x − 3x ⎞⎟ 2 2 ⎠ ⎝ = 4sin 3 x cos 2 x cos x
4x + 2x 4x − 2x sin + 2sin 3 x cos3 x 2 2 = 2cos3 x sin x + 2sin 3 x cos3 x = 2cos3 x(sin x + sin 3x) = 2cos3 x ⎛⎜ 2sin x + 3x cos x − 3x ⎞⎟ 2 2 ⎠ ⎝ = 2cos3 x(2sin 2 x cos x) = 4cos3 x sin 2 x cos x
2
2
2cos9 x cos x = 2cos9 x sin x = cot x
92.
10 x + 2 x 10 x − 2 x sin10 x + sin 2 x = 2sin 2 cos 2 cos10 x + cos 2 x 2cos 10 x + 2 x cos 10 x − 2 x 2
= sin 6 x cos 4 x cos 6 x cos 4 x = tan 6 x = 2 tan 3 x 1 − tan 2 3 x
Copyright © Houghton Mifflin Company. All rights reserved.
2
Section 6.4
93.
94.
399
sin 2 x + sin 4 x + sin 6 x sin 2 x + sin 6 x + sin 4 x = cos 2 x + cos 4 x + cos 6 x cos 2 x + cos6 x + cos 4 x 2x + 6x 2x − 6x + sin 4 x 2sin cos 2 2 = 2x + 6x 2x − 6x + cos 4 x 2cos cos 2 2 2sin 4 x cos 2 x + sin 4 x = 2cos 4 x cos 2 x + cos 4 x sin 4 x(2cos 2 x + 1) = cos 4 x(2cos 2 x + 1) sin 4 x = cos 4 x = tan 4 x 2 x +6 x 2 x −6 x sin 2 x + sin 6 x = 2sin 2 cos 2 cos 6 x − cos 2 x −2sin 6 x + 2 x sin 6 x − 2 x 2
= − 2sin 4 x cos 2 x 2sin 4 x sin 2 x = − cos 2 x sin 2 x = − cot 2 x
2
95.
cos 2 x − sin 2 x = cos x ⋅ cos x − sin x ⋅ sin x = 1 [ cos( x + x) + cos( x − x) ] − 1 [cos( x − x) − cos( x + x)] 2 2 = 1 cos 2 x + 1 cos 0 − 1 cos 0 + 1 cos 2 x 2 2 2 2 = cos 2 x
96.
2sin x cos x = 2 ⋅ 1 [sin( x + x) + sin( x − x)] 2 = sin 2 x + sin 0 = sin 2 x
97.
Let k = a 2 + b 2 , tan α =
a b
a 2 + b 2 ( a sin x + b cos x) a2 + b2 ⎛ ⎞ a b sin x + cos x ⎟ = a2 + b2 ⎜ ⎜ 2 ⎟ 2 2 2 a +b ⎝ a +b ⎠
a sin x + b cos x =
= k (sin α sin x + cos α cos x) because sinα = = k (cos x cos α + sin x sin α ) = k cos( x − α )
98.
Let k = a 2 + b 2 , tan α =
a 2
a +b
2
and cosα =
b 2
a + b2
b a
a 2 + b 2 (a sin cx + b cos cx) a 2 + b2 ⎛ ⎞ a b sin cx + cos cx ⎟ = a 2 + b2 ⎜ ⎜ 2 ⎟ 2 a 2 + b2 ⎝ a +b ⎠
a sin cx + b cos cx =
= k (cos α sin cx + sin α cos cx ) because cosα = = k (sin cx cosα + cos cx sin α ) = k sin(cx + α )
a a 2 + b2
and sinα =
b a2 + b2
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400
Chapter 6: Trigonometric Identities and Equations
.......................................................
Prepare for Section 6.5
PS1. A one-to-one function is a function for which each range value (y-value) is paired with one and only one domain value (x-value). PS3.
PS2. If every horizontal line intersects the graph of a function at most once, then the function is a one-to-one function.
f [ g ( x)] = f ⎡⎢ 1 x − 2 ⎤⎥ ⎣2 ⎦ ⎛ ⎞ 1 = 2⎜ x − 2⎟ + 4 ⎝2 ⎠ = x−4+4 =x
PS4.
PS5. The graph of f –1 is the reflection of the graph of f across the line given by y = x.
f [ f −1 ( x)] = x
PS6. No, it does not pass the horizontal line test.
Section 6.5 1.
y = sin −1 1
2.
sin y = 1 with − y=
4.
7.
π
≤ y≤
π 2
3 3
y=
3 3
5.
2 2
2 2
0< y <π
≤ y≤
2
π
cos y = − 3 2 5 π y= 6
2
4
y = tan −1(1)
y=−
8.
π
π
−
π
π 2
< y<
π
tan y = 3
2
y=
4 9.
0≤ y ≤π
y = tan −1 3
6.
y = cot −11 cot y = 1 0< y <π y=
⎛ ⎞ y = cos−1 ⎜ − 3 ⎟ 2 ⎝ ⎠
3.
−
tan y = −1
π
−
π 2
π
< y<
π 2
3
y = sec−1 2 sec y = 2 0 ≤ y ≤π
π
y=
4
π 3
3
y = sec−1 2 3 sec y = 3 y=
sin y =
2
y = cot −1
y=
13.
2
y = cos−1 ⎛⎜ − 1 ⎞⎟ ⎝ 2⎠ 1 cos y = − 0≤ y ≤π 2 y = 2π 3
cot y =
10.
π
y = sin −1
2 3 3
11.
(
y = csc−1 − 2 csc y = − 2
0≤ y ≤π
y=−
π 6
⎛ ⎞ y = sin −1 ⎜ − 3 ⎟ ⎝ 2 ⎠ sin y = − 3 2 y = −π 3
−π ≤ y≤π 2 2
14.
π
−
) π 2
≤y≤
π
csc y = −2
2 y=−
4
y = sin −1 1 2 sin y = 1 −π ≤ y≤π 2 2 2 y=π 6
y = csc−1(−2)
12.
15.
π
−
π 2
≤ y≤
6
y = cos −1 ⎛⎜ − 1 ⎞⎟ ⎝ 2⎠ cos y = − 1 0≤ y ≤π 2 y = 2π 3
Copyright © Houghton Mifflin Company. All rights reserved.
π 2
Section 6.5
401
y = cos −1
16.
cos y = y=
19.
21.
23.
3 2
3 2
y = tan −1
17.
0 ≤ y ≤π
tan y =
π
y=
6
a.
sin −1 (0.8422) ≈ 1.0014
b.
tan −1 (0.2385) ≈ 0.2341
a.
sec −1 (2.2500) = cos −1
b.
cot −1 (3.4545) = tan −1
()
cos θ = x or θ = cos −1 x 7 7
3 3
3 3
−
π 2
< y<
π
tan y = 1
2
π
y=
π 6
20.
1 ( 2.2500 ) ≈ 1.1102 1 ( 3.4545 ) ≈ 0.2818 24.
y = tan −1(1)
18.
22.
a.
cos −1 ( −0.0356) ≈ 1.6064
b.
tan −1 (3.7555) ≈ 1.3106
a.
csc −1 (1.3465) = sin −1
b.
cot −1 (0.1274) = tan −1
()
tan θ = 5 or θ = tan −1 5 x x
25.
−
π 2
< y<
4
1 (1.3465 ) ≈ 0.8370 1 ( 0.1274 ) ≈ 1.4441
1⎞ ⎛ y = cos ⎜ cos −1 ⎟ 2⎠ ⎝ y = cos
π 3
1 y= 2 26.
y = cos(cos −1 2)
27.
32.
35.
3⎞ ⎛ y = sin ⎜ tan −1 ⎟ 4⎠ ⎝ 3 y= 5 ⎡ ⎛ 3 ⎞⎤ y = sin ⎢cos −1 ⎜⎜ − ⎟⎟ ⎥ ⎢⎣ ⎝ 2 ⎠ ⎥⎦ 1 y= 2
⎛ π⎞ y = sin −1 ⎜ sin ⎟ 6⎠ ⎝ 1 = sin −1 2
y=
38.
30.
33.
36.
π
5⎞ ⎛ y = cos ⎜ sin −1 ⎟ 13 ⎝ ⎠ 12 y= 13
y = cos(sec−1 2) 1 y= 2 ⎛ 5π ⎞ y = sin −1 ⎜ sin ⎟ 6 ⎠ ⎝ 1 = sin −1 2
y=
6
5π ⎞ ⎛ y = cos −1 ⎜ cos ⎟ 4 ⎠ ⎝ ⎛ 2⎞ = cos −1 ⎜⎜ − ⎟⎟ ⎝ 2 ⎠ 3π y= 4
28.
y=2
y is not defined.
29.
y = tan(tan −1 2)
39.
31.
34.
37.
y = sin −1(sin 2)
⎛ π⎞ y = cos −1 ⎜ sin ⎟ 4⎠ ⎝ 2 2
= cos −1
y=
6
= sin −1 3 y is not defined.
⎛ 2⎞ y = tan ⎜⎜ sin −1 ⎟ 2 ⎟⎠ ⎝ y =1
≈ sin −1(0.9093) y ≈ 1.1416
π
π⎞ ⎛ y = sin −1 ⎜ tan ⎟ 3⎠ ⎝
y = tan ⎛⎜ tan −1 1 ⎞⎟ 2⎠ ⎝ 1 y= 2
40.
π 4
2π ⎞ ⎛ y = cos −1 ⎜ tan ⎟ 3 ⎠ ⎝
(
= cos −1 − 3
)
y is not defined.
Copyright © Houghton Mifflin Company. All rights reserved.
π 2
402
41.
45.
Chapter 6: Trigonometric Identities and Equations
⎛ π⎞ y = tan −1 ⎜ sin ⎟ 6⎠ ⎝ 1 = tan −1 2 y ≈ 0.4636
2π ⎞ ⎛ y = cot −1 ⎜ cos ⎟ 3 ⎠ ⎝ ⎛ −1 ⎞ = tan −1 ⎜ ⎟ +π ⎝ 0.5 ⎠ y ≈ −1.1071 + π y ≈ 2.0344
42.
1 and find y = tanθ . 2 π π 1 Then sinθ = and − ≤ θ ≤ . 2 2 2 Let θ = sin −1
43.
y=−
46.
48.
50.
(
= cos −1 − 3
Let θ = cos −1 3 and find y = cscθ . 4 Then cosθ = 3 , and 0 ≤ θ ≤ π . 4
Let θ = cos −1 3 and find y = tan θ . 5 Then cosθ = 3 , and 0 ≤ θ ≤ π . 5
Thus tanθ = 4 . 3 y=4 3
Copyright © Houghton Mifflin Company. All rights reserved.
)
y is not defined.
Let θ = csc−1 2 and find y = cot θ . π 1 π Then cscθ = 2, sinθ = and − ≤ θ ≤ . 2 2 2
Thus cscθ = 4 = 4 7 . 7 7 4 7 y= 7
Let θ = sin −1 7 and find y = cosθ . 25 Then sinθ = 7 , and − π ≤ θ ≤ π . 25 2 2
⎡ ⎛ π ⎞⎤ y = cos −1 ⎢ tan ⎜ − ⎟ ⎥ ⎣ ⎝ 3 ⎠⎦
6
y= 3
Let θ = sin −1 1 and find y = secθ . 4 Then sinθ = 1 , and − π ≤ θ ≤ π . 4 2 2
Thus cosθ = 24 . 25 24 y= 25
π
Thus cotθ = 3.
Thus secθ = 4 = 4 15 . 15 15 4 15 y= 15
49.
44.
⎛ 1⎞ = sin −1 ⎜ − ⎟ ⎝ 2⎠
Thus tanθ = 1 = 3 . 3 3 3 y= 3
47.
⎛ ⎡ 2π ⎤ ⎞ y = sin −1 ⎜ cos ⎢ − ⎥ ⎟ ⎣ 3 ⎦⎠ ⎝
Section 6.5
51.
403
Let α = sin −1 2 , α = π , sin α = 2 , cos α = 2 . 2 4 2 2
52.
3
53.
2
=
2
4 4 3 ⎛4⎞ Let α = sin −1 , sin α = , cosα = 1 − ⎜ ⎟ = . 5 5 5 ⎝5⎠
54.
1 − tan 2 α
=
( 3) = 2 3 = 2 3 2 1− 3 −2 1− ( 3) 2
Let α = tan −11, α = −1
π 4
, sin α =
2 2 , cosα = . 2 2
1)
= cos 2α = cos 2 α − sin 2 α 2
2
⎛ 2⎞ ⎛ 2⎞ = ⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ = 0 ⎝ 2 ⎠ ⎝ 2 ⎠
2 1⎞ ⎛ y = sin ⎜ sin −1 + cos−1 ⎟ 3 2⎠ ⎝
56. 2
1
⎛1⎞
2
3 5⎞ ⎛ y = cos ⎜ sin −1 + cos −1 ⎟ 4 13 ⎝ ⎠ 2
2 2 5 ⎛2⎞ Let α = sin −1 , sin α = , cosα = 1 − ⎜ ⎟ = . 3 3 3 3 ⎝ ⎠
3 3 7 ⎛3⎞ Let α = sin −1 , sin α = , cosα = 1 − ⎜ ⎟ = . 4 4 4 4 ⎝ ⎠ 2
5 5 12 ⎛5⎞ , cos β = , sin β = 1 − ⎜ ⎟ = . 13 13 13 13 ⎝ ⎠ y = cos(α + β ) = cosα cos β − sin α sin β
3
β = cos−1 , cosβ = , sinβ = 1 − ⎜ ⎟ = . 2 2 2 ⎝2⎠ y = sin(α + β ) = sin α cos β + cosα sin β
57.
2 tan α
y = cos(2 tan
⎛ 4 ⎞⎛ 3 ⎞ 24 = 2 ⎜ ⎟⎜ ⎟ = ⎝ 5 ⎠⎝ 5 ⎠ 25
1
3.
=− 3
4⎞ ⎛ y = sin ⎜ 2sin −1 ⎟ 5⎠ ⎝ = sin 2α = 2sin α cosα
55.
3 , cosα = 1 , tan α = 2 2
⎛ 3⎞ y = tan ⎜⎜ 2sin −1 ⎟⎟ 2 ⎝ ⎠ = tan 2α
= cos 2 α − sin 2 α 2
2
Then, α = π , sin α =
⎛ 2⎞ y = cos ⎜⎜ 2sin −1 ⎟⎟ 2 ⎝ ⎠ = cos 2α
⎛ 2⎞ ⎛ 2⎞ = ⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ =0
Let α = sin −1 3 .
β = cos −1
=
2⎛ 1⎞ 5⎛ 3⎞ ⎜ ⎟ ⎜ ⎟+ 3 ⎝ 2 ⎠ 3 ⎜⎝ 2 ⎟⎠
=
=
1 15 2 + 15 + = 3 6 6
=
7 ⎛ 5 ⎞ 3 ⎛ 12 ⎞ ⎜ ⎟− ⎜ ⎟ 4 ⎝ 13 ⎠ 4 ⎝ 13 ⎠ 5 7 36 5 7 − 36 − = 52 52 52
1 3⎞ ⎛ y = tan ⎜ cos −1 − sin −1 ⎟ 2 4⎠ ⎝ 2
3
1 1 3 ⎛1⎞ Let α = cos −1 , cosα = , sin α = 1 − ⎜ ⎟ = , tan α = 2 = 3. 1 2 2 2 ⎝2⎠ 3
3
⎛3⎞
β = sin −1 , sin β = , cos β = 1 − ⎜ ⎟ 4 4 ⎝4⎠
2
2 3 7 3 3 7 = = , tan β = 4 = . 7 4 7 7 4
y = tan(α − β ) =
tan α − tan β 1 + tan α tan β 3−3 7
3 − 3 7 7 7 3 − 3 7 7 3 − 3 7 7 − 3 21 112 3 − 84 7 3 7 − 4 3 1 7 7 ⋅ = = = = ⋅ = = = 3 7 −4 3 −140 5 5 7 + 3 21 7 + 3 21 7 − 3 21 1+ 3 ⋅ 3 7 1+ 3 ⋅ 3 7 7 7 7
(
Copyright © Houghton Mifflin Company. All rights reserved.
)
404
58.
Chapter 6: Trigonometric Identities and Equations
2 2⎞ ⎛ y = sec ⎜ cos−1 + sin −1 ⎟ 3 3⎠ ⎝ 2
2 2 5 ⎛2⎞ Let α = cos−1 , cosα = , sin α = 1 − ⎜ ⎟ = . 3 3 3 3 ⎝ ⎠ 2
⎛2⎞
2
2
5
β = sin −1 , sin β = , cos β = 1 − ⎜ ⎟ = . 3 3 3 ⎝3⎠ y = sec(α + β ) 1
1 cos(α + β ) cosα cos β − sin α sin β 1 1 = = 2 5 2 5 2⋅ 5 − 5⋅2 − =
3
3
=
3
3
9
9
y is undefined.
59.
sin −1 x = cos −1 5
13 −1 sin(sin x) = sin cos −1 5 13 12 x= 13
(
60.
)
tan −1 x = sin −1 24
sin −1 ( x − 1) =
61.
25 −1 tan(tan x) = tan sin −1 24 25
) ( x = tan ( sin −1 24 ) 25
π 2
( x − 1) = sin
π 2
( x − 1) = 1
x=2
x = 24 7
62.
65.
1⎞ π ⎛ cos −1 ⎜ x − ⎟ = 2⎠ 3 ⎝ 1⎞ π ⎛ ⎜ x − ⎟ = cos 2⎠ 3 ⎝ 1 1 x= + 2 2 x =1
63.
π 4
sin −1( x − 2) = −
64.
x = 1− 2 2
2− 2 = 2
66.
− sin −1
3 5
Let α = sin −1 3 , sin α = 3 , cosα = 4 . 5
5
⎛π ⎞ x = cos ⎜ − α ⎟ ⎝4 ⎠
π
4 π = 5 6
sin −1 x =
3⎞ ⎛π x = cos ⎜ − sin −1 ⎟ 5⎠ ⎝4 5
sin −1 x + cos −1
π
6 ⎛ π⎞ ( x − 2) = sin ⎜ − ⎟ ⎝ 6⎠ 1 x=− +2 2 3 x= 2
⎛ x + 2 ⎞ = tan π ⎜ 2 ⎟⎠ 4 ⎝
3 π sin −1 + cos −1 x = 5 4
cos −1 x =
π tan −1 ⎛⎜ x + 2 ⎞⎟ = 2 ⎠ 4 ⎝
π 6
− cos −1
4 5
4⎞ ⎛π x = sin ⎜ − cos −1 ⎟ 5⎠ ⎝6 Let α = cos−1 4 , cosα = 4 , sin α = 3 . 5
5
5
⎛π ⎞ x = sin ⎜ − α ⎟ ⎝6 ⎠
π
x = cos cos α + sin sin α 4 4 2 4 2 3 ⋅ + ⋅ x= 2 5 2 5 4 2 3 2 7 2 = + = 10 10 10
π
π
cosα − cos sin α 6 6 1 4 3 3 ⋅ x= ⋅ − 2 5 2 5 4 3 3 4−3 3 = − = 10 10 10
x = sin
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.5
67.
405
sin −1
2 2π + cos −1 x = 2 3
68.
cos −1 x = 2π − sin −1 2 3 2 ⎛ 2π ⎞ x = cos ⎜ − sin −1 2 ⎟ 2 ⎠ ⎝ 3
cos −1 x + sin −1 cos −1 x =
2 2 2 , sin α = , cos α = . 2 2 2 x = cos ⎛⎜ 2π − α ⎞⎟ ⎝ 3 ⎠ π 2 x = cos cos α + sin 2π sin α 3 3 3 2 2 1 + ⋅ x=− ⋅ 2 2 2 2 = − 2 + 6 = 6 − 2 ≈ 0.2588 4 4 4
71.
Let α = sin −1x, sin α = x, cos α = 1 − x 2 .
3 3 1 , sin α = , cosα = . 2 2 2
⎛π ⎞ x = cos ⎜ − α ⎟ ⎝2 ⎠
π
π
x = cos cos α + sin sin α 2 2 1 3 3 = x = 0 ⋅ + 1⋅ 2 2 2
Note: Since sin α =
2 2 π and cosα = , then α = . 2 2 4 ⎛ 2π ⎞ ⎛ 2π π ⎞ ⎛ 5π ⎞ Thus, cos ⎜ − α ⎟ = cos ⎜ − ⎟ = cos ⎜ ⎟ ≈ 0.2588. ⎝ 3 ⎠ ⎝ 3 4⎠ ⎝ 12 ⎠ 1 − x2 x
3 2
− sin −1
Let α = sin −1
Note: Since sin α =
tan(cos −1 x) =
2
⎛π 3⎞ x = cos ⎜⎜ − sin −1 ⎟ 2 ⎟⎠ ⎝2
Let α = sin −1
69.
π
3 π = 2 2
3 1 π and cos α = , then α = . 2 2 3
π 3 ⎛π ⎞ ⎛π π ⎞ . Thus, cos ⎜ − α ⎟ = cos ⎜ − ⎟ = cos = 6 2 ⎝2 ⎠ ⎝2 3⎠ x2 − 1 x
70.
sin(sec−1 x) =
72.
Let α = cos −1x, cosα = x, sinα = 1 − x 2 .
Let β = sin −1( − x), sin β = − x, cos β = 1 − x 2 .
Let β = cos −1 (− x), cos β = − x, sin β = 1 − x 2 .
sin −1 x + sin −1(− x) = α + β
cos −1 x + cos −1 (− x) = α + β
= sin −1 [sin(α + β )] = sin −1(sin α cos β + cosα sin β ) = sin −1 ⎡⎢ x 1 − x 2 + 1 − x 2 (− x) ⎤⎥ ⎣ ⎦ = sin −1 0 =0
= cos −1 [ cos(α + β )] = cos −1 (cos α cos β − sin α sin β ) = cos −1 ⎡ x(− x) − 1 − x 2 ⋅ 1 − x 2 ⎤ ⎢⎣ ⎥⎦ −1 2 2 = cos − x − 1 + x
(
−1
= cos (−1) =π
Copyright © Houghton Mifflin Company. All rights reserved.
)
406
73.
Chapter 6: Trigonometric Identities and Equations
1 1 Let α = tan −1x, tanα = x, β = tan −1 , tanβ = . x x −1 −1 1 =α + β tan x + tan x = tan −1 [ tan(α + β )]
74.
⎡ tan α + tan β ⎤ = tan −1 ⎢ ⎥ ⎣1 − tan α tan β ⎦ ⎡ x+ 1 ⎤ x ⎥ = tan −1 ⎢ ⎢⎣1 − x ⋅ 1x ⎥⎦ = tan Thus x = π 2 75.
−1
x 2 +1 x ,
1−1
= sin −1 [sin α cos β + cos α sin β ] = sin −1 ⎡ 1 − x 2 ⋅ 1 − x 2 + x ⋅ x ⎤ ⎢⎣ ⎥⎦ −1 2 2 = sin 1 − x + x
(
which is undefined
= sin 1 =π 2
The graph of y = sin −1( x) + 2 (shown as a black graph)
76.
The graph of y = sin −1( x + 1) − 2 (shown as a black graph)
The graph of y = 2cos −1 x (shown as a black graph) is the graph of y = cos −1 x (shown as a gray graph) stretched.
The graph of y = cos −1( x − 1) (shown as a black graph) is the graph of y = cos −1 x (shown as a gray graph) moved one unit to the right.
78.
is the graph of y = sin −1 x (shown as a gray graph) moved one unit to the left and two units down.
79.
)
−1
is the graph of y = sin −1 x (shown as a gray graph) moved two units up.
77.
Let sec −1 1 = α , secα = 1 , cosα = x, sinα = 1 − x 2 . x x Let csc −1 1 = β , cscβ = 1 , sin β = x, cos β = 1 − x 2 . x x −1 1 −1 1 + csc =α + β sec x x = sin −1 [sin(α + β )]
The graph of y = tan −1( x − 1) + 2 (shown as a black graph) is the graph of y = tan −1 x (shown as a gray graph) moved one unit to the right and two units up.
80.
The graph of y = −2 tan −1 x (shown as a black graph) is the graph of y = tan −1 x (shown as a gray graph) stretched and reflected through the x-axis.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.5
407
The graph of y = tan −1( x + 1) − 2 (shown as a black graph)
81.
82.
is the graph of y = tan −1 x (shown as a gray graph) moved one unit to the left and two units down.
83.
a.
s = 3960θ
the graph of y = sin −1 x (shown as a gray graph) moved two units to the right and one unit up.
b.
3960 a + 3960 θ = cos −1 3960 a + 3960 s = 3960cos −1 3960 a + 3960 cos θ =
(
The graph of y = sin −1( x − 2) + 1 (shown as a black graph) is
(
)
( a 3960 + 3960 ) ( a +3960 3960 )
5500 = 3960cos−1
5500 = cos −1 3960 5500 cos = 3960 3960 a + 3960 a + 3960 = 3960 cos 5500 3960 3960 − 3960 a= cos 5500 3960 a ≈ 17,930 mi
( )
)
( ) ( )
84. a.
85.
(Make sure you are in “radian” mode.)
[Note: f ( x) = cos −1 x is neither odd nor even.
b. Although the water rises 0.1 ft in each case, there is more surface area, and thus more volume of water, at the 4.9 to 5foot level near the diameter of the cylinder than at the 0.1 to 0.2-foot level near the bottom.
g(x) = sin −1 1 − x 2 is an even function.]
No, f ( x) ≠ g ( x) on the interval [−1, 1].
c. V (4) ≈ 352.04 ft 3. 3 d. If V = 288 ft , x ≈ 3.45 ft.
86.
87.
y = csc−1 2 x csc y = 2 x
−
y = cos(cos −1 x)
y ( x) = cos −1(cos x)
[Note: The domain of cos(cos −1 x) is − 1 ≤ x ≤ 1.
π
≤ y≤
π
, y≠0 2 2 2 x ≤ −1 or 2 x ≥ 1 1 1 x ≤ − or x ≥ 2 2
The domain of cos −1(cos x ) is all the real numbers.]
Copyright © Houghton Mifflin Company. All rights reserved.
408
88.
Chapter 6: Trigonometric Identities and Equations
y = 0.5sec −1 x 2 −1 x 2 y = sec 2 x sec 2 y = 2
89.
y≠
0< y<π , y ≠π 2 4 x ≤ −1 x ≥1 2 2 x ≤ −2 x≥2
90.
π
2 x − 1 ≤ −1 x≤0
0 < 2y < π, y ≠ π 2
91.
y = sec−1( x − 1) sec y = x − 1 0< y <π
y = 2 tan −1 2 x y = tan −1 2 x 2 tan y = 2 x y −π < < π 2 2 2 −π < y < π −∞ < 2 x < ∞
y = sec−1( x + π ) sec y = x + π 0< y <π y≠
2 x + π ≤ −1 x ≤ −π − 1
x −1 ≥ 1 x≥2
92.
π
y = tan −1( x − 1) tan y = x − 1 −
π 2
π
2 x − 1 → Graph is displaced one unit to the right.
....................................................... 93.
Let α = sin −1 x sin α = x
(
94.
)
cos sin −1 x = cos α =
1 − x2 1
= 1− x
2
Let α = sin −1 x sin α = x
Connecting Concepts 95.
sec(sin −1 x ) = sec α 1 =
1 − x2
2 = 1 − x2 1− x
x +π ≥1 x ≥1−π
Let α = csc −1 x csc α = x
96.
tan(csc −1 x ) = tan α 1 =
x2 − 1
=
x2 − 1 x2 − 1
Copyright © Houghton Mifflin Company. All rights reserved.
Let α = cot −1 x cot α = x
sin(cot −1 x ) = sin α 1 =
x2 + 1
=
x2 + 1 x2 + 1
Section 6.6
97.
409
5 x = tan −1 3 y tan 5 x = 3 y 1 y = tan 5 x 3
1 2 x = sin −1 2 y 2
98.
99.
4 x = sin −1 2 y sin 4 x = 2 y 1 y = sin 4 x 2
π
= cos −1( y − 3) 3 π⎞ ⎛ cos ⎜ x − ⎟ = y − 3 3⎠ ⎝ π⎞ ⎛ y = 3 + cos ⎜ x − ⎟ 3⎠ ⎝ x−
100.
2 y = 1 + tan ⎛⎜ x + π ⎞⎟ 2⎠ ⎝ y = 1 + 1 tan ⎛⎜ x + π ⎞⎟ 2 2 2⎠ ⎝
....................................................... PS1. x =
5 ± (−5) 2 − 4(3)(−4) 5 ± 73 = 2(3) 6
x + π = tan −1 (2 y − 1) 2 ⎛ tan ⎜ x + π ⎞⎟ = 2 y − 1 2⎠ ⎝
Prepare for Section 6.6 PS2. sin 2 x + cos 2 x = 1 sin 2 x = 1 − cos 2 x
PS3. π + 2(1)π = 5 π 2 2 π + 2(2)π = 9 π 2 2 π + 2(3)π = 13 π 2 2
⎛ ⎞ ⎛ ⎞ PS4. x 2 − 3 x + x − 3 = x ⎜ x − 3 ⎟ + 1⎜ x − 3 ⎟ 2 2 2 2 ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ 3 = ( x + 1) ⎜ x − ⎟ 2 ⎠ ⎝
PS5.
PS6.
2x2 − 2x = 0 2 x( x − 1) = 0
x = 0 x −1 = 0 x =1 The solutions are 0, 1.
Section 6.6 1.
2.
sec x − 2 = 0
2sin x = 3
sec x = 2 x=
5.
sin x =
π 7π , 4 4
x=
3.
tan x = 3
3 2
x=
π 2π ,
π 4π , 3 3
2sin x cos x = 3 sin x
2sin x cos x − 2 cos x = 0
2sin x cos x − 3 sin x = 0
cos x(2sin x − 2) = 0
sin x(2cos x − 3) = 0
cos x = 0
sin x = 0
2sin x − 2 = 0
2 2 π 3π π 3π x= , x= , 2 2 4 4 π π 3π 3π . The solutions are , , , 4 2 4 2 sin x =
cos x − 1 = 0 cos x = 1 x=0
3 3 6.
2sin x cos x = 2 cos x
4.
tan x − 3 = 0
2cos x − 3 = 0
3 2 π 11π x = 0, π x= , 6 6 π 11π . The solutions are 0, , π , 6 6
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cos x =
410
7.
Chapter 6: Trigonometric Identities and Equations
sin 2 x − 1 = 0
8.
sin 2 x = 1
cos 2 x = 1
sin x = ± 1 sin x = ±1 π 3π x= , 2 2 9.
cos x = ± 1 cos x = ±1 x = 0,π
4sin x cos x − 2 3 sin x − 2 2 cos x + 6 = 0
10.
sec x(sec x + 3) − 2(sec x + 3) = 0
(2cos x − 3)(2sin x − 2) = 0
(sec x + 3)(sec x − 2) = 0
2cos x − 3 = 0
sec x + 3 = 0
2sin x − 2 = 0
3 2 π 11π x= , 6 6
The solutions are
π π 3π 11π 6
,
4
,
4
,
6
14.
sin x − 1 = 0 sin x = 1
sin x = 1 2 5π 6
The solutions are
x=
π π 5π 6
,
2
,
6
2cos 2 x + 1 = −3cos x
π
2π 4π , 3 3 2π 4π The solutions are , π, . 3 3 x=
2
16.
cos 2 x = 3
cos x + 1 = 0 cos x = −1 x =π
cos x = − 1 2
.
4cos 2 x − 3 = 0
2sin 2 x − 1 = 0 sin 2 x = 1
2
4
sin x = ± 2
cos x = ± 3 2
π
3cot x + 3 = 0
2cos x + 1 = 0
2
5π 7π 11π x= , , , 6 6 6 6
x=
π 4
,
3π 5π 7π , , 4 4 4
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,
7π . 4
2cos 2 x + 3cos x + 1 = 0 (2cos x + 1)(cos x + 1) = 0
2sin x − 1 = 0
15.
4
cot x = −
2sin 2 x − 3sin x + 1 = 0 (2sin x − 1)(sin x − 1) = 0
6
, 2.1863, 4.0960,
π
3 3 2π 5π x= , 3 3
, 4 4
,
4
x=
3cot x = − 3
π 3π
π
π
The solutions are
12.
2sin 2 x + 1 = 3sin x
x=
sec x = 2
x ≈ 2.1863, 4.0969
.
csc x − 2 = 0
x=
sec x − 2 = 0
sec x = − 3
2 2 π 3π x= , 4 4
sin x =
csc x = 2
13.
sec 2 x + 3 sec x − 2 sec x − 6 = 0
2sin x(2cos x − 3) − 2(2cos x − 3) = 0
cos x =
11.
cos 2 x − 1 = 0
7π 4
Section 6.6
17.
411
2sin 3 x = sin x
4cos3 x = 3cos x
18.
2sin 3 x − sin x = 0
4cos3 x − 3cos x = 0
sin x(2sin 2 x − 1) = 0
cos x(4cos 2 x − 3) = 0 2sin 2 x = 1
sin x = 0 x = 0, π
sin x = ± 2
x=
2
π
3π 5π 7π x= , , , 4 4 4 4 π 3π 5π 7π The solutions are 0, , , π, , . 4 4 4 4 19.
4sin 2 x + 2 3 sin x − 3 = 2sin x
2 2
2
5π 7π 11π , , 6 6 6 π π 5π 7π 3π 11π The solutions are , , , , , . 6 2 6 6 2 6 x=
tan x(tan x + 1) − 3(tan x + 1) = 0
(2sin x + 3)(2sin x − 1) = 0
(tan x + 1)(tan x − 3) = 0
π 6
,
x=
π
6
,
5π 6
5π 4π 5π , , . 6 3 3
sin 4 x = sin 2 x
,
cos4 x = cos2 x
22.
sin 4 x − sin 2 x = 0
cos 4 x − cos 2 x = 0
sin 2 x(sin 2 x − 1) = 0 sin 2 x = 0 sin x = 0
6
tan x − 3 = 0 tan x + 1 = 0 tan x = −1 tan x = 3 3π 7π π 4π x= , x= , 4 4 3 3 π 3π 4π 7π The solutions are , , , . 3 4 3 4
2sin x − 1 = 0 1 sin x = 2
π
tan 2 x + tan x − 3 = 3 tan x
20.
2sin x(2sin x + 3) − (2sin x + 3) = 0
The solutions are
cos 2 x(cos 2 x − 1) = 0 sin 2 x − 1 = 0 sin x = ±1
cos 2 x = 0 cos x = 0
π
3π x = 0, π x= , 2 2 3π π The solutions are 0, , π , . 2 2
cos x − 0.75 = 0
24.
cos x = 0.75 x ≈ 41.4°, 318.6°
26.
cos x = ± 3
,
tan 2 x + tan x − 3 tan x − 3 = 0
3 sin x = − 2 4π 5π , x= 3 3
23.
4
π 3π
4sin 2 x + 2 3 sin x − 2sin x − 3 = 0
2sin x + 3 = 0
21.
cos 2 x = 3
cos x = 0
4cos x − 1 = 0 4cos x = 1 1 cos x = 4 x ≈ 75.5°, 284.5°
cos 2 x − 1 = 0 cos x = ±1 x = 0, π
π
3π x= , 2 2 The solutions are 0,
sin x + 0.432 = 0
25.
sin x = −0.432 x ≈ 205.6°, 334.4°
27.
3sec x − 8 = 0
π 2
, π,
3π . 2
3sin x − 5 = 0 3sin x = 5 sin x = 5 3 no solution
28.
4csc x + 9 = 0
3sec x = 8 sec x = 8 3 1 =8 cos x 3 cos x = 3 8 x ≈ 68.0°, 292.0°
Copyright © Houghton Mifflin Company. All rights reserved.
9 4 4 sin x = − 9 x ≈ 206.4°, 333.6°
csc x = −
412
Chapter 6: Trigonometric Identities and Equations
29.
cos x + 3 = 0 cos x = −3 no solution
30.
32.
4cos x − 5 = cos x − 3
33.
3cos x = 2 2 cos x = 3 x ≈ 48.2°, 311.8°
35.
3tan 2 x − 2 tan x = 0
sin x − 4 = 0 sin x = 4 no solution
1 2 3 3 sin x + = sin x + 2 3 4 5 1 3 2 − sin x = − 4 5 3 1 1 − sin x = − 4 15 4 sin x = 15 x ≈ 15.5°, 164.5° 36.
tan x(3tan x − 2) = 0 3tan x − 2 = 0 2 tan x = 3 x ≈ 33.7°, 213.7° The solutions are 0°, 33.7°, 180°, 213.7°. tan x = 0 x = 0, 180°
37.
3 cos x + sec x = 0 3 cos x +
38.
1 =0 cos x
3 cos 2 x + 1 = 0 cos 2 x = −
1 3
no solution
40.
csc2 x − 1 = 3 cot 2 x + 2 cot 2 x = 3 cot 2 x + 2 −2 cot 2 x = 2 cot 2 x = −1 no solution.
3 − 5sin x = 4sin x + 1 −9sin x = −2 2 sin x = 9 x ≈ 12.8°, 167.2°
31.
2 1 1 1 cos x − = − cos x 5 2 3 2 9 1 1 5 cos x = + = 10 3 2 6 25 cos x = 27 x ≈ 22.2°, 337.8°
34.
4cot 2 x + 3cot x = 0 cot x(4cot x + 3) = 0 cot x = 0
4cot x + 3 = 0
3 4 x = 90°, 270° x ≈ 126.9°, 306.9° The solutions are 90°, 126.9°, 270°, 306.9° cot x = 0
cot x = −
5sin x − csc x = 0 5sin x − 1 = 0 sin x 5sin 2 x − 1 = 0 sin 2 x = 1 5
tan 2 x = 3(1 + tan 2 x) − 2 tan 2 x = 3 + 3 tan 2 x − 2 −2 tan 2 x = 1 tan 2 x = −
sin x = ± 1 5 x = 26.6°, 153.4°, 206.6°, 333.4° 41.
tan 2 x = 3 sec 2 x − 2
39.
no solution
2 sin 2 x = 1 − cos x 2(1 − cos 2 x) = 1 − cos x 2 − 2 cos 2 x = 1 − cos x 0 = 2 cos 2 x − cos x − 1 0 = (2 cos x + 1)(cos x − 1) 2cos x + 1 = 0 1 cos x = − 2 x = 120°, 240°
cos x − 1 = 0 cos x = 1 x = 0°
The solutions are 0°, 120°, 240°.
Copyright © Houghton Mifflin Company. All rights reserved.
1 2
Section 6.6
42.
413
cos 2 x + 4 = 2 sin x − 3
43.
1 − sin 2 x + 4 = 2 sin x − 3
−5 ± 52 − 4(3)( −2) 2⋅3 − 5 ± 49 = = −5 ± 7 6 6 1 cos x = cos x = −2 3 x ≈ 70.5°, 289.5° no solution The solutions are 70.5°, 289.5°. cos x =
0 = sin 2 x + 2 sin x − 8 0 = (sin x + 4)(sin x − 2) sin x + 4 = 0 sin x − 2 = 0 sin x = −4 sin x = 2 no solution no solution There is no solution. 44.
2sin 2 x + 5sin x + 3 = 0 sin x =
45.
2 tan 2 x − tan x − 10 = 0 (tan x + 2)(2 tan x − 5) = 0
−5 ± 52 − 4(2)(3)
tan x + 2 = 0
2⋅2 −5 ± 1 −5 ± 1 = = 4 4
2 tan x − 5 = 0
5 2 x ≈ 116.6°, 296.6° x ≈ 68.2°, 248.2°
tan x = −2
sin x = − 3 2 no solution
sin x = −1 x = 270°
3 cos 2 x + 5 cos x − 2 = 0
tan x =
The solutions are 68.2°, 116.6°, 248.2°, 296.6°.
The solution is 270°. 46.
2cot 2 x − 7 cot x + 3 = 0
47.
3sin x cos x − cos x = 0 cos x(3sin x − 1) = 0 cos x = 0 3sin x − 1 = 0 x = 90°, 270° 1 sin x = 3 x ≈ 19.5°, 160.5° The solutions are 19.5°, 90°, 160.5°, 270°. 2 sin x cos x − sin x − 2 cos x + 1 = 0 sin x(2 cos x − 1) − (2 cos x − 1) = 0 (2 cos x − 1)(sin x − 1) = 0 2cos x − 1 = 0 sin x − 1 = 0 sin x = 1 cos x = 1 2 x = 90° x = 60°, 300° The solutions are 60°, 90°, 300°.
(2cot x − 1)(cot x − 3) = 0 2cot x − 1 = 0 1 cot x = 2 x ≈ 63.4°, 243.4°
cot x − 3 = 0 cot x = 3 x ≈ 18.4°, 198.4°
The solutions are 18.4°, 63.4°, 198.4°, 243.4°.
48.
tan x sin x − sin x = 0 sin x(tan x − 1) = 0 sin x = 0 tan x − 1 = 0 x = 0°, 180° x =1 x = 45°, 225° The solutions are 0°, 45°, 180°, 225°.
49.
50.
6 cos x sin x − 3 cos x − 4 sin x + 2 = 0 3 cos x(2 sin x − 1) − 2(2 sin x − 1) = 0 (2 sin x − 1)(3 cos x − 2) = 0
51.
2sin x − cos x = 1 2sin x − 1 = cos x (2sin x − 1)2 = (cos x )2 2
2sin x − 1 = 0 3cos x − 2 = 0 1 2 sin x = cos x = 2 3 x = 30°, 150° x ≈ 48.2°, 311.8° The solutions are 30°, 48.2°, 150°, 311.8°.
4sin x − 4sin x + 1 = cos2 x 4sin 2 x − 4sin x + 1 = 1 − sin 2 x 5sin 2 x − 4sin x = 0 sin x (5sin x − 4) = 0 sin x = 0 x = 180°
126.9° does not check. The solutions are 53.1°, 180°.
Copyright © Houghton Mifflin Company. All rights reserved.
5sin x − 4 = 0 sin x = 4 5 x ≈ 53.1° or 126.9°
414
52.
Chapter 6: Trigonometric Identities and Equations
sin x + 2cos x = 1
53.
sin x = 1 − 2cos x (sin x) 2 = (1 − 2cos x) 2
2sin x − 3cos x = 1 2sin x = 3cos x + 1 (2sin x )2 = (3cos x + 1)2 4sin 2 x = 9cos2 x + 6cos x + 1
sin 2 x = 1 − 4cos x + 4cos 2 x
4(1 − cos2 x ) = 9 cos2 x + 6cos x + 1
1 − cos 2 x = 1 − 4cos x + 4cos 2 x
0 = 13cos2 x + 6cos x − 3
0 = 5cos 2 x − 4cos x 0 = cos x(5cos x − 4) 5cos x − 4 = 0 4 cos x = 5 x ≈ 323.1° or 36.9°
cos x = 0 x = 90°
= −6 ± 192 26 cos x ≈ 0.3022 x ≈ 72.4° or 287.6°
36.9° does not check. The solutions are 90°, 323.1°.
54.
55.
3 sin x = 1 − cos x ( 3 sin x) = (1 − cos x)
cos x ≈ −0.7637 x ≈ 139.8° or 220.2°
287.6° and 139.8° do not check. The solutions are 72.4°, 220.2°.
3 sin x + cos x = 1 2
−6 ± 62 − 4(13)( −3) 2(13)
cos x =
3sin 2 x − sin x − 1 = 0 sin x =
2
3sin 2 x = 1 − 2 cos x + cos 2 x
1 ± ( −1)2 − 4(3)(−1) 2(3)
1 ± 13 6 sin x = 0.7676 =
3(1 − cos 2 x) = 1 − 2 cos x + cos 2 x 0 = 4 cos 2 x − 2 cos x − 2
sin x = −0.4343
x = 50.1°, 129.9°
The solutions are 50.1°, 129.9°, 205.7°, 334.3°.
0 = 2(2cos2 x − cos x − 1) 0 = 2(2cos x + 1)(cos x − 1) 2cos x + 1 = 0 cos x − 1 = 0 1 cos x = − cos x = 1 2 x = 120° or 240° x = 0° 240° does not check. The solutions are 0°, 120°. 56.
2cos 2 x − 5cos x − 5 = 0 cos x =
57.
5 ± ( −5)2 − 4(2)( −5) 2(2)
5 ± 65 4 cos x = 3.26 =
no solution
2cos x − 1 + 3sec x = 0 2cos x − 1 + 3 = 0 cos x 2cos 2 x − cos x + 3 = 0
1 ± (−1)2 − 4(2)(3) 2(2) = 1 ± −23 4 no solution
cos x = cos x = −0.7656 x = 140.0°, 220.0°
The solutions are 140.0°, 220.0°.
x = 205.7°, 334.3°
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.6
58.
415
3sin x − 5 + csc x = 0 1 3sin x − 5 + =0 sin x
59.
cos 2 x − 3sin x + 2sin 2 x = 0 1 − sin 2 x − 3sin x + 2sin 2 x = 0 sin 2 x − 3sin x + 1 = 0
2
3sin x − 5 sin x + 1 = 0
5 ± (−5)2 − 4(3)(1) sin x = 2(3)
sin x = =
5 ± 13 6 sin x = 1.4343 sin x = 0.2324 no solution x = 13.4°, 166.6° The solutions are 13.4°, 166.6°. =
60.
sin 2 x = 2cos x + 3cos 2 x 2
3± 5 2
sin x = 0.3820 x = 22.5°, 157.5° The solutions are 22.5°, 157.5°.
sin x = 2.6180 no solution
61.
2
1 − cos x = 2cos x + 3cos x 0 = 4cos 2 x + 2cos x − 1
tan 2 x − 1 = 0 tan 2 x = 1 2x =
−2 ± 22 − 4(4)(−1) cos x = 2(4) = −2 ± 20 8 cos x = 0.3090 x = 72°, 288°
3 ± (−3)2 − 4(1)(1) 2(1)
x=
π 4
π 8
+ kπ
+
kπ , where k is an integer 2
cos x = −0.8090 x = 144°, 216°
The solutions are 72.0°, 144.0°, 216.0°, 288.0°. 62.
2 3 =0 3 2 3 sec3x = 3
sec3x −
64.
+ 2kπ
cos 4 x = −
x=
3x =
or
2 2
3π + 2kπ 4 3π 1 x= + kπ 16 2
65.
5π + 2kπ 4 5π 1 x= + kπ 16 2 3π 1 5π 1 The solutions are + kπ , + kπ where k is an integer. 16 2 16 2 4x =
sin 5 x = 1 5x =
11π + 2kπ 6 6 11π 2 π 2 + kπ x = + kπ or x= 18 3 18 3 π 2 11π 2 The solutions are + kπ , + kπ , where k is an 18 3 18 3 integer. 3x =
π
63.
or
4x =
π 2
π
10
+ 2kπ 2 + kπ , where k is an integer 5
sin 2 x − sin x = 0 2 sin x cos x − sin x = 0 sin x(2 cos x − 1) = 0 sin x = 0 x = 0 + 2kπ or x = π + 2kπ
The solutions are 0 + 2kπ ,
where k is an integer. Copyright © Houghton Mifflin Company. All rights reserved.
2 cos x − 1 = 0 cos x = 1 2
x = π + 2 kπ 3 or x = 5π + 2kπ 3
π 3
+ 2kπ , π + 2kπ ,
5π + 2kπ 3
416
66.
Chapter 6: Trigonometric Identities and Equations
cos 2 x = − 2x = x=
3 2
67.
5π + 2kπ 6
or
5π + kπ 12
2x =
7π + 2kπ 6
7π + kπ 12
x=
π⎞ 1 ⎛ sin ⎜ 2 x + ⎟ = − 6 2 ⎝ ⎠ π 7π + 2kπ 2x + = 6 6 2 x = π + 2k π x=
5π 7π The solutions are + kπ , + kπ where k is an integer. 12 12
π 2
+ kπ
The solutions are
68.
π⎞ 2 ⎛ cos ⎜ 2 x − ⎟ = − 4⎠ 2 ⎝ π 3π + 2kπ or 2x − = 4 4 2 x = π + 2kπ
x=
π 2
+ kπ
The solutions are 70.
cos 2
π 2
+ kπ ,
69. 2x −
π
=
3π + kπ where k is an integer. 4
⎛ 1 + cos x ⎞ ⎜⎜ ± ⎟⎟ − cos x = 1 2 ⎝ ⎠ 1 + cos x − cos x = 1 2 1 + cos x − 2cos x = 2 − cos x = 1 cos x = −1 x = π + 2kπ where k is an integer 73.
cos 2 x = 1 − 3sin x
2 x = 0 + 2kπ or 2 x = π + 2kπ
72.
2cos 2 x − 1 = 2cos x − 1
2
2cos 2 x − 2cos x = 0 2cos x(cos x − 1) = 0
0 = sin x(2sin x − 3) sin x = 0
2sin x − 3 = 0 3 x = 0, π sin x = 2 no solution. The solutions are 0, π . 74.
2cos 2 x − 1 = 0 1 cos 2 x = 2 2x =
π
3
+ 2kπ or
5π 2x = + 2kπ 3 π π 5π x = 0 + kπ , + kπ , + kπ , + kπ 2 6 6 7π 3π 11π π π 5π The solutions are 0, , , , π, , , . 6 2 6 6 2 6
cos 2 x = 2cos x − 1
2
0 = 2sin x − 3sin x
sin 4 x − sin 2 x = 0 2sin 2 x cos 2 x − sin 2 x = 0 sin 2 x(2cos 2 x − 1) = 0 sin 2 x = 0
5π + kπ where k is an integer. 6
x + cos x = 1 2
1 − 2sin x = 1 − 3sin x
2
2
+ kπ ,
11π = + 2kπ 6 6 5π + 2kπ 2x = 3 5π + kπ x= 6
⎛ 1 − cos x ⎞ ⎜⎜ ± ⎟⎟ + cos x = 1 2 ⎝ ⎠ 1 − cos x + cos x = 1 2 1 − cos x + 2cos x = 2 cos x = 1 x = 0 + 2kπ where k is an integer
4 3π + 2kπ 2x = 2 3π + kπ x= 4
71.
sin 2
π
π
2x +
2
5π + 2kπ 4
x − cos x = 1 2
or
cos x = 0
x=
cos x − 1 = 0
π 3π 2
,
cos x = 1
2
x=0 3π The solutions are 0, , . 2 2
π
sin 4 x − cos 2 x = 0 2 sin 2 x cos 2 x − cos 2 x = 0 cos 2 x(2 sin 2 x − 1) = 0 cos2x = 0 2sin 2 x − 1 = 0 π 1 2 x = + 2kπ sin 2 x = 2 2 or π 2 x = + 2kπ 6 3π 2x = + 2kπ or 2 5π 2x = + 2kπ 6 π π 5π 3π x = + kπ , + kπ , + kπ , + kπ 12 4 12 4 π π 5π 3π 13π 5π 17π 7π The solutions are , , , , , , , . 12 4 12 4 12 4 12 4
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.6
75.
417
π
= sin x 2 1 − cos x = sin x sin x tan
76.
π
= 1 − cos x 2 sin x = 1 − cos x 1 + cos x tan
1 − cos x = sin 2 x
sin x = 1 − cos 2 x
1 − cos x = 1 − cos 2 x
sin x = sin 2 x
cos 2 x − cos x = 0
0 = sin 2 x − sin x 0 = sin x(sin x − 1) sin x = 0 sin x − 1 = 0 x = 0, π sin x = 1
cos x(cos x − 1) = 0 cos x = 0
x=
cos x = 1
π 3π 2
,
x=π 2
x=0
2
The solutions are 0,
π 2
,
π does not check.
3π . 2
The solutions are 0, π . 2
77.
sin 2 x cos x + cos 2 x sin x = 0 sin(2 x + x) = 0 sin 3 x = 0 3x = 0 + 2kπ or 3 x = π + 2kπ π 2 2 x = 0 + kπ or x = + kπ 3 3 3 π 2 4 5 The solutions are 0, , π , π , π π . 3 3 3 3
79.
sin x cos 2 x − cos x sin 2 x =
3 2 3 sin( x − 2 x) = 2 3 sin(− x) = 2
78.
cos 2 x cos x − sin 2 x sin x = 0 cos(2 x + x ) = 0 cos3 x = 0 π 3x = + 2kπ or 3 x = 3π + 2kπ 2 2 π π 2 x = + kπ or x = + 2 kπ 6 3 2 3 The solutions are π , π , 5π , 7π , 3π , 11π . 6 2 6 6 2 6
80.
cos 2 x cos x + sin 2 x sin x = −1 cos(2 x − x) = −1 cos x = −1 x =π
82.
cos 3 x + cos x = 0
3 2 4π 5π , x= 3 3
sin x = −
81.
sin 3 x − sin x = 0 3x + x 3x − 2 sin =0 2 2 2 cos 2 x sin x = 0
3x + x 3x − x =0 cos 2 2 2 cos 2 x cos x = 0
2 cos
2 cos
2(1 − 2 sin 2 x) sin x = 0
2(2 cos 2 x − 1) cos x = 0
sin x = 0 x = 0, π
1 − 2sin 2 x = 0 sin 2 x =
1 2
sin x = ±
3π , 4 4 π 3π 5π The solutions are 0, , , π, , 4 4 4
x=
π
,
2cos 2 x − 1 = 0
cos x = 0
x=
π 3π
cos x = ±
, 2 2
2 2
5π 7π , 4 4 7π . 4
x=
The solutions are
π 4
,
π 3π 5π 7π
2 2
, , , 4 4 4 4
3π π 5π 3π 7π , , , , . 4 2 4 2 4
Copyright © Houghton Mifflin Company. All rights reserved.
418
83.
Chapter 6: Trigonometric Identities and Equations
2sin x cos x + 2sin x − cos x − 1 = 0 2sin x(cos x + 1) − (cos x + 1) = 0 (cos x + 1)(2sin x − 1) = 0 cos x + 1 = 0 cos x = −1 x =π
84.
2sin x cos x − 2 2 sin x − 3 cos x + 6 = 0 2sin x(cos x − 2) − 3(cos x − 2) = 0 (cos x − 2)(2sin x − 3) = 0
2sin x − 1 = 0 sin x =
x=
1 2
π
6
,
cos x = 2 no solution
5π 6
π
5π The solutions are , , π. 6 6
sin x =
x=
The solutions are
π 3
,
3 2
π
3
,
2π . 3
85.
0.7391
86.
0, 1.8955
87.
−3.2957, 3.2957
88.
4.9172
89.
1.16
90.
0.5
91.
2π 3
(288)2 sin θ cosθ and d = 1295 for 0° ≤ θ ≤ 90°, − 500 ≤ d ≤ 3000. 16 Use the TRACE or INTERSECT feature of your graphing utility to determine the intersection of the two graphs.
Set your graphing utility to “degree” mode and graph d =
Thus, d = 1295 for θ ≈ 14.99° and θ ≈ 75.01°. 92.
Set your graphing utility to “degree” mode and graph d =
(375)2 sin θ cosθ . Use the MAXIMUM feature of your graphing utility 16
to determine the maximum horizontal range.
The maximum horizontal range is about 4394.5 ft., and it is produced when θ = 45°. The sine regression functions in Exercises 93 –98 were obtained on a TI-83/TI-83 Plus/TI-84 Plus calculator by using an iteration value of 16. The use of a different iteration factor may produce a sine regression function that varies from the regression functions listed below. 93. a.
f(x) = y ≈ 1.1213sin(0.01595 x + 1.8362) + 6.6257 b.
Set your graphing utility to “radian” mode. f (71) ≈ 1.1213sin(0.01595(71) + 1.8362) + 6.6257 ≈ 6.8186 hours ≈ 6 + .8186(60) → 6 : 49
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.6
94.
419
a.
f ( x ) = y ≈ 1.5347sin(0.01624 x − 1.1316) + 18.4118 b.
Set your graphing utility to “radian” mode. f (73) ≈ 1.5347sin(0.01624(73) − 1.1316) + 18.4118 ≈ 18.495 hours ≈ 18 + 0.495(60) → 18 : 30
95.
a.
f ( x ) = y ≈ 49.2125sin(0.2130 x − 1.4576) + 48.0550 b.
96.
Set your graphing utility to “radian” mode. f (31) ≈ 49.2125sin(0.2130(31) − 1.4576) + 48.0550 ≈ 3%
a.
f ( x ) = y ≈ 2.1350sin(0.01676 x − 1.3109) + 12.1197 b.
Set your graphing utility to “radian” mode. f (132) ≈ 2.1350sin(0.01676(132) − 1.3109) + 12.1197 ≈ 13.794 hours ≈ 13 + 0.794(60) → 13 hours 48 minutes
97.
a.
f ( x) = y ≈ 35.185sin(0.30395 x − 2.1630) + 2.1515 b.
Set your graphing utility to “radian” mode.
( 60 )
( 60 )
f 9 25 ≈ 35.185sin(0.30395 9 25 − 2.1630) + 2.1515 ≈ 24.8°
Copyright © Houghton Mifflin Company. All rights reserved.
420
98.
Chapter 6: Trigonometric Identities and Equations
d (t ) ≥ 10.75 for 28 ≤ t ≤ 314, so New Orleans will have at least 10.75 hours of daylight about 314 – 28 = 286 days of the year.
Xmin = 0, Xmax = 365, Xscl = 31, Ymin = 10, Ymax = 14, Yscl = 1
99.
a.
y , so y = 3sin θ 3 cos θ = x , so x = 3cosθ 3 1 A = 2 bh + 3 y 2 = (3sin θ )(3cosθ ) + 3(3sin θ ) = 9sin θ cos θ + 9sin θ = 9sin θ (cosθ + 1)
sin θ =
( )
b.
Make sure your calculator is in degree mode.
c.
The solutions are 42o and 79o . Make sure your calculator is in degree mode.
The value of θ that would maximize the area is 60o .
Copyright © Houghton Mifflin Company. All rights reserved.
Section 6.6
421
100. a.
b.
A B
tan π = 1 6 3 1 = 6d 3 d 2 − 8.75 2 d − 8.75 = 6 3d
d 2 − 6 3d − 8.75 = 0 Solve for d using quadratic formula.
First separate θ into A + B, where A is the angle above the horizontal for the observer and B is the angle below the horizontal. tan A = 3.5 d 2.5 tan B = d tan θ = tan( A + B) = tan A + tan B 1 − tan A tan B 3.5 + 2.5 d d tan θ = 1 − 3.5 2.5 d d d d 3.5 2.5 + tan θ = 2 d − 8.75 ⎞ θ = tan −1 ⎛⎜ 2 6d ⎟ ⎝ d − 8.75 ⎠
6 3 ± (−6 3) 2 − 4(1)(−8.75) 2(1) = 6 3 ± 108 + 35 2 ≈ 11.2 ft
d=
( )( )
101. θ = 20°
102.
cosθ = 2( x − 4) /( x + 4)
x = (4 + 18cot 20°)2 + 100 − (4 + 18cot 20°)
cos −1 2(2− 4) /(2+ 4) ≤ θ ≤ cos −1 2(1− 4) /(1+ 4)
x ≈ 0.93 ft.
cos −1 2−1/ 3 ≤ θ ≤ cos −1 2−3 / 5 37.5° ≤ θ ≤ 48.7°
θ = 30° x = (4 + 18cot 30°)2 + 100 − (4 + 18cot 30°) x ≈ 1.39 ft.
....................................................... 103.
104.
3 sin x + cos x = 3
a = 3, b = 1, k = ( 3)2 + 1 = 2, α is in first quadrant
α=
sin x − cos x = 1
k = 12 + 12 = 2 tan β =
1 3
tan β =
β=
Connecting Concepts
β=
π
−1 =1 1
π 4
α =−
6
π
x+π =π 6 3 x=π 6
4
fourth quadrant
π⎞ ⎛ 2 sin ⎜ x − ⎟ = 1 4⎠ ⎝
6
2sin ⎛⎜ x + π ⎞⎟ = 3 6⎠ ⎝ ⎛ π sin ⎜ x + ⎞⎟ = 3 6⎠ 2 ⎝
π
2 π⎞ ⎛ sin ⎜ x − ⎟ = 4⎠ 2 ⎝
x + π = 2π 6 3 x=π 2
x−
π 4
=
x=
π 4
π
2
Copyright © Houghton Mifflin Company. All rights reserved.
x−
π
3π = 4 4 x =π
422
105.
Chapter 6: Trigonometric Identities and Equations
106.
− sin x + 3 cos x = 3
k=
( −1)2 + (
3
)
2
tan β =
=2
β=
3 = 3 −1
(− 3)
k= tan β =
π
3 2π second quadrant α= 3
β=
6 third quadrant
⎞ ⎟ =1 ⎠ 5π ⎞ 1 ⎛ sin ⎜ x − ⎟= 6 ⎠ 2 ⎝ 5π π x− = 6 6
x=0
− π + 2π = 5π 3 3 The solutions are 0 and
2
+ ( −1) = 2
π
5π 6 5π ⎛ 2sin ⎜ x − 6 ⎝
x + 2π = 2π 3 3
x = −π 3
2
−1 1 = − 3 3
α =−
2sin ⎛⎜ x + 2π ⎞⎟ = 3 3 ⎠ ⎝ ⎛ ⎞= 3 2 π sin ⎜ x + ⎟ 3 ⎠ 2 ⎝ x + 2π = π 3 3
107.
− 3 sin x − cos x = 1
x =π
5π . 3
cos5 x − cos3 x = 0 5 x + 3x 5 x − 3x −2sin sin =0 2 2 −2sin 4 x sin x = 0 sin 4 x = 0 sin x = 0 4 x = 0 + 2kπ 4 x = π + 2kπ x = 0, π π 1 1 x = 0 + kπ x = + kπ 2 4 2 π π 3π 5π 7π 3π x = 0, , π , x= , , , 2 2 4 4 4 4 The solutions are 0,
π 4
,
π 2
,
3π 5π 3π 7π , π, , , . 4 4 2 4
cos5 x − cos x − sin 3 x = 0
108.
5x + x 5x − x −2sin sin − sin 3 x = 0 2 2 −2sin 3 x sin 2 x − sin 3 x = 0 sin 3x(−2sin 2 x − 1) = 0
sin 3x = 0
sin2x = −
3x = 0 + 2kπ
3 x = π + 2kπ
2 x = 0 + kπ 3 2 4π x = 0, π , 3 3 The solutions are 0,
π
2 + kπ 3 3 5π π x = , π, 3 3
x=
π 3
,
1 2
7π + 2kπ 6 7π x= + kπ 12 7π 19π , x= 12 12
2x =
11π + 2kπ 6 11π x= + kπ 12 11π 23π , x= 12 12
2x =
7π 2π 11π 4π 19π 5π 23π , , , π, , , , . 12 3 12 3 12 3 12
Copyright © Houghton Mifflin Company. All rights reserved.
x−
5π 5π = 6 6 5π x= 3
Assessing Concepts
423
.......................................................
Exploring Concepts with Technology
Approximate an Inverse Trigonometric Function with Polynomials 1.
y = f1 = x +
x3 2⋅3
y = f3 = x +
x3 1 ⋅ 3 x5 1 ⋅ 3 ⋅ 5 x 7 + + 2⋅3 2⋅4⋅5 2⋅4⋅6⋅7
where − 1 ≤ x ≤ 1
y = f2 = x +
f3 ( x ) − sin −1 x < 0.001 for − 0.6552 < x < 0.6552
4.
f6 ( x ) = x +
5.
6.
where − 1 ≤ x ≤ 1
x 3 1 ⋅ 3 x5 1 ⋅ 3 ⋅ 5 x 7 1 ⋅ 3 ⋅ 5 ⋅ 7 x9 + + + 2⋅3 2⋅ 4⋅5 2⋅ 4⋅6⋅7 2⋅ 4⋅6⋅8⋅9 where − 1 ≤ x ≤ 1
where − 1 ≤ x ≤ 1
2.
x3 1 ⋅ 3 x 5 + 2⋅3 2⋅4⋅5
y = f4 = x +
3.
f 4 ( x ) − sin −1 x < 0.001 for − 0.7186 < x < 0.7186
x3 1 ⋅ 3 x5 1 ⋅ 3 ⋅ 5 x7 1 ⋅ 3 ⋅ 5 ⋅ 7 x9 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 x11 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 ⋅ 11x13 + + + + + 2 ⋅ 3 2 ⋅ 4 ⋅ 5 2 ⋅ 4 ⋅ 6 ⋅ 7 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 9 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 11 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 ⋅ 13
1 3 5 35 63 231 + + + + + 6 40 112 1152 2816 13312 ≈ 1 + 0.167 + 0.075 + 0.045 + 0.030 + 0.022 + 0.017 Each term is much smaller than the previous term.
f6 (1) = 1 +
The largest-degree term in f10 ( x) is
1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 ⋅ 11 ⋅ 13 ⋅ 15 ⋅ 17 ⋅ 19 x 21 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 ⋅ 14 ⋅ 16 ⋅ 18 ⋅ 20 ⋅ 21
.......................................................
Assessing Concepts
1.
True
2.
False. cos −1 [cos(3π / 2) ] = cos−1 ( 0 ) = π / 2 ≠ 3π / 2.
3.
False. cos ( cos −1 2 ) ≠ 2 because cos −1 2 is undefined.
4.
True
5.
4
6.
Domain is − 1 ≤ x ≤ 1 . 2 2
Copyright © Houghton Mifflin Company. All rights reserved.
424
7.
9.
10.
Chapter 6: Trigonometric Identities and Equations
Range is 0 ≤ y ≤ π .
8.
( )
⎛ ⎞ sin −1 ⎜⎛ sin 7π ⎟⎞ = sin −1 ⎜ 3 ⎟ = π 3 ⎠ ⎝ 2 ⎠ 3 ⎝
2
tan 75° = tan ( 45° + 30° ) = tan 45° + tan 30° = 1 − tan 45° tan 30°
1+ 3 3 = 3+ 3 ⋅ 3+ 3 = 9 + 6 3 + 3 = 2 + 3 9−3 ⎛ 3 ⎞ 3− 3 3+ 3 1 − 1⎜ ⎟ ⎝ 3 ⎠
....................................................... 1.
cos(45° + 30°) = cos 45° cos30° − sin 45° sin 30° [6.2]
Chapter Review 2.
tan(210° − 45°) =
2 3 2 1 ⋅ − ⋅ 2 2 2 2 6 2 = − 4 4 6− 2 = 4 =
3.
2π π 2π π ⎛ 2π π ⎞ sin ⎜ cos + cos sin + ⎟ = sin 3 4 3 4 ⎝ 3 4⎠
4.
[6.2]
1 −1 1 −1 3 3 = 3 ⋅ 3 1 + 1 ⋅1 1 + 1 3 3
=
1− 3 1− 3 1− 2 3 + 3 ⋅ = 1− 3 3 +1 1− 3
=
4−2 3 = 3−2 −2
(3
3 2 1 2 ⋅ − ⋅ 2 2 2 2 6 2 = − 4 4 6− 2 = 4
=
3⎛ 2⎞ 1 2 ⎜⎜ − ⎟− ⋅ 2 ⎝ 2 ⎟⎠ 2 2
=−
6 2 6+ 2 − =− 4 4 4
[6.2]
1 3
4
3
4
1 −4 = ⋅ − 1 ⋅ 2 + ⎛⎜ − 3 ⎞⎟ ⋅ 2 −4 2 2 ⎝ 2 ⎠ 2
=
=
4)
cos 4π cos π + sin 4π sin π
2− 6 −4 ⋅ 2+ 6 2− 6
=
sin(60° − 135°) = sin 60° cos135° − cos 60° sin135°
[6.2]
=
1 ⎛ 4π π ⎞ sec ⎜ − ⎟= ⎝ 3 4 ⎠ cos 4π − π
=
5.
tan 210° − tan 45° 1 + tan 210° tan 45°
[6.2]
6.
⎛ 5π 7π − cos ⎜ 4 ⎝ 3
−4
(
2− 6 2−6
)=
2− 6
5π 7π 5π 7π ⎞ + sin sin ⎟ = cos cos 3 4 3 4 ⎠ 1 2 ⎛ 3 ⎞⎛ 2⎞ = ⋅ + ⎜− ⎟⎜ − ⎟⎟ 2 2 ⎜⎝ 2 ⎟⎜ 2 ⎠⎝ ⎠ =
2 6 2+ 6 + = 4 4 4
Copyright © Houghton Mifflin Company. All rights reserved.
[6.2]
Chapter Review
7.
425
sin 22.5° = sin 45° 2 1 − cos 45° = 2
8.
cos105° = cos 210° 2 1 cos 210° + =− 2
1− 2
=−
2
=
9.
[6.3]
2 2− 2 4
=− 2− 3 4
=
2− 2 2
=−
tan 67.5° = tan 135° 2 − ° 1 cos135 = sin135°
=
[6.3]
10.
( )
1− −
1+
1 2
= 2+ 2 4
1 2
=
3 3 − =0 4 4 2 tan α
1 − tan 2 α 2 3 2 ⎛⎜ 3 ⎞⎟ 3 3 = ⎝ ⎠ = 3 ⋅ 2 1− 1 3 3 1 − ⎛⎜ 3 ⎞⎟ ⎝ 3 ⎠
=
c.
sin
β 2
= =
=
1 − cos β 2 1− 1
1 = 4 1 = 2
2
3 ⎛ 1 ⎞ ⎛ 1 ⎞⎛ 3⎞ ⋅ ⎜ − ⎟ + ⎜ − ⎟ ⎜⎜ − ⎟ 2 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎟⎠
3 3 + =0 4 4 1 1 sec 2 β = = cos 2 β cos 2 β − sin 2 β 1 = 2 2 − 1 − ⎛⎜ − 3 ⎞⎟ 2 ⎝ 2 ⎠ 1 1 = = = −2 1 − 3 −2 =−
b.
( )
2 3 = 3 3 −1
2
2+ 2 2
[6.2/6.3] sin α = 3 ,cos α = − 1 ,quadrant II 2 2 cos β = − 1 ,sin β = − 3 ,quadrant III 2 2 a. sin(α + β ) = sin α cos β + cosα sin β
=
tan 2α =
2 2
2
3 1 1⎛ 3⎞ ⋅ + ⎜− ⎟ 2 2 2 ⎜⎝ 2 ⎟⎠
=
[6.3]
( )
1− −
=
= 2 +1
b.
2− 3 2
1 2
12.
)
sin112.5° = sin 225° 2 1 cos 225° − = 2
1 2
[6.2/6.3] sin α = 1 ,cos α = 3 ,quadrant I, tan α = 3 2 2 3 cos β = 1 ,sin β = − 3 ,quadrant IV, tan β = − 3 2 2 a. cos (α − β ) = cosα cos β + sin α sin β
3 2
2
=
=
11.
(
1+ −
[6.3]
4
c.
cos
α 2
= =
4
4
1 + cosα 2
( 2) =
1+ − 1 2
1 = 4 1 = 2 Copyright © Houghton Mifflin Company. All rights reserved.
1 2
2
426
13.
Chapter 6: Trigonometric Identities and Equations
1 3 3 [6.2/6.3] 14. sin α = − , cosα = , quadrant IV, tan α = − 2 2 3 3 1 cos β = − , sin β = − , quadrant III 2 2 a. sin(α − β ) = sin α cos β − cos α sin β
2 2 ,cos α = , quadrant I 2 2 3 1 3 cos β = ,sin β = − , quadrant IV, tan β = − 2 2 3 a. cos (α − β ) = cosα cos β + sin α sin β sin α =
1⎛ 3⎞ 3 ⎛ 1⎞ = − ⎜⎜ − ⎟− ⎜− ⎟ 2 ⎝ 2 ⎟⎠ 2 ⎝ 2 ⎠ 3 3 2 3 + = 4 4 4 3 = 2 2 tan α =
b.
tan 2α =
b.
2
1 − tan α
2 ⎛⎜ − 3 ⎞⎟ ⎝ 3 ⎠ = 2 1 − ⎛⎜ − 3 ⎞⎟ 3 ⎝ ⎠ =
β
−2 3 3 3 ⋅ 1− 1 3
2
=−
=−
2− 3 2
c.
17.
sin 4 x cos x − cos 4 x sin x = sin(4 x − x) [6.2] = sin 3 x
[6.1]
2
2 2 ⋅ 2 2
=1
2sin 3 x cos3 x = sin 2(3 x) = sin 6 x
sin 2θ = tan 2θ cos 2θ
2 ⎛⎜ − 3 ⎞⎟ ⎝ 3 ⎠
sin 2α = 2sin α cosα =2
15.
19.
1 − tan 2 β
3
1 + ⎛⎜ − 3 ⎞⎟ ⎝ 2 ⎠ =− 2 2− 3 4
6 2 6− 2 − = 4 4 4
2 3 = 3 −1 =− 3
1 + cos β 2
=−
=
1 − ⎛⎜ − 3 ⎞⎟ ⎝ 3 ⎠ −2 ⎛⎜ 3 ⎞⎟ 3 3 = ⎝ ⎠⋅ 3 1− 1
3
cos
2 3 2⎛ 1⎞ ⋅ + ⎜− ⎟ 2 2 2 ⎝ 2⎠
2 tan β
=
−2 3 = 3 −1 =− 3
c.
tan 2 β =
=
[6.3]
16.
18.
20.
tan 2 x + tan x = tan(2 x + x) 1 − tan 2 x tan x = tan 3 x cos 2 2θ − sin 2 2θ = cos 2(2θ ) = cos 4θ
[6.2]
[6.3]
2 2 1 − cos 2θ = 1 − (cos θ − sin θ ) sin 2θ 2sin θ cosθ 2 2 2 2 = sin θ + cos θ − cos θ + sin θ 2sin θ cos θ
2sin 2 θ 2sin θ cos θ = sin θ cos θ = tan θ =
Copyright © Houghton Mifflin Company. All rights reserved.
[6.3]
[6.2/6.3]
Chapter Review
21.
427
2θ + 4θ 2θ − 4θ sin 2 2 = −2sin 3θ sin(−θ )
cos 2θ − cos 4θ = −2sin
[6.4]
22.
= 2sin 3θ sin θ
23.
25.
3θ + 5θ 3θ − 5θ sin 2 2 = 2cos 4θ sin(−θ )
sin 3θ − sin 5θ = 2cos
[6.4]
= −2cos 4θ sin θ
6θ + 2θ 6θ − 2θ cos 2 2 = 2sin 4θ cos 2θ
sin 6θ + sin 2θ = 2sin
[6.4]
1 1 (sin x + 1) + (sin x − 1) + = sin x − 1 sin x + 1 (sin x − 1)(sin x + 1) 2sin x = sin 2 x − 1 2sin x = − cos 2 x = −2 tan x sec x
24.
26.
5θ + θ 5θ − θ sin 2 2 = 2cos3θ sin 2θ
sin 5θ − sin θ = 2cos
[6.4]
sin x = sin x(1 + cos x) 1 − cos x (1 − cos x)(1 + cos x) = sin x + sin x cos x 1 − cos 2 x x + sin x cos x sin = sin 2 x = sin x + sin x cos x sin 2 x sin 2 x
= csc x + cot x, 0 < x < π 2
27.
1 + sin x 2
cos x
=
1 2
cos x
+
sin x
28.
2
cos x
2
= sec x + tan x sec x
= cos 2 x − sin 2 x − sin 2 x
= tan 2 x + 1 + tan x sec x 29.
30.
31.
32.
cos 2 2 x − sin 2 2 x (cos 2 x − sin 2 x)(cos 2 x + sin 2 x) = cos 2 x + sin 2 x cos 2 x + sin 2 x = cos 2 x − sin 2 x
1 − cos x = 1 − cos 2 x cos x cos x 2 x sin = cos x = tan x sin x sin(270° − θ ) − cos(270° − θ ) = sin 270° cosθ − cos 270° sin θ − cos 270° cosθ − sin 270° sin θ = (−1) cos θ − 0 − 0 − (−1)sin θ = − cos θ + sin θ = sin θ − cosθ sin ⎛⎜ π − α ⎞⎟ = sin π cos α − cos π sin α 4 4 ⎝4 ⎠ 2 2 = cos α − sin α 2 2 = 2 (cos α − sin α ) 2
sin(180° − α + β ) = sin [180° − (α − β )] = sin180° cos(α − β ) − cos180° sin(α − β ) = 0 [ cos(α − β )] − (−1) [sin α cos β − cosα sin β ] = sin α cos β − cosα sin β
33.
2cos 4 x + 2 x sin 4 x − 2 x sin 4 x − sin 2 x 2 2 = cos 4 x − cos 2 x −2sin 4 x + 2 x sin 4 x − 2 x 2
cos3 x sin x =− sin 3 x sin x = − cot 3 x
2
34.
2sin x sin 3x = cos( x − 3x ) − cos( x + 3x ) = cos 2 x − cos 4 x = cos 2 x − (2cos2 2 x − 1) = 1 + cos 2 x − 2cos2 2 x = (1 − cos 2 x )(1 + 2cos 2 x )
Copyright © Houghton Mifflin Company. All rights reserved.
428
35.
Chapter 6: Trigonometric Identities and Equations
sin x − cos 2 x = sin x − (1 − 2sin 2 x)
36.
cos 4 x = 1 − 2sin 2 2 x
= 2sin 2 x + sin x − 1 = (2sin x − 1)(sin x + 1)
= 1 − 2(2sin x cos x)2 = 1 − 2(4sin 2 x cos 2 x) = 1 − 8sin 2 x cos 2 x = 1 − 8sin 2 x(1 − sin 2 x) = 1 − 8sin 2 x + 8sin 4 x
37.
tan 4 x =
2 tan 2 x
38.
2
1 − tan 2 x
2x + x 2x − x sin 2 x − sin x 2cos 2 sin 2 = cos 2 x + cos x 2cos 2 x + x cos 2 x − x 2
⎛ ⎞ 2 ⎜ 2 tan x ⎟ 1 − tan 2 x ⎠ ⎝ = 2 ⎛ ⎞ 1 − ⎜ 2 tan x ⎟ 2 ⎝ 1 − tan x ⎠
=
= = 39.
41.
2
cos π
2
= tan
π
2 1 − cos x = sin x
4 tan x (1 − tan 2 x)2 − 1 tan 2 x = ⋅ 2 2 2 (1 − tan x ) − (2 tan x ) (1 − tan 2 x) 2 (1 − tan 2 x )2
=
sin π
4 tan x(1 − tan 2 x) (1 − tan 2 x)2 − 4 tan 2 x 4 tan x − 4 tan 3 x 1 − 2 tan 2 x + tan 4 x − 4 tan 2 x 4 tan x − 4 tan 3 x 1 − 6 tan 2 x + tan 4 x
2sin 3 x cos3 x − 2sin x cos x = sin 6 x − sin 2 x 6x + 2x 6x − 2x = 2cos sin 2 2 = 2cos 4 x sin 2 x
40.
2sin x sin 2 x = 2sin x(2sin x cos x) = 4cos x sin 2 x
cos( x + y ) cos( x − y ) = 1 [cos( x + y + x − y ) + cos( x + y − x + y ) ] 2 = 1 (cos 2 x + cos 2 y ) 2 = 1 ⎡⎣ 2 cos2 x − 1 + 2 cos2 y − 1⎤⎦ 2 = cos2 x + cos2 y − 1
42.
cos( x + y )sin( x − y ) = 1 [sin( x + y + x − y ) − sin( x + y − x + y )] 2 1 = (sin 2 x − sin 2 y ) 2 = 1 [2sin x cos x − 2sin y cos y ] 2 = sin x cos x − sin y cos y
43.
y = sec ⎛⎜ sin −1 12 ⎞⎟ , α = sin −1 12 , sin α = 12 , cos α = 5 , sec α = 13 [6.5] 13 ⎠ 13 13 13 5 ⎝ 13 y = sec α = 5
Copyright © Houghton Mifflin Company. All rights reserved.
2
Chapter Review
44.
429
3⎞ ⎛ y = cos ⎜ sin −1 ⎟ 5 ⎝ ⎠ 3 4 −1 3 , sin α = , y = cosα = α = sin 5 5 5
45.
[6.5]
α = sin −1 ⎛⎜ − 3 ⎞⎟
β = cos −1 5
⎝ 5⎠
13
cos β = 5 13 sin β = 12 13
sin α = − 3 5 4 cos α = 5
[6.5]
⎡ ⎤ y = cos ⎢sin −1 ⎛⎜ − 3 ⎞⎟ + cos −1 5 ⎥ 13 ⎦ ⎝ 5⎠ ⎣ = cos(α + β ) = cos α cos β − sin α sin β = 4 ⋅ 5 − ⎛⎜ − 3 ⎞⎟ ⋅ 12 5 13 ⎝ 5 ⎠ 13 = 20 + 36 65 65 56 = 65
46.
y = cos ⎛⎜ 2sin −1 3 ⎞⎟ 5⎠ ⎝ y = cos 2α 2
2
= cos α − sin α 2
= ⎛⎜ 4 ⎞⎟ − ⎛⎜ 3 ⎞⎟ ⎝ 5⎠ ⎝5⎠ = 16 − 9 25 25 = 7 25
48.
2
α = sin −1
3 5
[6.5]
47.
sin −1 x + α = π 2
sin −1 x = π − α 2 x = sin ⎛⎜ π − α ⎞⎟ ⎝2 ⎠
x − 1 = sin π 6 1 x −1 = 2 x=3 2
α = cos −1 4 5
cos α = 4 5 3 sin α = 5
[6.6]
49.
4sin 2 x + 2 3 sin x − 2sin x − 3 = 0 [6.6]
(
The solutions are 0°, 45°, 135°.
1 2 x = 30°,150°
sin x =
The solutions are 30°, 150°, 240°, 300°.
[6.6]
51.
3cos 2 x + sin x = 1
cos x − 1 = 0 cos x = 1 x = 0°
[6.6]
3(1 − sin 2 x) + sin x = 1 0 = 3sin 2 x − sin x − 2 0 = (3sin x + 2)(sin x − 1)
(2sin x − 2)(cos x − 1) = 0 2 2 x = 45°,135°
2sin x − 1 = 0
2sin x + 3 = 0
cos x(2sin x − 2) − (2sin x − 2) = 0
sin x =
)
( 2sin x + 3 ) ( 2sin x − 1) = 0
3 sin x = − 2 x = 240°,300°
2sin x cos x − 2 cos x − 2sin x + 2 = 0
2sin x − 2 = 0
) (
2sin x 2sin x + 3 − 2sin x + 3 = 0
= sin π cos α − cos π sin α 2 2 = 1⋅ 4 − 0 ⋅ 3 5 5 =4 5
50.
[6.6]
sin −1 ( x − 1) = π 6
3 5 4 cos α = 5 sin α =
sin −1 x + cos −1 4 = π 5 2
2sin −1 ( x − 1) = π 3
3sin x + 2 = 0 2 sin x = − 3
sin x − 1 = 0
x = 3.8713 or 5.553
x=π 2
sin x = 1
The solutions are π + 2kπ , 3.8713 + 2kπ , 5.553 + 2kπ 2 where k is an integer.
Copyright © Houghton Mifflin Company. All rights reserved.
430
52.
Chapter 6: Trigonometric Identities and Equations
tan 2 x − 2 tan x − 3 = 0 [6.6] (tan x + 1)(tan x − 3) = 0 tan x = −1 tan x = 3 x=−
π 4
53.
The solutions are −
π 4
2x = x=
π 6
5π + 2kπ 6 5π x= + kπ 12
+ 2kπ
π 12
2x =
+ kπ
The solutions are
54.
π⎞ 3 ⎛ cos ⎜ 2 x − ⎟ = − 3⎠ 2 ⎝ π 5π 2x − = + 2kπ 3 6 7π 2x = + 2kπ 6 7π x= + kπ 12 The solutions are
56.
55.
[6.6] 2x −
π 5π 13π 17π . , , , 12 12 12 12
f ( x) = 3 sin x + cos x [6.4]
π⎞ ⎛ f ( x) = 2sin ⎜ x + ⎟ 6⎠ ⎝ amplitude = 2
7π + 2kπ 6
phase shift = −
π 6
7π 19π 3π 7π , , , . 12 12 4 4
f ( x) = −2sin x − 2cos x
[6.4]
57.
f ( x) = − sin x − 3 cos x
58.
[6.4]
4π ⎞ ⎛ f ( x) = 2sin ⎜ x + ⎟ 3 ⎠ ⎝ amplitude = 2 4π phase shift = − 3
amplitude = 2 2 5π phase shift = − 4
f ( x) = 2cos −1 x
=
3 3π 2x = + 2kπ 2 3π x= + kπ 4
5π ⎞ ⎛ f ( x) = 2 2 sin ⎜ x + ⎟ 4 ⎠ ⎝
59.
π
[6.6]
1 2
sin 2 x =
+ kπ , 1.2490 + kπ where k is an integer.
1 2
1 2
sin ( 3 x − x ) =
x = 1.2490 + kπ
+ kπ
sin 3 x cos x − cox3x sin x =
60.
f ( x ) = sin −1( x − 1)
61.
f ( x) =
3 1 sin x − cos x 2 2
(
)
[6.4]
(
f ( x ) = sin x + 11π or sin x − π 6
amplitude = 1 phase shift = π
6
f ( x) = sin −1
x 2
Copyright © Houghton Mifflin Company. All rights reserved.
62.
f ( x ) = sec −1 2 x
6
)
Quantitative Reasoning
63.
431
a.
b.
y ≈ 1.1835sin(0.01600 x + 1.8497) + 6.4394 y ≈ 1.1835sin(0.01600(104) + 1.8497) + 6.4394 ≈ 6.009 = 6 : 01
[6.6]
....................................................... QR1. a.
s=
( ) (
(28)2 cos α sec π sin α − π 18 18 16
Quantitative Reasoning
) α ≈ 0.87266
25
0.1745
1.35
–8 π
b.
π + 18 = π + π = 9π + π = 5π ≈ 0.8727 4
2
4
36
36
18
QR2. cos α sec β sin(α − β ) = ⎛⎜ cos α ⋅ 1 ⎞⎟ ( sin α cos β − cos α sin β ) cos β ⎠ ⎝ sin β = cos α sin α − cos2 α ⋅ cos β = cos α ⋅ cos α sin α − cos2 α ⋅ tan β cos α = cos2 α ⋅ sin α − cos2 α tan β cos α = cos2 α ( tan α − tan β )
s=
v 2 cos α sec β sin (α − β ) 16
s=
v 2 cos2 α ( tan α − tan β ) 16
16s = v 2 cos2 α ( tan α − tan β ) v2 = v=
16s cos2 α ( tan α − tan β ) 16s cos2 α ( tan α − tan β )
v = 4sec α
a.
v = 4sec α
s tan α − tan β
15
tan α − tan π 18
α ≈ 0.87266
35
0.1745
1.35
–8 π
b.
π + 18 = π + π = 9π + π = 5π ≈ 0.8727 4
2
4
36
36
18
Copyright © Houghton Mifflin Company. All rights reserved.
432
Chapter 6: Trigonometric Identities and Equations
....................................................... 1.
1 + sin 2 x sec2 x = 1 + sin 2 x
1
Chapter Test 2.
2
cos x
1 1 sec x + tan x − sec x + tan x − = sec x − tan x sec x + tan x sec2 x − tan 2 x 2 tan x 1 = 2 tan x
= 1 + tan 2 x
=
2
= sec x
1 cos x − sin x sin x 1 − cos x = sin x
3.
cos3 x + cos x sin 2 x = cos x(cos 2 x + sin 2 x) = cos x
4.
csc x − cot x =
5.
sin195° = sin(150° + 45°) [6.2] = sin150° cos 45° + cos150° sin 45°
6.
3 4 2 2 sin α = − ,cos α = − ,cos β = − ,sin β 5 5 2 2 sin(α + β ) = sin α cos β + cosα sin β
=
1⎛ 2⎞ ⎛ 3 ⎞⎛ 2 ⎞ ⎜ ⎟ + ⎜− ⎟⎜ ⎟ 2 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
=
7.
3π ⎛ sin ⎜θ − 2 ⎝
[6.2]
2 ⎞ ⎛ 4 ⎞⎛ 2 ⎞ ⎛ 3 ⎞⎛ = ⎜ − ⎟ ⎜⎜ − ⎟ + ⎜ − ⎟⎜ ⎟ ⎝ 5 ⎠ ⎝ 2 ⎟⎠ ⎝ 5 ⎠ ⎜⎝ 2 ⎟⎠
2− 6 4
3 2 −4 2 10 2 =− 10 =
3π 3π ⎞ − cos θ sin ⎟ = sin θ cos 2 2 ⎠ = sin θ ( 0 ) − cos θ ( −1)
8.
cos 6 x sin 3 x + sin 6 x cos 3x = sin ( 6 x + 3x ) [6.2] = sin 9 x
= cos θ 9.
sin θ =
4 3 , cos θ = − 5 5
[6.3]
2
2
4
= 4 sin x cos x + 4 cos x = 4 cos 2 x(sin 2 x + cos 2 x) = 4 cos 2 x
cos θ 1 − cos θ cos θ = + sin θ sin θ sin θ 1 − cos θ + cos θ = sin θ = csc θ
12.
sin15° cos 75° = 1 [sin(15° + 75°) + sin(15° − 75°)] 2 = 1 (sin 90° − sin 60°) 2 ⎛ ⎞ 1 = ⎜1 − 3 ⎟ 2⎝ 2 ⎠ 3 1 = − 2 4 = 2− 3 4
2
⎛ 3⎞ = 2 ⎜ − ⎟ −1 ⎝ 5⎠ 18 7 = −1 = − 25 25 sin 2 2 x + 4 cos 4 x = (2 sin x cos x) 2 + 4 cos 4 x
+
tan
cos 2θ = 2 cos 2 θ − 1
11.
θ
10.
2
Copyright © Houghton Mifflin Company. All rights reserved.
[6.4]
Chapter Test
13.
433
y=−
3 1 sin x + cos x 2 2
a=−
3 1 , b= 2 2
14.
[6.4]
θ = cos−1(0.7644) θ = 0.701
12 ⎞ ⎛ sin ⎜ cos −1 ⎟ [6.5] 13 ⎠ ⎝ 12 Let θ = cos−1 and find sinθ . 13 12 Then cosθ = and 0 ≤ θ ≤ π . 13 5 sin θ = 13 12 ⎞ 5 ⎛ sin ⎜ cos −1 ⎟ = 13 ⎠ 13 ⎝
15.
[6.5]
2
2 ⎛ 3⎞ ⎛1⎞ +⎜ ⎟ =1 k = ⎜− ⎟ ⎜ 2 ⎟ ⎝2⎠ ⎝ ⎠ 1
1 sin β = 2 = 1 2
β=
α =π −
π 6
π 6
5π = 6 5π ⎞ ⎛ y = sin ⎜ x + ⎟ 6 ⎠ ⎝ 16.
The graph of y = sin −1( x + 2) is the graph of y = sin −1 x moved two units to the left.
17.
3sin x − 2 = 0 2 sin x = 3 x = 41.8°, 138.2°
18.
sin x cos x − 3 sin x = 0 [6.6] 2 ⎛ ⎞ sin x ⎜ cos x − 3 ⎟ = 0 2 ⎠ ⎝
19.
sin 2 x + sin x − 2cos x − 1 = 0 2sin x cos x + sin x − 2cos x − 1 = 0 sin x(2cos x + 1) − (2cos x + 1) = 0 (2cos x + 1)(sin x − 1) = 0
cos x − 3 = 0 2
sin x = 0 x = 0, π
cos x = 3 2
x = π , 11π 6 6
π
11π The solutions are 0, , π , . 6 6 20.
cos x = − x=
1 2
sin x = 1
2π 4π , 3 3
The solutions are
x=
π 2
,
2π 4π , . 3 3
a.
b.
[6.6]
f ( x ) = y ≈ 1.7569sin(0.0168 x − 1.3056) + 12.1100 f (75) ≈ 1.7569sin(0.01675(75) − 1.3056) + 12.1100 [6.6] ≈ 12.0233 ≈ 12 hr 1 min
Copyright © Houghton Mifflin Company. All rights reserved.
π 2
434
Chapter 6: Trigonometric Identities and Equations
.......................................................
Cumulative Review
1.
x3 − y 3 = ( x − y )( x 2 + xy + y 2 ) [P.4]
2.
x −5 =3 [1.1] x − 5 = −3 x − 5 = 3 x=2 x =8
3.
Shift the graph of y = f (x) horizontally 1 unit to the left and up 2 units. [2.5]
4.
Reflect the graph of y = f (x) across the x-axis. [2.5]
5.
x − 2 = 0 [3.5] x=2
8.
11.
6.
x = 25 [4.3] log 2 x = 5 5π = 5π ⎛ 180o ⎞ = 300o ⎜ ⎟ 3 3 ⎝ π ⎠
[5.1]
f (− x) = − x − sin(− x) [2.5/5.5] = − x + sin x = −( x − sin x) = − f ( x) odd function
9.
log10 1000 = log10 103 [4.3] = 3log10 10 = 3
12.
sin θ =
opp 2 = hyp 3
[5.2]
7.
10.
13.
f ( x) = 5 x x −1 y = 5x x −1 5y x= y −1 x( y − 1) = 5 y xy − 5 y = x y ( x − 5) = x y= x x−5 f −1 ( x) = x x−5
[4.1]
⎛ ⎞ 240o = 240o ⎜ π o ⎟ = 4π ⎝ 180 ⎠ 3
[5.1]
cot θ > 0 in quadrant III Positive [5.3]
adjacent side = 32 − 22 = 9−4 = 5 opp = 2 =2 5 tan θ = adj 5 5 14.
θ = 310o
[5.3] o
15. o
Since 270 < θ < 360 ,
θ = θ ′ = 360o θ ′ = 50o
θ = 5π 3
[5.3]
Since 3π < θ < 2π , 2 θ = θ ′ = 2π
θ′ = π
3
16.
t=π 3
[5.4]
y = sin t = sin π = 3 3 2 x = cos t = cos π = 1 3 2 The point on the unit circle corresponding to t = π is ⎛⎜ 1 , 3 ⎞⎟ . 3 ⎝2 2 ⎠
Copyright © Houghton Mifflin Company. All rights reserved.
Cumulative Review
17.
435
y = 0.43cos ⎛⎜ 2 x − π ⎞⎟ [5.7] 6⎠ ⎝ amplitude: 0.43
18.
0 ≤ 2 x − π ≤ 2π 6 π ≤ 2 x ≤ 13π 6 6 π ≤ x ≤ 13π 12 12
y = sin −1 1 2 sin y = 1 −π ≤ y≤π 2 2 2 y=π 6
period = π , phase shift = π 12
19.
Domain: [–1, 1].
[6.5]
20.
(
Range: − π , π 2 2
)
[6.5]
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[6.5]
Chapter 7
Applications of Trigonometry Section 7.1 1.
2.
o
o
A = 180o − 125o − 25o
o
C = 180 − 42 − 61
A = 30o
C = 77 o
b a = sin B sin A 12 b = o sin 61 sin 42o b=
12sin 61o
sin 42o b ≈ 16
c b = sin C sin B c 5.0 = o sin125 sin 25o
c a = sin C sin A 12 c = o sin 77 sin 42o c=
12sin 77o
5.0sin125o
a=
o
5.0sin 30o
sin 25o a ≈ 5.9
sin 25 c ≈ 9.7
sin 42o c ≈ 17
4.
3.
B = 180o − 110o − 32o
A = 180o − 78o − 28o
B = 38o
A = 74o
b a = sin B sin A a 12 = o sin 38 sin110o a=
12sin110o
sin 38o a ≈ 18
5.
c=
a b = sin A sin B a 5.0 = o sin 30 sin 25o
b c = sin B sin C b 44 = sin 28o sin 78o
c b = sin C sin B c 12 = o sin 32 sin 38o c=
12sin 32o
b=
a c = sin A sin C a 44 = sin 74o sin 78o
44sin 28o
a=
sin 78o b ≈ 21
sin 38o c ≈ 10
a b = sin A sin B 22 16 = o sin B sin132
C ≈ 180o − 33o − 132o C ≈ 15o
16sin132o 22 sin B ≈ 0.5405 sin B =
B ≈ 33o
Copyright © Houghton Mifflin Company. All rights reserved.
44sin 74o
sin 78o a ≈ 43
a c = sin A sin C c 22 ≈ o sin132 sin15o c≈
22sin15o
sin132o c ≈ 7.7
Section 7.1
437
6.
b c = sin B sin C 6.0 3.0 = o sin C sin 82
a b = sin A sin B a 6.0 ≈ o sin 68.3 sin 82.0o
A ≈ 180.0o − 82.0o − 29.7o A ≈ 68.3o
3sin 82.0o 6.0 sin C ≈ 0.4951
a=
sin C =
6.0sin 68.3o
sin82.0o a ≈ 5.6
C ≈ 29.7o
7.
8.
C = 180o − 22.5o − 112.4 o
C = 180o − 21.5o − 104.2 o
C = 45.1o
C = 54.3o
b = a sin B sin A b = 16.3 sin112.4o sin 22.5o
c = a sin C sin A c = 16.3 sin 45.1o sin 22.5o
b = 16.3sin112.4 sin 22.5o b ≈ 39.4
o
a = c sin A sin C a = 57.4 sin 21.5o sin 54.3o o
b = c sin B sin C b = 57.4 sin104.2o sin 54.3o o
c = 16.3sin 45.1 sin 22.5o c ≈ 30.2
a = 57.4sin 21.5 sin 54.3o a ≈ 25.9
o
b = 57.4sin104.2 sin 54.3o b ≈ 68.5
10.
9.
C = 180o − 65.4o − 82.0o A = 180o − 72.6o − 54.8o
C = 32.6o c b = sin C sin B c 36.5 = o sin 32.6 sin 65.4o c=
a b = sin A sin B a 36.5 = o sin 82.0 sin 65.4o
36.5 sin 32.6o
sin 65.4 c ≈ 21.6
o
a=
36.5 sin 82.0o
sin 65.4o a ≈ 39.8
A = 52.6o b a = sin B sin A b 14.4 = 54.8o sin 52.6o b=
14.4 sin 54.8o
sin 52.6o b ≈ 14.8
Copyright © Houghton Mifflin Company. All rights reserved.
c a = sin C sin A c 14.4 = sin 72.6o sin 52.6o a=
14.4 sin 72.6o
sin 52.6o a ≈ 17.3
438
Chapter 7: Applications of Trigonometry
12.
11.
B = 180o − 98.5o − 33.8o
A = 180o − 69.2o − 36.9o
B = 47.7 o
A = 73.9o
a c = sin A sin C a 102 = sin 33.8o sin 98.5o a=
b c = sin B sin C b 102 = sin 47.7 o sin 98.5o
102 sin 33.8o
b=
sin 98.5o a ≈ 57.4
b a = sin B sin A b 166 = o sin 36.9 sin 73.9o
102 sin 47.7 o
b=
sin 98.5o b ≈ 76.3
c a = sin C sin A c 166 = o sin 69.2 sin 73.9o
166sin 36.9o
c=
sin 73.9o b ≈ 104
166sin 69.2o
sin 73.9o c ≈ 162
14.
13.
c b = sin C sin B 87.2 12.1 = o sin B sin114.2 sin B =
a c = sin A sin C 24.42 16.92 = o sin C sin 54.32
12.1sin114.2o ≈ 0.1266 87.2
sin C =
16.92sin 54.32o ≈ 0.5628 24.42
B ≈ 7.3o
C ≈ 34.25o
A ≈ 180o − 114.2o − 7.3o
B = 180o − 54.32o − 34.25o
A ≈ 58.5o
B = 91.43o
c a = sin C sin A 87.2 a ≈ sin114.2o sin 58.5o
b a = sin B sin A b 24.42 ≈ o sin 91.43 sin 54.32o
a=
87.2sin 58.5o sin114.2o
b=
≈ 81.5
15.
24.42sin 91.43o sin 54.32o
≈ 30.05
sin 37 o = h 40 h = 40sin 37 o h ≈ 24 Since h < 28, two triangles exist.
c = a sin C sin A 40 = 28 sin C sin 37o o sin C = 40sin 37 = 0.8597 28
C ≈ 59o or 121o
C = 59o B = 180o − 37o − 59o = 84o b = 28 sin84o sin 37o
C = 121o B = 180o − 121o − 37o = 22o b = 28 sin 22o sin 37o
o b = 28sin84 = 46 o sin 37
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o b = 28sin 22 ≈ 17 o sin 37
Section 7.1
439
16.
sin 32o =
h 14
h = 14sin 32o h ≈ 7.4
Since h < 9.0, two triangles exist. b c = sin B sin C 9.0 14 = o sin C sin 32 sin C =
C = 56o
C = 124o
A = 180o − 32o − 56o = 92o 9.0 a = o sin 92 sin 32o
14sin 32o = 0.8243 9
a=
C = 56o or 124o 17.
9sin 92o o
sin 32
≈ 17
18.
sin 65o =
h 10
h 18
h = 18sin 42o
h ≈ 9.06 Since h > 8, no triangle is formed.
20.
a=
9sin 24o sin 32o
≈ 6.9
19.
sin 42o =
h = 10sin 65o
A = 180o − 124o − 32o = 24o 9.0 a = o sin 24 sin 32o
h ≈ 12.04 Since h > 12, no triangle is formed.
sin 22.6o =
sin 30o =
h 2.4
h = 2.4sin 30o h ≈ 1.2 Since h > 1, no triangle is formed.
h 13.8
h = 13.8sin 22.6o h ≈ 5.30
Since h < 5.55, two solutions exist. a b = sin A sin B 13.8 5.55 = sin A sin 22.6o
A = 107.1o
C = 180o − 72.9o − 22.6o
C = 180o − 107.1o − 22.6o
C = 84.5o c 5.55 = sin 84.5o sin 22.6o
13.8sin 22.6o 5.55 sin A = 0.9555
sin A =
o
A = 72.9o
o
A ≈ 72.9 or 107.1
c=
C = 50.3o c 5.55 = sin 50.3o sin 22.6o
5.55sin 84.5o
sin 22.6o c ≈ 14.4
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c=
5.55sin 50.3o
sin 22.6o c ≈ 11.1
440
Chapter 7: Applications of Trigonometry
21.
sin14.8o = h 6.35 h = 6.35sin14.8o h ≈ 1.62
Since h < 4.80, two solutions exist. c a = sin C sin A 6.35 4.80 = sin C sin14.8o
C = 160.2o
B = 180o − 19.8o − 14.8o
B = 180o − 160.2o − 14.8o
= 145.4o b 4.80 = o sin145.4 sin14.8o
6.35sin14.8o 4.80 sin C = 0.3379 sin C =
o
C = 19.8o
b=
o
C ≈ 19.8 or 160.2
= 5.0o b 4.80 = o sin 5.0 sin14.8o
4.80sin145.4o
b=
sin14.8o b ≈ 10.7
22.
4.80sin 5.0o
sin14.8o b ≈ 1.64
h 3.50 h = 3.50sin 37.9°
sin 37.9° =
h ≈ 2.15
Since h < 2.84, two solutions exist.
b c = sin B sin C 3.50 2.84 = sin B sin 37.9° 3.50sin 37.9° sin B = 2.84 sin B = 0.7570 B ≈ 49.2° or 130.8° 23.
B = 49.2°
B = 130.8°
A = 180° − 49.2° − 37.9°
A = 180° − 130.8° − 37.9°
A = 92.9°
A = 11.3°
a 2.84 = sin 92.9° sin 37.9° 2.84sin 92.9° a= sin 37.9° a ≈ 4.62
a 2.84 = sin11.3° sin 37.9° 2.84sin11.3° a= sin 37.9° a ≈ 0.906
24.
h 8.25 h = 8.25sin 47.2°
h 16.3 h = 16.3sin 57.2°
sin 47.2° =
sin 52.7° =
h ≈ 13.0
h ≈ 6.05
Since h > 5.80, no triangle is formed.
Since h >12.3, no triangle is formed.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.1
441
25.
26.
Since b > a, one triangle exists. a = b sin A sin B 15.05 = 67.25 sin A sin117.32° sin A = 15.04sin117.32° ≈ 0.1988 67.25 A = 11.47°
h 24.62 h = 24.62 sin 49.22° h ≈ 18.64
sin 49.22° =
Since h > 16.92, no triangle is formed.
C = 180° − 11.47° − 117.32° C = 51.21°
c = b sin C sin B c = 67.25 sin 51.21° sin117.32° c = 67.25sin 51.21° sin117.32° c ≈ 59.00 27.
sin 20.5o = h 14.1 h = 14.1sin 20.5o h ≈ 4.9
Since h < 10.3, two solutions exist. a = c sin A sin C 10.3 = 14.1 sin 20.5o sin C o
sin C = 14.1sin 20.5 10.3 sin C = 0.4794
C ≈ 28.6o or 151.4o
28.
A = 20.5o
A = 20.5o
B = 180o − 28.6o − 20.5o
B = 180o − 151.4o − 20.5o
B = 130.9o b = 10.3 sin130.9o sin 20.5o b = 10.3sin130.9 sin 20.5o b ≈ 22.2
B = 8.1o b = 10.3 sin8.1o sin 20.5o o
o
b = 10.3sin8.1 sin 20.5o b ≈ 4.1
sin 41.2o = h 31.5 h = 31.5sin 41.2o h ≈ 20.7
Since h < 21.6, two solutions exist. a = b sin A sin B 31.5 = 21.6 sin A sin 41.2o sin A = 31.5sin 41.2 21.6 sin A = 0.9606 o
A = 73.9o
A = 106.1o
C = 180o − 73.9o − 41.2o
C = 180o − 106.1o − 41.2o
C = 64.9o c = 21.6 sin 64.9o sin 41.2o
o
o
A ≈ 73.9 or 106.1
C = 32.7o c = 21.6 sin 32.7o sin 41.2o
o c = 21.6sin 64.9 o sin 41.2 c ≈ 29.7
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o c = 21.6sin 32.7 o sin 41.2 c ≈ 17.7
442
Chapter 7: Applications of Trigonometry
29.
30.
∠C = 180° − (59.0° + 77.2°) ∠C = 43.8° ∠B = 180° − (39.4° + 64.9°) ∠B = 75.7°
b c = sin B sin C 7620 x = sin 77.2° sin 43.8° 7620sin 77.2° x= sin 43.8° x ≈ 10,700 feet
a b = sin A sin B 105 a = sin 39.4° sin 75.7° 105sin 39.4° a= sin 75.7° a ≈ 68.8 miles 31.
a = 155 yd, c = 165 yd, A = 42.0o
32.
b = 365 yd, A = 11.2o , C = 22.9o
c = a sin C sin A 165 = 155 sin C sin 42.0o 155 sin C = 165sin 42.0o 155 ⎞ = 45.4o C = sin −1 ⎛⎜ ⎟ ⎝ 165sin 42.0o ⎠
B = 180o − 11.2o − 22.9o = 145.9o c = b a. sin C sin B c = 365 sin 22.9o sin145.9o o = 253 yd c = 365sin 22.9 o sin145.9
B = 180o − 42.0o − 45.4o = 92.6o
b.
b = a sin B sin A b = 155 sin 92.6o sin 42.0o o b = 155sin 92.6o = 231 yd 165sin 42.0 33.
a = b sin A sin B a = 365 sin11.2o sin145.9o o = 126 yd a = 365sin11.2 o sin145.9
34.
B = 180° − 67° − 31° B = 82°
c b = sin C sin B 220 c = sin 31° sin82° 220sin 31° c= sin 82° c ≈ 110 feet
CAB = 180° − 27.2° = 152.8° ACB = 180° − 152.8° − 23.9° = 3.3° AC = AB CD = AC sin ABC sin ACB sin CAD sin ADC AC = 17.5 CD = 123.2 sin 23.9° sin 3.3° sin 27.2° sin 90° AC = 17.5sin 23.9° CD = 123.2sin 27.2° sin 3.3° sin 90° ≈ 123.2 m ≈ 56.3 m Responses will vary.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.1
443
35.
36.
ABC = 180° − 3° = 177° ABC = 180° − 78° = 102° ACB = 180° − 102° − 62° = 16°
AC = BC sin ABC sin CAB AC = 3550 sin177° sin 2.2° AC = 3550sin177° ≈ 4840 ft sin 2.2°
37.
AC AB = sin ABC sin ACB AC 30 = sin102° sin16° 30sin102° AC = sin16°
A = 32° − 11° A = 21° B = 180° − 90° − 32° B = 58° C = 180° − 58° − 21° C = 101°
38.
C = 180° − 54° − 47° C = 79°
b a = sin B sin A b 320 = sin 47° sin 54° 320sin 47° b= sin 54° b ≈ 290 feet
39.
A = 5° B = 180° − 90° − 75° B = 15° C = 180° − 15° − 5° C = 160°
b a = sin B sin A b 12 = sin15° sin 5° 12sin15° b= sin 5°
40.
α = 65° B = 65° + 8° B = 73° A = 180° − 50° − 65° A = 65° C = 180° − 65° − 73° C = 42°
h AC h = AC sin 62°
sin 62° =
30sin102° sin 62° sin16° h ≈ 94 feet h=
c a = sin C sin A c 35 = sin101° sin 21° 35sin101° c= sin 21° c ≈ 96 feet c a = sin C sin A c 320 = sin 79° sin 54° 320sin 79° c= sin 54° c ≈ 390 feet h b h = b sin 70°
sin 70° =
⎛ 12sin15° ⎞ h=⎜ ⎟ sin 70° ⎝ sin 5° ⎠ h ≈ 33 feet
b c = sin B sin C b 20 = sin 73° sin 42° 20sin 73° b= sin 42° b ≈ 29 miles
Copyright © Houghton Mifflin Company. All rights reserved.
444
Chapter 7: Applications of Trigonometry
φ = 360° − 332° φ = 28° α = 28°
41.
C = 82° − 28° C = 54° A = 28° + 36° A = 64° B = 180° − 64° − 54° B = 62°
A = 90° − 55°
42.
A = 35° C = 90° − 25° C = 65° B = 180° − 35° − 65° B = 80°
c b = sin C sin B c 8 = sin 65° sin 80° 8sin 65° c= sin 80°
A = 120° − 65° A = 55° α = 65° B = 38° + 65° B = 103° C = 180° − 103° − 55° C = 22°
43.
a b = sin A sin B a 8.0 = sin 64° sin 62° 8.0sin 64° a= sin 62° a ≈ 8.1 miles h c h = c sin 35°
sin 35° =
88sin 65° sin 35° sin 80° h ≈ 4.2 miles h=
b c = sin B sin C 450 b = sin103° sin 22° 450sin103° b= sin 22° b ≈ 1200 miles
44.
A = 40° − 15° A = 25° B = 180° − 90° − 40° B = 50° C = 180° − 25° − 50°
d 12 = sin A sin C 12sin 25° d= sin105° d ≈ 5.3 feet
C = 105°
....................................................... 45.
Connecting Concepts 46.
A = 180° − 67° − 68° A = 45° B = 67° + 11° B = 78° C = 180° − 45° − 78° C = 57°
c b = sin C sin B c 300 = sin 57° sin 78° 300sin 57° c= sin 78° c ≈ 260 meters
a b = sin A siinB a sin A = b sin B a sin A −1= −1 b sin B a − b sin A − sin B = b sin B
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.2
47.
445
a b = sin A sin B a sin A = b sin B sin A a +1= +1 sin b B a + b sin A + sin B = b sin B
48.
Use the results of Problems 46 and 47. sin A−sin B a −b sin B b = sin A+sin B a +b sin B b
a − b sin A − sin B = a + b sin A + sin B
49.
The graph of L is shown. The minimum value of L is approximately 11.19 m.
3 5 = cosθ , = sin θ d1 d2 3 5 , d2 = cosθ sin θ L = d1 + d 2
d1 =
L(θ ) =
3 5 + cosθ sin θ
....................................................... PS1.
(10.0) 2 + (15.0)2 − 2(10.0)(15.0) cos110.0o ≈ 20.7
PS3.
PS2.
c 2 = a 2 + b2 − 2ab cos C c 2 − a 2 − b2 = −2ab cos C 2 2 2 cos C = c − a − b −2ab 2 2 2 2 2 2 −1 ⎛ c − a − b ⎞ −1 ⎛ a + b − c ⎞ C = cos ⎜ ⎟ = cos ⎜ ⎟ 2ab −2ab ⎝ ⎠ ⎝ ⎠
PS5. s = a + b + c = 3 + 4 + 5 = 6 2 2
Prepare for Section 7.2 A = 1 bh = 1 (6)(8.5) = 22.5 in.2 2 2
PS4. P = 6 + 9 + 10 = 25 semiperimeter = 1 (25) = 12.5 m 2
PS6. c 2 = a 2 + b 2
6(6 − 3)(6 − 4)(6 − 5) = 6(3)(2)(1) = 6
Section 7.2 1.
c 2 = a 2 + b2 − 2ab cos C
2.
c 2 = 122 + 182 − 2(12)(18) cos 44°
a 2 = 302 + 242 − 2(30)(24) cos120°
c 2 = 468 − 432cos 44°
a 2 = 1476 − 1440cos120°
c = 468 − 432cos 44° c ≈ 13 3.
a 2 = b2 + c 2 − bc cos A
a = 1476 − 1440cos120° a ≈ 47
b2 = a 2 + c 2 − 2ac cos B
4.
c 2 = a 2 + b2 − 2ab cos C
b2 = 1202 + 1802 − 2(120)(180) cos56°
c 2 = 4002 + 6202 − 2(400)(620)cos116°
b2 = 46,800 − 43,200cos56°
c 2 = 544,400 − 496,000cos116°
b = 46,800 − 43,200cos56°
c = 544,400 − 496,000cos116°
b ≈ 150
c ≈ 870
Copyright © Houghton Mifflin Company. All rights reserved.
446
5.
7.
Chapter 7: Applications of Trigonometry
a 2 = b2 + c 2 − 2bc cos A
6.
a 2 = 602 + 842 − 2(60)(84) cos13°
b2 = 1222 + 1442 − 2(122)(144) cos 48°
a 2 = 10,656 − 10,080cos13°
b2 = 35,620 − 35,136cos 48°
a = 10,656 − 10,080cos13°
b = 35,620 − 35,136cos 48°
a ≈ 29
b ≈ 110
c 2 = a 2 + b2 − 2ab cos C
8.
a 2 = 122 + 222 − 2(12)(22) cos55°
c 2 = 130 − 126cos72°
a 2 = 628 − 528cos55° a = 628 − 528cos55° a ≈ 18
c 2 = a 2 + b2 − 2ab cos C
10.
a 2 = 12.32 + 14.52 − 2(12.3)(14.5)cos6.5°
c 2 = 73 − 66.24cos124°
a 2 = 361.54 − 356.7cos6.5° a = 361.54 − 356.7cos6.5° a ≈ 2.67
b2 = a 2 + c 2 − 2ac cos B
12.
15.
17.
c 2 = a 2 + b2 − 2ab cos C
b2 = 25.92 + 33.42 − 2(25.9)(33.4) cos84.0°
c 2 = 14.22 + 9.302 − 2(14.2)(9.30) cos9.20°
b2 = 1786.37 − 1730.12cos84.0°
c 2 = 288.13 − 264.12cos9.20°
b = 1786.37 − 1730cos84.0° b ≈ 40.1 13.
a 2 = b2 + c 2 − 2bc cos A
c 2 = 4.62 + 7.22 − 2(4.6)(7.2) cos124° c = 73 − 66.24cos124° c ≈ 10 11.
a 2 = b2 + c 2 − 2bc cos A
c 2 = 9.02 + 7.02 − 2(9.0)(7.0) cos72° c = 130 − 126cos72° c ≈ 9.5 9.
b2 = a 2 + c 2 − 2ac cos B
c = 288.13 − 264.12 cos9.20° c ≈ 5.24
b2 = a 2 + c 2 − 2ac cos B
14.
a 2 = b2 + c 2 − 2bc cos A
b2 = 1222 + 55.92 − 2(122)(55.9) cos 44.2°
a 2 = 444.82 + 389.62 − 2(444.8)(389.6) cos78.44°
b2 = 18,008.81 − 13,639.6cos 44.2°
a 2 = 349,635.2 − 346,588.16cos78.44°
b = 18,008.81 − 13,639.6cos 44.2°
a = 349,635.2 − 346,588.16cos78.44°
b ≈ 90.7
a ≈ 529.3
2 2 2 cos A = b + c − a 2bc 2 + 32 402 − 252 cos A = 2(32)(40) 1999 cos A = 2560 A = cos −1 ⎛⎜ 1999 ⎞⎟ ≈ 39° ⎝ 2560 ⎠
2 2 2 cos C = a + b − c 2ab 2 + 9.02 − 122 8.0 cos C = 2(8.0)(9.0) cos C = 1 144 C = cos −1 ⎛⎜ 1 ⎞⎟ ≈ 90° ⎝ 144 ⎠
16.
18.
2 2 2 cos B = a + c − b 2ac 2 + 60 1202 − 882 cos B = 2(60)(120) 10256 cos B = 14400 B = cos −1 ⎛⎜ 10256 ⎞⎟ ≈ 45° ⎝ 14400 ⎠ 2 2 2 cos A = b + c − a 2bc 2 + 1602 − 1082 132 cos A = 2(132)(160) 31,360 cos A = 42, 240 A = cos −1 ⎜⎛ 31360 ⎟⎞ ≈ 42.1° ⎝ 42240 ⎠
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.2
19.
447
2 2 2 cos B = a + c − b 2ac
2 2 2 cos B = a + c − b 2ac
20.
2 2 2 cos B = 80 + 124 − 92 2(80)(124) 13,312 cos B = 19,840
2 2 2 cos B = 166 + 139 − 124 2(166)(139) 31,501 cos B = 46,148
B = cos −1 ⎛⎜ 13312 ⎞⎟ ≈ 47.9° ⎝ 19840 ⎠ 21.
B = cos −1 ⎜⎛ 31501 ⎟⎞ ≈ 47.0° ⎝ 46148 ⎠
2 2 2 cos C = a + b − c 2ab 2
2 2 2 cos A = b + c − a 2bc
22. 2
cos C = 1025 + 625 − 1420 2(1025)(625) −575,150 cos C = 1, 281,250
2
2 2 2 cos A = 3.2 + 5.9 − 4.7 2(3.2)(5.9) 22.96 cos A = 37.76 A = cos −1 ⎜⎛ 22.96 ⎟⎞ ≈ 53° ⎝ 37.76 ⎠
⎛ −575,150 ⎞ C = cos −1 ⎜ ⎟ ≈ 116.67° ⎝ 1,281,250 ⎠
23.
25.
2 2 2 cos B = a + c − b 2ac 2 + 29.62 − 40.12 32.5 cos B = 2(32.5)(29.6) 324.4 cos B = 1924 B = cos −1 ⎛⎜ 324.4 ⎞⎟ ≈ 80.3° ⎝ 1924 ⎠
2
a ≈ 11.13860218 a ≈ 11.1
2 2 2 cos B = a + c − b 2ac
c 2 = a 2 + b2 − 2ab cos C
2
2
2
cos B = 11.13860218 + 17.2 − 15.5 2(11.13860218)(17.2)
2
2
2
2
2 2 2 cos B = 141 + 179.4509034 − 92.3 2(141)(179.4509034)
⎛ 20,840.91673 ⎞ A = cos −1 ⎜ ⎟ ⎝ 33,126.64242 ⎠ A ≈ 51.0°
2 2 2 cos A = 144 + 98.1 − 83.6 2(144)(98.1)
cos B = 83.6 + 98.1 − 144 2(83.6)(98.1)
2
⎛ 43,564.33673 ⎞ B = cos −1 ⎜ ⎟ ⎝ 50,605.15476 ⎠ B ≈ 30.6°
2 2 2 cos C = a + b − c 2ab 2
⎛ ⎞ B = cos −1 ⎜ −4123.43 ⎟ 16,402.32 ⎝ ⎠ B ≈ 104.6°
2
2
2 2 2 cos B = a + c − b 2ac
cos A = 92.3 + 179.4509034 − 141 2(92.3)(179.4509034)
2 2 2 cos B = a + c − b 2ac
2
cos C = 11.13860218 + 15.5 − 17.2 2(11.13860218)(15.5) C = cos −1 ⎛⎜ 68.47845852 ⎞⎟ ⎝ 345.2966676 ⎠ C ≈ 78.6°
2 2 2 cos A = b + c − a 2bc
2 2 2 cos A = b + c − a 2bc
⎛ 23,370.65 ⎞ A = cos −1 ⎜ ⎟ ⎝ 28,252.8 ⎠ A ≈ 34.2°
2 2 2 cos C = a + b − c 2ab
B = cos −1 ⎛⎜ 179.6584585 ⎞⎟ ⎝ 383.167915 ⎠ B ≈ 62.0°
c = 1412 + 92.32 − 2(141)(92.3) cos98.4°
27.
2
C ≈ 75.87°
a 2 = b2 + c 2 − 2bc cos A
c ≈ 179.4509034 c ≈ 179
2
cos C = 112.4 + 96.80 − 129.2 2(112.40)(96.80) cos C ≈ 0.2441
a = 15.52 + 17.22 − 2(15.5)(17.2) cos39.4°
26.
2 2 2 cos C = a + b − c 2ab
24.
2
2
2
cos C = 83.6 + 144 − 98.1 2(83.6)(144) ⎛ 18,101.35 ⎞ C = cos −1 ⎜ ⎟ ⎝ 24,076.8 ⎠ C ≈ 41.3°
Copyright © Houghton Mifflin Company. All rights reserved.
448
28.
Chapter 7: Applications of Trigonometry 2 2 2 cos A = b + c − a 2bc 2
2
cos A = 36.3 + 38.2 − 25.4 2(36.3)(38.2)
2
2 2 2 cos B = a + c − b 2ac
2 2 2 cos C = a + b − c 2ab
2 2 2 cos B = 25.4 + 38.2 − 36.3 2(25.4)(38.2)
2 2 2 cos C = 25.4 + 36.3 − 38.2 2(25.4)(36.3)
A = cos −1 ⎛⎜ 2131.77 ⎞⎟ ⎝ 2773.32 ⎠ A ≈ 39.8° 29.
K = 1 bc sin A 2 1 K = (12)(24)sin105° 2 K ≈ 140 square units
31.
C = 180° − 42° − 76° C = 62°
B = cos −1 ⎛⎜ 786.71 ⎞⎟ ⎝ 1940.56 ⎠ B ≈ 66.1°
K=
c 2 sin A sin B 2sin C
C = cos −1 ⎛⎜ 503.61 ⎞⎟ ⎝ 1844.04 ⎠ C ≈ 74.2°
30.
K = 1 ac sin B 2 1 K = (32)(25)sin127° 2 K ≈ 320 square units
32.
A = 180° − 102° − 27° A = 51°
122 sin 42° sin 76° 2sin 62° K ≈ 53 square units
K=
8.52 sin102° sin 27° 2sin 51° K ≈ 21 square units
K=
33.
35.
1 s = (a + b + c) 2 1 s = (16 + 12 + 14 ) 2 s = 21
K=
34.
1 (a + b + c) 2 1 s = ( 24 + 32 + 36 ) 2 s = 46 s=
K = s ( s − a )( s − b)( s − c)
K = s ( s − a )( s − b)( s − c)
K = 21(21 − 16)(21 − 12)(21 − 14)
K = 46(46 − 24)(46 − 32)(46 − 36)
K = 21(5)(9)(7)
K = 46(22)(14)(10)
K ≈ 81 square units
K ≈ 380 square units
a b = sin A sin B 22.4 26.9 = sin A sin 54.3° 22.4sin 54.3° sin A = 26.9 sin A ≈ 0.6762 A ≈ 42.5° C = 180° − 42.5° − 54.3° C = 83.2°
36.
1 ab sin C 2 1 K = (9.84)(13.4)sin18.2° 2 K ≈ 20.6 square units
K=
a 2 sin B sin C 2sin A
37.
C = 180° − 116° − 34° C = 30°
K=
c 2 sin A sin B 2sin C
8.52 sin116° sin 34° 2sin 30° K ≈ 36 square units K=
1 ab sin C 2 1 K = (22.4)(26.9)sin 83.2° 2 K ≈ 299 square units K=
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.2
38.
449
A = 180° − 76.3° − 42.8° A = 60.9° K=
39.
c 2 sin A sin B 2sin C
17.92 sin 60.9° sin 42.8° 2sin 76.3° K ≈ 97.9 square units
1 (a + b + c) 2 1 s = ( 3.6 + 4.2 + 4.8 ) 2 s = 6.3 s=
K = s ( s − a )( s − b)( s − c )
K=
K = 6.3(6.3 − 3.6)(6.3 − 4.2)(6.3 − 4.8) K = 6.3(2.7)(2.1)(1.5) K ≈ 7.3 square units
40.
s = 1 (a + b + c) 2 1 s = 1 (10.2 + 13.3 + 15.4 ) s = (a + b + c ) 2 2 s = 19.45
41.
K = s ( s − a)( s − b)( s − c) K = 19.45(19.45 − 10.2)(19.45 − 13.3)(19.45 − 15.4)
α = 32° β = 72°
K = 19.45(9.25)(6.15)(4.05) K ≈ 66.9 square units
B = 72o + 32° B = 104° b 2 = a 2 + c 2 − 2ac cos104° b 2 = 3202 + 5602 − 2(320)(560)cos104° b 2 = 416,000 − 358, 400cos104° b = 416,000 − 358, 400cos104° b ≈ 710 miles
42.
43.
a 2 = b 2 + c 2 − 2bc cos A
a 2 = b 2 + c 2 − 2bc cos A a 2 = 3002 + 4162 − 2(300)(416)cos 72° a 2 = 236,056 − 249,600cos 72° a = 236,056 − 249,600cos 72° a ≈ 430 feet
a 2 = 262 + 902 − 2(26)(90)cos 45° a 2 = 8776 − 4680cos 45° a = 8776 − 4680cos 45° a ≈ 74 feet
Copyright © Houghton Mifflin Company. All rights reserved.
450
Chapter 7: Applications of Trigonometry
44.
45.
b 2 = a 2 + c 2 − 2ac cos B b 2 = (105.6) 2 + (105.6) 2 − 2(105.6)(105.6) cos109.05° b 2 = 22302.72 − 22302.72 cos109.05° Let a = the length of the diagonal on the front of the box. Let b = the length of the diagonal on the right side of the box. Let c = the length of the diagonal on the top of the box.
b = 22302.72 − 22302.72 cos109.05° b ≈ 172.0 feet
a 2 = (4.75) 2 + (6.50) 2 = 64.8125 a = 64.8125 2
b = (3.25) 2 + (4.75) 2 = 33.125 b = 33.125 2
c = (6.50) 2 + (3.25) 2 = 52.8125 θ =C cos C = cosθ = cosθ =
a2 + b2 − c2 2ab 64.8125 + 33.125 − 52.8125 2 64.8125 33.125 45.125 2 64.8125 33.125 ⎛
⎞ 45.125 ⎟ ⎟ 2 64 . 8125 33 . 125 ⎝ ⎠
θ = cos −1 ⎜⎜ θ ≈ 60.9° 46.
47.
cos A =
b2 + c2 − a2 2bc
cos A =
(615) 2 + (499) 2 − (629) 2 231585 = 2(615)(499) 613770
⎛ 231585 ⎞ A = cos −1 ⎜ ⎟ ≈ 67.8° ⎝ 613770 ⎠ x sin A = 499 x sin 67.8° = 499 499 sin 67.8° = x x ≈ 462 feet
b = (18 mph)(10 hours) = 180 miles c = (22 mph)(10 hours) = 220 miles A = 318° − 198° A = 120° a 2 = b 2 + c 2 − 2bc cos A a 2 = 1802 + 2202 − 2(180)(220)cos120° a 2 = 120, 400 a ≈ 350 miles
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.2
48.
451
d 2 = 1362 + 1622 − 2(136)(162) cos 78°
49.
2
d = 44,740 − 44,064cos 78° d = 44,740 − 44,064cos 78° d ≈ 189 miles
360° 6 A = 60°
a 2 = 402 + 402 − 2(40)(40)cos 60°
A=
50.
a 2 = 1600 a = 40 cm
2242 + 1822 − 1652 2(224)(182) cos B ≈ 0.6877 B ≈ 46.5° cos B =
51.
52.
C = 90° + 14° C = 104° 180(5280) a= ⋅ 10 3600 a = 2640 feet
c 2 = 26402 + 4002 − 2(2640)(400)(cos104°)
α = 270° − 254° α = 16°
a 2 = b 2 + c 2 − 2bc cos A
A = 16° + 90° + 32° A = 138° b = 4 ⋅ 16 = 64 miles
c 2 ≈ 7,640,539 c ≈ 2800 feet
a 2 = 642 + 662 − 2(64)(66)cos138° a 2 = 8452 − 8448cos138° a ≈ 120 miles
c = 3 ⋅ 22 = 66 miles
53.
b2 = a 2 + c 2 − 2ac cos B b2 = 2252 + 1812 − 2(225)(181) cos163.9° b2 = 83,386 − 81,450cos163.9° b = 83,386 − 81,450cos163.9° b ≈ 402.046592 b ≈ 402 mi
2 2 2 cos A = b + c − a 2bc 2 2 2 cos A = 402.046592 + 181 − 225 2(402.046592)(181)
⎛ 143,777.4621 ⎞ A = cos −1 ⎜ ⎟ ⎝ 145,540.8663 ⎠ A ≈ 8.9°
α = 180 − (108.5 + 8.9) α = 62.6°
B = 108.5 + (180 − 124.6) B = 163.9°
The distance is 402 mi and the bearing is S62.6°E.
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452
Chapter 7: Applications of Trigonometry
54.
a.
162 = 42 + c 2 − 2(4)( c ) cos A
b.
162 = 42 + c 2 − 2(4)( c )cos A 0 = c 2 − (8cos A)c − 240 c=
8cos A ± ( −8cos A)2 − 4(1)( −240) 2(1)
2 c = 8cos A + 64cos A + 960 2
c.
8cos(55) + 64cos2 (55) + 960 2 c ≈ 18 cm c=
d.
a = b sin A sin B 16 = 4 sin 55 sin B
(
)
B = sin −1 4sin 55 ≈ 11.8° 16 C = 180 − 55 − 11.8 = 113.2° a = c sin A sin C 16 = c sin 55 sin113.2 c = 16sin113.2 ≈ 18 cm sin 55 They are the same. 55.
56.
57.
1 s = (a + b + c) 2 1 s = (236 + 620 + 814) 2 s = 835
K = s ( s − a )( s − b)( s − c )
1 (8 + 10 + 12 ) 2 s = 15
K = 2 s ( s − a )( s − b)( s − c)
s=
K = 835(835 − 236)(835 − 620)(835 − 814) K = 835(599)(215)(21) K ≈ 2, 258, 240,000 K ≈ 47,500 square meters
K = 2 15(15 − 8)(15 − 10)(15 − 12) K = 30 7 square feet
α = 90° ⎡1 ⎤ K = 4 ⎢ (9)(9)sin 90°⎥ ⎣2 ⎦ K = 162 in 2
58.
a = 24 1 s = ( 24 + 24 + 24 ) 2 s = 36
K = 6 36(36 − 24)(36 − 24)(36 − 24) K = 6(144) 3 K = 864 3 cm 2
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.2
453
59.
60.
1 (185 + 212 + 240 ) 2 s = 318.5
K = 318.5(318.5 − 185)(318.5 − 212)(318.5 − 240)
s=
1 ( 324 + 412 + 516 ) 2 s = 626 s=
K ≈ 18,854 ft 2 cost = 2.20(18,854) cost ≈ $41,000
K = 626(626 − 324)(626 − 412)(626 − 516) K ≈ 66,710 ft 2 cost = 4.15(66,710) cost ≈ $277,000
61.
1 ( 680 + 800 + 1020 ) 2 s = 1250
K = 1250(1250 − 680)(1250 − 800)(1250 − 1020)
s=
1 (420 + 500 + 540) 2 s = 730
K ≈ 271,558 ft 2 271,558 Acres = 43,560 Acres ≈ 6.23
62.
s=
63.
For ABC, 13.0 − 16.1 ≈ −0.2981 10.4 sin 53.5 − 86.5 2 ≈ −0.3022 40.0 cos 2 Triangle ABC has correct dimensions.
For DEF, 17.2 − 21.3 ≈ −0.1798 22.8 sin 52.1 − 59.9 2 ≈ −0.0820 68.0 cos 2 Triangle DEF has an incorrect dimension.
For ABC, 9.23 − 15.1 ≈ −0.3623 16.2 sin 34.1 − 66.2 2 ≈ −0.3601 cos 79.7 2 Triangle ABC has correct dimensions.
For DEF, 13.6 − 16.0 ≈ −0.1270 18.9 sin 45.0 − 56.2 2 ≈ −0.1263 cos 78.8 2 Triangle DEF has correct dimensions.
(
64.
(
( )
( )
K = 730(730 − 420)(730 − 500)(730 − 540) K ≈ 99, 445 99, 445 Acres = 4840 Acres ≈ 20.5
)
)
(
(
( )
( )
)
)
....................................................... 65.
d ( P1 , P2 ) =
[2 − (−2)]2 + (1 − 4) 2
Connecting Concepts =5
d ( P1 , P3 ) = (−2 − 4) 2 + (4 − (−3)) 2 = 85 d ( P2 , P3 ) = (2 − 4) 2 + (1 − ( −3)) 2 = 2 5
cosθ =
52 +
(
85
) − (2 5 ) 2
2 ⋅ 5 ⋅ 85 cosθ ≈ 0.9762 θ ≈ 12.5°
Copyright © Houghton Mifflin Company. All rights reserved.
2
454
Chapter 7: Applications of Trigonometry
66.
a = (b cos A − c)2 + (b sin A − 0)2 a 2 = b 2 cos 2 A − 2bc cos A + c 2 + b 2 sin 2 A a 2 = b 2 (cos 2 A + sin 2 A) + c 2 − 2bc cos A a 2 = b 2 + c 2 − 2bc cos A
67.
cos A =
2 2 2bc ( b + c − a )( b + c + a ) b 2 + c 2 − a 2 b 2 + 2bc + c 2 − a 2 − 2bc ( b + c ) − a = = − = −1 2bc 2bc 2bc 2bc 2bc
68.
1 xy sin A 2 K I = K II
KI =
K = 2K I K = xy sin A
69.
V = 18K , where K = area of triangular base 1 V = (4)(4)(sin 72°)(18) 2 3
V ≈ 140 in 70.
K = K BOC + K AOC + K AOB
1 ar 2 1 K AOC = br 2 1 K AOB = cr 2 K BOC =
1 1 1 ar + br + cr 2 2 2 1 K = r (a + b + c) 2 1 1 K = rs where s = ( a + b + c) 2 2 K=
....................................................... PS1.
( 53 ) + ( − 54 ) 2
2
=
9 + 16 = 25 25
PS3. tan α = − 3 3 3 tan α = 3
25 = 1 25
Prepare for Section 7.3 PS2. 10cos 228o ≈ −6.691 PS4. cos α = −17 338
α = cos −1 ⎛⎜ −17 ⎞⎟ ≈ 157.6o ⎝ 338 ⎠
⎛ ⎞ α = tan −1 ⎜ 3 ⎟ = 30o ⎝ 3 ⎠
PS5.
1 ⋅ 5= 5 5 5 5
PS6.
28 = 28 = 14 ⋅ 17 = 14 17 17 68 2 17 17 17
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Section 7.3
455
Section 7.3 1.
a = 5 −1 = 4 b=4−2=2
2.
A vector equivalent to P1P2 is v = 4, 2 .
3.
A vector equivalent to P1P2 is v = −1, −4 .
a = −3 − 2 = −5 b = 5 −1 = 4
4.
A vector equivalent to P1P2 is v = −5, 4 .
5.
a = 4 − (−3) = 7 b = −1 − 0 = −1
6.
a = −3 − 4 = −7 b = −3 − 2 = −5
8.
a = 2−2=0 b = 3 − (−5) = 8
10.
v = (−3)2 + 42 α = tan −1 v = 9 + 16 v =5
u=
α θ θ θ
4 = tan −1 4 −3 3
12.
≈ 53.1° = 180° − α ≈ 180° − 53.1° ≈ 126.9°
α = tan −1
v = 400 + 1600
α θ θ θ
20 −40 , = 20 5 20 5
v = 36 + 100
α ≈ 59.0° θ = 59.0°
u=
v = 202 + (−40)2
u=
α = tan −1
−40 = tan −1 2 20
5 −2 5 , 5 5
34
14.
v = (−50)2 + 302
α = tan −1
v = 2500 + 900
α θ θ θ
v = 3400 = 10 34 ≈ 58.3
u=
5 , −2 5 . 5 5
6 10 3 34 5 34 , = , 34 34 2 34 2 34
A unit vector in the direction of v is u = 3 34 , 5 34 .
5
≈ 63.4° = 360° − α ≈ 360° − 63.4° ≈ 296.6°
A unit vector in the direction of v is u =
10 5 = tan −1 6 3
≈ 11.7
5
≈ 44.7
v = 62 + 102 v = 2 34
−3 4 , 5 5
v = 2000 = 20 5
a = 3−3= 0 b = 0 − (−2) = 2 A vector equivalent to P1P2 is v = 0, 2 .
A unit vector in the direction of v is u = − 3 , 4 .
13.
a =0−0=0 b = 4 − (−3) = 7 A vector equivalent to P1P2 is v = 0, 7 .
A vector equivalent to P1P2 is v = 0, 8 .
11.
a = 3 − 5 = −2 b = 1 − (−1) = 2 A vector equivalent to P1P2 is v = −2, 2 .
A vector equivalent to P1P2 is v = −7, −5 .
9.
a = 3 − ( −1) = 4 b = 3 − 4 = −1 A vector equivalent to P1P2 is v = 4, −1 .
A vector equivalent to P1P2 is v = 7, −1 .
7.
a = 3 − 4 = −1 b = −2 − 2 = −4
34
30 3 = tan −1 −50 5
≈ 31.0° = 180° − α ≈ 180° − 31° ≈ 149°
30 5 34 3 34 −50 , , = − 34 34 10 34 10 34
A unit vector in the direction of v is u = − 5 34 , 3 34 .
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34
34
456
15.
Chapter 7: Applications of Trigonometry
v = 22 + (−4)2 α = tan −1 v = 4 + 16 v = 20 = 2 5
≈ 4.5
−4 = tan −1 2 2
16.
v = 25 + 36
α ≈ 63.4° θ = 360° − α θ ≈ 360° − 63.4°
α θ θ θ
v = 61
≈ 7.8
≈ 296.6° u=
−4 = 2 5 2 5 2
,
5 2 5 , − 5 5
u=
A unit vector in the direction of v is u =
17.
α = tan −1
v = 1764 + 324
α θ θ θ
v = 2088 = 6 58 ≈ 45.7
u=
−18 3 = tan −1 18. 42 7
6 5 61 6 61 , = − 61 61 61
v = (−22)2 + (−32) 2
α = tan −1
v = 484 + 1024
α θ θ θ
≈ 38.8
u=
−22 −32 , 2 377 2 377
= −
61
− 32 16 = tan −1 −22 11
≈ 55.5° = 180° − α ≈ 180° + 55.5° ≈ 235.5°
11 377 16 377 , − 377 377
A unit vector in the direction of v is
58
u = − 11 377 , − 16 377 . 377
20.
≈ 129.8°
= 2 377
42 7 58 3 58 −18 , , − = 58 58 6 58 6 58
3u = 3 −2, 4 = −6, 12
≈ 180° − 50.2°
v = 1508
A unit vector in the direction of v is u = 7 58 , − 3 58 .
19.
= 180° − α
61
≈ 23.2° = 360° − α ≈ 360° − 23.2° ≈ 336.8°
58
−5 , 61
≈ 50.2°
A unit vector in the direction of v is u = − 5 61 , 6 61 .
5 , −2 5 . 5 5
v = 422 + (−18)2
6 6 = tan −1 −5 5
v = (−5)2 + 62 α = tan −1
377
21.
−4 v = −4 −3, −2 = 12, 8
2u − v = 2 −2, 4 − −3, −2 = −4, 8 − −3, −2 = −1, 10
22.
23.
4 v − 2u = 4 −3, −2 − 2 −2, 4 = −12, −8 − −4, 8 = −8, −16
2 1 2 1 u + v = −2, 4 + −3, −2 3 6 3 6 4 8 1 1 = − , + − , − 3 3 2 3 = −
25.
u = (−2)2 + 42 = 20 = 2 5
24.
3 3 u − 2 v = −2, 4 − 2 −3, −2 4 4 2 = − , 3 − −6, −4 3
11 7 , 6 3 26.
=
9 , 7 2
v + 2u = −3, −2 + 2 −2, 4
= −3, −2 + −4, 8 = −7, 6 v + 2u = (−7)2 + 62 = 49 + 36 = 85 27.
3u − 4 v = 3 −2, 4 − 4 −3, −2 = −6, 12 − −12, −8
28.
−2u = −2(3i − 2 j) = −6i + 4 j
= 6, 20 3u − 4 v = 62 + 202 = 436 = 2 109
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.3
457
29.
4 v = 4(−2i + 3 j) = −8i + 12 j
30.
3u + 2 v = 3(3i − 2 j) + 2(−2i + 3 j) = (9i − 6 j) + (−4i + 6 j) = (9 − 4)i + ( −6 + 6) j = 5i + 0 j = 5i
31.
6u + 2 v = 6(3i − 2 j) + 2(−2i + 3 j) = (18i − 12 j) + ( −4i + 6 j) = (18 − 4)i + (−12 + 6) j = 14i − 6 j
32.
1 3 1 3 u − v = (3i − 2 j) − (−2i + 3 j) 2 4 2 4 ⎛3 ⎞ ⎛ 3 9 ⎞ = ⎜ i − j⎟ − ⎜ − i + j⎟ 4 ⎠ ⎝2 ⎠ ⎝ 2
2 3 2 3 v + u = ( −2i + 3j) + (3i − 2 j) 3 4 3 4 3 ⎞ ⎛ 4 ⎞ ⎛9 = ⎜ − i + 2j⎟ + ⎜ i − j⎟ ⎝ 3 ⎠ ⎝4 2 ⎠
34.
33.
9⎞ ⎛3 3⎞ ⎛ = ⎜ + ⎟ i + ⎜ −1 − ⎟ j 4⎠ ⎝2 2⎠ ⎝ 13 = 3i − j 4 v = (−2)2 + 32 = 13
3⎞ ⎛ 4 9⎞ ⎛ = ⎜ − + ⎟i + ⎜ 2 − ⎟ j 2⎠ ⎝ 3 4⎠ ⎝ 11 1 = i+ j 12 2 35.
u − 2 v = (3i − 2 j) − 2(−2i + 3 j) = (3i − 2 j) − ( −4i + 6 j) = (3 + 4)i + ( −2 − 6) j = 7i − 8 j
36.
u + 2 v = 52 + 0 2 = 5
u − 2 v = 7 2 + (−8)2 = 113 37.
a1 = 5cos 27° ≈ 4.5
38.
a2 = 5sin 27° ≈ 2.3
a1 = 4cos a2 = 4sin
π 4
v = a1i + a2 j ≈ −2.4i + 3.2 j
≈ 2.8
40.
π
≈ 2.8 4 v = a1i + a2 j ≈ 2.8i + 2.8 j
41.
heading = 124° ⇒ direction angle = −34°
a1 = 4cos127° ≈ −2.4 a2 = 4sin127° ≈ 3.2
v = a1i + a2 j ≈ 4.5i + 2.3 j
39.
2 v + 3u = 2(−2i + 3 j) + 3(3i − 2 j) = (−4i + 6 j) + (9i − 6 j) = (−4 + 9)i + (6 − 6) j = 5i
wind from the west ⇒ direction angle = 0°
8π ≈ −1.8 7 8π ≈ −0.9 a2 = 2sin 7 v = a1i + a2 j ≈ −1.8i − 0.9 j
a1 = 2cos
AB = 45i AD = 340cos(−34°)i + 340sin(−34°) j AD ≈ 281.9i − 190.1j AC = AB + AD AC = 45i + 281.9i − 190.1j AC ≈ 327i − 190 j AC = 327 2 + (−190) 2 AC ≈ 380 mph
The ground speed of the plane is approximately 380 mph.
Copyright © Houghton Mifflin Company. All rights reserved.
458
Chapter 7: Applications of Trigonometry
42.
α = sin −1
heading = θ = 90° − α θ ≈ 90° − 17.9° θ ≈ 72.1°
0.8 0.8 = sin −1 2.6 2.6
α ≈ 17.9° 43.
heading = 96° ⇒ direction angle = −6°
heading = 37° ⇒ direction angle = 53°
16.4 253.9 16.4 − 1 = tan 253.9 α ≈ 4°
α = tan −1
AB = 50cos53°i + 50sin 53° j ≈ 30.1i + 39.9 j AD = 225cos( −6°)i + 225sin(−6°) j ≈ 223.8i − 23.5 j AC = AB + AD ≈ 30.1i + 39.9 j + 223.8i − 23.5 j ≈ 253.9i + 16.4 j
θ = 90° − α θ ≈ 90° − 4° θ ≈ 86°
AC = (253.9)2 + (16.4)2 ≈ 250
The ground speed of the plane is about 250 mph at a heading of approximately 86° . 44.
heading = 327° ⇒ direction angle = 123°
heading = 60° ⇒ direction angle = 30°
17.1 −6.3 17.1 = tan −1 6.3 α ≈ 70° θ = 270° + 70° θ = 340°
α = tan −1
AB = 18cos123°i + 18sin123° j AB ≈ −9.8i + 15.1j AD = 4cos30°i + 4sin 30° j AD = 3.5i + 2 j AC = AB + AD = −9.8i + 15.1j + 3.5i + 2 j = −6.3i + 17.1j
AC = (−6.3)2 + (17.1)2 ≈ 18
The course of the boat is about 18 mph at a heading of approximately 340°. 45.
F 3000 F = 3000sin 5.6°
sin 5.6° =
46.
F ≈ 293 lb
Copyright © Houghton Mifflin Company. All rights reserved.
α =θ 120 800 α ≈ 8.6°
sin α =
Section 7.3
47.
a.
459
sin 22.4° =
F1
b.
345 F1 = 345sin 22.4°
a.
sin 31.8° =
F1
b.
345 F1 = 811sin 31.8° F1 ≈ 427 lb
49.
F1 + F2 + F3 = (18i + 13.1j) + ( −12.4i + 3.8 j) + ( −5.8i − 16.9 j)
50.
= (18 − 12.4 − 5.8)i + (13.1 + 3.8 − 16.9) j =0 The forces are in equilibrium. 51.
345 F2 = 345cos 22.4°
F1 + F2 + F3 = (155i − 257 j) + ( −124i + 149 j) + ( −31i + 98 j)
52.
F2
345 F2 = 811cos31.8° F2 ≈ 689 lb
F1 + F2 + F3 = ( −4.6i + 5.3j) + (6.2i + 4.9 j) + ( −1.6i − 10.2 j)
F1 + F2 + F3 = (23.5i + 18.9 j) + ( −18.7i + 2.5 j) + ( −5.6i − 15.6 j) = (23.5 − 18.7 − 5.6)i + (18.9 + 2.5 − 15.6) j = −0.8i + 5.8 j
The forces are not in equilibrium. F4 = 0i + 10 j F1 + F2 + F3 = (189.3i + 235.7 j) + (45.8i − 205.6 j)
cos31.8° =
= ( −4.6 + 6.2 − 1.6)i + (5.3 + 4.9 − 10.2) j =0 The forces are in equilibrium.
= (155 − 124 − 31)i + ( −257 + 149 + 98) j = 0i − 10 j
53.
F2
F2 ≈ 319 lb
F1 ≈ 131 lb 48.
cos 22.4° =
The forces are not in equilibrium. F4 = 0.8i − 5.8 j 54.
+ ( −175.2i − 37.7 j) + ( −59.9i + 7.6 j) = (189.3 + 45.8 − 175.2 − 59.9)i + (235.7 − 205.6 − 37.7 + 7.6) j =0 The forces are in equilibrium.
F1 = ( F1 cos144°) i + ( F1 sin144°) j
= ( 6223cos144°) i + ( 6223sin144°) j ≈ −5034i + 3658 j F1 + F2 + F3 = 0 F2 = − F1 − F3
= −(5034i + 3658 j) − (0i − 9450 j) = 5034i + 5792 j F2 = F2 cos49° + F2 sin 49° F2 = 5034cos49° + 5034sin 49° ≈ 3300 + 4370 ≈ 7670 lbs
55.
v ⋅ w = 3, −2 ⋅ 1, 3
56.
= 3(1) + (−2)3 = 3−6 = −3 58.
v ⋅ w = 2, −3 ⋅ 3, 2
= 2(3) + ( −3)2 =6−6 =0
v ⋅ w = 2, 4 ⋅ 0, 2
57.
v ⋅ w = 4, 1 ⋅ −1, 4
= 2(0) + (4)2 = 0+8 =8 59.
v ⋅ w = (i + 2 j) ⋅ (−i + j)
= 4(−1) + 1(4) = −4 + 4 =0 60.
v ⋅ w = (5i + 3 j) ⋅ (4i − 2 j)
= 1(−1) + 2(1) = −1 + 2 =1
Copyright © Houghton Mifflin Company. All rights reserved.
= 5(4) + 3(−2) = 20 − 6 = 14
460
Chapter 7: Applications of Trigonometry
61.
v ⋅ w = (6i − 4 j) ⋅ (−2i − 3 j) = 6(−2) + (−4)(−3) = −12 + 12 =0
64.
cosθ =
v⋅w v w
62.
v ⋅ w = (−4i + 2 j) ⋅ (−2i − 4 j) = (−4)(−2) + 2(−4) =8−8 =0
65.
cosθ =
1, − 5 ⋅ −2, 3
cosθ =
2
1 + ( −5)
2
2
cosθ =
2
(−2) + 3
1(−2) + (−5)(3) 26 13 −17 ≈ −0.9247 cosθ = 26 13 θ = 157.6° cos θ = v ⋅ w v w (5i − 2 j) ⋅ (2i + 5 j) cos θ = 52 + (−2) 2 22 + 52 5(2) + (−2)(5) cos θ = 29 29 0 cos θ = =0 29 29 θ = 90o
cosθ =
68.
2
3 + (−4)
2
cosθ =
2
6 + ( −12)
v⋅w v w
−1, 7 ⋅ 3, −2 (−1)2 + 7 2 32 + (−2)2
( −1)3 + 7(−2) 50 13 −17 = −0.6668 cosθ = 5 2 13 cosθ =
69.
cosθ = cosθ =
projw v = projw v =
2
v⋅w v w (5i + 2 j) ⋅ (−5i − 2 j) 2
5 + 22 (−5) 2 + ( −2)2
5(−5) + 2(−2) 29 29 −29 = −1 cosθ = 29 29
cosθ =
θ = 180o
Thus, the vectors are orthogonal. 71.
2
cosθ =
θ = 131.8o
(3i − 4 j) ⋅ (6i − 12 j) 2
2
cos θ = v ⋅ w v w (8i + j) ⋅ (−i + 8 j) cos θ = 82 + 12 (−1)2 + 82 8(−1) + (1)(8) cos θ = 65 65 0 cos θ = =0 65 65 θ = 90o
v⋅w v w
cosθ =
66.
θ = 45o
Thus, the vectors are orthogonal. 70.
cos θ = v ⋅ w v w 2, −1 ⋅ 3, 4 cos θ = 2 2 + (−1)2 32 + 42 2(3) + ( −1)4 cos θ = 5 25 2 cos θ = ≈ 0.1789 5 5 θ ≈ 79.7o
0, 3 ⋅ 2, 2 2
0 +3 2 +2 0(2) + 3(2) cosθ = 9 8 6 cosθ = ≈ 0.7071 6 2
cosθ =
67.
v⋅w v w
63.
v⋅w w 6, 7 ⋅ 3, 4 2
3 +4
2
=
18 + 28 46 = 5 25
3(6) + (−4)(−12) cosθ = 25 180 66 = 0.9839 cosθ = 5 180
θ = 10.3o 72.
projw v = projw v =
v⋅w w
73.
−7, 5 ⋅ −4, 1
( −4 )
2
2
+1
=
28 + 5 33 33 17 = = ≈ 8.0 17 17 17
projw v = projw v =
v⋅w w
−3, 4 ⋅ 2, 5 2
2 +5
2
Copyright © Houghton Mifflin Company. All rights reserved.
=
−6 + 20 14 14 29 = = ≈ 2.6 29 29 29
Section 7.3
74.
461
projw v = projw v =
76.
projw v = projw v =
v⋅w w
75.
2, 4 ⋅ −1, 5
( −1)
2
+5
2
=
18 9 26 −2 + 20 = = ≈ 3.5 13 26 26
v⋅w w
( −5)
projw v =
77.
( 5i + 2 j) ⋅ ( −5i + −2 j) = −25 − 4 = 2
+ ( −2 )
2
29
projw v =
−29 29
projw v = projw v =
= − 29 ≈ −5.4
78.
projw v = projw v =
80.
2
6 +3
20
5 = 5 ≈ 2.2 5
v⋅w w
( 3i − 4 j) ⋅ ( 3i − 4 j) = −18 − 48 = − 11 ( −6 )2 + 122
180
5
11 5 ≈ −4.9 5
W = F ⋅s W= F
−6 5
45
2
s cos α
W = (75)(15)(cos32°) W ≈ 954 foot-pounds
6 5 ≈ −2.7 5
W = F⋅s W= F
79.
( 2i + 2 j) ⋅ ( −4i − 2 j) = −8 − 4 =
=−
( 2i + j) ⋅ ( 6i + 3j) = 12 + 3 =
=−
v⋅w w
( −4 )2 + ( −2 )2
v⋅w w
82.
81. s cos α
W = (100)(25)(cos 42°) W ≈ 1858 foot-pounds
W = F ⋅s W= F
s cos α
W = F ⋅s W= F
W ≈ 779 foot-pounds
W = (50)(6)(cos 48°) W ≈ 201 foot-pounds
.......................................................
Connecting Concepts 84.
83.
Thus, the sum is 6, 9 .
s cos α
W = (75)(12)(cos30°)
Thus, u + v − w = 5, − 4 .
Copyright © Houghton Mifflin Company. All rights reserved.
462
Chapter 7: Applications of Trigonometry
86.
85.
87.
v ⋅ w = (2i − 5 j) ⋅ (5i + 2 j) = 10 − 10 =0
The vector from P1(3, −1) to
P2(5, −4) is equivalent to 2i − 3 j. The vector from P1(−2, 4) to P2(−3, 7) is equivalent to − 1, 3 .
88.
89.
90.
v = −2, 7
The vectors are not perpendicular.
92.
( u ⋅ v ) ⋅ w = ⎡⎣( ai + bj) ⋅ ( ci + dj)⎤⎦ ⋅ ( ei + fj) . = ( ac + bd ) ⋅ ( ei + fj)
Let v = a, b
Therefore, v ⋅ w = w ⋅ v
cv = ca, ab
ca, ab ⋅ d , e = cad + cbe
Therefore, c ( v ⋅ w ) = ( cv ) ⋅ w.
95.
Let θ be the angle between vectors v and w. v ⋅ w = v w cosθ
v ⋅ w is positive if cos θ is positive. cos θ is positive when 0° < θ < 90°. This is an acute angle. v ⋅ w is negative if cos θ is negative. cos θ is negative when 90° < θ < 180°. This is an obtuse angle.
c ( v ⋅ w ) = c a, b ⋅ d , e = c ( ad + be ) = cad + cbe
( cv ⋅ w ) =
Let v = a, b and w = c, d .
w ⋅ v = c, d ⋅ a, b = ca + db = ac + bd
94.
and w = d , e
Thus, u = −1, 4 is one example.
v ⋅ w = a, b ⋅ c, d = ac + bd
ac + bd is a scalar quantity. The product of a scalar and a vector is not defined. Therefore, no, ( u ⋅ v ) ⋅ w does not equal u ⋅ ( v ⋅ w ) .
93.
a = −1b 4 Let b = 4 a = −1
Thus, u = 7, 2 is one example.
Let u = ci + bj, v = ci + dj, and w = ei + fj.
w = 4i + j 4, 1 ⋅ a , b = 0 4a + b = 0
−2, 7 ⋅ a, b = 0 −2a + 7b = 0 a = 7b 2 Let b = 2 a=7
v ⋅ w = 5, 6 ⋅ 6, 5 = 30 + 30 = 60 ≠ 0
91.
The two vectors are perpendicular.
Neither. If the force and the distance are the same, the work will be the same.
.......................................................
Prepare for Section 7.4 2 + i ⋅ 3 + i = 6 + 5i + i 2 = 5 + 5i = 1 + 1 i 3−i 3+i 10 2 2 9 − i2
PS1. (1 + i )(2 + i ) = 2 + 3i + i 2 = 1 + 3i
PS2.
PS3. 2 – 3i
PS4. 3 + 5i
PS5. x =
−1 ± 12 − 4(1)(1) −1 ± −3 = = −1 ± 3i 2(1) 2 2 2
PS6. x 2 + 9 = 0 x 2 = −9 x = ± 3i
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.4
463
Section 7.4 1.
2.
z =
( −2 )2 + ( −2 )2
3.
z = 42 + ( −4 )
= 8=2 2
z =
2
=4 2
( 3 )2
z = 0 2 + (− 2 )2 = 2
=2
7.
+ ( −1)
2
6.
5.
z = 12 +
2
= 3 +1 = 4 =2
= 32
4.
( 3)
8.
z =
9.
(− 5)2 + 0 2
=5
r = 12 + (−1)2 r= 2
α = tan −1 −1 1
z = 3 2 + ( −5 )
−1
2
z =
= 34
10.
r=
+ ( −4 )
= tan 1 = 45° θ = 360° − 45° = 315°
2
= 41
(− 4)2 + (− 4)2
11.
r=
z = 2 cis 315°
( 3 ) 2 + ( −1) 2
r = 32
r=2
r=4 2
α = tan −1 −4
= tan −1 1 = 45o
α = 360o − 30o = 330o
z = 4 2 cis 225o
z = 2 cis 330o
14.
r = 02 + ( −2 )
r = 12 + ( 3 )
α = tan −1 = tan −1
3 1 3 = 60o
θ = 60o z = 2 cis 60o
2
15.
r=
( −5)2 + 02
r=2
r =5
θ = 90
θ = 270o
θ = 180o
z = 3 cis 90o
z = 3 cis 270o
z = 5 cis 180o
o
2
r=2
θ = 180o + 45o = 225o
r = 02 + 32 r =3
12.
−1 3 1 = tan −1 = 30o 3
α = tan −1 −4
13.
( −5 )
2
Copyright © Houghton Mifflin Company. All rights reserved.
464
16.
Chapter 7: Applications of Trigonometry
r = 32 + 02 r =3
17.
2
20.
−2 3 −2
2 2 −2 2
= tan −1 1 = 45o
3 = 60o
θ = 180o − 45o = 135o
z = 16 cis 120o
z = 4 cis 135o
(
r=
2 ) + (− 2 ) 2
2
21.
z = 2(cos 45o + i sin 45o ) ⎛ 2 2 ⎞ + z = 2 ⎜⎜ i⎟ 2 ⎟⎠ ⎝ 2
− 2 2
z = 2 +i 2
= tan −1 1 = 45o
3 = 60o
θ = 180o + 60o = 240o
θ = 360o − 45o = 315o
z = 4 cis 240o
z = 2 cis 315o
z = 3(cos 240o + i sin 240o )
23.
z = cos 315o + i sin 315o
24.
z = 5(cos 120o + i sin 120o )
⎛ 1 3 ⎞ z = 5 ⎜⎜ − + i ⎟⎟ ⎝ 2 2 ⎠
z= 2 − 2i 2 2
3 3 3 z=− − i 2 2
5 5 3 z=− + i 2 2
z = 6 cis 135o o
( −2 2 ) 2 + ( 2 2 ) 2
θ = 180o − 60o = 120o
α = tan −1
⎛ 1 3 ⎞ z = 3 ⎜⎜ − + i ⎟⎟ ⎝ 2 2 ⎠
25.
r=
α = tan −1
r=2
r=4
22.
8 3 −8
= tan −1
2 r = ( −2 ) + ( −2 3 )
= tan −1
18.
r=4
α = tan −1
z = 3 cis 0o
α = tan −1
2
r = 16
θ = 0o
19.
2 r = ( −8 ) + ( 8 3 )
26. o
z = 6(cos 135 + i sin 135 ) ⎛ ⎞ z = 6 ⎜− 2 + 2 i⎟ 2 ⎠ ⎝ 2
z = cis 315° z = cos 315° + i sin 315° z=
27.
z = 8 cis 0° z = 8(cos 0o + i sin 0o ) z = 8(1 + 0i ) z =8
2 2 − i 2 2
z = −3 2 + 3i 2
28.
31.
z = 5 cis 90° z = 5(cos 90° + i sin 90°) z = 5(0 + i ) z = 5i
z = 3 ⎛⎜ cos 3π + i sin 3π ⎞⎟ 2 2 ⎠ ⎝ z = 3(0 − i ) z = −3i
29.
32.
z = 2 ⎛⎜ cos 5π + i sin 5π ⎞⎟ 6 6 ⎠ ⎝
30.
z = 4 ⎛⎜ cos 5π + i sin 5π ⎞⎟ 3 3 ⎠ ⎝
⎛ ⎞ z = 2 ⎜− 3 + 1 i⎟ 2 2 ⎝ ⎠
⎛ ⎞ z = 4 ⎜ 1 − 3 i⎟ 2 2 ⎝ ⎠
z = − 3+i
z = 2 − 2i 3
z = 5(cos π + i sin π )
33.
3π 4 3π 3π ⎞ ⎛ = 8 ⎜ cos + i sin ⎟ 4 4 ⎠ ⎝
z = 8 cis
z = 5( −1 + 0i ) z = −5
⎛ 2 i 2⎞ + z = 8 ⎜⎜ − ⎟ 2 2 ⎟⎠ ⎝ z = −4 2 + 4i 2
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.4
34.
37.
465
z = 9 cis 4π 3 ⎛ = 9 ⎜ cos 4π + i sin 4π ⎞⎟ 3 3 ⎠ ⎝ ⎛ 1 i 3⎞ z = 9 ⎜− + ⎟ 2 ⎠ ⎝ 2 z = −9 − 9 3i 2 2
35.
z = 9 cis 11π 6 ⎛ 11 z = 9 ⎜ cos π + i sin 11π ⎞⎟ 6 6 ⎠ ⎝ ⎛ 3 1 ⎞ − i⎟ z =9 ⎜ ⎝ 2 2 ⎠ z = 9 3 − 9i 2 2
z = 2 cis 2 z = 2(cos 2 + i sin 2) z ≈ 2( −0.4161 + 0.9093i )
z1z2 = 2 cis 30° ⋅ 3 cis 225°
z = 5 cis 4 z = 5(cos 4 + i sin 4) z ≈ 5( −0.6536 − 0.7568i ) z ≈ −3.268 − 3.784i
40.
z1z2 z1z2 z1z2 z1z2
42.
z1z2 = 8(cos 88° + i sin 88°) ⋅ 12(cos 112° + i sin 112°) = 96[cos(88° + 112°) + i sin(88° + 112°)] = 96[cos 200° + i sin 200°] = 96 cis 200°
z1z2 = 6 cis(30° + 225°) z1z2 = 6 cis 255°
41.
43.
45.
47.
z1z2 z1z2 z1z2 z1z2
= 3(cos 122° + i sin 122°) ⋅ 4(cos 213° + i sin 213°) = 12[cos(122° + 213°) + i sin(122° + 213°)] = 12(cos 335° + i sin 335°) = 12 cis 335°
z1z2 = 5 ⎛⎜ cos 2π + i sin 2π ⎞⎟ ⋅ 2 ⎛⎜ cos 2π + i sin 2π ⎞⎟ 3 3 ⎠ ⎝ 5 5 ⎠ ⎝ ⎡ ⎛ 2π 2π ⎞ ⎤ ⎛ ⎞ π π 2 2 + + z1z2 = 10 ⎢cos ⎜ ⎟ + i sin ⎜ ⎟ 5 ⎠ 5 ⎠ ⎥⎦ ⎝ 3 ⎣ ⎝ 3 z1z2 = 10 ⎛⎜ cos 16π + i sin 16π ⎞⎟ 15 15 ⎠ ⎝ π 16 z1z2 = 10 cis 15 z1z2 = 4 cis 2.4 ⋅ 6 cis 4.1
z = cis 3π 2 3 z = cos π + i sin 3π 2 2 z =0−i z = −i
38.
z ≈ −0.832 + 1.819i
39.
36.
44.
= 4 cis 120° ⋅ 6 cis 315° = 24 cis(120° + 315°) = 24 cis 435° = 24 cis 75°
z1z2 = 5 cis 11π ⋅ 3 cis 12 z1z2 = 15 cis ⎛⎜ 11π + 4π 3 ⎝ 4 49 π z1z2 = 15 cis 12
4π 3 ⎞ ⎟ ⎠
z1z2 = 15 cis π 12
46.
z1 z 2 = 7 cis 0.88 ⋅ 5 cis 1.32
z1z2 = 24 cis (2.4 + 4.1)
z1 z 2 = 35 cis (0.88 + 1.32)
z1z2 = 24 cis 6.5
z1 z 2 = 35 cis 2.2
z1 32 cis 30o = z2 4 cis 150o z1 = 8 cis(30° − 150°) z2
48.
z1 = 8 cis( − 120°) z2 z1 = 8 (cos 120° − i sin 120°) z2 z1 =8 z2
⎛ 1 i 3⎞ ⎜⎜ − − ⎟ = −4 − 4i 3 2 ⎟⎠ ⎝ 2
z1 15 cis 240o = z2 3 cis 135o z1 = 5 cis (240o − 135o ) z2 z1 = 5 cis 105o z2 z1 = 5 (cos 105o + i sin 105o ) z2 z1 ≈ 5 ( − 0.2588 + 0.9659i ) z2 z1 ≈ −1.294 + 4.830i z2
Copyright © Houghton Mifflin Company. All rights reserved.
466
49.
Chapter 7: Applications of Trigonometry
z1 27(cos 315o + i sin 315o ) = z2 9(cos 225o + i sin 225o ) z1 = 3 [cos(315o − 225o ) + i sin(315o − 225o )] z2
50.
z1 = 3(cos 90o + i sin 90o ) = 3(0 + i ) = 3i z2
51.
(
) 6 )
2π 2π z1 12 cos 3 + i sin 3 = z2 4 cos 11π + i sin 11π
(
6
52.
55.
⎞⎤ ⎟⎥ ⎠⎦
z1 25 cis 3.5 = 5 cis 1.5 z2
54.
= 5 cis 2 = 5 (cos 2 + i sin 2) ≈ 5 ( − 0.4161 + 0.9093i ) ≈ −2.081 + 4.546i
z1 = 1 − i 3
r1 = 2
( 3)
3(cos 175o + i sin 175o )
= 3 [cos(25o − 175o ) + i sin(25o − 175o )] = 3(cos 150o − i sin 150o ) ⎛ ⎞ = 3⎜ − 3 − 1 i ⎟ = − 3 3 − 3 i 2 2 ⎝ 2 2 ⎠
4
⎡ ⎤ = 2 ⎢cos ⎛⎜ π − π ⎞⎟ + i sin ⎛⎜ π − π ⎞⎟ ⎥ ⎝3 4⎠ ⎝ 3 4 ⎠⎦ ⎣ = 2 ⎛⎜ cos π + i sin π ⎞⎟ 12 12 ⎠ ⎝ ≈ 2(0.9659 + 0.2588i ) ≈ 1.932 + 0.518i
z1 18 cis 0.56 = 6 cis 1.22 z2 z1 z2 z1 z2 z1 z2 z1 z2 z1 z2
= 5 cis (3.5 − 1.5)
r1 = 12 +
9(cos 25o + i sin 25o )
π π z1 10(cos 3 + i sin 3 ) = z2 5(cos π + i sin π )
z1 z2 z1 z2 z1 z2 z1 z2
z1 ⎡ ⎤ = 3 ⎢ − 3 − ⎛⎜ − 1 i ⎞⎟ ⎥ z2 ⎣ 2 ⎝ 2 ⎠⎦ z1 = − 3 3 + 3i 2 2 z2
z1 z2 z1 z2 z1 z2 z1 z2 z1 z2
=
4
z1 ⎡ = 3 ⎢ cos ⎛⎜ 2π − 11π ⎞⎟ + i sin ⎛⎜ 2π − 11π 6 ⎠ 6 z2 ⎝ 3 ⎝ 3 ⎣ z1 ⎛ ⎞ = 3 ⎜ cos 7π − i sin 7π ⎟ 6 6 ⎠ z2 ⎝
53.
z1 z2 z1 z2 z1 z2 z1 z2
= 3 cis (0.56 − 1.22) = 3 cis ( − 0.66) = 3 (cos 0.66 − i sin 0.66) ≈ 3 (0.7900 − 0.6131i ) ≈ 2.370 − 1.839i
z2 = 1 + i 2
α = tan −1
− 3 = 60o 1
θ1 = 300o
r2 = 12 + 12 r2 = 2
α = tan −1 θ 2 = 45o
z2 = 2(cos 45° + i sin 45°)
z1 = 2(cos300° + i sin 300°)
z1z2 = 2(cos 300° + i sin 300°) ⋅ 2(cos 45° + i sin 45°) z1z2 = 2 2[cos(300° + 45°) + i sin(300° + 45°)] z1z2 = 2 2(cos 345° + i sin 345°) z1z2 ≈ 2.732 − 0.732i
Copyright © Houghton Mifflin Company. All rights reserved.
1 = 45o 1
Section 7.4
56.
467
z1 = 3 − i
z2 = 1 + i 3
α = tan −1
r1 = ( 3)2 + (−1) 2 r1 = 2
−1 = 30° 3
θ1 = 330°
z1 = 2(cos 330o + i sin 330o )
r2 = 12 +
( 3)
2
r2 = 2
α = tan −1
3 = 60° 1
θ 2 = 60°
z2 = 2(cos 60° + i sin 60°)
z1z2 = 2(cos 330° + i sin 330°) ⋅ 2(cos 60° + i sin 60°)
z1z2 = 4[cos(330° + 60°) + i sin(330° + 60°)] z1z2 = 4(cos 390° + i sin 390°) ⎡ 3 i⎤ z1z2 = 4 ⎢ + ⎥ = 2 3 + 2i ⎣⎢ 2 2 ⎥⎦
57.
z1 = 3 − 3i
z2 = 1 + i
r1 = 32 + (−3)2
α = tan −1
r1 = 3 2
θ1 = 315°
−3 = 45° 3
r2 = 2
z1 = 3 2(cos 315° + i sin 315°) z1z2 z1z2 z1z2 z1z2 z1z2
58.
α = tan −1
r2 = 12 + 12
1 = 45° 1
θ 2 = 45°
z2 = 2(cos 45° + i sin 45°)
= 3 2(cos 315° + i sin 315°) ⋅ 2(cos 45° + i sin 45°) = 6[cos(315° + 45°) + i sin(315° + 45°)] = 6 ( cos 360° + i sin 360° ) = 6 + 0i =6
z1 = 2 + 2i
z2 = 3 − i
r1 = 22 + 22
α = tan −1 2 = 45o
r2 =
r1 = 2 2
θ1 = 45o
r2 = 2
2
z1 = 2 2(cos 45° + i sin 45°)
( 3)
2
+ (−1)2
α = tan −1 −1 = 30o 3
o
θ 2 = 330
z2 = 2(cos 330° + i sin 330°)
z1z2 = 2 2(cos 45° + i sin 45°) ⋅ 2(cos 330° + i sin 330°) z1z2 = 4 2[cos(45° + 330°) + i sin(45° + 330°)] z1z2 = 4 2(cos 375° + i sin 375°) z1z2 ≈ 5.4641 + 1.4641i
59.
z1 = 1 + i 3
r1 = 12 + ( 3)2 r1 = 2
z2 = 1 − i 3
α1 = tan −1 3 = 60o 1
θ1 = 60o
z1 = 2(cos 60° + i sin 60°)
r2 = 12 + ( 3)2 r2 = 2 z2 = 2(cos 300° + i sin 300°)
z1 2(cos 60° + i sin 60°) = z2 2(cos 300° + i sin 300°) z1 = cos(60° − 300°) + i sin(60° − 300°) z2 z1 = cos 240° − i sin 240° = − 1 + 3 i 2 2 z2
Copyright © Houghton Mifflin Company. All rights reserved.
α = tan −1 − 3 = 60o 1
θ 2 = 300o
468
60.
Chapter 7: Applications of Trigonometry
z1 = 1 + i
z2 = 1 − i
α1 = tan −1 1 = 45o
r1 = 12 + 12
1
r1 = 2
o
θ1 = 45
z1 = 2(cos 45° + i sin 45°)
r2 = 12 + ( −1)2
α 2 = tan −1 −1 = 45°
r2 = 2
θ 2 = 315°
1
z2 = 2(cos 315° + i sin 315°)
z1 2(cos 45° + i sin 45°) = z2 2(cos 315° + i sin 315°)
z1 = cos(45° − 315°) + i sin(45° − 315°) z2 z1 = cos 270° − i sin 270° = 0 − i(−1) = 0 + 1i = i z2
61.
z1 = 2 − i 2
z2 = 1 + i
r1 = ( 2)2 + ( 2)2 r1 = 2
α1 = tan −1 − 2 = 45o 2
o
θ1 = 315
r2 = 12 + 12
α 2 = tan −1 1 = 45°
r2 = 2
θ 2 = 45°
1
z2 = 2(cos 45° + i sin 45°)
z1 = 2(cos 315° + i sin 315°)
z1 2(cos 315° + i sin 315°) = z2 2(cos 45° + i sin 45°) z1 = 2[cos(315° − 45°) + i sin(315° − 45°)] z2 z1 = 2(cos 270° + i sin 270°) z2 z1 = 2[0 + i (−1)] = 2(0 − 1i ) = 0 − 2i = − 2i or − i 2 z2
62.
z1 = 1 + i 3
r1 = 12 + ( 3)2 r1 = 2
z2 = 4 − 4i
3 = 60o 1
α1 = tan −1 θ1 = 60o
r2 = 42 + ( −4)2
α 2 = tan −1
r2 = 4 2
θ 2 = 315°
z2 = 4 2(cos 315° + i sin 315°)
z1 = 2(cos 60° + i sin 60°) z1 2(cos 60° + i sin 60°) = z2 4 2(cos 315° + i sin 315°)
z1 2 [cos(60° − 315°) + i sin(60° − 315°)] = z2 4 z1 2 [cos( − 255°) + i sin( − 255°)] = z2 4 z1 ≈ −0.0915 + 0.3415i z2
Copyright © Houghton Mifflin Company. All rights reserved.
−4 = 45° 4
Section 7.4
469
....................................................... 63.
Connecting Concepts
z1 = 3 − 1
z2 = 2 + 2i
z3 = 2 − 2i 3
r1 = ( 3)2 + (−1) 2
r2 = 22 + 22
r3 = 22 + (−2 3)2
r1 = 2
r2 = 2 2
r3 = 4
α1 = tan −1
−1 = 30° 3
α 2 = tan −1
θ1 = 330° z1 = 2(cos 330° + i sin 330°)
2 = 45o 2
α3 = tan −1
−2 3 = 60° 2
θ 2 = 45o
θ 3 = 300°
z2 = 2 2(cos 45° + i sin 45°)
z3 = 4(cos 300° + i sin 300°)
z1z2 z3 = 2(cos 330° + i sin 330°) ⋅ 2 2(cos 45° + i sin 45°) ⋅ 4(cos 300° + i sin 300°) z1z2 z3 = 16 2[cos(330° + 45° + 300°) + i sin(330° + 45° + 300°)] z1z2 z3 = 16 2(cos 675° + i sin 675°) z1z2 z3 = 16 2(cos 315° + i sin 315°) 1 ⎞ ⎛ 1 − z1z2 z3 = 16 2 ⎜ i ⎟ = 16 − 16i 2 ⎠ ⎝ 2
64.
z1 = 1 − i
z2 = 1 + i 3
z3 = 3 − i
r1 = 12 + (−1)2
r2 = 12 + ( 3)2
r3 = ( 3)2 + (−1)2
r1 = 2
r2 = 2
r3 = 2
α1 = tan −1
−1 = 45o 1
α 2 = tan −1
3 = 60° 1
α3 = tan −1
−1 = 30° 3
θ1 = 315o
θ 2 = 60°
θ 3 = 330°
z1 = 2(cos 315° + i sin 315°)
z2 = 2(cos 60° + i sin 60°)
z3 = 2(cos 330° + i sin 330°)
z1z2 z3 = 2(cos 315° + i sin 315°) ⋅ 2(cos 60° + i sin 60°) ⋅ 2(cos 330° + i sin 330°) z1z2 z3 = 4 2[cos(315° + 60° + 330°) + i sin(315° + 60° + 330°)] z1z2 z3 = 4 2(cos 705° + i sin 705°) z1z2 z3 = 4 2(cos 345° + i sin 345°) z1z2 z3 ≈ 5.4641 − 1.4641i
Copyright © Houghton Mifflin Company. All rights reserved.
470
65.
Chapter 7: Applications of Trigonometry
z2 = 1 − i 3
z1 = 3 + i 3 2
r1 = ( 3) + ( 3)
2
z3 = 2 − 2i
2
r2 = 1 + ( − 3)
2
r3 = 22 + (−2)2
r2 = 2
r1 = 6 3 = 45o 3
α1 = tan −1
α 2 = tan −1
r3 = 2 2 − 3 = 60° 1
α3 = tan −1 θ 3 = 315°
θ 2 = 300°
θ1 = 45o z1 = 6(cos 45° + i sin 45°)
−2 = 45° 2
z2 = 2(cos 300° + i sin 300°)
z3 = 2 2(cos 315° + i sin 315°)
z1 6(cos 45° + i sin 45°) = z2 z3 2(cos 300° + i sin 300°) ⋅ 2 2(cos 315° + i sin 315°) z1 6(cos 45° + i sin 45°) = z2 z3 4 2[cos(300° + 315°) + i sin(300° + 315°)] z1 6(cos 45° + i sin 45°) = z2 z3 4 2(cos 255° + i sin 255°) z1 = 3 [cos(45° − 255°) + i sin(45° − 255°)] 4 z2 z3 z1 ⎛ ⎞ = 3 (cos 210° − i sin 210°) = 3 ⎜ − 3 + i ⎟ = − 3 + 3 i z2 z3 4 4 ⎝ 2 2⎠ 8 8
66.
z1 = 2 − 2i 3
z2 = 1 − i 3
z3 = 4 3 + 4i
r1 = 22 + ( −2 3)2
r2 = 12 + ( − 3) 2
r3 = (4 3)2 + 42
r1 = 4
r2 = 2
r3 = 8
α1 = tan −1
−2 3 = 60° 2
α 2 = tan −1
− 3 = 60° 1
α3 = tan −1
1 = 30° 3
θ1 = 300°
θ 2 = 300°
θ 3 = 30°
z1 = 4(cos 300° + i sin 300°)
z2 = 2(cos 300° + i sin 300°)
z3 = 8(cos 30° + i sin 30°)
z1 4(cos 300° + i sin 300°) ⋅ 2(cos 300° + i sin 300°) = 8(cos 30° + i sin 30°) z2 z3 z1 = 4 ⋅ 2 [cos(300° + 300° − 30°) + i sin(300° + 300° − 30°)] 8 z2 z3 z1 = (cos 210° + i sin 210°) = − 3 − i 2 2 z2 z3
67.
z1 = 1 − 3i
z2 = 2 + 3i
z3 = 4 + 5i
r1 = 12 + (−3)2
r2 = 22 + 32
r3 = 42 + 52
r1 = 10
r2 = 13
r3 = 41
α1 = tan −1 −3 ≈ 71.57 o
α 2 = tan −1 3 ≈ 56.31o
α 3 = tan1 5 ≈ 51.34°
θ1 = 288.43o
θ 2 = 56.31o
θ 3 = 51.34°
z1 = 10(cos 288.4° + i sin 288.4°)
z2 = 13(cos56.3° + i sin 56.3°)
z3 = 41(cos51.3° + i sin 51.3°)
2
1
z1z2 z3 = 10(cos 288.4° + i sin 288.4°) ⋅ 13(cos56.3° + i sin 56.3°) ⋅ 41(cos51.3° + i sin 51.3°) z1z2 z3 = 10 ⋅ 13 ⋅ 41[cos(288.43° + 56.31° + 51.34°) + i sin(288.43° + 56.31° + 51.34°)] z1z2 z3 ≈ 73.0(cos396.08° + i sin 396.08°) z1z2 z3 = 73.0(cos36.08° + i sin 36.08°) z1z2 z3 ≈ 59.0 + 43.0i Copyright © Houghton Mifflin Company. All rights reserved.
4
Section 7.4
68.
471
z1 = 2 − 5i
z2 = 1 − 6i
z3 = 3 + 4i
r1 = 22 + (−5) 2
r2 = 12 + ( −6) 2
r3 = 32 + 42
r1 = 29
r2 = 37
r3 = 5
−5 ≈ 68.1986° 2 θ1 = 291.8014°
α 2 = tan −1
θ 2 = 279.4623o
θ 3 = 53.1301o
z1 = 29 cis 291.8014°
z2 = 37 cis 279.4623°
z3 = 5 cis 53.1301°
α1 = tan −1
−6 ≈ 80.5377o 1
α3 = tan −1
4 ≈ 53.1301o 3
z1z2 29 cis 291.8014° ⋅ 37 cis 279.4623° = 5 cis 53.1301° z3 29 ⋅ 37 cis (291.8014° + 279.4623° − 53.1301°) 5 29 ⋅ 37 (cos518.1336° + i sin 518.1336°) = 5 ≈ −6.0800 + 2.4400i =
69.
z = r (cosθ + i sin θ )
z = r (cosθ − i sin θ )
70.
z ⋅ z = r (cosθ + i sin θ ) ⋅ r (cosθ − i sin θ ) z ⋅ z = r (cosθ + i sin θ ) ⋅ r[cos(−θ ) + i sin(−θ )]
z = r (cosθ + i sin θ )
z = r (cosθ − i sin θ )
z r (cosθ + i sin θ ) = z r (cosθ − i sin θ ) cosθ + i sin θ = cos(−θ ) + i sin(−θ ) = cos(θ + θ ) + i sin(θ + θ ) = cos 2θ + i sin 2θ
z ⋅ z = r 2[cos(θ − θ ) + i sin(θ − θ )] z ⋅ z = r 2 (cos 0 + i sin 0) z ⋅ z = r 2 or a 2 + b 2
....................................................... 2
Prepare for Section 7.5
PS1. ⎛ 2 + 2 i ⎞ = 2 + 2 2 i + 2 i 2 = i ⎜ 2 2 ⎟⎠ 4 4 4 ⎝
PS2. x3 − 8 = ( x − 2)( x 2 + 2 x + 4)
PS3. x5 − 243 = ( x − 3)(3 x 4 + 3 x3 + 9 x 2 + 27 x + 81)
PS4. r = 22 + 22 = 2 2
( x 2 + 2 x + 4) yields 2 complex solutions ( x − 2) yields 1 real solution The real root is 2.
(3 x 4 + 3 x3 + 9 x 2 + 27 x + 81) yields 4 complex solutions ( x − 3) yields 1 real solution The real root is 3.
α = tan −1
2 2
= tan −1 1 = 45o
θ = 45o z = 2 2 cis 45o or 2 2 cis
⎛ 3 + 1 i⎞ = − 3 + i o o PS5. 2(cos150 + i sin150 ) = 2 ⎜ − ⎟ 2 2 ⎠ ⎝
2
PS6.
⎛ 2⎞ ⎛ 2⎞ z = ⎜⎜ ⎟⎟ + ⎜⎜ − ⎟⎟ 2 2 ⎝ ⎠ ⎝ ⎠ =
2
2 2 + 4 4
=1 Copyright © Houghton Mifflin Company. All rights reserved.
π 4
472
Chapter 7: Applications of Trigonometry
Section 7.5 1.
[2(cos30° + i sin 30°)]8 = 28[cos(8 ⋅ 30°) + i sin(8 ⋅ 30°)] = 256(cos 240° + i sin 240°)
2.
(cos 240° + i sin 240°)12 = cos(12 ⋅ 240°) + i sin(12 ⋅ 240°) = cos 2880° + i sin 2880° = cos 0° + i sin 0° = 1 + 0i =1
4.
[2(cos 45° + i sin 45°)]10 = 210[cos(10 ⋅ 45°) + i sin(10 ⋅ 45°)] = 1024(cos 450° + i sin 450°) = 1024(cos90° + i sin 90°) = 0 + 1024i = 1024i
6.
[2cis(330°)]4 = 24 cis(4 ⋅ 330°)
= −128 − 128i 3
3.
[2(cos 240° + i sin 240°)]5 = 25[cos(5 ⋅ 240°) + i sin(5 ⋅ 240°)] o
o
= 32[cos1200 + i sin1200 ] = 32(cos120o + i sin120o ) = −16 + 16i 3
5.
7.
[2cis(225°)]5 = 25 cis(5 ⋅ 225°) = 32(cos1125° + i sin1125°) = 32(cos 45° + i sin 45°)
= 16(cos1320° + i sin1320°) = 16(cos 240° + i sin 240°)
= 16 2 + 16i 2
= −8 − 8i 3
[2cis (120°)]6 = 26 cis ( 6 ⋅ 2π 3)
8.
[4cis(150°)]3 = 43 cis(3 ⋅ 5π 6) = 64(cos 450° + i sin 450°) = 64(cos90° + i sin 90°) = 64(0 + 1i ) = 0 + 64i = 64i
10.
z =1+ i 3
= 64(cos 720° + i sin 720°) = 64(cos 0° + i sin 0°) = 64
9.
z =1− i r = 12 + (−1)2 r= 2
α = tan −1
−1 = 45o 1
r = 12 + ( 3)2 r=2
θ = 315o z = 2(cos315° + i sin 315°)
10
(1 − i )
= [ 2(cos315° + i sin 315°)]
z =1+ i r = 1 +1 r= 2
= 28[cos(8 ⋅ 60°) + i sin(8 ⋅ 60°)] = 256(cos 480° + i sin 480°) = 256(cos120° + i sin120°) = −128 + 128i 3
12. 2
α = tan
−1 1
1
o
= 45
θ = 45o
z = 2(cos 45° + i sin 45°) 4
(1 + i ) = [ 2(cos 45° + i sin 45°)]4 = ( 2 )4 [cos(4 ⋅ 45°) + i sin(4 ⋅ 45°)] = 4(cos180° + i sin180°) = −4 + 0i = −4
θ = 60o
(1 + i 3)8 = [2(cos 60° + i sin 60°)]8
= ( 2) [cos(10 ⋅ 315°) + i sin(10 ⋅ 315°)] = 32(cos3150° + i sin 3150°) = 32(cos 270° + i sin 270°) = 0 − 32i = −32i
2
3 = 60o 1
z = 2(cos 60° + i sin 60°) 10
10
11.
α = tan −1
z = 2 − 2i 3
r = 22 + ( −2 3)2 r=4
α = tan −1
−2 3 = 60o 2
θ = 300o z = 4(cos300° + i sin 300°)
(2 − 2i 3)3 = [4(cos300° + i sin 300°)]3 = 43[cos(3 ⋅ 300°) + i sin(3 ⋅ 300°)] = 64(cos900° + i sin 900°) = 64(cos180° + i sin180°) = −64 + 0i = −64
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.5
13.
473
z = 2 + 2i 2
r = 2 +2
14. 2
α = tan
r=2 2
−1 2
2
o
z = 2 3 − 2i
r=4
θ = 45o
= 45[cos(5 ⋅ 330°) + i sin(5 ⋅ 330°)] = 1024(cos1650° + i sin1650°) = 1024(cos 210° + i sin 210°)
= 1024 2[cos(7 ⋅ 45°) + i sin(7 ⋅ 45°)] = 1024 2(cos315° + i sin 315°) = 1024 − 1024i
z=
= −512 3 − 512i
2 2 +i 2 2
16.
r = ( 2 2)2 + ( 2 2) 2 r =1
θ = 330o
(2 3 − 2i )5 = [4(cos330° + i sin 330°)]5
7
(2 + 2i ) = [2 2(cos 45° + i sin 45°)]
15.
α = tan −1
2 2 = 45° 2 2
z=−
2 2 +i 2 2
r = (− 2 2)2 + ( 2 2)2 r =1
θ = 45°
α = tan −1 θ = 135o
12
⎛ 2 −i 2 ⎞ ⎜− ⎟ 2 ⎠ ⎝ 2
6
⎛ 2 2⎞ 6 −i ⎜⎜ ⎟ = (cos 45° + i sin 45°) 2 ⎟⎠ ⎝ 2 = cos(6 ⋅ 45°) + i sin(6 ⋅ 45°) = cos 270° + i sin 270° = 0 − 1i = −i 9 = 9(cos 0 o + i sin 0 o )
18.
1 ⎛ 0o + 360o k 0o + 360o k ⎞ ⎟ + i sin wk = 9 2 ⎜ cos ⎜ ⎟ 2 2 ⎝ ⎠
w0 = 3(cos 0o + i sin 0o )
k = 0,1
= (cos135° + i sin135°)12 = cos(12 ⋅ 135°) + i sin(12 ⋅ 135°) = cos1620° + i sin1620° = cos180° + i sin180° = −1 + 0i = −1
16 = 16(cos 0o + i sin 0o ) 1 ⎛ 0o + 360o k 0o + 360o k ⎞ ⎟ + i sin wk = 16 2 ⎜ cos ⎜ ⎟ 2 2 ⎝ ⎠
w0 = 4(cos 0o + i sin 0o )
w0 = 3 + 0i = 3
w0 = 4 + 0i = 4
⎛ cos 0 + 360 0 + 360 ⎟ + i sin w1 = 3 ⎜ ⎜ ⎟ 2 2 ⎝ ⎠
⎛ 0o + 360o 0o + 360o ⎞ ⎟ + i sin w1 = 4 ⎜ cos ⎜ ⎟ 2 2 ⎝ ⎠
w1 = 3(cos180o + i sin180o )
w1 = 4(cos180o + i sin180o )
w1 = −3 + 0i = −3
w1 = −4 + 0i = −4
o
o
o
2 2 = 45o − 2 2
z = cos135° + i sin135°
z = cos 45° + i sin 45°
17.
−2 = 30o 2 3
z = 4(cos330° + i sin 330°)
z = 2 2(cos 45° + i sin 45°) 7
α = tan −1
r = (2 3)2 + (−2)2
= 45
o⎞
Copyright © Houghton Mifflin Company. All rights reserved.
k = 0,1
474
19.
Chapter 7: Applications of Trigonometry
64 = 64(cos 0° + i sin 0°) 0° + 360°k 0° + 360°k ⎞ ⎛ wk = 641/ 6 ⎜ cos + i sin ⎟ 6 6 ⎝ ⎠
k = 0, 1, 2, 3, 4, 5 0 + 360° 0 + 360° ⎞ ⎛ + i sin w1 = 2 ⎜ cos ⎟ 6 6 ⎝ ⎠ w1 = 2(cos 60° + i sin 60°)
w0 = 2(cos 0° + i sin 0°) w0 = 2 + 0i = 2
⎛1 3⎞ w1 = 2 ⎜⎜ + i ⎟⎟ 2 2 ⎝ ⎠
w1 = 1 + i 3 0° + 360° ⋅ 2 0° + 360° ⋅ 2 ⎞ ⎛ + i sin w2 = 2 ⎜ cos ⎟ 6 6 ⎝ ⎠ w2 = 2(cos120° + i sin120°)
0° + 360° ⋅ 3 0° + 360° ⋅ 3 ⎞ ⎛ + i sin w3 = 2 ⎜ cos ⎟ 6 6 ⎝ ⎠ w3 = 2(cos180° + i sin180°)
⎛ 1 3⎞ w2 = 2 ⎜⎜ − + i ⎟ 2 ⎟⎠ ⎝ 2
w3 = 2(−1 + 0i ) w3 = −2 + 0i = −2
w2 = −1 + i 3
20.
0° + 360° ⋅ 4 0° + 360° ⋅ 4 ⎞ ⎛ w4 = 2 ⎜ cos + i sin ⎟ 6 6 ⎝ ⎠ w4 = 2(cos 240° + i sin 240°)
0° + 360° ⋅ 5 0° + 360° ⋅ 5 ⎞ ⎛ w5 = 2 ⎜ cos + i sin ⎟ 6 6 ⎝ ⎠ w5 = 2(cos300° + i sin 300°)
⎛ 1 3⎞ w4 = 2 ⎜⎜ − − i ⎟ 2 ⎟⎠ ⎝ 2
⎛1 3⎞ w5 = 2 ⎜⎜ − i ⎟ 2 ⎟⎠ ⎝2
w4 = −1 − i 3
w5 = 1 − i 3
(
32 = 32 cos 0o + i sin 0o
)
0° + 360°k 0° + 360°k ⎞ ⎛ + i sin wk = 321/ 5 ⎜ cos ⎟ 5 5 ⎝ ⎠
k = 0, 1, 2, 3, 4 0° + 360° 0° + 360° ⎞ ⎛ + i sin w1 = 2 ⎜ cos ⎟ 5 5 ⎝ ⎠ w1 = 2(cos 72° + i sin 72°)
w0 = 2(cos 0° + i sin 0°) w0 = 2 + 0i = 2
w1 ≈ 2(0.3090 + 0.9511i ) w1 = 0.6180 + 1.9021i 0° + 360° ⋅ 2 0° + 360° ⋅ 2 ⎞ ⎛ + i sin w2 = 2 ⎜ cos ⎟ 5 5 ⎝ ⎠ w2 = 2(cos144° + i sin144°)
0° + 360°° ⋅ 3 0° + 360° ⋅ 3 ⎞ ⎛ w3 = 2 ⎜ cos + i sin ⎟ 5 5 ⎝ ⎠ w3 = 2(cos 216° + i sin 216°)
w2 ≈ 2( −0.8090 + 0.5878i )
w3 ≈ 2(−0.8090 − 0.5878i )
w2 = −1.6180 + 1.1756i
w3 = −1.6180 − 1.1756i
0° + 360° ⋅ 4 0° + 360° ⋅ 4 ⎞ ⎛ + i sin w4 = 2 ⎜ cos ⎟ 5 5 ⎝ ⎠ w4 = 2(cos 288° + i sin 288°) w4 ≈ 2(0.3090 − 0.9511i ) w4 = 0.6180 − 1.9021i
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.5
21.
475
−1 = 1(cos180° + i sin180°) 180° + 360°k 180° + 360°k ⎞ ⎛ wk = 11/ 5 ⎜ cos + i sin ⎟ 5 5 ⎝ ⎠
k = 0, 1, 2, 3, 4
180° + 360° 180° + 360° + i sin 5 5 w1 = cos108° + i sin108°
w0 = 1(cos36° + i sin 36°)
w1 = cos
w0 ≈ 0.809 + 0.588i
w1 ≈ −0.309 + 0.951i 180° + 360° ⋅ 3 180° + 360° ⋅ 3 + i sin 5 5 w3 = cos 252° + i sin 252°
180° + 360° ⋅ 2 180° + 360° ⋅ 2 + i sin 5 5 w2 = cos180° + i sin180° w2 = cos
w3 = cos
w2 = −1 + 0i = −1
w3 ≈ −0.309 − 0.951i
180° + 360° ⋅ 4 180° + 360° ⋅ 4 + i sin 5 5 w4 = cos324° + i sin 324° w4 = cos
w4 ≈ 0.809 − 0.588i
22.
−16 = 16(cos180° + i sin180°) 180° + 360°k 180° + 360°k ⎞ ⎛ + i sin wk = 161/ 4 ⎜ cos ⎟ 4 4 ⎝ ⎠
k = 0, 1, 2, 3 180° + 360° 180° + 360° ⎞ ⎛ + i sin w1 = 2 ⎜ cos ⎟ 4 4 ⎝ ⎠ w1 = 2(cos135° + i sin135°)
w0 = 2(cos 45° + i sin 45°) w0 = 2 ⎛⎜ 2 + i 2 ⎞⎟ 2 ⎠ ⎝ 2
w1 = 2 ⎛⎜ − 2 + i 2 ⎞⎟ 2 ⎠ ⎝ 2
w0 = 2 + i 2
w1 = − 2 + i 2
23.
180° + 360° ⋅ 2 180° + 360° ⋅ 2 ⎞ ⎛ + i sin w2 = 2 ⎜ cos ⎟ 4 4 ⎝ ⎠ w2 = 2(cos 225° + i sin 225°)
180° + 360° ⋅ 3 180° + 360° ⋅ 3 ⎞ ⎛ w3 = 2 ⎜ cos + i sin ⎟ 4 4 ⎝ ⎠ w3 = 2(cos315° + i sin 315°)
w2 = 2 ⎛⎜ − 2 − i 2 ⎞⎟ 2 ⎠ ⎝ 2
w3 ≈ 2 ⎛⎜ 2 − i 2 ⎞⎟ 2 ⎠ ⎝ 2
w2 = − 2 − i 2
w3 = 2 − i 2
1 = cos 0° + i sin 0° 0° + 360°k 0° + 360°k wk = cos + i sin 3 3 0° 0° + i sin 3 3 w0 = cos 0° + i sin 0°
k = 0, 1, 2
0° + 360° 0° + 360° + i sin 3 3 w1 = cos120° + i sin120°
w0 = cos
w1 = cos
w0 = 1 + 0i = 1
w1 = − 1 + 3 i 2
2
0o + 360o ⋅ 2 0o + 360o ⋅ 2 + i sin 3 3 w2 = cos 240° + i sin 240° w2 = cos
w2 = − 1 − 3 i
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2
2
476
24.
Chapter 7: Applications of Trigonometry
i = cos90° + i sin 90° 90° + 360°k 90° + 360°k wk = cos + i sin 3 3 90o 90o + i sin 3 3 w0 = cos30° + i sin 30°
90o + 360o 90o + 360o + i sin 3 3 w1 = cos150° + i sin150°
w0 = cos
w1 = cos
25.
2
2
2
1 + i = 2(cos 45° + i sin 45°) wk =
( 2)
1/ 4 ⎛
⎜ cos ⎝
45° + 360°k 45° + 360°k ⎞ + i sin ⎟ 4 4 ⎠
k = 0, 1, 2, 3
45° 45° ⎞ ⎛ + i sin w0 = 21/ 8 ⎜ cos ⎟ 4 4 ⎠ ⎝ w0 = 21/ 8 (cos11.25° + i sin11.25°) w0 ≈ 1.070 + 0.213i
26.
90o + 360o ⋅ 2 90o + 360o ⋅ 2 + i sin 3 3 w2 = cos 270° + i sin 270° w2 = cos
w2 = 0 − i = −i
w1 = − 3 + 1 i
w0 = 3 + 1 i 2
k = 0, 1, 2
⎛ 45o + 360o 45o + 360o ⎞ ⎟ w1 = 21/ 8 ⎜ cos + i sin ⎜ ⎟ 4 4 ⎝ ⎠ w1 = 21/ 8 (cos101.25° = i sin101.25°) w1 ≈ −0.213 − 1.070i
⎛ 45o + 360o ⋅ 2 45o + 360o ⋅ 2 ⎞ ⎟ w2 = 21/ 8 ⎜ cos + i sin ⎜ ⎟ 4 4 ⎝ ⎠
⎛ 45o + 360o ⋅ 3 45o + 360o ⋅ 3 ⎞ ⎟ + i sin w3 = 21/ 8 ⎜ cos ⎜ ⎟ 4 4 ⎝ ⎠
w2 = 21/ 8 (cos191.25° + i sin191.25°)
w3 = 21/ 8 (cos 281.25° + i sin 281.25°)
w2 ≈ −1.070 − 0.213i
w3 ≈ 0.213 − 1.070i
−1 + i = 2(cos135° + i sin135°) wk =
( 2)
1/ 5 ⎛
⎜ cos ⎝
135° + 360°k 135° + 360°k ⎞ + i sin ⎟ 5 5 ⎠
k = 0, 1, 2, 3, 4
⎛ 135o 135o ⎞ ⎟ w0 = 21/10 ⎜ cos + i sin ⎜ 5 5 ⎟⎠ ⎝
⎛ 135o + 360o 135o + 360o ⎞ ⎟ w1 = 21/10 ⎜ cos + i sin ⎜ ⎟ 5 5 ⎝ ⎠
w0 = 21/10 (cos 27° + i sin 27°)
w1 = 21/10 (cos99° + i sin 99°)
w0 ≈ 0.955 + 0.487i
w1 ≈ −0.168 + 1.059i
⎛ 135o + 360o ⋅ 2 135o + 360o ⋅ 2 ⎞ ⎟ w2 = 21/10 ⎜ cos + i sin ⎜ ⎟ 5 5 ⎝ ⎠
135° + 360° ⋅ 3 135° + 360° ⋅ 3 ⎞ ⎛ + i sin w3 = 21/10 ⎜ cos ⎟ 5 5 ⎝ ⎠
w2 = 21/10 (cos171° + i sin171°) w2 ≈ −1.059 + 0.168i
w3 = 21/10 (cos 243° + i sin 243°) w3 ≈ −0.487 − 0.955i
⎛ 135o + 360o ⋅ 4 135o + 360o ⋅ 4 ⎞ ⎟ w4 = 21/10 ⎜ cos + i sin ⎜ ⎟ 5 5 ⎝ ⎠ w4 = 21/10 (cos315° + i sin 315°) w4 ≈ 0.758 − 0.758i
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.5
27.
477
2 − 2i 3 = 4(cos300° + i sin 300°)
k = 0, 1, 2
300° + 360°k 300° + 360°k ⎞ ⎛ + i sin wk = 41/ 3 ⎜ cos ⎟ 3 3 ⎝ ⎠ 300° 300° ⎞ ⎛ + i sin w0 = 41/ 3 ⎜ cos ⎟ 3 3 ⎠ ⎝ w0 = 41/ 3 (cos100° + i sin100°)
⎛ 300o + 360o 300o + 360o ⎞ ⎟ w1 = 41/ 3 ⎜ cos + i sin ⎜ ⎟ 3 3 ⎝ ⎠ w1 = 41/ 3 (cos 220° + i sin 220°)
w0 ≈ −0.276 + 1.563i
w1 ≈ −1.216 − 1.020i
300° + 360° ⋅ 2 300° + 360° ⋅ 2 ⎞ ⎛ + i sin w2 = 41/ 3 ⎜ cos ⎟ 3 3 ⎝ ⎠ w2 = 41/ 3 (cos340° + i sin 340°) w2 ≈ 1.492 − 0.543i
28.
−2 + 2i 3 = 4(cos120° + i sin120°) 120° + 360°k 120° + 360°k ⎞ ⎛ + i sin wk = 41/ 3 ⎜ cos ⎟ 3 3 ⎝ ⎠ o o⎞ ⎛ 120 120 ⎟ w0 = 41/ 3 ⎜ cos + i sin ⎜ 3 3 ⎟⎠ ⎝
k = 0, 1, 2 120° + 360° 120° + 360° ⎞ ⎛ w1 = 41/ 3 ⎜ cos + i sin ⎟ 3 3 ⎝ ⎠ w1 = 41/ 3 (cos160° + i sin160°)
w0 = 41/ 3 (cos 40° + i sin 40°)
w1 ≈ −1.492 + 0.543i
w0 ≈ 1.216 + 1.020i 120° + 360° ⋅ 2 120° + 360° ⋅ 2 ⎞ ⎛ w2 = 41/ 3 ⎜ cos + i sin ⎟ 3 3 ⎝ ⎠ w2 = 41/ 3 (cos 280° + i sin 280°) w2 ≈ 0.276 − 1.563i
29.
−16 + 16i 3 = 32(cos120° + i sin120°) 120° + 360°k 120° + 360°k ⎞ ⎛ wk = 321/ 2 ⎜ cos + i sin ⎟ 2 2 ⎝ ⎠
k = 0, 1 o o o o⎞ ⎛ w1 = 4 2 ⎜ cos 120 + 360 + i sin 120 + 360 ⎟ 2 2 ⎝ ⎠ w1 = 4 2(cos 240° + i sin 240°)
w0 = 4 2 ⎛⎜ cos 120° + i sin 120° ⎞⎟ 2 2 ⎠ ⎝ w0 = 4 2(cos 60° + i sin 60°) w0 ≈ 2 2 + 2i 6
30.
w1 ≈ −2 2 − 2i 6
−1 + 3i = 2(cos120° + i sin120°) 120° + 360°k 120° + 360°k ⎞ ⎛ wk = 21/ 2 ⎜ cos + i sin ⎟ 2 2 ⎝ ⎠
k = 0, 1
w0 = 21/ 2 ⎜⎛ cos 120° + i sin 120° ⎟⎞ 2 2 ⎠ ⎝
w1 = 21/ 2 ⎜⎛ cos 120° + 360° + i sin 120° + 360° ⎟⎞ ⎝ ⎠ 2 2
w0 = 21/ 2 (cos 60° + i sin 60°)
w1 = 21/ 2 (cos 240° + i sin 240°)
w0 = 2 + 6 i 2 2
w1 = − 2 − 6 i 2 2
Copyright © Houghton Mifflin Company. All rights reserved.
478
31.
Chapter 7: Applications of Trigonometry
x3 + 8 = 0 x3 = −8
Find the three cube roots of –8. −8 = 8(cos180° + i sin180°) ⎛ 180o + 360°k 180o + 360°k ⎞ ⎟ + i sin xk = 81/ 3 ⎜ cos ⎜ ⎟ 3 3 ⎝ ⎠
k = 0,1, 2
o o⎞ ⎛ w0 = 2 ⎜ cos 180 + i sin 180 ⎟ 3 3 ⎠ ⎝ w0 = 2(cos 60° + i sin 60°) w0 = 2 cis 60°
w1 = 2 ⎜⎛ cos 180° + 360° + i sin 180° + 360° ⎟⎞ 3 3 ⎝ ⎠ w1 = 2(cos180° + i sin180°) w1 = 2 cis 180°
w2 = 2 ⎜⎛ cos 180° + 360° ⋅ 2 + i sin 180° + 360° ⋅ 2 ⎟⎞ 3 3 ⎝ ⎠ w2 = 2(cos300° + i sin 300°) w2 = 2 cis 300°
32.
x5 − 32 = 0 x5 = 32
Find the five fifth roots of 32. 32 = 32(cos 0° + i sin 0°) 0° + 360°k 0° + 360°k ⎞ ⎛ wk = 321/ 5 ⎜ cos + i sin ⎟ 5 5 ⎝ ⎠ 0° 5 w0 = 2 cis 0° w0 = 2 cis
0o + 360o 5
w1 = 2 cis
w1 = 2 cis 72o
0o + 360o ⋅ 3 5 w3 = 2 cis 216° w3 = 2 cis
33.
k = 0, 1, 2, 3, 4 0o + 360o ⋅ 2 5 w2 = 2 cis 144° w2 = 2 cis
0o + 360o ⋅ 4 5 w4 = 2 cis 288° w4 = 2 cis
x4 + i = 0 x 4 = −i
Find the four fourth roots of –i. −i = (cos 270o + i sin 270o ) 270° + 360°k 270° + 360°k wk = cos + i sin 4 4 270° 4 w0 = cis 67.5° w0 = cis
270° + 360° 4 w1 = cis 157.5° w1 = cis
k = 0, 1, 2, 3 270° + 360° ⋅ 2 4 w2 = cis 247.5° w2 = cis
270° + 360° ⋅ 3 4 w3 = cis 337.5° w3 = cis
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.5
34.
479
x3 − 2i = 0 x3 = 2i
Find the three cube roots of 2i. 2i = 2(cos90° + i sin 90°) 90° + 360°k 90° + 360°k ⎞ 3 ⎛ 90° + 360°k 90° + 360°k ⎞ ⎛ wk = 21/ 3 ⎜ cos + i sin + i sin ⎟ = 2 ⎜ cos ⎟ 3 3 3 3 ⎝ ⎠ ⎝ ⎠ w0 = 3 2 cis
90° 3
w1 = 3 2 cis
w2 = 3 2 cis
w1 = 3 2 cis 150°
w0 = 3 2 cis 30°
35.
90° + 360° 3
k = 0, 1, 2
90°+360° ⋅ 2 3
w2 = 3 2 cis 270°
x3 − 27 = 0 x3 = 27
Find the three cube roots of 27. 27 = 27(cos 0° + i sin 0°) 0° + 360°k 0° + 360°k ⎞ ⎛ wk = 3 ⎜ cos + i sin ⎟ 3 3 ⎝ ⎠ 0° 3 w0 = 3 cis 0°
0° + 360° 3 w1 = 3 cis 120°
w0 = 3 cis
36.
k = 0, 1, 2 0° + 360° ⋅ 2 3 w2 = 3 cis 240°
w1 = 3 cis
w2 = 3 cis
x5 + 32i = 0 x5 = −32i
Find the five fifth roots of –32i. −32i = 32(cos 270° + i sin 270°) 270° + 360°k 270° + 360°k ⎞ ⎛ wk = 321/ 5 ⎜ cos + i sin ⎟ 5 5 ⎝ ⎠ 270° 5 w0 = 2 cis 54°
270° + 360° 5 w1 = 2 cis 126°
w0 = 2 cis
270° + 360° ⋅ 2 5 w2 = 2 cis 198°
w1 = 2 cis
270° + 360° ⋅ 3 5 w3 = 2 cis 270° w3 = 2 cis
37.
k = 0, 1, 2, 3, 4 w2 = 2 cis
270° + 360° ⋅ 4 5 w4 = 2 cis 342° w4 = 2 cis
x 4 + 81 = 0 x 4 = −81
Find the four fourth roots of –81. −81 = 81(cos180° + i sin180°) 180° + 360°k 180° + 360°k ⎞ ⎛ wk = 811/ 4 ⎜ cos + i sin ⎟ 4 4 ⎝ ⎠ 180° 4 w0 = 3 cis 45° w0 = 3 cis
180° + 360° 4 w1 = 3 cis 135° w1 = 3 cis
k = 0, 1, 2, 3 0° + 360° ⋅ 2 5 w2 = 2 cis 225° w2 = 2 cis
Copyright © Houghton Mifflin Company. All rights reserved.
180° + 360° ⋅ 3 4 w3 = 3 cis 315° w3 = 3 cis
480
38.
Chapter 7: Applications of Trigonometry
x3 − 64i = 0 x3 = 64i
Find the three cube roots of 64i. 64i = 64(cos90° + i sin 90°) 90° + 360°k 90° + 360°k ⎞ ⎛ wk = 641/ 3 ⎜ cos + i sin ⎟ 3 3 ⎝ ⎠ 90° 3 w0 = 4 cis 30° w0 = 4 cis
39.
k = 0, 1, 2
90° + 360° 3 w1 = 4 cis 150°
90° + 360° ⋅ 2 3 w2 = 4 cis 270°
w1 = 4 cis
w2 = 4 cis
x 4 − (1 − i 3) = 0 x4 = 1 − i 3
Find the four fourth roots of 1 − i 3. 1 − i 3 = 2(cos300° + i sin 300°) ⎛ 300o + 360o k 300o + 360o k ⎞⎟ wk = 21/ 4 ⎜ cos + i sin ⎜ ⎟ 4 4 ⎝ ⎠ w0 = 4 2 cis
300° 4
w0 = 4 2 cis 75°
40.
w1 = 4 2 cis
k = 0, 1, 2, 3
300° + 360° 4
w2 = 4 2 cis
w1 = 4 2 cis 165°
300° + 360° ⋅ 2 4
w2 = 4 2 cis 255°
x3 = −2 3 + 2i
Find the three cube roots of −2 3 + 2i. −2 3 + 2i = 4(cos150° + i sin150°)
150° 3 w0 = 3 4 cis 50° w0 = 3 4 cis
41.
k = 0, 1, 2
150° + 360° 3 w1 = 3 4 cis 170°
150° + 360° ⋅ 2 3 w2 = 3 4 cis 290°
w1 = 3 4 cis
w2 = 3 4 cis
x3 + (1 + i 3) = 0 x3 = −1 − i 3 Find the three cube roots of −1 − i 3. −1 − i 3 = 2(cos 240° + i sin 240°) 240° + 360°k 240° + 360°k ⎞ ⎛ wk = 21/ 3 ⎜ cos + i sin ⎟ 3 3 ⎝ ⎠ w0 = 3 2 cis
240° 3
w0 = 3 2 cis 80°
w1 = 3 2 cis
k = 0, 1, 2
240° + 360° 3
w1 = 3 2 cis 200°
w2 = 3 2 cis
300° + 360° ⋅ 3 4
w3 = 4 2 cis 345°
x3 + (2 3 − 2i ) = 0
150° + 360°k 150° + 360°k ⎞ ⎛ wk = 41/ 3 ⎜ cos + i sin ⎟ 3 3 ⎝ ⎠
w3 = 4 2 cis
240° + 360° ⋅ 2 3
w2 = 3 2 cis 320°
Copyright © Houghton Mifflin Company. All rights reserved.
Section 7.5
42.
481
x6 − ( 4 − 4i ) = 0 x6 = 4 − 4i Find the six sixth roots of 4 − 4i. 4 − 4i = 4 2(cos315° + i sin 315°) 315° + 360°k 315° + 360°k ⎞ + i sin ⎟ 6 6 ⎠
wk = 4 2
(
)
w0 = 4 2
(
)
cis
315° 6
w1 = 4 2
(
)
cis
w0 = 4 2
(
)
cis 52.5°
w1 = 4 2
(
)
cis 112.5°
w3 = 4 2
(
)
cis
w4 = 4 2
(
)
cis
(
)
cis 232.5°
(
)
cis 292.5°
w3 = 4 2
1/ 6 ⎛
1/ 6 1/ 6
1/ 6 1/ 6
⎜ cos ⎝
315° + 360° ⋅ 3 6
1/ 6
w4 = 4 2
1/ 6
1/ 6 1/ 6
k = 0, 1, 2, 3, 4, 5
315° + 360° 6
315° + 360° ⋅ 4 6
(
)
cis
w2 = 4 2
(
)
cis 172.5°
w5 = 4 2
(
)
cis
(
)
cis 352.5°
w5 = 4 2
1/ 6 1/ 6
1/ 6 1/ 6
....................................................... 43.
Let z = a + bi. Then z = a − bi by definition. Substitute a = r cosθ and b = r sin θ . Thus z = rcosθ − ri sin θ = r (cosθ − i sin θ )
z = r (cos θ + i sin θ ) 2
2
44.
w5 = 1.33 ( cos352.5o + i sin 352.5o ) 1= 1 cos θ − i sin θ = z r (cos θ + i sin θ ) r (cos θ + i sin θ )(cosθ − i sin θ ) cos θ − i sin θ = = cosθ − i sin θ r (cos 2 θ + i 2 sin 2 θ ) r (cos 2 θ + sin 2 θ )
46.
z = r (cos 2θ + i sin 2θ ) 1 = 1 z 2 r 2 (cos 2θ + i sin 2θ ) cos 2θ − i sin 2θ = 2 r (cos 2θ + i sin 2θ )(cos 2θ − i sin 2θ ) 2θ = 2 cos22θ − i sin = cos 2θ − i sin 2θ r (cos 2θ − i 2 sin 2 2θ ) r 2 (cos2 2θ + sin 2 2θ ) z −2 = r −2 (cos 2θ − i sin 2θ )
315° + 360° ⋅ 5 6
Connecting Concepts
z −1 = r −1 (cos θ − i sin θ )
45.
315° + 360° ⋅ 2 6
w2 = 4 2
Exercises 42 and 43 imply that z − n = r − n (cos nθ − i sin nθ ).
z = 1 − i 3 = 2(cos300° + i sin 300°) z −4 = 2−4[cos 4(300°) − i sin 4(300°)] z −4 = 1 (cos1200° − i sin1200°)
16 z = 1 (cos120° − i sin120°) 16 z −4 = 1 [cos(−120°) + i sin(−120°)] 16 −4 z = 1 cis ( − 120°) 16 −4
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482
47.
Chapter 7: Applications of Trigonometry
For n = 2, the two square roots of 1 are 1 and –1. The sum of these roots is 1 + (–1) = 0.
48.
For n = 3, the three cube roots of 1 are (from exercise 23)
For n = 2, the two square roots of 1 are 1 and –1. The product of these roots is 1 ⋅ ( −1) = −1 . For n = 3, the three cube roots of 1 are (from exercise 23)
1, − 1 + 3 i and − 1 − 3 i . 2 2 2 2
1, − 1 + 3 i and − 1 − 3 i .
The sum of these roots is
The product of these roots is
1− 1 + 3 i − 1 − 3 i = 0 .
1⋅ − 1 + 3 i ⋅ − 1 − 3 i = 1 .
2
2
2
2
(
2
2
2
2
2
)(
2
2
2
)
For n = 4, the four fourth roots of 1 are 1, –1, i, and –i. The sum of these roots is 1−1+ i − i = 0
For n = 4, the four fourth roots of 1 are 1, –1, i, and –i. The product of these roots is 1 ⋅ ( −1) ⋅ ( i ) ⋅ ( −i ) = −1
For n = 5, the five fifth roots of 1 are 1, cis 72° , cis 144° , cis 216° , cis 288° The sum of these roots is 1 + cis 72° + cis 144° + cis 216° + cis 288° = 0
For n = 5, the five fifth roots of 1 are 1, cis 72° , cis 144° , cis 216° , cis 288° The product of these roots is 1 ⋅ ( cis 72° ) ⋅ ( cis 144° ) ⋅ ( cis 216° ) ⋅ ( cis 288° ) = 1
For n = 6, the six sixth roots of 1 are
For n = 6, the six sixth roots of 1 are
1, –1, − 1 + 3 i , − 1 − 3 i , 1 + 3 i , 1 − 3 i
1, –1, − 1 + 3 i , − 1 − 3 i , 1 + 3 i , 1 − 3 i
The sum of these roots is
The product of these roots is
2
1 − 1 − 12 +
2
3 i − 21 2
2
−
3 i 2
2
+ 12 +
2
2
3 i + 21 2
2
−
3 i 2
2
=0
For n ≥ 2 , the sum of the nth roots of 1 is 0.
2
(
2
1 ⋅ ( −1) ⋅ − 12 +
2
3 i 2
) ⋅ (−
2
1 2
−
2
3 i 2
)⋅(
2
1 2
+
2
3 i 2
)⋅(
1 2
−
3 i 2
) = −1
For n ≥ 2 , the sum of the nth roots of 1 is –1 if n is even and 1 if n is odd.
.......................................................
Exploring Concepts with Technology
Optimal Branching of Arteries The following graph is the graph of R as given in Equation (2) with a = 8 cm, b = 4 cm, r1 = 0.4 cm, and r2 = 0.3 cm.
The following graph is a close-up of the graph of R, for 71o ≤ θ ≤ 74 o .
According to this graph R is a minimum when θ = 72 o (to the nearest degree). Using Equation (3) yields ⎛ 3 r1 ⎞ cos θ = ⎜⎜ 4 ⎟⎟ ⎝ r1 ⎠ 81 cos θ = 256
2
4
81 ≈ 72 o θ = cos −1 ( 256 )
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Chapter Review
483
.......................................................
Assessing Concepts
1.
An oblique triangle that does not contain a right angle.
2.
The Law of Cosines
3.
SSA
4.
A scalar
5.
A scalar
6.
Yes
7.
Yes
8.
One
9.
Five
10.
Yes
11.
No
12.
3 + 5i
....................................................... 1.
B = 180° − 92° − 37° B = 51°
Chapter Review a 4 a = 14 tan 37°
tan A =
a ≈ 11
2.
A = 180° − 94.0° − 77.4° A = 8.6°
b 11.8 b = 11.8 sin 77.4°
sin B =
b ≈ 11.5
3.
4.
5.
cos B =
122 + 202 − 152 2 (12 )( 20 )
cos C =
122 + 152 − 202 2 (12 )(15 )
cos B ≈ 0.6646
cos C ≈ −0.0861
B ≈ 48°
C ≈ 95°
242 + 322 − 282 2(24)(32) cos C ≈ 0.5313 C ≈ 58° cos C =
322 + 282 − 242 2(32)(28) cos A ≈ 0.6875 A ≈ 47° cos A =
c 2 = 222 + 182 − 2 ( 22 )(18 ) cos35° c 2 ≈ 159 c = 159 c ≈ 13
18 159 = sin A sin 35° 18sin 35° sin A = 159 sin A ≈ 0.8188 A ≈ 55°
Copyright © Houghton Mifflin Company. All rights reserved.
cos A =
14 c
[7.1]
14 cos 37° c ≈ 18 c=
a [7.1] 11.8 a = 11.8 cos 77.4°
cos B =
a ≈ 2.57 A = 180° − 48° − 95° [7.2] A = 37°
B ≈ 180° − 58° − 47° [7.2] B ≈ 75°
B ≈ 180 − 35° − 55° [7.2] B ≈ 90°
484
Chapter 7: Applications of Trigonometry
6.
a 2 = 1022 + 1502 − 2 (102 )(150 ) cos82° a 2 ≈ 28645 a = 28645 a ≈ 169
150 28645 = sin C sin 82° 150sin 82° sin C = 28645 sin C ≈ 0.8776
B ≈ 180° − 61° − 82° [7.2] B ≈ 37°
C ≈ 61°
7.
10 8 [7.1] = sin C sin105° 10sin105° sin C = 8 sin C ≈ 1.207 No triangle is formed.
8.
110 = 80 [7.1] sin B sin 55° sin B = 110sin 55° 80 sin B ≈ 1.1263 No triangle is formed.
9.
C = 180° − 80° − 55° C = 45°
25 a = sin 45° sin 55° 25sin 55° a= sin 45° a ≈ 29
25 b [7.1] = sin45° sin80° 25sin 80° b= sin 45° b ≈ 35
10.
A = 180° − 40° − 25° A = 115°
40 a = sin115° sin 40° 40sin115° a= sin 40° a ≈ 56
40 b [7.1] = sin 25° sin 40° 40sin 25° b= sin 40° b ≈ 26
11.
1 (a + b + c) [7.2] 2 1 s = ( 24 + 30 + 36 ) 2 s = 45
12.
s=
K = s ( s − a )( s − b )( s − c )
K = s ( s − a )( s − b )( s − c )
K = 45 ( 45 − 24 )( 45 − 30 )( 45 − 36 )
K = 14 (14 − 9.0 )(14 − 7.0 )(14 − 12 )
K = 127,575
K = 980 K ≈ 31 square units
K ≈ 360 square units
13.
14.
1 (a + b + c) [7.2] 2 1 s = ( 9.0 + 7.0 + 12 ) 2 s = 14 s=
1 ab sin C 2 1 K = ( 60 )( 44 ) sin 44° 2 K ≈ 920 square units
[7.2]
1 K = bc sin A 2 1 K = ( 8.0 )(12 ) sin 75° 2 K ≈ 46 square units
[7.2]
K=
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Review
485
15.
50 75 = sin B sin15° 50sin15° sin B = 75 sin B ≈ 0.1725
A ≈ 180° − 10° − 15° A ≈ 155°
K≈
C ≈ 180° − 42° − 68° C ≈ 70°
K≈
C ≈ 180° − 110° − 26° C ≈ 44°
K≈
A ≈ 180° − 45° − 35° A ≈ 100°
K≈
1 ( 50 )( 75) sin155° [7.2] 2 K ≈ 790 square units
B ≈ 10°
16.
18 25 = sin B sin 68° 18sin 68° sin B = 25 sin B ≈ 0.6676
1 (18)( 25) sin 70° [7.2] 2 K ≈ 210 square units
B ≈ 42°
17.
15 32 = sin B sin110° 15sin110° sin B = 32 sin B ≈ 0.4405
1 (15)( 32 ) sin 44° [7.2] 2 K ≈ 170 square units
B ≈ 26°
18.
18 22 = sin B sin 45° 18sin 45° sin B ≈ 22 sin B ≈ 0.5785
1 (18)( 22 ) sin100° [7.2] 2 K ≈ 190 square units
B ≈ 35°
19.
21.
Let P1P2 = a1i + a2 j. [7.3] a1 = 3 − ( −2 ) = 5
a1 = −3 − ( −4 ) = 1 a2 = 6 − 0 = 6
A vector equivalent to P1P2 is v = 5, 3 .
A vector equivalent to P1P2 is v = 1, 6 .
v =
( −4 )2 + 22
v ≈ 4.5
u =
( −2 )2 + 32
u = 4+9 u ≈ 3.6
25.
Let P1P2 = a1i + a2 j. [7.3]
a2 = 7 − 4 = 3
v = 16 + 4
23.
20.
w =
( −8)2 + 52
w = 89
α ≈ tan −1
2 1 = tan −1 [7.3] 22. 4 2 −
α ≈ 26.6° θ ≈ 180° − 26.6° θ ≈ 153.4° α = tan −1
3 3 [7.3] 24. = tan −1 2 −2
−8 , 89
2
v = 36 + 9 v ≈ 6. 7
α ≈ 56.3° θ ≈ 180° − 56.3° θ ≈ 123.7°
u =
v = 6 2 + ( −3 )
u =
( −4 )2 + ( −7 )2
u = 16 + 49 u ≈ 8.1
5 8 89 5 89 = − , 89 89 89
[7.3]
A unit vector in the direction of w is u = − 8 89 , 5 89 . 89
89
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α ≈ tan −1
−3 1 = tan −1 [7.3] 6 2
α ≈ 26.6° θ ≈ 360° − 26.6° θ ≈ 333.4° α = tan −1
−7 7 [7.3] = tan −1 −4 4
α ≈ 60.3° θ ≈ 180° + 60.3° θ ≈ 240.3°
486
26.
Chapter 7: Applications of Trigonometry
w = 7 2 + ( −12 )
2
−12 7 7 193 12 193 = , , − 193 193 193 193
u=
w = 193
[7.3]
A unit vector in the direction of w is u = 7 193 , − 12 193 . 193
27.
v = 52 + 12
u=
v = 26
193
5 1 5 26 26 i+ j= i+ j [7.3] 26 26 26 26
A unit vector in the direction of v is u = 5 26 i + 26 j. 26
28.
v = 32 + ( −5 )
2
u=
v = 34
26
3 5 3 34 5 34 i− j= i− j [7.3] 34 34 34 34
A unit vector in the direction of v is u = 3 34 i − 5 34 j. 34
29.
v − u = −4, −1 − 3, 2
34
30.
[7.3]
2u − 3v = 2 3, 2 − 3 −4, −1
= −7, −3
[7.3]
= 6, 4 − −12, −3 = 18,7
31.
33.
−u +
1 1 v = − (10i + 6 j) + ( 8i − 5 j) [7.3] 2 2 5 ⎞ ⎛ = ( −10i − 6 j) + ⎜ 4i − j ⎟ 2 ⎠ ⎝ 5⎞ ⎛ = ( −10 + 4 ) i + ⎜ −6 − ⎟ j 2⎠ ⎝ 17 = −6i − j 2
v = 400sin 204°i + 400cos 204° j v ≈ −162.7i − 365.4 j w = −45i R=v+w R ≈ −162.7i − 365.4 j − 45i R ≈ −207.7i − 365.4 j
32.
2 3 2 3 v − u = ( 8i − 5 j) − (10i + 6 j) [7.3] 3 4 3 4 ⎛ 16 10 ⎞ ⎛ 15 9 ⎞ = ⎜ i − j⎟ − ⎜ i + j⎟ 3 ⎠ ⎝ 2 2 ⎠ ⎝ 3 ⎛ 16 15 ⎞ ⎛ 10 9 ⎞ = ⎜ − ⎟i + ⎜ − − ⎟ j ⎝ 3 2 ⎠ ⎝ 3 2⎠ 32 − 45 −20 = 27 = i+ j 6 6 13 47 =− i− j 6 6
R ≈
α = tan −1
( −207.7 )2 + ( −365.4 )2
R ≈ 420 mph
−365.4 365.4 = tan −1 −207.7 207.7
α ≈ 60° θ ≈ 180° + 60° θ ≈ 240°
The ground speed is approximately 420 mph at a heading of 240° [7.3]
34.
θ = sin −1 40 [7.3] θ ≈ 7°
320
35.
u ⋅ v = 3, 7 ⋅ −1, 3
[7.3]
36.
v ⋅ u = −8, 5 ⋅ 2, − 1
= ( 3)( −1) + ( 7 )( 3)
= ( −8 )( 2 ) + ( 5 )( −1)
= 18
= −21
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[7.3]
Chapter Review
37.
487
v ⋅ u = ( −4i − j) ⋅ ( 2i + j) [7.3]
38.
u ⋅ v = ( −3i + 7 j) ⋅ ( −2i + 2 j) [7.3]
= ( −4 )( 2 ) + ( −1)(1)
= ( −3)( −2 ) + ( 7 )( 2 )
= −9
= 20
39.
cos α =
cosα =
7, −4 ⋅ 2,3 7 2 + ( −4 )
2
22 + 32
14 + (−12 )
65 13 cosα ≈ 0.0688 α ≈ 86°
40.
−5, 2 ⋅ 2, −4
cos α =
( −5)2 + 22
2 2 + ( −4 )
2
41.
−10 − 8 29 20 cosα ≈ −0.7474 [7.3] α = 138° projw v = projw v =
v⋅w [7.3] w
46.
44.
−2,5 ⋅ 5, 4
2
projw v =
r = 13
α α θ θ
−3 3 = tan −1 = tan −1 2 2 ≈ 56° ≈ 360° − 56° ≈ 304°
v⋅w [7.3] w
cos α = cosα =
( i − 5 j) ⋅ ( i + 5 j) 12 + ( −5 )
2
12 + 52
1 − 25
26 26 cosα ≈ −0.9231 [7.3] α ≈ 157°
45.
w= F
S cosθ
[7.3]
w = 60 ⋅ 14cos38° w ≈ 662 foot-pounds
( 4i − 7 j) ⋅ ( −2i − 5 j) ( −2 )2 + ( −5)2
−8 + 35 29 27 = 29 =
=
[7.4]
22 + 42
12 − 44
projw v =
10 41 41
r = 22 + ( −3)
2
42.
157 20 [7.3] cosα ≈ −0.5711 cosα ≈ 125°
52 + 4 2 −10 + 20 = 41 10 = 41 =
62 + ( −11)
cosα =
cos α =
43.
( 6i − 11j) ⋅ ( 2i − 4 j)
cos α =
47.
r=
27 29 29
( −5 )2 + (
3
)
2
[7.4]
r = 28 ≈ 5.29
α = tan −1
3 3 = tan −1 −5 5
α ≈ 19° θ ≈ 180° − 19° θ ≈ 161°
48.
z = 2 − 2i [7.4] r = 22 + ( −2 )
2
r= 8=2 2
α = tan −1
−2 = tan −11 2
α = 45° θ = 360° − 45° θ = 315° z = 2 2 cis 315°
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[7.3]
488
49.
Chapter 7: Applications of Trigonometry
50.
z = − 3 + 3i [7.4] r=
(− 3)
2
4π 4π ⎞ ⎛ z = 6⎜ cos + i sin ⎟ [7.4] 3 3 ⎠ ⎝
51.
⎛ 2 i 2⎞ z = 5 ⎜⎜ − ⎟ 2 ⎟⎠ ⎝ 2
2
+3
r = 12 = 2 3
α = tan −1
z = 5 ( cos 315° + i sin 315° ) [7.4]
z=
⎛ 1 i 3⎞ ⎟ z = 6⎜ − − ⎜ 2 2 ⎟⎠ ⎝
5 2 5 2 i − 2 2
z = −3 − 3i 3
3 3 = tan −1 − 3 3
α = 60° θ = 180° − 60° θ = 120° z = 2 3 cis 120°
52.
z1z2 = 5 cis 162° ⋅ 2 cis 63° [7.4]
53.
z1z2 = 3 cis 12° ⋅ 4 cis 126° [7.4]
z1z2 = 10 cis (162° + 63° )
z1z2 = 12 cis (12° + 126° )
z1z2 = 10 cis 225°
z1z2 = 12 cis 138°
z1z2 = 10 ⎛⎜ − 2 − 2 i ⎞⎟ 2 ⎠ ⎝ 2
z1z2 ≈ −8.918 + 8.030i
z1z2
z1z2 = 12 ( cos138° + i sin138° )
z1z2 = 10 ( cos 225° + i sin 225° )
z1z2 z1z2
z1z2 = −5 2 − 5 2i
z1z2
or − 5 2 − 5i 2
55.
z1z2 = (3 cis 1.8) ⋅ (5 cis 2.5) [7.4]
56.
z1z2 = 15 cis (1.8 + 2.5 )
[7.4]
z1 30 cis 165° = z2 10 cis 55°
57.
[7.4]
z1z2 = 15 cis 4.3
z1 = 3 cis 9 (50° − 150°) z2
z1 = 3 cis (165° − 55° ) z2
z1z2 ≈ −6.012 − 13.742i
z1 = 3 cis ( − 100°) or 3 cis 260° z2
z1 = 3 cis 110° z2
z1z2 = 15 ( cos 4.3 + i sin 4.3)
58.
z1 6 cis 50° = z2 2 cis 150°
2π π [7.4] ⋅ 4 cis 3 4 ⎛ 2π π ⎞ = 28 cis ⎜ + ⎟ 4⎠ ⎝ 3 11π = 28 cis 12 11π 11π ⎞ ⎛ = 28 ⎜ cos + i sin ⎟ 12 12 ⎠ ⎝ ≈ −27.046 + 7.247i
z1z2 = 7 cis
54.
z1 40 cis 66° = = 5 cis ( 66° − 125° ) [7.4] z2 8 cis 125° z1 = 5 cis ( −59° ) or 5 cis 301° z2
59.
z1 = 3 −i 1+ i z2 2 cis 330° = 2 cis 45° = 2 cis ( 330° − 45° )
3 − i = 2 cis 330°
1 + i = 2 cis 45° [7.4]
= 2 cis 285°
60.
( 3 cis 45° )6 = 36 cis 6 ⋅ 45°
[7.5]
= 729 cis 270° = 729 ( cos 270° + i sin 270° ) = 729 ( 0 − 1i ) = 0 − 729i
61.
(cis 116π )
8
= cis 8 ⋅ 11π = cis 44π = cis 2π [7.5] 6 3 π π 2 2 = cos + i sin 3 3
3
= − 1 + 3 i or − 1 + i 3 2
2
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2
2
Chapter Review
62.
489
(1 − i 3 )
7
= ( 2 cis 300° ) = 27 cis 7 ⋅ 300° 7
= 128 cis 2100° = 128 cis 300° =128 ( cos300° + i sin 300° ) = 128
63.
[7.5]
(
1 2
−
3 i 2
( −2 − 2i )10 = ( 2
90° + 360°k 3
90° 3 w0 = 3 cis 30°
)
= 32,768i
k = 0, 1, 2 90° + 300° ⋅ 2 3 w2 = 3 cis 270°
w1 = 3 cis
w2 = 3 cis
8i = 8 cis 90° [7.5] wk = 81/ 4 cis
90° + 360°k 4 90° + 360°k = 8 cis 4 4
w0 = 4 8 cis
90° 4
w0 = 4 8 cis 22.5° wk = 2561/ 4 cis
120° + 360°k 4
120° 4 w0 = 4 cis 30°
( 12 ) + ⎛⎜⎝ 23 ⎞⎟⎠ 2
2
w1 = 4 8 cis
k = 0, 1, 2, 3
90° + 360° 4
w1 = 4 8 cis 112.5°
w2 = 4 8 cis
90° + 360° ⋅ 2 4
w2 = 4 8 cis 202.5°
1/ 5
cis
120° + 360° ⋅ 2 4 w2 = 4 cis 210°
w1 = 4 cis
w2 = 4 cis
= 1 + 3 = 1 [7.5] 4 4
60° + 360°k 60° + 360°k = 1 cis 5 5
k = 0, 1, 2, 3, 4
w0 = cis 60°
w1 = cis 60°+360°
w2 = cis 60°+360°⋅2
w0 = cis 12°
w1 = cis 84°
w2 = cis 156°
5
5
w3 = cis 60°+360°⋅3
w4 = cis 60°+360°⋅4
w3 = cis 228°
w4 = cis 300°
5
90° + 360° ⋅ 3 4
w3 = 4 8 cis 292.5°
z = 1 cis 60o wk = (1)
w3 = 4 8 cis
k = 0, 1, 2, 3 [7.5] 120° + 360° 4 w1 = 4 cis 120°
w0 = 4 cis
67.
= 32,768 cis 10 ⋅ 225° [7.5]
= 32,768 cis 2250° = 32,768 ( cos 2250° + i sin 2250° )
90° + 360° 3 w1 = 3 cis 150°
w0 = 3 cis
66.
10
27i = 27 cis 90° [7.5] wk = 271/ 3 cis
65.
)
= 0 + 32,768i
= 64 − 64i 3
64.
2 cis 225°
5
5
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120°+360° ⋅ 3 4 w3 = 4 cis 300° w3 = 4 cis
490
Chapter 7: Applications of Trigonometry
.......................................................
Quantitative Reasoning
QR1. The distance between MCO and LAX is d = cos −1 [cos(lat1) cos(lat2) cos(lon1 − lon2) + sin(lat1)sin(lat2)] d = cos −1 [cos(0.496187) cos(0.592409) cos( −1.419110 − ( −2.066611)) + sin(0.496187)sin(0.592409) ] d ≈ 0.559146 radian The great circle distance between MCO and LAX is d ≈ 0.559146 × 3960 d ≈ 2210 mi
QR2. Since sin(lon2 − lon1) = sin( −2.066611 − ( −1.419110)) < 0 , we use Formula (3) to find the initial heading. sin(lat2) − sin(lat1)cos( d ) ⎤ h1 = 2π − cos −1 ⎡⎢ ⎥ sin( d ) cos(lat1) ⎣ ⎦ −1 ⎡ sin(0.592409) − sin(0.496187)cos(0.559146) ⎤ = 2π − cos ⎢ ⎥ sin(0.559146) cos(0.496187) ⎣ ⎦ ≈ 5.050614 radians o⎞ ⎛ = 5.050614 ⎜ 180 ⎟ ⎝ π ⎠ ≈ 289o
QR3. The distance between JFK and DEN is d = cos −1 [cos(lat1) cos(lat2) cos(lon1 − lon2) + sin(lat1)sin(lat2)] d = cos −1 [cos(0.709476) cos(0.695717) cos( −1.287756 − ( −1.826892)) + sin(0.709476)sin(0.695717)] d ≈ 0.409558 radian The great circle distance between JFK and DEN is d ≈ 0.409558 × 3960 d ≈ 1620 mi
QR4. Flying from DEN to JFK, sin(lon2 − lon1) = sin( −1.287756 − ( −1.826892)) > 0 , we use Formula (2) to find the initial heading. sin(lat2) − sin(lat1) cos( d ) ⎤ h1 = cos −1 ⎡⎢ ⎥ sin( d ) cos(lat1) ⎣ ⎦ −1 ⎡ sin(0.709476) − sin(0.695717) cos(0.409558) ⎤ = cos ⎢ ⎥ sin(0.409558) cos(0.695717) ⎣ ⎦ ≈ 1.361499 radians o⎞ ⎛ = 1.361499 ⎜ 180 ⎟ ⎝ π ⎠ ≈ 78o
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Test
491
....................................................... 1.
Chapter Test 2.
B = 180° − 70° − 16° B = 94°
b a = sin B sin A
c a = sin C sin A 14 sin 70° a= sin 16° a ≈ 48
⎛ b sin A ⎞ B = sin −1 ⎜ ⎟ [7.1] ⎝ a ⎠ ⎛ 13 sin 140° ⎞ B = sin −1 ⎜ ⎟ 45 ⎝ ⎠ B ≈ 11°
c b [7.1] = sin C sin B 14 sin 94° b= sin16° b ≈ 51
3.
4.
a 2 + c2 − b2 2ac ⎛ 322 + 182 − 242 ⎞ ⎟ [7.2] B = cos −1 ⎜ ⎜ ⎟ 2(32)(18) ⎝ ⎠ B ≈ 48°
c 2 = a 2 + b 2 − 2ab cos C
cos B =
c 2 = 202 + 122 − 2(20)(12) cos 42° [7.2] c ≈ 14
5.
1 ab sin C 2 1 K = (7)(12)(sin110°) 2 K ≈ 39 square units [7.2] K=
7.
A = 180° − 42° − 75° A = 63°
6.
2 K = b sin A sin C 2sin B 2 K = 12 sin 63° sin 75° 2sin 42° K ≈ 93 square units [7.2]
8.
a1 = 12cos 220° ≈ −9.2 [7.3] a2 = 12sin 220° ≈ −7.7
v = a1i + a2 j
1 s = (a + b + c) 2 1 s = (17 + 55 + 42 ) = 57 2 K = s ( s − a )( s − b)( s − c)
v = −9.2i − 7.7 j
K = 57(57 − 17)(57 − 55)(57 − 42) K ≈ 260 square units
[7.2]
Copyright © Houghton Mifflin Company. All rights reserved.
492
Chapter 7: Applications of Trigonometry
9.
3u − 5v = 3(2i − 3 j) − 5(5i + 4 j) [7.3] = (6i − 9 j) − (25i + 20 j) = (6 − 25)i + (−9 − 20) j = −19i − 29 j
11.
cosθ =
3,5 ⋅ −6, 2 u⋅v = u v 32 + 52 (−6) 2 + 22
10.
[7.3]
12.
−18 + 10 −8 cosθ = = 34 40 34 40 θ ≈ 103°
u ⋅ v = (−2i + 3 j) ⋅ (5i + 3 j) [7.3] = (−2 ⋅ 5) + (3 ⋅ 3) = −10 + 9 = −1 3 2 = tan −1 2 −3 2
z = −3 2 + 3i
α =tan −1
z = (−3 2) 2 + 32
α ≈ 35° θ ≈ 180° − 35° θ ≈ 145°
z =3 3 z ≈ 3 3 cis 145° [7.4]
13.
z = 5 cis 315° [7.4] z = 5(cos315° + i sin 315°)
14.
z = 1 + 3 i [7.5] 2 2 r = (1 2)2 + ( 3 2)2
z = 5 ⎛⎜ 2 − 2 i ⎞⎟ 2 ⎠ ⎝ 2
r =1
z=5 2 −5 2i 2
α = tan −1
3 2 = 60o 12
z = cos 60° + i sin 60°
2
3
⎛1 3 ⎞ = (cos 60° + i sin 60°)3 ⎜ +i ⎟ 2 ⎠ ⎝2 = cos(3 ⋅ 60°) + i sin(3 ⋅ 600°) = cos180° + i sin180° = −1 + 0i = −1
15.
z1 25 cis 115° = [7.4] z2 10 cis 210°
16.
z = 2 −i z =
z1 = 2.5 cis (115° − 210°) z2
2
2 + (−1)2
z = 3
z1 = 2.5 cis ( − 95°) or 2.5 cis 265° z2
−1 2 = tan −1 2 2 α ≈ 35.2644°
α = tan −1
θ ≈ 360° − 35.2644° θ ≈ 324.7356°
z ≈ 3 cis 324.7356° 5
z ≈ ( 3)5 cis (5 ⋅ 324.7356°) z 5 ≈ 9 3(cos 1623.678° + i sin1623.678°) z 5 ≈ −15.556 − 1.000i
17.
27i = 27(cos90° + i sin 90°) = 27 cis 90° 90° + 360°k wk = 271/ 3 cis 3
90° = 3 cis 30° 3 3 3 3 + i w0 = 2 2
[7.5]
18.
k = 0, 1, 2
w0 = 3 cis
w1 = 3 cis 150° w1 = −
A = 142° − 65° = 77° R 2 = 242 + 182 − 2(24)(18) cos 77° R ≈ 27 miles
3 3 3 + i 2 2
w2 = 3 cis 270° w2 = 0 − 3i or − 3i [7.5]
Copyright © Houghton Mifflin Company. All rights reserved.
[7.3]
Cumulative Review
19.
493
2 2 + i = 1(cos 45° + i sin 45°) 2 2 wk = cos 45°+ 360°k + isin 45°+ 360°k 5 5 w0 w0 w0
20. k = 0, 1, 2, 3, 4
1 (112 + 165 + 140 ) = 208.5 2 K = 208.5(208.5 − 112)(208.5 − 165)(208.5 − 140) S=
o o = cos 45 + i sin 45 5 5 = (cos9° + i sin 9°) = cis 9°
K ≈ 7743 cost ≈ 8.50(7743)
o o o o w1 = cos 45 + 360 + i sin 45 + 360 5 5 w1 = (cos81° + i sin81°) w1 = cis 81°
cost ≈ $66,000 [7.2]
o o o o w2 = cos 45 + 360 ⋅ 2 + i sin 45 + 360 ⋅ 2 5 5 w2 = cos153° + i sin153° w2 = cis 153° o o o o w3 = cos 45 + 360 ⋅ 3 + i sin 45 + 360 ⋅ 3 5 5 w3 = cos 225° + i sin 225° w3 = cis 225° o o o o w4 = cos 45 + 360 ⋅ 4 + i sin 45 + 360 ⋅ 4 5 5 w4 = cos 297° + i sin 297° w4 = cis 297°
....................................................... 1.
( f o g )( x) = f [ g ( x)]
= f [ x + 1]
[2.6]
Cumulative Review 2.
f ( x) = 2 x + 8 y = 2x + 8 x = 2y + 8 x − 8 = 2y x −8 = y 2 f −1 ( x) = 1 x − 4 2
[4.1]
4.
hyp = 32 + 42 = sin θ = 3 , cosθ = 5
25 = 5 [5.2] 4 , tan θ = 3 5 4
6.
y = 3sin π x
2
= cos( x 2 + 1)
⎛ 180o ⎞ o ⎜ ⎟ = 270 [5.1] ⎝ π ⎠
3.
3π 2
5.
[5.2] cos 26.0o = 15.0 c c = 15.0 o ≈ 16.7 cm cos 26.0
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494
7.
Chapter 7: Applications of Trigonometry
(
y = 4cos 2 x − π 2
)
8.
[5.7]
0 ≤ 2 x − π ≤ 2π 2 π ≤ 2 x ≤ 5π 2 2 π ≤ x ≤ 5π 4 4
k = 12 + (−1) 2 = 2
sin α = 1 , cos α = −1 , α = 7π or − π 4 4 2 2 7 π sin x − cos x = 2 sin x + or 2 sin x − π 4 4
(
amplitude = 4, period = π , phase shift = π 4
9.
y = sin x is an odd function. [2.5]
11.
tan ⎛⎜ sin −1 12 ⎞⎟ = tan ( 67.38o ) = 12 13 ⎠ 5 ⎝
13.
( )
10.
[6.5]
2cos 2 x + sin x − 1 = 0 [6.6]
1 − sin x = 1 − sin 2 x sin x sin x 2 x cos = sin x = cos x cos x sin x = cos x cot x
sin x (2cos x − 3) = 0
sin x = 0 x = 0, π
sin x − 1 = 0 sin x = 1
2
v = 9 + 16
α θ θ θ
4 = tan −1 4 3 −3
≈ 53.1° = 180° − α v =5 ≈ 180° − 53.1° ≈ 126.9° magnitude: 5, angle: 126.9°
2cos x − 3 = 0 cos x = 3 2
x = π , 11π 6 6 π 11 π The solutions are 0, , π , . 6 6
x=π 2
The solutions are π , 7π , 11π . 2 6 6
v = (−3)2 + 42 α = tan −1
[6.1]
sin 2 x = 3 sin x [6.6] 2sin x cos x − 3 sin x = 0
x = 7π , 11π 6 6
[7.3]
)
14.
2sin 2 − sin x + 1 = 0 (2sin x + 1)(sin x − 1) = 0 sin x = − 1
(
[6.2] sin 2 x cos3 x − cos 2 x sin 3 x = sin(2 x − 3 x) = sin( − x) or − sin x
2(1 − sin x) + sin x − 1 = 0
2sin x + 1 = 0
)
12.
2
15.
sin x − cos x [6.4] a = 1, b = −1
16.
cos θ = v ⋅ w [7.3] v w 2, − 3 ⋅ −3, 4 cos θ = 2 2 + ( −3)2 (−3)2 + 42 2( −3) + (−3)(4) cos θ = 13 25 cos θ = −18 ≈ −0.9985 5 13 θ = 176.8°
Copyright © Houghton Mifflin Company. All rights reserved.
Cumulative Review
17.
495
AB = 415(cos 42i + sin 42 j) ≈ 308.4i + 277.6 j [7.3] AD = 55[cos(−25°)i + sin(−25°) j] ≈ 49.8i − 23.2 j AC = AB + AD AC = 308.4i + 277.6 j + 49.8i − 23.2 j AC ≈ 358.2i + 254.4 j
18.
a = b [7.1] sin A sin B sin A = a sin B b 42sin 32o = 0.4364041 = 51 A = sin −1 (0.4364041) ≈ 26o
20.
i = cos90° + i sin 90° [7.5] wk = cos 90° + 360°k + i sin 90° + 360°k 2 2
AC = 358.22 + (254.4) 2 AC ≈ 439 mph
( 358.2 )
α = 90o − θ = 90o − tan −1 254.4 ≈ 54.6o 19.
z = 1 − i [7.5] r = 12 + (−1)2 r= 2
−1 = 45o α = tan −1 1
θ = 315o
z = 2(cos315° + i sin 315°)
(1 − i )8 = [ 2(cos315° + i sin 315°)]8 = ( 2)8[cos(8 ⋅ 315°) + i sin(8 ⋅ 315°)]
o o w0 = cos 90 + isin 90 2 2 w0 = cos45°+ isin45°
w0 = 2 + 2 i 2
2
= 16(cos 2520° + i sin 2520°) = 16(cos 0° + i sin 0°) = 16 − 0i = 16
Copyright © Houghton Mifflin Company. All rights reserved.
k = 0, 1
o o o o w1 = cos 90 + 360 + isin 90 + 360 2 2 w1 = cos225°+ i sin225°
w1 = − 2 − 2 i 2
2
Chapter 8
Topics in Analytic Geometry Section 8.1 1.
a. iii
b. i
3.
x 2 = −4 y 4 p = −4
c. iv
d. ii
2. y2 = 1 x
4.
d. iv 6.
x2 = − 1 y 4
4p = − 1
4p = 1 3 1 p= 12
vertex = ( 0, 0 )
4
p=−1
16
vertex = ( 0, 0 )
vertex = ( 0, 0 )
focus = 1 , 0
focus = 1 , 0
focus = 0, − 1
directix: x = − 1 8
directrix: x = − 1
(8 )
8.
4p = 8 p = 2 (h, k + p ) = (2, − 3 + 2) = (2, − 1) focus = (2, − 1) k − p = −3 − 2 = −5 directrix : y = −5
c. i
3
4p = 1 2 p=1 8
(x − 2)2 = 8( y + 3) vertex = (2, − 3)
b. iii
y2 = 1 x
5.
2
p = −1 vertex = (0, 0) focus = (0, − 1) directrix: y = 1
7.
a. ii
(
( 12 )
16 directrix: y = 1 16
12
( y + 1)2 = 6(x − 1) vertex = (1, − 1) ,
(
9.
4p = 6 p = 3
) (2
2
( y + 4)2 = −4(x − 2) vertex = (2, − 4 )
)
( h + p, k ) = 1 + 3 , − 1 = 5 , − 1
(
2
)
)
focus = 5 , − 1 2 h − p = 1− 3 = − 1 2 2 directrix: x = − 1 2
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4 p = −4 p = −1
(h + p, k ) = (2 − 1, − 4) = (1, − 4) focus = (1, − 4 ) h − p = 2 +1 = 3 directrix : x = 3
Section 8.1
10.
497
( x − 3) 2 = − ( y + 2 ) vertex = (3, − 2 ) 4 p = −1 p = −
11.
1 4
4p = 2 p =
(h, k + p ) = ⎛⎜ 3, − 2 − 1 ⎞⎟ = ⎛⎜ 3, − 9 ⎞⎟ 4⎠
⎝
( 2 x − 4 )2 = 8 y − 16
12.
(x + 2)2 = 3( y − 2) vertex = (−2, 2 )
1 2
4p = 3 p =
2 ⎠ ⎝ ⎛ 7 ⎞ focus = ⎜ − , 1⎟ ⎝ 2 ⎠ 1 9 h − p = −4 − = − 2 2 9 directrix : x = − 2
14.
⎝ 2
3 4
(h, k + p ) = ⎛⎜ − 2, 2 + 3 ⎞⎟ = ⎛⎜ − 2, 11 ⎞⎟
(h + p, k ) = ⎛⎜ − 4 + 1 , 1⎞⎟ = ⎛⎜ − 7 , 1⎞⎟
4⎠
⎝
9⎞ ⎛ focus = ⎜ 3, − ⎟ 4⎠ ⎝ 1 7 k − p = −2 + = − 4 4 7 directrix : y = − 4
13.
( y − 1)2 = 2(x + 4) vertex = (−4, 1)
4⎠
⎝
⎠
11 ⎞ ⎛ focus = ⎜ − 2, ⎟ 4⎠ ⎝ 3 5 k − p = 2− = 4 4 5 directrix : y = 4
(3x + 6)2 = 18 y − 36
15.
x 2 + 8x − y + 6 = 0
4 ( x − 2 ) = 8( y − 2)
9 ( x + 2 ) = 18 ( y − 2 )
x2 + 8x = y − 6
( x − 2 )2 = 2 ( y − 2 )
x 2 + 8 x + 16 = y − 6 + 16
vertex = ( 2, 2 )
( x + 2 )2 = 2 ( y − 2 ) vertex = ( −2, 2 )
4p = 2 p = 1
4p = 2 p = 1
2
2
( h, k + p ) = ( 2,
(
focus = 2, 5
)
2
2) (
2)
2 + 1 = 2, 5
2 1 k − p = 2− = 3 2 2 directrix: y = 3 2
(
2
( x + 4 )2 = y + 10 vertex = ( −4, − 10 ) 2) (
2)
( h, k + p ) = −2, 2 + 1 = −2, 5
(
2)
focus = −2, 5
k − p = 2− 1 = 3 2
2
directrix: y = 3 2
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4 p = 1, p = 1
(
4
focus = −4, − 39 4 41 directrix: y = − 4
)
⎝
4⎠
498
16.
Chapter 8: Topics in Analytic Geometry
x 2 − 6 x + y + 10 = 0
17.
18.
x − y2 − 4y + 9 = 0
x 2 − 6 x = − y − 10
y2 − 3 y = − x − 4
− y2 − 4 y = − x − 9
x 2 − 6 x + 9 = − y − 10 + 9
y2 − 3 y + 9 / 4 = − x − 4 + 9 / 4
y2 + 4 y = x + 9
( x − 3)3 = − ( y + 1) vertex = ( 3, − 1)
( y − 32 )
y2 + 4 y + 4 = x + 9 + 4
(
focus = 3, − 5
4
)
2
(
(
directrix: y = − 3 4
20.
4
)
4
)
( y + 2 )2 = ( x + 13) vertex = ( −13, − 2 )
2 4 p = −1, p = − 1 4 3 focus = −2, 2 directrix: x = − 3 2
4
2x − y 2 − 6 y +1 = 0
(
= − x+ 7
vertex = − 7 , 3
4 p = −1, p = − 1
19.
x + y 2 − 3y + 4 = 0
4 p = 1, p = 1
4 51 focus = − , − 2 4 directrix: x = − 53 4
)
3x + y 2 + 8 y + 4 = 0
(
21.
)
x 2 + 3x + 3 y − 1 = 0
− y 2 − 6 y = −2 x − 1
y 2 + 8 y = −3x − 4
x 2 + 3 x = −3 y + 1
y2 + 6 y = 2x + 1
y 2 + 8 y + 16 = −3x − 4 + 16
x 2 + 3x + 9 / 4 = −3 y + 1 + 9 / 4
y2 + 6 y + 9 = 2x + 1 + 9
( y + 4)2 = −3(x − 4)
( x + 32 )
( y + 3)
2
= 2 ( x + 5)
vertex = ( −5, − 3) 4 p = 2, p = 1
(
2
focus = − 9 , − 3 2
directrix: x = − 11
)
3 vertex = (4, − 4 ), 4 p = −3, p = − 4 ⎞ ⎛ 13 focus = ⎜ , − 4 ⎟ ⎠ ⎝ 4 19 directrix : x = 4
2
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2
(
(
= −3 y − 13
vertex − 3 , 13 2
12
)
4 p = −3, p = − 3
(
focus = − 3 , 1 2 3 directrix: y = 11 6
4
)
12
)
Section 8.1
22.
499
x 2 + 5x − 4 y − 1 = 0
23.
x2 + 5x = 4 y + 1 x 2 + 5 x + 25 = 4 y + 1 + 25 4
( x + 52 )
4
2 x2 − 8x − 4 y + 3 = 0
−3 y 2 − 12 y = −6 x − 4
2( x 2 − 4 x + 4) = 4 y − 3 + 8
−3( y 2 − 4 y ) = −6 x − 4
( x − 2)2 = 2 y +
5 2
−3( y − 2)2 = −6 x − 16
26.
(
)
⎝
focus = 2, − 3 4 directrix y = − 7 4
4 p = 2, p = 1
)
4 x 2 − 12 x + 12 y + 7 = 0 4( x 2 − 3x) = −12 y − 7
( y + 1)
(2
2 4 1 = − x− 9 2 2
2
)
vertex = 9 , − 1
4p = − 1, p = −1 2
(
8
)
focus = 35 , − 1 8 directrix x = 37 8
(
(
)
27.
)
3
)
)
3x 2 − 6 x − 9 y + 4 = 0
)
4 x 2 − 3x + 9 = −12 y − 7 + 9 4 3 2
( 2 ) = −12 y + 2 2 ( x − 32 ) = −3 y + 12 2 ( x − 32 ) = −3( y − 16 ) vertex = ( 3 , 1 ) , 4 p = −3, p = − 3 2 6 4 3 7 focus = ( , − ) 2 12 4 x−
3
2 13 focus = − , − 2 6 directrix x = − 19 6
(
(
3
= 2 x+8
vertex = − 8 , − 2
4( y 2 + 2 y ) = −2 x + 5
( y + 1) 2 = − 1 x + 9
(
2
vertex = 2, − 5 , 4 p = 2, p = 1 4 2
4 x 2 − 12 x = −12 y − 7
4( y + 1) 2 = −2 x + 9
( y + 2)
4⎠
4 y 2 + 8 y = −2 x + 5 4( y 2 + 2 y + 1) = −2 x + 5 + 4
( y − 2)2 = 2 x + 16
(x − 2)2 = 2 ⎛⎜ y + 5 ⎞⎟
(
2x + 4 y2 + 8 y − 5 = 0
−3( y 2 − 4 y + 4) = −6 x − 4 − 12
2( x − 2)2 = 4 y + 5
(
= 4 y + 29
directrix: y = − 45 16
25.
6 x − 3 y 2 − 12 y + 4 = 0
2( x 2 − 4 x) = 4 y − 3
16 ) vertex = ( − 5 , − 29 ) , 4 p = 4, p = 1 2 16 5 13 focus = ( − , − ) 2 16 2
24.
directrix y = 11 12
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3( x2 − 2 x ) = 9 y − 4 3 ( x 2 − 2 x + 1) = 9 y − 1
( x − 1)2 = 3 ⎛⎜ y − 1 ⎞⎟ 9⎠ ⎝ 3 ⎛ 1⎞ vertex = ⎜1, ⎟ , 4 p = 3, p = 4 ⎝ 9⎠ ⎛ 31 ⎞ focus = ⎜1, ⎟ ⎝ 36 ⎠ 23 directrix y = − 36
500
28.
Chapter 8: Topics in Analytic Geometry
( y − 32 )
) vertex = ( − 47 , 3 ) , 4 p = 2 , p = 1 8 2 3 6 137 3 focus = ( − , 24 2 ) 2
(
= 2 x + 47 3
8
29.
vertex (0, 0), focus (0, − 4 )
directrix x = − 145 24
31.
( x − h)
x = 4 py p = −4 since focus is (0, p )
y 2 = 4 px p = 5 since focus is ( p, 0)
x 2 = 4(− 4 ) y
y 2 = 4(5)x
x 2 = −16 y
y 2 = 20 x
32.
= 4 p( y − k)
h = 2, k = −3. Since the focus is (h + p, k), 2 + p = 0 and p = −2.
( x + 1)2 = 4 (1)( y − 2 )
( y + 3)2 = 4 ( −2 )( x − 2 )
( x + 1)2 = 4 ( y − 2 )
( y + 3)2 = −8 ( x − 2 )
focus (3, − 3), directrix y = −5 The vertex is the midpoint of the line segment joining (3,−3) and the point (3, −5) on the directrix. ⎛ 3 + 2 −3 + ( −5 ) ⎞ , ( h, k ) = ⎜ ⎟ = ( 3, − 4 ) 2 ⎝ 2 ⎠
34.
The distance p from the vertex to the focus is 1. 4 p = 4 (1) = 4
35.
vertex ( 2, − 3) , focus ( 0, − 3)
( y − k )2 = 4 p ( x − h )
h = −1, k = 2. The distance p from the vertex to the focus is 1.
33.
vertex (0, 0 ), focus (5, 0 )
2
vertex (−1, 2 ), focus (−1, 3) 2
30.
focus (−2, 4 ), directrix x = 4 The vertex is the midpoint of the line segment joining (−2,4) and the point (4, 4) on the directrix. −2 + 4 4 + 4 ⎞ , ( h, k ) = ⎛⎜ ⎟ = (1, 4 ) 2 ⎠ ⎝ 2 The distance p from the vertex to the focus is −3. 4 p = 4 ( −3) = −12
( x − h )2 = 4 p ( y − k )
( y − k )2 = 4 p ( x − h )
( x − 3)2 = 4 ( y + 4 )
( y − 4 )2 = −12 ( x − 1)
vertex = ( −4, 1) , point: ( −2, 2 ) on the parabola.
36.
vertex = ( 3, − 5 ) and the point ( 4, 3) is on the parabola. The equation of the parabola in standard form must be
Axis of symmetry x = −4.
( y + 5 )2 = 4 p ( x − 3)
2
If P1 = ( −2, 2 ) ( x + 4 ) = 4 p ( y − 1) . Since (−2, 2) is on the curve, we get
( 3 + 5 )2 = 4 p ( 4 − 3)
( −2 + 4 )2 = 4 p ( 2 − 1)
82 = 4 p (1)
4 = 4p ⇒ p =1
Thus, the equation in standard form is
( x + 4 )2 = 4 ( y − 1)
Since ( 4, 3) is on the curve, we get
64 = 4 p p = 16 Thus, the equation of the parabola in standard form is
( y + 5)2 = 64 ( x − 3)
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.1
37.
501
Find the vertex. x = −0.325 y 2 + 13 y + 120 x − 120 = −0.325 y 2 + 13 y x − 120 = −0.325( y 2 − 40 y ) x − 120 − 130 = −0.325( y 2 − 40 y + 400) x − 250 = −0.325( y − 20) 2 − 40 ( x − 250 ) = ( y − 20) 2 13 The vertex is (250, 20).
38.
To find where the fountains intersect, set the equations equal and solve for x. −0.25 x 2 + 2 x = −0.25 x 2 + 4.5 x − 16.25 2 x = 4.5 x − 16.25 −2.5 x = −16.25 x = 6.5 Substitute the value of x into one equation and solve for y.
y = −0.25 x 2 + 2 x y = −0.25(6.5)2 + 2(6.5) y = 2.4375 The fountains of water intersect 2.4375 feet above the base.
Find the focus. 4 p = − 40 , p = − 10 13 13 The focus is ⎛⎜ 3240 , 20 ⎞⎟ . ⎝ 13 ⎠ 39.
Place the satellite dish on an xy-coordinate system with its vertex at (0, −1) as shown.
The equation of the parabola is x 2 = 4 p ( y + 1)
40.
The focus of the parabola is (0, p) where x 2 = 4 py. Half of 81 feet = 40.5 feet. Therefore, the point (40.5, 16) is on the parabola. −1 ≤ y ≤ 0
Because (4, 0) is a point on this graph, (4, 0) must be a solution of the equation of the parabola. Thus, 16 = 4 p ( 0 + 1) 16 = 4 p
(40.5)2 = 4 p (16) 1640.25 = 64 p 1640.25 =p 64 p ≈ 25.6 feet
4= p Because p is the distance from the vertex to the focus, the focus is on the x-axis 4 feet above the vertex.
Copyright © Houghton Mifflin Company. All rights reserved.
502
Chapter 8: Topics in Analytic Geometry
41. The focus of the parabola is (p, 0) where y 2 = 4 px. Half of 18.75 inches is 9.375 inches. Therefore, the point (3.66, 9.375) is on the parabola.
42.
a.
The focus of the parabola is (0, 75).
x 2 = 4 py x 2 = 4(75) y x 2 = 300 y or y = 1 x 2 300
b.
Half of 250 feet is 125 feet. Therefore, (125, y) is a point on the parabola, where y is the depth of the dish.
x 2 = 300 y (125)2 = 300 y 15625 = 300 y 15625 =y 300 y ≈ 52 feet
(9.375)2 = 4 p (3.66) 87.890625 = 14.64 p 87.890625 =p 14.64 p ≈ 6.0 inches
43.
S=
a.
)
(
⎤ πr ⎡ 2 2 3/ 2 − r3 ⎥ ⎢ r + 4d
6d 2 ⎣ r = 40.5 feet d = 16 feet
S=
44.
⎦
x 2 = 4 py − 100 ≤ x ≤ 100
)
( ⎣
π (40.5) ⎡
3/ 2 ⎤ [40.5]2 + 4[16]2 − (40.5)3 ⎥ 2 ⎢
6(16) 40.5π ⎡ = ( 266.25)3 / 2 − 66430.125⎤⎥⎦ 1536 ⎢⎣ 40.5π = [137518.9228 − 66430.125] 1536 40.5π = [71088.79775] 1536 ≈ 5900 square feet
b.
The equation of the mirror is
⎦
Because (100, 3.75375) is a point on the parabola,
(100, 3.75375) must be 100 2 = 4 p (3.75375)
10000 = 15.015 p 666 ≈ p The focus is approximately 666 inches above the vertex.
r = 125 feet d = 52 feet
π (125) ⎡
(
)
3/ 2 ⎤ [125]2 + 4[52]2 − (125)3 ⎥ ⎢ 2 6(52) ⎣ ⎦ 125π ⎡ 3/ 2 = ( 26441) − 1953125⎤⎦⎥ 16224 ⎣⎢
S=
a solution of the equation. Thus
125π [ 4299488.724 − 1953125] 16224 125π = [ 2346363.724] 16224 ≈ 56,800 square feet =
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.1
45.
503
The equation of the mirror is given by
46.
Place the headlight on a coordinate grid as shown.
x 2 = 4 py − 60 ≤ x ≤ 60 Because p is the distance from the vertex to the focus and the coordinates of the focus are (0, 600), p = 600. Therefore, x 2 = 4 ( 600 ) y x 2 = 2400 y To determine a, substitute (60, a) into the equation
The equation of the parabola is y 2 = 4 px. Because (6, 3) is on the graph of the parabola, the coordinates
x 2 = 2400 y and solve for a.
must be a solution of the equation y 2 = 4 px. Thus,
2
x = 2400 y
y 2 = 4 px 32 = 4 p ( 6 ) 9 = 24 p p = 0.375 The value p is the distance from the vertex to the focus. Therefore, the focus is 0.375 inches to the right of the vertex.
602 = 2400a 3600 = 2400a 1.5 = a. The concave depth of the mirror is 1.5 inches.
47.
a.
The equation of the parabola is
b.
x = 4 p ( y − 32 ) 2
To find the width use x = 900, and solve for y and then multiply by 2 2⎞ ⎛ x 2 = 4 ⎜ 800 ⎟ ( y − 32 ) ⎝ 84 ⎠
Because (–800, 53) is a point on this graph, (–800, 53) must be a solution of the equation of the parabola. Thus,
2⎞ ⎛ 9002 = 4 ⎜ 800 ⎟ ( y − 32 ) ⎝ 84 ⎠
( −800 )2 = 4 p ( 53 − 32 ) 8002 = 84 p
⎛ 9002 ⎞ 84 ⎜ ⎟ + 32 = y ⎝ 8002 ⎠ 4 y ≈ 58.58
2
800 = p 84 2⎞ ⎛ The equation of the parabola is x 2 = 4 ⎜ 800 ⎟ ( y − 32 ) . ⎝ 84 ⎠
48.
2 y ≈ 117 The width of the ski is 117 mm.
An infinite number of parabolas pass through the points (2, 3) and (–2, 3).
....................................................... 49.
x2 = 4 y 4p = 4 p =1
Connecting Concepts 50.
focus = (0, 1) Substituting the vertical coordinate of the focus for y to obtain x-coordinates of endpoints ( x1, y1 ) , ( x2 , y2 ) , we have
y 2 = −8 x 4 p = −8 p = −2 focus = ( −2, 0 ) Substituting the horizontal coordinate of the focus for x to obtain the y-coordinates of the endpoints ( x1, y1 ) , ( x2 , y2 ) , we have
x 2 = 4 (1) , or x 2 = 4
y 2 = −8 ( −2 ) = 16 y = ± 16 y1 = −4 y2 = 4
x=± 4 x1 = −2 x2 = 2
Length of latus rectum = x2 − x1
= 2 − ( −2 ) = 4.
Length of latus rectum = y2 − y1 = 4 − ( −4 ) = 8
Copyright © Houghton Mifflin Company. All rights reserved.
504
51.
Chapter 8: Topics in Analytic Geometry
( x − h )2 = 4 p ( y − k ) focus = ( h, k + p ) Substituting the vertical coordinate of the focus for y to obtain x-coordinates of endpoints ( x1, y1 ) , ( x2 , y2 ) ,
Substituting the horizontal coordinate of the focus for x to obtain the y-coordinates of the endpoints ( x1, y1 ) , ( x2 , y2 ) ,
we have
we have
( y − k )2 = 4 p ( h + p − h ) ( y − k )2 = 4 p 2
2
( x − h) = 4 p (k + p − k ) ( x − h )2 = 4 p 2
y − k = ±2p y1 = k − 2 p y2 = k + 2 p
x − h = ±2p x1 = h − 2 p x2 = h + 2 p
Solving for x2 − x1 , we obtain Δx = x2 − x1 = h + 2 p − h + 2 p = 4 p
or
( y − k )2 = 4 p ( x − h )
Thus, the length of the latus rectum is 4 p .
focus = (h + p, k) 52.
53.
4p = 2 p=1 2
focus
4 p = −1 p = −1
(3,
one point: one point:
54.
Solving for y2 − y1 , we obtain Δy = y2 − y1 = k + 2p − k + 2p =4 p
−1
2
)
( h + 2 p, ( h − 2 p,
4 3, −4 4 one point: 3 , k + 2 p = 3 , − 9 4 4 2 one point: 3 , k − 2 p = 3 , − 7 4 4 2
focus
) ( ) (
− 1 = 4, − 1
2 2 − 1 = 2, − 1 2 2
) )
By definition, the point (x, y ) on the curve must be equidistant from the focus (–c, 0) and the directrix (x = c). So,
( x + c 2 ) + ( y − 0 )2 =
55.
(
( (
)
) ( ) (
Graph y = 7 + 1 x x . 4
4
( x − c )2
( x + c )2 + ( y − 0 )2 = ( x − c )2 x 2 + 2cx + c 2 + y 2 = x 2 − 2cx + c 2 y 2 = −4cx
Copyright © Houghton Mifflin Company. All rights reserved.
) )
Section 8.1
56.
505
Since the axis of symmetry passes through the vertex (0, 0) and focus (1, 1), its equation is given by y = x. Because the directrix is perpendicular to the axis of symmetry, its slope m must be − 1 or –1. 1
Since the vertex (0, 0) is the midpoint of the line segment connecting the focus (1, 1) and the directrix, the distance from the vertex ⎛ ⎞ 2 2 to the focus ⎜ (1 − 0 ) + (1 − 0 ) or 2 ⎟ must equal the distance from the vertex to the directrix [at point ( x1, y1 ) ] along the axis ⎝ ⎠ of symmetry. Therefore, y1 = x1 (since the point is also on the axis of symmetry) and
( x1 − 0 )2 + ( y1 − 0 )2
= 2.
Thus, by substituting y1 for x , we obtain: 1
( y1 − 0 )2 + ( y1 − 0 )2
= 2
y12 + y12 = 2 2 y12 = 2 Thus, y12 = 1 and y1 = ±1. If y1 = 1 Thus, y1 = −1, and x1 = −1.
( and
x1 = 1) , the directrix would pass through the focus, which is an impossibility.
The equation of the directrix is derived by substituting y1, x1, and m in the point slope form of the straight line. y = y1 = m ( x − x1 ) y − ( −1) = ( −1) ( x − ( −1) ) y + 1 = − ( x + 1) y +1 = −x −1 y = −x − 2
57.
By definition, any point on the curve (x, y) will be equidistant from both the focus (1, 1) and the directrix, ( y2 = − x2 − 2 ) . If we let d1 equal the distance from the focus to the point (x, y), we get d1 =
( x − 1)2 + ( y − 1)2
To determine the distance d2 from the point (x, y) to the line y = −x –2, draw a line segment from (x, y) to the directrix so as to meet the directrix at a 90° angle. Now drop a line segment parallel to the y-axis from (x, y) to the directrix. This segment will meet the directrix at a 45° angle, thus forming a right isosceles triangle with the directrix and the line segment perpendicular to the directrix from (x, y). The length of this segment, which is the hypotenuse of the triangle, is the difference between y and the y-value of the directrix at x, or –x – 2. Thus, the y+x+2 hypotenuse has a length of y + x + 2, and since the right triangle is also isosceles, each leg has a length of . 2 y+x+2 But since d 2 is the length of the leg drawn from ( x, y ) to the directrix, d 2 = . 2 x+ y+2 2 2 Thus, d1 = ( x − 1) + ( y − 1) and d 2 = . 2 By definition, d1 = d 2 . So, by substitution,
2
x+ y+2 2
( x − 1)2 + ( y − 1)2
=
( x − 1)2 + ( y − 1)2
= x+ y+2
2 ⎡( x − 1) + ( y − 1) = x 2 + y 2 + 4 x + 4 y + 2 xy + 4 ⎣⎢ ⎦⎥ 2
2⎤
)
(
2 x 2 − 2 x + 1 + y 2 − 2 y + 1 = x 2 + y 2 + 4 x + 4 y + 2 xy + 4 2
2
2 x − 4 x + 2 y − 4 y + 4 = x 2 + y 2 + 4 x + 4 y + 2 xy + 4 x 2 + y 2 − 8 x − 8 y − 2 xy = 0
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506
Chapter 8: Topics in Analytic Geometry
....................................................... x +x PS1. midpoint: 1 2 2 5 + −1 =2 2 The midpoint is (2, 3).
y1 + y2 2 1+ 5 =3 2
Prepare for Section 8.2 PS2.
x+8= 0 x−2=0 x = −8 x=2 The solutions are –8, 2.
( x2 − x1)2 + ( y2 − y1)2
length:
x 2 + 6 x − 16 = 0 ( x + 8)( x − 2) = 0
(−1 − 5)2 + (5 − 1)2 = 36 + 16 = 52 = 2 13 The length is 2 13 .
x2 − 2 x = 2
PS3.
PS4. x 2 − 8 x + 16 = ( x − 4)2
2
x − 2x + 1 = 2 + 1 ( x − 1)2 = 3 x −1 = ± 3 x =1± 3
PS5. ( x − 2)2 + y 2 = 4
( x − 2)2 + ( y + 3)2 = 16 Center: (2, –3), radius 4
PS6.
y 2 = 4 − ( x − 2) 2 y = ± 4 − ( x − 2)2
Section 8.2 1. 3.
a. iv
b. i
c. ii
x2 y2 + =1 16 25
d. iii
2. 4.
x2 y2 + =1 49 36
5.
a. iii
b. i
c. iv
d. ii
x2 y2 + =1 9 4
6.
x2 y2 + =1 64 25
a 2 = 25 → a = 5
a 2 = 49 → a = 7
a2 = 9 → a = 3
a 2 = 64 → a = 8
b 2 = 16 → b = 4
b 2 = 36 → b = 6
b2 = 4 → b = 2
b 2 = 25 → b = 5
c = a2 − b2
c = a 2 − b2
c = a2 − b2
c = a 2 − b2
= 25 − 16
= 49 − 36
= 9−4
= 9
= 13
= 5
=3
( 0, 0 ) Vertices ( 0, ± 5 ) Foci ( 0, ± 3)
Center
( 0, 0 ) Vertices ( ±7, 0 )
Center
Foci
Foci
Center
(±
13, 0
)
= 64 − 25
( 0, 0 ) ( ±3, 0 )
Vertices
(±
5, 0
)
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= 39 Center ( 0, 0 ) Vertices Foci
(±
( ±8, 0 ) 39, 0
)
Section 8.2
7.
507
2 x2 + y = 1 9 7
8.
x2 y2 + =1 5 4
9.
4x 2 y 2 + =1 9 16
10.
a2 = 9 → a = 3
a2 = 5 → a = 5
Rewrite as
Rewrite as
b2 = 7 → b = 7
b2 = 4 → b = 2
c = a 2 − b2
c = a 2 − b2
x2 y2 + =1 9 / 4 16
x2 y2 + =1 9 16 / 9
= 9−7
= 5−4
a 2 = 16 → a = 4
a2 = 9 → a = 3
= 2
= 1 =1
b 2 = 9 / 4 → b = 3/ 2
b 2 = 16 / 9 → b = 4 / 3
( 0, 0) Vertices ( 0, ± 3)
Center Foci
( 0,
± 2
Center
Vertices
)
Foci
c = a 2 − b2
( 0, 0 )
(±
5, 0
)
c = a 2 − b2
= 16 − 9 / 4
= 9 − 16 / 9
= 55 / 2
( ±1, 0 )
Center
= 65 / 3 Center ( 0, 0 )
( 0, 0 ) ( 0, ± 4 )
Vertices
⎛ Foci ⎜⎜ 0, ± ⎝
11.
x2 9y2 + =1 9 16
( x − 3) 2 + ( y + 2 ) 2
25 16 Center ( 3, − 2 )
Vertices Foci
( 3 ± 5,
( 3 ± 3,
12.
=1
55 ⎞ ⎟ 2 ⎟⎠
(x + 3)2 + ( y + 1)2
− 2 ) = ( 6, − 2 ) , ( 0, − 2 )
( ±3, 0 )
⎛ Foci ⎜⎜ ± ⎝
65 , 3
=1
9 16 Center (−3, − 1)
− 2 ) = ( 8, − 2 ) , ( −2, − 2 )
Vertices
Vertices (− 3, − 1, ± 4 ) = (− 3, 3), (− 3, − 5)
(
Foci − 3, − 1 ± 7
)
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⎞ 0 ⎟⎟ ⎠
508
13.
Chapter 8: Topics in Analytic Geometry
( x + 2 )2
y2 =1 9 25 Center (−2, 0 )
14.
+
Vertices (− 2, 5), (− 2, − 5) Foci (− 2, 4), (− 2, − 4 )
15.
(x − 1)2 + ( y − 3)2
21 4 Center (1, 3)
Vertices Foci
16.
=1
)
( 0, 2 ± 9 ) = ( 0, 11) , ( 0,
( 0, 2 ±
) (
56 = 0, 2 ± 2 14
(x + 5)2 + ( y − 3)2
9 7 Center (−5, 3)
)
− 7)
=1
Vertices 1 ± 21, 3
Vertices (− 5 ± 3, 3) = (− 2, 3), (− 8, 3)
Foci 1 ± 17 , 3
Foci − 5 ± 2 , 3
(
17.
(
x 2 ( y − 2 )2 + =1 25 81 Center ( 0, 2 )
)
9(x − 1)2 ( y + 1)2 + =1 16 9 Center (1, − 1)
(
18.
Vertices (1, − 1 ± 3) = (1, 2 ), (1, − 4 )
⎛ 65 ⎞⎟ Foci ⎜1, − 1 ± ⎜ 3 ⎟⎠ ⎝
( x + 6 )2
)
25 y 2 =1 25 144 Center (−6, 0 ) +
Vertices (− 6 ± 5, 0 ) = (− 1, 0), (− 11, 0 )
⎛ 481 ⎞⎟ ,0 Foci ⎜ − 6 ± ⎜ ⎟ 5 ⎝ ⎠
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Section 8.2
19.
509
3x 2 + 4 y 2 = 12
20.
x2 y 2 + =1 4 3
( 0, 0 ) Vertices ( ±2, 0 ) Foci ( ±1, 0 )
Center
Foci
23.
x2 y 2 + =1 12 25 Center ( 0, 0 ) Foci
25.
( 0,
( 0,
)
)(
) 4(x 2 − 6 x + 9)+ (y 2 − 8 y + 16 ) = −48 + 36 + 16 4 x 2 − 6 x + y 2 − 8 y = −48 4(x − 3)3 + ( y − 4 )2 = 4
(x − 3)2 + ( y − 2)2 1
4
=1
Vertices (3, 4 ± 2 ) = (3, 6), (3, 2)
(
Foci 3, 4 ± 3
)
( 0, 0 ) Vertices ( 0, ± 5 ) Foci ( 0, ± 3)
Center
( 0, ± 5 ) ± 1)
64 x 2 + 25 y 2 = 400 25 4
4 x 2 + y 2 − 24 x − 8 y + 48 = 0
(
( 0,
x2
± 5)
± 13
Center (3, 4 )
25 x 2 + 16 y 2 = 400
x2 y 2 + =1 16 25
( 0, 0 )
Vertices
25 x 2 + 12 y 2 = 300
Vertices
21.
x2 y 2 + =1 4 5
Center
22.
5 x 2 + 4 y 2 = 20
+
24.
9 x 2 + 64 y 2 = 144
y2 =1 16
x2 y 2 + =1 9 16 4
Center
( 0, 0 ) Vertices ( 0, ± 4 )
Center
Foci ⎛⎜ 0, ± 39 ⎞⎟ 2 ⎠ ⎝
Foci ⎛⎜ ± 55 , 0 ⎞⎟ ⎝ 2 ⎠
26.
( 0, 0 ) Vertices ( ±4, 0 )
x 2 + 9 y 2 + 6 x − 36 y + 36 = 0
(x 2 + 6 x)+ 9(y 2 − 4 y ) = −36 (x 2 + 6x + 9)+ 9(y 2 − 4 y + 4) = −36 + 9 + 36 (x + 3)2 + 9( y − 2)2 = 9 (x + 3)2 + ( y − 2)2 = 1
9 Center (−3, 2 )
1
Vertices (− 3 ± 3, 2 ) = (0, 2 ), (− 6, 2)
(
Foci − 3 ± 2 2 , 2
)
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510
27.
Chapter 8: Topics in Analytic Geometry
5 x 2 + 9 y 2 − 20 x + 54 y + 56 = 0
) ( ) 2 5(x − 4 x + 4)+ 9(y 2 + 6 y + 9) = −56 + 20 + 81 (
28.
5 x 2 − 4 x + 9 y 2 + 6 y = −56
9
5
Center (2, − 3)
2
( 2 ) = 144 1 2 ( x + 2 )2 + ( y − 2 ) = 1 2
9 ( x + 2 ) + 16 y − 1
=1
Vertices (2 ± 3, − 3) = (− 1, − 3), (5, − 3) Foci (2 ± 2, 3) = (0, − 3), (4, − 3)
29.
) ( ) 9 ( x + 4 x + 4 ) + 16 ( y 2 − y + 1 ) = 104 + 36 + 4 4 (
9 x 2 + 4 x + 16 y 2 − y = 104
5(x − 2)2 + 9( y + 3)2 = 45
(x − 2)2 + ( y + 3)2
9 x 2 + 16 y 2 + 36 x − 16 y − 104 = 0
16 x 2 + 9 y 2 − 64 x − 80 = 0
) 16(x 2 − 4 x + 4)+ 9 y 2 = 80 + 64 (
16 x 2 − 4 x + 9 y 2 = 80
16(x − 2)2 + 9 y 2 = 144
16
2
9
( −2, 12 ) Vertices ( −2 ± 4, 1 ) = ( 2, 1 ) , ( −6, 1 ) 2 2 2 1 Foci ( −2 ± 7, ) 2
Center
30.
16 x 2 + 9 y 2 + 36 y − 108 = 0
) 16 x 2 + 9(y 2 + 4 y + 4) = 108 + 36 (
16 x 2 + 9 y 2 + 4 y = 108
16 x 2 + 9( y − +2)2 = 144
Vertices (2, ± 4) = (2, 4 ), (2, − 4 )
x 2 ( y + 2 )2 + =1 9 16 Center (0, − 2 ) Vertices (0, − 2 ± 4 ) = (0, 2), (0, − 6 )
Foci 2 ± 7
Foci 0, − 2 ± 7
( x − 2 )2 9 Center (2, 0)
(
)
+
y2 =1 16
(
)
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Section 8.2
31.
511
25 x 2 + 16 y 2 + 50 x − 32 y − 359 = 0
32.
) ( ) 2 25(x + 2 x + 1)+ 16(y 2 − 2 y + 1) = 359 + 25 + 16 (
25 x 2 + 2 x + 16 y 2 − 2 y = 359
16 x 2 + 9 y 2 − 64 x − 54 y + 1 = 0
) ( ) 16(x − 4 x + 4 )+ 9(y 2 − 6 y + 9) = −1 + 64 + 81 (
16 x 2 − 4 x + 9 y 2 − 6 y = −1
2
25(x + 1)2 + 16( y − 1)2 = 400
(x + 1)2 + ( y − 1)2 Center (−1, 1)
16
25
16(x − 2 )2 + 9( y − 3)2 = 144
(x − 2)2 + ( y − 3)2
=1
Center (2, 3)
(
Foci 2, 3 ± 7
8 x 2 + 25 y 2 − 48 x + 50 y + 47 = 0
34.
) ( ) 8(x 2 − 6 x + 9 )+ 25(y 2 + 2 y + 1) = −47 + 72 + 25 (
8 x 2 − 6 x + 25 y 2 + 2 y = −47
25 / 4 Center: ( 3, − 1)
35.
2
=1
)
4 x 2 + 9 y 2 + 24 x + 18 y + 44 = 0
) ( ) 4(x 2 + 6 x + 9)+ 9(y 2 + 2 y + 1) = −44 + 36 + 9 (
4 x 2 + 6 x + 9 y 2 + 2 y = −44
8(x − 3)2 + 25( y + 1)2 = 50
(x − 3)2 + ( y + 1)2
16
Vertices (2, 3 ± 4) = (2, 7 ), (2, − 1)
Vertices (− 1, 1 ± 5) = (− 1, 6 ), (− 1, − 4 ) Foci (− 1, 1 ± 3) = (− 1, 4), (− 1, − 2 )
33.
9
4(x + 3)2 + 9( y + 1)2 = 1
(x + 3)2 + ( y + 1)2
=1
1/ 4 Center (−3, − 1)
1/ 9
=1
5 ⎛ ⎞ ⎛ 11 ⎞ ⎛1 ⎞ Vertices: ⎜ 3 ± , − 1⎟ = ⎜ , − 1⎟ , ⎜ , − 1⎟ 2 2 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
1 ⎛ ⎞ ⎛ 5 ⎞ ⎛ 7 ⎞ Vertices ⎜ − 3 ± , − 1⎟ = ⎜ − , − 1⎟, ⎜ − , − 1⎟ 2 2 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ 17 Foci: ⎜⎜ 3 ± , − 1⎟⎟ 2 ⎝ ⎠
⎞ ⎛ 5 , − 1⎟ Foci ⎜ − 3 ± ⎟ ⎜ 6 ⎠ ⎝
2a = 10 a=5
36.
2
a = 25 c=4 c2 = a2 − b2
16 = 25 − b 2
b =9 x2 y2 + =1 25 9
2
2b = 6 b=3
37.
a=6 2
38.
a=7
2
b =9
a = 36 b=4
a 2 = 49 b=5
c=4
b 2 = 16
b 2 = 25
c2 = a2 − b2 2
16 = a − 9
x2 y2 + =1 36 16
25 = a 2 x2 y2 + =1 9 25
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x2 y2 + =1 49 25
512
Chapter 8: Topics in Analytic Geometry
2a = 12 a=6
39.
2a = 8 a=4
40.
a 2 = 16
a 2 = 36
2 x2 + y = 1 16 b2
x2 y2 + =1 36 b2
( −2 )2 ( 2 )2
( 2 )2 ( −3)2 36
+
=1 b2 4 9 + =1 36 b2 9 8 = b2 9
16
2
8b = 81 b2 = 81 8
c=3 2a = 8 a = 16
c=3 2b = 4 2b = 4 b=2
c2 = a2 − b2
b2 = 4
a=4 2
9 = 16 − b 2 2
b =7
( x + 2 )2 + ( y − 4 )2 16
7
( −2 ) 3 ( 2 ) 2
=1
b2 4 + 4 =1 16 b2 4 =3 b2 4
b2
+
=1 16 4 + 4 =1 b2 16 4 =3 b2 4
3b2 = 16
3b2 = 16
b2 = 16 3
b2 = 16 3
2 2 x 2 + y = 1 or x 2 + y = 1 16 16 / 3 16 / 3 16
x2 y2 + =1 36 81/ 8 41.
+
2 x2 + y = 1 b2 16
=1
42.
c2 = a2 − b2
9 = a2 − 4 a 2 = 13 x 2 ( y − 3)2 + =1 4 13
43.
2a = 10 a=5 a 2 = 25 Since the center of the ellipse is (2, 4) and the point (3, 3) is on the ellipse, we have
( x − 2 )2 + ( y − 4 )2 b2
(3− 2)2 (3−4)2 b2
=1
a2
+
=1
25
1 b2
= 1− 1
25
2
b = 25 24 2
( x − 2 )2 + ( y − 4 ) 25 24
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25
=1
Section 8.2
513
2b = 8 b=4
44.
center (5, 1)
45.
46.
c=4 2a = 12 a=6
a 2 = 25
a 2 = 36
c2 = a2 − b2
c2 = a2 − b2
9 = 25 − b 2
16 = 36 − b 2
2
b = 16
( x + 4 )2
+
a2
( y − 1)2
=1
16
(0 + 4)2 + (4 − 1)2 a2
=1
16
16 a
9 =1 16
+
2
16 a
=
2
center (−1, − 1)
c=3 2a = 10 a=5
b 2 = 16
(x − 5)2 + ( y − 1)2
7 16
16
25
b 2 = 20
(x + 1)2 + ( y + 1)2
=1
36
20
=1
7 a 2 = 256 256 a2 = 7
(x + 4)2 + ( y − 1)2 256 / 7
=1
16
2a = 10 a=5
47.
48.
a 2 = 25 c 2 = a 5 c 2 = 5 5 c=2 2
2
c = a −b 4 = 25 − b
2
2
y x + =1 25 21
center (1, 3)
c2 = a 2 − b2
4 = 25 − b 2 b 2 = 21
( x − 1) 2 ( y − 3) 2 + =1 25 21
50.
c2 = a 2 − b2
16 = 36 − b 2
81 = 144 − b
center
( 0, 0 ) c=3 c 1 = a 4 3 1 = a 4 a = 12
2
2
b = 63
52.
c=2 c 2 = a 5 2 2 = a 5 a=5
( 0, 0 )
center
c=4 c 2 = a 3 4 2 = a 3 a=6
c2 = a 2 − b2
9 = 144 − b 2
b 2 = 20
x2 y2 + =1 144 63
b = 21
51.
49.
c2 = a2 − b2 2
2
2
center (0, 0 ) c=9 c 3 = a 4 9 3 = a 4 a = 12
b 2 = 135
x2 y 2 + =1 20 36
center ( − 2, 1)
x2 y2 + =1 135 144
2a = 24 a = 12 c 2 = a 3 c 2 = 12 3 c =8
53.
c=3 c 1 = a 4 3 1 = a 4 a = 12
c2 = a2 − b2
c2 = a 2 − b2
64 = 144 − b 2
9 = 144 − b 2
b 2 = 80
b 2 = 135
( x + 2)2 ( y − 1) 2 + =1 135 144
x2 y2 + =1 80 144
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514
Chapter 8: Topics in Analytic Geometry
2a = 15
54.
55.
a = 15
2 225 a = 4 2
c 3 = a 5 2c 3 = 15 5
484 = 64 + c 2 c 2 = 420 c = 20.494 2c = 40.9878 ≈ 41 The emitter should be placed 41 cm away.
c=9
2
c2 = a2 − b2 81 = 225 − b 2 4 4 b 2 = 144 = 36 4 2
x2 y + =1 225 / 4 36
56.
a.
b.
4= h 5 4.5 h = 18 = 3.6 5 The value of h is 3.6 in.
Major axis: 2
From part b, 2a = 5.763 2b = 4.5 a = 2.88 b = 2.25 a 2 = 8.3025 b 2 = 5.0625
c. 2
(4.5) + (3.6) = 5.76 in. Minor axis: Diameter of vent pipe = 4.50 in.
The equation is 57.
Aphelion = 2a − perihelion 934.34 = 2a − 835.14 a = 884.74 million miles Aphelion = a + c = 934.34 884.74 + c = 934.34 c = 49.6 million miles
58.
The mean distance is a = 67.08 million miles.
b = a −c
a = semimajor axis = 50 feet b = height = 30 feet c2 = a 2 − b2
Aphelion = a + c = 67.58 million miles. Thus c = 67.58 − a = 0.50 million miles.
2
59.
c 2 = 502 − 302 c = 1600 = 40 The foci are located 40 feet to the right and to the left of center.
2
b = a2 − c2
= 884.74 2 − 49.6 2 ≈ 883.35 million miles
= 67.082 − 0.502 ≈ 67.078
An equation of the orbit of Saturn is x2 884.742
+
y2 883.352
=1
An equation of the orbit of Venus is x2 67.08
2
+
y2 67.0782
2 x2 + y =1 . 8.3025 5.0625
=1
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.2
60.
515
The length of the semimajor axis is 50 feet. Thus
61.
c2 = a2 − b2
2b = 9
2a = 36 a = 18 2
b=9
2
2
c = a −b
32 2 = 50 2 − b 2
2
(2)
2 c 2 = 182 − 9
b 2 = 50 2 − 32 2
c 2 = 324 − 81
b = 50 2 − 32 2 b ≈ 38.4 feet
4
2
c = 1215 4 c = 9 15 2
Since one focus is at (0, 0), the center of the ellipse is at (9 15 / 2, 0) (17.43, 0). The equation of the path of Halley’s Comet in astronomical units is
(x − 9
15 / 2
)
2
324 62.
The reflective property of an ellipse.
63.
+
y2 =1 81/ 4
2 x2 + y = 1 752 342 Solve for y, where x = 55. 2 552 + y = 1 752 342
y2
2
= 1 − 552 34 75 2⎞ ⎛ y 2 = 342 ⎜ 1 − 552 ⎟ ⎝ 75 ⎠ 2
2⎞ ⎛ y = 342 ⎜ 1 − 552 ⎟ ⎝ 75 ⎠ y ≈ 23 ft h = y + 1 = 23 + 1 = 24 ft
64.
The gear on the left speeds up and slows down twice as the gear on the right makes one complete revolution at a constant angular speed.
65.
a.
c 2 = a 2 − b2 c 2 = 42 − 32 c2 = 7 c= 7 7 ft to the right and left of O.
b.
2a = 2(4) = 8 ft
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516
66.
Chapter 8: Topics in Analytic Geometry
2a = 3.04 ⇒ a = 1.52 2b = 2.99 ⇒ b = 1.495
67.
) ( 2 (1.522 + 1.4952 )
y=
p = π 2 a 2 + b2 =π
= π 2(2.3104 + 2.235025) = π 2(4.545425) = π 9.09085 AU ⋅ (92.96 million miles per AU) ≈ 881 million miles
68.
9 y 2 + 36 y + 16 x 2 − 108 = 0
25 y 2 + 50 y + 8 x 2 − 48 x + 47 = 0
69.
−36 ± 362 − 4(9)(16 x 2 − 108) 2(9)
=
−36 ± 1296 − 36(16 x 2 − 108) 18
=
−36 ± 1296 − 576 x 2 + 3888 18
=
−36 ± −576 x 2 + 5184 18
=
−36 ± 576(− x 2 + 9) 18
=
−36 ± 24 (− x 2 + 9) 18
=
−6 ± 4 (− x 2 + 9) 3
9 y 2 + 18 y + 4 x 2 + 24 x + 44 = 0
y=
−50 ± 502 − 4(25)(8 x 2 − 48 x + 47) 2(25)
y=
−18 ± 182 − 4(9)(4 x 2 + 24 x + 44) 2(9)
=
−50 ± 2500 − 100(8 x 2 − 48 x + 47) 50
=
−18 ± 324 − 36(4 x 2 + 24 x + 44) 18
=
−50 ± 2500 − 800 x 2 + 4800 x − 4700 50
=
−18 ± 324 − 144 x 2 − 864 x − 1584 18
=
−50 ± −800 x 2 + 4800 x − 2200 50
=
−18 ± −144 x 2 − 864 x − 1260 18
=
−50 ± 100( −8 x 2 + 48 x − 22) 50
=
−18 ± 36(−4 x 2 − 24 x − 35) 18
=
−50 ± 10 −8 x 2 + 48 x − 22 50
=
−18 ± 6 −4 x 2 − 24 x − 35 18
=
−5 ± −8 x 2 + 48 x − 22 5
=
−3 ± −4 x 2 − 24 x − 35 3
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Section 8.2
517
....................................................... 70.
4 x 2 + 9 y − 16 x − 2 = 0
Connecting Concepts 71.
This is not the equation of an ellipse because there is no y 2 term. It is a quadratic equation.
The sum of the distances between the two foci and a point on the ellipse is 2a. 2
2
⎛9 ⎞ ⎛9 ⎞ 2 2 2 a = ⎜ − 0 ⎟ + ( 3 − 3 ) + ⎜ − 0 ⎟ + ( 3 + 3) ⎝2 ⎠ ⎝2 ⎠
9 y = −4 x 2 + 16 x + 2
2
225 ⎛9⎞ = ⎜ ⎟ + 4 ⎝2⎠ 9 15 = + 2 2 = 12 a=6
c=3 2
c = a2 − b2 9 = 36 − b 2 b 2 = 27 x2 y2 + =1 36 27 72.
The sum of the distances between the two foci and a point on the ellipse is 2a.
2a =
(4 − 4)2 + ⎛⎜ 9 − 0 ⎞⎟ ⎝5
2
1681 ⎛9⎞ = ⎜ ⎟ + 25 ⎝5⎠ 9 41 = + 5 5 = 10
⎠
2
+
(4 + 4)2 + ⎛⎜ 9 − 0 ⎞⎟ ⎝5
⎠
2
73.
The sum of the distances between the two foci and a point on the ellipse is 2a. 2a =
( 5 − 2 )2 + ( 3 + 1)2 + ( 5 − 2 )2 + ( 3 − 3)2
= 25 + 32 = 5+3 =8 a=4 c=2 c2 = a 2 − b2 4 = 16 − b 2
5=a c=4 c2 = a2 − b2 16 = 25 − b 2
b 2 = 12
( x − 1)2 + ( y − 2 )2 16
12
=1
b2 = 9 x2 y2 + =1 9 25
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518
Chapter 8: Topics in Analytic Geometry
74.
The sum of the distances between the two foci and a point on the ellipse is 2a. a=4 2 2 ⎛3 ⎞ ⎛3 ⎞ c=3 2a = (1 − 7 )2 + ⎜ + 1⎟ + (1 − 1)2 + ⎜ + 1⎟ ⎝4 ⎠ ⎝4 ⎠ 2 c = a2 − b2 2 625 ⎛7⎞ 9 = 16 − b 2 = + ⎜ ⎟ 16 ⎝4⎠ b2 = 7 25 7 = + (x + 1)2 + ( y − 4)2 = 1 4 4 =8 7 16
75.
Center (1, − 1) c2 = a 2 − b2 c 2 = 16 − 9 c2 = 7 c= 7
The latus rectum is on the graph of y = −1 + 7, or y = −1 − 7
( x − 1)2 + ( y + 1)2 9
16
( x − 1)2 + ( −1 + 9
)
7 +1
2
= 1 or
16
( x − 1)2 + 9
=1
( x − 1)2 + ( −1 − 9
)
7 +1 16
2
=1
7 =1 16
( x − 1)2 9
=
9 16
2
16 ( x − 1) = 81
( x − 1)2 =
81 16
81 16 9 x −1 = ± 4 13 5 x= and − 4 4 x −1 = ±
The x-coordinates of the endpoints of the latus rectum are
13 5 and − . 4 4
13 ⎛ 5 ⎞ 9 −⎜− ⎟ = 4 ⎝ 4⎠ 2
The length of the latus rectum is
9 . 2
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Section 8.2
76.
519
9 x 2 + 16 y − 36 x + 96 y + 36 = 0
9 x 2 − 36 x + 16 y 2 + 96 y + 36 = 0
) ( ) 9(x − 4 x + 4 )+ 16(y 2 + 6 y + 9) = −36 + 36 + 144 (
9 x 2 − 4 x + 16 y 2 + 6 y = −36
2
9(x − 2 )2 + 16( y + 3)2 = 144
( x − 2 ) 2 + ( y + 3) 2 16
( 2,
Center 2
2
9
=1
− 3)
c = a −b
2
c 2 = 16 − 9 c2 = 7 c= 7
The latus rectum is on the graph of x = 2 + 7 or x = 2 − 7.
(x − 2)2 + ( y + 3)2
(2 +
16
9
7 −2 16
=1
)2 + (y + 3)2 = 1 or (2 − 9
)2
7 (y + 3 + 16 9
( y + 3)2 9
7 −2 16
)2 + ( y + 3)2 = 1 9
=1 =
9 16
16( y + 3)2 = 81
( y + 3)2 81 16
81 16 9 y+3= ± 4 3 21 y = − and − 4 4 y+3= ±
The y-coordinates of the endpoints of the latus rectum are − 3 and − 21 . 4 4 3 ⎛ 21 ⎞ 9 − −⎜− ⎟ = 4 ⎝ 4⎠ 2
The length of the latus rectum is 9 . 2
Copyright © Houghton Mifflin Company. All rights reserved.
520
77.
Chapter 8: Topics in Analytic Geometry
Let us transform the general equation of an ellipse into an x′y′ - coordinate system where the center is at the origin by replacing ( x − h) by x′ and (y − k ) by y′. x′2
(c ) 2 a2
+
y ′2 b2
+
y ′2
= 1. a 2 b2 Letting x′ = c and solving for y′ yields We have
=1
b 2c 2 + a 2 y′2 = a 2b 2 a 2 y′2 = a 2b 2 − b 2c 2 a 2 y ′2 = b 2 ( a 2 − c 2 ) But since c 2 = a 2 − b 2 , b 2 = a 2 − c 2 , we can substitute to obtain a 2 y ′2 = b2 (b2 ) 4 y ′2 = b
a2
2 4 y′ = ± b = ± b 2 a a 2⎞ 2⎞ ⎛ ⎛ The endpoints of the latus rectum, then, are ⎜ c, b ⎟ and ⎜ c, − b ⎟ . a ⎠ a ⎠ ⎝ ⎝ 2 The distance between these points is 2b . a
78.
Let P(x, y ) be a point on the ellipse and let F1 (0, c ) and F2 (0, − c ) be the foci. By the definition of an ellipse,
d ( P, F1 ) + d ( P, F2 ) = 2a ( x − 0)2 + ( y − c )2 + ( x − 0)2 + ( y + c )2 = 2a x 2 + ( y + c ) 2 = 2a − x 2 + ( y − c ) 2 x 2 + ( y + c ) 2 = 4a 2 − 4 a x 2 + ( y − c ) 2 + x 2 + ( y − c ) 2 x 2 + y 2 + 2cy + c 2 = 4a 2 − 4a x 2 + ( y − c ) + x 2 + y 2 − 2cy + c 2 4cy = 4a 2 − 4a x 2 + ( y − c )2 cy = a 2 − a x 2 + ( y − c )2 cy − a 2 = − a x 2 + ( y − c)2 c 2 y 2 − 2cya 2 + a 4 = a 2 ( x 2 + ( y − c)2 ) 2 2
2
4
2 2
2 2
Square each side. 2
c y − 2cya + a = a x + a y − 2cya + a 2 c 2 a 4 − a 2c2 = a2 x2 + a 2 y 2 − c2 y 2 a 2 (a 2 − c2 ) = a 2 x 2 + (a 2 − c2 ) y 2 a 2 b2 = a 2 x 2 + b2 y 2 1=
x2 b2
+
Let b2 = a 2 − c 2 .
y2 a2
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Square each side.
Section 8.3
521
....................................................... PS1.
4 + −2 = 1 2 −3 + 1 = −1 2 Midpoint: (1, –1)
PS2.
( x − 1)( x + 3) = 5 x2 + 2 x − 3 = 5 x2 + 2 x − 8 = 0 ( x + 4)( x − 2) = 0
4 =4 8 =8 2 = 2 8 8 8
PS3.
x+4 = 0 x−2 = 0 x = −4 x=2
(−2 − 4) 2 + (1−−3) 2 = 52 = 2 13 Length: 2 13
PS4. 4 x 2 + 24 x = 4( x 2 + 6 x ) = 4( x 2 + 6 x + 9) = 4( x + 3)2
Prepare for Section 8.3
PS5.
2 x2 − y = 1 4 9 2 y2 − = 1− x 9 4 2 y2 = 9x − 9 4 2 y = ± 9x − 9 4 y = ± 3 x2 − 4 2
PS6.
Section 8.3 1. 3.
a. iii
b. ii
c. i
x2 y2 − =1 16 25
( 0, 0 ) Vertices ( ±4, 0 )
(±
41, 0
2. 4.
Center Foci
d. iv
)
5 Asymptotes y = ± x 4
x2 y2 − =1 16 9
5.
a. iii
b. i
c. iv
y2 x2 − =1 4 25
6.
( 0, 0 ) Vertices ( 0, ± 2 )
( 0, 0 ) Vertices ( ±4, 0 ) Foci ( ±5, 0 )
Center
3 Asymptotes y = ± x 4
2 Asymptotes y = ± x 5
Center
d. ii
Foci
( 0,
± 29
)
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y2 x2 − =1 25 36
( 0, 0 ) Vertices ( 0, ± 5)
Center Foci
( 0,
± 61
)
5 Asymptotes y = ± x 6
522
7.
Chapter 8: Topics in Analytic Geometry
x2 y2 − =1 7 9 Center
( 0, 0 )
Vertices Foci
8.
(±
x2 y2 − =1 5 4 Center
7, 0
)
Foci
Asymptotes y = ±
( 0, 0 )
Vertices
( ±4, 0 ) 3 7 7
x
9.
(±
4x 2 y 2 − =1 9 16 Center
5, 0
) 2 5 5
Center
⎛ 97 ⎞ Foci ⎜⎜ ± , 0 ⎟⎟ 3 ⎝ ⎠
⎛ 73 ⎞ Foci ⎜⎜ ± , 0 ⎟⎟ 2 ⎝ ⎠
x
4 Asymptotes y = ± x 9
8 Asymptotes y = ± x 3
11.
( x − 3) 2 − ( y + 4 ) 2 16
9
12.
=1
Foci
( 3 ± 4,
( 3 ± 5,
− 4 ) = ( 7, − 4 ) , ( −1, − 4 )
− 4 ) = ( 8, − 4 ) , ( −2, − 4 )
( y + 2)2 − (x − 1)2 4
16
Foci
( −3 ±
) (
) ( −3 −
29, 0 = −3 + 29, 0 ,
29, 0
)
2 Asymptotes y = ± ( x + 3) 5
14.
=1
( y − 2)2 − (x + 1)2 36
Center (1, − 2 )
49
=1
( −1, 2 ) Vertices ( −1, 2 ± 6 ) = ( −1, 8 ) , ( −1,
Center
Vertices (1, − 2 ± 2 ) = (1, 0 ) , (1, − 4 )
(
y2 =1 4
Center
3 Asymptotes y + 4 = ± ( x − 3) 4
13.
25
−
( −3, 0 ) Vertices ( −3 ± 5, 0 ) = ( 2, 0 ) , ( −8, 0 )
Center ( 3, − 4 ) Vertices
(x + 3)2
x2 9y2 − =1 9 16
( 0, 0 ) Vertices ( ±3, 0 )
( 0, 0 )
⎛ 3 ⎞ Vertices ⎜ ± , 0 ⎟ ⎝ 2 ⎠
( ±3, 0 )
Asymptotes y = ±
10.
) (
) (
Foci 1, − 2 ± 2 5 = 1, − 2 + 2 5 , 1, − 2 − 2 5 Asymptotes y + 2 = ±
1 ( x − 1) 2
)
Foci
( −1, 2 ±
Asymptotes
) (
− 4)
) (
85, 0 = −1, 2 + 85 , −1, 2 − 85 6 7
( y − 2 ) = ± ( x + 1)
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)
Section 8.3
15.
523
( x + 2 )2
−
9 Center
y2 =1 25
16.
( −2, 0 ) ( −2 ± 3, 0 ) = (1, 0 ) , ( −5, 0 )
( 0, 2 ) Vertices ( 0 ± 5, 2 ) = ( 5, 2 ) , ( −5, 2 )
Center
Vertices Foci
( −2 ±
34, 0
Asymptotes y = ±
17.
x 2 ( y − 2 )2 − =1 25 81
)
Foci
5 ( x + 2) 3
(±
Asymptotes
9(x − 1)2 ( y + 1)2 − =1 16 9
18.
25
( x + 6 )2 −
Center (1, − 1)
Center
9
25
⎛ 4 ⎞ ⎛7 ⎞ ⎛ 1 ⎞ Vertices ⎜1 ± , − 1⎟ = ⎜ , − 1⎟ , ⎜ − , − 1⎟ 3 3 3 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ 97 Foci ⎜⎜ 1 ± , − 1⎟⎟ 3 ⎝ ⎠ Asymptotes
9 4
( y + 1) = ± ( x − 1)
144
) (
106, 2 , − 106, 2
( y − 2) = ±
( x + 6 )2 − 25 y 2
( x − 1)2 − ( y + 1)2 16 / 9
) (
106, 2 =
9 x 5
=1
y2 =1 144 / 25
( −6, 0 ) Vertices ( −6 ± 5, 0 ) = ( −1, 0 ) , ( −11, 0 )
⎛ 769 ⎞ Foci ⎜⎜ −6 ± , 0 ⎟⎟ 5 ⎝ ⎠ 12 Asymptotes y = ± ( x + 6 ) 25
Copyright © Houghton Mifflin Company. All rights reserved.
)
524
19.
Chapter 8: Topics in Analytic Geometry
x2 − y2 = 9
20.
21.
16 y 2 − 9 x 2 = 144
x2 y2 − =1 9 9
x2 y2 − =1 4 16
y2 x2 − =1 9 16
( 0, 0 ) Vertices ( ±3, 0 )
Center
( 0, 0 ) Vertices ( ±2, 0 )
Center
Foci
Foci
Center
( ±3
2, 0
)
Asymptotes y = ± x
22.
4 x 2 − y 2 = 16
9 y 2 − 25 x 2 = 225
( ±2
5, 0
23.
9 y 2 − 36 x 2 = 4
y2 x2 − =1 4 / 9 1/ 9
( 0, 0 ) Vertices ( 0, ± 5 )
Center
Foci
( 0,
± 34
)
5 Asymptotes y = ± x 3
)
3 Asymptotes y = ± x 4
Asymptotes y = ±2 x
y2 x2 − =1 25 9 Center
( 0, 0) Vertices ( 0, ± 3) Foci ( 0 ± 5 )
( 0, 0 )
24.
16 x 2 − 25 y 2 = 9
y2 x2 − =1 9 / 16 9 / 25 Center
( 0, 0 )
2⎞ ⎛ Vertices ⎜ 0, ± ⎟ 3⎠ ⎝
⎛ 3 Vertices ⎜ ± , ⎝ 4
⎛ 5⎞ Foci ⎜⎜ 0, ± ⎟ 3 ⎟⎠ ⎝ Asymptotes y = ±2 x
⎛ 3 41 ⎞ Foci ⎜⎜ ± , 0 ⎟⎟ ⎝ 20 ⎠ 4 Asymptotes y = ± x 5
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⎞ 0⎟ ⎠
Section 8.3
25.
525
x 2 − y 2 − 6x + 8 y = 3
26.
(x 2 − 6x)− (y 2 − 8 y ) = 3 (x 2 − 6x + 9)− (y 2 − 8 y + 16) = 3 + 9 − 16
4 x 2 − 25 y 2 + 16 x + 50 y − 109 = 0
) ( ) 4(x + 4 x + 4)− 25(y 2 − 2 y + 1) = 109 + 16 − 25 (
4 x 2 + 4 x − 25 y 2 − 2 y = 109
2
4(x + 2)2 − 25( y − 1)2 = 100
(x − 3)2 − ( y − 4)2 = −4 ( y − 4)2 − (x − 3)2 = 1 4
(x + 2)2 − ( y − 1)2 25
4
) (
)(
Foci 3, 4 ± 2 2 = 3, 4 + 2 2 , 3, 4 − 2 2 Asymptotes y − 4 = ± (x − 3)
27.
Vertices (− 2 ± 5, 1) = (− 7, 1), (3, 1)
(
)
28.
) ( ) 9(x + 4 x + 4 )− 4(y 2 + 2 y + 1) = −68 + 36 − 4 9 x + 4 x − 4 y + 2 y = −68
( y + 3)2 − (x − 1)2
9(x + 2 )2 − 4( y + 1)2 = −36 4
16
=1
) (
9
=1
Center (1, − 3)
Center (−2, − 1) Vertices (− 2, − 1 ± 3) = (− 2, 2 ), (− 2, − 4)
(
) ( ) 16(x − 2 x + 1)− 9(y 2 + 6 y + 9 ) = −79 + 16 − 81 (
16(x − 1)2 − 9( y + 3)2 = −144
2
9
16 x 2 − 9 y 2 − 32 x − 54 y + 79 = 0 2
2
( y + 1)2 − (x + 2)2
)(
16 x 2 − 2 x − 9 y 2 + 6 y = −79
9 x 2 + 36 x − 4 y 2 − 8 y = −68
(
) (
)(
Foci − 2, − 1 ± 13 = − 2, − 1 + 13 , − 2, − 1 − 13 3 Asymptotes y + 1 = ± (x + 2 ) 2
)
Foci − 2 ± 29 , 1 = − 2 + 29 , 1 , − 2 − 29 , 1 2 Asymptotes y − 1 = ± (x + 2) 5
9 x 2 − 4 y 2 + 36 x − 8 y + 68 = 0 2
=1
Center (−2, 1)
Center (3, 4 ) Vertices (3, 4 ± 2 ) = (3, 6), (3, 2 )
(
4
Vertices (1, − 3 ± 4 ) = (1, 1), (1, − 7 )
)
Foci (1, − 3 ± 5) = (1, 2), (1, − 8)
Asymptotes ( y + 3) = ±
4 (x − 1) 3
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526
29.
Chapter 8: Topics in Analytic Geometry
y=
= =
(
−6 ± 62 − 4 ( −1) 4 x 2 + 32 x + 39 2 ( −1)
(
−6 ± 36 + 4 4 x 2 + 32 x + 39
)
30.
)
−2 2
−6 ± 16 x + 128 x + 192 −2
=
=
−6 ± 4 x 2 + 8 x + 12 −2
=
2
= 3 ± 2 x + 8 x + 12
31.
y=
y=
=
−6 ± 16( x 2 + 8 x + 12) = −2 =
64 ±
64 ±
(
( −64 )2 − 4 ( −16 ) 9 x 2 − 36 x + 116 2 ( −16 )
)
32.
y=
( −64 )2 − 4 ( −16 ) 2 ( −16 )
( x2 + 8x + 16)
(
64 ± 4096 + 64 x 2 + 8 x + 16 −32
(
2
64 ± 64 x + 8 x + 16 + 64
)
)
−32
64 ± 8
−8 ±
( x2 + 8x + 80) (x
−32 2
+ 8 x + 80
)
4
18 ± (−18)2 − 4(−9)(2 x 2 + 12 x + 18) 2(−9)
=
64 ± 4096 + 64(9 x 2 − 36 x + 116) −32
=
18 ± 324 + 36(2 x 2 + 12 x + 18) −18
=
64 ± 64(9 x 2 − 36 x + 116 + 64) −32
=
18 ± 36(2 x 2 + 12 x + 18 + 9) −18
=
64 ± 8 (9 x 2 − 36 x + 180) −32
=
−3 ± 2 x 2 + 12 x + 27 3
=
64 ± 8 9( x 2 − 4 x + 20) −32
=
64 ± 24 x 2 − 4 x + 20 −32
=
−8 ± 3 x 2 − 4 x + 20 4
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Section 8.3
33.
527
y=
=
=
=
=
35.
( −18)2 − 4 ( −9 ) 2 ( −9 )
18 ±
( 4 x2 + 8x − 6)
(
18 ± 324 + 36 4 x 2 + 8 x − 6 −18
(
2
18 ± 36 4 x + 8 x − 6 + 9
34.
)
=
)
=
−18
( 4 x 2 + 8 x + 3)
18 ± 6
−18
−3 ±
( 4x
2
+ 8x + 3
y=
=
)
=
3
vertices (3, 0) and (−3, 0), foci (4, 0 ) and (−4, 0 )
36.
(
−36 ± 362 − 4 ( −9 ) 2 x 2 − 8 x − 46 2 ( −9 )
(
−36 ± 1296 + 36 2 x 2 − 8 x − 46
(
−18 2
−36 ± 36 2 x − 8 x − 46 + 36
)
)
)
−18
( 2 x2 − 8x − 10)
−36 ± 6
−18 2
6 ± 2 x − 8 x − 10 3
vertices
( 0, 2 ) and ( 0,
− 2 ) , foci
( 0, 3) and ( 0,
− 3)
Traverse axis is on x-axis. For a standard hyperbola, the vertices are at (h + a, k) and (h – a, k), h + a = 3, h – a = −3, and k = 0..
Transverse axis is on y -axis. Since vertices are at ( h, k + a ) and ( h, k − a ) , k + a = 2, k − a = −2, and h = 0.
If h + a = 3 and h – a = −3, then h = 0 and a = 3.
Therefore, k = 0 and a = 2.
The foci are located at (4, 0) and (−4, 0). Thus, h = 0 and c = 4.
The foci are located at ( h, k + c ) and specifically at ( 0, 3) and ( 0, − 3) .
Since c 2 = a 2 + b 2 , b 2 = c 2 − a 2 b 2 = ( 4 ) − ( 3) = 16 − 9 = 7
Since k = 0, c = 3.
( x − h )2 − ( y − k )2
b 2 = c 2 − a 2 = ( 3) − ( 2 ) = 9 − 4 = 5
2
a
2
2
( x − 0) ( 3)2
b
2
−
2
( y − 0 )2 7
=1 =1
x2 y 2 − =1 9 7
2
( y − k ) 2 ( x − h) 2 a2
−
b2
2
=1
y2 x2 − =1 4 5
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( h, k − c ) ,
or
528
37.
Chapter 8: Topics in Analytic Geometry
foci ( 0, 5) and ( 0, − 5 ) , asymptotes y = 2 x and y = −2 x Transverse axis is on y -axis. Since foci are at ( h, k + c ) and
( h, k − c ) ,
k + c = 5, k − c = −5, and h = 0.
Therefore, k = 0 and c = 5. Since one of the asymptotes is y =
a a = 2 and x, b b
a = 2b. 2
38.
foci
( 4, 0 )
and
( −4, 0 ) ,
Transverse axis is on x-axis. Since foci are at and
2
a + b = c ; then substituting a = 2b and c = 5 yields
( 2b )2 + b2 = ( 5)2 ,
or 5b 2 = 25.
Therefore, h = 0 and c = 4. Since the asymptotes are b b b y = x, and y = − x, = 1 and b = a. a a a
a 2 + a 2 = 42 , or 2a 2 = 16. Therefore, a 2 = 8 and a = 2 2. Since b = a, b = 2 2.
Therefore, b 2 = 5 and b = 5. Since a = 2b, a = 2
)2
−
a2
y
( x − h )2 b2
2
(2 5 )
2
−
( 5) = 2
5.
( x − h )2 − ( y − k )2 a2
b2
x2 − 8
=1
2
x =1 5
y2
(2 2 )2
=1 =1
x2 y2 − =1 8 8
y2 x2 − =1 20 5 39.
vertices (0, 3) and (0, − 3) , point (2, 4) The distance between the two vertices is the length of the transverse axis, which is 2a. 2a = 3 − (− 3) = 6 or a = 3. Since the midpoint of the transverse axis is the center of the hyperbola, the center is given by ⎛ 0 + 0 3 + (− 3) ⎞ , ⎜ ⎟ , or (0, 0) 2 ⎝ 2 ⎠ Since both vertices lie on the y-axis, the transverse axis must be on the y-axis. Taking the standard form of the hyperbola, we have y2
−
( h + c, k )
( h − c, k ) , h + c = 4, h − c = −4, and k = 0.
a 2 + b 2 = c 2 ; then substituting b = a and c = 4 yields
2
(y − k
asymptotes y = x and y = − x
x2
=1 a2 b2 Substituting the point (2, 4) for x and y, and 3 for a, we have 16 4 − =1 9 b2 Solving for b 2 yields b 2 = 36 . 7
Therefore, the equation is y2 x2 − =1 9 36 / 7
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Section 8.3
40.
529
vertices (5, 0) and (−5, 0) , point (−1, 3) The length of the transverse axis, 5 − (−5) , or 10, is equal to 2a. Therefore, a = 5. The midpoint of the transverse axis, or the center of the hyperbola, is given by ⎛ 5 + (− 5) 0 + 0 ⎞ , ⎜ ⎟ , or (0, 0) 2 2 ⎠ ⎝ Since both vertices are on the x-axis, the transverse axis must lie on the x-axis. Therefore, we have x2
−
y2
=1 a2 b2 Substituting the point (−1, 3) for x and y, and 5 for a, we have 1 9 − =1 25 b 2 Solving for b 2 yields b 2 = − 225 . 24
However, b 41.
2
must be positive. Therefore, no such hyperbola exists.
vertices (0, 4 ) and (0, − 4 ) , asymptotes y = 1 x and y = − 1 x. 2
2
The length of the transverse axis, or the distance between the vertices, is equal to 2a . 2a = 4 − (−4 ) = 8 , or a = 4 The center of the hyperbola, or the midpoint of the line segment joining the vertices, is ⎛ 0 + 0 4 + (− 4 ) ⎞ , ⎜ ⎟ , or (0, 0) 2 ⎝ 2 ⎠ Since both vertices lie on the y-axis, the transverse axis must lie on the y-axis. Therefore, the asymptotes are given by y = a x and y = − a x. One asymptote is y = 1 x. Thus a = 1 or b = 2a. b
b
2
b
2
Since b = 2a and a = 4, b = 2(4) = 8. Thus, the equation is y2 42 42.
−
x2 82
= 1 or
y2 x2 − =1 16 64
vertices (6, 0) and (−6, 0) , asymptotes y = 2 x and y = − 2 x 3
3
Length of transverse axis = distance between vertices 2a = 6 − (−6 ) 2a = 12 a=6 The center of the hyperbola is given by the midpoint of the line segment joining the vertices, or ⎛ 6 + (− 6 ) 0 + 0 ⎞ , ⎜ ⎟ , which is (0, 0). 2 2 ⎠ ⎝ Since both vertices lie on the x-axis, the transverse axis must lie on the x-axis. Thus, the asymptotes must be given by y = b x and y = − b x .
a a 2 2 Since y = x and y = − x define the actual asymptotes, b = 2 and b = 2a . 3 3 a 3 3
2(6) =4. 3 If a = 6 and b = 4.
Since a = 6, b = x2 62
−
y2 42
=1
x2 y2 − =1 36 16
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530
43.
Chapter 8: Topics in Analytic Geometry
vertices (6, 3) and (2, 3), foci (7, 3) and (1, 3) Length of transverse axis = distance between vertices 2a = 6 − 2 a=2 ⎛ 6 + 2 3+ 3⎞ , The center of the hyperbola (h, k) is the midpoint of the line segment joining the vertices, or the point ⎜ ⎟. 2 ⎠ ⎝ 2 6+2 3+3 Thus, h = , or 4, and k = , or 3 . 2 3 Since both vertices lie on the horizontal line y = 3, the transverse axis is parallel to the x-axis. The location of the foci is given by (h + c, k) and (h – c, k), or specifically (7, 3) and (1, 3). Thus h + c = 7, h – c = 1, and k = 3. Solving for h and c simultaneously yields h = 4 and c = 3. Since c 2 = a 2 + b 2 , b 2 = c 2 − a 2 . Substituting, we have b 2 = 3 2 − 2 2 = 9 − 4 = 5. Substituting a = 2, b 2 = 5, h = 4, and k = 3 in the standard equation
44.
(x − h)2 − ( y − k )2 a2
b2
= 1 yields
(x − 4)2 − ( y − 3)2 4
5
vertices (−1, 5) and (−1, − 1) , foci (−1, 7 ) and (−1, − 3) Length of transverse axis = distance between vertices 2a = 5 − (− 1) = 6 a=3 Center (midpoint of transverse axis) is ⎛ −1 − 1 5 − 1 ⎞ , ⎜ ⎟ , or (-1, 2) 2 ⎠ ⎝ 2 Thus, h = −1 and k = 2. The foci, (h, k + c ) and (h, k − c ) , are (−1, 7 ) and (−1, − 3) . Thus, since k + c = 7 and k = 2, c = 5. b 2 = c 2 − a 2 = 5 2 − 3 2 = 25 − 9 = 16 Thus, we obtain
45.
( y − 2)2 − (x + 1)2 9
16
=1
foci (1, −2) and (7, −2) , slope of an asymptote = 5
4
Both foci lie on the horizontal line y = −2 ; therefore, the transverse axis is parallel to the x-axis. The foci are given by (h + c, k) and (h −c, k ) . Thus, h − c = 1, h + c = 7, and k = −2 . Solving simultaneously for h and c yields h = 4 and c = 3. Since y − k = b (x − h ) is the equation for an asymptote, and the slope of an asymptote is given as 5 , b = 5 , b = 5a , 4 a
a
2 and b 2 = 25a .
16
Because a 2 + b 2 = c 2 , substituting c = 3 and b 2 = Therefore, b 2 =
144 25a 2 . yields a 2 = 41 16
3600 225 = . 656 41
Substituting in the standard equation for a hyperbola yields
( x − 4 )2 − ( y + 2 )2 144 / 41
225 / 41
=1
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4
4
=1 .
Section 8.3
46.
531
foci (−3, − 6 ) and (−3, − 2 ) , slope of an asymptote =1 Both foci are on the vertical line x = −3 ; therefore, the transverse axis is parallel to the y-axis. The foci are given by (h, k + c) and (h, k −c) . Thus, k − c = −6, k + c = −2, and h = −3 . Solving simultaneously for k and c yields k = − 4 and c = 2 . Since y − k = a (x − h ) is the equation for one asymptote, and the slope of an asymptote is given as 1, a = 1, b = a, and b 2 = a 2 . b
b
2
2
2
2
2
2
2
Because a + b = c , substituting c = 2 and b = a yields a = 2 ; therefore, b = 2 . Substituting in the standard equation of a hyperbola yields 47.
( y + 4)2 − (x + 3)2 2
2
=1
Because the transverse axis is parallel to the y-axis and the center is (7, 2), the equation of the hyperbola is
( y − 2)2 − (x − 7)2
=1 a2 b2 Because (9, 4) is a point on the hyperbola,
(4 − 2)2 − (9 − 7)2 a2
b2
=1
The slope of the asymptote is 1 . Therefore 1 = a or b = 2a . 2
b
2
Substituting, we have 4 4 − =1 2 a 4a 2 4 1 − = 1 , or a 2 = 3 a2 a2 Since b = 2a, b 2 = 4a 2 , or b 2 = 12 . The equation is 48.
( y − 2)2 − (x − 7)2 3
12
=1.
Because the transverse axis is parallel to the x-axis, and the center is (3, 3), the equation of the hyperbola is
(x − 3)2 − ( y − 3)2
=1 a2 b2 Because (6, 1) is a point on the hyperbola,
(6 − 3)2 − (1 − 3)2 a2
b2
=1
The slope of an asymptote is 2. Therefore, 2 =
b , or 2a = b . a
Substituting, we have 9 4 9 1 8 − = − = =1 2 2 2 2 a 4a a a a2 Thus, a 2 = 8 . Since 2a = b, 4a 2 = b 2 , or b 2 = 32 . The equation of the hyperbola is
(x − 3)2 − ( y − 3)2 8
32
= 1.
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532
49.
Chapter 8: Topics in Analytic Geometry
vertices (1, 6) and (1, 8), eccentricity = 2 Length of transverse axis = distance between vertices 2a = 6 − 8 = 2 a = 1 and a 2 = 1 ⎛1+1 6 + 8 ⎞ , Center (midpoint of transverse axis) is ⎜ ⎟ , or (1, 7 ) . 2 ⎠ ⎝ 2 Therefore, h = 1 and k = 7. Since both vertices lie on the vertical line x = 1, the transverse axis is parallel to the y-axis.
Since e = c , c = ae = (1)(2 ) = 2 . a 2
Because b = c 2 − a 2 , b 2 = (2 )2 − (1)2 = 4 − 1 = 3 . Substituting h, k, a 2 , and b 2 into the standard equation yields
50.
( y − 7 )2 − (x − 1)2 1
3
=1
vertices (2, 3) and (− 2, 3) , eccentricity = 5
2
Length of transverse axis = distance between vertices 2a = 2 − (− 2 ) = 4 a = 2 and a 2 = 4 ⎛ 2 − 2 3+ 3⎞ Center (midpoint of transverse axis) is ⎜ , ⎟ , or (0, 3) . 2 ⎠ ⎝ 2 Thus, h = 0 and k = 3. Since both vertices lie on the horizontal line y = 3, the transverse axis is parallel to the x-axis. c ⎛5⎞ Since e = , c = ae = (2) ⎜ ⎟ = 5 . a ⎝2⎠ Because b 2 = c 2 − a 2 , b 2 = 5 2 − 2 2 = 25 − 4 = 21 . Substituting h, k, a 2 , and b 2 into the standard equation yields 51.
x 2 ( y − 3)2 − =1 4 21
foci (4, 0) and (−4, 0) , eccentricity = 2
⎛ 4 + (− 4 ) 0 + 0 ⎞ , Center (midpoint of line segment joining foci) is ⎜ ⎟ , or (0, 0) 2 ⎠ 2 ⎝ Thus, h = 0 and k = 0. Since both foci lie on the horizontal line y = 0, the transverse axis is parallel to the x-axis. The locations of the foci are given by (h + c, k) and (h − c, k ) , or specifically (4, 0) and (-4, 0) Since h = 0, c = 4. c 4 c Because e = , a = = = 2 and a 2 = 4 . e 2 a Because b 2 = c 2 − a 2 , b 2 = 4 2 − 2 2 = 16 − 4 = 12 . Substituting h, k, a 2 and b 2 into the standard formula for a hyperbola yields
x2 y2 − =1 4 12
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Section 8.3
52.
533
foci (0, 6) and (0, − 6 ) , eccentricity = 4 3
Center (midpoint of the line segment joining foci) is ⎛0+0 6−6⎞ , ⎜ ⎟ , or (0, 0) 2 ⎠ ⎝ 2 Thus, h = 0 and k = 0. Since both foci lie on the vertical line x = 0, the transverse axis is parallel to the y-axis. The location of the foci are given by (h, k + c) and (h, k − c ) , or specifically (0, 6) and (0, − 6) . Since k = 0, c = 6. 81 c 6 9 c Because e = , a = = = and a 2 = . 4 e 43 2 a 2
81 63 ⎛9⎞ . = Because b 2 = c 2 − a 2 , b 2 = 6 2 − ⎜ ⎟ = 36 − 4 4 ⎝2⎠ Substituting h, k, a 2 , and b 2 into the standard formula for a hyperbola yields y2 x2 − =1 81/ 4 63/ 4 53.
conjugate axis length = 4, center (4, 1), eccentricity = 4 3
2b = conjugate axis length = 4 b = 2 and b 2 = 4 Since e=
c 4 4a 16a 2 16a 2 = , c= and c 2 = . Since a 2 + b 2 = c 2 , substituting b 2 = 4 and c 2 = and solving for a 2 yields a 2 = 36 . 7 a 3 3 9 9
Substituting into the two standard equations of a hyperbola yields 54.
( x − 4 )2 − ( y − 1)2 36 / 7
4
= 1 and
( y − 1)2 − ( x − 4 )2 36 / 7
4
=1
conjugate axis length = 6, center (−3 , − 3) , eccentricity = 2 2b = conjugate axis length = 6 b = 3 and b 2 = 9 c Since e = = 2 , c = 2a and c 2 = 4a 2 . Since a 2 + b 2 = c 2 , substituting b 2 = 9 and c 2 = 4a 2 and solving for a 2 yields a 2 = 3 . a Substituting into the two standard equations for a hyperbola yields
55.
a.
Because the transmitters are 250 miles apart, 2a = rate × time
b.
3
9
9
x = 100
− y2 ≈ − 3.6248121 13, 462.75
2a = 0.186 × 500 = 93 Thus, a = 46.5 miles. 2
3
10,000 y2 − =1 2,162.25 13, 462.75
2c = 250 and c = 125.
2
(x + 3)2 − ( y + 3) 2 = 1 and ( y + 3)2 − (x + 3)2
2
2
b = c − a = 125 − 46.5 = 13, 462.75 miles The ship is located on the hyperbola given by x2 y2 − =1 2,162.25 13, 462.75
y 2 ≈ 48,799.939 y ≈ 221 The ship is 221 miles from the coastline.
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=1
534
56.
Chapter 8: Topics in Analytic Geometry
a.
Because the transmitters are 300 miles apart,
b.
x2
2c = 300 and c = 150.
74.42
2a = rate × time
57.
130.25 2
=1 =1
x = 74.4 miles The ship reaches the coastline 74.4 miles to the left of the origin at the point (−74.4, 0).
The ship is located on the hyperbola given by 74.4 2
130.252
x 2 = 74.42
b = c 2 − a 2 = 150 2 − 74.4 2 ≈ 130.25 miles y2
02
74.42
Thus, a = 74.4 miles.
−
−
x2
2a = 0.186 × 800 = 148.8 miles
x2
The ship will reach the coastline when y = 0. Thus,
=1
When the wave hits Earth, z = 0.
58.
a.
At the top of the tower, y = 380. 2
y 2 = x 2 + ( z −10,000)2 y 2 = x 2 + (0 −10,000)2 2 y − x 2 =10,0002 It is a hyperbola.
x 2 − (380 − 220) = 1 802 1802 x 2 = 1 + 1602 802 1802 2⎞ ⎛ x 2 = 802 ⎜ 1 + 1602 ⎟ ⎝ 180 ⎠ 2⎞ ⎛ x = 802 ⎜ 1 + 1602 ⎟ ⎝ 180 ⎠ x ≈ 107 ft
At the bottom of the tower, y = 0. 2
x 2 − (0 − 220) = 1 802 1802 x 2 = 1 + 2202 802 1802 2⎞ ⎛ x 2 = 802 ⎜ 1 + 2202 ⎟ ⎝ 180 ⎠ 2⎞ ⎛ x = 802 ⎜ 1 + 2202 ⎟ ⎝ 180 ⎠ x ≈ 126 ft
b. 59.
a.
Using the eccentricity, and a = 2, c = 17 ⇒ c = 17 2 4 2 Solve for b. 2
2
a +b =c
2
b2 = c 2 − a 2 2
⎛ ⎞ b2 = ⎜ 17 ⎟ − 22 ⎝ 2 ⎠ b2 = 17 − 16 4 4 2 1 b = 4 b = 1 = 0.5 2
b.
From the equation, a = 80 ft. For FG, y = 0.6. x 2 − 0.62 = 1 22 0.52 x 2 = 1 + 0.62 22 0.52 2⎞ ⎛ x 2 = 22 ⎜ 1 + 0.62 ⎟ ⎝ 0.5 ⎠ 2⎞ ⎛ x = 22 ⎜ 1 + 0.62 ⎟ ⎝ 0.5 ⎠ x ≈ 3.1241 FG = 2 x ≈ 6.25 in.
2 x2 − y = 1 2 2 0.52
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Section 8.3
60.
535
a.
hyperbola
b.
Foci: (–2, 0), (2, 0) Center (0, 0) c=2 2a = F1P − F2 P F1P − F2 P = 2 a =1 2
2
61.
2
4(x − 2 )2 + 9( y − 2)2 = 36
(x − 2)2 + ( y − 2)2 9
2
vertices (2 ± 3, 2 ) = (5, 2) , (−1, 2 )
(
) (
)(
foci 2 ± 5 , 2 = 2 + 5 , 2 , 2 − 5 , 2
) 2 2(x − 4 x + 4 ) = −3 y − 2 + 8 (
=1
center (2, 2)
2
2 x 2 − 4 x = −3 y − 2
4
ellipse
x 2 − y =1 1 3
2 x 2 + 3 y − 8x + 2 = 0
) ( ) 4(x − 4 x + 4 ) + 9(y 2 − 4 y + 4) = −16 + 16 + 36 (
4 x 2 − 4 x + 9 y 2 − 4 y = −16
c = a +b 22 =12 + b 2 b2 = 3
62.
4 x 2 + 9 y 2 − 16 x − 36 y + 16 = 0
63.
5 x − 4 y 2 + 24 y − 11 = 0
) − 4(y − 6 y + 9) = −5 x + 11 − 36 (
− 4 y 2 − 6 y = −5 x + 11 2
2(x − 2 )2 = −3 y + 6
− 4( y − 3)2 = −5(x − 25)
2(x − 2 )2 = −3( y − 2 ) (x − 2)2 = − 3 ( y − 2) 2
− 4( y − 3)2 = −5(x + 5) ( y − 3)2 = 5 (x + 5) 4
parabola vertex (2, 2) 3 ⎞ ⎛ 13 ⎞ ⎛ focus ⎜ 2, 2 − ⎟ = ⎜ 2, ⎟ 8 ⎝ ⎠ ⎝ 8⎠ 19 3 directrix y = 2 + , or y = 8 8
parabola vertex
(−5, 3)
5 ⎞ ⎛ 75 ⎞ ⎛ , 3⎟ focus ⎜ − 5 + , 3 ⎟ = ⎜ − 16 ⎠ ⎝ 16 ⎠ ⎝ 5 −85 directrix x = −5 − , or x = 16 16
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)
536
64.
Chapter 8: Topics in Analytic Geometry
9 x 2 − 25 y 2 − 18 x + 50 y = 0
65.
) ( ) 2 9(x − 2 x + 1) − 25(y 2 − 2 y + 1) = 9 − 25 (
9 x 2 − 2 x − 25 y 2 − 2 y = 0
x 2 − 8 x = −2 y x 2 − 8 x + 16 = −2 y + 16
(x − 4)2 = −2( y − 8)
9(x − 1)2 − 25( y − 1)2 = −16
( y − 1)2 − (x − 1)2 16 25
=1
16 9
x 2 + 2 y − 8x = 0
parabola vertex (4, 8) 1 ⎞ ⎛ 15 ⎞ ⎛ foci ⎜ 4, 8 − ⎟ = ⎜ 4, ⎟ 2⎠ ⎝ 2 ⎠ ⎝
hyperbola center (1, 1) 4⎞ ⎛ 9⎞ ⎛ 1⎞ ⎛ vertices ⎜1, 1 ± ⎟ = ⎜1, ⎟ , ⎜1, ⎟ 5⎠ ⎝ 5⎠ ⎝ 5⎠ ⎝
directrix y = 8 +
17 1 , or y = 2 2
⎛ 4 34 ⎞⎟ 4 34 ⎞⎟ ⎛⎜ 4 34 ⎞⎟ ⎛⎜ = 1, 1 + , 1, 1 − foci ⎜1, 1 ± ⎜ ⎟ ⎜ ⎟ ⎜ 15 ⎟⎠ 15 ⎠ ⎝ 15 ⎠ ⎝ ⎝ 3 asymptotes y − 1 = ± (x − 1) 5
66.
9 x 2 + 16 y 2 + 36 x − 64 y − 44 = 0
) ( ) 2 9(x + 4 x + 4 ) + 16(y 2 − 4 y + 4) = 44 + 36 + 64 (
9 x 2 + 4 x + 16 y 2 − 4 y = 44
9(x + 2 )2 + 16( y − 2)2 = 144
(x + 2)2 + ( y − 2)2 16
9
67.
25 x 2 + 9 y 2 − 50 x − 72 y − 56 = 0
) ( ) 2 25 ( x − 2 x + 1) + 9 ( y − 8 y + 16 ) = 56 + 25 + 144 (
25 x 2 − 2 x + 9 y 2 − 8 y = 56
2
2
=1
( x −1)2 9
ellipse
ellipse
center (−2, 2 )
center (1, 4)
vertices (−2 ± 4, 2 ) = ( 2, 2 ) , (−6, 2 )
(
) (
2
25 ( x − 1) + 9 ( y − 4 ) = 225
)(
foci − 2 ± 7 , 2 = − 2 + 7 , 2 , − 2 − 7 , 2
+
( y − 4 )2 25
vertices (1, 4 ± 5) = (1, 9 ) , (1, − 1)
)
foci (1, 4 ± 4 ) = (1, 8) , (1, 0)
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=1
Section 8.3
68.
537
(x − 3)2 + ( y − 4)2 = (x + 1)2 x 2 − 6 x + 9 + y 2 − 8 y + 16 = x 2 + 2 x + 1 − 8 x + 8 = − y 2 + 8 y − 16 − 8(x − 1) = −( y − 4 )2 8(x − 1) = ( y − 4 )2 parabola vertex (1, 4) focus (1 + 2, 4 ) = (3, 4 ) directrix x = 1 − 2 , or x = −1
....................................................... 69.
Connecting Concepts
foci F1 (2, 0 ) , F2 (−2, 0) passing through P1 (2, 3) d (P1 , F2 ) − d (P1 , F1 ) =
(2 + 2)2 + 32
−
(2 − 2)2 + 3 2
=5−3=2
Let P(x, y) be any point on the hyperbola. Since the difference between F1 P and F2 P is the same as the difference between F1 P1 and F2 P1 , we have
(x − 2)2 + y 2
−
(x + 2)2 + y 2
=2
( x − 2)2 + y 2 = 2 + ( x + 2)2 + y 2 x 2 − 4 x + 4 + y 2 = 4 + 4 ( x + 2)2 + y 2 + x 2 + 4 x + 4 + y 2 −8 x − 4 = 4 ( x + 2)2 + y 2 −2 x − 1 = ( x + 2) 2 + y 2 4 x2 + 4 x + 1 = x2 + 4 x + 4 + y2 3x 2 − y 2 = 3 2 x2 − y = 1 1 3
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538
70.
Chapter 8: Topics in Analytic Geometry
foci ( 0, 3) and ( 0, − 3) , point ⎜⎛ 5 , 3 ⎞⎟ ⎝2 ⎠ Difference of distances from (x, y) to foci = difference of distances from ⎛⎜ 5 , 3 ⎞⎟ to foci ⎝2 ⎠
( x − 0 ) 2 + ( y − 3) 2 − ( x − 0 ) 2 + ( y + 3) 2
=
( 25 − 0)
2
+ ( 3 − 3) − 2
( 25 − 0)
2
+ ( 3 + 3)
2
x 2 + y 2 − 6 y + 9 − x 2 + y 2 + 6 y + 9 = 5 − 13 = − 4 2 2 x2 + y2 − 6 y + 9 = x2 + y2 + 6 y + 9 = − 4 x 2 + y 2 − 6 y + 9 = x 2 + y 2 + 6 y + 9 − 8 x 2 + y 2 + 6 y + 9 + 16 −12 y − 16 = −8 x 2 + y 2 + 6 y + 9 3 y + 4 = 2 x2 + y2 + 6 y + 9 9 y 2 + 24 y + 16 = 4 x 2 + 4 y 2 + 24 y + 36 5 y 2 − 4 x 2 = 20 y2 x2 − =1 4 5
71.
⎛7 ⎞ foci (0, 4 ) and (0, − 4) , point ⎜ , 4 ⎟ ⎝3 ⎠ ⎛7 ⎞ Difference in distances from (x, y) to foci = difference of distances from ⎜ , 4 ⎟ to foci ⎝3 ⎠ 2
2
⎛7 ⎞ ⎛7 ⎞ 2 2 = ⎜ − 0 ⎟ + ( 4 − 4) − ⎜ − 0 ⎟ + ( 4 + 4) ⎝3 ⎠ ⎝3 ⎠ 7 25 x 2 + y 2 − 8 y + 16 − x 2 + y 2 + 8 y + 16 = − =−6 3 3
( x − 0 )2 + ( y − 4 ) 2 − ( x − 0 )2 + ( y + 4 )2
x 2 + y 2 − 8 y + 16 =
x 2 + y 2 + 8 y + 16 − 6
x 2 + y 2 − 8 y + 16 = x 2 + y 2 + 8 y + 16 − 12 x 2 + y 2 + 8 y + 16 + 36 −16 y − 36 = −12 x 2 + y 2 + 8 y + 16 4 y + 9 = 3 x 2 + y 2 + 8 y + 16 16 y 2 + 72 y + 81 = 9 x 2 + 9 y 2 + 72 y + 144 7 y 2 − 9 x 2 = 63 y 2 x2 − =1 9 7
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Section 8.3
72.
539
⎛ 9⎞ foci (5, 0 ) and (− 5, 0 ) , point ⎜ 5, ⎟ ⎝ 4⎠ ⎛ 9⎞ Difference of distances from (x, y) to foci = difference of distances from ⎜ 5, ⎟ to foci ⎝ 4⎠
( x − 5 )2 + ( y − 0 )2 − ( x + 5 )2 + ( y − 0 )2
=
( 5 − 5)2 + ⎛⎜
2
9 ⎞ − 0⎟ − 4 ⎝ ⎠
( 5 + 5)2 + ⎛⎜
9 ⎞ − 0⎟ 4 ⎝ ⎠
2
9 41 = −8 x 2 − 10 x + 25 + y 2 − x 2 + 10 x + 25 + y 2 = − 4 4 x 2 − 10 x + 25 + y 2 =
x 2 + 10 x + 25 + y 2 − 8
x 2 + 10 x + 25 + y 2 = x 2 + 10 x + 25 + y 2 − 16 x 2 + 10 x + 25 + y 2 + 64 −20 x − 64 = −16 x 2 + 10 x + 25 + y 2 5 x + 16 = 4 x 2 + 10 x + 25 + y 2 25 x 2 + 160 x + 256 = 16 x 2 + 160 x + 400 + 16 y 2 9 x 2 − 16 y 2 = 144 x2 y 2 − =1 16 9 73.
74.
The hyperbola in a. has the larger eccentricity.
.......................................................
Prepare for Section 8.4
PS1. cos(α + β ) = cos α cos β − sin α sin β
PS2. sin(α + β ) = sin α cos β + cos α sin β
PS3. cot 2α = 3 3 tan 2α = 3 3
PS4. sin α = 1 , α = 30o or 150o 2 cos α = − 3 α = 150o or 210o 2 α = 150o
2α = tan −1 ⎛⎜ 3 ⎞⎟ = π ⎝ 3⎠ 3
α =π
6
PS6.
PS5. 4 x 2 − 6 y 2 + 9 x + 16 y − 8 = 0 A = 4, B = 0, C = −6
Since B 2 − 4 AC = 02 − 4 ( 4 )( −6 ) = 96 > 0, the graph is a hyperbola.
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540
Chapter 8: Topics in Analytic Geometry
Section 8.4 1.
xy = 3 A = 0, B = 1, C = 0
2.
A−C B 0−0 = 1 =0 = 90° = 45°
cot 2α = A − C B 5 − ( −5) cot 2α = −3 cot 2α = − 10 3 2α ≈ 163.3 α ≈ 81.7°
cot 2α = cot 2α cot 2α 2α
α 3.
9 x 2 − 24 xy + 16 y 2 − 320 x − 240 y = 0 A = 9, B = −24, C = 16
4.
cot 2α = A − C B cot 2α = 9 − 16 −24 − cot 2α = 7 −24 cot 2α = 7 24 2α ≈ 73.74° α ≈ 36.9° 5.
cot 2a = A − C B 5 − ( −11) cot 2a = −6 3 cot 2a = 5 + 11 −6 3 cot 2a = 16 −6 3 cot 2a = − 8 3 3 8 cot 2a = − 3 9 2α ≈ 147° α ≈ 73.5°
x 2 + 4 xy + 4 y 2 − 6 x − 5 = 0 A = 1, B = 4, C = 4
A−C B 1− 4 cot 2a = 4 3 cot 2a = − 4 2α ≈ 126.9° α ≈ 63.4° cot 2a =
5 x 2 − 6 3xy − 11y 2 + 4 x − 3 y + 2 = 0
A = 5, B = −6 3, C = −11
5 x 2 − 3 xy − 5 y 2 − 1 = 0 A = 5, B = −3, C = −5
6.
5 x 2 + 4 xy + 8 y 2 − 6 x + 3 y − 12 = 0 A = 5, B = 4, C = 8
A−C B 5−8 cot 2a = 4 3 cot 2a = − 4 2α ≈ 126.9° α ≈ 63.4° cot 2a =
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Section 8.4
7.
541
2 x 2 + xy + y 2 − 4 = 0 A = 2, B = 1, C = 1
8.
A = −2, B = 3, C = −3
A−C cot 2a = B 2 −1 cot 2a = 1 cot 2a = 1 2α ≈ 45° α ≈ 22.5° 9.
−2 x 2 + 3xy − 3 y 2 + 2 x + 6 y + 36 = 0 cot 2a = A − C B −2 − ( −3) cot 2a = 3 cot 2a = 3 3 2α ≈ 60.0° α ≈ 30.0°
xy = 4 xy − 4 = 0 A = 0, B = 1, C = 0, F = −4 cot 2α = A − C = 0 − 0 = 0 B 1
csc2 2α = cot 2 2α + 1 csc2 2α = 02 + 1 = 1 csc 2α = +1 (2α is in the first quadrant.) sin 2α =
1 1 = =1 csc 2α 1
sin 2 2α + cos2 2α = 1 cos2 2α = 1 − sin 2 2α 2 cos2 2α = 1 − (1)
cos2 2α = 0 cos 2α = 0 sin α =
1 − (0) = 2 2 2
cos α =
1 + (0) = 2 2 2
α = 45° 2
2
2
2
⎛ 2⎞ ⎛ 2 ⎞⎛ 2 ⎞ ⎛ 2⎞ 1 A ' = A cos2 α + B cos α sin α + C sin 2 α = 0 ⎜ ⎟ + 1⎜ ⎟⎜ ⎟ + 0⎜ ⎟ = 2 ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 2⎞ ⎛ 2 ⎞⎛ 2 ⎞ ⎛ 2⎞ 1 C ' = A sin 2 α − B cos α sin α + C cos2 α = 0 ⎜ ⎟ − 1⎜ ⎟⎜ ⎟ + 0⎜ ⎟ =− 2 ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ F ' = F = −4 2 1 x '2 − 1 y '2 − 4 = 0 or x '2 − y ' = 1 2 2 8 8
Copyright © Houghton Mifflin Company. All rights reserved.
542
10.
Chapter 8: Topics in Analytic Geometry
xy = −10 xy + 10 = 0 A = 0, B = 1, C = 0, F = 10
cot 2α =
A−C 0−0 = =0 B 1
csc2 2α = cot 2 2α + 1 csc2 2α = 02 + 1 = 1 csc 2α = +1 (2α is in the first quadrant.) sin 2α =
1 1 = =1 csc 2α 1
sin 2 2α + cos 2 2α = 1 cos 2 2α = 1 − sin 2 2α cos 2 2α = 1 − (1)2 cos 2 2α = 0 cos 2α = 0
sin α =
1 − (0) 2 = 2 2
cosα =
1 + (0) 2 = 2 2
α = 45° 2
2
2
2
⎛ 2⎞ ⎛ 2 ⎞⎛ 2 ⎞ ⎛ 2⎞ 1 A ' = A cos 2 α + B cos α sin α + C sin 2 α = 0 ⎜⎜ ⎟⎜ ⎟⎟ + 1⎜⎜ 2 ⎟⎜ ⎟⎟ + 0 ⎜⎜ 2 ⎟⎟ = 2 2 2 ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎛ 2⎞ ⎛ 2 ⎞⎛ 2 ⎞ ⎛ 2⎞ 1 C ' = A sin 2 α − B cos α sin α + C cos 2 α = 0 ⎜⎜ ⎟⎟ − 1⎜⎜ ⎟⎜ ⎟⎟ + 0 ⎜⎜ ⎟⎟ = − ⎟⎜ 2 ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ F ' = F = 10
1 2 1 2 ( y ')2 ( x ')2 x ' − y ' + 10 = 0 or − =1 2 2 20 20
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.4
11.
543
6 x 2 − 6 xy + 14 y 2 − 45 = 0 A = 6, B = −6, C = 14, F = −45
A − C 6 − 14 4 = = B −6 3
cot 2α =
csc2 2α = cot 2 2α + 1 2
25 ⎛4⎞ csc2 2α = ⎜ ⎟ + 1 = 3 9 ⎝ ⎠ 25 5 = 9 3
csc 2α = +
sin 2α =
( 2α
is in the first quadrant.)
1 3 = csc 2α 5
sin 2 α + cos 2 2α = 1 cos 2 2α = 1 − sin 2 2α 2
16 ⎛3⎞ cos 2 2α = 1 − ⎜ ⎟ = 25 ⎝5⎠ cos 2α = +
sin α =
(5) =
1− 4 2
16 4 = 25 5
10 10
( 2α is in the first quadrant.)
cos α =
( 5 ) = 3 10
1+ 4 2
10
α = 18.4° 2
2
⎛ 3 10 ⎞ ⎛ 3 10 ⎞⎛ 10 ⎞ ⎛ 10 ⎞ A ' = A cos 2 α + B cos α sin α + C sin 2 α = 6 ⎜⎜ ⎟⎟ − 6 ⎜⎜ ⎟⎜ ⎟⎟ + 14 ⎜⎜ ⎟⎟ = 5 ⎟⎜ ⎝ 10 ⎠ ⎝ 10 ⎠⎝ 10 ⎠ ⎝ 10 ⎠ 2
2
⎛ 10 ⎞ ⎛ 3 10 ⎞⎛ 10 ⎞ ⎛ 3 10 ⎞ C ' = A sin 2 α − B cos α sin α + C cos 2 α = 6 ⎜⎜ ⎟⎟ + 6 ⎜⎜ ⎟⎜ ⎟⎟ + 14 ⎜⎜ ⎟⎟ = 15 ⎟⎜ ⎝ 10 ⎠ ⎝ 10 ⎠⎝ 10 ⎠ ⎝ 10 ⎠ F ' = F = −45
5 x '2 + 15 y '2 − 45 = 0 or
( x ') 2 ( y ')2 + =1 9 3
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544
12.
Chapter 8: Topics in Analytic Geometry
11x 2 − 10 3 xy + y 2 − 20 = 0
A = 11, B = −10 3, C = 1, F = −20
cot 2α =
A − C 11 − 1 3 = = 3 B −10 3
csc2 2α = cot 2 2α + 1 2
⎛ 3⎞ 4 csc2 2α = ⎜⎜ − ⎟⎟ + 1 = 3 3 ⎝ ⎠ csc 2α = +
sin 2α =
4 2 3 = 3 3
( 2α is in second quadrant.)
1 3 3 = = csc 2α 2 3 2
sin 2 2α + cos2 2α = 1 cos2 2α = 1 − sin 2 2α 2
⎛ 3⎞ 1 cos2 2α = 1 − ⎜⎜ ⎟⎟ = 2 4 ⎝ ⎠ cos 2α = −
sin α =
( 2) =
1− − 1 2
1 1 =− 4 2
3 2
( 2α is in second quadrant.)
cos α =
( 2) = 1
1+ − 1 2
2
α = 60° 2
2 ⎛1⎞ ⎛ 1 ⎞⎛ 3 ⎞ ⎛ 3 ⎞ A ' = A cos 2 α + B cos α sin α + C sin 2 α = 11⎜ ⎟ − 10 3 ⎜ ⎟ ⎜⎜ ⎟ + 1⎜ ⎟ = −4 ⎝2⎠ ⎝ 2 ⎠ ⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ 2
2 ⎛ 3⎞ ⎛ 1 ⎞⎛ 3 ⎞ ⎛ 1 ⎞ C ' = A sin 2 α − B cos α sin α + C cos 2 α = 11⎜⎜ ⎟⎟ + 10 3 ⎜ ⎟ ⎜⎜ ⎟ + 1⎜ ⎟ = 16 ⎟ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ F ' = F = −20
−4 x '2 + 16 y '2 − 20 = 0 or
4( y ')2 ( x ')2 − =1 5 5
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.4
13.
545
x 2 − 4 xy + 2 y 2 − 1 = 0 A = 1, B = 4, C = −2, F = −1 A − C 1 − ( −2) 3 = = 4 4 B
cot 2α =
csc2 2α = cot 2 2α + 1 2
25 ⎛3⎞ csc2 2α = ⎜ ⎟ + 1 = 16 ⎝4⎠ 25 5 = 16 4
csc 2α = +
sin 2α =
(2α is in the first quadrant.)
1 4 = csc 2α 5
sin 2 α + cos2 2α = 1 cos2 2α = 1 − sin 2 α 2
9 ⎛4⎞ cos2 2α = 1 − ⎜ ⎟ = 25 ⎝5⎠ cos 2α = +
sin α =
(5) =
1− 3 2
5 5
9 3 = 25 5
( 2α
is in first quadrant. )
cosα =
(5) = 2
1+ 3 2
5 5
α ≈ 26.6° 2
2
⎛2 5⎞ ⎛ 2 5 ⎞⎛ 5 ⎞ ⎛ 5⎞ A ' = A cos 2 α + B cos α sin α + C sin 2 α = 1⎜⎜ ⎟⎟ + 4 ⎜⎜ ⎟⎜ ⎟⎜ 5 ⎟⎟ − 2 ⎜⎜ 5 ⎟⎟ = 2 5 5 ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ 2
2
⎛ 5⎞ ⎛ 2 5 ⎞⎛ 5 ⎞ ⎛2 5⎞ C ' = A sin 2 α − B cos α sin α + C cos 2 α = 1⎜⎜ ⎟⎟ − 4 ⎜⎜ ⎟⎜ ⎟⎟ − 2 ⎜⎜ ⎟⎟ = −3 ⎟⎜ ⎝ 5 ⎠ ⎝ 5 ⎠⎝ 5 ⎠ ⎝ 5 ⎠ F ' = F = −1 2( x ') 2 + 3( y ')2 = 1
Copyright © Houghton Mifflin Company. All rights reserved.
546
14.
Chapter 8: Topics in Analytic Geometry
9 x 2 − 24 xy + 16 y 2 + 100 = 0 A = 9, B = −24, C = 16, F = 100 A − C 9 − 16 7 = = −24 24 B
cot 2α =
csc2 2α = cot 2 2α + 1 2
625 ⎛ 7 ⎞ csc2 2α = ⎜ ⎟ + 1 = 576 ⎝ 24 ⎠ csc 2α = +
sin 2α =
625 25 = 576 24
( 2α is in the first quadrant.)
1 24 = csc 2α 25
sin 2 2α + cos 2 2α = 1 cos 2 2α = 1 − sin 2 2α 2
49 ⎛ 24 ⎞ cos 2 2α = 1 − ⎜ ⎟ = 625 ⎝ 25 ⎠ cos 2α = +
sin α =
( 25 ) = 3
1− 7 2
5
49 7 = 625 25
(2α is in the first quadrant.)
cos α =
( 25 ) = 4
1+ 7 2
5
α ≈ 36.9° 2
2
2
2
⎛4⎞ ⎛ 3 ⎞⎛ 4 ⎞ ⎛ 3⎞ A ' = A cos2 α + B cos α sin α + C sin 2 α = 9 ⎜ ⎟ − 24 ⎜ ⎟⎜ ⎟ + 16 ⎜ ⎟ = 0 ⎝5⎠ ⎝ 5 ⎠⎝ 5 ⎠ ⎝5⎠ ⎛3⎞ ⎛ 3 ⎞⎛ 4 ⎞ ⎛ 4⎞ C ' = A sin 2 α − B cos α sin α + C cos 2 α = 9 ⎜ ⎟ + 24 ⎜ ⎟⎜ ⎟ + 16 ⎜ ⎟ = 25 5 5 5 ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝5⎠ F ' = F = 100 25( y ')2 + 100 = 0 ( y ')2 = −4
This equation has no real solutions, so there is no graph.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.4
15.
547
3x 2 + 2 3 xy + y 2 + 2 x − 2 3 y + 16 = 0
A = 3, B = 2 3, C = 1, D = 2, E = −2 3, F = 16
cot 2α =
A − C 3 −1 3 = = 3 B 2 3
csc2 2α = cot 2 2α + 1 2
⎛ 3⎞ 4 csc2 2α = ⎜⎜ ⎟⎟ + 1 = 3 3 ⎝ ⎠ 4 2 3 = 3 3
csc 2α = +
sin 2α =
(2α is in the first quadrant.)
1 3 = csc 2α 2
sin 2 α + cos 2 2α = 1 cos 2 2α = 1 − sin 2 2α 2
⎛ 3⎞ 1 cos 2 2α = 1 − ⎜⎜ ⎟⎟ = 2 4 ⎝ ⎠ cos 2α = +
sin α =
1 1 = 4 2
(2α is in the first quadrant.)
(2) = 1
1− 1 2
cosα =
2
(2) =
1+ 1 2
3 2
α = 30° 2
⎛ 3⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞2 A ' = A cos 2 α + B cos α sin α + C sin 2 α = 3 ⎜⎜ ⎟ + 2 3 ⎜⎜ ⎟ ⎟⎟ ⎜ ⎟ + 1⎜ ⎟ = 4 ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ 2
2 ⎛ 3 ⎞⎛ 1 ⎞ ⎛ 3 ⎞ ⎛1⎞ C ' = A sin 2 α − B cos α sin α + C cos 2 α = 3 ⎜ ⎟ − 2 3 ⎜⎜ ⎟⎟ ⎜ ⎟ + 1⎜⎜ ⎟⎟ = 0 ⎝2⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 3⎞ ⎛1⎞ D ' = D cosα + E sin α = 2 ⎜⎜ ⎟⎟ − 2 3 ⎜ ⎟ = 0 2 2⎠ ⎝ ⎝ ⎠ ⎛ 3⎞ ⎛1⎞ E ' = − D sin α + E cosα = −2 ⎜ ⎟ − 2 3 ⎜⎜ ⎟⎟ = −4 ⎝2⎠ ⎝ 2 ⎠ F ' = F = 16
4( x ')2 − 4 y '+ 16 = 0 or y ' = ( x ') 2 + 4
Copyright © Houghton Mifflin Company. All rights reserved.
548
16.
Chapter 8: Topics in Analytic Geometry
x 2 + 2 xy + y 2 + 2 2 x − 2 2 y = 0
A = 1, B = 2, C = 1, D = 8, E = − 8 cot 2α =
A − C 1−1 = =0 B 2
csc2 2α = cot 2 2α + 1 csc2 2α = 02 + 1 = 1 csc 2α = 1 sin 2α =
1 1 = =1 csc 2α 1
sin 2 α + cos 2 2α = 1 cos 2 2α = 1 − sin 2 2α cos 2 2α = 1 − (1)2 = 0 cos 2α = 0
sin α =
1 − (0) 2 = 2 2
cosα =
1 + (0) 2 = 2 2
α = 45° 2
2
2
2
⎛ 2⎞ ⎛ 2 ⎞⎛ 2 ⎞ ⎛ 2 ⎞ A ' = A cos 2 α + B cos α sin α + C sin 2 α = 1⎜⎜ ⎟⎟ + 2 ⎜⎜ ⎟⎜ ⎟⎟ + 1⎜⎜ ⎟⎟ = 2 ⎟⎜ 2 ⎝ ⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 2⎞ ⎛ 2 ⎞⎛ 2 ⎞ ⎛ 2 ⎞ C ' = A sin 2 α − B cos α sin α + C cos 2 α = 1⎜⎜ ⎟⎟ − 2 ⎜⎜ ⎟⎜ ⎟⎟ + 1⎜⎜ ⎟⎟ = 0 ⎟⎜ ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 2⎞ ⎛ 2⎞ D ' = D cosα + E sin α = 8 ⎜⎜ ⎟⎟ − 8 ⎜⎜ ⎟⎟ = 0 2 ⎝ ⎠ ⎝ 2 ⎠ ⎛ 2⎞ ⎛ 2⎞ E ' = − D sin α + E cosα = − 8 ⎜⎜ ⎟⎟ − 8 ⎜⎜ ⎟⎟ = −4 2 ⎝ ⎠ ⎝ 2 ⎠ 1 2( x ') 2 − 4 y ' = 0 or y ' = ( x ') 2 2
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.4
17.
549
9 x 2 + 24 xy + 16 y 2 − 40 x − 30 y + 100 = 0 A = 9, B = −24, C = 16, D = −40, E = −30 F = 100 A − C 9 − 16 7 = = −24 24 B
cot 2α =
csc2 2α = cot 2 2α + 1 2
625 ⎛ 7 ⎞ csc2 2α = ⎜ ⎟ + 1 = 576 ⎝ 24 ⎠ csc 2α = +
sin 2α =
625 25 = 576 24
(2α is in the first quadrant.)
1 24 = csc 2α 25
sin 2 α + cos2 2α = 1 cos 2 2α = 1 − sin 2 2α 2
49 ⎛ 24 ⎞ cos 2 2α = 1 − ⎜ ⎟ = 625 ⎝ 25 ⎠ cos 2α = +
sin α =
( 25 ) = 3
1− 7 2
5
49 7 = 625 25
cosα =
(2α is in first quadrant.)
( 25 ) = 4
1+ 7 2
5
α ≈ 36.9° 2
2
2
2
⎛4⎞ ⎛ 4 ⎞⎛ 3 ⎞ ⎛ 3⎞ A ' = A cos 2 α + B cos α sin α + C sin 2 α = 9 ⎜ ⎟ − 24 ⎜ ⎟⎜ ⎟ + 16 ⎜ ⎟ = 0 ⎝5⎠ ⎝ 5 ⎠⎝ 5 ⎠ ⎝5⎠ ⎛3⎞ ⎛ 4 ⎞⎛ 3 ⎞ ⎛ 4⎞ C ' = A sin 2 α − B cos α sin α + C cos 2 α = 9 ⎜ ⎟ + 24 ⎜ ⎟⎜ ⎟ + 16 ⎜ ⎟ = 25 ⎝5⎠ ⎝ 5 ⎠⎝ 5 ⎠ ⎝5⎠ ⎛4⎞ ⎛3⎞ D ' = D cosα + E sin α = −40 ⎜ ⎟ − 30 ⎜ ⎟ = −50 ⎝5⎠ ⎝5⎠ 3 ⎛ ⎞ ⎛ 4⎞ E ' = − D sin α + E cosα = 40 ⎜ ⎟ − 30 ⎜ ⎟ = 0 ⎝5⎠ ⎝5⎠ F ' = F = 100 25( x ') 2 − 50 x '+ 100 = 0 or (y ') 2 = 2 ( x − 2 )
9
Copyright © Houghton Mifflin Company. All rights reserved.
550
18.
Chapter 8: Topics in Analytic Geometry
24 x 2 + 16 3 xy + 8 y 2 − x + 3 y − 8 = 0
A = 24, B = 16 3, C = 8, D = −1, E = 3, F = −8 cot 2α =
3 A − C 24 − 8 = = 3 B 16 3
csc2 2α = cot 2 2α + 1 2
⎛ 3⎞ 4 csc2 2α = ⎜⎜ ⎟⎟ + 1 = 3 3 ⎝ ⎠ 4 2 3 = 3 3
csc 2α = + sin 2α =
(2α is in the first quadrant.)
1 3 = csc 2α 2
sin 2 2α + cos 2 2α = 1 cos 2 2α = 1 − sin 2 2α 2
⎛ 3⎞ 1 cos 2 2α = 1 − ⎜⎜ ⎟⎟ = 2 4 ⎝ ⎠ cos 2α = +
sin α =
(2) = 1
1− 1 2
2
1 1 = 4 2
(2α is in the first quadrant.)
cosα =
(2) =
1+ 1 2
3 2
α = 30° 2
⎛ 3⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞2 A ' = A cos 2 α + B cos α sin α + C sin 2 α = 24 ⎜⎜ ⎟⎟ + 16 3 ⎜⎜ ⎟⎟ ⎜ ⎟ + 8 ⎜ ⎟ = 32 ⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ 2
2 ⎛ 3 ⎞⎛ 1 ⎞ ⎛ 3 ⎞ ⎛1⎞ C ' = A sin 2 α − B cos α sin α + C cos 2 α = 24 ⎜ ⎟ − 16 3 ⎜⎜ ⎟⎟ ⎜ ⎟ + 8 ⎜⎜ ⎟⎟ = 0 ⎝2⎠ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 3⎞ ⎛1⎞ D ' = D cosα + E sin α = −1⎜⎜ ⎟⎟ + 3 ⎜ ⎟ = 0 2 ⎝2⎠ ⎝ ⎠ ⎛ 3⎞ ⎛1⎞ E ' = − D sin α + E cosα = 1⎜ ⎟ + 3 ⎜⎜ ⎟⎟ = 2 ⎝2⎠ ⎝ 2 ⎠ F ' = F = −8
32( x ')2 + 2 y '− 8 = 0 or y ' = −16( x ')2 + 4
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.4
19.
551
6 x 2 + 24 xy − y 2 − 12 x + 26 y + 11 = 0 A = 6, B = 24, C = −1, D = −12, E = 26 F = 11
A − C 6 − (−1) 7 = = B 24 24
cot 2α =
csc2 2α = cot 2 2α + 1 2
625 ⎛ 7 ⎞ csc2 2α = ⎜ ⎟ + 1 = 576 ⎝ 24 ⎠ 625 25 = 576 24 1 24 sin 2α = = csc 2α 25
csc 2α = +
(2α is in the first quadrant.)
sin 2 2α + cos 2 2α = 1 cos 2 2α = 1 − sin 2 2α 2
49 ⎛ 24 ⎞ cos 2 2α = 1 − ⎜ ⎟ = 25 625 ⎝ ⎠ cos 2α = +
sin α =
( 25 ) = 3
1− 7 2
5
49 7 = 625 25
(2α is in the first quadrant.)
cosα =
( 25 ) = 4
1+ 7 2
5
α ≈ 36.9° 2
2
2
2
⎛4⎞ ⎛ 4 ⎞⎛ 3 ⎞ ⎛ 3 ⎞ A ' = A cos 2 α + B cos α sin α + C sin 2 α = 6 ⎜ ⎟ + 24 ⎜ ⎟⎜ ⎟ − 1⎜ ⎟ = 15 5 ⎝ ⎠ ⎝ 5 ⎠⎝ 5 ⎠ ⎝ 5 ⎠ ⎛3⎞ ⎛ 4 ⎞⎛ 3 ⎞ ⎛ 4 ⎞ C ' = A sin 2 α − B cos α sin α + C cos 2 α = 6 ⎜ ⎟ − 24 ⎜ ⎟⎜ ⎟ − 1⎜ ⎟ = −10 ⎝5⎠ ⎝ 5 ⎠⎝ 5 ⎠ ⎝ 5 ⎠ 4 3 ⎛ ⎞ ⎛ ⎞ D ' = D cosα + E sin α = −12 ⎜ ⎟ + 26 ⎜ ⎟ = 6 ⎝5⎠ ⎝5⎠ ⎛3⎞ ⎛4⎞ E ' = − D sin α + E cosα = 12 ⎜ ⎟ + 26 ⎜ ⎟ = 28 ⎝5⎠ ⎝5⎠ F ' = F = 11 15( x ')2 − 10( y ')2 + 6 x '+ 28 y '+ 11 = 0
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552
20.
Chapter 8: Topics in Analytic Geometry
x 2 + 4 xy + 4 y 2 − 2 5 x + 5 y = 0
A = 1, B = 4, C = 4, D = −2 5, E = 5, F = 0 cot 2α =
A − C 1− 4 3 = =− B 4 4
csc2 2α = cot 2 2α + 1 2
25 ⎛ 3⎞ csc2 2α = ⎜ − ⎟ + 1 = 16 ⎝ 4⎠ 25 5 = 16 4 1 4 sin 2α = = csc 2α 5
csc 2α = +
(2α is in the second quadrant.)
sin 2 2α + cos 2 2α = 1 cos 2 2α = 1 − sin 2 2α 2
9 ⎛4⎞ cos 2 2α = 1 − ⎜ ⎟ = 5 25 ⎝ ⎠ 9 3 =− 25 5
cos 2α = −
sin α =
( 5) = 2
1− − 3 2
5 5
(2α is in second quadrant.)
cosα =
( 5) =
1+ − 3 2
5 5
α ≈ 63.43° 2
2
⎛ 5⎞ ⎛ 5 ⎞⎛ 2 5 ⎞ ⎛2 5⎞ A ' = A cos α + B cos α sin α + C sin α = 1⎜⎜ ⎟⎟ + 4 ⎜⎜ ⎟⎜ ⎟⎜ 5 ⎟⎟ + 4 ⎜⎜ 5 ⎟⎟ = 5 5 5 ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ 2
2
2
2
⎛2 5⎞ ⎛ 5 ⎞⎛ 2 5 ⎞ ⎛ 5⎞ C ' = A sin 2 α − B cos α sin α + C cos 2 α = 1⎜⎜ ⎟⎟ − 4 ⎜⎜ ⎟⎜ ⎟⎟ + 4 ⎜⎜ ⎟⎟ = 0 ⎟⎜ ⎝ 5 ⎠ ⎝ 5 ⎠⎝ 5 ⎠ ⎝ 5 ⎠ ⎛ 5⎞ ⎛2 5⎞ D ' = D cosα + E sin α = −2 5 ⎜⎜ ⎟⎟ + 5 ⎜⎜ ⎟⎟ = 0 5 ⎝ ⎠ ⎝ 5 ⎠ ⎛2 5⎞ ⎛ 5⎞ E ' = − D sin α + E cosα = 2 5 ⎜⎜ ⎟⎟ − 5 ⎜⎜ 5 ⎟⎟ = 5 5 ⎝ ⎠ ⎝ ⎠ 5( x ')2 + 5 y ' = 0 or y ' = −(x') 2
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.4
21.
553
A = 6, B = −1, C = 2, D = 4, E = −12, F = 7
22.
−( − x − 12) ± (− x − 12)2 − 8(6 x 2 + 4 x + 7) 4 The graph will appear disconnected at the endpoints of the minor axes on a graphing utility. Graph y =
23.
A = 1, B = −6, C = 1, D = −2, E = −5, F = 4
Graph y =
25.
Graph y =
Graph y =
24.
−(−6 x − 5) ± (−6 x − 5)2 − 4( x 2 − 2 x + 4) 2
A = 3, B = −6, C = 3, D = 10, E = −8, F = −2
−( −6 x − 8) ± (−6 x − 8) 2 − 12(3 x 2 + 10 x − 2) 6
A = 5, B = −2, C = 10, D = −6, E = −9, F = −20
A = 2, B = −10, C = 3, D = −1, E = −8, F = −7
Graph y =
26.
−( −2 x − 9) ± ( −2 x − 9)2 − 40(5 x 2 − 6 x − 20) 20
−(−10 x − 8) ± ( −10 x − 8) 2 − 12(2 x 2 − x − 7) 6
A = 2, B = −8, C = 8, D = 20, E = −24, F = −3
Graph y =
−( −8 x − 24) ± (−8 x − 24)2 − 32(2 x 2 + 20 x − 3) 16
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554
27.
Chapter 8: Topics in Analytic Geometry
2 x '2 3 y '2 − =1 1 1
sin α =
5 5
cosα =
2 5 5
a2 = 1 ; a = 2 2 2 2 1 b = ; b= 3 3 3
b 6 Asymptotes: y ' = ± x ' or y ' = ± x' a 3 Using the transformation formulas for x ' and y ' yields 6 ( x cos α + y sin α ) 3 2 5 5 6⎛2 5 5 ⎞ y− x=± x+ y⎟ ⎜ 5 5 3 ⎜⎝ 5 5 ⎟⎠ ⎛ 2 30 2 5 5 30 ⎞ y− x = ± ⎜⎜ y ⎟⎟ + 5 5 15 15 ⎝ ⎠
y cos α − x sin α = ±
Multiplying both sides of the equation by 15 / 5 yields
(
6y − 3 x = ± 2 6 x + 6 y
6y − 3 x = 2 6 x + 6 y 6y − 6 y = 3x + 2 6 x y = 3+ 2 6 x 6− 6
) 6y − 3 x = −(2 6 x + 6 y ) 6y + 6 y = 3 x − 2 6 x y = 3−2 6 x 6+ 6
and
Rationalizing the denominators, we obtain y= 28.
2+ 6 x 2
and
y=
2− 6 x 2
( 2)
From Exercise 16, y ' = 1 x ' 2 . Relative to x ' y '-coordinates, the focus is 0, 1 and α = 45° and the directrix is y ' = − 1 . 2
Using the transformation equation x = x ' cos α − y ' sin α y = y 'cosα + x 'sin α we have 1 2 x = 0 ⋅ cos 45° − sin 45° = − 2 4 1 2 y = cos 45° + 0 ⋅ sin 45° = 2 4 The coordinates of the focus in the xy-coordinate system are ⎛⎜ − 2 , 2 ⎞⎟ . 4 ⎠ ⎝ 4 The equation of the directrix is given by y' = − 1 2 y cos 45° − x sin 45° = − 1 2 y 2 − x 2 = −1 2 2 2 2 y − 2 x = −1 or 2 x − 2 y = 1. The equation of the directrix is 2 x − 2 y = 1
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2
Section 8.4
29.
555
From Exercise 11,
x '2 y '2 10 3 10 2 + = 1, sin α = , cos α = , a = 9, b2 = 3, c 2 = 9 − 3 = 6. 9 3 10 10
Thus c = 6.
Foci in x ' y '-coordinates are ( ± 6, 0). Thus x ' = ± 6, y ' = 0. x = x 'cos α − y 'sin α
y = y 'cosα + x 'sin α
⎛ 3 10 ⎞ 10 = ± 6 ⎜⎜ ⎟⎟ − 0 ⋅ 10 ⎝ 10 ⎠ =±
⎛ 10 ⎞ = 0 ⋅ cosα ± 6 ⎜⎜ ⎟⎟ ⎝ 10 ⎠
3 15 5
Foci in the xy -coordinate system are ⎛⎜ 3 15 , ⎝ 5
30.
15 5
=±
15 ⎞ and ⎛ − 3 15 , − 15 ⎞ . ⎜ 5 ⎟⎠ 5 5 ⎟⎠ ⎝
xy = 4 xy − 4 = 0 A = 0, B = 1, C = 0
31.
Since B 2 − 4 AC = 12 − 4 (1)( −1) = 5 > 0, the graph is a hyperbola.
Since B 2 − 4 AC = 12 − 4(0)(0) = 1 > 0, the graph is a hyperbola. 32.
11x 2 − 10 3 xy + y 2 − 20 = 0
33.
A = 11, B = −10 3, C = 1 2
Since B 2 − 4 AC = (2 3) 2 − 4(3)(1) = 0, the graph is a parabola.
Since B − 4 AC = (−10 3) − 4(11)(1) = 256 > 0, the graph is a hyperbola. 9 x 2 − 24 xy + 16 y 2 + 8 x − 12 y − 20 = 0 A = 9, B = −24, C = 16
35.
Since B 2 − 4 AC = (−24) 2 − 4(9)(16) = 0, the graph is a parabola. 36.
5 x 2 − 4 xy + 8 y 2 − 6 x + 3 y − 12 = 0 A = 5, B = 4, C = 8
6 x 2 − 6 xy + 14 y 2 − 14 x + 12 y − 60 = 0 A = 6, B = −6, C = 14 2
2
Since B − 4 AC = ( −6) − 4(6)(14) = −300 < 0, the graph is an ellipse.
40.
4 x 2 − 4 xy + y 2 − 12 y − 20 = 0 A = 4, B = −4, C = 1
Since B 2 − 4 AC = (−4) 2 − 4(4)(1) = 0, the graph is a parabola.
37.
5 x 2 − 6 3xy − 11y 2 + 4 x − 3 y + 2 = 0
A = 5, B = −6 3, C = −11
Since B 2 − 4 AC = (4) 2 − 4(5)(8) = −144 < 0, the graph is an ellipse. 38.
3x 2 + 2 3xy + y 2 − 3x + 2 y + 20 = 0
A = 3, B = 2 3, C = 1
2
34.
x 2 + xy − y 2 − 40 = 0 A = 1, B = 1, C = −1
Since B 2 − 4 AC = (−6 3) 2 − 4(5)(−11) = 328 > 0, the graph is a hyperbola.
39.
6 x 2 + 2 3xy + 5 y 2 − 3x + 2 y − 20 = 0 A = 6, B = 2 3, C = 5 Since B 2 − 4 AC = (2 3)2 − 4(6)(5) = −108 < 0, the graph is an ellipse.
5 x 2 − 2 3xy + 3 y 2 − x + y − 12 = 0 A = 5, B = −2 3, C = 3 Since B 2 − 4 AC = ( −2 3 ) − 4(5)(3) = −48 < 0, the graph is an ellipse. 2
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556
Chapter 8: Topics in Analytic Geometry
....................................................... 41.
Connecting Concepts
x2 + y 2 = r 2 Substitute x = x ' cos α − y ' sin α and y = y 'cos α + x 'sin α .
( x 'cos α − y 'sin α ) 2 + ( y 'cos α + x 'sin α ) 2 = r 2 2
2
2
2
x ' cos α − 2 x ' y 'cos α sin α + y ' sin α + y 'cos 2 α + 2 x ' y 'cosα sin α + x '2 sin 2 α = r 2 x '2 (cos 2 α + sin 2 α ) + x ' y '(2cos α sin α − 2cos α sin α ) + y '2 (sin 2 α + cos 2 α ) = r 2 x '2 (1) + x ' y '(0) + y '2 (1) = r 2 x '2 + y '2 = r 2 42.
Because the vertices are (1, 1) and ( − 1, − 1), the transverse axis is on the line y = x. Consider an x ' y '-coordinate system rotated 45° (tan α = 1 implies α = 45°) from an xy-coordinate system. The equation of the hyperbola is x' 2 y ' 2 − =1 2 2
(1)
This follows from the fact that (1, 1) in xy -coordinates is Therefore, a = 2. Also,
(
)
(
)
2,0 in x ' y '-coordinates.
2 in x ' y '-coordinates is ( 2,0 ) in x ' y '-coordinates.
2,
Therefore, c = 2. Since a = 2 and c = 2, then b = 2. Now let x ' =
2 2 2 2 x+ y and y ' = y− x and substitute into Equation (1): 2 2 2 2 2
2
⎛ 2 x + 2 y⎞ ⎛ 2 y − 2 x⎞ ⎜ 2 ⎜ 2 2 ⎟⎠ 2 ⎟⎠ ⎝ −⎝ =1 2 2 1 x 2 + xy + 1 y 2 1 x 2 − xy + 1 y 2 ) ) =1 ( ( 2 2 2 2 − Simplifying, we have
2
2
or xy = 1.
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Section 8.4
43.
557
Vertices (2, 4) and ( − 2, − 4) imply that the major axis is on the line y = 2 x. Consider an x ' y '-coordinate system rotated an angle
α,
where tan α = 2. From this equation, using identities, cos α =
5 2 5 and sin α = . 5 5
The equation of the ellipse in the x ' y '-coordinate system is x '2 y '2 + =1 20 10
(1)
This follows from the fact that ( 2, 4 ) in xy -coordinates is Therefore, α = 20. Also,
(
)
(
2, 2 2 in xy -coordinates is
)
20, 0 in x ' y '-coordinates.
(
)
10, 0 in x ' y '-coordinates.
Therefore, c = 10. Thus b = 10.
Now let x ' =
5 2 5 5 2 5 x+ y and y ' = y− x and substitute into Equation (1) : 5 5 5 5 2
2
⎛ 5 x + 2 5 y⎞ ⎛ 5 y − 2 5 x⎞ ⎜ 5 ⎟ ⎜ ⎟ 5 5 ⎝ ⎠ +⎝ 5 ⎠ =1 20 10 Simplifying, we have
( 15 x2 + 54 xy + 54 y2 ) + ( 54 x2 − 54 xy + 15 y2 ) = 1 20
10
9 x 2 − 4 xy + 6 y 2 5 5 5 =1
20
9 x 2 − 4 xy + 6 y 2 =1 100 or 9 x 2 − 4 xy + 6 y 2 = 100
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558
44.
Chapter 8: Topics in Analytic Geometry
Vertex:
( 0, 0)
Focus: (1, 3)
Since both the vertex and the focus lie on the axis of symmetry, we can find the equation of the line by using these two points. We obtain y = 3x The slope of the line m is equal to 3. But θ , the angle the line makes with the x-axis, is equal to the arctangent of m. In other words, tan θ = m = 3. But rotating the axes upward through θ (α = θ ) , we can place the vertex and focus on the x '-axis. Since tan α = 3, sin α =
3 10 10 and cosα = . 10 10
Vertex: ( 0, 0 ) Since the origins for both the xy - and x ' y '-systems are coincident, the x ' y ' vertex is ( 0, 0 ) . Focus: (1, 3) x = 1, y = 3 x ' = x cosα + y sin α y ' = y cos α − x sin α ⎛ 10 ⎞ ⎛ 3 10 ⎞ x ' = 1⎜⎜ ⎟⎟ + 3 ⎜⎜ ⎟⎟ = 10 ⎝ 10 ⎠ ⎝ 10 ⎠ ⎛ 10 ⎞ ⎛ 3 10 ⎞ y ' = 3 ⎜⎜ ⎟⎟ − 1⎜⎜ ⎟⎟ = 0 ⎝ 10 ⎠ ⎝ 10 ⎠ x' y'
Focus: ( 10,0)
Since the vertex is at the origin at the focus at ( 10, 0), p = 10. Therefore, y '2 = 4 10 x '. Substituting x' = x cos α + y sin α and y ' = y cosα − x sin α , we have ( y cosα − x sin α ) 2 = 4 10 ( x cosα + y sin α ) Substituting sinα =
3 10 10 and cos α = yields 10 10 2
⎛ 10 ⎛ 10 3 10 ⎞ 3 10 y− x ⎟⎟ = 4 10 ⎜⎜ x+ ⎜⎜ 10 10 10 10 ⎝ ⎠ ⎝ 1 2 3 9 y − xy + x 2 = 4 x + 12 y 10 5 10
⎞ y ⎟⎟ ⎠
9 x 2 − 6 xy + y 2 − 40 x − 120 y = 0 45.
A '+ C ' = A cos 2 α + B cosα sin α + C sin 2 α + A sin 2 α − B cosα sin α + C cos 2 α = A(cos 2 α + sin 2 α ) + B (cosα sin α − cosα sin α ) + C (sin 2 α + cos 2 α ) = A+C
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Section 8.4
46.
559
Begin by using double-angle formula to rewrite A ', B ', and C '. Recall cos2α = sin 2 a =
1 + 2cos 2α , 2
1 − 2cos 2α , and sin2α = 2cosα sin α . 2
A ' = A cos 2 α + B cosα sin α + C sin 2 α ⎛ 1 + cos 2α ⎞ B ⎛ 1 − cos 2α ⎞ = A⎜ ⎟ + sin 2α + C ⎜ ⎟ 2 2 ⎝ ⎠ 2 ⎝ ⎠ 1 = [( A + C ) + B sin 2α + ( A − C ) cos 2α ] 2 B ' = B(cos 2 α − sin 2 α ) + (C − A)2sin α cosα = B cos 2α − ( A − C )sin 2α C ' = A sin 2 α − B cosα sin α + C cos 2 α ⎛ 1 − cos 2α ⎞ B ⎛ 1 + cos 2α ⎞ = A⎜ ⎟ ⎟ − sin 2α + C ⎜ 2 2 2 ⎝ ⎠ ⎝ ⎠ 1 = [( A + C ) − B sin 2α − ( A − C )cos 2α ] 2 Using these expressions, we have ⎛1 ⎞ B '2 − 4 A ' C ' = [ B cos 2α − ( A − C )sin 2α ]2 − 4 ⎜ [( A + C ) + B sin 2α + ( A − C ) cos 2α ] ⎟ ⎝2 ⎠ ⎛1 ⎞ × ⎜ [( A + C ) − B sin 2α − ( A − C ) cos 2α ] ⎟ ⎝2 ⎠ = [ B 2 cos 2 2α − 2 B( A − C ) cos 2α sin 2α + ( A − C )2 sin 2 2α ] − [( A + C )2 − B 2 sin 2 α − ( A − C )2 cos 2 2α − 2 B( A − C ) cos 2α sin 2α ] = B 2 cos 2 2α − 2 B( A − C ) cos 2α sin 2α + ( A − C )2 sin 2 2α − ( A + C )2 + B 2 sin 2 2α + ( A − C )2 cos2 2α + 2 B ( A − C ) cos 2α sin 2α = B 2 (cos 2 2α + sin 2 2α ) + ( A − C )2 (sin 2 2α + cos 2 2α ) − ( A + C ) 2 2
= B2 + ( A − C ) − ( A + C )
2
= B 2 + A2 − 2 AC + C 2 − A2 − 2 AC − C 2 = B 2 − 4 AC
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560
47.
Chapter 8: Topics in Analytic Geometry
Ellipse with major axis parallel to x -axis: ( x − h) 2 a
2
+
( y − k )2 b2
=1
b 2 ( x − h ) 2 + a 2 ( y − k ) 2 = a 2b 2 b ( x − 2hx + h 2 ) + a 2 ( y 2 − 2ky + k 2 ) = a 2b 2 2 2 b x − 2b 2 hx + b 2 h 2 + a 2 y 2 − 2a 2 ky + a 2 k 2 = a 2b 2 2 2 b x + a 2 y 2 − 2b 2 hx − 2a 2 ky + b 2 h 2 + a 2 k 2 − a 2b 2 = 0 2
2
A = b 2 , B = 0, C = a 2 B 2 − 4 AC = 02 − 4b 2 a 2 = 4a 2b 2 < 0 B 2 − 4 AC < 0 for an ellipse whose major axis is parallel to the x-axis. Ellipse with major axis parallel to y -axis: ( y − k ) 2 ( x − h) 2 + =1 a2 b2 b 2 ( y − k ) 2 + a 2 ( x − h ) 2 = a 2b 2 b ( y − 2ky + k 2 ) + a 2 ( x 2 − 2hx + h 2 ) = a 2b 2 2 2 b y − 2b 2 ky + b 2 k 2 + a 2 x 2 − 2a 2 hx + a 2 h 2 = a 2b 2 2 2 b y + a 2 x 2 − 2b 2 ky − 2a 2 hx + b 2 k 2 + a 2 h 2 − a 2b 2 = 0 2
2
A = b 2 , B = 0, C = a 2 B 2 − 4 AC = 02 − 4b 2 a 2 = −4a 2b 2 < 0 B 2 − 4 AC < 0 for an ellipse whose major axis is parallel to the y -axis.
Parabola with axis of symmetry parallel to y-axis:
( x − h )2 = 4 p ( y − k ) 2
x − 2hx + h 2 = 4 py − 4 pk x 2 − 2hx − 4 py + h 2 + 4 pk = 0 A = 1, B = 0, C = 0 B 2 − 4 AC = 02 − 4(1)(0) = 0 B 2 − 4 AC = 0 for a parabola with axis of symmetry parallel to the y -axis. Parabola with axis of symmetry parallel to x-axis:
( y − k )2 = 4 p ( x − h ) y 2 − 2ky + k 2 = 4 px − 4 ph y − 2ky − 4 px + k 2 + 4 ph = 0 A = 1, B = 0, C = 0 B 2 − 4 AC = 02 − 4 (1)( 0 ) = 0 2
B 2 − 4 AC = 0 for a parabola with axis of symmetry parallel to the x-axis. Hyperbola with the transverse axis parallel to x-axis:
( x − h) 2 a2
−
( y − k )2 b2
=1
b 2 ( x − h ) 2 − a 2 ( y − k ) 2 = a 2b 2 b ( x − 2hx + h 2 ) − a 2 ( y 2 − 2ky + k 2 ) = a 2b 2 b 2 x 2 − 2b 2 hx + b 2 h 2 − a 2 y 2 + 2a 2 ky − a 2 k 2 = a 2b 2 2 2 b x − a 2 y 2 − 2b 2 hx + 2a 2 ky + b 2 h 2 − a 2 k 2 − a 2b 2 = 0 2
2
A = b 2 , B = 0, C = − a 2 B 2 − 4 AC = 02 − 4b 2 (− a 2 ) = 4a 2b 2 > 0 B 2 − 4 AC > 0 for a hyperbola whose transverse axis is parallel to the x-axis.
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Section 8.4
561
Hyperbola with transverse axis parallel to y -axis: ( y − k )2 a2
−
( x − h)2 b2
=1
b 2 ( y − k ) 2 − a 2 ( x − h ) 2 = a 2b 2 b ( y − 2ky + k 2 ) − a 2 ( x 2 − 2hx + h 2 ) = a 2b 2 b 2 y 2 − 2b 2 ky + b 2 k 2 − a 2 x 2 + 2a 2 hx − a 2 h 2 = a 2b 2 2 2 b y − a 2 x 2 − 2b 2 ky + 2a 2 hx + b 2 k 2 − a 2 h 2 − a 2b 2 = 0 2
2
A = − a 2 , B = 0, C = b 2 B 2 − 4 AC = 02 − 4b 2 (− a 2 ) = 4a 2b 2 > 0 B 2 − 4 AC > 0 for a hyperbola whose transverse axis is parallel to the y -axis. 48.
x = x 'cosα − y 'sin α y = y 'cos α + x 'sin α If this represents a system where the xy-axes have been rotated through α to create x’y’ axes, then we can rotate the new system backward through α (that is, the angle of rotation is − α ) and create an x”y” system that is consistent with the original xy-system. Thus, we can use the original formulas with x replaced by x’, y replaced by y’, x’ replaced by x”, y’ replaced by y”, and α replaced by −α to rotate the x’y’-system backward through α to an x”y”-system. x ' = x ''cos(−α ) − y ''sin(−α ) y ' = y ''cos(−α ) + x ''sin(−α ) Since cos(−α ) = cos(α ) and sin(−α ) = − sin(α ), we can say x ' = x ''cos(α ) + y ''sin(α ) y ' = y ''cos(α ) − x ''sin(α ) But since the x '' y ''-system is coincident with the xy -system, x '' can be replaced by x and y '' by y. Hence x ' = x cosα + y sin α y ' = y cos α − x sin α
.......................................................
Prepare for Section 8.5
PS1. sin(− x) = − sin x odd function
PS2. cos(− x) = cos x even function
PS3. tan α = − 3 α = 2π , 5π 3 3
PS4. sin α = − 3 , α = 240o or 300o 2 cos α = − 1 α = 120o or 240o 2 α = 240o
PS5. (r cosθ )2 + ( r sin θ ) 2 = r 2 cos 2 θ + r 2 sin 2 θ = r 2 (cos 2 θ + sin 2 θ ) = r2
PS6.
y 5 cos32o = x 5 (4.2, 2.6)
sin 32o =
y = 5sin 32o ≈ 2.6 x = 5cos32o ≈ 4.2
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562
Chapter 8: Topics in Analytic Geometry
Section 8.5 1.
2.
3.
4.
5.
6.
7.
8.
12.
9.
0 ≤ θ ≤ 2π
10.
0 ≤ θ ≤ 2π
11.
13.
0 ≤θ ≤π
14.
0 ≤θ ≤π
15.
0 ≤ θ ≤ 2π
16.
0 ≤θ ≤π
17.
0 ≤θ ≤π
18.
0 ≤θ ≤π
19.
0 ≤ θ ≤ 2π
20.
0 ≤ θ ≤ 2π
21.
0 ≤ θ ≤ 2π
22.
0 ≤ θ ≤ 2π
23.
0 ≤ θ ≤ 2π
24.
0 ≤ θ ≤ 2π
25.
Graph for 0 ≤ θ ≤ 2π .
26.
See Exercise 24. Graph for 0 ≤ θ ≤ 2π .
27.
Graph for 0 ≤ θ ≤ π .
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Section 8.5
563
28.
Graph for 0 ≤ θ ≤ 2π .
29.
Graph for 0 ≤ θ ≤ π .
30.
Graph for 0 ≤ θ ≤ π .
31.
Graph for 0 ≤ θ ≤ π .
32.
Graph for 0 ≤ θ ≤ π .
33.
Graph for 0 ≤ θ ≤ 4π .
34.
Graph for 0 ≤ θ ≤ 8π .
35.
Graph for 0 ≤ θ ≤ 6π .
36.
Graph for 0 ≤ θ ≤ 6π .
37.
Graph for 0 ≤ θ ≤ 2π .
38.
Graph for 0 ≤ θ ≤ 4π .
39.
Graph for 0 ≤ θ ≤ 2π with θ step = π 200. (Some graphing utilities may draw a false asymptote in “connected” mode.)
40.
Graph for 0 ≤ θ ≤ 2π with θ step = π 200.
41.
r = x2 + y 2
42.
r = x2 + y 2
= 12 + (− 3)2 = 1+ 3 = 4 =2
θ = tan −1
y x
= tan −1 ⎛⎜ − 3 ⎞⎟ ⎝ 1 ⎠ = tan −1 ⎛⎜ − 3 ⎞⎟ ⎝ 1 ⎠ = −60o
The polar coordinates of the point are (2, − 60o ).
= (−2 3)2 + (2)2 = 12 + 4 = 16 =4
θ = tan −1
y x
= tan −1 ⎛⎜ 2 ⎞⎟ ⎝ −2 3 ⎠ = tan −1 ⎛⎜ − 1 ⎞⎟ ⎝ 3⎠ = 150o
The polar coordinates of the point are (4, 150o ).
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564
43.
Chapter 8: Topics in Analytic Geometry
x = r cosθ
2π ⎞ ⎛ = (−3) ⎜ cos ⎟ 3 ⎠ ⎝ ⎛ 1⎞ = (−3) ⎜ − ⎟ ⎝ 2⎠ 3 = 2
y = r sin θ
44.
2π ⎞ ⎛ = ( − 3) ⎜ sin ⎟ 3 ⎠ ⎝ ⎛ 3⎞ = ( −3) ⎜⎜ ⎟⎟ ⎝ 2 ⎠ =−
⎡ = (2) ⎢cos ⎣ ⎛1⎞ = (2) ⎜ ⎟ ⎝2⎠ =1
3 3 2
x = r cosθ
⎛ π⎞ = 0cos ⎜ − ⎟ ⎝ 2⎠ =0
y = r sin θ
⎛ π ⎞⎤
⎡ = (2) ⎢sin ⎣ ⎛ = (2) ⎜⎜ − ⎝
⎜ − ⎟⎥ ⎝ 3 ⎠⎦
⎛ π ⎞⎤ ⎜ − ⎟⎥ ⎝ 3 ⎠⎦ 3⎞ ⎟ 2 ⎟⎠
=− 3
(
)
The rectangular coordinates of the point are 1, − 3 .
⎛ ⎞ The rectangular coordinates of the point are ⎜ 3 , − 3 3 ⎟ . 2 2 ⎝ ⎠ 45.
y = r sin θ
x = r cosθ
46.
⎛ π⎞ = 0sin ⎜ − ⎟ ⎝ 2⎠ =0
y = r sin θ
x = r cosθ
⎛ 5π ⎞ = (3) ⎜ sin ⎟ 6 ⎠ ⎝ ⎛1⎞ = (3) ⎜ ⎟ ⎝2⎠ 3 = 2
5π ⎞ ⎛ = (3) ⎜ cos ⎟ 6 ⎠ ⎝ ⎛ 3⎞ = (3) ⎜⎜ − ⎟⎟ 2 ⎝ ⎠
The rectangular coordinates of the point are (0, 0)
=−
3 3 2
⎛ The rectangular coordinates of the point are ⎜ − 3 3 , ⎝ 2 47.
r = x2 + y 2
=
( 3)2 + ( 4 )2
= 9 + 16
y x = tan −1 4 3 o
θ = tan −1
48.
=
= 169 = 13
The approximate polar coordinates of the point are (5, 53.1°).
52.
r = 3cosθ r − 3cosθ = 0
50.
r = 2sin θ r − 2sin θ = 0
r 2 − 3r cosθ = 0
r 2 − 2r sin θ = 0
x 2 + y 2 − 3x = 0
x2 + y 2 − 2 y = 0
r = 4cscθ 4 sin θ r sin θ = 4 y=4 r=
( ) ( )
(12 )2 + ( −5)2
= 144 + 25
≈ 53.1
= 25 =5
49.
y x = tan −1 −5 12 = tan −1 − 5 12
θ = tan −1
r = x2 + y 2
53.
r=4 x2 + y 2 = 4 x 2 + y 2 = 16
≈ 337.4°
The approximate polar coordinates of the point are (13, 337.4°). 51.
54.
r = 3secθ 3 r= cosθ r cosθ = 3 x=3
θ=
π
4 y tan θ = x π y tan = 4 x y 1= x y=x
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3 ⎞⎟ . 2⎠
Section 8.5
55.
565
θ =π
56.
6 y tan θ = x y tan π = 6 x 3= y 3 x
r cos θ = −4 x = −4
57.
r = tan θ r = sin θ cosθ r cosθ = sin θ r cos θ − sin θ = 0 r 2 cosθ − r sin θ = 0 x2 + y 2 ( x) − y = 0 y x2 + y2 = x y2 2 2 x +y = 2 x x4 − y2 + x2 y2 = 0
y= 3x 3
58.
r = cot θ r = cosθ sin θ r sin θ = cosθ r sin θ − cosθ = 0 2 r sin θ − r cosθ = 0
59.
x2 + y 2 − y = 2 x2 + y 2 = 2 + y x2 + y 2 = 4 + 4 y + y 2 x − 4y − 4 = 0 x2 = 4 y + 4
x2 + y 2 = 4 − 4 x + x2 y + 4x − 4 = 0
2
2
r (sin θ − 2 cosθ ) = 6 r sin θ − 2r cosθ = 6 y − 2x = 6
62.
r (2 cosθ + sin θ ) = 3 2r cosθ + r sin θ = 3 2x + y = 3
r sin θ = 2 r = 2 csc θ
x = −4
65.
66.
y = 3x r sin θ = 3 ( r cos θ ) tan θ = 3 3
x=3
68.
r cos θ = 3 r = 3sec θ
y = x2 y =x x tan θ = r cos θ tan θ sec θ = r
θ =π
70.
y=2
63.
y = −2 x + 3
r cos θ = −4 r = −4sec θ
67.
r=
x2 + y 2 = 2 − x
y = 2x + 6 64.
2 1 − sin θ r − r sin θ = 2
60.
x2 + y 2 + x = 2
x2 + y 2 ( y) − x = 0 x2 + y 2 = x y 2 2 2 x x +y = 2 y 4 2 2 2 y −x +x y =0 61.
2 1 + cos θ r + r cos θ = 2 r=
xy = 4 ( r cosθ ) ( r sin θ ) = 4
69.
x2 + y2 = 4 r2 = 4
2
r sin θ cos θ = 4
2x − 3y = 6
71.
2 r cos θ − 3r sin θ = 6 r (2 cos θ − 3sin θ ) = 6
r=
r=2
x2 = 8 y r 2 cos 2 θ = 8r sin θ
6 2 cos θ − 3sin θ
r cos 2 θ = 8 sin θ
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566
72.
Chapter 8: Topics in Analytic Geometry
y2 = 4 y
x 2 − y 2 = 25
73.
r 2 sin 2 θ = 4r sin θ r 2 sin 2 θ − 4r sin θ = 0 r sin θ (r sin θ − 4) = 0
x 2 + 4 y 2 = 16
74.
r 2 cos 2 θ − r 2 sin 2 θ = 25
r 2 cos2 θ + 4 r 2 sin 2 θ = 16
r 2 (cos 2 θ − sin 2 θ = 25
r 2 (cos2 θ + 4sin 2 θ ) = 16
r 2 (cos 2θ ) = 25
r 2 (cos2 θ + sin 2 θ + 3sin 2 θ = 16 r 2 (1 + 3sin 2 θ ) = 16 r 2 + 3r 2 sin 2 θ = 16 3r 2 sin 2 θ + r 2 = 16
75.
76.
77.
78.
79.
80.
81.
82.
....................................................... 83.
cosθ = ± cos 2 θ is not an identity.
84.
cos 2θ = 2 cos 2 θ − 1 is an identity.
Connecting Concepts 85.
Enter as r = 4cos 2θ and r = − 4cos 2θ for 0 ≤ θ ≤ 4π .
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.5
86.
567
87.
Enter as r = −2sin 2θ and r = − −2sin 2θ for 0 ≤ θ ≤ π .
Graph for 0 ≤ θ ≤ 2π with θ step = π 200.
88.
Graph for π 2 ≤ θ ≤ 3π 2 . (Some graphing utilities may produce a false asymptote in “connected” mode.)
91.
Graph for −30 ≤ θ ≤ 30.
b.
0 ≤ θ ≤ 20π
(Some graphing utilities may produce a false asymptote in “connected” mode.)
Graph for 0 ≤ θ ≤ 2π .
89.
Graph r = 2 θ for − 4π < θ < 4π .
90.
92.
Graph for 0 < θ ≤ 10π .
93. a. 0 ≤ θ ≤ 5π
....................................................... PS1.
2 x2 + y = 1 25 16 a 2 = 25, a = 5
b 2 = 16
Prepare for Section 8.6
y2 = 4x 4p = 4 p =1 directrix: x = –1
PS3.
PS5. For a hyperbola, e > 1.
PS6.
PS2.
c 2 = a 2 − b 2 = 25 − 16 = 9 c=3 e= c =3 a 5 The eccentricity is 3 . 5 PS4. 1 + sin x = 0 sin x = −1 x = 3π 2
y = 2(1 + yx) y = 2 + 2 yx y − 2 yx = 2 y (1 − 2 x) = 2 y= 2 1 − 2x
4 1 4sec x = cos x 2sec x − 1 2 1 − 1 cos x
4 1 x cos cos x = ⋅ cos x 2 1 − 1 cos x 4 = 2 − cos x
Copyright © Houghton Mifflin Company. All rights reserved.
568
Chapter 8: Topics in Analytic Geometry
Section 8.6 1.
12 4 = 3 − 6 cosθ 1 − 2 cosθ e = 2 The graph is a hyperbola. The transverse axis is on the polar axis.
r=
2.
Let θ = 0. 12 r= = 12 = −4 3 − 6cos 0 3 − 6 Let θ = π . 12 r= = 12 = 4 3 − 6cos π 3 + 6 3
Let θ = 0. 8 = 8 = −4 2 − 4cos 0 2 − 4 Let θ = π . 8 r= = 8 =4 2 − 4cos π 2 + 4 3 r=
(3 )
The vertices are at (−4, 0) and 4 , π .
3.
r=
8 4 = 2 − 4 cosθ 1 − 2 cosθ e = 2 The graph is a hyperbola. The transverse axis is on the polar axis.
r=
(3 )
The vertices are at (−4, 0) and 4 , π .
2 8 = 4 + 3 sin θ 1 + 3 sin θ
4.
r=
2 6 = 3 + 2 cosθ 1 + 2 cosθ 3
4
e = 3 The graph is an ellipse.
e = 2 The graph is an ellipse.
The major axis is on the line θ = π .
The major axis is on the polar axis.
3
4
2
Let θ = r=
π 2
Let θ = 0. 6 6 6 = = 3 + 2 cos 0 3 + 2 5 Let θ = π . 6 6 r= = =6 3 + 2 cos π 3 − 2
.
r=
8 4 + 3 sin π
2
=
8 8 = 4+3 7
3π Let θ = . 2
(5 )
Vertices on major axis are at 6 , 0 and (6, π ) .
8
8 r= = =8 4 + 3 sin 3π 4 − 3 2
Let θ = π . 2
(7 2)
Vertices on major axis are at 8 , π
and
( 8, 32π ) .
Let θ = 0. 8 8 = =2 4 + 3 sin 0 4 + 0 Let θ = π . 8 8 r= = =2 4 + 3 sin π 4 + 0 The curve also goes through (2, 0) and (2, π ).
r=
r=
6 3 + 2 cos π
=
2
6 =2 3+ 0
3π Let θ = . 2 r=
6 3 + 2 cos 3π 2
=
6 =2 3+0
( 2) ( 2 )
The curve also goes through 2, π and 2, 3π .
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.6
5.
r=
569 5
9 3 = 3 − 3 sin θ 1 − sin θ
6.
e = 1 The graph is a parabola.
The axis of symmetry is θ = π .
2
2
When θ = π , r is undefined.
When θ = π , r is undefined.
2
r=
2
3π . 2
Let θ = 9
3 − 3 sin 3π 2
=
9 3 = 3+3 2
r=
r=
3π . 2 5 2 − 2 sin 3π 2
=
5 5 = 2+2 4
(4 2 )
(2 2 )
Vertex is at 5 , 3π .
Vertex is at 3 , 3π .
7.
5 2 = 2 − 2 sin θ 1 − sin θ
e = 1 The graph is a parabola.
The axis of symmetry is θ = π .
Let θ =
r=
2 10 = 5 + 6 cosθ 1 + 6 cosθ
8.
5
e = 6 The graph is a hyperbola. 5
The transverse axis is on the polar axis. Let θ = 0. 10 10 10 r= = = 5 + 6 cos 0 5 + 6 11 Let θ = π . 10 10 = = −10 r= 5 + 6 cos π 5 − 6
( )
The vertices are at 10 , 0 and ( −10, π ) . 11
r=
8 4 = 2 + 4 cosθ 1 + 2 cosθ
e = 2 The graph is a hyperbola. The transverse axis is on the polar axis. Let θ = 0. 8 8 4 = = 2 + 4 cos 0 2 + 4 3 Let θ = π . 8 8 r= = = −4 2 + 4 cosπ 2 − 4
r=
(3 )
The vertices are at 4 , 0 and (−4,π ) .
Copyright © Houghton Mifflin Company. All rights reserved.
570
9.
Chapter 8: Topics in Analytic Geometry
r= = =
4 secθ 2 secθ − 1
10.
4 4 cosθ = 2 − 1 2 − cosθ cosθ
r=
=
3 secθ 2 secθ + 2 3 3 cosθ = 2 + 2 2 + 2 cosθ cosθ
e = 1 The graph is a parabola. The axis of symmetry is the polar axis.
2 1 − 1 cosθ 2
e = 1 The graph is an ellipse.
Let θ = 0 .
2
The major axis is on the polar axis. However, the original equation is undefined at π and at 3π . 2
Thus, the ellipse will have holes at those angles. Let θ = 0 . r=
4 4 = =4 2 − cos 0 2 − 1 4 4 4 = = 2 − cos π 2 + 1 3
θ=
2
π 2
. 3
2 + 2 cos π
=
3 3 = 2+0 2
(3 )
r=
3 2 + 2 cos 3π 2
=
2 − cos π
=
2
3 2
Thus, the equation r = 3secθ
( 32 , π2 ) and ( 23 , 32π ) . 4
2secθ +1
4 =2 2−0
3π Let θ = . 2 r=
4 2 − cos 3π 2
=
4 =2 2−0
Vertices on minor axis of
(2, π2 ) and (2, 32π ) .
2 are at 1− 12 cosθ
Thus, the equation r = 4 secθ will have holes at
( 2) ( 2 )
⎛3 ⎞ is at ⎜ , 0 ⎟ . ⎝4 ⎠
3 3 = 2 + 2 cos 0 4
2
.
r=
Let θ =
1 + cosθ
3π Let θ = . 2
Vertices on major axis are at (4, 0) and 4 , π .
π
2
r=
Let θ = π . r=
r=
Vertex of
3 2
2 secθ −1
2, π and 2, 3π
Copyright © Houghton Mifflin Company. All rights reserved.
has holes at
Section 8.6
11.
r= = =
571
12 cscθ 6 cscθ − 2
12.
12 12 sin θ = 6 − 2 6 − 2 sin θ sin θ
3 3 sin θ = 2 + 2 2 + 2 sin θ sin θ 3 2 =
2
1 + sin θ
e = 1 The graph is a parabola.
e = 1 The graph is an ellipse. 3
The axis of symmetry is θ = π .
The major axis is on θ = π .
2
2
r=
π 2
Let θ =
. 12
=
6 − 2 sin π
2
r=
12 =3 6−2
6 − 2 sin 3π 2
2
. 3
2 + 2sin π
=
3 3 = 2+2 4
(4 2 )
Vertex is at 3 , π .
2
12
π
2
Let θ = 3π . r=
3 cscθ 2 cscθ + 2
=
1 − 1 sin θ 3
Let θ =
r=
=
(2 )
( 2)
(2 2 )
Vertices on major axis are at 3, π and 3 , 3π . The equation r = 12 cscθ
6 cscθ − 2
(2 )
The parabola has holes at 3 , 0 and 3 , π .
12 3 = 6+2 2
has holes at (2, 0) and (2, π ).
Copyright © Houghton Mifflin Company. All rights reserved.
572
13.
Chapter 8: Topics in Analytic Geometry
3 cosθ − 1 −3 = 1 − cosθ
r=
14.
=
1 + 1 sin θ
e = 1 The graph is an ellipse. 2
The major axis is on θ = π .
Let θ = π . −3 −3 −3 3 r= = = =− 1 − cos π 1 − (−1) 1 + 1 2
2
Let θ = π . 2
)
r=
Vertex is at − 3 , π . 2
2 sin θ + 2 1 2
e = 1 The graph is a parabola. The axis of symmetry is the polar axis.
(
r=
2 sin π + 2
=
2
2 2 = 1+ 2 3
Let θ = 3π . 2 r=
2 sin 3π + 2 2
=
2 =2 −1+ 2
(3 2)
Vertices of major axis are at 2 , π
and
( 2, 32π ) .
Let θ = 0. r=
2 2 = =1 sin θ + 2 0 + 2
Let θ = π . r=
2 2 = =1 sin π + 2 0 + 2
The curve also goes through (1, 0) and (1, π ).
12 3 − 6cosθ r (3 − 6cos θ ) = 12 3r − 6r cosθ = 12
r=
15.
8 2 − 4cosθ r (2 − 4cos θ ) = 8 2r − 4r cosθ = 8 r=
16.
2 x2 + y 2 − 4 x = 8
3 x 2 + y 2 − 6 x = 12
2 x2 + y 2 = 4 x + 8
3 x 2 + y 2 = 6 x + 12
x2 + y 2 = 2 x + 4
x2 + y 2 = 2 x + 4
x 2 + y 2 = 4 x 2 + 16 x + 16 3x 2 − y 2 + 16 x + 16 = 0
x 2 + y 2 = 4 x 2 + 16 x + 16 3x 2 − y 2 + 16 x + 16 = 0
8 4 + 3sin θ r (4 + 3sin θ ) = 8 4r + 3r sin θ = 8 r=
17.
6 3 + 2cosθ r (3 + 2cosθ ) = 6 3r + 2r cosθ = 6 r=
18.
4 x2 + y 2 + 3 y = 8
3 x2 + y 2 + 2 x = 6
4 x 2 + y 2 = −3 y + 8 2
2
3 x 2 + y 2 = −2 x + 6
2
16 x + 16 y = 9 y − 48 y + 64 2
2
16 x + 7 y + 48 y − 64 = 0
9 x 2 + 9 y 2 = 4 x 2 − 24 x + 36 2
2
5 x + 9 y + 24 x − 36 = 0 Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.6
573
9 3 − 3 sin θ r (3 − 3sin θ ) = 9
r=
19.
5 2 − 2 sin θ r ( 2 − 2 sin θ ) = 5
r=
20.
3r − 3r sin θ = 9
2r − 2r sin θ = 5
3 x2 + y 2 − 3 y = 9
2 x2 + y2 − 2 y = 5
3 x2 + y 2 = 3 y + 9
2 x2 + y2 = 2 y + 5
3 x 2 + y 2 = 3( y + 3)
4 x 2 + 4 y 2 = 4 y 2 + 20 y + 25 4 x 2 − 20 y − 25 = 0
x2 + y 2 = y + 3 x2 + y 2 = y 2 + 6 y + 9 x2 − 6 y − 9 = 0
21.
e = 2, r cosθ = −1 ,
22.
d = −1 = 1 ed 1 − e cosθ (2) (1) = 1 − (2) cosθ 2 = 1 − 2 cosθ
r=
23.
e = 1, r sin θ = 2, d = 2 = 2
24.
2 , r sin θ = −4, 3 d = −4 = 4
e=
r= =
ed 1 − e sin θ
( 23 ) (4) 1 − ( 2 ) sin θ 3
8 3 = 1 − 2 sin θ 3
=
8 3 − 2sin θ
=
( 32 ) (1) 1 + ( 3 ) sin θ 2
=
3 2 1 + 3 sin θ 2
=
3 2 + 3sin θ
e = 1, r cosθ = −2, d = −2 = 2
ed 1 + e sin θ (1) (2) = 1 + (1) sin θ 2 = 1 + sin θ
r=
25.
3 , r sin θ = 1 , d = 1 = 1 2 ed r= 1 + e sin θ
e=
ed 1 − e cosθ (1) (2) = 1 − (1) cosθ 2 = 1 − cosθ
r=
26.
1 , r cosθ = 2, d = 2 = 2 2 ed r= 1 + e cosθ
e=
= =
( 12 ) (2) 1 + ( 1 ) cosθ 2 1 1 + 1 cosθ 2
=
2 2 + sin θ
Copyright © Houghton Mifflin Company. All rights reserved.
574
27.
Chapter 8: Topics in Analytic Geometry
3 , r = 2 secθ 2 2 r= cosθ r cosθ = 2, d = 2 = 2
e=
r= = =
28.
ed 1 + e cosθ 3 (2)
r=
(2) 1 + ( 3 ) cos θ 2
=
3 1 + 3 cos θ
=
6 2 + 3cos θ
=
2
=
29.
3 , r = 2 cscθ 4 2 r= sin θ r sin θ = 2, d = 2 = 2
e=
vertex: (2, π ), curve: parabola ed e = 1 (by definition of a parabola) r= 1 − e cosθ When θ = π , r = 2. Substituting into ed , r= 1 − e cosθ we have 1⋅ d d 2= = 1 − 1 ⋅ cos(π ) 2 Therefore, d = 4. Substituting e = 1 and d = 4 yields (1) (4) 4 r= or r = 1 − (1)cosθ 1 − cosθ
30.
ed 1 + e sin θ 3 (2)
(4) 1 + ( 3 ) sin θ 4 3 2 1 + 3 sin θ 4
6 4 + 3sin θ
vertex: (4, 0), e =
1 2
ed 1 + e cosθ When θ = 0, r = 4. Substituting into ed r= , 1 + e cosθ we have 1d 1d 2 2 = d 4= = 3 3 1 + 12 ⋅ cos(0) 2 r=
Therefore, d = 12. Substituting e = r=
( 12 ) (12) 1 + ( 1 ) cos θ 2
=
6 1 + 1 cos θ
=
12 2 + cosθ
1 and d = 12 yields 2
2
31.
vertex: (1, 3π / 2), e = 2 ed r= 1 − e sin θ 3π , r = 1. Substituting into When θ = 2 ed , r= 1 − e sin θ we have 2d 1= = 2d 3 1 − 2sin 3π
(2)
Therefore d = 3 . Substituting e = 2 and d = 3 yields 2 2 r=
(3)
(2) 2
1 − (2)sin θ 3 = 1 − 2sin θ
32.
vertex: (2, 3π / 2), e = 2 3
ed r= 1 − e sin θ
When θ = 3π , r = 2. Substituting into 2
ed r= , 1 − e sin θ we have (2 / 3)d 2d / 3 = 2= 1 − (2 / 3)sin(3π / 2) 5 / 3
Therefore, d = 5. Substituting e = 2 / 3 and d = 5 yields (2 / 3) ( 5 ) r= 1 − (2 / 3)sin θ (10 / 3) = 1 − (2 / 3)sin θ 10 = 3 − 2sin θ
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.6
575
33.
34.
Rotate the graph of Exercise 1
π 6
radians counterclockwise
Rotate the graph of Exercise 2
π 2
radians counterclockwise
about the pole.
about the pole. 35.
36.
Rotate the graph of Exercise 3 π radians counterclockwise about the pole.
Rotate the graph of Exercise 4
π 3
radians counterclockwise
about the pole. 37.
38.
Rotate the graph of Exercise 5 π radians clockwise
Rotate the graph of Exercise 6 π radians clockwise
about the pole.
about the pole.
2
6
39.
40.
Rotate the graph of Exercise 7 π radians clockwise about the pole.
Rotate the graph of Exercise 8 π radians clockwise 3
about the pole.
....................................................... 41.
42.
44.
45.
Connecting Concepts 43.
0 ≤ θ ≤ 12π
46.
0 ≤ θ ≤ 8π
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576
47.
Chapter 8: Topics in Analytic Geometry
Convert the equations for the conic and the directrix to rectangular form. ed 1 − e cosθ r (1 − e cosθ ) = ed
d = − r cosθ (by definition)
r=
= −x x = −d
r − er cosθ = ed x 2 + y 2 − ex = ed x 2 + y 2 = ex + ed = e( x + d ) x 2 + y 2 = e 2 ( x 2 + 2dx + d 2 ) x + y − e x − 2e dx − e2 d 2 = 0 (1 − e 2 ) x 2 + y 2 − (2e 2 d ) x − (e2 d 2 ) = 0 2
2
2 2
2
Solving for y 2 yields y 2 = (e 2 − 1) x 2 + (2ed ) x + (e 2 d 2 ). Now, with y 2 = (e 2 − 1) x 2 + (2ed ) x + (e 2 d 2 ) and a directrix of x = − d , let k =
d ( P, F ) , where the focus is at the origin (by d ( P, D)
definition). k=
d ( P, F ) = d ( P, D )
x2 + y2 x+d
We can substitute y 2 = (e 2 − 1) x 2 + (2ed ) x + (e 2 d 2 ) to obtain k=
x 2 + (e 2 − 1) x 2 + (2ed ) x + (e 2 d 2 ) = x+d
Solving for k 2 gives us k 2 =
e 2 x 2 + 2e 2 dx + e 2 d 2 x+d
e 2 x 2 + 2e 2 dx + e 2 d 2 2
x + 2dx + d
Since k 2 = e 2 , k = ± e 2 = ± e . But, since k = Therefore, k = e, or 48.
2
=
e 2 ( x 2 + 2dx + d 2 ) x 2 + 2dx + d 2
= e2.
d ( P, F ) , and the ratio of two distances must be positive, k cannot be negative. d ( P, D )
d ( P, F ) = e. d ( P, D )
Convert the polar equation of the directrix into the xy-coordinate system: r sin θ = d y=d Since we have shown (in Exercise 47) that
d ( P, F ) = e and since the focus is at the origin (or pole), we can say d ( P, D )
PF =e PD x2 + y2 =e y−d x2 + y2 ( y − d )2
= e2
x 2 + y 2 = e 2 (d − y ) 2 Converting back into polar coordinates yields r 2 = e 2 (d − r sin θ ) 2 r = e( d − r sin θ ) r = ed − er sin θ r + er sin θ = ed r (1 + e sin θ ) = ed ed r= 1 + e sin θ
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.7
577
....................................................... PS1.
( ) = ( y + 32 )
y2 + 3y + 3 2
2
2
PS2.
y = x 2 = (2t + 1) 2 = 4t 2 + 4t + 1
PS4. x 2 + y 2 = (sin t ) 2 + (cos t ) 2 = 1
PS3. ellipse PS5.
Prepare for Section 8.7
y = ln t
PS6. Domain: (−∞, ∞) Range: [–3, 3]
ey = t
Section 8.7 1.
A table of five arbitrarily chosen values of t and the corresponding values of x and y are shown in the table below.
2.
t
x = 2t
y = −t
( x, y )
t
x = −3t
y = 6t
( x, y )
−2 −1 0 1 2
−4 −2 0 2 4
2 1 0 −1 −2
(−4, 2) (−2, 1) (0, 0) (2, −1) (4, −2)
−2 −1 0 1 2
6 3 0 −3 −6
−12 −6 0 6 12
(6, −12) (3, −6) (0, 0) (−3, 6) (−6, 12)
Plotting points for several values of t yields the following graph.
3.
A table of five arbitrarily chosen values of t and the corresponding values of x and y are shown the table below.
A table of five arbitrarily chosen values of t and the corresponding values of x and y are shown in the table below.
t −2 −1 0 1 2
x = −t 2 1 0 −1 −2
y = t2 −1
( x, y )
3 0 −1 0 3
(2, 3) (1, 0) (0, −1) (−1, 0) (−2, 3)
Plotting points for several values of t yields the following graph.
Plotting points for several values of t yields the following graph.
4.
A table of five arbitrarily chosen values of t and the corresponding values of x and y are shown the table below.
t −2 −1 0 1 2
x = 2t
y = 2t 2 − t + 1
( x, y )
−4 −2 0 2 4
11 4 1 2 7
(−4, 11) (−2, 4) (0, 1) (2, 2) (4, 7)
Plotting points for several values of t yields the following graph.
Copyright © Houghton Mifflin Company. All rights reserved.
578
5.
Chapter 8: Topics in Analytic Geometry
A table of five arbitrarily chosen values of t and the corresponding values of x and y are shown the table below.
t −2 −1 0 1 2
x = t2 4 1 0 1 4
y = t3
( x, y )
−8 −1 0 1 8
(4, −8) (1, −1) (0, 0) (1, 1) (4, 8)
6.
A table of eight values of t in the specified interval and the corresponding values of x and y are shown the table below.
t 0 π /4
8.
( x, y )
t
0
(2, 0)
2 0
3 2/2 3
− 2
3 2/2 0
(− 2, 3 2 / 2)
0 π /4 π /2 3π / 4
−3 2 / 2
(− 2, − 3 2 / 2)
−3
(0, −3)
−3 2 / 2
( 2, − 3 2 / 2)
−2 − 2 0
3π / 2 7π / 4
2
( 2, 3 2 / 2) (0, 3)
π
(−2, 0)
5π / 4 3π / 2 7π / 4
t −2 −1 0 1 2
x = 2t 1/4 1/2 1 2 4
y = 2t +1
1/2 1 2 4 8
( x, y ) (1/4, 1/2) (1/2, 1) (1, 2) (2, 4) (4, 8)
Plotting points for several values of t yields the following graph.
3 0 −1 0 3
(5, 3) (2, 0) (1, −1) (2, 0) (5, 3)
x = 1 − sin t 1 ≈ 0.29 0 ≈ −0.29 1 ≈ 1.7 2 ≈ 1.7
y = 1 + cos t
( x, y )
2 ≈ 1.7 1 ≈ 0.29 0 ≈ 0.29 1 ≈ 1.7
(1, 2) (≈ 0.29, ≈ 1.7) (0, 1) (≈ −0.29, ≈ 0.29) (1, 0) (≈ 1.7, ≈ 0.29) (0, −3) (≈ 1.7, ≈ 1.7)
Plotting points for several values of t yields the following graph.
Plotting points for several values of t yields the following graph.
A table of five arbitrarily chosen values of t and the corresponding values of x and y are shown the table below.
( x, y )
A table of eight values of t in the specified interval and the corresponding values of x and y are shown the table below.
y = 3sin t
π 5π / 4
y = t2 −1
Plotting points for several values of t yields the following graph.
x = 2cos t 2
π /2 3π / 4
9.
x = t2 +1 5 2 1 2 5
t −2 −1 0 1 2
Plotting points for several values of t yields the following graph.
7.
A table of five arbitrarily chosen values of t and the corresponding values of x and y are shown the table below.
10.
A table of five values of t in the domain of the parameter and the corresponding values of x and y are shown the table below.
t 1 2 3 4 5
x = t2 1 4 9 16 25
y = 2log 2 t
( x, y )
0 2 ≈ 3.17 4 ≈ 4.64
(1, 0) (4, 2) (9, ≈ 3.17) (16, 4) (25, ≈ 4.64)
Plotting points for several values of t yields the following graph.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 8.7
11.
579
x = sec t
−
π 2
π
12.
2
y = tan t 2
0 ≤ t < 2π
x = 3 + 2cos t y = −1 − 3sin t
13.
y = 3 + 2t
x−3 2 y +1 sin t = 3
cos t =
2
tan t + 1 = sec t
y 2 + 1 = x2 x2 − y 2 − 1 = 0
x ≥ 1, y ∈ R
2
Because x = 2 − t 2 and t 2 ≥ 0 for all real numbers t , x ≤ 2 for all t. Similarly, y ≥ 3 for all t.
2
⎛ x − 3 ⎞ ⎛ y +1⎞ ⎜ ⎟ +⎜− ⎟ =1 3 ⎠ ⎝ 2 ⎠ ⎝
( x − 3)2 + ( y + 1)2 4
x = 1 + t2 y = 2−t
15.
t∈R
9
x = cos3 t
=1
0 ≤ t < 2π
16.
3
2
y = sin t
−t
2
cos t = x
y = 2 − ( x − 1) y = −x + 3
x = t +1 y=t
t ≥ −1
x=
y +1
x≥0
2
y ≥ −1
y = x −1
t
t ∈ R, x > 0 y>0
t
(t − t )
e ⋅e = e = e0 = 1 xy = 1 for x > 0 and y > 0
23
sin 2 t = y 2 3 cos 2 t + sin 2 t = 1
Because x = 1 + t 2 and t 2 ≥ 0 for all real numbers t , x ≥ 1 for all t. Similarly, y ≤ 2 for all t.
17.
x = e −t y=e
2
x = 1+ t → t2 = x −1
t∈R 2
x = 2 − t2 → t2 = 2 − x y = 3 + 2(2 − x) y = −2 x + 7
cos 2 t + sin 2 t = 1
14.
x = 2 − t2
x
18.
23
+y
23
=1
−1 ≤ x ≤ 1 −1 ≤ y ≤ 1
x= t y = 2t − 1
t ≥ 0, x ≥ 0 y ≥ −1
y = 2t − 1 → t =
y +1 2
x=
y +1 2
19.
x = t3 y = 3 ln t
t > 0, x > 0 y∈R
x = t 3 → t = x1/ 3 y = 3 ln x1/ 3 y = ln x for x > 0 and y ∈ R
y = 2 x 2 − 1 for x ≥ 0 and y ≥ −1
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580
20.
Chapter 8: Topics in Analytic Geometry
x = et y = e2t t
21.
t ∈ R, x > 0
y = 3 + 2sin t y − 3 = 2sin t y−3 = sin t 2
x = 2 + 3cos t x − 2 = 3cos t
y>0
x − 2 = cos t 3
x = e → x = e2t → x 2 = y 2
y = x 2 for x > 0 and y > 0
( x −3 2 ) ( x −3 2 )
2
2
⎛ y − 3⎞ 2 2 +⎜ ⎟ = cos t + sin t ⎝ 2 ⎠
2
⎛ y − 3⎞ +⎜ ⎟ =1 ⎝ 2 ⎠
2
At t = 0 , x = 2 + 3cos0 = 5 At t = π , x = 2 + 3cos π = −1
y = 3 + 2sin 0 = 3 y = 3 + 2sin π = 3
The point traces the top half of the ellipse
( x −3 2 ) + ⎛⎜⎝ y 2− 3 ⎞⎟⎠ 2
2
= 1 , as shown in the figure. The point
starts at (5, 3) and moves counterclockwise along the ellipse until it reaches the point (–1, 3) at time t = π .
22.
x = sin t
y = − cos t
23.
x 2 + y 2 = sin 2 t + cos2 t x2 + y2 = 1
At t = 0 , x = sin 0 = 0 At t = 3π , 2 x = sin 3π = −1 2
y = − cos0 = −1
y = − cos 3π = 0 2
The point traces a portion of the circle x 2 + y 2 = 1 , as shown in the figure. The point starts at (0, –1) and moves counter clockwise along the circle until it reaches the point (–1, 0) at time t = 3π . 2
y = t +1 ⇒ t = y −1 x = 2t − 1 x = 2( y − 1) − 1 x = 2y − 3 x + 3 = 2y y = 1 x+3 2 At t = 0 , x = 2(0) − 1 = −1 At t = 3 , x = 2(3) − 1 = 5
y = 0 +1=1 y = 3+1 = 4
The point traces a line segment, as shown in the figure. The point starts at (–1, 1) and moves along the line segment until it reaches the point (5, 4) at time t = 3.
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Section 8.7
24.
581
x = t +1 ⇒ t = x −1
25.
y= t y = x − 1 or y 2 = x − 1
( )
x = tan π − t 4
y2 − x2
At t = 0 ,
x = 0 +1=1 At t = 4 ,
y= 0 =0
x = 4 +1= 5
y= 4 =2
( ) = sec ( π − t ) − tan ( π − t ) 4 4
y2 − x2 = 1 At t = 0 ,
y = sec π − t 4
2
2
Since 1 + tan 2 θ = sec2 θ
( )
y = sec π − 0 = 2 4
(
y = sec π − π = 2 4 2
x = tan π − 0 = 1 4
2
At t = π , 2
The point traces a portion of the parabola y = x − 1 , as shown in the figure. The point starts at (1, 0) and moves along the parabola until it reaches the point (5, 2) at time t = 4.
)
x = tan π − π = −1 4 2
( ) (
)
The point traces a portion of the top branch of the hyperbola y 2 − x 2 = 1 , as shown in the figure. The point starts at (1,
2 ) and moves along the hyperbola until it
reaches the point ( −1,
26.
x =1− t ⇒ t = 1− x y=t
27.
2 2
y = (1 − x ) or y = ( x − 1)
x = 1 − 2 = −1
C1 : x = 2 + t 2 y = 1 − 2t 2
2
y = (0)2 = 0
x = 2 + t2 → t2 = x − 2 y = 1 − 2( x − 2) y = −2 x + 5 x ≥ 2, y ≤ 1
y = (2)2 = 4
C2 : x = 2 + t
At t = 0 ,
x =1− 0 =1 At t = 2 ,
2 ) at time t = π . 2
The point traces a portion of the parabola given by
y = ( x − 1)2 , as shown in the figure. The point starts at (1, 0) and moves along the parabola until it reaches (–1, 4) at time t = 2.
y = 1− 2t x = 2+t →t = x−2 y = 1 − 2( x − 2) y = −2 x + 5 x ∈ R, y ∈ R The graph of C1 is a ray beginning at (2, 1) with slope −2. The graph of C 2 is a line passing through (2, 1) with slope –2.
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582
28.
Chapter 8: Topics in Analytic Geometry
29.
C1 : x = sec2 t y = tan 2 t
x = sin t y = csc t 1 sin t 1 y= x
csc t =
tan 2 t + 1 = sec 2 t y +1 = x y = x −1
Because 0 ≤ t < π , 2 1 ≤ sec2 t < ∞ and 0 ≤ tan 2 t < ∞ 1 ≤ x < ∞ and 0 ≤ y <∞ Thus, x ≥ 1 and y ≥ 0.
C2 : x = 1 + t 2 y = t2 x =1+ y y = x −1
Because 0 < t <
Range for graph 1 : 0 < t ≤
π
2 0 < x ≤1 y ≥1
,
2
2
π
2
0 ≤ t 2 < π and 1 ≤ t 2 + 1 < 1 + π 4 4
2 2 Thus, 1 ≤ x < 1 + π and 0 ≤ y < π .
4
3π 2 −1 ≤ x ≤ 0 y ≤ −1
Range for graph 2 : π ≤ t ≤
.
4
2 2⎞ ⎛ C1 is a ray from (1, 0) in the direction of ⎜1 + π , π ⎟ . 4 4 ⎝ ⎠ C 2 is the points on the line y = x − 1 between (1, 0) and
⎛ π2 π2 ⎞ ⎛ π2 π2 ⎞ ⎜1 + 4 , 4 ⎟ . C 2 includes (1, 0) but not ⎜1 + 4 , 4 ⎟ . ⎝ ⎠ ⎝ ⎠
C1 : x = cos t
30.
y = cos 2 t
0 ≤ t ≤ π , −1 ≤ x ≤ 1
C2 : x = sin t y = sin 2 t
0 ≤ y ≤1
(cos t )2 = cos 2 t
0 ≤ t ≤ π, 0 ≤ x ≤1 0 ≤ y ≤1
(sin t )2 = sin 2 t
2
( x)2 = y
( x) = y
y = x 2 for − 1 ≤ x ≤ 1 and 0 ≤ y ≤ 1
y = x 2 for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1
C1 is the graph of the parabola y = x 2 for − 1 ≤ x ≤ 1, while C2 is the graph of the parabola y = x 2 for 0 ≤ x ≤ 1.
31.
33.
32.
a.
b.
For the Hummer, x=6 y = 60t for t ≥ 0 Using the graphing calculator in SIMUL and PAR mode, the Hummer is the first to reach the intersection.
34.
a.
For the Learjet, x = 300 − 420t y = 200 for t ≥ 0
b.
Using the graphing calculator in SIMUL and PAR mode, the Piper Seneca is the first to reach the intersection point.
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Section 8.7
583
35.
36.
Maximum height (to the nearest foot) of 462 feet is attained when t ≈ 5.38 seconds.
Maximum height (to the nearest foot) of 195 feet is attained when t ≈ 3.50 seconds.
The projectile has a range (to the nearest foot) of 1295 feet and hits the ground in about 10.75 seconds.
The projectile has a range (to the nearest foot) of 1117 feet and hits the ground in about 6.99 seconds.
37.
38.
Maximum height: 963 ft at time t ≈ 7.76 sec Range: 3009 ft at time t ≈ 15.51 sec
Maximum height (to the nearest foot) of 694 feet is attained when t ≈ 6.59 seconds. The projectile has a range (to the nearest foot) of 3084 feet and hits the ground in about 13.17 seconds. 39.
40.
41.
42.
....................................................... 43.
Let P1 ( x1 , y1 ) and P2 ( x2 , y 2 ) be two distinct points on a line. 44. If P(x, y) is any other point on the line, then y − y1 y2 − y1 = (Slope is constant along entire line.) This x − x1 x2 − x1 equation can be rewritten as y − y1 x − x1 = Let this value equal t. y2 − y1 x2 − x1
x − x1 y − y1 = t and = t. x2 − x1 y2 − y1 Solving for x and y, respectively, we have x = x1 + t ( x2 − x1) and y = y1 + t ( y 2 − y1 )
Thus,
Connecting Concepts a > 0, x ∈ R, 0 ≤ t < 2π b > 0, y ∈ R, 0 ≤ t < 2π x = h + a sin t → sin t = x − h a y−k y = k + b cos t → cos t = b
x = h + a sin t y = k + b cos t
sin 2 t + cos 2 t = 1 2
2 ⎛ x − h ⎞ + ⎛ y − k ⎞ =1 ⎜ ⎟ ⎜ ⎟ ⎝ a ⎠ ⎝ b ⎠
(x − h )2 + ( y − k )2
= 1, which is the standard equation a2 b2 for an ellipse at (h, k).
Copyright © Houghton Mifflin Company. All rights reserved.
584
45.
Chapter 8: Topics in Analytic Geometry
radius = a, θ = ∠TOR
46.
The x-coordinate of P(x, y) is given by x = OR + QP. The y-coordinate is given by y = TR − TQ.
Let α = ∠COP. Because the smaller circle does not slip, b bθ = aα or α = θ . a
From the figure,
OR = a cosθ and QP = aθ sin θ . Thus, x = a cosθ + aθ sin θ TR = a sin θ and TQ = aθ cosθ . Thus, y = a sin θ − aθ cosθ The parametric equations are
x = a cosθ + aθ sin θ y = a sin θ − aθ cosθ
The coordinates of P(x, y) are given by
x = BC + DP y = OB − OD
π⎞ ⎛ a+b Thus, x = (a + b) cosθ + a sin ⎜ θ− ⎟ 2⎠ ⎝ a ⎛a+b ⎞ θ⎟ = (a + b) cosθ − a cos ⎜ ⎝ a ⎠ π⎞ ⎛a+b θ− ⎟ y = (a + b)sin θ − a cos ⎜ 2⎠ ⎝ a ⎛a+b ⎞ θ⎟ = (a + b)sin θ − a cos ⎜ ⎝ a ⎠ The parametric equations are ⎛ a+b ⎞ x = (a + b) cosθ − a cos⎜ θ⎟ ⎠ ⎝ a ⎛ a+b ⎞ y = (a + b) sin θ − a sin ⎜ θ⎟ ⎠ ⎝ a
47.
Because the circle moves without slipping, bθ = aα . bθ Therefore, α = . Let P( x, y ) be the coordinates of the moving point. a π ⎛b−a⎞ Angle φ = − ⎜ ⎟θ 2 ⎝ a ⎠ ⎡π ⎛ b − a ⎞ ⎤ Thus, x = (b − a ) cosθ + a sin ⎢ − ⎜ ⎟θ ⎥ ⎣2 ⎝ a ⎠ ⎦ ⎡π ⎛ b − a ⎞ ⎤ y = (b − a )sin θ − a cos ⎢ − ⎜ ⎟θ⎥ ⎣2 ⎝ a ⎠ ⎦
Simplifying, we have ⎛b−a ⎞ x = (b − a) cosθ + a cos⎜ θ⎟ ⎠ ⎝ a ⎛b−a ⎞ y = (b − a ) sin θ − a sin ⎜ θ⎟ ⎠ ⎝ a
Copyright © Houghton Mifflin Company. All rights reserved.
Exploring Concepts with Technology
585
.......................................................
Exploring Concepts with Technology
Using a Graphing Calculator to Find the nth Roots of z 1.
The procedure for a TI-83 calculator is illustrated below. z = −27 = 27(cos180° + i sin180°), thus, r = 27, θ = 180°. cube roots ⇒ n = 3
Use the TRACE feature and the arrow keys to display and move to each of the vertices of the polygon.
2.
The procedure for a TI-83 calculator is illustrated below. z = 32i = 32(cos90° + i sin 90°), thus, r = 32, θ = 90°. fifth roots ⇒ n = 5
Use the TRACE feature and the arrow keys to display and move to each of the vertices of the polygon.
Thus, the three cube roots of –27 are 1.5 + 2.598076i, 1.5 − 2.598076i, and − 3.
Thus, the five fifth roots of 32i are 1.902113 + 0.61803399i, −1.902113 + 0.61803399i, −1.1755705 – 1.618034i, 1.1755705 – 1.618034i, and 2i.
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586
3.
Chapter 8: Topics in Analytic Geometry
Here is the procedure for a TI-83 graphing calculator. Be sure the calculator is in parametric and degree mode. In the WINDOW menu, set Tmin=0, Tmax=360, and, since 360 / 4 = 90, set Tstep=90. Set Xmin, Xmax, Ymin, and Ymax to appropriate values that will allow the roots to be seen. Since z = 8 + 8i = 4(cos 45° + i sin 45°), in the Y- menu, enter X1T=4^(1/4)cos(45/4+T) and Y1T=4^(1/4)sin(45/4+T).Press GRAPH to display a polygon. The x- and y-coordinates of each vertex represent a root of z in the rectangular form x + yi. Use the TRACE feature and the arrow keys to display and move to each of the vertices of the polygon. The fourth roots of 8 + 8i are 1.38704 + 0.2758994i, −0.2758994 + 1.38704i, −1.38704 − 0.2758994i, and 0.2758994 − 1.38704i.
4.
Here is the procedure for a TI-83 graphing calculator. Be sure the calculator is in parametric and degree mode. In the WINDOW menu, set Tmin=0, Tmax=360, and, since 360 / 6 = 60, set Tstep=60. Set Xmin, Xmax, Ymin, and Ymax to appropriate values that will allow the roots to be seen. Since z = −64i = 64(cos 270° + i sin 270°), in the Y- menu, enter X1T=64^(1/6)cos(270/6+T) and Y1T=64^(1/6)sin(270/6+T).Press GRAPH to display a polygon. The x- and ycoordinates of each vertex represent a root of z in the rectangular form x + yi. Use the TRACE feature and the arrow keys to display and move to each of the vertices of the polygon. The sixth roots of −64i are 1.414214 + 1.414214i, −0.5176381 + 1.931852i, −1.931852 + 0.5176381i, −1.414214 − 1.414214i, 0.5176381 − 1.931852i, and 1.931852 − 0.5176381i.
.......................................................
Assessing Concepts
1.
d
2.
b
3.
e
4.
c
5.
a
6.
f
7.
g
8.
i
9.
h
10.
j
11.
k
12.
k
....................................................... 1.
x2 − y2 = 4
[8.3]
x2 y2 − =1 4 4
hyperbola center: (0, 0) vertices: (±2, 0)
Chapter Review 2.
y 2 = 16 x
[8.1]
parabola vertex: (0, 0) focus: (4, 0) directrix: x = −4
foci: (±2 2, 0) asymptotes: y = ± x
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Chapter Review
587
x2 + 4 y 2 − 6 x + 8 y − 3 = 0
3.
2
[8.2]
4.
3 x 2 − 4 y 2 + 12 x − 24 y − 36 = 0
2
2
x − 6 x + 4( y + 2 y ) = 3 2
3( x + 4 x) − 4( y + 6 y ) = 36
2
2
3( x + 4 x + 4) − 4( y 2 + 6 y + 9) = 36 + 12 − 36
( x − 6 x + 9) + 4( y + 2 y + 1) = 3 + 9 + 4 ( x − 3)2 + 4( y + 1)2 = 16
3( x + 2)2 − 4( y + 3)2 = 12
( x − 3) 2 ( y + 1)2 + =1 16 4
( x + 2)2 ( y + 3)2 − =1 4 3
ellipse center: (3, − 1) vertices: (3 ± 4, − 1) = (7, − 1) , (−1, − 1)
hyperbola center: (−2, − 3) vertices: (−2 ± 2, − 3) = (0, − 3), ( − 4, − 3)
foci: (3 ± 2 3, − 1) = (3 + 2 3, − 1), (3 − 2 3, − 1)
foci: (−2 ± 7, − 3) = (−2 + 7, − 3), ( − 2 − 7, − 3) asymptotes: y + 3 = ± 3 ( x + 2) 2
5.
[8.3]
2
3x − 4 y 2 + 8 y + 2 = 0
[8.1]
2
6.
3x + 2 y 2 − 4 y − 7 = 0 2
−4( y − 2 y ) = −3 x − 2
2( y − 2 y ) = −3 x + 7
2
2( y 2 − 2 y + 1) = −3 x + 7 + 2
−4( y − 2 y + 1) = −3 x − 2 − 4 −4( y − 1)2 = −3( x + 2)
2( y − 1)2 = −3( x − 3)
3 ( y − 1)2 = ( x + 2) 4
3 ( y − 1)2 = − ( x − 3) 2
parabola vertex: (−2, 1)
parabola vertex: (3, 1)
3 ⎞ ⎛ 29 ⎞ ⎛ focus: ⎜ − 2 + , 1⎟ = ⎜ − , 1⎟ 16 ⎠ ⎝ 16 ⎠ ⎝ 3 35 directrix: x = −2 − , or x = − 16 16
3 ⎞ ⎛ 21 ⎞ ⎛ focus: ⎜ 3 − , 1⎟ = ⎜ , 1⎟ 8 ⎠ ⎝ 8 ⎠ ⎝ 3 27 directrix: x = 3 + , or x = 8 8
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[8.1]
588
Chapter 8: Topics in Analytic Geometry
9 x 2 + 4 y 2 + 36 x − 8 y + 4 = 0
7.
2
[8.2]
8.
11x 2 − 25 y 2 − 44 x − 50 y − 256 = 0
2
2
9( x + 4 x) + 4( y − 2 y ) = −4 2
11( x − 4 x) − 25( y + 2 y ) = 256
2
2
9( x + 4 x + 4) + 4( y − 2 y + 1) = −4 + 36 + 4
11( x − 4 x + 4) − 25( y 2 + 2 y + 1) = 256 + 44 − 25
9( x + 2) 2 + 4( y − 1)2 = 36
11( x − 2)2 − 25( y + 1)2 = 275
( x + 2)2 ( y − 1)2 + =1 4 9
( x − 2)2 ( y + 1)2 − =1 25 11
ellipse center: (−2, 1) vertices: (−2, 1 ± 3) = (−2, 4), ( − 2, − 2)
hyperbola center: (2, − 1) vertices: (2 ± 5, − 1) = (7, − 1), ( − 3, − 1) foci: (2 ± 6, − 1) = (8, − 1), ( − 4, − 1)
foci: (−2, 1 ± 5) = (−2, 1 + 5), ( − 2, 1 − 5)
asymptotes: y + 1 = ±
9.
4 x 2 − 9 y 2 − 8 x + 12 y − 144 = 0
[8.3]
4 4( x − 2 x) − 9( y − y ) = 144 3 4 4⎞ ⎛ 4( x 2 − 2 x + 1) − ⎜ 9 y 2 − y + ⎟ = 144 + 4 − 4 3 9⎠ ⎝ 2
10.
2
( x − 1)2 − 36
1⎞ ⎛ 9( x 2 + 4 x + 4) + 16 ⎜ y 2 − y + ⎟ = 104 + 36 + 4 4⎠ ⎝ 2
1⎞ ⎛ 9( x + 2)2 + 16 ⎜ y − ⎟ = 144 2⎠ ⎝
2
(
16
)
[8.2]
2
9( x + 4 x) + 16( y − y ) = 104
2⎞ ⎛ 4( x − 1) − 9 ⎜ y − ⎟ = 144 3⎠ ⎝ y−2 3
11 ( x − 2) 5
9 x 2 + 16 y 2 + 36 x − 16 y − 104 = 0
2
2
[8.3]
2
(
y−1 ( x + 2)2 2 + 16 9
2
=1
)
2
=1
ellipse
hyperbola ⎛ 2⎞ center: ⎜1, ⎟ ⎝ 3⎠
1⎞ ⎛ center: ⎜ − 2, ⎟ 2⎠ ⎝
2⎞ ⎛ 2⎞ ⎛ 2⎞ ⎛ vertices: ⎜1 ± 6, ⎟ = ⎜ 7, ⎟ , ⎜ −5, ⎟ 3⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ 2⎞ ⎛ 2⎞ ⎛ 2⎞ ⎛ foci: ⎜1 ± 2 13, ⎟ = ⎜ 1 + 2 13, ⎟ , ⎜1 − 2 13, ⎟ 3 3 3⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 2 2 asymptotes: y − = ± ( x − 1) 3 3
1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ vertices: ⎜ −2 ± 4, ⎟ = ⎜ 2, ⎟ , ⎜ −6, ⎟ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ foci: ⎜ −2 ± 7, ⎟ = ⎜ −2 + 7, ⎟ , ⎜ −2 − 7, ⎟ 2 2 2⎠ ⎝ ⎠ ⎝ ⎠ ⎝
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Chapter Review
11.
589
4 x 2 + 28 x + 32 y + 81 = 0
x 2 − 6 x − 9 y + 27 = 0
12.
[8.1]
[8.1]
2
2
4( x + 7 x) = −32 y − 81
x − 6 x = 9 y − 27 2
49 ⎞ ⎛ 4 ⎜ x 2 + 7 x + ⎟ = −32 y − 81 + 49 4 ⎠ ⎝
x − 6 x + 9 = 9 y − 27 + 9 ( x − 3)2 = 9( y − 2)
2
7⎞ ⎛ 4 ⎜ x + ⎟ = −32( y + 1) 2⎠ ⎝
parabola
2
4p = 9 ⇒ p =
7⎞ ⎛ ⎜ x + ⎟ = −8( y + 1) 2⎠ ⎝
9 4
vertex: (3, 2) 9 ⎞ ⎛ 17 ⎞ ⎛ focus: ⎜ 3, 2 + ⎟ = ⎜ 3, ⎟ 4⎠ ⎝ 4⎠ ⎝ 9 1 directrix: y = 3 − , or y = − 4 4
parabola 4p = −8 ⇒ p = −2 ⎛ 7 ⎞ vertex: ⎜ − , − 1⎟ ⎝ 2 ⎠ ⎛ 7 ⎞ ⎛ 7 ⎞ focus: ⎜ − , − 1 − 2 ⎟ = ⎜ − , − 3 ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ directrix: y = 1
13.
14.
2a = 7 − (−3) = 10 [8.2]
a=5
2a = 4 − (−2) = 6
[8.3]
15.
a=3 a2 = 9 e= c = 4 a 3 c=4 3 3 c=4 c2 = a 2 + b2 16 = 9 + b 2 b2 = 7
2
a = 25 2b = 8 b=4 b 2 = 16 center (2, 3) ( x − 2)2 ( y − 3)2 + =1 25 16
center (−2, 2), c = 3 2a = 4 a=2
[8.3]
a2 = 4 c2 = a2 + b2 9 = 4 + b2 b2 = 5
( x + 2)2 ( y − 2)2 − =1 4 5
center (1, 1) ( x − 1) 2 ( y − 1)2 − =1 9 7
16.
⎛ 6 + 2 −3 − 3 ⎞ (h, k ) = ⎜ , ⎟ = (4, − 3) 2 ⎠ ⎝ 2 p = 2−4 p = −2 4 p = −8 ( y + 3)2 = −8( x − 4)
[8.1]
17.
( x − h) 2 = 4 p ( y − k )
or
(3 − 0)2 = 4 p(4 + 2) 9 = 4 p(6) 3 p= 8
( y − k ) 2 = 4 p ( x − h) (4 + 2)2 = 4 p(3 − 0) 36 = 4 p(3) p=3
Thus, there are two parabolas that satisfy the given conditions: 3 x 2 = ( y + 2) or (y + 2)2 = 12 x 2
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[8.1]
590
18.
Chapter 8: Topics in Analytic Geometry
center (−2, − 1)
a = 6 and the transverse axis is on the x-axis.
19.
c=2 c 2 e= = a 3 2 2 = a 3 a=3
±
c2 = a 2 − b2 4 = 9 − b2
( x − h) 2 = 4 p ( y − k ) (1 − h)2 = 4 p (0 − k ) (2 − h)2 = 4 p (1 − k ) (0 − h) 2 = 4 p(1 − k )
20.
b 1 =± a 9 b 1 = 6 9 2 b= 3 2
In the last two equations, by substitution: (2 − h) 2 = (0 − h)2 4 − 4h + h 2 = h 2
2
x y − = 1 [8.3] 36 4 / 9
b2 = 5
4 − 4h = 0 4h = 4 h =1
( x + 2)2 ( y + 1)2 + = 1 [8.2] 9 5
Thus: (1 − 1)2 = 4 p(0 − k ) 0 = −4 pk k =0 (2 − 1)2 = 4 p (1 − k ) 1 = 4 p(1) 1 p= 4 The equation is y = ( x − 1)2
21.
A = 11, B = −6, C = 19, D = 0, 11 − 19 −8 4 cot 2α = = = −6 −6 3 16 = csc 2 2α = 1 + cot 2 2α = 1 + 9 5 csc 2α = 3
E = 0, F = −40
22.
5
2
25 9
1 + cot 2 2α = csc2 2α
cosα =
1 = csc2 2α csc 2α = 1 Thus, sin 2α = 1 so 2α = 90° or α = 45°.
3 10 10
2
Therefore, cos α = 2 , sin α = 2 . 2
⎛ 3 10 ⎞ ⎛ 10 ⎞ ⎛ 3 10 ⎞ A′ = 11⎜⎜ ⎟⎟ − 6 ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ + 19 10 ⎝ ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ 99 18 19 100 = − + = = 10 10 10 10 10 B′ = 0 2
⎛ 10 ⎞ ⎛ 10 ⎞ ⎛ 3 10 ⎞ C ′ = 11⎜⎜ ⎟⎟ + 6 ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ + 19 10 ⎝ ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ 11 18 171 200 = + + = = 20 10 10 10 10 F′ = F
⎛ 10 ⎞ ⎜⎜ ⎟⎟ ⎝ 10 ⎠
⎛ 3 10 ⎞ ⎜⎜ ⎟⎟ ⎝ 10 ⎠
10( x′)2 + 20( y′)2 − 40 = 0 or (x′) 2 + 2( y′)2 − 4 = 0
The graph is an ellipse.
[8.4]
3−3 =0 6
cot 2α =
5
10
A = 3, B = 6, C = 3, D = −4, E = 5, F = −12
(2α is in quadrant I.)
Thus, sin 2α = 3 and cos 2α = 4 . 1− 4 5 = 10 sin α =
[8.1]
2
2
⎛ 2⎞ ⎛ 2⎞ 3 6 3 ⎜⎜ ⎟⎟ + 3 ⎜⎜ ⎟⎟ = + + = 6 2 2 2 2 2 ⎝ ⎠ ⎝ ⎠
2
2⎞ ⎛ 2⎞ ⎛ 2⎞ 3 6 3 ⎟ ⎜ ⎟ + 3⎜ ⎟ = − + =0 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ 2 2 2 ⎛ 2⎞ 2 ⎜⎜ ⎟⎟ = 2 2 ⎝ ⎠
⎛ 2⎞ ⎛ 2⎞ A′ = 3 ⎜⎜ ⎟⎟ + 6 ⎜⎜ ⎟⎟ 2 ⎝ ⎠ ⎝ 2 ⎠ B′ = 0 ⎛ 2⎞ ⎛ C ′ = 3 ⎜⎜ ⎟⎟ − 6 ⎜⎜ 2 ⎝ ⎠ ⎝ ⎛ 2⎞ D′ = −4 ⎜⎜ ⎟⎟ + 5 ⎝ 2 ⎠
⎛ 2⎞ E′ = 4 ⎜⎜ ⎟⎟ + 5 ⎝ 2 ⎠ F ′ = −12 6( x′)2 +
2
2
⎛ 2⎞ 9 2 ⎜⎜ ⎟⎟ = 2 ⎝ 2 ⎠
2 9 2 x′ + y′ − 12 = 0 2 2
The graph is a parabola. [8.4] Copyright © Houghton Mifflin Company. All rights reserved.
2
2
Chapter Review
23.
591
A = 1, B = 2 3 , C = 3, D = 8 3 , E = −8, F = 32
cot 2α =
24.
1− 3 1 =− . Thus 90º < 2α < 180º. 2 3 3
0−0 = 0. Thus α = 45o. 1 2 2 Therefore, sin α = and cos α = . 2 2 cot 2α =
cot 2 2α + 1 = csc 2 2α 1 + 1 = csc 2 2α 3 4 = csc 2 2α , or csc 2α = 2 3 3
⎛ 2⎞ ⎛ 2⎞ ⎟ ⎜ ⎟+0 = 1 A′ = 0 + 1 ⎜ ⎜ 2 ⎟ ⎜ 2 ⎟ 2 ⎝ ⎠ ⎝ ⎠ ′ B =0
Therefore, sin 2α = 3 and cos 2α = − 1 . 2
2
Since 2α is in quadrant II, cos α =
1 + (−1/ 2) 1 = 2 2
()
2 A′ = 1 1 + 2 3 2
() 1 2
C′ = 0 −
1 − (−1/ 2) 3 = 2 2
sinα =
1 2 2 +0 = − 2 2 2
⎛ 2⎞ ⎛ 2⎞ D′ = −1 ⎜⎜ ⎟⎟ − 1 ⎜⎜ 2 ⎟⎟ = − 2 2 ⎝ ⎠ ⎝ ⎠
2
⎛ 3⎞ ⎛ 3⎞ 1 6 9 ⎜ 2 ⎟ + 3⎜ 2 ⎟ = 4 + 4 + 4 = 4 ⎝ ⎠ ⎝ ⎠
2 2 + ( −1) =0 2 2 F ′ = −1 E′ =
B′ = 0 2 2 C ′ = 1 ⎛⎜ 3 ⎞⎟ − 2 3 1 ⎛⎜ 3 ⎞⎟ + 3 1 = 3 − 6 + 3 = 0 2 2 2 2 4 4 4 ⎝ ⎠ ⎝ ⎠
()
A = 0, B = 1, C = 0, D = −1, E = −1, F = −1
()
()
1 1 ( x′) 2 − ( y′) 2 − 2 x′ − 1 = 0 2 2
D′ = 8 3 1 − 8 ⎛⎜ 3 ⎞⎟ = 0 2 ⎝ 2 ⎠ E′ = 8 3 ⎜⎛ 3 ⎟⎞ − 8 1 = −12 − 4 = −16 2 ⎝ 2 ⎠ ′ F = 32
()
The equation is a hyperbola. [8.4]
4( x′) 2 − 16 y′ + 32 = 0 or (x′) 2 − 4 y′ + 8 = 0
The graph is a parabola. 25.
[8.4] 26.
27.
[8.5] 29.
28.
[8.5] 30.
vertical line through (3, 0).
[8.5] 31.
[8.5]
horizontal line through (0, 4)
32.
[8.5]
[8.5] [8.5]
33.
[8.5]
34.
y 2 = 16 x
35.
[8.5]
( r sin θ ) 2 = 16( r cos θ ) r 2 sin 2 θ = 16r cos θ
[8.5] x2 + y2 + 4 x + 3 y = 0
36. 2
r sin 2 θ = 16 cos θ
[8.5]
2
( r cos θ ) + (r sin θ ) + 4( r cos θ ) + 3( r sin θ ) = 0
[8.5]
37.
3x – 2y = 6 [8.6] 3r cos θ – 2r sin θ = 6
r 2 (cos 2 θ + sin 2 θ ) + 4r cos θ + 3r sin θ = 0 r + 4 cos θ + 3sin θ = 0
Copyright © Houghton Mifflin Company. All rights reserved.
592
38.
Chapter 8: Topics in Analytic Geometry
xy = 4 [8.5] ( r cos θ )( r sin θ ) = 4
39.
r 2cosθ sinθ = 4
4 1 − cos θ r − r cos θ = 4 r=
[8.5]
x2 + y 2 − x = 4
r 2 (2)cosθ sinθ = 4(2)
x2 + y 2 = x + 4
r 2 sin 2θ = 8
x 2 + y 2 = x 2 + 8 x + 16 y 2 = 8 x + 16
r 2 = 3r cos θ − 4r sin θ
40. 2
r 2 = cos 2θ
41.
[8.5]
2
2
[8.5]
r = cos θ − sin 2θ
x + y = 3x − 4 y x 2 − 3x + y 2 + 4 y = 0
2
r 4 = r 2cos2θ − r 2sin 2θ (r 2 )2 = r 2cos 2θ − r 2sin 2θ ( x 2 + y 2 )2 = x 2 − y 2 x4 + 2 x2 y 2 + y 4 = x2 − y 2 x4 + y 4 + 2 x2 y 2 − x2 + y 2 = 0
42.
θ = 1 [8.5] tan θ = tan 1 y ≈ 1.5574 x y = 1.5574 x
43.
r=
4 3 − 6 sin θ
44.
r=
[8.6]
46.
r=
6 4 + 3 sin θ
47.
45.
r=
2 2 − cos θ
[8.6]
x = 4t − 2, y = 3t + 1, t ∈ R 4t = x + 2
x+2 t= 4 y = 3t + 1
[8.6]
2 1 + cos θ
48.
[8.6]
x = 1 − t 2 , y = 3 − 2t 2 , t ∈ R
t2 = −x + 1 y = 3 − 2t 2 y = 3 − 2( − x + 1) y = 3 + 2x − 2
⎛ x+2⎞ y=3 ⎜ ⎟ +1 ⎝ 4 ⎠ 3 5 y= x+ 4 2
y = 2 x + 1, x ≤ 1
[8.7]
[8.7]
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Review
49.
593
y = 3 cos t
x = 4 sin t
0 ≤ t < 2π
50.
x 2 = sec2 t
1 y = cos t 3
1 x = sin t 4
x = sec t
1 2 y = cos t 2 9
1 2 x = sin 2 t 16
Using the trignometric identity sin 2 t + cos 2 t = 1, we have 1 2 1 2 x + y =1 16 9
y = 4 tan t 1 y = tan t 4 2 ⎛ 1 y ⎞ = tan 2 t ⎜ ⎟ ⎝4 ⎠ 1 y 2 = tan 2 t 16
−
π 2
π 2
Using the trignometric identity 1 + tan 2 t = sec 2 t , we have y2 = x2 16 2 x2 − y = 1 1 16 1+
x2 y 2 + =1 19 9
[8.7] [8.7] 51.
x=
2 t ⎛1⎞ y = −2 ⎜ ⎟ ⎝t⎠ y = −2 x , x > 0
1 t
y=−
t>0
52.
y = 2 − sin t 2 − y = sin t −( y − 2) = sin t
x = 1 + cos t x − 1 = cos t ( x − 1)2 = cos 2 t
0 ≤ t < 2π
( −( y − 2) )2 = sin 2 t ( y − 2) 2 = sin 2 t
Using the trigonometric identity cos 2 t + sin 2 t = 1 we have ( x − 1) 2 + ( y − 2) 2 = 1
[8.7] [8.7] 53.
x = t , y = 2 −t , t ≥ 0
t=x
54.
2 2
y = 2− x , x ≥ 0 [8.7]
[8.7] 55.
Graph y =
−(4 x + 5) ± (4 x + 5) 2 − 8( x 2 − 2 x + 1) . 4
56.
[8.5] [8.4] Copyright © Houghton Mifflin Company. All rights reserved.
594
Chapter 8: Topics in Analytic Geometry
57.
Graph in parametric mode. Use the TRACE feature to determine that the maximum height (to the nearest foot) of 278 feet is attained when t ≈ 4.17 seconds. [8.7]
....................................................... QR1. They appear to be the same.
Quantitative Reasoning QR2. They appear to be the same.
QR3. 5.2 units
....................................................... 1.
3.
vertex: (0, 0) [8.1]
y = 1 x2 8 x2 = 8 y 4p = 8 p=2
Chapter Test 2.
focus: (0, 2) directrix: y = −2 [8.2]
25 x 2 − 150 x + 9 y 2 + 18 y + 9 = 0
[8.2]
2b = 6
4.
b = 3
25( x 2 − 6 x + 9) + 9( y 2 + 2 y + 1) = −9 + 255 + 9 25( x − 3)2 + 9( y + 1) 2 = 225
a 2 = 9 + 36 = 45
( x − 3) 2 ( y + 1)2 + =1 9 25
center = (0, −3) x 2 ( y + 3)2 + =1 45 9
a=5 b=3 c=4 vertices: (3, 4), (3, −6) foci: (3, 3), (3, −5) 5.
c = 6 [8.2]
6.
x2 y2 − =1 36 64
7.
( y + 1)2 ( x + 3)2 − =1 4 16
vertices: (6, 0), (−6, 0) [8.3]
foci: (10, 0), (−10, 0) asymptotes: y = ±
8.
x 2 − 4 xy − 5 y 2 + 3x − 5 y − 20 = 0 [8.4] A = 1 B = −4 C = −5 D = 3 E = −5 F = −20 1 − (−5) cot 2α = A − C = =−3 2α is in quadrant II. B −4 2 tan 2α = − 2 3
9.
4 x 3
[8.3] [8.3] A=8
B=5 2
C=2 2
D = −10
E =5
F =4
Since B − 4 AC = (5) − 4(8)(2) = −39 < 0, the graph is an ellipse. [8.4]
( 3)
2α = tan −1 − 2
2α ≈ (−33.69° + 180°) α ≈ 73.15° Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Test
10.
595
r = x2 + y 2
P (1, − 3)
r sin θ = y 2sin θ = − 3 sin θ = − 3 2
r cosθ = x 2cosθ = 1 cosθ = 1 2
r = 12 + (− 3)2 r=2
θ is in quadrant IV. θ = 300°
P(1, − 3) = P(2, 300°) [8.5] 11.
r = 4 cos θ
12.
r = 3(1 − sin θ )
13.
[8.5] 14.
17.
y = r sin θ x = r cos θ 7π 7π y = 5sin x = 5cos 3 3 5 5 3 x= y= 2 2 The rectangular coordinates of the point are (5 / 2, 5 3 / 2). [8.5]
r − r cos x = 4
15.
16.
r=
x2 + y 2 = x + 4 y 2 − 8 x − 16 = 0
18.
4 [8.6] 1 + sin θ r + r sin θ = 4
x2 + y 2 + y = 4
x 2 + y 2 = x 2 + 8 x + 16
( x + 3)2 = t 2
x 2 + y 2 = 16 − 8 y + y 2 x 2 + 8 y − 16 = 0
x = 4sin θ sin θ = x / 4
y = cosθ + 2 cosθ = y − 2
sin 2 θ + cos 2 θ = 1
2
2
⎛x⎞ 2 ⎜ ⎟ + ( y − 2) = 1 ⎝4⎠
2
2( x + 3) = y ( x + 3)2 =
[8.5]
x2 + y2 − x = 4
x+3=t 2( x + 3) = 2t
[8.5]
[8.5]
x =t −3
2
r = 2 sin 4θ
1 y 2
x 2 (y − 2) 2 + =1 16 1
[8.7]
[8.7] 19.
20.
[8.7]
The projectile will travel 256 3 feet ≈ 443 feet. [8.7]
Copyright © Houghton Mifflin Company. All rights reserved.
596
Chapter 8: Topics in Analytic Geometry
....................................................... 1.
x 4 − 2 x 2 − 8 = 0 [1.4]
Cumulative Review 2.
Let u = x 2 . u 2 − 2u − 8 = 0 (u − 4)(u + 2) = 0 u=4 or 2
u = −2 2
2 3 2( x + 2) − 3( x − 1) − = [P.5] x −1 x + 2 ( x − 1)( x + 2) 2 x + 4 − 3x + 3 = ( x − 1)( x + 2) −x + 7 = ( x − 1)( x + 2)
x =4 x = −2 x=±2 x = ±i 2 The solutions are 2, − 2, i 2, − i 2.
3.
2⎤ ⎡ 2⎤ ⎡ f (2 + h) − f (2) ⎣⎢1− (2 + h) ⎦⎥ − ⎣⎢1− (2) ⎦⎥ = h h
=
( f o g )( x) = f [ g ( x)]
[2.6]
= f [2 − x 2 ] = 3(2 − x 2 ) + 2 = 6 − 3x2 + 2
2
= −3 x 2 + 8 ( f o g )(−3) = −3(−3)2 + 8 = −27 + 8 = −19
By the Linear Factor Theorem, since the polynomial is of degree 6, there are 6 complex number solutions to
6.
x 6 + 2 x 4 − 3 x3 − x 2 + 5 x − 7 = 0 . [3.4]
8.
4.
1− 4 − 4h − (h)2 −1+ 4 h
= −4h − h h = −4 − h
5.
[2.6]
d = (−5 − (−3)) 2 + (4 − (−1)) 2
2 − (−4) 6 3 [2.3] = =− 2 −3 − 1 −4 3 y − ( −4) = − ( x − 1) 2 3 3 y+4=− x+ 2 2 3 5 y=− x− 2 2
m=
9.
7.
x = –3, y = 2 [3.5]
10.
log 2 ( x + 3) − log 2 ( x) = 2 [4.5] ( x + 3) log 2 =2 x ( x + 3) 22 = x 4x = x +3 3x = 3 x =1
= (−2) 2 + (5) 2 = 4 + 25 = 29 [2.1] [4.2] 11.
12.
f ( x) = 2 x − 8 x = 2y − 8 x + 8 = 2y 1x+4= y
[4.1]
2
f −1( x) = 1 x + 4 2
[2.5]
Copyright © Houghton Mifflin Company. All rights reserved.
Cumulative Review
13.
597
2i
–2i
1
1
1
2i 1 + 2i
–8 –4 + 2i –12 + 2i
1 + 2i –2i 1
–12 + 2i –2i –12
1 1
4 –4 – 24i –24i
–48 48 0
–24i 24i 0
x 2 + x − 12 = ( x − 3)( x + 4) = 0 x = 3, x = −4 The remaining zeros are −2i and − 4. [3.4]
14.
17.
3( − x)
= −3x = − f ( x) (− x) + 1 x 2 + 1 odd [2.5] f (− x) =
2
15.
120o = 120o ⎛⎜ π o ⎞⎟ = 2π ⎝ 180 ⎠ 3
f ( x) = 3cos 4 x [5.5]
(
cos x = 0 x = π , 3π 2 2
[6.6]
)
sin x − 1 = 0 2 sin x = 1 2
tan 40o = a [5.2] 15 a = 15sin 40o a = 12.6 cm
sin 2α tan α = (2sin α cos α ) tan α = 2sin α cos α sin α cos α = 2sin 2 α
20.
v ⋅ w = (3i − 4 j) ⋅ (4i + j) [7.3] = 3(4) + (−4)(1) = 12 − 4 =8
2 amplitude: 3 sin x cos x − 1 cos x = 0 2 cos x sin x − 1 = 0 2
16.
18.
period: π
19.
[5.1]
x = π , 5π 6 6
Copyright © Houghton Mifflin Company. All rights reserved.
[6.3]
Chapter 9
Systems of Equations and Inequalities Section 9.1 1.
2.
⎧2 x − 3 y = 16 ⎨ x=2 ⎩
3.
⎧3 x − 2 y = −11 ⎨ y= 1 ⎩
⎧3x + 4 y = 18 ⎨ y = −2 x + 3 ⎩
3x + 4( −2 x + 3) = 18
2( 2) − 3 y = 16 − 3 y = 12 y = −4
3x − 2(1) = −11
The solution is (2, −4).
The solution is (−3, 1).
3x − 8 x + 12 = 18
3 x = −9 x = −3
−5 x = 6 x=−
6 5
⎛ 6⎞ y = −2 ⎜ − ⎟ + 3 ⎝ 5⎠ 27 y= 5
⎛ 6 27 ⎞ The solution is ⎜ − , ⎟. ⎝ 5 5 ⎠ 4.
5.
⎧5 x − 4 y = −22 ⎨ y = 5x − 2 ⎩
6.
⎧−2 x + 3 y = 6 ⎨ x = 2y − 5 ⎩
8(3 y + 15) + 3 y = −7
−2(2 y − 5) + 3 y = 6 − 4 y + 10 + 3 y = 6 − y = −4 y=4
5 x − 4(5 x − 2) = −22 5 x − 20 x + 8 = −22 − 15 x = −30 x=2 y = 5(2) − 2 y=8
24 y + 120 + 3 y = −7 27 y = −127 y=−
127 27
⎛ 127 ⎞ x = 3 ⎜− ⎟ + 15 ⎝ 27 ⎠ 8 x= 9
x = 2( 4) − 5 x=3
The solution is (2, 8).
⎧8 x + 3 y = −7 ⎨ x = 3 y + 15 ⎩
The solution is (3, 4).
127 ⎞ ⎛8 The solution is ⎜ , − ⎟. 27 ⎠ ⎝9 7.
⎧6 x + 5 y = 1 ⎨ ⎩ x − 3y = 4
(1) ( 2)
8.
⎧−3 x + 7 y = 14 ⎨ ⎩ 2 x − y = −13
(1) ( 2)
Solve (2) for x : x = 3 y + 4
Solve (2) for y : y = 2 x + 13
6(3 y + 4) + 5 y = 1 18 y + 24 + 5 y = 1
x =1
−3 x + 7(2 x + 13) = 14 − 3x + 14 x + 91 = 14 11x = −77 x = −7 y = 2( −7) + 13 y = −1
The solution is (1, −1).
The solution is (−7, −1).
23 y = −23 y = −1 x = 3( −1) + 4
9.
⎧7 x + 6 y = −3 ⎪ 2 ⎨ y = x−6 ⎪ 3 ⎩
(1) (2)
⎛2 ⎞ 7 x + 6⎜ x − 6 ⎟ = −3 ⎝3 ⎠ 7 x + 4 x − 36 = −3 11x = 33 x=3 2 y = (3) − 6 3 y = −4 The solution is (3, −4).
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.1
10.
599
11.
(1) ⎧9 x − 4 y = 3 ⎪ 4 ⎨ x = y + 3 (2) ⎪ 3 ⎩ ⎛4 ⎞ 9 ⎜ y + 3⎟ − 4 y = 3 ⎝3 ⎠ 12 y + 27 − 4 y = 3 8 y = −24 y = −3
12.
⎧ y = 4x − 3 ⎨ ⎩ y = 3x − 1 4 x − 3 = 3x − 1
5x + 1 = 4x − 2
x=2
x = −3
y = 4(2) − 3 y=5
4 ( −3) + 3 3 x = −1
y = 5(−3) + 1 y = −14 The solution is (−3, −14).
The solution is (2, 5).
x=
⎧ y = 5x + 1 ⎨ ⎩ y = 4x − 2
The solution is (−1, −3). 13.
14.
⎧ y = 5x + 4 ⎨ ⎩ x = −3 y − 4
y y 16 y y
= 5(−3 y − 4) + 4 = −15 y − 20 + 4 = −16 = −1
15.
⎧ y = −2 x − 6 ⎨ ⎩ x = −2 y − 2
Solve (1) for x and substitute into (2). 3x = 4 y + 2 4y + 2 x= 3
y = −2(−2 y − 2) − 6 y = 4y + 4 − 6 − 3 y = −2 y=
x = −3( −1) − 4 x = −1
2 3
⎛ 4y + 2 ⎞ 4⎜ ⎟ + 3 y = 14 ⎝ 3 ⎠ 16 y + 8 + 9 y = 42 25 y = 34 34 y= 25 4 ⎛ 34 ⎞ 2 x = ⎜ ⎟+ 3 ⎝ 25 ⎠ 3 62 x= 25 ⎛ 62 34 ⎞ The solution is ⎜ , ⎟. ⎝ 25 25 ⎠
2 x = −2( ) − 2 3 10 x=− 3
The solution is (−1, −1).
⎛ 10 2 ⎞ The solution is ⎜ − , ⎟ . ⎝ 3 3⎠
16.
⎧6 x + 7 y = −4 ⎨ ⎩2 x + 5 y = 4
17.
(1) ( 2)
Solve (2) for x and substitute into (1). 2x + 5 y = 4 4 − 5y x= 2 ⎛ 4 − 5y ⎞ 6⎜ ⎟ + 7 y = −4 ⎝ 2 ⎠ 3(4 − 5 y ) + 7 y = −4 12 − 15 y + 7 y = −4 − 8 y = −16 y=2 4 − 5(2) 2 x = −3 x=
⎧3 x − 3 y = 5 ⎨ ⎩4 x − 4 y = 9
(1) ( 2)
⎧3 x − 4 y = 2 (1) ⎨ ⎩4 x +3 y = 14 (2)
18.
⎧3 x − 4 y = 8 ⎨ ⎩6 x − 8 y = 9
(1) ( 2)
Solve (1) for x and substitute into (2). 3x − 3 y = 5 3y + 5 x= 3 ⎛ 3y + 5 ⎞ 4⎜ ⎟ − 4y = 9 ⎝ 3 ⎠ 12 y + 20 − 12 y = 27 20 = 27
Solve (1) for x and substitute into (2). 3x − 4 y = 8 4y + 8 x= 3 ⎛ 4y + 8 ⎞ 6⎜ ⎟ − 8y = 9 ⎝ 3 ⎠ 8 y + 16 − 8 y = 9 16 = 9
The system of equations is inconsistent and has no solution.
The system of equations is inconsistent and has no solution.
The solution is (−3, 2).
Copyright © Houghton Mifflin Company. All rights reserved.
600
19.
22.
Chapter 9: Systems of Equations and Inequalities
20.
⎧4 x + 3 y = 6 ⎪ 4 ⎨ y = − x+2 ⎪ 3 ⎩
⎛ 5 ⎞ 5 x + 2⎜ − x + 1⎟ = 2 ⎝ 2 ⎠ 5x − 5x + 2 = 2 0=0
The system of equations is dependent. 4 Let x = c and y = − c + 2. 3 4 ⎛ ⎞ The solutions are ⎜ c, − c + 2 ⎟ . 3 ⎝ ⎠
The system of equations is dependent. 5 Let x = c and y = − c + 1. 2 5 ⎛ ⎞ The solutions are ⎜ c, − c + 1⎟ . 2 ⎝ ⎠
⎧3 x + 4 y = −5 ⎨ ⎩ x − 5 y = −8
23.
(1) ( 2)
13 x
⎧4 x + 7 y = 21 ⎨ ⎩ 5 x − 4 y = −12
(1)
19 y = 19 y =1
x=0 The solution is (0, 3).
−10 x + 6 y = 0 10 x − 6 y = 0
26.
(1) ( 2)
− 2 times (1) (2)
0=0 5 x − 3c = 0 3c x= 5
⎛ 3c ⎞ The solution is ⎜ , c ⎟ . ⎝ 5 ⎠
⎧3 x + 2 y = 0 ⎨ ⎩2 x + 3 y = 0
3 x − 8 y = −6 −10 x + 8 y = 20
(1) 2 times (2)
( 2)
6x + 4 y = 0
2 times (1)
−6 x − 9 y = 0
− 3 times (2)
x=0 The solution is (0, 0).
(1)
= 14 x = −2
3(−2) − 8 y = −6 −8 y = 0 y=0 The solution is (−2, 0). (1)
−5 y = 0 y=0 3x + 2(0) = 0
= 26 x=2
( 2)
−7 x
x = −3 The solution is (−3, 1).
3 times (1) (2)
⎧ 3x − 8 y = −6 ⎨ ⎩− 5 x + 4 y = 10
51y = 153 y=3 4 x + 7(3) = 21
⎧ 5x − 3 y = 0 ⎨ ⎩10 x − 6 y = 0
24.
( 2)
x − 5(1) = −8
(1) (2)
3(2) − y = 10 6 − y = 10 y = −4 The solution is (2, −4).
20 x + 35 y = 105 5 times (1) − 20 x + 16 y = 48 −4 times (2)
(1) − 3 times (2)
⎧ 3 x − y = 10 ⎨ ⎩ 4 x + 3 y = −4 9 x − 3 y = 30 4 x + 3 y = −4
⎛ 4 ⎞ 4 x + 3⎜ − x + 2 ⎟ = 6 ⎝ 3 ⎠ 4x − 4x + 6 = 6 0=0
3 x + 4 y = −5 −3 x + 15 y = 24
25.
21.
⎧5 x + 2 y = 2 ⎪ 5 ⎨ y = − x +1 ⎪ 2 ⎩
27.
⎧6 x + 6 y = 1 ⎨ ⎩4 x + 9 y = 4 12 x + 12 y = 2
(1) ( 2) 2 times (1)
−12 x − 27 y = −12 −3 times (2)
−15y = −10 2 y= 3 ⎛2⎞ 6x + 6⎜ ⎟ = 1 ⎝3⎠ 6 x = −3 1 x=− 2 ⎛ 1 2⎞ The solution is ⎜ − , ⎟ . ⎝ 2 3⎠
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.1
28.
601
⎧ 4x + 5 y = 2 ⎨ ⎩8 x − 15 y = 9 12 x + 15 y = 6 8 x − 15 y = 9
20x
29.
(1) ( 2) 3 times (1) (2)
= 15 3 x= 4
30.
⎧3 x + 6 y = 11 ⎨ ⎩2 x + 4 y = 9
( 2)
⎧4 x − 2 y = 9 (1) ⎨ ⎩ 2 x − y = 3 ( 2)
6 x + 12 y = 22 −6 x − 12 y = −27
2 times (1) − 3 times (2)
4x − 2 y = 9 −4 x + 2 y = 3
(1)
0=3
0 = −5
The system of equations is inconsistent and has no solution.
The system of equations is inconsistent and has no solution.
⎛3⎞ 4 ⎜ ⎟ + 5y = 2 ⎝4⎠ 5 y = −1 1 y=− 5
(1) − 2 times (2)
1⎞ ⎛3 The solution is ⎜ , − ⎟ . 4 5⎠ ⎝ 31.
1 ⎧5 ⎪⎪ 6 x − 3 y = −6 ⎨ ⎪1 x + 2 y = 1 ⎪⎩ 6 3
32.
(1) (2)
5 2 x − y = −12 3 3 1 2 x+ y = 1 6 3 11 x = −11 6 x = −6 1 2 x + y =1 6 3 2 1 (−6) + y = 1 3 6 2 y=2 3 y=3
The solution is (−6, 3).
2 times (1) ( 2)
2 ⎧3 ⎪⎪ 4 x + 5 y = 1 ⎨ ⎪ 1 x − 3 y = −1 ⎪⎩ 2 5 ⎧15 x + 8 y = 20 ⎨ ⎩ 5 x − 6 y = −10
(1) (2) 20 times (1) 10 times (2)
15 x + 8 y = 20 − 15 x + 18 y = 30 26 y = 50 25 y= 13
33.
1 ⎧3 ⎪⎪ 4 x + 3 y = 1 ⎨ ⎪1 x + 2 y = 0 ⎪⎩ 2 3 ⎧9 x + 4 y = 12 ⎨ ⎩3 x + 4 y = 0 9 x + 4 y = 12 − 3x − 4 y = 0 6x
5 x = −10 + 150 13 20 5x = 13 x= 4 13 ⎛ 4 25 ⎞ The solution is ⎜ , ⎟. ⎝ 13 13 ⎠
(1) −1 times (2)
= 12
x=2 3(2) + 4 y = 0 4 y = −6
⎛ ⎞ 5 x − 6 ⎜ 25 ⎟ = −10 ⎝ 13 ⎠ 5 x = −10 + 150 13
(1) ( 2)
y=−
3 2
3⎞ ⎛ The solution is ⎜ 2,− ⎟ . 2⎠ ⎝
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602
34.
Chapter 9: Systems of Equations and Inequalities
⎧3 x− 2 y =7 ⎪5 3 ⎨2 5 ⎪ x− y =7 ⎩5 6 ⎧ 9 x − 10 y = 105 ⎨ ⎩12 x − 25 y = 210
35.
6 3x − 9 y =
(1) ( 2)
36 x − 40 y = 420
y=
3
2 3 x − 3(3) = 3 3x = 6 x=2 3
The solution is ( 2 3, 3) . 37.
(1) (2) 4 times (1)
9 x + 12 5 y = 51
3 times (2)
{ ππ
3 x−4 y =6 2 x +3 y = 5
(1) (2)
6π x − 8 y = 12 −6π x − 9 y = −15
2 times (1) −3 times (2)
−17 y = −3
= −25 x = −1 3(−1) + 4 5 y = 17 25 x
y=
4 5 y = 20 5y = 5 5 y= 5
3 17
9π x − 12 y = 18
3 times (1)
8π x + 12 y = 20
4 times (2)
17π x
= 38 x=
y= 5
38.
3 times (1)
2 3 x = 12
16 x − 12 5 y = −76
The solution is ( −1,
9
− 13 y = −39
4 times (1)
35 y = −210 y = −6 9 x − 10(−6) = 105 9 x = 45 x=5 The solution is (5, −6). ⎪⎧4 x − 3 5 y = −19 ⎨ ⎩⎪3x + 4 5 y =17
(1) (2)
− 6 3 x − 4 y = −48 −2 times (2)
−36 x + 75 y = −630 − 3 times (2)
36.
⎪⎧2 3x − 3 y = 3 ⎨ ⎩⎪3 3 x + 2 y = 24
38 17π
The solution is ⎛⎜ 38 , 3 ⎞⎟ . ⎝ 17π 17 ⎠
5) .
⎧⎪3 2 x − 4 3 y = −6 ⎨ ⎪⎩2 2 x + 3 3 y =13
⎧2 x − 5πy = 3 ⎨ ⎩3x + 4πy = 2
(1)
8 x − 20π y = 12
4 times (1)
6 2 x − 8 3 y = −12
2 times (1)
15 x + 20π y = 10
5 times (2)
−6 2 x − 9 3 y = −39
− 3 times (2)
= 22 22 x= 23 6 x − 15π y = 9
39.
( 2)
23 x
−6 x − 8π y = −4 −23π y = 5
3 times (1) −2 times (2)
5 23π 5 ⎛ 22 The solution is ⎜ , − 23π ⎝ 23
(1) (2)
−17 3 y = −51 3 y= 3 y= 3
9 2 x − 12 3 y = −18 8 2 x + 12 3 y = 52
y=−
17 2 x ⎞ ⎟. ⎠
= 34 2 x= 2 x= 2
The solution is
(
2,
3) .
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Section 9.1
40.
603
⎧⎪2 2 x + 3 5 y = 7 (1) ⎨ ⎪⎩3 2 x − 5 y = −17 (2)
41.
6 2 x + 9 5 y = 21
3 times (1)
−6 2 x + 2 5 y = 34
− 2 times (2)
Solve the system by substitution. 20 p − 2000 = −4 p +1000 24 p = 3000 p =125 The solution is $125.
11 5 y = 55 5 y= 5 y= 5 2 2x + 3 5 y = 7
(1)
9 2 x − 3 5 y = −51
3 times (2)
11 2 x = −44 −4 x= 2 x = −2 2
(
The solution is −2 2, 42.
)
5 .
Solve the system by substitution.
43.
25 p − 500 = −7 p +1100 32 p =1600 p = 50
Rate of plane with the wind: r + w Rate of plane against the wind: r − w r ⋅t = d ⎧(r + w) ⋅ 3 = 450 ⎨ ⎩(r − w) ⋅ 5 = 450
The solution is $50.
r + w = 150 r − w = 90 2r
= 240
r = 120 120 + w = 150 w = 30
Rate of plane = 120 mph. Rate of wind = 30mph. 44.
Rate of plane with the wind: r + w Rate of plane against the wind: r − w r ⋅t = d ⎧(r + w) ⋅ 4 = 800 ⎨ ⎩(r − w) ⋅ 5 = 800 r + w = 200 r − w = 160 2r
= 360
r = 180 180 + w = 200 w = 20
Rate of plane = 180 mph. Rate of wind = 20 mph.
45.
Rate of boat with the current: r + w Rate of boat against the wind: r − w r ⋅t = d
⎧(r + w) ⋅ 4 = 120 ⎨ ⎩(r − w) ⋅ 6 = 120 r + w = 30 r − w = 20 2r
= 50
r = 25 25 + w = 30
w=5
Rate of boat = 25 mph. Rate of current = 5 mph.
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604
46.
Chapter 9: Systems of Equations and Inequalities
Rate of canoeist with the current: r + w Rate of canoeist against the current: r − w
47.
r ⋅t = d
⎧30 x + 45 y = 1080 ⎨ ⎩15 x + 12 y = 372
⎧(r + w) ⋅ 2 = 12 ⎨ ⎩(r − w) ⋅ 4 = 12 r+w=6 r−w= 3 2r
(1) − 2 times (2)
21 y = 336 y = 16 15 x + 12(16) = 372 15 x = 180 x = 12
=9
r = 4.5 w = 1.5
Rate of canoeist = 4.5 mph. Rate of current = 1.5 mph.
Cost of iron alloy: $12 per kilogram Cost of lead alloy: $16 per kilogram
x = cost of hydrochloric acid y = cost of silver nitrate
49.
⎧10 x + 15 y = 14.10 (1) ⎨ ⎩12 x + 20 y = 18.16 (2)
40 x + 60 y = 56.40 −36 x − 60 y = −54.48
(1) ( 2)
⎪⎧30 x + 45 y = 1080 ⎨−30 x − 24 y = −744 ⎪⎩
4. 5 + w = 6
48.
x = cost per kilogram of iron alloy y = cost per kilogram of lead alloy
4 times (1) −3 times (2)
= 1.92 x = 0.48 10(0.48) + 15 y = 14.10 15 y = 9.30 y = 0.62
x = amount of 40% gold y = amount of 60% gold x + y = 20 (1) ⎧ ⎨ + = 0.40 0.60 (0.52)(20) (2) x y ⎩ −0.40 x − 0.40 y = −8 −0.40 times (1)
0.40 + 0.60 y = 10.4 0.20 y = 2.4 y = 12
4x
x + 12 = 20 x=8
Cost of hydrochloric acid: $0.48/liter Cost of silver nitrate: $0.62/liter
Amount of 40% gold: 8 g Amount of 60% gold: 12 g
Copyright © Houghton Mifflin Company. All rights reserved.
(2)
Section 9.1
50.
605
x = amount of 70% solution y = amount of 30% solution x + y = 20 ⎧ ⎨ + x y = 0.40(20) 0.70 0.30 ⎩ −0.30 x − 0.30 y = −6
51.
Sketch a graph to visualize the right triangle.
(1) (2)
−0.30 times (1)
0.70 x + 0.30 y = 8 0.40 x
=2 x=5 To find the coordinates of point A, solve the system ⎧⎪ y = 0 ⎨y = 1 x ⎪⎩ 2
5 + y = 20 y = 15 Amount of 70% solution: 5 liters. Amount of 30% solution: 15 liters.
By substitution, 1 x = 0 2
x = 0 Thus A is (0, 0). To find the coordinates of point B, solve the system ⎧y = 0 ⎨ ⎩ y = −2 x + 6 By substitution, −2 x + 6 = 0 −2 x = −6 x=3 Thus B is (3, 0). To find the coordinates of the point C, solve the system (1) ⎧⎪ y = −2 x + 6 ⎨y = 1 x (2) ⎪⎩ 2 By substitution, 1 x = −2 x + 6 2 5x=6 2 x = 12 5
Substituting 12 for x in Equation (2), we have
( )
5
( 5 5)
y = 1 12 = 6 . 2 5
Thus C is 12 , 6 .
5
From the graph, ∠C is the right angle. Use the distance formula to find AC and BC. AC =
( 125 − 0) + ( 65 − 0) 2
2
= 144 + 36 = 180 = 6 5 25
BC = =
25
25
5
(3 − 125 ) ( ) 2 2 ( 53 ) + ( − 56 ) = 4525 = 53 2
2 + 0− 6 5
Area = 1 (base)(height) 2
(
=1 6
)(
5 3 5 2 5 5 = 9 (5) 25 = 9 square units 5
)
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5
606
52.
Chapter 9: Systems of Equations and Inequalities
Solve the system (1) ⎧2 x + 3 y = 1 ⎨ − = 3 x 4 y 10 (2) ⎩ 6x + 9 y = 3 −6 x + 8 y = −20
17 y = −17 y = −1 2 x + 3(−1) = 1 2x − 3 = 1 2x = 4
53.
3 times (1) − 2 times (2)
(1)
54.
⎧Z + 6 = Y ⎨ 7=X ⎩ X +Y = 7+ Z +6
⎧Z − 4 = Y ⎨ 8= X ⎩ X +Y =8+ Z − 4
X + Y = Z + 13
X +Y = Z + 4
If Z + 13 is divisible by 3, then Z = 2, 5, or 8. If Z + 4 is divisible by 3, then Z = 2, 5, or 8.
4 x + ky = 5. 4(2) + k (−1) = 5 8−k =5 − k = −3 k =3
2x + 5 = 6x + k = 4x − 7
In both cases, the largest digit Z can be is 8.
55.
2 x + 5 = 4 x − 7 ⇒ −2 x = −12 ⇒ x = 6 2(6) + 5 = 6(6) + k 12 + 5 = 36 + k 17 = 36 + k −19 = k
Case 2: Z + 5 + 1 > 9 ⎧ Z + 5 + 1 = 10 + Y ⎨ ⎩ 5 + 2 +1 = X
XY3 is divisible by 3 ⇒ X + Y is divisible by 3.
x=2 Thus the point is (2, −1). Substitute this point into
5Z 7 + 256 XY 3 Case 1: Z + 5 + 1 ≤ 9 ⎧Z + 5 + 1 = Y ⎨ ⎩ 5+2 = X
14 = c − b 126 = c + b
18 = c − b 98 = c + b
140 = 2c 70 = c, b = 56
116 = 2c 58 = c, b = 40
294 = c − b 6 = c +b
2 = c −b 882 = c + b
300 = 2c 150 = c, b =144
884 = 2c 442 = c, b = 440
The Pythagorean triples are: 42, 56, 70; 42, 40, 58; 42, 144, 150; 42, 440, 442.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.1
56.
607
If a = 30, then a2 = 900
57.
18 = c − b 50 = c + b
90 = c − b 10 = c + b
68 = 2c 34 = c, b =16
100 = 2c 50 = c, b = 40
150 = c − b 6 = c+b
2 = c −b 450 = c + b
156 = 2c 78 = c, b = 72
452 = 2c 226 = c, b = 224
The Pythagorean triples are: 30, 16, 34; 30, 40, 50; 30, 72, 78; 30, 224, 226.
x = people who like lip balm but do not like skin cream y = people who like lip balm and skin cream z = people who do not like lip balm but do like skin cream w = people who do not like lip balm nor skin cream x + y + z + w =100 0.80( y + z ) = y 0.50( x + w) = w x + y = 77 Rewrite the system by solving eq (2) for z, eq (3) for w, and eq (4) for x. x + y + z + w =100 z = 0.25 y w= x x = − y + 77 Substitute the values from equations (2), (3), and (4) into equation (1) and solve for y. (− y + 77) + y + 0.25 y + (− y + 77) =100 −0.75 y +154 =100 −0.75 y = −54 y = 72 z = 0.25(72) = 18 x = –72 + 77 = 5 w=5 Find the number of people who like skin cream (y + z) y + z = 72 +18 = 90
90 people liked the skin cream.
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608
58.
Chapter 9: Systems of Equations and Inequalities
x = people who pass the fire exam but not the chemical exam 59. y = people who pass the fire exam and the chemical exam z = people who did not pass the fire exam but did pass the chemical exam w = people who did not pass either fire exam nor chemical exam x + y + z + w = 200 0.75( x + y ) = y 0.25( z + w) = z y + z =120
⎧ 1 S − 1 A = 8750 ⎪ 2 2 ⎨ ⎪ 3 S − 2 3 A = 11,250 4 ⎩4
( )
S − A = 17,500 S − 2 A = 15,000 A = 2500 S = 20,000
Rewrite the system by solving eq (2) for x, eq (3) for w, and eq (4) for z. x + y + z + w =100 x=1 y
(1) (2)
2 times (1) 4 times (2) 3
The supply pump can pump 20,000 gal/h. The outlet pump can pump 2500 gal/h.
3
w = 3z z = − y +120 Substitute the values from equations (2), (3), and (4) into equation (1) and solve for y. 1 3
S = supply pump A = outlet pump
y + y + (− y +120) + 3(− y +120) = 200 − 8 y + 480 = 200 3
− 8 y = −280 3
y =105 z = –105 + 120 = 15 w = 3(15) = 45 x = 1 (105) = 35 3
Find the number of people who passed the fire exam (x + y) x + y = 35 +105 =140 140 people passed the basic fire science exam. 60.
a.
x + y = 40
b.
V = π r2h + 1 π r2h 3 2 1 π (2) y + π (2)2 y = 477.5 3 The system is ⎧ x + y = 40 ⎪ ⎨ 2 2 1 ⎪⎩π (2) y + 3 π (2) y = 477.5
y = 40 − x 4π x + 4 π (40 − x ) = 477.5 3 4 4π x + π 40 − 4 π x = 477.5 3 3 x 4π − 4 π = 477.5 − 160 π 3 3 160 477.5 − π 3 x= 4π − 4 π 3 x ≈ 37.0 ft y = 40 − x ≈ 40 − 37.0 = 3.0 ft
(
)
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Section 9.1
609
....................................................... 61.
(3 + 2i ) x + (4 − 3i ) y = 2 − 16i 3x + 2 xi + 4 y − 3 yi = 2 − 16i (3 x + 4 y ) + (2 x − 3 y )i = 2 − 16i ⎧2 x + 4 y = 2 ⎨ ⎩2 x − 3 y = −16
9 x + 12 y = 6 8 x − 12 y = −64
Connecting Concepts 62.
⎧ 4 x + 5 y = 11 ⎨ ⎩−3 x + 2 y = 9 12 x + 15 y = 33 −12 x + 8 y = 36
(1) ( 2)
3 times (1) 4 times (2)
(1) (2)
3 times (1) 4 times (2)
23 y = 69
= −58 58 x=− 17 3x + 4 y = 2
17 x
y=3 4 x + 5(3) = 11 4 x = −4 x = −1
⎛ 58 ⎞ 4 y = 2 − 3⎜ − ⎟ ⎝ 17 ⎠ 208 4y = 17 52 y= 17
63.
(4 − 3i) x + (5 + 2i ) y = 11 + 9i 4 x − 3xi + 5 y + 2 yi = 11 + 9i 4 x + 5 y + (−3x + 2 y )i = 11 + 9i
(2 + 6i ) x + (4 − 5i ) y = −8 − 7i 2 x + 6 xi + 4 y − 5 yi = −8 − 7i (2 x + 4 y ) + (6 x − 5 y )i = −8 − 7i ⎧ 2 x + 4 y = −8 ⎨ ⎩ 6 x − 5 y = −7
−6 x − 12 y = 24 6 x − 5 y = −7
(1) (2)
−3 times (1) (2)
− 17 y = 17 y = −1 2 x + 4( −1) = −8 2 x = −4 x = −2
....................................................... PS1. 2 x − 5 y =15 −5 y = −2 x +15 y = 2 x −3 5
Prepare for Section 9.2 PS2.
x = 2c +1 y = −c + 3 z = 2x +5 y − 4 z = 2(2c +1) + 5(−c + 3) − 4 = 4c + 2 − 5c +15 − 4 = −c +13
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610
Chapter 9: Systems of Equations and Inequalities
PS3. ⎧5 x − 2 y =10 ⎨ 2 y =8 ⎩
PS4. ⎧ 3 x − y =11 (1) ⎨ x y + = − 2 3 11 (2) ⎩
y=4
Solve (1) for y : y = 3 x −11
5 x − 2(4) =10 5 x =18 x = 18 5
2 x + 3(3x −11) = −11 2 x + 9 x − 33 = −11 11x = 22 x=2 y = 3(2) −11 y = −5
⎛ ⎞ The solution is ⎜ 18 , 4 ⎟ . ⎝5 ⎠
The solution is (2, −5). PS5. ⎧ y = 3 x − 4 ⎨ ⎩ y = 4x −2
PS6. ⎧ 4 x + y = 9 (1) ⎨ 8 x 2 y 18 (2) − − = − ⎩
3x − 4 = 4 x − 2 x = −2
Solve (1) for y : y = −4 x + 9
−8 x − 2(−4 x + 9) = −18 −8 x + 8 x −18 = −18 0=0 The system of equations is dependent. Let x = c and y = –4c + 9. The solutions are (c, –4c + 9).
y = 3(−2) − 4 y = −10
The solution is (–2, –10).
Section 9.2 1.
⎧2 x − y + z = 8 ⎪ 2 y − 3 z = −11 ⎨ ⎪ y + 2z = 3 3 ⎩ 6 y − 9 z = −33 −6 y − 4 z = −6
(2) (3) 3 times (2) − 2 times (3)
−13 z = −39 z =3
2.
(1)
⎧3x + y + 2 z = −4 ⎪ ⎨ − 3 y − 2 z = −5 ⎪ 2 y + 5 z = −4 ⎩ −6 y − 4 z = −10 6 y + 15 z = −12
(1) ( 2) (3) 2 times (2) 3 times (3)
11z = −22 (4)
z = −2
⎧2 x − y + z = 8 ⎪ ⎨ 2 y − 3 z = −11 ⎪ z=3 ( 4) ⎩
⎧3x + y + 2 z = −4 ⎪ ⎨ − 3 y − 2 z = −5 ⎪ z = −2 ⎩
2 y − 3(3) = −11 y = −1
−3 y − 2(−2) = −5 y=3 3x + 3 + 2(−2) = −4
2 x − (−1) + 3 = 8 x=2
The solution is (2, − 1, 3).
(4)
( 4)
x = −1 The solution is (−1, 3, − 2).
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Section 9.2
3.
611
4.
⎧ x + 3 y − 2 z = 8 (1) ⎪ ⎨ 2 x − y + z = 1 (2) ⎪3x + 2 y − 3 z = 15 (3) ⎩
−2 x − 6 y + 4 z = −6 2x − y + z = 1
⎧ x − 2 y + 3z = 5 ⎪ ⎨ 3x − 3 y + z = 9 ⎪5 x + y − 3z = 3 ⎩
(1) ( 2) (3)
− 2 times (1) (2)
−3 x + 6 y − 9 z = −15 3x − 3 y + z = 9
−7 y + 5 z = −15 (4) −3 x − 9 y + 6 z = −24 − 3 times (1) 3 x + 2 y − 3z = 15 (3)
3 y − 8z = − 6
− 7 y + 3 z = −9 (5) ⎧x + 3y − 2z = 8 ⎪ ⎨ −7 y + 5 z = −15 ⎪ −7 y + 3 z = −9 ⎩
−7 y + 5 z = −15 7 y − 3z = 9
5x +
y − 3z =
33 y − 88 z = −66
− 1 times (5)
− 33 y + 54 z = 66 − 34 z = 0 z=0
(6)
−7 y + 5(−3) = −15 y=0
−5 x + 10 y − 15 z = −25
⎧ x − 2 y + 3z = 5 ⎪ ⎨ 3 y − 8 z = −6 ⎪ 11 y − 18 z = −22 ⎩
(4) (5)
(4)
⎧x + 3 y − 2z = 8 ⎪ ⎨ − 7 y + 5 z = −15 ⎪ z = −3 ⎩
(4) − 5 times (1)
3
(3)
11y − 18 z = −22
2 z = −6 z = −3
− 3 times (1) (2)
⎧ x − 2 y + 3z = 5 ⎪ ⎨ 3 y − 8 z = −6 ⎪ z=0 ⎩
(6)
x + 3(0) − 2(3) = 8 x=2
The solution is (2, 0, − 3).
3 y − 8(0) = −6 y = −2
(5)
(4) (5) 11 times (4) − 3 times (5) (6)
(6)
x − 2(−2) + 3(0) = 5 x =1
The solution is (1, − 2, 0).
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612
5.
Chapter 9: Systems of Equations and Inequalities
⎧3x + 4 y − z = −7 ⎪ ⎨ x − 5 y + 2 z = 19 ⎪5 x + y − 2 z = 5 ⎩
6.
(1) (2) (3)
3x + 4 y − z = − 7
(1)
−3 x + 15 y − 6 z = −57
− 3 times (2)
19 y − 7 z = −64
(4)
−5 x + 25 y − 10 z = −95 5x + y − 2 z = 5
− 5 times (2) (3)
26 y − 12 z = −90 ⎧3x + 4 y − z = −7 ⎪ ⎨ 19 y − 7 z = −64 ⎪ 26 y − 12 z = −90 ⎩
(5)
(4) (5)
494 y − 182 z = −1664
26 times (4)
−494 y + 228 z = 1710 −19 times (3) 46z = 46
(6)
z =1 ⎧3x + 4 y − z = −7 ⎪ ⎨ 19 y − 7 z = −64 ⎪ z =1 ⎩
(6)
19 y − 7(1) = −64 y = −3 3x + 4(−3) − 1 = −7 x=2
The solution is (2, − 3, 1).
⎧ 2 x − 3 y − 2 z = 12 (1) ⎪ ⎨ x + 4 y + z = −9 (2) ⎪4 x + 2 y − 3 z = 6 (3) ⎩ 2 x − 3 y − 2 z = 12
(1)
−2 x − 8 y − 2 z = 18 − 2 times (2)
−11y − 4 z = 30
(4)
−4 x − 16 y − 4 z = 36 4 x + 2 y − 3z = 6
− 4 times (2) (3)
−14 y − 7 z = 42 − 2 y − z = 6 (5) ⎧2 x − 3 y − 2 z = 12 ⎪ ⎨ − 11 y − 4 z = 30 (4) ⎪ − 2 y − z = 6 (5) ⎩ −22 y − 8 z = 60 2 times (4) 22 y + 11z = −66 − 11 times (5)
3z = −6 z = −2
(6)
⎧2 x − 3 y − 2 z = 12 ⎪ ⎨ −11 y − 4 z = 30 ⎪ z = −2 ⎩
(6)
−11 y − 4(−2) = 30 y = −2 2 x − 3(−2) − 2(−2) = 12 x =1
The solution is (1, − 2, − 2).
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.2
7.
613
⎧2 x − 5 y + 3 z = −18 ⎪ ⎨ 3 x + 2 y − z = −12 ⎪ x − 3 y − 4 z = −4 ⎩
3x + 2 y −
8.
(1) (2) (3)
z = −12
(2) −2 times (3)
−3 x + 9 y + 12 z = 12
11 y + 11z = 0 y+z=0
(4)
y + 11z = −10
−10 z = 10 z = −1
(2)
2 x + 3 y − 3z = −13 − 2 x − 10 y − 2 z = −14 − 7 y − 5 z = −27
(2) − 2 times (2) (5)
(2)
⎧2 x − 5 y + 3 z = −18 ⎪ y+z =0 ⎨ ⎪ y + 11z = −10 ⎩
− y − 11z = 10
4 x − y + 2 z = −1 (1) − 4 x − 6 y + 6 z = 26 − 2 times (2) − 7 y + 8 z = 25
2 x − 5 y + 3 z = −18 (1) − 2 x + 6 y + 8z = 8 − 2 times (3)
y+ z=0
⎧ 4 x − y + 2 z = −1 (1) ⎪ ⎨2 x + 3 y − 3 z = −13 (2) ⎪ x + 5y + z = 7 (3) ⎩
(4) (5)
⎧4 x − y + 2 z = −1 ⎪ ⎨ − 7 y + 8 z = 25 ⎪ − 7 y − 5 z = 27 ⎩
(4) (5)
−7 y + 8 z = 25
(4) − 1 times (5)
7 y + 5 z = 27 13z = 52 z=4
(6)
(6)
⎧2 x − 5 y + 3z = −18 ⎪ y+z =0 ⎨ ⎪ z = −1 (6) ⎩
y −1 = 0 y =1 2 x − 5(1) + 3(−1) = −18 x = −5
⎧4 x − y + 2 z = −1 ⎪ ⎨ − 7 y + 8 z = 25 ⎪ z = 4 (6) ⎩
−7 y + 8(4) = 25 y =1 4 x − 1 + 2(4) = −1 x = −2
The solution is (−2, 1, 4).
The solution is (−5, 1, − 1).
Copyright © Houghton Mifflin Company. All rights reserved.
614
9.
Chapter 9: Systems of Equations and Inequalities
10.
⎧ x + 2 y − 3 z = −7 ⎪ ⎨ 2 x − y + 4 z = 11 ⎪4 x + 3 y − 4 z = −3 ⎩
(2) (3)
⎧ x − 3 y + 2 z = −11 ⎪ ⎨ 3x + y + 4 z = 4 ⎪5 x − 5 y + 8 z = −18 ⎩
−2 x − 4 y + 6 z = 14 2 x − y + 4 z = 11
− 2 times (1) (2)
−3 x + 9 y − 6 z = 33 3x + y + 4 z = 4
− 5 y + 10 z = 25
(1)
− 5 y + 8 z = 25
− 4 times (1) (3) (5)
⎧ x + 2 y − 3 z = −7 ⎪ ⎨ − y + 2 z = 5 (4) ⎪ − 5 y + 8 z = 25 (5) ⎩ 5 y − 10 z = −25
− 1 times (4)
− 5 y + 8 z = 25
(5)
− 2z = 0
(2) (3)
− 3 times (1) (2)
10 y − 2 z = 37
(4)
−4 x − 8 y + 12 z = 28 4 x + 3 y − 4 z = −3
(1)
(4)
−5 x + 15 y − 10 z = 55 − 5 times (1) 5 x − 5 y + 8 z = −18 (3) 10 y − 2 z = 37 ⎧ x − 3 y + 2 z = −11 ⎪ ⎨ 10 y − 2 z = 37 ⎪ 10 y − 2 z = 37 ⎩ 10 y − 2 z = 37 −10 y + 2 z = −37
(5)
( 4) (5) (4) − 1 times (5)
0=0
z=0
(6)
⎧ x + 2 y − 3 z = −7 ⎪ ⎨ − y + 2z = 5 ⎪ z = 0 (6) ⎩
− y + 2(0) = 5 y = −5 x + 2(−5) − 3(0) = −7 x=3
The solution is (3, − 5, 0).
The system is dependent. Let z = c.
10 y − 2c = 37 37 + 2c y= 10 ⎛ 37 + 2c ⎞ x − 3⎜ ⎟ + 2c = −11 ⎝ 10 ⎠ 111 6c − = −11 − 2c x− 10 10 111 110 20c 6c − − + x= 10 10 10 10 1 − 14c x= 10
⎛ 1 − 14c 37 + 2c The solution is ⎜ , , 10 ⎝ 10
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⎞ c ⎟. ⎠
Section 9.2
11.
615
12.
⎧2 x − 5 y + 2 z = −4 (1) ⎪ ⎨ 3x + 2 y + 3z = 13 (2) ⎪ 5 x − 3 y − 4 z = −18 (3) ⎩
6 x − 15 y + 6 z = −12 3 times (1) −6 x − 4 y − 6 z = −26 −2 times (2) −19 y y
= −38 =2
(4)
10 x − 25 y + 10 z = −20
5 times (1)
−10 x + 6 y + 8 z = 36
− 2 times (3)
−19 y + 18 z = 16
(5)
⎧2 x − 5 y + 2 z = −4 ⎪ y = 2 (4) ⎨ ⎪ − 19 y + 18 z = 16 (5) ⎩ = 38 19 y −19 y − 18 z = 16
19 times (4) (5)
15 x + 10 y − 25 z = 30 −15 x + 12 y − 9 z = 36 22 y − 34 z = 66 11 y − 17 z = 33 12 x + 8 y − 20 z = 24 −12 x − 15 y + 6 z = −45 −7 y − 14 z = −21 y + 2z = 3 ⎧3x + 2 y − 5 z = 6 ⎪ ⎨ 11 y − 17 z = 33 ⎪ y + 2z = 3 ⎩ 11y − 17 z = 33
5 times (1) − 3 times (2) (4) 4 times (1) − 3 times (3) (5)
(4) (5) (4)
− 11 y − 22 z = −33 − 11 times (5)
18 z = 54 z =3
(1) ⎧ 3x + 2 y − 5 z = 6 ⎪ − + = − 12 (2) ⎨ 5 x 4 y 3z ⎪4 x + 5 y − 2 z = 15 (3) ⎩
(6)
⎧2 x − 5 y + 2 z = −4 ⎪ y=2 ⎨ ⎪ z = 3 (6) ⎩ 2 x − 5(2) + 2(3) = −4 x=0
The solution is (0, 2, 3).
− 39 z = 0 z=0
(6)
⎧3x + 2 y − 5 z = 6 ⎪ ⎨ 11 y − 17 z = 33 ⎪ z=0 ⎩ 11y − 17(0) = 33 y=3 3x + 2(3) − 5(0) = 6 x=0
The solution is (0, 3, 0).
Copyright © Houghton Mifflin Company. All rights reserved.
616
13.
Chapter 9: Systems of Equations and Inequalities
⎧ 2 x + y − z = −2 ⎪ ⎨3x + 2 y + 3z = 21 ⎪ 7 x + 4 y + z = 17 ⎩
14.
(1) (2) (3)
6 x + 3 y − 3 z = −6 3 times (1) − 6 x − 4 y − 6 z = −42 − 2 times (2)
⎧ 3x + y + 2 z = 2 (1) ⎪ ⎨ 4 x − 2 y + z = −4 (2) ⎪11x − 3 y + 4 z = −6 (3) ⎩ 12 x + 4 y + 8 z = 8
− y − 9 z = −48 (4) 14 x + 7 y − 7 z = −14 − 14 x − 8 y − 2 z = −34 − y − 9 z = −48
4 times (1)
− 12 x + 6 y − 3 z = 12 − 3 times (2) 10 y + 5 z = 20 2y + z = 4
(4)
7 times (1) − 2 times (3)
33 x + 11 y + 22 z = 22
(5)
− 33 x + 9 y − 12 z = 18
11 times (1) − 3 times (3)
20 y + 10 z = 40 ⎧2 x + y − z = −2 ⎪ ⎨ − y − 9 z = −48 ⎪ − y − 9 z = −48 ⎩
− y − 9 z = −48 y + 9 z = 48 0=0
2y + z = 4 (4) (5)
⎧3 x + y + 2 z = 2 ⎪ 2y + z = 4 ⎨ ⎪ 2y + z = 4 ⎩
(4) − 1 times (5)
(5)
(4) (5)
2 y + z = 4 (4) − 2 y − z = −4 − 1 times (5)
(6)
⎧2 x + y − z = −2 ⎪ ⎨ − y − 9 z = −48 ⎪ 0 = 0 (6) ⎩
0=0
The system of equations is dependent. Let z = c.
− y − 9c = −48 y = 48 − 9c 2 x + (48 − 9c) − c = −2 x = 5c − 25
The solution is (5c − 25, 48 − 9c, c).
(6)
⎧3 x + y + 2 z = 2 ⎪ 2y + z = 4 ⎨ ⎪ 0=0 ⎩
(6)
The system of equations is dependent. Let z = c
2y + c = 4 y=
3x +
4−c + 2c = 2 2 3 x = 2 − 2c − x=−
4−c 2
c 2
⎛ c 4−c ⎞ , c ⎟. The solution is ⎜ − , 2 ⎝ 2 ⎠
Copyright © Houghton Mifflin Company. All rights reserved.
4−c 2
Section 9.2
15.
617
16.
⎧3x − 2 y + 3z = 11 (1) ⎪ ⎨ 2 x + 3 y + z = 3 (2) ⎪5 x + 14 y − z = 1 (3) ⎩
6 x − 4 y + 6 z = 22 − 6 x − 9 y − 3z = −9
− 3 times (3)
(1) − 2 times (2) (4)
x − 3 y + 4 z = 4 (2) − x + 12 y − 6 z = 2 (3) 9 y − 2z = 6
(5)
⎧2 x + 3 y + 2 z = 14 ⎪ 3y − 2z = 2 ⎨ ⎪ 9 y − 2z = 6 ⎩
(4) (5) − 2 times (4) (5)
−9 y + 6 z = −6
(6)
⎧3 x − 2 y + 3z = 11 ⎪ ⎨ − 13 y + 3z = 13 ⎪ z = 0 (6) ⎩
−31 y + 3(0) = 13 y = −1
(2) (3)
9 y − 6z = 6 3y − 2z = 2
− 52 y + 18 z = 52 − 26 y + 9 z = 26
z=0
(1)
5 times (1)
− 15 x − 42 y + 3 z = −3
26 y − 6 z = −26 − 26 y + 9 z = 26
− 2 x + 6 y − 8 z = −8
(4)
15 x − 10 y + 15 z = 55
⎧3x − 2 y + 3 z = 11 ⎪ ⎨ − 13 y + 3 z = 13 ⎪ − 26 y + 9 z = 26 ⎩
2 x + 3 y + 2 z = 14
2 times (1) − 3 times (2)
13 y + 3 z = 13
⎧ 2 x + 3 y + 2 z = 14 ⎪ ⎨ x − 3y + 4z = 4 ⎪− x + 12 y − 6 z = 2 ⎩
3x − 2(−1) + 3(0) = 11 x=3
The solution is (3, − 1, 0).
(5)
(4) (5)
− 3 times (4)
9 y − 2z = 6
(5)
4z = 0 z=0
(6)
⎧2 x + 3 y + 2 z = 14 ⎪ 3y − 2z = 2 ⎨ ⎪ z=0 ⎩ 3 y − 20 = 2 y=
2 3
⎛ 2 ⎞ The solution is ⎜ 6, , 0 ⎟ . ⎝ 3 ⎠
Copyright © Houghton Mifflin Company. All rights reserved.
⎛2⎞ 2 x + 3 ⎜ ⎟ + 2(0) = 14 ⎝3⎠ x=6
618
17.
Chapter 9: Systems of Equations and Inequalities
⎧2 x − 3 y + 6 z = 3 ⎪ ⎨ x + 2 y − 4z = 5 ⎪3x + 4 y − 8 z = 7 ⎩
18.
(1) (2) (3)
2x − 3y + 6z = 3
(4)
−3 x − 6 y + 12 z = −15
(3)
− 19 y
(5)
(4) (5)
y=2 − y = −2 0=0
2 x − 3 y + 5 z =14 x + 4 y − 3z = −2
2 x − 3 y + 5 z = 14 − 2x − 8 y + 6z = 4 − 11y + 11z = 18
{
2 x − 3 y + 5 z =14 −11y +11z =18
Let z = c.
(4) −1 times (5) (6)
(6)
The system of equations is dependent. Let z = c.
The system of equations is inconsistent and has no solution.
{
(5)
(4) (5)
⎧2 x + 3 y − 6 z = 5 ⎪ y =2 ⎨ ⎪⎩ 0=0
(6)
⎧2 x − 3 y + 6 z = 3 ⎪ − y + 2 z = −1 ⎨ ⎪ 0 = 3 (6) ⎩
19.
= −38
⎧2 x + 3 y − 6 z = 5 ⎪ y =2 ⎨ ⎪⎩ =2 y
(4) − 1 times (5)
0=3
− 3 times (3)
y=2
− y + 2 z = −4
− y + 2 z = −1 y − 2z = 4
3 times (2)
− 6 x − 15 y + 18 z = −24
− 2 y + 4 z = −8
⎧2 x − 3 y + 6 z = 3 ⎪ − y + 2 z = −1 ⎨ ⎪ − y + 2 z = −4 ⎩
(2) (3)
6 x − 4 y − 18 z = −14
− 3 times (2)
3x + 4 y − 8 z = 7
(1)
6 x + 9 y −18 z =12 3 times (1) −6 x + 4 y +18 z =14 − 2 times (2) 13 y = 26 (4) y=2
(1)
− 2 x − 4 y + 8 z = −10 − 2 times (2) − 7 y + 14 z = −7 − y + 2 z = −1
⎧2 x + 3 y − 6 z = 4 ⎪ ⎨3x − 2 y − 9 z = −7 ⎪2 x + 5 y − 6 z = 8 ⎩
(1) − 2 times (2) (3)
⎧ x − 3y + 4z = 9 ⎨ ⎩3 x − 8 y − 2 z = 4
(1) (2)
−3 x + 9 y − 12 z = −27 3x − 8 y − 2 z = 4
− 3 times (1) (2)
y − 14 z = −23
(3)
−11 y + 11c = 18 18 − 11c y= −11 11c − 18 y= 11 c − 11 18 ⎛ ⎞ 2 x − 3⎜ ⎟ + 5c = 14 ⎝ 11 ⎠ 33c − 54 2 x = 14 − 5c + 11 154 − 55c + 33c − 54 2x = 11 50 − 11c x= 11 ⎛ 50 − 11c 11c − 18 ⎞ The solution is ⎜ , , c ⎟. 11 ⎝ 11 ⎠
2 x + 3(2) − 6c = 4 x = 3c − 1
The solution is (3c − 1, 2, c). 20.
(1) (2)
y=2
⎧x − 3 y + 4z = 9 ⎨ y − 14 z = −23 ⎩
Let z = c.
(3)
(3)
y − 14c = −23 y = 14c − 23
x − 3(14c − 23) + 4c = 9 x = 9 − 4c + 42c − 69 x = 38c − 60
The solution is (38c − 60, 14c − 23, c).
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.2
21.
619
⎧6 x − 9 y + 6 z = 7 ⎨ ⎩4 x − 6 y + 4 z = 9
22.
(1) (2)
⎧4 x − 2 y + 6 z = 5 ⎨ ⎩ 2 x − y + 3z = 2
0 = −26
0 =1 ⎧4 x − 2 y + 6 z = 5 ⎨ 0 =1 ⎩
(3)
The system of equations is inconsistent and has no solution. 23.
⎧5 x + 3 y + 2 z = 10 ⎨ ⎩3x − 4 y − 4 z = −5
15 x + 9 y + 6 z = 30 3 times (1) − 15 x + 20 y + 20 z = 25 − 5 times (2)
⎧5 x + 3 y + 2 z = 10 ⎨ ⎩ 29 y + 26 z = 55
Let z = c.
(3)
(3)
The system of equations is inconsistent and has no solution. 24.
(1) (2)
29 y + 26 z = 55
(2)
4x − 2 y + 6z = 5 (1) − 4 x + 2 y − 6 z = −4 − 2 times (2)
24 x − 36 y + 24 z = 28 4 times (1) − 24 x + 36 y − 24 z = −54 − 6 times (2)
⎧6 x − 9 y + 6 z = 7 ⎨ 0 = −26 ⎩
(1)
⎧3x − 4 y − 7 z = −5 ⎨ ⎩2 x + 3 y − 5z = 2
(1) (2)
6 x − 8 y − 14 z = −10 − 6 x − 9 y + 15 z = −6 − 17 y + z = −16
(3)
2 times (1) − 3 times (2) (3)
⎧3 x − 4 y − 7 z = −5 ⎨ ⎩ − 17 y + z = −16
(3) 29 y + 26c = 55 55 − 26c y= 29
Let z = c
(3) −17 y + c = −16 16 + c y= 17
3 x − 4 y − 7 z = −5
⎛ 55 − 26c ⎞ 5 x + 3⎜ ⎟ + 2c = 10 ⎝ 29 ⎠ 165 − 78c 5 x = 10 − 2c − 29 290 − 58c − 165 + 78c 5x = 29 25 + 4c x= 29 ⎛ 25 + 4c 55 − 26c The solution is ⎜ , , 29 ⎝ 29
⎞ c ⎟. ⎠
⎛ 16 + c ⎞ 3x − 4⎜ ⎟ − 7c = −5 ⎝ 17 ⎠ 3 x = 7c − 5 +
64 + 4c 17
119c − 85 + 64 + 4c 17 41c − 7 x= 17
3x =
⎛ 41c − 7 16 + c , , The solution is ⎜ 17 ⎝ 17
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⎞ c ⎟. ⎠
620
25.
Chapter 9: Systems of Equations and Inequalities
⎧ x + 3y − 4z = 0 ⎪ ⎨ 2x + 7 y + z = 0 ⎪3 x − 5 y − 2 z = 0 ⎩
(1)
−2 x − 6 y + 8 z = 0 2x + 7 y + z = 0 y + 9z = 0 −3 x − 9 y + 12 z = 0
(1)
(2) (3)
⎧ x − 2 y + 3z = 0 ⎪ ⎨3x − 7 y − 4 z = 0 ⎪ 4x − 4 y + z = 0 ⎩
− 2 times (1) (2)
−3 x + 6 y − 9 z = 0 3x − 7 y − 4 z = 0
− 3 times (1) (2)
− 3 times (1) (3)
−4 x + 8 y − 12 z = 0 4x − 4 y + z = 0
− 4 times (1) (3) (5)
(5)
(4) (5)
⎧ x − 2 y + 3z = 0 ⎪ ⎨ − y − 13 z = 0 ⎪ 4 y − 11z = 0 ⎩
(4) (5)
(6)
7 y + 63 z = 0
7 times (4)
− 7 y + 5z = 0
(5)
⎧ x − 2 y + 3z = 0 ⎪ ⎨ − y − 13 z = 0 ⎪ z=0 ⎩
68 z = 0 z=0
(6)
−4 y − 42 z = 0
y + 9(0) = 0 y=0
(4)
4 y − 11z = 0
− 14 y + 10 z = 0 − 7 y + 5z = 0
⎧x + 3 y − 4z = 0 ⎪ y + 9z = 0 ⎨ ⎪ z=0 ⎩
(2) (3)
− y − 13 z = 0
(4)
3x − 5 y − 2 z = 0
⎧x + 3 y − 4z = 0 ⎪ y + 9z = 0 ⎨ ⎪ − 7 y + 5z = 0 ⎩
26.
4 times (4)
4 y − 11z = 0
(5)
− 53z = 0 z=0
(6)
z = 0, y = 0, x = 0. The solution is (0, 0, 0).
(6) x + 3(0) − 4(0) = 0 x=0
The solution is (0, 0, 0).
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.2
27.
621
⎧ 2x − 3 y + z = 0 ⎪ ⎨2 x + 4 y − 3 z = 0 ⎪ 6x − 2 y − z = 0 ⎩
28.
(1) (2) (3)
−2 x + 3 y − z = 0
− 1 times (1)
2 x + 4 y − 3z = 0
(2)
7 y − 4z = 0
(4)
−6 x + 9 y − 3 z = 0 6x − 2 y − z = 0 7 y − 4z = 0 ⎧2 x − 3 y + z = 0 ⎪ ⎨ 7 y − 4z = 0 ⎪ 7 y − 4z = 0 ⎩
−7 y + 4 z = 0 7 y − 4z = 0
⎧5 x − 4 y − 3z = 0 ⎪ ⎨ 2x + y + 2z = 0 ⎪ x − 6 y − 7z = 0 ⎩
(1) (2) (3)
10 x − 8 y − 6 z = 0 2 times (1) − 10 x − 5 y − 10 z = 0 − 5 times (2) − 13 y − 16 z = 0
− 3 times (1) (3)
(4)
2 x + y + 2 z = 0 (2) − 2 x + 12 y + 14 z = 0 − 2 times (3) 13 y + 16 z = 0
(5)
⎧5 x − 4 y − 3z = 0 ⎪ ⎨ − 13 y − 16 z = 0 ⎪ 13 y + 16 z = 0 ⎩
(4) (5)
(5)
(4)
−13 y − 16 z = 0 13 y + 16 z = 0
− 1 times (4) (5)
0=0
0=0
(6)
⎧2 x − 3 y + z = 0 ⎪ ⎨ 7 y − 4z = 0 ⎪ 0=0 ⎩
⎧5 x − 4 y − 3z = 0 ⎪ ⎨ − 13 y − 16 z = 0 ⎪ 0=0 ⎩
(6)
4 Let z = c. Then 7 y = 4c or y = c. Substitute for y and z in 7 Eq. (1) and solve for x.
Let z = c.
−13 y − 16c = 0
⎛4 ⎞ 2 x − 3⎜ c ⎟ + c = 0 ⎝7 ⎠ 5 2x = c 7 5 x= c 14
⎛ 16c ⎞ 5 x − 4⎜ − ⎟ − 3c = 0 ⎝ 13 ⎠
4 ⎛ 5 ⎞ The solution is ⎜ c, c, c ⎟ . 14 7 ⎝ ⎠
y=−
64c 13 25c 5x = − 13 5 x=− c 13 5 x = 3c −
16 ⎛ 5 The solution is ⎜ − c, − c, 13 ⎝ 13
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⎞ c ⎟. ⎠
16c 13
622
29.
Chapter 9: Systems of Equations and Inequalities
⎧ 3x − 5 y + 3z = 0 ⎪ ⎨ 2 x −3 y + 4 z = 0 ⎪⎩7 x −11y +11z = 0 −6 x + 10 y − 6 z = 0 6 x − 9 y + 12 z = 0
− 2 times (1) 3 times (2)
y + 6z = 0
⎧3 x − 5 y + 3 z = 0 ⎪ y + 6z = 0 ⎨ ⎪ 2 y + 12 z = 0 ⎩
0=0
− y + 11z = 0
(5)
⎧5 x − 2 y − 3 z = 0 ⎪ ⎨ − y +11z = 0 ⎪⎩ − y +11z = 0
(1) (4) (5)
− y + 11z = 0 y − 11z = 0
(6)
⎧5 x − 2 y − 3 z = 0 ⎪ ⎨ − y +11z = 0 ⎪⎩ 0=0
(1) (4) (6)
Let z = c.
3x − 5(−6 z ) + 3z = 0 3 x = −33 z x = −11z
5 x − 2(11c) − 3c = 0
4x − 7 y − 2z = 0 − 4x − 8 y − 6z = 0 − 15 y − 8 z = 0
(1) − 2 times (2)
⎧4 x − 7 y − 2 z = 0 ⎪ ⎨ −15 y − 8 z = 0 ⎪⎩ 16 y +19 z = 0
The solution is (5c, 11c, c ). ⎧ 5 x + 2 y + 3z = 0 ⎪ ⎨ 3x + y − 2 z = 0 ⎪⎩4 x − 7 y + 5 z = 0
(1) (2) (3)
15 x + 6 y + 9 z = 0 −15 x − 5 y + 10 z = 0
3 times (1) − 5 times (2)
y + 19 z = 0
(4)
(4)
12 x + 4 y − 8 z = 0 −12 x + 21 y − 15 z = 0 25 y − 23 z = 0
(5)
⎧5 x + 2 y + 3 z = 0 ⎪ y + 19 z = 0 ⎨ ⎪ 25 y − 23 z = 0 ⎩
(4) (5)
−240 y − 128 z = 0
16 times (4)
240 y + 285 z = 0
15 times (5)
157z = 0 z=0
− y + 11c = 0 y = 11c 5 x = 22c + 3c x = 5c
32.
(1) (2) (3)
6 x + 12 y + 9 z = 0 3 times (2) − 6 x + 4 y + 10 z = 0 − 2 times (3) 16 y + 19 z = 0
(5)
(4) (5)
From Eq. (4), y = −6z. Substitute into Eq. (1).
⎧4 x − 7 y − 2 z = 0 ⎪ ⎨ 2 x + 4 y + 3z = 0 ⎪⎩ 3 x − 2 y − 5 z = 0
(4)
(4) − 1 times (5)
0=0
(6)
Let z be any real number c, then the solutions are (−11c, − 6c, c). 31.
3 times (1) − 5 times (2)
12 x − 4 y − 16 z = 0 4 times (2) − 12 x + 3 y + 27 z = 0 − 3 times (3)
− 2 times (4) (5)
⎧3x − 5 y + 3 z = 0 ⎪ y +6z = 0 ⎨ ⎪⎩ 0=0
(1) (2) (3)
− y + 11z = 0
− 7 times (1) 3 times (3)
2 y + 12 z = 0
⎧5 x − 2 y − 3 z = 0 ⎪ ⎨ 3x − y − 4 z = 0 ⎪⎩ 4 x − y − 9 z = 0
15 x − 6 y − 9 z = 0 −15 x + 5 y + 20 z = 0
(4)
−21x + 35 y − 21z = 0 21x − 33 y + 33 z = 0
−2 y − 12 z = 0 2 y + 12 z = 0
30.
(1) (2) (3)
(6)
⎧4 x − 7 y − 2 z = 0 ⎪ ⎨ −15 y − 8 z = 0 ⎪⎩ (6) z =0 z = 0, y = 0, x = 0. The solution is (0, 0, 0).
4 times (2) − 3 times (3) (5)
(4) (5)
−25 y − 475 z = 0 25 y − 23 z = 0 −498 z = 0 z =0
− 25 times (4) (5) (6)
⎧5 x + 2 y + 3 z = 0 ⎪ y +19 z = 0 ⎨ ⎪⎩ z =0
(6)
z = 0, y = 0, x = 0. The solution is (0, 0, 0).
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Section 9.2
33.
623
34.
y = ax 2 + bx + c
y = ax 2 + bx + c
3 = a (2)2 + b(2) + c
−2 = a(1)2 + b(1) + c
7 = a( −2)2 + b( −2) + c
−4 = a(3)2 + b(3) + c
−2 = a (1) 2 + b(1) + c
−2 = a(2)2 + b(2) + c
⎧4a + 2b + c = 3 (1) ⎪ ⎨4a − 2b + c = 7 (2) ⎪ a + b + c = −2 (3) ⎩
⎧ a + b + c = −2 ⎪ ⎨ 9a + 3b + c = −4 ⎪4a + 2b + c = −2 ⎩
4a + 2b + c = 3 (1) − 4a + 2b − c = −7 − 1 times (2)
Eliminate c from Eq. (2) by multiplying Eq. (1) by –1 and adding to Eq. (2), and eliminate c from Eq. (3) by multiply-ing Eq. (1) by –1 and adding to Eq. (3).
4b = −4 4a + 2b + c = 3 − 4a − 4b − 4c = 8
(4) (1) − 4 times (3)
− 2b − 3c = 11 (5) ⎧4a + 2b + c = 3 ⎪ 4b = −4 ⎨ ⎪ − 2b − 3c = 11 ⎩
⎧a + b + c = −2 ⎪ ⎨ 8a + 2b = −2 ⎪ 3a + b = 0 ⎩
(1) (2) (3)
(4) (5)
Eliminate b from Eq. (5) by multiplying Eq. (4) by − (4) (5)
1 2
and adding to Eq. (5). ⎧a + b + c = −2 ⎪ ⎨ 8a + 2b = −2 ⎪ a = −1 ⎩
From (4) : 4b = −4 b = −1 From (5) : − 2(−1) − 3c = 11
a = −1
c = −3
(6)
8(−1) + 2b = −2 b=3
From (1) : 4a + 2(−1) − 3 = 3 a=2
a + b + c = −2 −1 + 3 + c = −2 c = −4
The equation whose graph passes through the three points is
The equation whose graph passes through the three points is
y = − x 2 + 3 x − 4.
y = 2 x 2 − x − 3.
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624
Chapter 9: Systems of Equations and Inequalities
35.
x 2 + y 2 + ax + by + c = 0
52 + 32 + a (5) + b(3) + c = 0 (−1)2 + (−5)2 + a( −1) + b(−5) + c = 0 (−2)2 + 22 + a (−2) + b(2) + c = 0 ⎧ 5a + 3b + c = −34 ⎪ ⎨ − a − 5b + c = −26 ⎪− 2a + 2b + c = −8 ⎩
5a + 3b + c = −34 a + 5b − c = 26 6a + 8b = −8 3a + 4b = −4 5a + 3b + c = −34 2a − 2b − c = 8 7 a + b = −26 ⎧5a + 3b + c = −34 ⎪ ⎨ 3a + 4b = 4 ⎪ 7 a + b = −26 ⎩
3a + 4b = −4 − 28a − 4b = 104 − 25a
= 100 a = −4
⎧5a + 3b + c = −34 ⎪ = −4 ⎨3a + 4b ⎪ a = −4 ⎩ 3(−4) + 4b = −4
36.
x 2 + y 2 + ax + by + c = 0 0 + 36 + a(0) + b(6) + c = 0 1 + 25 + a (1) + b(5) + c = 0 49 + 1 + a(−7) + b(−1) + c = 0 6b + c = −36 (1) ⎧ ⎪ ⎨ a + 5b + c = 26 (2) ⎪− 7 a − b + c = −50 (3) ⎩
(1) (2) (3)
7 a + 35b + 7c = −182 − 7 a − b + c = −50
(1) − 1 times (2)
7 times (2) (3)
34b + 8c = −232 17b + 4c = −116 (4)
(4)
6b + c = −36 ⎧ ⎪ 17 + 4c = −116 b ⎨ ⎪− 7 a − b + c = −50 ⎩
(1) − 1 times (3) (5)
(4)
−24b − 4c = 144 − 4 times (1) 17b + 4c = −116 (4)
(4) (5)
− 7b
(4) − 4 times (5)
= 28 b = −4
= −4 b ⎧ ⎪ ⎨ 17b + 4c = −116 ⎪− 7 a − b + c = −50 ⎩
(6)
(5) (5)
17(−4) + 4c = −116 c = −12
a=6
The equation whose graph passes through the three points is
(6) 5(−4) + 3(2) + c = −34
b=2
−7 a − (−4) − 12 = −50
x 2 + y 2 + 6 x − 4 y − 12 = 0.
c = −20
The equation whose graph passes through the three points is x 2 + y 2 − 4 x + 2 y − 20 = 0.
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Section 9.2
625
37.
x 2 + y 2 + ax + by + c = 0
38.
x 2 + y 2 + ax + by + c = 0
(−2)2 + 102 + a(−2) + b(10) + c = 0
22 + 52 + a (2) + b(5) + c = 0
(−12)2 + (−14)2 + a (−12) + b( −14) + c = 0
(−4) 2 + (−3) 2 + a (−4) + b(−3) + c = 0
52 + 32 + a (5) + b(3) + c = 0
32 + 42 + a (3) + b(4) + c = 0
⎧ − 2a + 10b + c = −104 ⎪ ⎨− 12a − 14b + c = −340 ⎪ 5a + 3b + c = −34 ⎩
−2a + 10b + c = −104 12a + 14b − c = 340 10a + 24b 5a + 12b
= 236 = 118
⎧ 2a + 5b + c = −29 ⎪ ⎨− 4a − 3b + c = −25 ⎪ 3a + 4b + c = −25 ⎩
(1) (2) (3)
(1) (2) (3)
(1) − 1 times (2)
2a + 5b + c = −29 4a + 3b − c = 25
(1) − 1 times (2)
(4)
6a + 8b 3a + 4b
(4)
= −4 = −2
−2a + 10b + c = −104 − 5a − 3b − c = 34
(1) − 1 times (3)
−4a − 3b + c = −25 − 3a − 4b − c = 25
(2) − 1 times (3)
− 7 a + 7b −a+b
(5)
− 7 a − 7b a+b
(5)
= −70 = −10
⎧− 2a + 10b + c = −104 ⎪ = 118 ⎨ 5a + 12b ⎪ −a+b = −10 ⎩
⎧2a + 5b + c = −29 ⎪ = −2 (4) ⎨ 3a + 4b ⎪ a+b =0 (5) ⎩
(1) (4) (5)
5a + 12b = 118 (4) − 5a + 5b = −50 5 times (5) 17b = 68 b=4
3a + 4b = −2 − 4a − 4b = 0 −a
(6)
⎧− 2a + 10b + c = −104 ⎪ = 118 ⎨ 5a + 12b ⎪ =4 (6) b ⎩ 5a + 12( 4) = 118
=0 =0
(4) − 4 times (5)
= −2 a = 2 (6)
⎧2a + 5b + c = −29 ⎪ = −2 ⎨ 3a + 4b ⎪ a =2 (6) ⎩
−2(14) + 10(4) + c = −104
3(2) + 4b = −2
2(2) + 5(−2) + c = −29
c = −116
b = −2
c = −23
a = 14
The equation whose graph passes through the three points
The equation whose graph passes through the three points
is x 2 + y 2 + 14 x + 4 y − 116 = 0.
is x 2 + y 2 + 2 x − 2 y − 23 = 0.
( x 2 + 14 x + 49) + ( y 2 + 4 y + 4) = 116 + 49 + 4
( x 2 + 2 x + 1) + ( y 2 − 2 y + 1) = 23 + 1 + 1
( x + 7) 2 + ( y + 2) 2 = 169 The center is (−7, − 2) and radius is 13.
( x + 1) 2 + ( y − 1) 2 = 25 The center is (−1, 1) and radius is 5.
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626
39.
Chapter 9: Systems of Equations and Inequalities
For intersection A, 275 + 225 = x1 + x2 x1 + x2 = 500
40.
For intersection A, 200 + x3 =165 + x1 x1 − x3 = 35
For intersection B, x2 + 90 = x3 +150 x2 − x3 = 60
For intersection B, 50 + x1 =100 + x2 x1 − x2 = 50
For intersection C, x1 + x3 = 240 + 200 x1 + x3 = 440
For intersection C, 200 + x2 =185 + x3 x2 − x3 =15
⎧ x1 + x2 = 500 (1) ⎪ ⎨ x2 − x3 = 60 (2) ⎪ x + x = 440 (3) ⎩ 1 3
⎧ x1 − x3 = 35 (1) ⎪ ⎨ x1 − x2 = 50 (2) ⎪ x − x =15 (3) ⎩ 2 3
The equations are dependent. Solve equation (2) for x3 and substitute the inequality for x2. x3 = x2 − 60
The equations are dependent. Solve equation (1) for x3 and substitute the inequality for x1. x3 = x1 − 35
Because 150 ≤ x2 ≤ 250 , then 90 ≤ x3 ≤190 . The flow between B and C is 90 to 190 cars per hour.
Solve equation (2) for x2 and substitute the inequality for x1. x2 = x1 − 50 Because 60 ≤ x1 ≤ 80 , .then 10 ≤ x2 ≤ 30 and 25 ≤ x3 ≤ 45 . The flow between C and A is 25 to 45 cars per hour and the flow between B and C is 10 to 30 cars per hour.
41.
For intersection A, 256 + x4 = 389 + x1 x1 − x4 = −133
42.
For intersection A, 75 + x4 = 60 + x1 x1 − x4 = −15
For intersection B, 437 + x1 = x2 + 300 x1 − x2 = −137
For intersection B, 50 + x1 = x2 +100 x1 − x2 = 50
For intersection C, 298 + x3 = 249 + x4 x3 − x4 = −49
For intersection C, 45 + x2 = 50 + x3 x2 − x3 = 5
For intersection D, 314 + x2 = 367 + x3 x2 − x3 = 53
For intersection D, 80 + x3 = 40 + x4 x3 − x4 = −40
⎧ x1 − x4 = −133 ⎪ x − x = −137 ⎪ 1 2 ⎨ ⎪ x3 − x4 = −49 ⎪⎩ x2 − x3 = 53
⎧ x1 − x4 =15 ⎪ x − x = 50 ⎪ 1 2 ⎨ ⎪ x2 − x3 = 5 ⎪⎩ x3 − x4 = −40
(1) (2) (3) (4)
The equations are dependent. Solving the system gives x1 = x4 −133 x2 = x4 + 4 x3 = x4 − 49 Because 125 ≤ x1 ≤175 , then 125 ≤ x4 −133 ≤175 and 258 ≤ x2 − 4 ≤ 308 and 258 ≤ x3 + 49 ≤ 308 258 ≤ x4 ≤ 308 262 ≤ x2 ≤ 312 209 ≤ x3 ≤ 259
(1) (2) (3) (4)
The equations are dependent. Solving the system gives x1 = x2 + 50 x3 = x2 − 5 x4 = x2 + 35 Since there cannot be a negative number of cars per hour in an intersection, then x3 ≥ 0 and therefore x2 ≥ 5 . The minimum number of cars traveling between B and C is 5 cars per hour.
The flow between C and A is 258 to 308 cars per hour The flow between B and D is 262 to 312 cars per hour. The flow between D and C is 209 to 259 cars per hour.
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Section 9.2
43.
627
w1d1 + w2 d 2 = w3d3 and w1 = 2, w2 = 6, w3 = 9 2d1 + 6d 2 = 9d3 (1) From the words in the exercise, d1 + d3 = 13 (2) d 2 = 1 d1 (3) 3
2d1 + 6d 2 − 9d3 = 0 2d1 − 6d 2 =0 4d1 − 9d3 = 0
4d1 − 9d3 = 0 9d1 + 9d3 = 0 13d1 = 117 d1 = 9
(1) 6 times (3) (4)
(4) 9 times (2) (5)
44.
d3 + d 4 = 20 ⎧ ⎪ d1 + d 2 = 10 ⎪ d5 + d 6 = 8 ⎪⎪ ⎨ 6d1 = 4d 2 ⎪ ⎪ 5d5 = 3d6 ⎪ ⎪⎩(2 + 6 + 4) d3 = (10 + 5 + 3)d 4
4d1 + 4d 2 = 40 6d1 − 4d 2 = 0 = 40 10d1 d1 = 4
(1) (2) (3) (4) (5) (6)
4 times (2) (4)
4d 2 = 6(4) d2 = 6
Substitute into equation (3) and (2) d 2 = 1 (9) = 3 3
9 + d3 = 13 d3 = 4 Therefore d1 = 9 in., d2 = 3 in., and d3 = 4 in. d2 + d3 = 3 + 4 = 7 in. d1 – d2 = 9 – 3 = 6 in. So the middle chime is 7 in. from the 9 ounce chime and 6 in. from the 2 ounce chime.
3d5 + 3d 6 = 24 5d5 − 3d 6 = 0 = 24 8d 5 d5 = 3
3 times (3) (5)
3d 6 = 5(3) d6 = 5 18d3 + 18d 4 = 360 12d3 − 18d 4 = 0 = 360 30d3 d3 = 12
18 times (1) (6)
12 + d 4 = 20 d4 = 8 The lengths are: d1 = 4 in., d2 = 6 in., and d3 = 12 in., d4 = 8 in., d5 = 3 in., and d6 = 5 in.
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628
Chapter 9: Systems of Equations and Inequalities
....................................................... 45.
⎧ 2 x + y − 3 z + 2 w = −1 ⎪ 2 y − 5 z − 3w = 9 ⎪ ⎨ 3 y − 8 z + w = −4 ⎪ ⎪⎩ 2 y − 2 z + 3w = −3
6 y − 15 z − 9w = 27 − 6 y + 16 z − 2w = 8
2 y − 5 z − 3w = 9
46.
(1) (2) (3) (4)
3 times (2) − 2 times (3)
z − 11w = 35
⎧3x − y + 2 z − 3w = 5 ⎪ ⎪ 2 y − 5 z + 2 w = −7 ⎨ 4 y − 9 z + w = −19 ⎪ ⎪⎩ 3 y + z − 2w = −12
− 1 times (4)
(2) (3) (4)
− 2 times (2) 4 y − 9 z + w = −19 (3) (5)
6 y − 15 z + 6 w = −21 − 6 y − 2 z + 4 w = 24 − 17 z + 10w = 3
− 3 z − 6 w = 12
(1)
−4 y + 10 z − 4 w = 14
z − 3w = −5
(5) (2)
− 2 y + 2 z − 3w = 8
Connecting Concepts
3 times (2) − 2 times (4) (6)
z + 2 w = −4 (6) ⎧ 2 x + y − 3 z + 2 w = −1 ⎪ 2 y − 5 z − 3w = 9 ⎪ ⎨ z − 11w = 35 ⎪ ⎪⎩ z + 2 w = −4
z − 11w = 35 − z − 2w = 4
(5) (6)
(5) − 1 times (6)
− 13w = 39 w = −3
⎧3 x − y + 2 z − 3 w = 5 ⎪ 2 y − 5 z + 2 w = −7 ⎪ ⎨ z − 3w = −5 (5) ⎪ ⎪⎩ −17z + 10w = 3 (6)
17 z − 51w = −85 − 17 z + 10w = 3
17 times (5) (6)
− 41w = −82 w=2
(7)
(7)
⎧ 2 x + y − 3 z + 2 w = −1 ⎪ 2 y − 5 z − 3w = 9 ⎪ ⎨ z − 11w = 35 ⎪ ⎪⎩ w = −3
⎧3 x − y + 2 z − 3 w = 5 ⎪ 2 y − 5 z + 2 w = −7 ⎪ ⎨ z − 3w = − 5 ⎪ ⎪⎩ w = 2 (7) (7) z − 3(2) = −5 z =1
z − 11(−3) = 35 z=2
2 y − 5(2) − 3( −3) = 9 2 y = 10 y =5
2 y − 5(1) + 2(2) = −7 2 y = −6 y = −3
3x − (−3) + 2(1) − 3(2) = 5 2 x + 5 − 3(2) + 2( −3) = −1 2x = 6 x=3
3x = 6 x=2
The solution is (2, − 3, 1, 2).
The solution is (3, 5, 2, − 3).
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Section 9.2
47.
629
48.
⎧ x − 3y + 2z − w = 2 ⎪ ⎪ 2 x − 5 y − 3z + 2 w = 21 ⎨ ⎪ 3 x − 8 y − 2 z − 3w = 12 ⎪⎩− 2 x + 8 y + z + 2 w = −13
(1)
−2 x + 6 y − 4 z + 2w = −4 2 x − 5 y − 3 z + 2w = 21
− 2 times (1) (2)
(2) (3) (4)
y − 7 z + 4w = 17
(5)
−3 x + 9 y − 6 z + 3w = −6 3 x − 8 y − 2 z − 3w = 12 y − 8z
− 3 times (1) (3)
=6
3 y − 2 z + 4w = 8
−2 x + 4 y − 6 z − 4 w = −16 2 x − 5 y + 2 z − w = 19 − y − 4 z − 5w = 3
4 x − 8 y + 3 z + 2 w = 29 3z + 2w = 1
⎧ x − 2 y + 3z + 2w = 8 ⎪⎪ − y −11z − 3w = −6 ⎨ ⎪ − y − 4 z − 5w = 3 3 z + 2 w =1 ⎩⎪
(6) (7)
(5) − 1 times (7)
− 2 y − 5z
(8)
− y − 11z − 3w = −6 y + 4 z + 5w = −3
− 2 times (1) (3)
− 4 times (1) (4)
(5) (6) (7)
(8)
⎧ x − 2 y + 3z + 2w = 8 ⎪⎪ − y −11z − 3w = −6 ⎨ −7 z + 2 w = −9 ⎪ 3 z + 2 w =1 ⎪⎩
(6) (8)
−7 z + 2w = −9 − 3z − 2 w = −1
(8)
− 10 z
− 21z = 21 z = −1 (9)
(7)
(5) − 1 times (6)
− 7 z + 2 w = −9
2 times (6)
⎧x − 3 y + 2z − w = 2 ⎪ ⎪ y − 7 z + 4 w = 17 ⎨ =6 ⎪ y − 8z ⎪⎩ = −1 z
(5)
− 9 z − 6 w = −3
(5)
⎧x − 3 y + 2z − w = 2 ⎪ ⎪ y − 7 z + 4 w = 17 ⎨ =6 ⎪ y − 8z ⎪⎩ − 2 y − 5 z =9
− 3 times (1) (2)
(6)
−4 x + 8 y − 12 z − 8w = −32
y − 7 z + 4 w = 17 − 3 y + 2 z − 4 w = −8
− 2 y − 5z = 9
− y − 11z − 3w = −6
(7)
⎧x − 3 y + 2z − w = 2 ⎪ ⎪ y − 7 z + 4 w = 17 ⎨ =6 ⎪ y − 8z ⎪⎩ 3 y − 2 z + 4 w = 8
2 y − 16 z = 12
−3 x + 6 y − 9 z − 6 w = −24 3x − 7 y − 2 z + 3w = 18
(6)
2 x − 5 y − 3 z + 2 w = 21 (2) − 2 x + 8 y + z + 2 w = −13 (4)
=9
(1) ⎧ x − 2 y + 3z + 2w = 8 ⎪3 x − 7 y − 2 z + 3w =18 (2) ⎨ 2 x − 5 y + 2 z − w =19 (3) ⎪ ⎩4 x −8 y + 3 z + 2 w = 29 (4)
(8) (7)
(8) − 1 times (7)
= −10 z =1
⎧ x − 2 y + 3z + 2w = 8 ⎪⎪ − y −11z − 3w = −6 ⎨ −7 z + 2 w = −9 ⎪ z =1 ⎪⎩ (9)
−7(1) + 2w = −9 2w = −2 w = −1
y − 8(−1) = 6 y = −2
− y − 11(1) − 3(−1) = −6 −y=2
(−2) − 7(−1) + 4 w = 17 4 w = 12
y = −2
w=3
x − 2(−2) + 3(1) + 2( −1) = 8 x=3
x − 3( −2) + 2(−1) − 3 = 2 x =1
The solution is (1, − 2, − 1, 3).
The solution is (3, − 2, 1, − 1).
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630
49.
Chapter 9: Systems of Equations and Inequalities
50.
⎧ x + 2 y − 2 z + 3w = 2 ⎪ ⎪ 2 x + 5 y + 2 z + 4w = 9 ⎨ ⎪4 x + 9 y − 2 z + 10 w = 13 ⎪⎩ − x − y + 8 z − 5w = 3
(1)
(4)
⎧ x − 2 y + 3 z − 2 w = −1 ⎪ ⎪3x − 7 y − 2 z − 3w = −19 ⎨ ⎪ 2 x − 5 y + 2 z − w = −11 ⎪⎩ − x + 3 y − 2 z − w = 3
−2 x − 4 y + 4 z − 6 w = −4 2 x + 5 y + 2 z + 4w = 9
− 2 times (1) (2)
−3 x + 6 y − 9 z + 6 w = 3 − 3 times (1) 3 x − 7 y − 2 z − 3w = −19 (2)
(2) (3)
y + 6 z − 2w = 5
y + 6 z − 2w = 5
− 4 times (1) (3)
⎧ x + 2 y − 2 z + 3w = 2 ⎪ y + 6 z − 2w = 5 ⎪ ⎨ y + 6 z − 2w = 5 ⎪ ⎪⎩ y + 6 z − 2w = 5
(3) (4)
−2 x + 4 y − 6 z + 4w = 2 − 2 times (1) 2 x − 5 y + 2 z − w = −11 (3) − y − 4 z + 3w = −9
(6)
x − 2 y + 3 z − 2 w = −1 − x + 3y − 2z − w = 3
x + 2 y − 2 z + 3w = 2 (1) − x − y + 8 z − 5w = 3 (4) y + 6 z − 2w = 5
(2)
− y − 11z + 3w = −16 (5)
(5)
−4 x − 8 y + 8 z − 12w = −8 4 x + 9 y − 2 z + 10w = 13
(1)
(1) (4)
y + z − 3w = 2
(7)
(6)
(7)
⎧ x − 2 y + 3 z − 2w = −1 ⎪ ⎪ − y − 11z + 3w = −16 ⎨ ⎪ − y − 4 z + 3w = −9 ⎪⎩ y + z − 3w = 2
(5) (6) (7)
y + 11z − 3w = 16
⎧ x + 2 y − 2 z + 3w = 2 ⎪ y + 6 z − 2w = 5 ⎪ ⎨ 0=0 ⎪ ⎪⎩ 0=0
− y − 4 z + 3w = −9 7z
− y − 4 z + 3w = −9 y + z − 3w = 2
y + 6a − 2b = 5 y = −6a + 2b + 5
− 3z
x + 2(−6a + 2b + 5) − 2a + 3b = 2 x − 12a + 4b + 10 − 2a + 3b = 2 x = 14a − 7b − 8
The solutions are (14a − 7b − 8, − 6a + 2b + 5, a, b).
(6) (7)
− 1 times (5) (6)
=7 z =1
Let z = a, w = b.
(5)
= −7 7 z= 3
(8)
(6) (7)
(9)
⎧ x − 2 y + 3 z − 2w = −1 ⎪ ⎪⎪ − y − 11z + 3w = −16 =1 (8) z ⎨ ⎪ 7 ⎪ = (9) z 3 ⎩⎪ ⎧ x − 2 y + 3z − 2 w = −1 ⎪ − y − 11z + 3w = −16 ⎪⎪ =1 z ⎨ ⎪ 4 ⎪ 0=− ⎪⎩ 3 The system of equations is inconsistent and has no solutions.
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Section 9.2
631
51-52. ⎧ x − 3y − 2z = A2 ⎪⎪ ⎨2 x − 5 y + Az = 9 ⎪ 2 x − 8 y + z = 18 ⎪⎩
53-55. ⎧ x + 2 y + z = A2 (1) ⎪ − − x 2 3 y + Az =1 (2) ⎨ ⎪7 x +12 y + A2 z = 4 A2 − 3 (3) ⎩
(1) (2) (3)
Multiply Eq. (1) by 2 and add to Eq. (2). Then multiply Eq. (1) by –7 and add to Eq. (3). The resulting system is
Multiply Eq. (1) by –2 and add to Eq. (2). Now multiply Eq. (1) by –2 and add to Eq. (3). The resulting system is ⎧ x − 3y − 2z = A2 ⎪ ⎪ 2 ⎨ y + (4 + A) z = −2 A + 9 ⎪ 2 ⎪⎩ − 2 y + 5 z = −2 A + 18
(5)
⎧ x + 2 y + z = A2 (4) ⎪ 2 y + ( A + 2) z = 2 A +1 (5) ⎨ ⎪−2 y + ( A2 − 7) z = −3 A2 − 3 (6) ⎩
(6)
Multiply Eq. (5) by 2 and add to Eq. (6).
(4)
⎧ x + 2 y + z = A2 ⎪ 2 ⎨ y + ( A + 2) z = 2 A +1 ⎪( A2 + 2 A − 3) z = A2 −1 (7) ⎩
Multiply Eq. (5) by 2 and add to Eq. (6). We now have ⎧ x − 3y − 2z = A2 ⎪ ⎪ 2 ⎨ y + (4 + A) z = −2 A + 9 ⎪ 2 ⎪⎩ (2 A + 13) z = −6 A + 36 (7)
In Exercise 53, the system of equations will have a unique solution when ( A 2 + 2 A − 3) ≠ 0 in Eq. (7). That is, ( A + 3)( A − 1) ≠ 0, or A ≠ −3, A ≠ 1
For Exercise 51, the system of equations has no solution 13 when 2 A + 13 = 0 or A = − . 2
In Exercise 54, the system will have an infinite number of solutions when A 2 + 2 A − 3 = 0 and A 2 − 1 = 0. This occurs when A = 1. In Exercise 55, the system of equations will have no solution
For Exercise 52, the system of equations has a unique solution 13 when 2 A + 13 ≠ 0 or A ≠ − . 2 56.
z = ax + by + c
57.
⎧ 2a + b + c =1 (1) ⎪ ⎨− a + 2b + c =12 (2) ⎪⎩ 3a + 2b + c = 0 (3) 2a + b + c = 1 −2a + 4b + 2c = 24 2 times (2) 5b + 3c = 25 (4) −3a + 6b + 3c = 36 3a + 2b + c = 0 8b + 4c = 36
⎧ 2 a + b + c =1 ⎪ ⎨ 5b + 3c = 25 ⎪⎩ 8b + 4c = 36
when A 2 + 2 A − 3 = 0 and A 2 − 1 ≠ 0. This occurs when A = −3.
3 times (2) (5)
⎧ a − b + c = 5 (1) ⎪ ⎨ 2a − 2b + c = 9 (2) ⎪− 3a − b + c = −1 (3) ⎩ 3a − 3b + 3c = 15
3 times (1)
− 3a − b + c = −1
(3)
− 4b + 4c = 14
−2a + 2b − 2c = −10 2a − 2b + c = 9
(4) (5)
40b + 24c = 200 −40b − 20c = −180 4c = 20 c=5
z = ax + by + c
− 2 times (1) (2)
− c = −1 8 times (4) − 5 times (5) (6)
⎧ 2 a + b + c =1 ⎪ ⎨ 5b + 3c = 25 ⎪⎩ c = 5 (6) 2 a + 2 + 5 =1 5b + 3(5) = 25 5b =10 2a = −6 a = −3 b=2 Thus the equation of the plane is z = −3 x + 2 y + 5.
c =1 −4b + 4(1) = 14 5 b=− 2 ⎛ 5⎞ a − ⎜ − ⎟ +1 = 5 ⎝ 2⎠ 3 a= 2 Thus the equation of the plane is z = 3x − 5 y − 2 z = −2.
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3 5 x − y + 1 or 2 2
632
Chapter 9: Systems of Equations and Inequalities
....................................................... PS1. x 2 + 2 x − 2 = 0
Prepare for Section 9.3 PS2. ⎧ x + 4 y = −11 (1) ⎨ (2) ⎩3x − 2 y = 9
x 2 + 2 x +1= 2 +1 ( x +1) 2 = 3
Solve equation (1) for x: x = −4 y −11
x +1= ± 3
3(−4 y −11) − 2 y = 9 −12 y − 33 − 2 y = 9 −14 y = 42 y = −3 x = −4(−3) −11 x =1
x = −1± 3
The solution is (1, –3). PS3. parabola
PS4. hyperbola
PS5. 2
PS6. 4
Section 9.3 1.
⎧⎪ y = x 2 − x (1) ⎨ ⎪⎩ y = 2 x − 2 (2) Set the expressions for y equal to each other.
2.
x2 − x = 2 x − 2 x − 3x + 2 = 0 ( x − 2)( x −1) = 0 x−2 = 0 x −1 = 0
x 2 + 2 x − 3 = x −1 x2 + x − 2 = 0 ( x + 2)( x −1) = 0 x+2=0 x = −2
2
x=2
x =1
When x = 2, y = 22 − 2 = 2 (From Eq. (1))
x −1 = 0 x =1
When x = 1, y = 1 − 1 = 0
When x = −2, y = −2 − 1 = −3 (From Eq. (2)) When x = 1, y = 1 − 1 = 0
The solutions are (1, 0) and (2, 2).
The solutions are (−2, −3) and (1, 0).
2
3.
⎧⎪ y = x 2 + 2 x − 3 (1) ⎨ ⎪⎩ y = x − 1 (2) Set the expressions for y equal to each other.
⎧ y = 2 x 2 − 3 x − 3 (1) ⎨ (2) ⎩ y = x−4 Set the expressions for y equal to each other.
2 x2 − 3x − 3 = x − 4 2 x2 − 4 x + 1 = 0 4 ± 16 − 4(2)(1) (Quadratic Formula) 2⋅2 = 4± 8 = 4±2 2 = 2± 2 4 4 2
x=
Substitute for x in (1) and solve for y. When x = 2 + 2 , y = 2 + 2 − 4 = −6 + 2 . 2 2 2 When x = 2 − 2 , y = 2 − 2 − 4 = −6 − 2 . 2 2 2 The solutions are ⎛ 2− 2 −6− 2 ⎞ ⎛ 2 + 2 −6 + 2 ⎞ ⎟. ⎜ , ⎜ ⎟ and ⎜ 2 , ⎟ 2 2 ⎝ 2 ⎠ ⎠ ⎝
4.
⎧ y = −x 2 + 2x − 4 ⎪ ⎨ 1 ⎪y = x +1 2 ⎩ Set the expressions for y equal to each other.
−x + 2x − 4 = − x2 +
1 x +1 2
3 x−5= 0 2
2 x 2 − 3x + 10 = 0 3 ± 9 − 80 (Quadratic Formula) 2⋅2 3 ± −71 x= 4 x=
Because the solutions are not real numbers, the system of equations has no real number solutions.
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Section 9.3
5.
633
⎧⎪ y = x 2 − 2 x + 3 (1) ⎨ ⎪⎩ y = x 2 − x − 2 (2) Set the expressions for y equal to each other.
6.
2 x2 − x + 1 = x2 + 2 x + 5
x 2 − 2x + 3 = x 2 − x − 2 − x = −5 x=5 Substitute for x in Eq. (1).
x 2 − 3x − 4 = 0 ( x − 4)( x + 1) = 0 x = 4 or x = −1 Substitute for x in Eq. (2).
y = 5 2 − 2(5) + 3
y = (4) 2 + 2(4) + 5 y = 29
y = 18 The solution is (5, 18). 7.
⎧ x + y = 10 ⎨ ⎩ xy = 24
8.
(1) (2)
2 y 2 + 3y +1 = 0 (2 y + 1)( y + 1) = 0
0 = x 2 − 10 x + 24 0 = ( x − 4)( x − 6)
y=−
x = 4 or x = 6
The solutions are (4, 6) and (6, 4).
Solve Eq. (1) for y. (3)
Substitute into Eq. (2). x(2 x − 1) = 6
⎧ x − 3 y = 7 (1) ⎨ xy = −4 (2) ⎩ Solve Eq. (1) for x. x = 3 y + 7 (3) Substitute into Eq. (2). (3 y + 7) y = −4
3 y 2 + 7 y = −4 3y 2 + 7 y + 4 = 0 (3 y + 4)( y + 1) = 0
2x 2 − x = 6 2x 2 − x − 6 = 0 (2 x + 3)( x − 2) = 0
3 y + 4 = 0, or y + 1 = 0 y = −4 / 3 y = −1
2 x + 3 = 0, or x − 2 = 0 3 2
The solutions are (2, −1/2) and (1, −1). 10.
⎧2 x − y = 1 (1) ⎨ xy = 6 (2) ⎩
1 or y = −1 2
Substitute for y in Eq. (1). ⎛ 1⎞ x − 2(−1) = 3 x − 2⎜ − ⎟ = 3 ⎝ 2⎠ x =1 x=2
Substitute for x in Eq. (1). 4 + y = 10 6 + y = 10 y=6 y=4
x=−
⎧ x − 2 y = 3 (1) ⎨ xy = −1 (2) ⎩
Substitute x from Eq. (1) into Eq. (2). ( 2 y + 3) y = −1
10 x − x 2 = 24
y = 2x −1
y = (−1) 2 + 2(−1) + 5 y=4
The solutions are (4, 29) and (−1, 4).
Substitute y from Eq. (1) into Eq. (2). x(10 − x) = 24
9.
⎧⎪ y = 2 x 2 − x +1 (1) ⎨ 2 ⎪⎩ y = x + 2 x + 5 (2) Set the expressions for y equal to each other.
x=2
Substitute for x in Eq. (3). 3 ⎛ 3⎞ When x = − , y = 2 ⎜ − ⎟ − 1 = −4. 2 ⎝ 2⎠ When x = 2, y = 2(2) − 1 = 3 .
Substitute for y in Eq. (1). ⎛ 4⎞ When y = −4 / 3, x = 3 ⎜ − ⎟ + 7 = 3 ⎝ 3⎠ When y = −1, x = 3( −1) + 7 = 4 The solutions are (3, −4/3) and (4, −1).
The solutions are (−3/2, −4) and (2, 3).
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634
11.
Chapter 9: Systems of Equations and Inequalities
⎧⎪3x 2 − 2 y 2 = 1 (1) ⎨ ⎪⎩ y = 4 x − 3 (2) Substitute y from Eq. (2) into Eq. (1).
12.
3x 2 − 2(4 x − 3) 2 = 1
(6 − 4 y ) 2 + 3 y 2 = 7
3x 2 − 32 x 2 + 48 x − 18 = 1
36 − 48 y + 16 y 2 + 3 y 2 = 7
29 x 2 − 48 x + 19 = 0 (29 x − 19)( x − 1) = 0
19 y 2 − 48 y + 29 = 0 (19 y − 29)( y − 1) = 0
x=
19 or x = 1 29
y=
Substitute for x in Eq. (2). ⎛ 19 ⎞ y = 4⎜ ⎟ − 3 ⎝ 29 ⎠ 76 87 − y= 29 29 11 y=− 29
29 19
y =1
Substitute for y in Eq. (2). x + 4(1) = 6 ⎛ 29 ⎞ x + 4⎜ ⎟ = 6 x=2 ⎝ 19 ⎠ 114 116 − x= 19 19 2 x=− 19 ⎛ 2 29 ⎞ The solutions are ⎜ − , ⎟ and (2, 1). ⎝ 19 19 ⎠
y = 4(1) − 3 y =1
The solutions are (19/29, −11/29) and (1, 1). 13.
⎧⎪ x 2 + 3 y 2 = 7 (1) ⎨ ⎪⎩ x + 4 y = 6 (2) Substitute x from Eq. (2) into Eq. (1).
⎧⎪ y = x 3 + 4 x 2 − 3 x − 5 (1) ⎨ ⎪⎩ y = 2 x 2 − 2 x − 3 (2) Set the expressions for y equal to each other. x 3 + 4 x 2 − 3x − 5 = 2 x 2 − 2 x − 3
14.
⎧⎪ y = x 3 − 2 x 2 + 5 x + 1 (1) ⎨ ⎪⎩ y = x 2 + 7 x − 5 (2) Set the expressions for y equal to each other. x 3 − 2 x 2 + 5x + 1 = x 2 + 7 x − 5
x 3 + 2x 2 − x − 2 = 0
x 3 − 3x 2 − 2 x + 6 = 0
x 2 ( x + 2) − ( x + 2) = 0
x 2 ( x − 3) − 2( x − 3) = 0
( x + 2)( x 2 − 1) = 0 ( x + 2)( x − 1)( x + 1) = 0
( x − 3)( x 2 − 2) = 0
x = 3, x = 2 , or x = − 2
x = −2, x = 1, or x = −1
Substitute for x in Eq. (2). Substitute for x in Eq. (2).
y = 3 2 + 7(3) − 5
y = 2(−2) 2 − 2(−2) − 3 y=9
y = 25 y=
y = 2(1) 2 − 2(1) − 3 y = −3
( 2 )2 + 7
2 −5
y = −3 + 7 2
( )2 + 7( 2 )− 5
y= − 2
y = 2(−1) 2 − 2(−1) − 3 y =1
y = −3 − 7 2
The solutions are (−2, 9), (1, −3) and (−1, 1).
The solutions are (3, 25), (− 2, − 3 − 7 2).
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(
)
2, − 3 + 7 2 and
Section 9.3
15.
635
⎧⎪2 x 2 + y 2 = 9 ⎨ ⎪⎩ x 2 − y 2 = 3
16.
(1) (2)
⎧⎪3 x 2 − 2 y 2 = 19 ⎨ ⎪⎩ x 2 − y 2 = 5
(1) (2)
2x 2 + y 2 = 9
3x 2 − 2 y 2 = 19
x2 − y2 = 3
−3 x 2 + 3 y 2 = −15
3 x 2 = 12 Add the equations.
− 3 times Eq. (2)
y2 = 4 y = ±2
2
x =4 x = ±2
Add the equations.
When y = −2, x 2 − (−2)2 = 5 From Eq. (1) 2
2
When x = −2, (−2) − y = 3 From Eq. (2)
x2 − 4 = 5
2
4− y =3
x2 = 9 x = ±3
− y 2 = −1
When y = 2, x 2 − 22 = 5
y2 = 1 y = ±1
x2 − 4 = 5 x2 = 9 x = ±3
When x = 2, (2)2 − y 2 = 3 4 − y2 = 3
The solutions are (3, − 2), (3, 2), ( −3, 2) and (−3, − 2).
− y 2 = −1 y2 = 1 y = ±1 The solutions are (−2, 1), (−2, − 1), (2, 1) and (2, − 1). 17.
18.
⎧⎪ x 2 − 2 y 2 = 8 (1) ⎨ ⎪⎩ x 2 + 3 y 2 = 28 (2) Use the elimination method to eliminate x 2 . x2 − 2y2 = 8 2
⎧⎪2 x 2 + 3 y 2 = 5 (1) ⎨ ⎪⎩ x 2 − 3 y 2 = 4 (2) Use the elimination method to eliminate y 2 . 2x 2 + 3y 2 = 5
2
2
− x − 3 y = −28
x − 3y = 4
2
3x 2
=9
2
2
=3
− 5 y = −20 y =4
x
y = ±2
x=± 3
Substitute for y in Eq. (1). x 2 − 2( 2) 2 = 8 x 2 = 16 x = ±4
(1)
2
x 2 − 2(−2) 2 = 8 x 2 = 16 x = ±4
The solutions are (4, 2), (−4, 2), (4, −2) and (−4, −2).
Substitute for y in Eq. (2).
( 3 )2 − 3 y 2 = 4
y2 = −
1 3
y 2 = −1 / 3 has no real number solutions. The graphs of the equations do not intersect.
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636
19.
Chapter 9: Systems of Equations and Inequalities
20.
⎧⎪2 x 2 + 4 y 2 = 5 (1) ⎨ ⎪⎩ 3x 2 + 8 y 2 = 14 (2) Use the elimination method to eliminate y 2 . − 4 x 2 − 8 y 2 = −10 2
−x
(1) (2)
Use the elimination method to eliminate y 2 . 4 x 2 + 6 y 2 = 22
− 2 times (1)
2
2
2
− 9 x − 6 y = −57
3x + 8 y = 14 2
⎧⎪2 x 2 + 3 y 2 = 11 ⎨ ⎪⎩3 x 2 + 2 y 2 = 19
− 5x
=4
2
2 times (1) − 3 times(2)
= −35
x 2 = −4
x=± 7
Substitute for x in Eq. (1).
x 2 = −4 has no real number solutions. The graphs of the equations do not intersect.
2
2( 7) + 3 y = 11 3 y 2 = −3 y 2 = −1 y 2 = −1 has no real number solutions. The graphs of the equations do not intersect.
21.
⎧⎪ x 2 − 2 x + y 2 = 1 (1) ⎨ 2 x + y = 5 (2) ⎪⎩ Substitute y from Eq. (2) into Eq. (1).
22.
⎧⎪ x 2 + y 2 + 3 y = 22 (1) ⎨ ⎪⎩ 2 x + y = −1 (2) Substitute y from Eq. (2) into Eq. (1).
x 2 − 2 x + (5 − 2 x) 2 = 1
x 2 + (−2 x − 1) 2 + 3(−2 x − 1) = 22
x 2 − 2 x + 25 − 20 x + 4 x 2 = 1
x 2 + 4 x 2 + 4 x + 1 − 6 x − 3 = 22
5 x 2 − 22 x + 24 = 0 (5 x − 12)( x − 2) = 0 x=
5 x 2 − 2 x − 24 = 0 (5 x − 12)( x + 2) = 0
12 or x = 2 5
x=
12 or x = −2 5
Substitute for x in Eq. (2).
Substitute for x in Eq. (2).
⎛ 12 ⎞ 2⎜ ⎟ + y = 5 ⎝ 5⎠ 1 y= 5
⎛ 12 ⎞ 2 ⎜ ⎟ + y = −1 ⎝ 5⎠ 29 y=− 5
2( 2) + y = 5 y =1
The solutions are (12/5, 1/5) and (2, 1).
2(−2) + y = −1 y =3
The solutions are (12/5, −29/5) and (−2, 3).
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Section 9.3
23.
637
⎧( x − 3) 2 + ( y + 1) 2 = 5 (1) ⎪⎪ x y − = 3 7 (2) ⎨ ⎪ x = 3y + 7 ⎪⎩ Substitute x from Eq. (2) into Eq. (1).
24.
⎧⎪( x + 2) 2 + ( y − 2) 2 = 13 (1) ⎨ ⎪⎩ 2x + y = 6 (2) Substitute y from Eq. (2) into Eq. (1). ( x + 2) 2 + (4 − 2 x) 2 = 13 x 2 + 4 x + 4 + 16 − 16 x + 4 x 2 = 13
(3 y + 4) 2 + ( y + 1) 2 = 5
5 x 2 − 12 x + 7 = 0 (5 x − 7)( x − 1) = 0
9 y 2 + 24 y + 16 + y 2 + 2 y + 1 = 5 10 y 2 + 26 y + 12 = 0 5 y 2 + 13 y + 6 = 0 (5 y + 3)( y + 2) = 0 y=−
x=
3 or y = −2 5
Substitute for x in Eq. (2). ⎛7⎞ 2⎜ ⎟ + y = 6 ⎝5⎠ 16 y= 5
Substitute for y in Eq. (3). ⎛ 3⎞ x = 3⎜ − ⎟ + 7 ⎝ 5⎠ 26 x= 5
7 or x = 1 5
x = 3(−2) + 7 x =1
2(1) + y = 6 y=4
The solutions are (7/5, 16/5) and (1, 4).
The solutions are (26/5, −3/5) and (1, −2). 25.
⎧⎪ x 2 − 3x + y 2 = 4 (1) ⎨ ⎪⎩ 3 x + y = 11 (2) Substitute y from Eq. (2) into Eq. (1). x 2 − 3x + (11 − 3 x) 2 = 4
2
10 x 2 − 69 x + 117 = 0 (10 x − 29)( x − 3) = 0 39 or x = 3 10
29 x 2 − 60 x + 4 = 0 (29 x − 2)( x − 2) = 0
Substitute for x in Eq. (2). ⎛ 39 ⎞ 3 ⎜ ⎟ + y = 11 ⎝ 10 ⎠ y=−
7 10
⎧⎪ x 2 + y 2 − 4 y = 4 (1) ⎨ ⎪⎩ 5 x − 2 y = 2 (2) Substitute y from Eq. (2) into Eq. (1). ⎞ ⎛5 ⎞ ⎛5 x 2 + ⎜ x − 1⎟ − 4⎜ x − 1⎟ = 4 ⎠ ⎝2 ⎠ ⎝2 25 x2 + x 2 − 5 x + 1 − 10 x + 4 = 4 4 29 2 x − 15 x + 1 = 0 4
x 2 − 3x + 121 − 66 x + 9 x 2 = 4
x=
26.
3(3) + y = 11 y=2
The solutions are (39/10, −7/10) and (3, 2).
2 x= or x = 2 29
Substitute for x in Eq. (2). ⎛ 2 ⎞ 5⎜ ⎟ − 2 y = 2 ⎝ 29 ⎠ y=−
24 29
5(2) − 2 y = 2 y=4
The solutions are (2/29, −24/29) and (2, 4).
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638
27.
Chapter 9: Systems of Equations and Inequalities
⎧⎪( x − 1) 2 + ( y + 2) 2 = 14 (1) ⎨ ⎪⎩( x + 2) 2 + ( y − 1) 2 = 2 (2) Expand the binomials and then subtract.
28.
⎧⎪( x + 2) 2 + ( y − 3) 2 = 10 (1) ⎨ ⎪⎩ ( x − 3) 2 + ( y + 1) 2 = 13 (2) Expand the binomials and then subtract.
x 2 − 2 x + 1 + y 2 + 4 y + 4 = 14
x 2 + 4 x + 4 + y 2 − 6 y + 9 = 10
x 2 + 4x + 4 + y 2 − 2 y +1 = 2
x 2 − 6 x + 9 + y 2 + 2 y + 1 = 13
− 6x − 3 − 6x
10 x − 5 10 x
+ 6 y + 3 = 12 + 6 y = 12 y = x+2
Substitute for y in Eq. (2).
− 8 y + 8 = −3 − 8y = −6 5x + 3 y= 4
Substitute for y in Eq. (1): y − 3 =
( x + 2) 2 + ( x + 2 − 1) 2 = 2 2
x 2 + 4x + 4 + x 2 + 2x + 1 = 2
⎛ 5x − 9 ⎞ ( x + 2) 2 + ⎜ ⎟ = 10 ⎝ 4 ⎠
2x 2 + 6x + 3 = 0 x= =
5x − 9 . 4
x 2 + 4x + 4 +
− 6 ± 36 − 4 ⋅ 2 ⋅ 3 4
25 x 2 − 90 x + 81 = 10 16
16 x 2 + 64 x + 64 + 25 x 2 − 90 x + 81 = 160
− 6 ± 12 − 3 ± 3 = 4 2
41x 2 − 26 x − 15 = 0 ( 41x + 15)( x − 1) = 0
Substitute for x in y = x + 2.
x=−
y=
−3+ 3 +2 2
y=
−3− 3 +2 2
y=
1+ 3 2
y=
1− 3 2
The solutions are ⎛ −3 + 3 1 + 3 ⎞ ⎛ −3 − 3 1 − 3 ⎞ , , ⎜⎜ ⎟⎟ and ⎜⎜ ⎟. 2 2 ⎠ 2 2 ⎟⎠ ⎝ ⎝
15 or x = 1 41
Substitute for x in y =
5x + 3 . 4
5 ⎛ 15 ⎞ 3 y = ⎜− ⎟ + 4 ⎝ 41 ⎠ 4 12 y= 41
5(1) + 3 4 y=2 y=
The solutions are (−15/41, 12/41) and (1, 2).
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Section 9.3
29.
639
⎧⎪( x + 3) 2 + ( y − 2) 2 = 20 (1) ⎨ ⎪⎩( x − 2) 2 + ( y − 3) 2 = 2 (2) Expand the binomials and then subtract. x 2 + 6 x + 9 + y 2 − 4 y + 4 = 20
30.
⎧⎪( x − 4) 2 + ( y − 5) 2 = 8 (1) ⎨ ⎪⎩ ( x + 1) 2 + ( y + 2) 2 = 34 (2) Expand the binomials and then subtract. x 2 − 8 x + 16 + y 2 − 10 y + 25 = 8
x 2 − 4x + 4 + y 2 − 6 y + 9 = 2 10 x + 5 10 x
x 2 + 2 x + 1 + y 2 + 4 y + 4 = 34
+ 2 y − 5 = 18 + 2y = 18 y = −5 x + 9
− 10 x + 15
− 14 y + 21 = −26
− 10 x
− 14 y
= −62
5x
+ 7y
= 31 y=
Substitute for y in Eq. (2): y − 3 = −5 x + 6. ( x − 2) 2 + (−5 x + 6) 2 = 2 x 2 − 4 x + 4 + 25 x 2 − 60 x + 36 = 2
Substitute for y in Eq. (1): y − 5 =
26 x 2 − 64 x + 38 = 0
2
13x − 32 x + 19 = 0 (13x − 19)( x − 1) = 0 19 or x = 1 13
x 2 − 8 x + 16 +
25 x 2 + 40 x + 16 =8 49
49 x 2 − 392 x + 784 + 25 x 2 + 40 x + 16 = 392 74 x 2 − 352 x + 408 = 0
Substitute for x in y = −5x + 9. ⎛ 19 ⎞ y = −5 ⎜ ⎟ + 9 ⎝ 13 ⎠ 22 y= 13
−5 x − 4 . 7
⎛ − 5x − 4 ⎞ ( x − 4) 2 + ⎜ ⎟ =8 7 ⎠ ⎝
2
x=
31 − 5 x 7
y = −5(1) + 9 y=4
37 x 2 − 176 x + 204 = 0 (37 x − 102)( x − 2) = 0 x=
102 or x = 2 37
Substitute for x in y = The solutions are (19/13, 22/13) and (1, 4).
31 − 5 x . 7
31 5 ⎛ 102 ⎞ − ⎜ ⎟ 7 7 ⎝ 37 ⎠ 91 y= 37 y=
31 − 5(2) 7 y=3 y=
The solutions are (102/37, 91/37) and (2, 3).
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640
31.
Chapter 9: Systems of Equations and Inequalities
⎧⎪ ( x −1)2 + ( y +1)2 = 2 (1) ⎨ 2 2 ⎪⎩( x + 2) + ( y − 3) = 3 (2) Expand the binomials and then subtract.
32.
⎧⎪ ( x +1) 2 + ( y − 3)2 = 4 (1) ⎨ 2 2 ⎪⎩( x − 3) + ( y + 2) = 2 (2) Expand the binomials and then subtract.
x 2 − 2x +1+ y 2 + 2 y +1 = 2
x 2 + 2x +1 + y 2 − 6 y + 9 = 4
x 2 + 4x + 4 + y 2 − 6 y + 9 = 3
x 2 − 6x + 9 + y 2 + 4 y + 4 = 2
− 6x − 3 − 6x
+ 8 y − 8 = −1 + 8y = 10 3x + 5 y= 4
Substitute for y in Eq. (1): y + 1 =
8x − 8 8x
3x + 9 . 4
− 10 y + 5 = 2 − 10 y =5 8x − 5 y= 10
Substitute for y in Eq. (1): y − 3 =
2
2
⎛ 3x + 9 ⎞ ( x − 1) 2 + ⎜ ⎟ =2 ⎝ 4 ⎠ x 2 − 2x +1+
⎛ 8 x − 35 ⎞ ( x + 1) 2 + ⎜ ⎟ =4 ⎝ 10 ⎠
9 x 2 + 54 x + 81 =2 16
x 2 + 2x + 1 +
16 x 2 − 32 x + 16 + 9 x 2 + 54 x + 81 = 32
33.
64 x 2 − 560 x + 1225 =4 100
100 x 2 + 200 x + 100 + 64 x 2 − 560 x + 1225 = 400
25 x 2 + 22 x + 65 = 0 x=
8x − 5 . 10
164 x 2 − 360 x + 925 = 0
− 22 ± 22 2 − 4(25)(65)
x=
2( 25)
360 ± 360 2 − 4(164)(925) 2(164)
− 22 ± − 6016 x= 50
360 ± − 477200 x= 2(164)
x is not a real number. There are no real solutions. The curves do not intersect.
x is not a real number. There are no real solutions. The curves do not intersect.
h = height w = weight
34.
V = lwh, l = w l 2 h = 121
2h + 2 w = 25
2l + 2h = 19 ⇒ l = 19 − 2h 2
wh = 37.5 ⇒ w = 37.5 h
(19 −2 2h ) h = 121
( )
2
2h + 2 37.5 = 25 h 2 2h + 75 = 25h
⎛ 361 − 76h + 4h 2 ⎞ ⎜ ⎟ h = 121 4 ⎝ ⎠
2h 2 − 25h + 75 = 0 (2h − 15)( h − 5) = 0 2h − 15 = 0 h−5=0 h = 7.5 h=5 Since the height is greater than the width, h = 7.5. w = 37.5 = 5 7.5 The width is 5 in. and the height is 7.5 in.
4h 3 − 76h 2 4 4 −76 16 4 −60
+ 361h − 484 = 0 361 −484 −240 484 121 0
h=4 19 − 2(4) = 5.5 l = 19 − 2h = 2 2 The height is 4 in. and the length and width is 5.5 in.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.3
35.
641
36.
x 2 + y 2 = 865 2
2
x − y = 703
h 2 + w2 = (25.0)2 h = 1.6w
2 x 2 = 1568 x 2 = 784 x = 28
(1.6w)2 + w2 = 625 3.56 w2 = 625 w2 = 625 3.56 w = 625 ≈ 13.2 3.56
282 + y 2 = 865 y 2 = 81 y=9
The small carpet is 9 ft by 9 ft and the large carpet is 28 ft by 28 ft.
37.
h = height w = weight
r = radius of the small globe R = radius of the large globe V = 4 π r3 3 4 π R 3 = 8 4 π r 3 ⇒ R 3 = 8r 3 3 3 4 π R 3 − 4 π r 3 = 15,012.63 3 3
(
)
− 4 π R 3 + 32 π r 3 = 0 3 3 4 π R 3 − 4 π r 3 = 15,012.63 3 3 28 π r 3 = 15,012.63 3 3(15,012.63) r3 = 28π 3(15,012.63) r=3 28π r ≈ 8.0
h = 1.6 w = 1.6(13.2) = 21.1 The width is 13.2 ft and the height is 21.1 ft 38.
2 x 2 + ( −16) = 1 472 252
( )
2
x 2 = 1 − 16 25 472 ⎛ x 2 = 472 ⎜ 1 − 16 25 ⎝ x=±
( ) ⎞⎟⎠ ⎛ ⎞ 47 ⎜ 1 − ( 16 ) ⎟ 25 ⎠ ⎝ 2
x ≈ −36.1 The point A is (–36.1, –16).
R 3 = 8r 3 R 3 = 8(8.0)3 R = 16.0 The radius of the large globe is 16.0 in. and the radius of the small globe is 8.0 in.
Copyright © Houghton Mifflin Company. All rights reserved.
2
2
642
39.
Chapter 9: Systems of Equations and Inequalities
40.
⎧⎪ x 2 = y (1) ⎨ (2) ⎪⎩18 x − 22 = 3 y + 5 Substitute for y in Eq. (2).
Let x = the original distance of the ladder from the wall and y = the length of the ladder.
( )
18 x − 22 = 3 x 2 + 5 0 = 3 x 2 − 18 x + 27 0 = 3( x 2 − 6 x + 9) 0 = 3( x − 3)2 x=3
By the Pythagorean Theorem, 2
72 + x2 = y 2
y=3 =9
49 + x 2 = y 2
P = x 2 + 3 y + 5 + y + 18 x − 22
and
52 + ( x + 1)2 = y 2 25 + x 2 + 2 x + 1 = y 2
(1)
x 2 + 2 x + 26 = y 2
= 32 + 3(9) + 5 + 9 + 18(3) − 22 = 82 units
Set the expressions for y 2 equal to each other. x 2 + 2 x + 26 = x 2 + 49 2 x = 23 x = 11.5 Substitute for x in Eq. (1). y 2 = 49 + (11.5)2 = 181.25 y = 181.25 ≈ 13.46 meters
41.
42.
⎧⎪ x 2 + y 2 = r 2 (1) ⎨ (2) y = 2x + 1 ⎪⎩ Substitute for y in Eq. (1)
x 2 + (2 x + 1)2 = r 2 2
2
x + 4x + 4x + 1 = r
2
5x2 + 4 x + 1 = r 2 Minimize r 2 by completing the square. 4 4 ⎞ 4 ⎛ r 2 = 5x2 + 4 x + 1 = 5 ⎜ x2 + x + ⎟ + 1 − 5 25 5 ⎝ ⎠ 2
2⎞ 1 ⎛ = 5⎜ x + ⎟ + 5⎠ 5 ⎝
ab = ( a − 3)(b + 2) ab = ab + 2a − 3b − 6 6 = 2a − 3b
(1)
ab = (a + 3)(b − 1) ab = ab − a + 3b − 3 3 = − a + 3b
(2)
Adding Eq. (1) and Eq. (2) we have 9 = a. Substitute 9 for a in Eq. (1).
⎛ 2 1⎞ Thus ⎜ − , ⎟ is the point on both x 2 + y 2 = r 2 and ⎝ 5 5⎠ y = 2 x + 1 for which x 2 + y 2 = r 2 has the smallest radius. Substitute for x in r 2 = 5 x 2 + 4 x + 1 2
1 ⎛ 2⎞ ⎛ 2⎞ r 2 = 5⎜ − ⎟ + 4⎜ − ⎟ + 1 = . 5 ⎝ 5⎠ ⎝ 5⎠ r= Therefore r ≥
ab = ( a − 3)(b + 2) = (a + 3)(b − 1)
1 or 5
6 = 2(9) − 3b 6 = 18 − 3b −12 = −3b 4=b Thus, the dimensions of the first rectangle are 9 and 4. Therefore the area is (9)(4) = 36 square units.
5 is the minimum radius. 5
5 . 5
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.3
43.
45.
643
⎧ p2 ⎪ x = − 20 ⎪ 5 ⎨ 17,710 ⎪x = p +1 ⎩⎪ Use a graphing calculator and INTERSECTION.
44.
on [0, 100] by [0, 600] The graphs intersect at (45, 385).
on [0, 100] by [0, 600] The graphs intersect at (66, 342).
The solution is $45.
The solution is $66. 46.
⎧⎪ y = 2 x ⎨ ⎪⎩ y = x + 1
49.
⎧ y = log 2 x ⎨ ⎩y = x − 3 Using a graphing calculator, graph the two equations on the same coordinate grid. Using the ZOOM feature, estimate the coordinates of the points where the graphs intersect. These coordinates are the solutions of the system of equations. For this system of equations, the solutions are approximately (0.1375, −2.8625) and (5.4449, 2.4449).
Using a graphing calculator, graph the two equations on the same coordinate grid. Using the ZOOM feature, estimate the coordinates of the points where the graphs intersect. These coordinates are the solutions of the system of equations. For this system of equations, the solutions are (0, 1) and (1, 2). 47.
⎧ p2 ⎪ x = − 384 ⎪ 6 ⎨ 22,914 ⎪x = p +1 ⎩⎪ Use a graphing calculator and INTERSECTION.
⎧⎪ y = e − x ⎨ ⎪⎩ y = x 2 Using a graphing calculator, graph the two equations on the same coordinate grid. Using the ZOOM feature, estimate the coordinates of the points where the graphs intersect. These coordinates are the solutions of the system of equations. For this system of equations, the solution is approximately (0.7035, 0.4949).
48.
⎧y = x ⎪ ⎨ 1 ⎪y = x −1 ⎩ Using a graphing calculator, graph the two equations on the same coordinate grid. Using the ZOOM feature, estimate the coordinates of the points where the graphs intersect. These coordinates are the solutions of the system of equations. For this system of equations, the solution is approximately (1.7549, 1.3247).
50.
⎧ y = ln x ⎨ ⎩ y = −x + 4
Using a graphing calculator, graph the two equations on the same coordinate grid. Using the ZOOM feature, estimate the coordinates of the points where the graphs intersect. These coordinates are the solutions of the system of equations. For this system of equations, the solution is approximately (2.9263, 1.0737). 6 ⎧ ⎪⎪ y = x + 1 ⎨ ⎪y = x x −1 ⎩⎪ Using a graphing calculator, graph the two equations on the same coordinate grid. Using the ZOOM feature, estimate the coordinates of the points where the graphs intersect. These coordinates are the solutions of the system of equations. For this system of equations, the solutions are (2, 2) and (3, 3/2).
Copyright © Houghton Mifflin Company. All rights reserved.
644
Chapter 9: Systems of Equations and Inequalities
....................................................... 51.
Connecting Concepts 52.
⎧⎪ y = x 2 + 4 ⎨ ⎪⎩ x = y 2 − 24 Solve by substitution.
⎧⎪ y = x 2 − 5 ⎨ ⎪⎩ x = y 2 − 13 Solve by substitution.
x = ( x + 4) − 24
x = ( x 2 − 5)2 − 13
x = x 4 + 8 x 2 + 16 − 24
x = x 4 − 10 x 2 + 25 − 13
2
2
0 = x 4 − 10 x 2 − x + 12 0 = x 4 + 8x 2 − x − 8
By the Rational Zero Theorem, the possible rational roots are ±1, ± 2, ± 3, ± 4, ± 6, ± 12. By Descartes’ rule of signs, there are 0 or 2 positive roots and 0 or 2 negative roots.
0 = ( x − 1)( x 3 + x 2 + 9 x + 8) x3 + x 2 + 9 x + 8 is not factorable over the rational numbers because the Rational Zero Theorem implies the only rational; zeros are ±1, ± 2, ± 4, ± 8. Thus, the only rational orderedpair solution is (1, 5).
Using synthetic division, we test possible rational roots. 3
1 1
0 3 3
−10 9 −1
−1 −3 −4
12 −12 0
Thus, 3 is a root of the equation. The rational ordered-pair solution is (3, 4). 53.
54.
x 2 − 3 xy + y 2 = 5
x 2 + 2 xy − y 2 = 1
x 2 − xy − 2 y 2 = 0 Factor the second equation.
x 2 + 3xy + 2 y 2 = 0 Factor the second equation.
( x − 2 y )( x + y ) = 0
( x − 2 y )( x + y ) = 0
Thus x = 2y or x =-y. Substituting each expression into the first equation, we have
Thus x = −y or x = −2y. Substitute each expression into the first equation and solve for y.
(2 y ) 2 − 3(2 y ) y + y 2 = 5 2
2
(− y ) 2 + 2(− y ) y − y 2 = 1
2
4y − 6y + y = 5
y2 − 2y2 − y2 =1
− y2 = 5
− 2y2 = 1
y 2 = −5
y2 = −
There are no rational solutions.
1 2
There are no rational solutions.
(− y ) 2 − 3(− y ) y + y 2 = 5 y 2 + 3y 2 + y 2 = 5
(−2 y ) 2 + 2(−2 y ) y − y 2 = 1
5y 2 = 5
4y2 − 4y2 − y2 = 1
y 2 = 1 ⇒ y = ±1 Substituting into x =-y, we have x = −1 or x = 1. The rational ordered-pair solutions are (−1, 1) and (1, −1).
− y2 =1 y 2 = −1 There are no rational solutions.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.3
55.
645
56.
⎧⎪2 x 2 − 4 xy − y 2 = 6 ⎨ ⎪⎩4 x 2 − 3 xy − y 2 = 6 Subtract the two equations.
⎧⎪3x 2 + 2 xy − 5 y 2 =11 ⎨ 2 2 ⎪⎩ x + 3 xy + y =11 Subtract the two equations. 2 x 2 − xy − 6 y 2 = 0 Factoring, we have (2 x + 3 y )( x − 2 y ) = 0. Then 2 x = −3 y or x = 2 y
− 2 x 2 − xy = 0 − x( 2 x + y ) = 0 x = 0 or y = −2 x
x=−
2
Substituting x = 0 into the first equation gives − y = 6 or y 2 = −6. There are no rational solutions.
3 2y
Substituting into the first equation, we have 2
⎛ ⎞ ⎛ ⎞ 3 ⎜ − 3 y ⎟ + 2 ⎜ − 3 y ⎟ y − 5 y 2 =11 ⎝ 2 ⎠ ⎝ 2 ⎠ 27 y 2 − 3 y 2 − 5 y 2 =11 4 − 5 y 2 =11 4 y 2 = − 44 5 There are no rational solutions.
Substituting y = −2 x into the first equation gives 2 x 2 + 8x 2 − 4x 2 = 6 6x 2 = 6 x2 =1 x = ±1
The rational ordered-pair solutions are (1, − 2) and (−1, 2).
3(2 y ) 2 + 2(2 y ) y − 5 y 2 =11 12 y 2 + 4 y 2 − 5 y 2 =11 11y 2 =11 y 2 =1
y = 1 or y = −1 Substituting each of these into the first equation and solving for x, we have (2, 1) and (−2, −1) as solutions.
.......................................................
Prepare for Section 9.4
PS1. x 4 +14 x 2 + 49 = ( x 2 + 7) 2
PS2.
PS3.
PS4. ⎧ 1= A + B (1) ⎨ ⎩11= −5 A + 3B (2)
2 7 − 6 + 10 = ( x −1) ⋅ 7 − x( x −1) ⋅ 6 + x ⋅ 10 2 x x −1 ( x −1) ( x −1) 2 x x( x −1) x −1 x ( x −1) 2 2 2 = 7 x −14 x2+ 7 − 6 x − 6 2x + 10 x 2 x( x −1) x( x −1) x( x −1) 2
= x + 2 x +27 x( x −1)
5 + 1 = x + 2 ⋅ 5 + x −1⋅ 1 x −1 x + 2 x + 2 x −1 x −1 x + 2 x −1 = 5 x +10 + ( x −1)( x + 2) ( x −1)( x + 2 = 6x +9 ( x −1)( x + 2)
Solve equation (1) for A and substitute into equation (2). 11= −5(1− B ) + 3B 11= −5 + 8 B 16 = 8 B 2= B A=1− 2 = −1 The solution is (–1, 2).
Copyright © Houghton Mifflin Company. All rights reserved.
646
Chapter 9: Systems of Equations and Inequalities
PS5. ⎧ 0 = A + B (1) ⎪ = − B + C 3 2 (2) ⎨ ⎪16 = 7 A − 2C (3) ⎩
PS6.
x3 − 4 x 2 −19 x − 35 x2 − 7 x Use long division. x +3 x 2 − 7 x x3 − 4 x 2 −19 x − 35
Solve equation (1) for A and substitute into equation (3). 16 = –7B –2C (4)
x3 − 7 x 2
Multiply equation (2) by 2 and add to equation (4). 6 = −4 B + 2C 16 = −7 B − 2C 22 = −11B −2 = B
3 x 2 −19 x 3 x 2 − 21x 2 x − 35 x3 − 4 x 2 −19 x − 35 = x + 3 + 2 x − 35 x2 − 7 x x2 − 7 x
A=2 C = 2B + 3 = 2(−2) + 3 = −1 The solution is (2, –2, –1).
Section 9.4 1.
x + 15 A B = + x( x − 5) x x − 5
2.
x +15 = A( x − 5) + Bx x +15 = ( A + B ) x − 5 A
{
1= A + B 15 = −5 A
3.
A = −3
5 x − 6 = A( x + 3) + Bx 5 x − 6 = ( A + B) x + 3 A
1 = A( x − 1) + B(2 x + 3) 1 = Ax − A + 2 Bx + 3B 1 = ( A + 2 B ) x + ( − A + 3B ) 0 = A + 2B 1 = − A + 3B
B=
5B 1 5
{
− 3 + B =1 B=4
1 A B = + ( 2 x + 3)( x − 1) 2 x + 3 x − 1
1=
5x − 6 A B = + x( x + 3) x x + 3
⎛1⎞ 0 = A + 2⎜ ⎟ ⎝5⎠ 2 A=− 5
5 = A+ B −6 = 3 A
4.
A = −2
− 2+ B = 5 B=7
6x − 5 A B = + ( x + 4)(3x + 2) x + 4 3 x + 2 6 x − 5 = A(3x + 2) + B( x + 4) 6 x − 5 = 3 Ax + 2 A + Bx + 4 B 6 x − 5 = (3 A + B) x + (2 A + 4 B)
{
6 = 3A+ B −5 = 2 A + 4 B
Solve the system. −24 = −12 A − 4 B −5 = 2 A + 4 B −29 = −10 A 29 =A 10
⎛ 29 ⎞ 6 = 3⎜ ⎟ + B ⎝ 10 ⎠ 27 B=− 10
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.4
5.
647
x+9 x( x − 3)
2
=
A B C + + x x − 3 ( x − 3) 2
2x − 7
6.
( x + 1)( x − 2)
=
2
2 x − 7 = A( x − 2) 2 + B( x +1)( x − 2) + C ( x +1) 2 x − 7 = Ax 2 − 4 Ax + 4 A + Bx 2 − Bx − 2 B + Cx + C 2 x − 7 + ( A + B) x 2 + (−4 A − B + C ) x + (4 A − 2 B + C )
x + 9 = A( x − 3) 2 + Bx( x − 3) + Cx x + 9 = Ax 2 − 6 Ax + 9 A + Bx 2 − 3Bx + Cx x + 9 = ( A + B ) x 2 + ( −6 A − 3B + C ) x + 9 A
(1) ⎧⎪ 0 = A + B ⎨ 2 = −4 A − B + C (2) ⎪⎩−7 = 4 A − 2 B + C (3) 2 = −4 A − B + C (2) 7 = −4 A + 2 B − C − 1 times (3)
⎧0 = A + B ⎪ ⎨1 = −6 A − 3B + C ⎪9 = 9 A ⎩
9 = −8 A + B
A =1
A+ B = 0 1+ B = 0 B = −1
A B C + + x + 1 x − 2 ( x − 2) 2
−6 A − 3B + C = 1 −6(1) − 3(−1) + C = 1 C=4
(4)
⎧⎪0 = A + B ⎨ 2 = −4 A − B + C ⎪⎩9 = −8 A + B
0 = A+ B −9 = 8 A − B 9=9A −1 = A
(4)
(1) −1 times (4) (5)
(5) ⎧⎪ −1 = A ⎨ 2 = −4 A − B + C ⎪⎩ 9 = −8 A + B
A = −1
7.
4x 2 + 3 2
( x − 1)( x + x + 5)
=
A Bx + C + x −1 x 2 + x + 5
8.
4 x 2 + 3 = ( A + B ) x 2 + ( A − B + C ) x + (5 A − C )
⎧1= A + C ⎪ ⎨1= −3 A + B ⎩⎪3 = −3B + 7C
(1) (2) (3)
3 = 3 A + 3C 1 = −3 A + B
From (3), C = 5A – 3. From (1), B = 4 − A. Substitute C and B into Eq. (2). 0 = A − 4 + A + 5A − 3 7 = 7A 1= A
2
C = 5(1) − 3 C=2
=
Ax + B 2
x +7
−4(−1) − 1 + C = 2 C = −1
+
C x−3
x 2 + x + 3 = ( Ax + B )( x − 3) + C ( x 2 + 7) x 2 + x + 3 = Ax 2 − 3 Ax + Bx − 3B + Cx 2 + 7C x 2 + x + 3 = ( A + C ) x 2 + ( − 3 A + B ) x + ( − 3 B + 7C )
4 x 2 + 3 = Ax 2 + Ax + 5 A + Bx 2 − Bx + Cx − C
0 = A − (4 − A) + 5 A − 3
x2 + x + 3 ( x + 7)( x − 3)
4 x 2 + 3 = A( x 2 + x + 5) + ( Bx + C )( x − 1)
⎧4 = A + B ⎪ ⎨0 = A − B + C ⎪3 = 5 A − C ⎩
9 = −8(−1) + B B =1
4 = 3C + B B = 4 −1 B=3
(1) (2) (3) 3 times (1) (2) (4)
⎧1 = A + C ⎪ (4) ⎨4 = B + 3C ⎪⎩3 = −3B + 7C 12 = 3B + 9C 3 times (4) 3 = −3B + 7C (3) 15 =
16C
(5)
⎧ 1= A + C ⎪ ⎨ 4 = B + 3C ⎪⎩15 =16C (5) C=
15 16
B + 3 ⎛⎜ 15 ⎞⎟ = 4 ⎝ 16 ⎠ B = 19 16
Copyright © Houghton Mifflin Company. All rights reserved.
A + 15 = 1 16 A= 1 16
648
9.
Chapter 9: Systems of Equations and Inequalities
x 3 + 2x ( x 2 + 1) 2
=
Ax + B
Cx + D
+
x 2 +1
( x 2 + 1) 2
x3 + 2 x = ( Ax + B )( x 2 +1) + (Cx + D) x3 + 2 x = Ax3 + Ax + Bx 2 + B + Cx + D x3 + 2 x = Ax3 + Bx 2 + ( A + C ) x + ( B + D)
⎧1 = A ⎪0 = B ⎨2 = A + C ⎪ ⎩0 = B + D 10.
A =1
3x 3 + x 2 − x − 5 3
( x + 2 x + 5)
2
=
1+ C = 2 C =1
B=0
Ax + B 2
x + 2x + 5
+
0+ D =0 D=0
Cx + D 2
( x + 2 x + 5) 2
3x3 + x 2 − x − 5 = ( Ax + B )( x 2 + 2 x + 5) + Cx + D 3x3 + x 2 − x − 5 = Ax3 + 2 Ax 2 + 5 Ax + Bx 2 + 2 Bx + 5 B + Cx + D 3x3 + x 2 − x − 5 = Ax3 + (2 A + B ) x 2 + (5 A + 2 B + C ) x + (5B + D) ⎧3 = A ⎪1 = 2 A + B ⎪ ⎨ ⎪ −1 = 5 A + 2 B + C ⎪⎩−5 = 5 B + D
11.
A=3
8 x + 12 A B = + x( x + 4) x x + 4
2(3) + B = 1 B = −5
5(3) + 2(−5) + C = −1 C = −6 12.
8 x + 12 = A( x + 4) + Bx 8 x + 12 = Ax + 4 A + Bx 8 x + 12 = ( A + B) x + 4 A ⎧ 8 = A+ B ⎨ ⎩12 = 4 A
A=3
3 x + 50
3+ B = 8
⎧ 1= A+ B ⎨ ⎩− 14 = −7 A
B=5
14.
=
3x + 50 = A( x + 2) + B( x − 9) 3x + 50 = Ax + 2 A + Bx − 9 B 3x + 50 = ( A + B ) x + (2 A − 9 B)
3 x + 50 2
x − 7 x − 18
=
2+ B =1 B = −1
3 = A + (−4) 7= A
7 x + 44
7 x + 44 x ( 4)( x + 6) + x + 10 x + 24 A B = + x+4 x+6 2
⎧ 7 = A+ B ⎨ ⎩44 = 6 A + 4 B −4 A − 4 B = −28
6 A + 4 B = 44 2A
7 −4 + x −9 x + 2
=
7 x + 44 = A( x + 6) + B( x + 4) 7 x + 44 = Ax + 6 A + Bx + 4 B 7 x + 44 = ( A + B) x + (6 A + 4 B )
⎧ 3 = A+ B ⎨ ⎩50 = 2 A − 9 B
−2 A − 2 B = −6 2 A − 9 B = 50 −11B = 44 B = −4
A=2
x − 14 2 −1 = + x( x − 7) x x − 7
3 x + 50 − x ( 9)( x + 2) x − 7 x − 18 A B = + x−9 x+2 2
x − 14 A B = + x ( x − 7) x x − 7 x − 14 = A( x − 7) + Bx x − 14 = Ax − 7 A + Bx x − 14 = ( A + B) x − 7 A
8 x + 12 3 5 = + x( x + 4) x x + 4 13.
5(−5) + D = −5 D = 20
= 16 A=8
7 x + 44 2
x + 10 x + 24
8+ B = 7 B = −1 =
8 −1 + x+4 x+6
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.4
15.
649
16 x + 34 4 x 2 + 16 x + 15
16 x + 34 (2 x + 3)(2 x + 5) A B = + 2x + 3 2x + 5
=
16.
−15 x + 37 (3x + 1)(3 x − 5) A B = + 3x + 1 3x − 5 =
16 x + 34 = A(2 x + 5) + B (2 x + 3) 16 x + 34 = 2 Ax + 5 A + 2 Bx + 3B
−15 x + 37 = A(3x − 5) + B(3 x + 1) − 15 x + 37 = 3 Ax − 5 A + 3Bx + b
16 x + 34 = (2 A + 2 B ) x + (5 A + 3B)
− 15 x + 37 = (3 A + 3B) x + (−5 A + B)
⎧16 = 2 A + 2 B ⎨ ⎩34 = 5 A + 3B
⎧−15 = 3 A + 3B ⎨ ⎩ 37 = −5 A + B
(1) (2)
6 A + 6 B = 48 3 times (1) − 10 A − 6 B = −68 − 2 times (2) − 4A
= −20
2(5) + 2 B = 16 B=3 16 x + 34
4 x 2 + 16 x + 15
=
x−5 A B = + (3x + 5)( x − 2) 3 x + 5 x − 2 x − 5 = A( x − 2) + B(3 x + 5) x − 5 = Ax − 2 A + 3Bx + 5B x − 5 = ( A + 3B) x + (−2 A + 5B)
2 A + 6B = 2 −2 A + 5B = −5 11B = −3 3 B=− 11
2
9 x − 12 x − 5 18.
⎧ 1 = A + 3B (1) ⎨ ⎩− 5 = −2 A + 5 B (2)
= 42 = −7
−15 x + 37
5 3 + 2x + 3 2x + 5
(2)
−
− 5 A + B = 37 − 6A A
(1)
1 times (1) 3 (2)
− A− B = 5
A=5
17.
−15 x + 37 9 x 2 − 12 x − 5
=
−7 2 + 3x + 1 3x − 5
1 A B = + ( x + 7)(2 x − 5) x + 7 2 x − 5 1 = A(2 x − 5) + B ( x + 7) 1 = 2 Ax − 5 A + Bx + 7 B 1 = (2 A + B) x + ( −5 A + 7 B) ⎧0 = 2 A + B (1) ⎨ ⎩1 = −5 A + 7 B (2)
−14 A − 7 B = 0 −5 A + 7 B = 1
2 times (1) (2) ⎛ 3⎞ A + 3⎜ − ⎟ = 1 ⎝ 11 ⎠ 20 A= 11
x−5 20 −3 = + (3 x + 5)( x − 2) 11(3 x + 5) 11( x − 2)
−19 A
− 7 times (1) (2)
=1 A=−
1 19
⎛ 1⎞ 2⎜ − ⎟ + B = 0 ⎝ 19 ⎠ 2 B= 19
1 −1 2 = + ( x + 7)(2 x − 5) 19( x + 7) 19(2 x − 5)
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650
19.
Chapter 9: Systems of Equations and Inequalities
x+3
20.
x 2 − 4 x 3 + 3x 2 − 4 x − 8 x3
x +1 x 2 − x − 12 x 3
x 3 − x 2 − 12 x
− 4x 3x 2 3x
2
− 13x − 9
−8
x2 −
x−9
− 12
2
x − 12
x −
4 3
2
x + 3x − 4 x − 8 2
x −4
= x +3+
3 3
x − 13 x − 9
3 = x +1+ ( x − 4)( x + 3) x 2 − x − 12 A B 3 = + ( x − 4)( x + 3) x − 4 x + 3 3 = A( x + 3) + B ( x − 4)
4 ( x − 2)( x + 2)
4 A B = + ( x − 2)( x + 2) x − 2 x + 2 4 = A( x + 20 + B( x − 2) 4 = Ax + 2 A + Bx − 2 B
3 = Ax + 3 A + Bx − 4 B 3 = ( A + B ) x + (3 A − 4 B )
4 = ( A + B) x + (2 A − 2 B) ⎧0 = A + B ⎨ ⎩4 = 2 A − 2 B
⎧0 = A + B ⎨ ⎩3 = 3 A − 4 B
(1) (2)
4 A + 4B = 0 3 A − 4B = 3
2 A + 2 B = 0 2 times (1) 2 A − 2 B = 4 (2) =4
4A
A =1 x 3 + 3x 2 − 4 x − 8 x2 − 4
21.
3 x 2 + 49 x ( x + 7)
2
=
7A
1+ B = 0 B = −1 = x +3+
(2)
4 times (1) (2)
=3
3 +B=0 7
3 7
B=−
A=
1 −1 + x−2 x+2
(1)
x 3 − 13x − 9 2
x − x − 12
= x +1+
3 −3 + 7( x − 4) 7( x + 3)
A B C + + x x + 7 ( x + 7) 2
3x 2 + 49 = A( x + 7) 2 + Bx( x + 7) + Cx 3x 2 + 49 = Ax 2 +14 Ax + 49 A + Bx 2 + 7 Bx + Cx 3x 2 + 49 = ( A + B ) x 2 + (14 A + 7 B + C ) x + 49 A ⎧ 3 = A+ B ⎪ ⎨ 0 = 14 A + 7 B + C ⎪49 = 49 A ⎩ 3 x 2 + 49 x( x + 7)
22.
2
x − 18 x( x − 3) 2
A =1
1+ B = 3
14(1) + 7(2) + C = 0
B=2
=
1 2 − 28 + + x x + 7 ( x + 7) 2
=
A B C + + x x − 3 ( x − 3) 2
C = −28
x −18 = A( x − 3) 2 + Bx( x − 3) + Cx x −18 = Ax 2 − 6 Ax + 9 A + Bx 2 − 3Bx + Cx x −18 = ( A + B ) x 2 + (−6 A − 3B + C ) x + 9 A 9 A = −18 ⎧ 0 = A+ B ⎪ 1 = − 6 A − 3 B + C A = −2 ⎨ ⎪− 18 = 9 A ⎩ −2 2 −5 x − 18 = + + 2 x x − 3 ( x − 3) 2 x( x − 3)
−2+ B = 0 B=2
3 7
− 6(−2) − 3(2) + C = 1 C = −5
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.4
23.
651
5 x 2 − 7 x + 2 = 5 x 2 − 7 x + 2 = A + Bx + C x3 − 3 x 2 + x x( x 2 − 3 x + 1) x x 2 − 3 x + 1 5 x 2 − 7 x + 2 = A( x 2 − 3 x +1) + ( Bx + C ) x 5 x 2 − 7 x + 2 = Ax 2 − 3 Ax + A + Bx 2 + Cx 5 x 2 − 7 x + 2 = ( A + B ) x 2 + ( −3 A + C ) x + A
⎧⎪ 5 = A + B ⎨ −7 = − 3 A + C ⎪⎩ 2 = A
5x 2 − 7 x + 2 3
2
x − 3x + x 24.
=
A=2
− 3(2) + C = −7 C = −1
2 3x − 1 + x x 2 − 3x + 1
9 x 2 − 3x + 49 3
2+ B =5 B=3
2
x − x + 10 x − 10 2
9 x 2 − 3x + 49
=
2
( x − 1)( x + 10)
=
A Bx + C + x − 1 x 2 + 10
2
9 x − 3 x + 49 = A( x +10) + ( Bx + C )( x −1) 9 x 2 − 3 x + 49 = Ax 2 +10 A + Bx 2 − Bx + Cx − C 9 x 2 − 3 x + 49 = ( A + B ) x 2 + (− B + C ) x + (10 A − C ) ⎧ 9= A+ B ⎪ ⎨−3 = − B + C ⎪⎩ 49 = 10 A − C 9 x 2 − 3 x + 49 3
2
x − x + 10 x − 10 25.
−3 = − B + C 49 = 10 A − C 46 = 10 A − B
5+ B =9 B=4
−4 + C = −3 C =1
5 4x +1 + x − 1 x 2 + 10
=
2 x 3 + 9 x 2 + 26 x + 41 2
10 A − B = 46 A+ B =9 11A = 55 A=5
2
( x + 3) ( x + 1)
=
A B Cx + D + + x + 3 ( x + 3) 2 x 2 +1
3
2 x + 9 x 2 + 26 x + 41 = A( x + 3)( x 2 + 1) + B( x 2 + 1) + (Cx + D)( x + 3) 2 2 x3 + 9 x 2 + 26 x + 41 = Ax3 + Ax + 3 Ax 2 + 3 A + Bx 2 + B + Cx3 + 6Cx 2 + 9Cx + Dx 2 + 6 Dx + 9 D 2 x3 + 9 x 2 + 26 x + 41 = ( A + C ) x3 + (3 A + B + 6C + D ) x 2 + ( A + 9C + 6 D) x + (3 A + B + 9 D) ⎧ 2 = A+C ⎪ 9 = 3 A + B + 6C + D ⎨26 = A + 9C + 6 D ⎪ ⎩ 41 = 3 A + B + 9 D
(1) (2) (3) (4)
3 A + B + 6C + D = 9 − 3 A − B − 9 D = −41
(3) − 1 times (4)
6C − 8D = −32 (5) A + 9C + 6 D = 26 (3) − A−C = −2 − 1 times (1) 8C + 6 D = 24
(6)
⎧ 2 = A+C ⎪ 9 = 3 A + B + 6C + D ⎨ 24 = 8C + 6 D (6) ⎪ (5) ⎩−32 = 6C − 8D 64C + 48D = 192 8 times (6) 36C − 48D = −192 6 times (5) 100C
C=0 3
=0 A+0 = 2
8(0) + 6 D = 24
A=2 2
2 x + 9 x + 26 x + 41 2
2
( x + 3) ( x + 1)
D=4 =
3(2) + B + 9(4) = 41 B = −1
2 −1 4 + + x + 3 ( x + 3) 2 x 2 + 1
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652
26.
Chapter 9: Systems of Equations and Inequalities
12 x 3 − 37 x 2 + 48 x − 36 ( x − 2) 2 ( x 2 + 4)
=
A B Cx + D + + x − 2 ( x − 2) 2 x 2 + 4
12 x 3 − 37 x 2 + 48 x − 36 = A( x − 2)( x 2 + 4) + B ( x 2 + 4) + (Cx + D )( x − 2) 2 12 x 3 − 37 x 2 + 48 x − 36 = Ax 3 + 4 Ax − 2 Ax 2 − 8 A + Bx 2 + 4 B + Cx 3 − 4Cx 2 + 4Cx + Dx 2 − 4 Dx + 4 D 12 x 3 − 37 x 2 + 48 x − 36 = ( A + C ) x 3 + (−2 A + B − 4C + D ) x 2 + (4 A + 4C − 4 D ) x + (−8 A + 4 B + 4 D) (1) ⎧ 12 = A + C ⎪ ⎪− 37 = −2 A + B − 4C + D (2) ⎨ (3) ⎪ 48 = 4 A + 4C − 4 D ⎪⎩− 36 = −8 A + 4 B + 4 D (4)
4 A + 4C = 48 − 4 A − 4C + 4 D = −48 4D = 0 D=0
4 times (1) − 1 times (3) (5)
⎧ 12 = A + C ⎪ ⎪− 37 = −2 A + B − 4C + D ⎨ (5) ⎪ 0=D ⎪⎩− 36 = −8 A + 4 B + 4 D
From (1) C = 12 − A From (4) 4 B = 8 A − 36 − 4 D B = 2A − 9 −2 A + (2 A − 9) − 4(12 − A) + 0 = −37 −2 A + 2 A − 9 − 48 + 4 A = −37 4 A = 20 A=5
Substitute in Eq. (2).
12 x 3 − 37 x 2 + 48 x − 36 2
2
( x − 2) ( x + 4)
27.
=
C = 12 − 5 C=7
B = 10 − 9 B =1
5 1 7x + + x − 2 ( x − 2) 2 x 2 + 4
3x − 7 A B = + ( x − 4) 2 x − 4 ( x − 4) 2 3x − 7 = A( x − 4) + B
28.
5 x − 53 A B = + ( x − 11) 2 x − 11 ( x − 11) 2 5 x − 53 = A( x − 11) + B
3x − 7 = Ax − 4 A + B 3x − 7 = Ax + (−4 A + B)
5 x − 53 = Ax − 11A + B 5 x − 53 = Ax + ( −11A + B )
⎧ 3= A B − 4(3) = −7 ⎨ − 7 = − 4 A + B B=5 ⎩ 3x − 7 3 5 = + ( x − 4) 2 x − 4 ( x − 4) 2
⎧ 5= A −11(5) + B = −53 ⎨ − 53 = − 11 A + B B=2 ⎩ 5 x − 53 5 2 = + ( x − 11) 2 x − 11 ( x − 11) 2
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.4
29.
653
3 x 3 − x 2 + 34 x − 10 2
( x + 10)
2
Ax + B 2
x + 10
+
Cx + D 2
( x + 10)
4
=
2
x + 14 x + 49
2 x3 + 9 x + 1 ( x 2 + 7) 2
=
Ax + B x2 + 7
+
Cx + D ( x 2 + 7)2
2 x3 + 9 x +1= ( Ax + B)( x 2 + 7) + Cx + D 2 x3 + 9 x +1= Ax3 + Bx 2 + 7 Ax + 7 B + Cx + D 2 x3 + 9 x +1= Ax3 + Bx 2 + (7 A + C ) x + (7 B + D)
⎧ 3= A ⎪ ⎪ −1 = B ⎨ ⎪ 34 = 1 − A + C ⎪⎩− 10 = 10 B + D
⎧2 = A ⎪ ⎪0 = B ⎨ ⎪9 = 7 A + C ⎪⎩1 = 7 B + D
2
( x + 10)
1 2
k −x
=
2
2
10(3) + C = 34
10(−1) + D = −10
C=4
=
3x − 1 2
x + 10
+
D=0
2
( x + 10)
7( 2) + C = 9 C = −5
2 x3 + 9 x + 1
4x
4
2
=
2
x + 14 x + 49
1 A B = + (k − x)(k + x) k − x k + x
32.
1= A(k + x) + B (k − x ) 1= Ak + Ax + Bk − Bx 1= ( A − B ) x + ( Ak + Bk )
2x 2
x +7
+
( x 2 + 7)2
1 = A+ B x( k + mx ) x k + mx 1 = A( k + mx) + Bx 1 = Ak + Amx + Bx 1 = ( Am + B) x + Ak 0 = Am + B 1 = Ak
{
(1) (2)
A=
Ak + Bk = 1 Ak − Bk = 0 2 Ak =1
(2) k times (1)
m k 1 1 −m = + x(k + mx) kx k (k + mx)
1 2
2
k −x
=
1 m+B=0 k
1 k
B=−
1 −B =0 2k
1 2k
B=
1 2k
1 1 + 2k ( k − x ) 2k ( k + x)
x 2
3
34.
2
x − x x − x − x −1
x +1 2
2 x + 3x − 2 2 x 3 + 5 x 2 + 3x − 8
x3 − x 2
2 x 3 + 3x 2 − 2 x − x −1
3
2
x − x − x −1 2
= x+
2x 2 + 5x − 8
− x −1
2 x 2 + 3x − 2
2
x −x x −x −x −1 A B = + x( x − 1) x x − 1
2x − 6 2 x3 + 5 x 2 + 3 x − 8 2 x2 +3 x − 2
− x − 1 = Ax − A + Bx − x − 1 = ( A + B) x − A ⎧−1 = A + B ⎨ ⎩− 1 = − A 3
2
x − x − x −1 2
x −x
7(0) + D = 1 D =1
−5 x + 1
⎧0 = A − B ⎨ ⎩1 = Ak + Bk
A=
33.
2 x3 + 9 x + 1
30. 2
3x3 − x 2 + 34 x −10 = ( Ax + B)( x 2 +10) + Cx + D 3x3 − x 2 + 34 x −10 = Ax3 + Bx 2 +10 Ax +10 B + Cx + D 3x3 − x 2 + 34 x −10 = Ax3 + Bx 2 + (10 A + C ) x + (10 B + D)
3 x 3 − x 2 + 34 x − 10
31.
=
2 x −6 (2 x −1)( x + 2)
A =1
= x+
1 + B = −1
1 −2 + x x −1
B = −2
= x +1+ =
2 x−6 2 x2 + 3 x − 2
A B + 2 x −1 x + 2
2 x − 6 = A( x + 2) + B(2 x − 1) 2 x − 6 = Ax + 2 A + 2 Bx − B 2 x − 6 = ( A + 2 B) x + (2 A − B ) 2 = A + 2B 2 = A + 2 B ⎧ − 12 = 4 A − 2 B ⎨ − 10 = 5 A ⎩− 6 = 2 A − B
−2 + 2 B = 2
−2 = A 3
2
2 x + 5 x + 3x − 8 2
2 x + 3x − 2
= x +1+
Copyright © Houghton Mifflin Company. All rights reserved.
−2 2 + 2x −1 x + 2
2B = 4 B=2
654
35.
Chapter 9: Systems of Equations and Inequalities
2x − 2
36.
x 2 − x −1 2x 3 − 4x 2
+5
2x 3 − 2x 2 − 2x − 2 x 2 + 2x + 5 − 2 x 2 + 2x + 2 3 3
2
2x − 4x + 5 x 2 − x −1
= 2x − 2 +
x +1 x 3 − 3 x 2 x 4 − 2 x3 − 2 x 2 − x + 3 x 4 − 3x3 x3 − 2 x 2 − x + 3 x3 − 3 x 2 x2 − x + 3 x 4 − 2x 3 − 2x 2 − x + 3 2
3
x ( x − 3)
x 2 − x −1
= x +1+
x2 − x + 3 x 2 ( x − 3)
x 2 − x + 3 = Ax( x − 3) + B ( x − 3) + Cx 2 x 2 − x + 3 = Ax 2 − 3 Ax + Bx − 3B + Cx 2 x 2 − x + 3 = ( A + C ) x 2 + ( −3 A + B ) x − 3 B
⎧ 1= A + C ⎪ ⎨−1= −3 A + B ⎪⎩ 3 = −3B
−3B = 3
−3 A − 1 = −1 A=0
B = −1
0+C =1 C =1
x 4 − 2 x 3 − 2 x 2 − x + 3 = x + 1 + −1 + 1 x 2 ( x − 3) x2 x − 3
....................................................... 37.
( x −1)( x +1) x 2 −1 = ( x −1)( x + 2)( x − 3) ( x −1)( x + 2)( x − 3) x +1 = = A + B ( x + 2)( x − 3) x + 2 x − 3 x + 1 = A( x − 3) + B ( x + 2) x + 1 = Ax − 3 A + Bx + 2 B x + 1 = ( A + B) x + ( −3 A + 2 B) A+ B =1 ⎧ ⎨ ⎩− 3 A + 2 B = 1 3 A + 3B = 3
5B = 4 4 B= 5
x2 + x
38.
=
2
x ( x + 1)
=
2
x ( x − 4) x ( x − 4) Cancel the common term. x +1 A B = + x( x − 4) x x − 4
x +1 x ( x − 4)
x +1= A( x − 4) + Bx x +1= Ax − 4 A + Bx x +1= ( A + B ) x − 4 A
{
1= A + B 1 = −4 A
− 1 + B =1 4 B=5 4
−4 A = 1 1 5
A=
A=−1 4
x 2 −1 1 4 = + ( x − 1)( x + 2)( x − 3) 5( x + 2) 5( x − 3) 39.
Connecting Concepts
− x 4 − 4 x 2 + 3x − 6 4
x ( x − 2)
=
x2 + x 2
x ( x − 4)
=
−1 5 + 4 x 4( x − 4)
A B C D E + + + + x x 2 x3 x 4 x − 2
4
− x − 4 x 2 + 3 x − 6 = Ax3 ( x − 2) + Bx 2 ( x − 2) + Cx( x − 2) + D( x − 2) + Ex 4 − x 4 − 4 x 2 + 3 x − 6 = Ax 4 − 2 Ax3 + Bx3 − 2 Bx 2 + Cx 2 − 2Cx + Dx − 2 D + Ex 4 − x 4 − 4 x 2 + 3 x − 6 = ( A + E ) x 4 + (−2 A + B ) x3 + (−2 B + C ) x 2 + (−2C + D) x + (−2 D) ⎧ −1= A + E ⎪ 0 = −2 A + B ⎪ ⎨−4 = −2 B + C ⎪ 3 = −2C + D ⎪−6 = −2 D ⎩
− x4 − 4 x2 + 3x − 6 4
x ( x − 2)
−2 D = −6 D=3 =
−2C + 3 = 3 C =0
−2 B + 0 = −4 B=2
−2 A + 2 = 0 A = −2
−2 1 2 3 + + + x x2 x4 x − 2
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1 + E = −1 E = −2
Section 9.4
40.
655
3x 2 − 2 x − 1 2
( x − 1)
2
=
3x2 − 2 x − 1
=
2
[( x − 1)( x + 1)]
3x 2 − 2 x − 1 ( x − 1)2 ( x + 1)2
=
A B C D + + + x − 1 ( x − 1)2 x + 1 ( x + 1)2
3x 2 − 2 x − 1 = A( x − 1)( x + 1)2 + B( x + 1) 2 + C ( x + 1)( x − 1)2 + D( x − 1)2 3x 2 − 2 x − 1 = A( x − 1)( x 2 + 2 x + 1) + B ( x 2 + 2 x + 1) + C ( x + 1)( x 2 − 2 x + 1) + D ( x 2 − 2 x + 1) 3x 2 − 2 x − 1 = A( x3 + x 2 − x − 1) + B( x 2 + 2 x + 1) + C ( x3 − x 2 − x + 1) + D( x 2 − 2 x + 1) 3x 2 − 2 x − 1 = Ax3 + Ax 2 − Ax − A + Bx 2 + 2 Bx + B + Cx3 − Cx 2 − Cx + C + Dx 2 − 2 Dx + D 3 x 2 − 2 x − 1 = ( A + C ) x3 + ( A + B − C + D ) x 2 + ( − A + 2 B − C − 2 D ) x + (− A + B + C + D )
⎧ 0= A+C ⎪ 3= A+ B−C + D ⎨−2 = − A + 2 B − C − 2 D ⎪ ⎩ −1 = − A + B + C + D A+ B−C + D = 3 A− B −C − D =1 2 A − 2C =4 A −C =2
(1) (2) (3) (4)
(2) − 1 times (4) (5)
−2 A + 2 B + 2C + 2 D = −2 + 3C + 4 D = 0
−1
3(−1) + 4 D = 0 − 4 + 4D = 0
(1)
A−C = 2 2A =2 A =1
(5)
A+ B−C + D = 3 1+ B +1+1 = 3 B=0
2 times (4)
A − 2B + C + 2D = 2 −A
A+C = 0
− 1 times (3)
1+ C = 0 C = −1
(2)
D =1 2
3x − 2 x − 1 2
( x − 1) 41.
2
2 x 2 + 3x − 1 3
( x − 1)
=
=
1 1 1 − + x − 1 x + 1 ( x + 1)2 2 x 2 + 3x − 1 2
( x − 1)( x + x + 1)
=
A Bx + C + x −1 x 2 + x +1
2
2 x + 3 x − 1 = A( x 2 + x + 1) + ( Bx + C )( x − 1) 2 x 2 + 3 x − 1 = Ax 2 + Ax + A + Bx 2 − Bx + Cx − C 2 x 2 + 3x − 1 = ( A + B) x 2 + ( A − B + C ) x + ( A − C ) (1) ⎧ 2 = A+ B ⎪ A B C 3 (2) = − + ⎨ ⎪− 1 = A − C (3) ⎩ Solve Eq. (1) for B and Eq. (3) for C and substitute into Eq. (2).
A − C = −1 C = A +1
A+ B = 2 B = 2− A A− B +C = 3 A − (2 − A) + ( A + 1) = 3 A − 2 + A + A +1 = 3 3A = 4 4 A= 3 2 x 2 + 3x − 1 3
x −1
=
B = 2− A 4 B = 2− 3 2 B= 3
C = A +1 4 +1 3 7 C= 3 C=
4 2x + 7 + 3( x − 1) 3( x 2 + x + 1)
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656
42.
Chapter 9: Systems of Equations and Inequalities
x3 − 2x 2 + x − 2 4
3
x − x + x −1
=
x 3 − 2x 2 + x − 2 ( x − 1)( x + 1)( x 2 − x + 1)
=
A B Cx + D + + x −1 x +1 x 2 − x +1
x 3 − 2 x 2 + x − 2 = A( x + 1)( x 2 − x + 1) + B ( x − 1)( x 2 − x + 1) + (Cx + D )( x − 1)( x + 1) x 3 − 2 x 2 + x − 2 = Ax 3 + A + Bx 3 − 2 Bx 2 + 2 Bx − B + Cx 3 − Cx + Dx 2 − D x 3 − 2 x 2 + x − 2 = ( A + B + C ) x 3 + (−2 B + D) x 2 + (2 B − C ) x + ( A − B − D ) ⎧ 1= A+ B +C ⎪−2 = −2 B + D ⎪ ⎨ ⎪ 1 = 2B − C ⎪⎩−2 = A − B − D
(1) (2) (3) (4)
−2 B + D = −2 (2) A − B − D = −2 (4) A − 3B = −4
(5)
⎧ 1= A+ B+C ⎪−2 = −2 B + D ⎪ ⎨ ⎪ 1 = 2B − C ⎪⎩−4 = A − 3B
(1) (2) (3) (5)
Solve Eq. (3) for C and Eq. (5) for A and substitute for C and A in Eq. (1). A+ B +C =1
2B − C = 1 C = 2B − 1
A − 3B = −4 A = 3B − 4
(3B − 4) + B + (2 B − 1) = 1 3B − 4 + B + 2 B − 1 = 1 6B = 6
C = 2(1) − 1 C =1
A = 3B − 4 A = −1
−2 B + D = −2 D=0
B =1 3
2
x − 2x + x − 2 4
3
x − x + x −1 43.
=
1 x −1 + + x −1 x +1 x 2 − x +1
1 1 (a − b)( p ( x) + b) + (b − a )( p ( x) + a) + = (b − a )( p ( x) + a ) (a − b)( p ( x ) + b) (b − a )(a − b)( p( x ) + a )( p ( x) + b) (a − b) p ( x) + (a − b)b + (b − a ) p ( x) + (b − a )a = (b − a)(a − b)( p ( x) + a )( p ( x) + b) (a − b) p ( x) + (a − b)b − (a − b) p ( x) − ( a − b)a = (b − a )(a − b)( p ( x) + a )( p ( x) + b) ( a − b )b − ( a − b ) a = (b − a )(a − b)( p( x ) + a )( p( x) + b) (a − b)(b − a ) = (b − a )(a − b)( p( x ) + a )( p( x) + b) 1 = ( p ( x) + a)( p( x) + b)
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.5
44.
657
a.
Let p ( x) = x 2 , a = 4, b = 1 (See Exercise 43.) 1 1 1 = + 2 2 2 2 ( x + 4)( x + 1) − 3( x + 4) 3( x + 1)
b.
Let p( x) = x 2 , a = 1, b = 9 1 1 1 = + ( x 2 + 1)( x 2 + 9) 8( x 2 + 1) −8( x 2 + 9)
c.
Let p( x) = x 2 + x, a = 1, b = 2 1 1 1 = − 2 2 2 2 ( x + x + 1)( x + x + 2) x + x + 1 x + x + 2
d.
Let p( x) = x 2 + 2 x, a = 4, b = 9 1 1 1 = − 2 2 2 2 ( x + 2 x + 4)( x + 2 x + 9) 5( x + 2 x + 4) 5( x + 2 x + 9)
....................................................... PS1.
PS2.
Prepare for Section 9.5 PS3.
PS5.
PS4.
PS6.
Section 9.5 1.
5.
2.
y > 2x−2 3
6.
y > −x – 2
y ≤ − 3 x +1 4
3.
7.
4.
y ≤ −4x+4 3
Copyright © Houghton Mifflin Company. All rights reserved.
8.
y > 5x−4 2
658
9.
Chapter 9: Systems of Equations and Inequalities
vertex (0,0)
10.
vertex (0,0)
13.
center (2, 1), r = 4
17.
4 x 2 + 9 y 2 − 8 x + 18 y ≥ 23 4( x − 2 x + 1) + 9( y 2 + 2 y + 1) ≥ 23 + 4 + 9 4( x − 1)2 + 9( y + 1)2 ≥ 36 ( x − 1)2 ( y + 1)2 + ≥1 9 4 center (1, −1), a = 3, b = 2
14.
center (−2, 3), r = 5
vertex (1,−4)
12.
15.
center (3, −1), a = 3, b = 4
16.
18.
25 x 2 − 16 y 2 − 100 x − 64 y < 64
11.
2
(4
vertex 1 , − 25 8
center (−1, 3), a = 5, b = 4
25( x 2 − 4 x + 4) − 16( y 2 + 4 y + 4) < 64 + 100 − 64 25( x − 2)2 − 16( y + 2) 2 < 100 ( x − 2)2 ( y + 2) 2 − <1 4 25 / 4 center (2, − 2), a = 2, b = 5 / 2
19.
20.
21.
22.
23.
24.
25.
26.
27.
The graphs of each of the 28. two inequalities are shown.
29.
30.
Because the solution sets of the inequalities do not intersect, the system has no solution and cannot be graphed. Copyright © Houghton Mifflin Company. All rights reserved.
)
Section 9.5
659
31.
32.
35.
no solution
33.
34.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
Substitute Ashley’s age, 35, in the first inequality to find the minimum value. y ≥ 0.55(220 − x ) y ≥ 0.55(220 − 35) y ≥ 0.55(185) y ≥ 101.75 Substitute Ashley’s age in the second inequality to find the maximum value. y ≤ 0.75(220 − x ) y ≤ 0.75(220 − 35) y ≤ 0.75(185) y ≤ 138.75 The minimum is 102 beats per minute and maximum is 139 beats per minute.
46.
Substitute the sprinter’s age, 26, in the first inequality to find the minimum value. y ≥ 0.80(220 − x ) y ≥ 0.80(220 − 26) y ≥ 0.80(194) y ≥ 155.2 Substitute the sprinter’s age in the second inequality to find the maximum value. y ≤ 0.88(220 − x ) y ≤ 0.88(220 − 26) y ≤ 0.88(194) y ≤ 164.9 The minimum is 155 beats per minute and maximum is 165 beats per minute.
....................................................... 47.
48.
no solution
Connecting Concepts 49.
Copyright © Houghton Mifflin Company. All rights reserved.
50.
660
Chapter 9: Systems of Equations and Inequalities
51.
53.
52.
If x is a negative number, then the inequality is reversed when multiplying both sides of the inequality by the 1 negative number . x
54.
If x is a negative number, then the inequality is reversed when multiplying both sides of the inequality by a negative number.
....................................................... PS1.
PS3.
Prepare for Section 9.6 PS2.
C = 3x + 4 y C (0, 5) = 3(0) + 4(5) = 20 C (2, 3) = 3(2) + 4(3) =18 C (6, 1) = 3(6) + 4(1) = 22 C (9, 0) = 3(9) + 4(0) = 27
PS4.
PS5. ⎧3x + y = 6 (1) ⎨ ⎩ x + y = 4 (2)
PS6. ⎧ 300 x + 100 y = 900 ⎨ ⎩400 x + 300 y = 2200
Solve equation (1) for y and substitute into equation (2). x + (−3 x + 6) = 4 −2 x + 6 = 4 −2 x = −2 x =1 y = −3(1) + 6 =3 The solution is (1, 3).
C = 6 x + 4 y +15 C (6, 0) = 6(0) + 4(20) +15 = 95 C (4, 18) = 6(4) + 4(18) +15 =111 C (10, 10) = 6(10) + 4(10) +15 =115 C (15, 0) = 6(15) + 4(0) +15 =105
−900 x − 300 y = −2700 400 x + 300 y = 2200
−500 x x =1
= −500
300(1) +100 y = 900 100 y = 600 y =6
The solution is (1, 6).
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(1) (2) − 3 times (1) (2)
Section 9.6
661
Section 9.6 1.
C ( x, y ) = 3x + 4 y C (0, 5) = 3(0) + 4(5) = 20 C (2, 3) = 3(2) + 4(3) = 18 C (6, 1) = 3(6) + 4(1) = 22 C (9, 0) = 3(9) + 4(0) = 27 The minimum of 18 occurs at (2, 3).
2.
C ( x, y ) = 12 x + 2 y + 48 C (0, 24) = 12(0) + 2(24) + 48 = 96 C (4, 16) = 12(4) + 2(16) + 48 = 128 C (32, 2) = 12(32) + 2(2) + 48 = 436 C (40, 0) = 12(40) + 2(0) + 48 = 528 The minimum of 96 occurs at (0, 24).
3.
C ( x, y ) = 2.5 x + 3 y + 5 C (0, 20) = 2.5(0) + 3(20) + 5 = 65 C (5, 19) = 2.5(5) + 3(19) + 5 = 74.5 C (20, 4) = 2.5(20) + 3(4) + 5 = 67 C (22.5, 0) = 2.5(22.5) + 3(0) + 5 = 61.25 The maximum of 74.5 occurs at (5, 19).
4.
C ( x, y ) = 6 x + 4 y + 15 C (0, 20) = 6(0) + 4(20) + 15 = 95 C (4, 18) = 6(4) + 4(18) + 15 = 111 C (10, 10) = 6(10) + 4(10) + 15 = 115 C (15, 0) = 6(15) + 4(0) + 15 = 105 The maximum of 115 occurs at (10, 10).
5.
C = 4x + 2 y
⎧ x+ y ≥7 ⎪ ⎪⎪4 x + 3 y ≥ 24 ⎨ ⎪ x ≤ 10, y ≤ 10 ⎪ ⎪⎩ x ≥ 0, y ≥ 0
C = 4x + 2y (0, 10) (0, 8) (3, 4) (7, 0) (10, 0) (10, 10)
20 16 20 28 40 60
C = 5x+4y (0, 15) (0, 8) (4, 5) (12, 3) (12, 15)
60 32 40 72 120
minimum
The minimum is 16 at (0, 8). 6.
C = 5x + 4 y
⎧3 x + 4 y ≥ 32 ⎪ ⎪⎪ x + 4 y ≥ 24 ⎨ ⎪ x ≤ 12, y ≤ 15 ⎪ ⎪⎩ x ≥ 0, y ≥ 0
minimum
The minimum is 32 at (0, 8). 7.
C = 6x + 7 y ⎧ x + 2 y ≤ 16 ⎪ ⎪5 x + 3 y ≤ 45 ⎨ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
C = 6x + 7 y (0, 8) (0, 0)
56 0
(9, 0)
54
(6, 5)
71
maximum
The maximum is 71 at (6, 5).
Copyright © Houghton Mifflin Company. All rights reserved.
662
8.
Chapter 9: Systems of Equations and Inequalities
C = 6x + 5 y ⎧2 x + 3 y ≤ 27 ⎪ ⎪7 x + 3 y ≤ 42 ⎨ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
C = 6x + 5 y (0, 9) (0, 0)
45 0
(6, 0)
36
(3, 7)
53
maximum
The maximum is 53 at (3, 7). 9.
C = x + 6y ⎧ 5 x + 8 y ≤ 120 ⎪ ⎪7 x + 16 y ≤ 192 ⎨ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
C = x + 6y (0, 12) (0, 0)
72 0
(24, 0)
24
(16, 5)
46
maximum
The maximum is 72 at (0, 12). 10.
C = 4x + 5y ⎧ x + 3 y ≥ 30 ⎪ ⎪3 x + 4 y ≥ 60 ⎨ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
C = 4x + 5y (0, 15) (12, 6)
75 78
(30, 0)
120
minimum
The minimum is 75 at (0, 15). 11.
C = 4x + y ⎧3 x + 5 y ≥ 120 ⎪ ⎪ x + y ≥ 32 ⎨ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
C = 4x + y (40, 0) (0, 32)
160 32
(20, 12)
92
minimum
The minimum is 32 at (0, 32). 12.
C = 7x + 2 y ⎧ x + 3 y ≤ 108 ⎪ ⎪7 x + 4 y ≤ 280 ⎨ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
C = 7x + 2 y (0, 36) (40, 0) (24, 28) (0, 0)
72 280 224 0
minimum
The maximum is 280 at (40, 0).
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.6
13.
663
C = 2x + 7 y
⎧ x + y ≤ 10 ⎪ ⎪⎪ x + 2 y ≤ 16 ⎨2 x + y ≤ 16 ⎪ ⎪ ⎩⎪ x ≥ 0, y ≥ 0
C = 2x + 7 y (0, 8) (0, 0)
56 0
(8, 0)
16
(6, 4)
40
(4, 6)
50
maximum
The maximum is 56 at (0, 8). 14.
C = 4x + 3y ⎧ 2x + y ≥ 8 ⎪ ⎪ 2 x + 3 y ≥ 16 ⎪⎪ x + 3 y ≥ 11 ⎨ ⎪ ⎪ x ≤ 20, y ≤ 20 ⎪ ⎪⎩ x ≥ 0, y ≥ 0
C = 4x + 3y (0, 20) (0, 8)
60 24
(2, 4)
20
(5, 2) (11, 0) (20, 0) (20, 20)
26 44 80 140
minimum
The minimum is 20 at (2, 4). 15.
C = 3x + 2 y ⎧ 3 x + y ≥ 12 ⎪ ⎪⎪2 x + 7 y ≥ 21 ⎨ x+ y ≥8 ⎪ ⎪ ⎩⎪ x ≥ 0, y ≥ 0
C = 3x + 2 y (0, 12) (2, 6)
24 18
(7, 1)
23
(10.5, 0)
31.5
minimum
The minimum is 18 at (2, 6). 16.
C = 2x + 6 y ⎧ x + y ≤ 12 ⎪ ⎪⎪3 x + 4 y ≤ 40 ⎨ x + 2 y ≤ 18 ⎪ ⎪ ⎩⎪ x ≥ 0, y ≥ 0
C = 2x + 6 y (0, 9) (4, 7) (8, 4) (12, 0) (0, 0)
54 50 40 24 0
maximum
The maximum is 54 at (0, 9). 17.
C = 3x + 4 y ⎧ 2 x + y ≤ 10 ⎪ ⎪⎪2 x + 3 y ≤ 18 ⎨ x− y ≤ 2 ⎪ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
C = 3x + 4 y (0, 6) (3, 4)
24 25
(4, 2)
20
(2, 0)
6
(0, 0)
0
maximum
The maximum is 25 at (3, 4).
Copyright © Houghton Mifflin Company. All rights reserved.
664
18.
Chapter 9: Systems of Equations and Inequalities
C = 3x + 7 y ⎧ x+ y≥9 ⎪3x + 4 y ≥ 32 ⎪⎪ ⎨ x + 2 y ≥ 12 ⎪ ⎪ ⎩⎪ x ≥ 0, y ≥ 0
C = 3x + 7 y (0, 9) (4, 5)
63 47
(8, 2)
38
(12, 0)
36
maximum
The minimum is 36 at (12, 0). 19.
C = 3x + 2 y ⎧ x + 2y ≥ 8 ⎪ 3x + y ≥ 9 ⎪⎪ ⎨ x + 4 y ≥ 12 ⎪ ⎪ ⎩⎪ x ≥ 0, y ≥ 0
C = 3x + 2 y (0, 9) (2, 3)
18 12
(4, 2)
16
(12, 0)
36
maximum
The minimum is 12 at (2, 3). 20.
C = 4x + 5y ⎧3x + 4 y ≤ 250 ⎪ x + y ≤ 75 ⎪⎪ ⎨ 2 x + 3 y ≤ 180 ⎪ ⎪ ⎩⎪ x ≥ 0, y ≤ 0
C = 4x + 5y (0, 60) (30, 40)
300 320
(50, 25)
325
(75, 0)
300
(0, 0)
0
maximum
The maximum is 325 at (50, 25). 21.
C = 6x + 7 y ⎧ x + 2 y ≤ 900 ⎪ x + y ≤ 500 ⎪⎪ ⎨3x + 2 y ≤ 1200 ⎪ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
C = 6x + 7 y (0, 450) (100, 400) (200, 300) (400, 0) (0, 0)
3150 3400 3300 2400 0
maximum
The maximum is 3400 at (100, 400). 22.
C = 11x + 16 y ⎧ x + 2 y ≥ 45 ⎪ x + y ≥ 40 ⎪⎪ ⎨2 x + y ≥ 45 ⎪ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
C = 11x + 16 y (0, 45) (5, 35)
720 615
(35, 5)
465
(45, 0)
495
minimum
The minimum is 465 at (35, 5).
Copyright © Houghton Mifflin Company. All rights reserved.
Section 9.6
665
23. x = number of cups of Oat Flakes y = number of cups of Crunchy O’s
C = 0.38 x + 0.32 y Constraints: ⎧ 6 x + 3 y ≥ 210 ⎪30 x + 40 y ≥ 1200 ⎪ ⎨ x≥0 ⎪ ⎪⎩ y≥0
0.38 x + 0.32 y = C (0, 70) (32, 6)
$22.40 $14.08
(40, 0)
$15.20
minimum
The minimum cost of $14.08 is achieved by mixing 32 cups of Oat Flakes and 6 cups of Crunchy O’s. 24. x = number of two-person tents y = number of family tents
P = 3 + x + 49 y Constraints: x ≤ 4y ⎧ ⎪2 x + 2 y ≤ 50 ⎪ ⎨2 x + 4 y ≤ 80 ⎪ x≥0 ⎪ y≥0 ⎩⎪
x + 49 y + 3 = P (0, 0) (0, 20)
0 $980
(10, 15)
$1075
(20, 5)
$925
Maximum
The maximum profit of $1075 is achieved by making 10 two-person tents and 15 family tents. 25. W = acres of wheat to plant B = acres of barley to plant
P = 50W + 70 B Constraints: ⎧ 4W + 3B ≤ 200 ⎪ W + 2 B ≤ 100 ⎪ ⎨ ⎪ ⎪⎩ W ≥ 0, B ≥ 0
50W + 70 B = P (0, 50) (20, 40)
3500 3800
(50, 0)
2500
(0, 0)
0
maximum
The maximum profit is achieved by planting 20 acres of wheat and 40 acres of barley. 26. x = hours of machine 1 use y = hours of machine 2 use
Cost = 28 x + 25 y Constraints: ⎧ 4 x + 3 y ≥ 60 ⎪ ⎪5 x + 10 y ≥ 100 ⎨ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
28 x + 25 y = Cost (0, 20) (12, 4)
500 436
(20, 0)
560
minimum
To achieve the minimum cost, use machine 1 for 12 hours and machine 2 for 4 hours.
Copyright © Houghton Mifflin Company. All rights reserved.
666
Chapter 9: Systems of Equations and Inequalities
27. S = Number of Starter sets P = Number of Pro sets
Profit = 35S + 55P Constraints: ⎧ 4S + 6 P ≤ 108 ⎪ ⎪ S + P ≤ 24 ⎨ ⎪ ⎪⎩ S ≥ 0, P ≥ 0
Profit = 35S + 55P (0, 18) (18, 6) (24, 0) (0, 0)
990 960 840 0
maximum
To maximize profit, produce zero starter sets and 18 pro sets. 28. S = Number of standard model D = Number of deluxe model
Profit = 25S + 35 D Constraints: ⎧ S + 3D ≤ 24 ⎪ ⎪⎪ S + D ≤ 10 ⎨ 2 S + D ≤ 16 ⎪ ⎪ ⎪⎩ S ≥ 0, D ≥ 0
Profit = 25S + 35D (0, 8) (3, 7) (6, 4)
280 320 290
(8, 0)
200
(0, 0)
0
maximum
To maximize profit, produce 3 standard models and 7 deluxe models.
.......................................................
Connecting Concepts
29. A = ounces of food group A B = ounces of food group B
Cost = 40 A + 10 B Constraints: ⎧ 3 A + B ≥ 24 ⎪ ⎪⎪ A + B ≥ 16 ⎨ A + 3B ≥ 30 ⎪ ⎪ ⎩⎪ A ≥ 0, B ≥ 0
Cost = 40 A + 10 B (0, 24) (4, 12)
240 280
(9, 7)
430
(30, 0)
1200
minimum
To minimize cost, use 24 ounces of food group B and zero ounces of food group A. The minimum cost is $2.40. 30. A = liters of Pymex A B = liters of Pymex B
Profit = 12 A + 9 B Constraints: ⎧ A + B ≤ 10 ⎪ ⎪⎪3 A + 4 B ≤ 36 ⎨3 A + 2 B ≤ 27 ⎪ ⎪ ⎪⎩ A ≥ 0, B ≥ 0
Profit = 12 A + 9 B (0, 9) 81 (4, 6) 102 (7, 3) 111 maximum (9, 0) 108 (0, 0) 0 To maximize profit, make 7 liters of Pymex A and 3 liters of Pymex B. The maximum profit is $111. Copyright © Houghton Mifflin Company. All rights reserved.
Assessing Concepts
667
31. x = number of 4-cylinder engines y = number of 6-cylinder engines
Profit = 150 x + 250 y Constraints: x+ y ≤9 ⎧ ⎪ 5 x + 10 y ≤ 80 ⎪⎪ ⎨ 3x + 2 y ≤ 24 ⎪ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
150 x + 250 y = Profit (0, 8) (2, 7) (6, 3)
2000 2050 1650
(8, 0)
1200
(0, 0)
0
maximum
To achieve the maximum profit of $2050, produce two 4-cylinder engines and seven 6-cylinder engines. 32. x = number of pounds of F1
y = number of pounds of F2 Cost = 450 x + 300 y Constraints: ⎧ 200 x + 100 y ≥ 5000 ⎪ ⎪⎪ 100 x + 200 y ≥ 7000 ⎨100 x + 400 y ≥ 10,000 ⎪ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
450 x + 300 y = Cost (0, 50) (10, 30)
15,000 13,500
(40, 15)
22,500
(100, 0)
45,000
minimum
To achieve the minimum cost, produce 10 pounds of F1 and 30 pounds of F2 . The minimum cost is $13,500.
.......................................................
Exploring Concepts with Technology
Ill-Conditioned Systems of Equations System 1 When the coefficients are approximated to: a. the nearest 0.01 the solution is (266.6666667, −1433.333333, 1366,666667). b. the nearest 0.001 the solution is (30.50502049, −209.9900227, 226.6966693). c. the nearest 0.0001 the solution is (27.32317303, −193.65719, 211.5374196). d. the limit of most calculators (nearest 0.00000000001) the solution is (27.00000003, −192.0000002, 210.0000002).
System 2 When the coefficients are approximated to: a. the nearest 0.01 the solution is (55.98194131, −295.2595937, 190.9706546, 118.510158). b. the nearest 0.001 the solution is (81.29464132, −708.9158728, 1335.912871, −682.7827636). c. the nearest 0.0001 the solution is (−90.6440169, 1188.443724, −3201.68217, 2258.171289). d. the limit of most calculators (nearest 0.00000000001) the solution is (−64.00000245, 900.0000269, −2520.000064, 1820.000042)
.......................................................
Assessing Concepts
1.
True
2.
False, a graph of a linear equation in three variables is a plane.
3.
False, a homogeneous system is one where the constant term in each equation is zero.
4.
True
5.
True
6.
Yes
Copyright © Houghton Mifflin Company. All rights reserved.
668
Chapter 9: Systems of Equations and Inequalities
7.
At a vertex of the set of feasible solutions.
8.
It is a nonsquare system of equations.
9.
A line
10.
(0, 0)
....................................................... 1.
⎧2 x − 4 y = − 3 ⎨ ⎩3 x + 8 y = −12
2.
(1) (2)
4 x − 8 y = −6
2 times (1)
3x + 8 y = −12
(2)
7 x = −18 x=−
2 −
3.
⎧4 x − 3 y = 15 ⎨ ⎩2 x = 5 y = −12 4x −
3 y = 15
(1)
−4 x − 10 y = 24
18 7
y=−
(1) ⎧⎪3x − 4 y = −5 ⎨ 2 y = x +1 (2) ⎪⎩ 3 Substitute y from Eq. (2) into Eq. (1).
(3 )
− 2 times (2)
− 13 y = 39 y = −3
( ) − 4 y = −3
18 7
Chapter Review
3x − 4 2 x + 1 = −5 3x − 8 x − 4 = −5
4 x − 3( −3) = 15
3
3 x= 2
15 28
1 x = −1 3
x = −3
⎛3 ⎞ The solution is ⎜ , − 3 ⎟ . ⎝2 ⎠ [9.1]
15 ⎞ ⎛ 18 The solution is ⎜ − , − ⎟ . [9.1] 28 ⎠ ⎝ 7
y = 2 ( −3) + 1 3
y = −1 The solution is ( −3, − 1) .
4.
5. (1) ⎪⎧7 x + 2 y = −14 ⎨ y = −5 x−3 (2) ⎪⎩ 2 Substitute y from Eq. (2) into Eq. (1).
(
6. (1) ⎧ y = 2x − 5 ⎨ x = 4 y − 1 (2) ⎩ Substitute x from Eq. (2) into Eq. (1).
2
(1) ⎧ y = 3x + 4 ⎨ (2) ⎩x = 4 y − 5 Substitute x from Eq. (2) into Eq. (1).
y = 2 ( 4 y − 1) − 5
)
7 x + 2 − 5 x − 3 = −14
y = 8y − 2 − 5
7 x − 5 x − 6 = −14 2 x = −8 x = −4
−7 y = −7 y =1
x = 4 (1) − 1 x=3
x = 4 (1) − 5
y = 12 y − 15 + 4
x = −1
The solution is ( −1, 1) .
[9.1]
2
y = 3( 4 y − 5) + 4 −11 y = −11 y =1
The solution is ( 3, 1) .
y = − 5 (−4) − 3
[9.1]
y=7 The solution is (−4, 7). [9.1] 7.
⎧ 6 x + 9 y = 15 ⎨ ⎩10 x + 15 y = 25
8.
(1)
4x − 8 y =
9
−4 x + 8 y = −10
( 2)
(1)
−2 times (2)
0 = −1
1 times (1) 2x + 3y = 5 3 2x + 3 y = 5 1 times (2) 0=0 5
The system of equations is inconsistent. The system has no solution. [9.1]
2 x + 3c = 5 5 − 3c x= 2 ⎛ 5 − 3c , The ordered-pair solutions are ⎜ ⎝ 2
[9.1]
Let y = c.
⎞ c ⎟ . [9.1] ⎠
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Review
9.
669
⎧ 2 x − 3 y + z = −9 ⎪ ⎨2 x + 5 y − 2 z = 18 ⎪ 4 x − y + 3z = − 4 ⎩
(3)
2 x − 3 y + z = −9
(1)
10.
(1) ( 2)
−2 x − 5 y + 2 z = −18 −1 times (2)
⎧ x − 3 y + 5 z = 1 (1) ⎪ ⎨ 2 x + 3 y − 5 z = 15 (2) ⎪ 3 x + 6 y + 5 z = 15 (3) ⎩
−2 x + 6 y − 10 z = −2 −2 times (1) 2 x + 3 y − 5z = 15 (2) 9 y − 15 z = 13 (4)
−8 y + 3z = −27 (4) −4 x + 6 y − 2 z = 18 4 x − y + 3z = −4 5 y + z = 14
−2 times (1) (3)
15 y − 10 z = 12 (5)
(5)
⎧ x − 3 y + 5z = 1 ⎪ ⎨ 9 y − 15 z = 13 ( 4) ⎪ 15 y − 10 z = 12 (5) ⎩
⎧2 x − 3 y + z = − 9 ⎪ ⎨ − 8 y + 3z = −27 (4) ⎪ 5 y + z = 14 (5) ⎩ − 8 y + 3 z = −27
−15 y − 3 z = −42 −23 y
18 y − 30 z = 26 2 times (4)
(4) − 3 times (5)
= −69
−45 y + 30 z = −36 −27 y
y=3
(6)
2 x − 3(3) − 1 = 9 2x = 1 x =1
⎛1 ⎞ The ordered-triple solution is ⎜ , 3, −1⎟ . [9.2] ⎝2 ⎠
−3 times (5)
= −10 y=
⎧2 x − 3 y + z = − 9 ⎪ ⎨ − 8 y + 3 z = −27 ⎪ y = 3 ( 6) ⎩
−8(3) + 3 z = −27 3z = −3 z = −1
−3x + 9 y − 15z = −3 −3 times (1) 3x + 6 y + 5z = 15 (3)
10 27
⎧ ⎪ x − 3 y + 5z = 1 ⎪ ⎨ 9 y − 15z = 13 ⎪ 10 y= ⎪ 27 ⎩
(6)
( 6)
−15z = 13 − 9 y
x = 3 y − 5z + 1
⎛ 10 ⎞ − 15z = 13 − 9⎜ ⎟ ⎝ 27 ⎠ 29 − 15 z = 3 29 z=− 45
⎛ 10 ⎞ ⎛ 29 ⎞ x = 3 ⎜ ⎟ −5 ⎜− ⎟ + 1 ⎝ 27 ⎠ ⎝ 45 ⎠ 16 x= 3
29 ⎞ ⎛ 16 10 , − ⎟ . [9.2] The solution is ⎜ , 45 ⎠ ⎝ 3 27
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670
11.
Chapter 9: Systems of Equations and Inequalities
⎧ x + 3 y − 5 z = −12 ⎪ ⎨3x − 2 y + z = 7 ⎪5 x + 4 y − 9 z = −17 ⎩ −3 x − 9 y + 15 z = 36
3x − 2 y +
12.
(1) (2) (3)
−3 times (1)
z= 7
(2)
−11 y + 16 z = 43
(4)
2 x − y + 2 z = 5 (1)
−2 x − 6 y + 6 z = −4 −2 times (2) −7 y + 8 z = 1 (4)
−5 x − 15 y + 25 z = 60 −5 times (1)
5 x + 4 y − 9 z = −17
(1) (2) (3)
⎧ 2x − y + 2z = 5 ⎪ ⎨ x + 3 y − 3z = 2 ⎪5 x − 9 y + 8 z = 13 ⎩
(3)
−5 x − 15 y + 15 z = −10 −5 times (2) 5 x − 9 y + 8 z = 13 (3) −24 y + 23 z =
−11 y + 16 z = 43 (5)
3
(5)
⎧ x + 3 y − 5 z = −12 ⎪ ⎨ − 11 y + 16 z = 43 (4) ⎪ − 11 y + 16 z = 43 (5) ⎩
⎧ 2x − y + 2z = 5 ⎪ − 7 y + 8 z = 1 (4 ) ⎨ ⎪ − 24 y + 23 z = 3 (5) ⎩
⎧ x + 3 y − 5 z = −12 ⎪ ⎨ − 11 y + 16 z = 43 ⎪ 0 = 0 ( 6) ⎩
−168 y + 192 z = 24 24 times (4)
168 y − 161z = −21 −7 times (2) 31z =
The system of equations is dependent. Let z = c. −11 y + 16c = 43 16c − 43 y= 11 ⎛ 16c − 43 ⎞ x+3 ⎜ ⎟ − 5c = −12 ⎝ 11 ⎠ 7c − 3 x= 11
z=
3
3 31
(6)
⎧ ⎪2 x − y + 2 z = 5 ⎪ ⎨ − 7 y + 8z = 1 ⎪ 3 z= (6) ⎪ 31 ⎩
⎛ 7c − 3 16c − 43 ⎞ The solution is ⎜ , , c ⎟ . [9.2] 11 ⎝ 11 ⎠
⎛ 3⎞ −7 y + 8 ⎜ ⎟ = 1 ⎝ 31 ⎠ 31 24 −7 y = − 31 31 1 y=− 31 ⎛ 1⎞ ⎛ 3⎞ 2x − ⎜ − ⎟ + 2⎜ ⎟ = 5 ⎝ 31 ⎠ ⎝ 31 ⎠ 1 6 155 − + 31 31 31 148 2x = 31 74 x= 31 2x = −
1 3⎞ ⎛ 74 The solution is ⎜ , − , ⎟ . [9.2] 31 31 31 ⎠ ⎝
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Chapter Review
13.
671
14.
⎧ 3 x + 4 y − 6 z = 10 (1) ⎪ ⎨ 2 x + 2 y − 3 z = 6 ( 2) ⎪ x − 6 y + 9 z = −4 (3) ⎩
Rearrange the equations so that the equation with the 1 as the coefficient of x is in the first row. ⎧ x − 6 y + 9 z = −4 (3) ⎪ ⎨ 2 x + 2 y − 3 z = 6 ( 2) ⎪ 3 x + 4 y − 6 z = 10 (1) ⎩
−2 x + 12 y − 18 z = 8 2 x + 2 y − 3z = 6
−3 x + 18 y − 27 z = 12 3x + 4 y − 6 z = 10
(4) –3 times (3) (1)
22 y − 33 z = 22 2 y − 3z = 2
x − 6 y + 9 z = −4 2 y − 3z = 2
(3) (4)
2 y − 3z = 2
(5)
(3)
5x − 3 y
(4)
=7
13x − 3 y
= 15
2 times (2) (3) (5)
⎧x − 6 y + 4 z = 6 ⎪ =7 ⎨5 x − 3 y ⎪13 x − 3 y = 15 ⎩
(5)
5x − 3 y = 7
(4)
− 13 x + 3 y = −15
(4)
− 1 times (5)
= −8 x =1
⎧x − 6 y + 4 z = 6 ⎪ =7 ⎨5 x − 3 y ⎪ x =1 ⎩
⎧ x − 6 y + 9 z = −4 ⎪ 2 y − 3z = 2 ⎨ ⎪ 0= 0 ⎩
(2)
(1) (2)
− 8x
(5)
(1)
x − 6 y + 4z = 6 4x + 3 y − 4z = 1
8x + 6 y − 8z = 2 5 x − 9 y + 8 z = 13 −2 times (3) (2)
14 y − 21z = 14 2 y − 3z = 2
⎧ x − 6 y + 4z = 6 ⎪ ⎨4 x + 3 y − 4 z = 1 ⎪ 5 x − 9 y + 8 z = 13 ⎩
5(1) − 3 y = 7
y=−
2 3
(6)
(6)
⎛ 2⎞ 1 − 6⎜ − ⎟ + 4z = 6 ⎝ 3⎠ 4z = 1
The system of equations is dependent. Let z = c. 2 y − 3c = 2 3c + 2 y= 2 ⎛ 3c + 2 ⎞ x−6 ⎜ ⎟ + 9c = −4 ⎝ 2 ⎠ x − 9c − 6 + 9c = −4 x=2 ⎛ 3c + 2 The solution is ⎜ 2, , 2 ⎝
z=
2 1⎞ ⎛ The solution is ⎜1,− , ⎟. [9.2] 3 4⎠ ⎝
⎞ c ⎟ . [9.2] ⎠
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672
15.
Chapter 9: Systems of Equations and Inequalities
⎧2 x + 3 y − 2 z = 0 ⎪ ⎨ 3x − y − 4 z = 0 ⎪⎩ 5 x +13 y − 4 z = 0
16.
(1) (2) (3)
6 x + 9 y − 6 z = 0 3 times (1) − 6 x + 2 y + 8 z = 0 − 2 times (2)
3x − 5 y + z = 0 −3 x − 12 y + 9 z = 0
11y + 2 z = 0 (4) 15 x − 5 y − 20 z = 0
⎧ 3 x − 5 y + z = 0 (1) ⎪ ⎨ x + 4 y − 3 z = 0 ( 2) ⎪2 x + y − 2 z = 0 (3) ⎩
− 17 y + 10 z = 0
5 times (2)
− 15 x − 39 y + 12 z = 0
− 3 times (3)
−2 x − 8 y + 6 z = 0 2x + y − 2z = 0
− 44 y − 8 z = 0 11y + 2z = 0
(5)
−34 y + 20 z = 0 35 y − 20 z = 0 y=0
Let z = c 11 y + 2c = 0 2 11
− 2c = 0 2x =
x=
The solution is 17.
{
(
14 c, 11
28c 11 14c 11
−
2 c, 11
(5)
(6)
(6)
x = 0, y = 0, z = 0. The solution is (0, 0, 0). [9.2]
)
c . [9.2] 18.
x − 2 y + z =1 (1) 3 x + 2 y − 3 z =1 (2)
−3 x + 6 y − 3z = −3 − 3 times (1) 3x + 2 y − 3z = 1 (2) 8 y − 6 z = −2 4 y − 3 z = −1 (3)
{
2 x − 3 y + z = 1 (1) 4 x + 2 y + 3 z = 21 (2)
−4 x + 6 y − 2 z = −2 4 x + 2 y + 3z = 21 8y + z =19
{
2 x − 3 y + z =1 8y + z =19
⎧x − 2 y + z = 1 ⎨ 4 y − 3 z = −1 (3) ⎩
Let z = c.
(4)
2 times (4) − 5 times (5)
⎧ 3x − 5 y + z = 0 ⎪ ⎨ −17 y + 10 z = 0 ⎪ y =0 ⎩
y=− c 2c 11
(5)
z=0 ⎧ 3x − 5 y + ⎪ − + y z=0 17 10 ⎨ ⎪ − 7 y + 4z = 0 ⎩
⎧2 x + 3 y − 2 z = 0 ⎪ 11 y + 2 z = 0 ⎨ ⎪ 0=0 ⎩
( )
(4) − 2 times (2) (3)
− 7 y + 4z = 0
⎧2 x + 3 y − 2 z = 0 (1) ⎪ ⎨ 11 y + 2 z = 0 (4) ⎪ 11 y + 2 z = 0 (5) ⎩
2x + 3 −
(1) − 3 times (2)
Let z = c.
4 y − 3c = −1 3c − 1 y= 4
− 2 times (1) (2) (3) (3) 8 y = 19 − c 19 − c y= 8
⎛ 3c − 1 ⎞ x − 2⎜ ⎟ + c =1 ⎝ 4 ⎠ c +1 x= 2
⎛ 19 − c ⎞ 2x − 3 ⎜ ⎟ + c =1 ⎝ 8 ⎠ 65 − 11c 2x = 8 65 − 11c x= 16
⎛ c + 1 3c − 1 ⎞ , , c ⎟ . [9.2] The solution is ⎜ 4 ⎝ 2 ⎠
⎛ 65 − 11c 19 − c ⎞ , , c ⎟ . [9.2] The solution is ⎜ 8 ⎝ 16 ⎠
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Chapter Review
19.
673
20.
⎧⎪ y = x 2 − 2 x − 3 ⎨ ⎪⎩ y = 2 x − 7
2 x2 + x = 2 x + 1
x 2 − 2x − 3 = 2x − 7
2 x2 − x − 1 = 0 (2 x + 1)(x − 1) = 0
x 2 − 4x + 4 = 0 ( x − 2)( x − 2) = 0
x=2
21.
x =−
1 2
x=1
( ) +1 1 2
y = 2(2) − 7 y = −3
y=2 −
The solution is (2, −3). [9.3]
The solutions are − , 0 and (1, 3). [9.3] 22.
or x = 1
()
3 2 2
3 2
⎧⎪( x − 1) 2 + ( y + 1) 2 = 5 ⎨ ⎪⎩ y = 2x − 3
x2 − 2x + 1 + 4x2 − 8x + 4 = 5
2
( x + 1) + (−2 x + 2) = 4
5 x 2 − 10 x = 0
2
x + 2 x + 1 + 4 x −8 x + 4 = 4
5x( x − 2) = 0
5x 2 − 6x + 1 = 0 (5 x − 1)( x − 1) = 0
x = 0 or x = 2 y = 2 (0) − 3
x = 1 or x = 1 5
y = 18
( , 3) and (1, − 1). [9.3]
The solutions are
( x − 1)2 + (2 x − 2)2 = 5
( x + 1) + (−2 x + 4 − 2) = 4
1 5
3 2
(x − 1)2 + (2 x − 3 + 1)2 = 5
2
( )+4
()−6
y = 2 (1)2 + 3(1) − 6 y=− 1
24.
(1) (2)
From Eq. (2), y = −2 x + 4. Substitute y in Eq. (1).
y = −2
+ 3
y=3
x has no real number solution. The system of equations is inconsistent. The graphs of the equations do not intersect. No real solution. [9.3]
2
3 2
y=2
3 ± 9 − 16 3 ± −7 = 4 4
2
⎪⎧ y = 4 x 2 − 2 x − 3 ⎨ 2 ⎪⎩ y = 2 x + 3x − 6
x=
2(2)
2
)
2 x 2 − 5x + 3 = 0 (2x − 3)( x − 1) = 0
− (−3) ± (−3) 2 − 4(2)(2)
( x + 1)2 + ( y − 2)2 = 4 2x + y = 4
1 2
4 x 2 − 2 x − 3 = 2 x 2 + 3x − 6
2 x2 − 3x + 2 = 0
x=
y=3
(
⎧⎪ y = 3x 2 − x + 1 ⎨ ⎪⎩ y = x 2 + 2 x − 1
x=
y = 2(1) + 1
y=0
3x2 − x + 1 = x 2 + 2 x − 1
23.
⎧⎪ y = 2 x 2 + x ⎨ ⎪⎩ y = 2 x + 1
y = −3
y = −2(1) + 4 y=2
y = 2(2) − 3 y =1
The solutions are ( 0, − 3) and (2, 1). [9.3]
5
The solutions are ⎛⎜ 1 , 18 ⎞⎟ and (1, 2). [9.3] ⎝5 5 ⎠ Copyright © Houghton Mifflin Company. All rights reserved.
674
25.
Chapter 9: Systems of Equations and Inequalities
⎧⎪( x − 2) 2 + ( y + 2) 2 = 4 (1) ⎨ ⎪⎩ ( x + 2) 2 + ( y + 1) 2 = 17 (2) Expand the binomials. Then subtract.
26.
⎧⎪ ( x + 1) 2 + ( y − 2) 2 = 1 (1) ⎨ ⎪⎩( x − 2) 2 + ( y + 2) 2 = 20 (2) Expand the binomials. Then subtract.
x2 − 4 x + 4 + y 2 + 4 y + 4 =
4
x 2 + 2x + 1 + y 2 − 4 y + 4 = 1
x2 + 4 x + 4 + y 2 + 2 y + 1 =
17
x 2 − 4 x + 4 + y 2 + 4 y + 4 = 20
−8 x − 8x
+ 2 y + 3 = − 13 + 2y = − 16
6x − 3 6x
− 8y − 8y
= −19 = −16
y = 4x − 8
y=
Substitute y into Eq. (1).
3x + 8 4
2
2
2
( x − 2) + (4 x − 8 + 2) = 4 ( x − 2) 2 + (4 x − 6) 2 = 4
⎞ ⎛ 3x + 8 ( x + 1) 2 + ⎜ − 2⎟ = 1 ⎠ ⎝ 4 2
⎛ 3x ⎞ ( x + 1) 2 + ⎜ ⎟ = 1 ⎝ 4 ⎠
x 2 − 4 x + 4 + 16 x 2 − 48 x + 36 = 4
17 x 2 − 52 x + 36 = 0 ( x − 2)(17 x − 18) = 0
y=0
9x 2 =1 16
16 x 2 + 32 x + 16 + 9 x 2 = 16
18 x = 2 or x = 17
y = 4(2) − 8
x 2 + 2x + 1 +
25 x 2 + 32 x = 0
x(25 x + 32) = 0
⎛ 18 ⎞ y = 4⎜ ⎟ − 8 ⎝ 17 ⎠ 64 y=− 17
64 ⎞ ⎛ 18 The solutions are (2, 0) and ⎜ , − ⎟ . [9.3] 17 ⎠ ⎝ 17
x=0
y=
or
3 (0) + 2 4
y=2
x=−
32 25
y=
3 ⎛ 32 ⎞ ⎜− ⎟ + 2 4 ⎝ 25 ⎠
y=
26 25
⎛ 32 26 ⎞ The solutions are (0, 2) and ⎜ − , ⎟ . [9.3] ⎝ 25 25 ⎠
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Chapter Review
27.
675
⎧⎪ x 2 − 3xy + y 2 = − 1 (1) ⎨ ⎪⎩ 3x 2 − 5 xy − 2 y 2 = 0 (2) Factor Eq. (2). ( 3x + y )( x − 2 y ) = 0 −y x= or x = 2 y 3 y x = − implies y = −3 x. Substitute for y in Eq. (1). 3 x 2 − 3 x(−3 x) + (−3x) 2 = −1 x 2 + 9 x 2 + 9 x 2 = −1
19 x 2 = −1 1 x2 = − 19 This equation yields no real solutions. Substituting x = 2 y in Eq. (1) yields (2 y ) 2 − 3(2 y ) y + y 2 = −1 4 y 2 − 6 y 2 + y 2 = −1 − y 2 = −1 y2 =1 y = ±1
The solutions are (2, 1) and (−2, − 1). [9.3] 28.
⎧⎪2 x 2 + 2 xy − y 2 = −1 ⎨ ⎪⎩6 x 2 + xy − y 2 = 0
(1) (2)
Factor Eq. (2). (3 x − y )(2 x + y ) = 0 y = 3x
y = −2 x
2 x 2 + 2 x(3 x) − (3 x)2 = −1
2x 2 + 2 x( −2 x) − (−2 x) 2 = −1
2 x 2 + 6 x 2 − 9 x 2 = −1
2 x 2 − 4 x 2 − 4 x 2 = −1 1 x2 = 6
x2 = 1 x = ±1
x=± y = 3(1) y=3
y = 3(−1) y = −3
⎛ 6⎞ y = −2 ⎜⎜ ⎟⎟ ⎝ 6 ⎠ y=−
6 3
6 6
⎛ 6⎞ y = −2 ⎜⎜ − ⎟⎟ ⎝ 6 ⎠ y=
6 3
⎛ 6 ⎛ 6⎞ 6 6⎞ , − , The solutions are (1, 3), (−1, − 3), ⎜⎜ ⎟⎟ and ⎜⎜ − ⎟ . [9.3] 3 ⎠ 3 ⎟⎠ ⎝ 6 ⎝ 6
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676
29.
Chapter 9: Systems of Equations and Inequalities
⎧⎪ 2 x 2 − 5 xy + 2 y 2 = 56 ⎨ ⎪⎩14 x 2 − 3xy − 2 y 2 = 56
30.
(1) (2)
⎧⎪2 x 2 + 7 xy + 6 y 2 = 1 ⎨ ⎪⎩6 x 2 + 7 xy + 2 y 2 = 1
(1) (2)
Subtract Eq. (1) From Eq. (2)
Subtract Eq. (2) from Eq. (1)
12 x 2 + 2 xy − 4 y 2 = 0
−4 x 2 + 4 y 2 = 0 − x2 + y 2 = 0 y = x or y = − x
Factor 2(3 x + 2 y )(2 x − y ) = 0. Thus 3x + 2 y = 0
2x − y = 0 y = 2x
or
y =−3 2
When y = x, we have, from Eq. (1), 2 x2 + 7 x2 + 6 x2 = 1
Substituting y = − 3 x into Eq. (1), we have
15 x 2 = 1
2
(
) (
x2 = 1
15
)
2 2 x − 5 x − 3 x + 2 − 3 x = 56 2 2 2 x 2 + 15 x 2 + 9 x 2 = 56 2 2 2 2
x = ± 15 15
Since y = x, y = ± 15 . The solutions are 15
14 x = 56
⎛ 15 , 15 ⎞ , ⎛ − 15 , − 15 ⎞ . ⎜ 15 15 ⎟⎠ ⎜⎝ 15 15 ⎟⎠ ⎝
x2 = 4 x = ±2
When y = − x, we have, from Eq. (1)
When x = 2, y = − 3 (2) = −3;
2 x2 − 7 x2 + 6 x2 = 1
2
When x = −2, y = − 3 (− 2 ) = 3.
x2 = 1 x = ±1
2
Two solutions are (2, −3) and (−2, 3). Substituting y =2x into Eq. (1) yields 0 = 56. Thus the only solutions of the system are (2, −3), and (−2, 3). [9.3]
Since y = − x, y = ±1. The solutions are (1, − 1) , ( −1, 1) . The solutions of the system of equations are ⎛ 15 , 15 ⎞ , ⎛ − 15 , − 15 ⎞ , 1, − 1 , −1, 1 . [9.3] ( )( ) ⎜ 15 15 ⎟ ⎜ 15 15 ⎟⎠ ⎝ ⎠ ⎝
31.
7x − 5 2
x −x−2
=
7x − 5 A B = + ( x − 2 )( x + 1) x − 2 x + 2
7 x − 5 = A ( x + 1) + B ( x − 2 )
1 = −A + B
5 = − A + 2B
7x − 5 2
x −x−2
x +1
A+4 = 7 A
=
=
A B + x − 1 ( x − 1)2
1= A
⎧ 7 = A+ B ⎨ ⎩− 5 = A − 2 B A+ B 7=
4=B
2
x + 1 = Ax − A + B
7 x − 5 = ( A + B ) x + ( A − 2B )
3B
x +1
( x − 1)
x + 1 = A(x − 1) + B
7 x − 5 = Ax + A + Bx − 2b
12 =
32.
( x − 2)
=3
2
=
−1+ B = 1 B=2
1 2 + [9.4] x − 1 ( x − 1)2
3 4 + [9.4] x − 2 x +1
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Chapter Review
33.
2x − 2
( x 2 + 1)( x
677
+ 2)
=
Ax + B x2 + 1
+
34.
C x+2
2 x − 2 = ( Ax + B ) ( x + 2 ) + C ( x 2 + 1)
5 x 2 − 10 x + 9
( x − 2)
2
( x + 1)
=
A B C + + x − 2 ( x − 2 )2 x + 1 2
2 x − 2 = Ax 2 + 2 Ax + Bx + 2 B + Cx 2 + C
5 x 2 − 10 x + 9 = A( x − 2)( x + 1) + B ( x + 1) + C ( x − 2)
2 x − 2 = ( A + C ) x2 + (2 A + B ) x + (2B + C )
5 x 2 − 10 x + 9 = Ax 2 − Ax − 2 A + Bx + B + Cx 2 − 4Cx + 4C 5 x 2 − 10 x + 9 = ( A + C ) x 2 + ( − A + B − 4C ) x + ( −2 A + B + 4C )
⎧ 0 = A + C (1) ⎪ ⎨ 2 = 2 A + B (2) ⎪− 2 = 2 B + C (3) ⎩ ⎧0 = A + C ⎪ ⎨2 = 2 A + B ⎪2 = A − 2 B ⎩
(1) ⎧ 5= A+C ⎪ ⎨− 10 = − A + B − 4C (2) ⎪ 9 = −2 A + B + 4C (3) ⎩
(4)
0= A +C 2 = − 2B − C
(1) − 1 times (3)
2 = A − 2B
(4)
4 = 4 A + 2B
2 times (2)
2 = A − 2B
(4)
6 = 5A
(5)
− 19 = A
(4)
24 =
()
2 6 +B=2 5
( x 2 + 1) ( x − 2 )
=
6x − 2
5 ( x 2 + 1)
− 8C
5 = A + C (1) 19 = − A − 8C − 1 times (4)
6 +C =0 5 C =−6 5
B=−2 5
2x − 2
(2) − 1 times (3)
⎧ 5= A+C ⎪ (4) ⎨− 19 = A − 8C ⎪ 9 = −2 A + B + 4C ⎩
6 = A 5
A= 6 5
−10 = − A + B − 4C − 9 = 2 A − B − 4C
+
−6 [9.4] 5( x + 2)
9C
8 =C 3
(5)
⎧5 = A + C ⎪8 ⎪ C ⎨ = ⎪3 ⎩⎪9 = −2 A + B + 4C
A+ 8 = 5 3
(5)
(3 )
(3 )
−2 7 +B+4 8 =9
A= 7
B = 9 + 14 − 32
3
3
3
B=3 5 x 2 − 10 x + 9 2
( x − 2 ) ( x + 1)
=
7 3 8 + + [9.4] 3 ( x − 2 ) ( x − 2 )2 3 ( x + 1)
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678
35.
Chapter 9: Systems of Equations and Inequalities
11x 2 − x − 2 3
x −x
=
11x 2 − x − 2 A B C = + + x( x − 1)( x + 1) x x − 1 x + 1
36.
x 4 + x3 + 4 x 2 + x + 3 2
( x + 1)
11x 2 − x − 2 = A( x − 1)( x + 1) + Bx( x + 1) + Cx ( x − 1) 11x 2 − x − 2 = ( A + B + C ) x 2 + ( B − C ) x + (− A) (2) (3)
⎧ 11 = A + B + C ⎪ ⎨− 1 = B − C ⎪ 2 = −A ⎩
(2) (4)
=
x 4 + x3 + 4 x 2 + x + 3
=1+
11x 2 − x − 2 = Ax 2 − A + Bx 2 + Bx + Cx 2 − Cx ⎧ 11 = A + B + C ⎪ ⎨ −1 = B − C ⎪− 2 = − A ⎩
2
=1+
(1)
=1+
(1)
=1+
x4 + 2 x2 + 1 x3 + 2 x 2 + x + 2 ( x 2 + 1)2 x 2 ( x + 2) + 1( x + 2) ( x 2 + 1)2 ( x 2 + 1)( x + 2) ( x 2 + 1)2 x+2
[9.4]
x2 + 1
B + C = 9 from (1) with A = 2 B − C = −1 (2) =8 B=4 4 − C = −1 C =5
2B
11x 2 − x − 2 3
x −x 37.
=
2 4 5 [9.4] + + x x −1 x +1 38.
[9.5]
39.
40.
49 ⎞ ⎛5 vertex ⎜ , − ⎟ 4 ⎠ ⎝2
[9.5]
[9.5]
[9.5] 41.
42.
center (−3, − 1), r = 3
43.
[9.5]
44.
[9.5]
[9.5] [9.5]
45.
46.
[9.5]
47.
[9.5]
48.
[9.5] [9.5]
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Chapter Review
49.
679
50.
51.
52.
[9.5] [9.5] [9.5] 53.
[9.5]
54.
55.
56.
[9.5]
57.
[9.5]
[9.5]
[9.5] 58.
59.
60.
[9.5] [9.5] 61.
P = 2 x + 2 y [9.6] ⎧ x + 2 y ≤ 14 ⎪ ⎨5 x + 2 y ≤ 30 ⎪ x ≤ 0, y ≤ 0 ⎩
[9.5]
[9.5] P = 2x + 2 y (0, 7) (6, 0) (4, 5) (0, 0)
14 12 18 0
maximum
The maximum is 18 at (4, 5). 62.
P = 4 x + 5 y [9.6] ⎧ 2 x + 3 y ≤ 24 ⎪ ⎪ 4 x + 3 y ≤ 36 ⎨ ⎪ ⎪⎩ x ≥ 0, y ≥ 0
4x + 5y = P (0, 8) (6, 4)
40 44
(9, 0)
36
(0, 0)
0
maximum
The maximum is 44 at (6, 4). 63.
P = 4 x + y [9.6] ⎧5 x + 2 y ≥ 16 ⎪x + 2 y ≥ 8 ⎪⎪ ⎨ ⎪0 ≤ x ≤ 20 ⎪ ⎩⎪0 ≤ y ≤ 20
P = 4x + y (0, 8) (2, 3)
8 11
(8, 0)
32
(20, 0) (20, 20) (0, 20)
80 100 20
minimum
The minimum is 8 at (0, 8).
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680
64.
Chapter 9: Systems of Equations and Inequalities
P = 2x + 7 y ⎧4 x + 3 y ≥ 24 ⎪4 x + 7 y ≥ 40 ⎪⎪ ⎨ ⎪0 ≤ x ≤ 10 ⎪ ⎩⎪0 ≤ y ≤ 20
[9.6] P = 2x + 7 y (0, 8)
56
(0, 10) (3, 4) (10, 0) (10, 10)
70 34 20 90
minimun
The minimum is 20 at (10, 0). 65.
P = 6 x + 3 y [9.6] ⎧5 x + 2 y ≥ 20 ⎪ x+ y ≥ 7 ⎪⎪ ⎨ x + 2 y ≥ 10 ⎪ ⎪ ⎩⎪0 ≤ x ≤ 15, 0 ≤ y ≤ 10
6x + 3y = P (0, 10) (2, 5)
30 27
(4, 3)
33
(10, 0)
60
(0, 15)
45
(15, 15)
135
(15, 0)
90
minimum
The minimum is 27 at (2, 5). 66.
P = 5x + 4 y ⎧ x + y ≤ 10 ⎪ 2 x + y ≤ 13 ⎪⎪ ⎨ 3x + y ≤ 18 ⎪ ⎪ ⎩⎪ x ≥ 0, y ≥ 0
[9.6] 5x + 4 y = P (0, 10) (3, 7)
40 43
(5, 3)
37
(6, 0) (0, 0)
30 0
maximum
The maximum is 43 at (3, 7).
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Chapter Review
67.
681
68.
y = ax 2 + bx + c
x 2 + y 2 + ax + by + c = 0 42 + 22 + 4a + 2b + c = 0
0 = a(1) 2 + b(1) + c
02 + 12 + 0a + 1b + c = 0
2
5 = a (−1) + b(−1) + c
32 + (− 1)2 + 3a + (− 1)b + c = 0
2
3 = a ( 2) + b ( 2) + c ⎧ a+b+c = 0 ⎪ ⎨ a−b+c = 5 ⎪4a + 2b + c = 3 ⎩
(2) (3)
a+b+c = 0 − a + b − c = −5
(1) − 1 times (2)
2b
⎧4a + 2b + c = −20 ⎪ b + c = −1 ⎨ ⎪ 3a − b + c = −10 ⎩
(1)
= −5
−12a − 6b + 3c = 60 12a − 4b + 4c = −40 − 10b + c = 20
(2) − 1 times (3)
− 3a − 3b
(5)
⎧ a+b+c = 0 ⎪ 2b = −5 ⎨ ⎪− 3a − 3b = 2 ⎩
− 3 times (1) 4 times (3) (4)
(4)
a−b+c = 5 − 4a − 2b − c = −3 =2
(1) (2) (3)
⎧4a + 2b + c = −20 ⎪ b + c = −1 ⎨ ⎪ − 10b + c = 20 ⎩
b + c = − 1 (2) 10b − c = −20 − 1 times (4)
(4) (5)
⎛ 5⎞ 2b = −5 − 3a − 3⎜ − ⎟ = 2 ⎝ 2⎠ 5 b=− 11 a= 2 6
11 5 − +c = 0 6 2 2 c= 3
The equation of the graph that passes through the three points is 11 2 5 2 y= x − x + . [9.2] 6 2 3
(4)
11b
= −21
b
= − 21 11
(5)
⎧4a + 2b + c = −20 ⎪ b + c = −1 ⎨ ⎪ 11 b = −21 ⎩
(5)
− 21 + c = −1
4a + 2 − 21 + 10 = −20
c = 10
4a = − 188
( 11 ) ( 11 )
11
11
11
a = − 47 11
The equation of the circle that passes through the three points is x 2 + y 2 − 47 x − 21 y + 10 = 0. [9.2] 11
11
11
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682
69.
Chapter 9: Systems of Equations and Inequalities
x = ax + by + c 2 = 2a + b + c 0 = 3a + b + c − 2 = −2a − 3b + c
70.
x = amount of 20% acid 0.20 x + 0.10(10) = 0.16( x + 10) 0.20 x + 1 = 0.16 x + 1.6 0.04 x = 0.6
⎧ 2a + b + c = 2 ⎪ ⎨ 3a + b + c = 0 ⎪− 2a − 3b + c = −2 ⎩
x = 15 liters
(1)
[9.1]
(2) (3)
6a + 3b + 3c = 6 3 times (1) − 6a − 2b − 2c = 0 - 2 times (2) b + c = 6 (4) 2a + b + c = 2 (1) − 2a − 3b + c = −2 (3) − 2b + 2c = 0 −b+ c = 0 ⎧2a + b + c = 2 ⎪ b+c=6 ⎨ ⎪ −b+c = 0 ⎩
b+ c =6 −b + c = 0 2c = 6 c=3 ⎧2a + b + c = 2 ⎪ b+c = 6 ⎨ ⎪ c=3 ⎩ b+3=6 b=3
(5)
(4) (5)
(4) (5) (6)
(6) 2a + 3 + 3 = 2 2 a = −4 a = −2
The equation of the graph that passes through the three points is z = −2 x + 3 y + 3. [9.2]
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Chapter Review
71.
683
Rate flying with the wind: r + w Rate flying against the wind: r − w ⎧⎪855 = ( r + w ) 5 ⎨ ⎪⎩575 = ( r − w ) 5 171 = 143 + w 28 = w
171 = r + w 115 = r − w 286 = 2r 143 = r
Rate of the wind is 28 mph. Rate of the plane in calm air is 143 mph. [9.1]
72.
x = number of nickels y = number of dimes z = number of quarters ⎧ x + y + z =10 ⎨ ⎩ 5 x +10 y + 25 z =125 ⎧ x + y + z =10 ⎨ y + 4 z =15 ⎩ −5 x − 5 y − 5 z = −50 5 x +10 y + 25 z =125
(1) (2)
− 5 times (1)
5 y + 20 z = 75 y + 4 z =15 Solving the last equation for y, we have y = −4 z +15. Subsitute this into Eq. (1) and solve for x.
x + (−4 z + 15) + z = 10 x − 3z + 15 = 10 x = 3x − 5 Since x, y, and z must all be positive integers, from x = 3 z − 5, we have x ≥ 2. For y = −4 x + 15, x ≤ 3. Thus z = 2 or z = 3. When z = 2, x = 1, y = 7, there could be 1 nickel, 7 dimes and 2 quarters. When z = 3, x = 4, y = 3, there could be 4 nickels, 3 dimes and 3 quarters. [9.2] 73.
Given (a, b, c) with ab = c, ac = b. [9.2] If ab = c, then b ⋅ ( ab ) = a and b 2 = 1, b = ±1, or a = 0. If bc = a, then bc ⋅ c = b and c 2 = 1, c = ±1, or b = 0 If ac = b, then a ⋅ ac = c and a 2 = 1, a ± 1, or c = 0 The ordered triples are (1, 1, 1) , (1, − 1, − 1) , ( −1, − 1, 1) , ( −1, 1, − 1) , ( 0, 0, 0 ) .
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684
Chapter 9: Systems of Equations and Inequalities
.......................................................
Quantitative Reasoning
QR1.
QR2.
QR3. Answers will vary.
QR4. Answers will vary.
....................................................... 1.
⎧⎪3 x + 2 y = −5 (1) ⎨ ⎪⎩2 x − 5 y = −16 (2)
15 x + 10 y = −25
5 times (1)
4 x − 10 y = −32
2 times (2)
19 x = −57 x = −3 3(−3) + 2 y = −5 2y = 4 y=2 The solution is (−3, 2). [9.1]
Chapter Test 2.
⎧ x− 1 y =3 ⎪ ⎨ 2 ⎪⎩2 x − y = 6 −2 x + y = −6 2 x − y = −6 0=0
(1) (2) − 2 times (1)
The system of equations is dependent. Let y = c. 2x −c = 6 x = 6+ c 2 ⎛6+c ⎞ The solution is ⎜ , c ⎟ . [9.1] ⎝ 2 ⎠
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Test
3.
685
⎧ x + 3y − z = 8 ⎪ ⎨2 x − 7 y + 2 z = 1 ⎪4 x − y + 3z = 13 ⎩
(2) (3)
−2 x − 6 y + 2 z = −16 2x − 7 y + 2z =
− 2 times (1)
1
(2)
− 13 y + 4 z = −15
(4)
−4 x − 12 y + 4 z = −32 4x −
4.
(1)
(3)
− 13 y + 7 z = −19
(5)
⎧ x + 3y − z = 8 ⎪ ⎨ − 13 y + 4 z = −15 ⎪ − 13 y + 7 z = 19 ⎩
(4)
−13 y + 4 z = −15
(4)
13 y − 7 z = 19
− 8 y + 7 z = −1
4 3
(1) (5)
(5)
⎧3x − 2 y + z = 2 ⎪ ⎨ − 8 y + 7 z = −1 ⎪ − 8 y + 7 z = −1 ⎩
(5)
8y − 7z =
− 1 times (5)
(5)
(5) (6)
⎧3x − 2 y + z = 2 ⎪ ⎨ − 8 y + 7 z = −1 ⎪ 0=0 ⎩
(6)
(4)
1 − 1 times (4)
− 8y + 7z = −1 0=0
⎧ 8 ⎪x + 3y − z = ⎪ ⎨ − 13 y + 4 z = −15 ⎪ 4 z= − ⎪⎩ 3 ⎛ 4⎞ −13 y + 4 ⎜ − ⎟ = −15 ⎝ 3⎠ 29 y= 39
(2) (3)
−4 x − 8 y + 8z = −4 − 4 times (2) 4x − z = − 3 (3) − 8 y + 7 z = −1
− 3z = 4 z=−
3x − 2 y + z = 2
(1)
− 3x − 6 y + 6 z = − 3 − 3 times (2)
− 4 times (1)
y + 3z = 13
⎧3 x − 2 y + z = 2 ⎪ ⎨ x + 2 y − 2z = 1 ⎪4 x −z =3 ⎩
(6)
The system of equations is dependent. Let z = c. ⎛ 29 ⎞ ⎛ 4 ⎞ x + 3⎜ ⎟ − ⎜ − ⎟ = 8 ⎝ 39 ⎠ ⎝ 3 ⎠ 173 x= 39
− 8 y + 7 ( c ) = −1 y=
7c + 1 8
⎛ c + 3 7c + 1 ⎞ The solution is ⎜ , , c⎟ . 8 ⎝ 4 ⎠ [9.2]
4⎞ ⎛ 173 29 The solution is ⎜ , , − ⎟ . [9.2] 3⎠ ⎝ 39 39
Copyright © Houghton Mifflin Company. All rights reserved.
4x − c = 3 x=
c+3 4
686
5.
Chapter 9: Systems of Equations and Inequalities
⎧2 x − 3 y + z = −1 ⎨ ⎩ x + 5 y − 2z = 5
6.
(1) (2)
2x − 3y + z = − 1
(1)
4x + 2 y + z = 0
− 2 x − 10 y + 4 z = −10 − 2 times (2) − 13 y + 5 z = −11 ⎧2 x − 3 y + z = −1 ⎨ ⎩ − 13 y + 5 z = − 11
(1) (2) (3) (1)
− 4 x + 12 y + 8 z = 0 − 4 times (2)
(3)
14 y + 9 z = 0
(4)
−3 x + 9 y + 6z = 0 − 3 times (2) 3 x + 5y + 3z = 0 (3)
(3)
14 y + 9 z = 0
The system of equations is dependent. Let z = c.
⎧4 x + 2 y + z = 0 ⎪ ⎨ x − 3y − 2z = 0 ⎪ 3x + 5 y + 3z = 0 ⎩
− 8 y + 7 ( c ) = −1 7c + 1 y= 8
⎛ c + 10 5c + 11 The solution is ⎜ , , 13 ⎝ 13
⎧4 x + 2 y + z = 0 ⎪ 14 y + 9 z = 0 ⎨ ⎪ 14 y + 9z = 0 ⎩
4x − c = 3 c+3 x= 4
14y + 9 z = 0 −14 y − 9z = 0
⎞ c ⎟ . [9.2] ⎠
0=0
(5)
(4) (5)
− 1 times (5) (6)
⎧4 x + 2 y + z = 0 ⎪ ⎨ 14 y + 9 z = 0 ⎪ 0=0 ⎩ The system of equations is dependent. Let z = c.
14y + 9c = 0 y=−
9c 14
⎛ 9c ⎞ x − 3 ⎜ − ⎟ − 2c = 0 ⎝ 14 ⎠ c x= 14
9c ⎞ ⎛ c The solution is ⎜ , − , c ⎟ . [9.2] 14 ⎠ ⎝ 14
7.
(1) ⎧⎪ y = x + 3 ⎨ 2 ⎪⎩ y = x + x − 1 (2)
8.
Set the expressions equal to each other.
⎧⎪ y = x 2 − x − 3 ⎨ 2 ⎪⎩ y = 2 x + 2 x − 1
(1) (2)
Set the expressions equal to each other.
2
x + x −1 = x + 3
x2 + 2 x − 1 = x2 − x − 3 x 2 + 3x + 2 = 0 ( x + 2 )( x + 1) = 0
2
x =4 x = ±2 Substitute for x in Eq. (1).
Substitute for x in Eq. (1).
y = 2+3
y = −2 + 3
y = (− 2 )2 − (− 2 ) − 3
y = (− 1)2 − (− 1) − 3
y=5
y =1
y=3
y = −1
The solutions are (2, 5) and (−2, 1). [9.3] 9.
The solutions are (−2, 3) and (−1, −1). [9.3]
10.
[9.5]
11.
[9.5]
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[9.5]
Chapter Test
12.
687
No solution. The solution set is empty. [9.5]
13.
14.
[9.5] 15.
3x − 5 2
x − 3x − 4
=
3x − 5 ( x − 4)( x + 1)
=
A B + x − 4 x +1
[9.5] 2x + 1
16.
x( x 2 + 1)
=
A Bx + C + x x 2 +1
2 x + 1 = A( x 2 + 1) + ( Bx + C ) x 2 x + 1 = Ax 2 + A + Bx 2 + Cx
3 x − 5 = A( x + 1) + B ( x − 4) ⎧0 = A + B ⎪ ⎨2 = C ⎪1 = A ⎩
3 x − 5 = Ax + A + Bx − 4 B ⎧ 3 = A+ B ⎨ ⎩− 5 = A − 4 B
2x + 1
3= A+ B 5 = − A + 4B
2
x( x + 1)
8 = 5B
3= A+
8 =B 5
7 =A 5
=
0 = 1+ B −1 = B 1 −x + 2 [9.4] + x x2 +1
8 5
3x − 5 7 8 [9.4] = + ( x − 4)( x + 1) 5( x − 4) 5( x + 1) 17.
x = 20 + y 2 x + π y = 554.16 Using substitution, 2(20 + y ) + π y = 554.16 40 + 2 y + π y = 554.16 y (2 + π ) = 514.16 y = 514.16 2 +π y ≈ 100
18.
− x − 6 y = −14.50 x + 8 y = 18.00 14 y = 3.50 y = 1.75 x + 6(1.75) = 14.50 x=4 The fee for the first hour is $4; the fee for each additional half-hour or portion of the half-hour is $1.75. [9.1]
x = 20 + y = 20 + 100 = 120 Length is 120 m and width is 100 m. [9.1]
19.
x = Acres of oats y = Acres of barley Constraints ⎧ x + y ≤ 160 ⎪15 x + 13 y ≤ 2200 ⎪ ⎨ ⎪15 x + 20 y ≤ 2600 ⎪⎩ x ≤ 0, y ≥ 0
x = rate of first hour y = rate of n additional half hour or portion of the half-hour x + 6 y = 14.50 x + 8 y = 18.00
maximize p =120x + 150y (0, 130) ⎛ 680 , 400 ⎞ ⎜ ⎟ 7 ⎠ ⎝ 7
$19,500 $20,228.57
maximum
⎛ 146 2 , 0 ⎞ ⎜ ⎟ $17,600 3 ⎠ ⎝ (0, 0) 0 680 400 To maximize profit, acres of oats and acres of barley must be planted. 7 7 [9.6]
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688
20.
Chapter 9: Systems of Equations and Inequalities
x 2 + y 2 + ax + by + c = 0 32 + 52 + 3a + 5b + c = 0 ( −3)2 + ( −3)2 − 3a − 3b + c = 0 4 2 + 4 2 + 4 a + 4b + c = 0 ⎧ 3a + 5b + c = −34 ⎪ ⎨ −3a − 3b + c = −18 ⎪ 4a + 4b + c = −32 ⎩
(1) (2) ( 3)
3a + 5b + c = −34 − 3a − 3b + c = −18 2b + 2c = −52 b + c = −26
(1) (2) (4)
−12a − 12b + 4c = −72 4 times ( 2 ) 12a + 12b + 3c = −96 3 times ( 3) 7c = −168 c = −24 ( 5) ⎧3a + 5b + c = −34 ⎪ b + c = −26 ⎨ ⎪ c = −24 ⎩
c = −24
(4) (5)
b + (−24) = −26 b = −2
3a + 5(−2) − 24 = −34 a=0
The equation of the circle is x 2 + y 2 − 2 y − 24 = 0 . [9.2]
....................................................... 1.
3.
Cumulative Review
[2.4] f ( x) = − x 2 + 2 x − 4 = −( x 2 − 2 x ) + 4 = −( x 2 − 2 x + 1) + 4 − 1 = −( x − 1) 2 + 3 Vertex (1, –3), parabola opens down. Range: {y | y < –3 }
2.
Vertex = (4, 2), point (–1, 1), axis of symmetry x = 4
4.
log 6 ( x − 5) + 3log 6 (2 x) = log 6 ( x − 5) + log 6 (2 x)3 [4.4] = log 6 ( x − 5)(2 x)3 = log[8 x3 ( x − 5)]
even [2.5]
2
( x − 4) = 4 p( y − 2) (−1 − 4) 2 = 4 p(1 − 2) 25 = 4 p (−1) p = − 25 4 ( x − 4) 2 = 4 ⎛⎜ − 25 ⎞⎟ ( y − 2) ⎝ 4 ⎠ [8.1] ( x − 4) 2 = −25( y − 2)
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Cumulative Review
5.
689
log x − log(2 x − 3) = 2 [4.5] x log =2 2x − 3 x = 102 2x − 3 x = 100(2 x − 3) x = 200 x − 300 −199 x = −300 300 x= 199
6.
vertices (2, 2) and (10, 2), eccentricity = 3 [8.3] Length of transverse axis = distance between vertices 2a = 10 − 2 = 8 a = 4 and a 2 = 16 Center (midpoint of transverse axis) is ⎜⎛ 2 +10 , 2 + 2 ⎟⎞ = (6,2) 2 ⎠ ⎝ 2 Therefore, h = 6 and k = 2. Since both vertices lie on the horizontal line x = 2, the transverse axis is parallel to the x-axis. Since e = c , c = ae = (4)(3) = 12 a Because b 2 = c 2 − a 2 , b 2 = 144 − 16 = 128 Substituting h, k, a2, b2 into the standard equation yields ( x − 6) 2 ( y − 2) 2 − =1. 16 128
7.
( )
g −
1 2
=
1 2
− −2 −
1 2
=
5 2 1 − 2
−
8.
f (−2) ⋅ g (−2) = [(−2) 2 − 1][(−2) 2 − 4( −2) − 2] [2.6] = [4 − 1][4 + 8 − 2] = [3][10] = 30
= 5 [2.2]
9.
y = x2 + 0.4x – 0.8 10.
( x + 2)( x − 3i)( x + 3i) = ( x + 2)( x 2 + 9) [3.4] = x3 + 2 x 2 + 9 x + 18
11.
2 [4.1] 1− r 2 r= 1− Q r (1 − Q) = 2 r − rQ = 2 −rQ = 2 − r 2−r Q= −r r−2 Q −1 (r ) = r
12.
2x + 1 x − x − 1 2x − x − 2 2 x3 − 2 x 2 x2 − 2 x2 − x − 1 x +1
13.
g[ f (1)] = g[21 ] [4.2] = g[2] = 32(2) = 34 = 81
2
3
2
Q(r ) =
3 2 H ( x) = 2 x 2 − x − 2 = 2 x + 1 + 2 x + 1 x − x −1 x − x −1 Slant asymptote: y = 2x + 1 [3.5]
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690
Chapter 9: Systems of Equations and Inequalities
14.
( )
15.
cos 11π = 3 6 2
17.
sin15o = sin(45o − 30o )
[5.4]
[4.2] 16.
b 2 = a 2 + c 2 − 2ac cos B [7.2] 2 2 2⎞ ⎛ B = cos −1 ⎜ b − a − c ⎟ −2ac ⎝ ⎠ 2 2 2 ⎛ −1 (25) − (30) − (35) ⎞ = cos ⎜ ⎟ −2(30)(35) ⎝ ⎠
o
v = 3i − 2 j, w = i + 4 j
cos α = v ⋅ w = v w
(
[7.3] (3i − 2 j) ⋅ (i + 4 j)
(3) 2 + (−2)2 = 3 − 8 = −5 13 17 221
)(
(1)2 + (4) 2
o
o
= sin 45 cos30 − cos 45 sin 30 2 ⋅ 3 − 2 ⋅1 2 2 2 2 6 2 − = 4 =
≈ 44o 18.
[6.2]
o
cos −1 x + tan −1 x = π 2
19.
)
α = cos −1 ⎛⎜ −5 ⎞⎟ ≈ 109.7o
[6.5]
()
cos [ cos −1 x + tan −1 x ] = cos π 2 cos(α + β ) = 0 cos α cos β − sin α sin β = 0 cos α = x, sin α = 1 + x 2
⎝ 221 ⎠
cos β = 1 − x 2 , sin β =
x
(
) (
1 − x2 −
)
1 + x2 ⋅
x
1 + x2 x
1 + x2
=0
( 1 − x2 ) − x = 0 x ( 1 − x 2 − 1) = 0
x
x=0
1 − x2 − 1 = 0 1 − x2 = 1 1 − x2 = 1 0 = x2 0=x
20.
(10, 54π ) x = 10cos 5π = −10 4 5 y = 10sin π = −10 4
2 = −5 2 2 2 = −5 2 2
The rectangular coordinates are (−5 2, − 5 2) . [8.5]
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Chapter 10
Matrices Section 10.1 1.
⎡ 2 − 3 1 1⎤ ⎡ 2 − 3 1⎤ ⎢ ⎥ ⎢ ⎥ ⎢ 3 − 2 3 0⎥ , ⎢ 3 − 2 3⎥ , ⎢⎣ 1 0 5 4⎥⎦ ⎢⎣ 1 0 5⎥⎦
3.
⎡ 2 −3 −4 ⎢ 2 1 ⎢0 ⎢ 1 −1 2 ⎢ ⎣⎢ 3 − 3 − 2
4.
3 −2⎤ ⎡ 1 −1 2 3⎤ ⎡ 1 −1 2 ⎥ ⎢ ⎥ ⎢ 1 −2 1⎥ ⎢ 2 0 1 − 2⎥ ⎢ 2 0 , , ⎢ 3 0 0 −2 3⎥ ⎢ 3 0 0 − 2⎥ ⎥ ⎢ ⎥ ⎢ 0 3⎥⎦ ⎢⎣ − 1 3 − 1 0⎦⎥ ⎣⎢ − 1 3 − 1
5.
−2 R1 + R2 ⎡ 1 −1 2 2⎤ 2⎤ ⎡ 2 − 1 3 − 2⎤ ⎡ 1 −1 2 ⎡ 1 −1 2 2 ⎤ −5 R 2 + R 3 ⎢ −3R1 + R3 ⎢ ⎥ ⎢ ⎥ ⎯⎯⎯⎯⎯ ⎢ 2 −1 3 −2 ⎥ ⎯⎯⎯⎯⎯→ ←⎯ → 0 1 −1 − 6 ⎥ R R − − − → − 2 1 1 6 1 1 2 2 2⎢ ⎥ 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 3 2 − 1 3⎥⎦ ⎣⎢ 3 2 −1 3⎦⎥ ⎣⎢0 0 −2 27 ⎦⎥ ⎣⎢ 0 5 −7 −3⎦⎥ 2⎤ ⎡ 1 −1 2 ( −1 2) R3 ⎢ ⎥ ⎯⎯⎯⎯⎯ → ⎢0 1 −1 −6⎥ ⎢⎣0 0 1 − 27 2 ⎥⎦
6.
−2 R1 + R2 ⎡ 1 2 1⎤ 4 1⎤ ⎡1 2 4 −3R1 + R3 ⎢ ⎢ ⎥ 3⎥ ⎯⎯⎯⎯⎯→ 0 −2 −1 1⎥ ⎢2 2 7 ⎢ ⎥ ⎢⎣ 3 6 8 −1⎥⎦ ⎢⎣0 0 −4 −4 ⎥⎦
1 2⎤ ⎥ 0 2⎥ , 0 4⎥ ⎥ 0 1⎦⎥
2.
⎡ 1⎤ ⎢ ⎥ ⎢0⎥ ⎢⎣4⎥⎦
⎡ 2 −3 −4 ⎢ 2 1 ⎢0 ⎢ 1 −1 2 ⎢ ⎣⎢ 3 − 3 − 2
1⎤ ⎥ 0⎥ , 0⎥ ⎥ 0⎦⎥
3⎤ ⎡0 − 3 2 ⎢ ⎥ ⎢ 2 − 1 0 − 1⎥, ⎢⎣ 3 − 2 3 4⎥⎦
⎡ 0 − 3 2⎤ ⎢ ⎥ ⎢ 2 − 1 0⎥, ⎢⎣ 3 − 2 3⎥⎦
⎡ 3⎤ ⎢ ⎥ ⎢−1⎥ ⎢⎣ 4⎥⎦
⎡2⎤ ⎢ ⎥ ⎢2⎥ ⎢4⎥ ⎢ ⎥ ⎣⎢ 1⎦⎥ ⎡−2⎤ ⎢ ⎥ ⎢ 1⎥ ⎢ 3⎥ ⎢ ⎥ ⎣⎢ 3⎦⎥
4 1⎤ ⎡1 2 ( −1 2) R 2 ⎢ 1 1 ⎯⎯⎯⎯⎯→ 0 1 − 2⎥ 2 ⎢ ⎥ ⎢⎣ 0 0 −4 −4 ⎥⎦
⎡1 2 ( −1 4) R3 ⎢ ⎯⎯⎯⎯⎯→ 0 1 ⎢ ⎢⎣ 0 0
4 1
2
1
−1
1⎤ ⎥ 2⎥ 1⎥⎦
7.
−3R1 + R 2 ⎡ 1 −2 −1 2⎤ 3⎤ 3⎤ ⎡4 − 5 − 1 ⎡ 1 −2 −1 3⎤ ⎡ 1 −2 −1 1 R −4 R1 + R3 ⎢ ⎢ ⎥ ⎢ ⎥ ⎥ ⎯⎯⎯⎯ 2 2 → ⎢0 11 ⎥ − − 3 4 1 2 R R 3 2 1 2 1 2 ←⎯ → − − ⎯⎯⎯⎯⎯ → − − 0 2 4 11 1 3⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎢⎣ 1 − 2 − 1 ⎢⎣ 0 ⎢⎣ 4 −5 −1 2 ⎥⎦ ⎢⎣ 0 3⎥⎦ 3 3 −10⎥⎦ 3 3 −10⎥⎦ 3⎤ 3⎤ ⎡ 1 − 2 −1 ⎡ 1 −2 −1 −3R 2 + R 3 ⎢ ⎥ ⎥ − 13 R 3 ⎢ 11 11 ⎯⎯⎯⎯⎯⎯ → ⎢0 1 2 − 2 ⎥ ⎯⎯⎯⎯→ ⎢ 0 1 2 − 2⎥ ⎢⎣ 0 ⎢⎣0 0 −3 13 2 ⎥⎦ 0 1 − 13 6 ⎥⎦
8.
⎡ − 2 1 − 1 3⎤ ⎢ ⎥ ⎢ 2 2 4 6⎥ ⎢⎣ 3 1 − 1 2⎥⎦
⎡ 2 2 4 6⎤ R1 ←⎯→ R 2 ⎢ −2 1 −1 3⎥ ⎢ ⎥ ⎣⎢ 3 1 −1 2 ⎥⎦
2 R1 + R2 1 2 3⎤ ⎡ 1 1 2 3⎤ ⎡1 −3R1 + R3 ⎢ (1/ 2) R1 ⎢ 3 3 9⎥ ⎯⎯⎯⎯⎯ → −2 1 −1 3⎥ ⎯⎯⎯⎯⎯ → 0 ⎢ ⎥ ⎢ ⎥ ⎢⎣0 −2 −7 −7 ⎦⎥ ⎣⎢ 3 1 −1 2 ⎥⎦
⎡1 1 2 1 2 3⎤ ⎡ 1 1 2 3⎤ ⎡1 2 R 2 + R3 ⎢ (1 3) R 2 ⎢ ( −1/ 5) R 3 ⎢ → ⎢0 1 1 0 1 1 3⎥ ⎯⎯⎯⎯⎯ → 0 1 1 3⎥ ⎯⎯⎯⎯⎯ ⎯⎯⎯⎯→ ⎢ ⎥ ⎢ ⎥ ⎢0 0 1 ⎢⎣0 0 −5 −1⎥⎦ ⎣⎢0 −2 −7 −7 ⎦⎥ ⎣
3⎤ ⎥ 3⎥ 1 ⎥ 5⎦
Copyright © Houghton Mifflin Company. All rights reserved.
692
9.
10.
11.
12.
13.
Chapter 10: Matrices
−3R1 + R 2 −4⎤ ⎡ 1 −2 3 ⎡ 1 −2 3 −4 ⎤ ⎡ 1 −2 3 −4 ⎤ −5R1 + R 3 ⎡ 1 −2 3 −4 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ 0 2 4 −1 1 2 − 12⎥ 0 0 1 −2 (1 2) R2 ⎢0 −2 R 1 + R 4 ⎢ 3 − 6 10 − 14⎥ ⎯⎯⎯⎯⎯⎯ ⎥ R 2 ←⎯→ R 3 ⎢ ⎥ ⎯⎯⎯⎯⎯ ⎥ →⎢ → ⎢ 5 − 8 19 − 21⎥ ⎢ 0 2 4 −1⎥ ⎢0 ⎢0 0 1 −2 ⎥ 0 1 −2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 1 −2 ⎦ 0 1 −2 ⎦ 0 1 −2 ⎦ ⎣0 ⎣0 ⎣0 ⎣⎢ 2 − 4 7 − 10⎥⎦ ⎡ 1 −2 3 −4 ⎤ ⎢ 1 2 − 12⎥ − R3 + R 4 ⎢0 ⎥ ⎯⎯⎯⎯⎯ → ⎢0 0 1 −2 ⎥ ⎢ ⎥ 0 0 0⎦ ⎣0 2⎤ ⎥ 3⎥ 2⎥ ⎥ 8⎥⎦ 1 R 2 + R3 ⎡⎢ 0 5R 2 + R 4 ⎢ ⎯⎯⎯⎯⎯ → ⎢0 ⎢ ⎣0
−2 R1 + R 2 3⎤ ⎡ 1 2 −1 3⎤ −3R1 + R 3 ⎢ 0 −5 5 −4 ⎥ ⎥ 2 −4 R 1 + R 4 ⎢ ⎥ ⎥ ⎯⎯⎯⎯⎯⎯→ ⎢ 0 −1 1 −7 ⎥ 2⎥ ⎢ ⎥ ⎥ 8⎦ ⎣ 0 −5 5 −4 ⎦ ⎡ 1 2 −1 3⎤ ⎢ 0 1 −1 4 ⎥ 5R − 31 3 5⎥ ⎯⎯⎯⎯ →⎢ ⎢ 0 0 0 1⎥ ⎢ ⎥ ⎣0 0 0 0⎦
⎡ 1 2 −1 ⎢ 2 −1 3 R1 ←⎯→ R 2 ⎢ ⎢ 3 5 −2 ⎢ 1 ⎣4 3
3 ⎡ 2 −1 ⎢ 1 2 1 − ⎢ ⎢3 5 − 2 ⎢ 1 ⎢⎣ 4 3
2 −1 1 −1 0 0
3⎤ ⎥ 5⎥ 0 − 31 5 ⎥ ⎥ 0 0⎦
4 2 1⎤ ⎡ 1 −3 ⎢ ⎥ 2 − 3 5 − 2 − 1⎥ ⎢ ⎢⎣ − 1 2 −3 1 3⎥⎦ ⎡ 1 −3 4 R2 + R3 ⎢ 1 −1 ⎯⎯⎯⎯ → 0 ⎢ ⎢⎣ 0 0 0
4
⎡ 1 2 −1 3⎤ ⎢ 1 −1 4 5 ⎥ (1 5) R 2 ⎢0 ⎥ ⎯⎯⎯⎯→ ⎢0 −1 1 −7 ⎥ ⎢ ⎥ ⎣0 −5 5 −4 ⎦
−2 R1 + R 2 ⎡ 1 −3 4 2 2 1⎤ 1⎤ ⎡ 1 −3 4 R1 + R3 ⎢ (1 3) R 2 ⎢ 0 1 −1 −2 −1⎥ ⎯⎯⎯⎯⎯ → 0 3 −3 −6 −3⎥ ⎯⎯⎯⎯→ ⎢ ⎥ ⎢ ⎥ ⎢⎣0 −1 1 3 4 ⎦⎥ ⎣⎢0 −1 1 3 4 ⎥⎦ 2 1⎤ −2 −1⎥ ⎥ 1 3⎥⎦
−2 R1 + R2 ⎡ 1 −2 2 1 −1⎤ 2 2⎤ 1 −1⎤ ⎡2 − 1 3 ⎡ 1 −2 2 −3R1 + R 2 ⎢ ⎢ ⎥ ⎢ 2 −1 3 2 2 ⎥ ⎯⎯⎯⎯⎯ 0 3 1 0 4⎥ 1 − 2 2 1 − 1 R R → − ←⎯ → 2 ⎢ ⎢ ⎥ 1 ⎢ ⎥ ⎥ ⎢⎣ 3 − 5 − 1 − 2 1 −7 −5 6⎦⎥ 3⎥⎦ ⎣⎢ 0 ⎣⎢ 3 −5 −1 −2 3⎦⎥ 2 1 −1⎤ 2 1 −1⎤ 2 1 −1⎤ ⎡ 1 −2 ⎡ 1 −2 ⎡ 1 −2 (1/ 20) R 3 ⎢ −3R 2 + R3 ⎢ ⎥ → ⎢0 1 −7 −5 6⎥ 1 −7 −5 6⎥ ⎯⎯⎯⎯⎯ 1 −7 −5 6⎥ ⎯⎯⎯⎯⎯ R 2 ←⎯→ R 3 ⎢0 → 0 ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 ⎢⎣ 0 0 20 15 −14 ⎥⎦ ⎢⎣0 0 1 3 4 − 7 10 ⎥⎦ 3 −1 0 4 ⎥⎦ −5R1 + R 2 ⎡ 1 2 −2 −2 ⎤ ⎡ 1 2 − 2 − 2⎤ −3R1 + R3 ⎢ ⎥ ⎢ ⎥ ⎢5 9 − 4 − 3⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 −1 6 7 ⎥ ⎢⎣3 4 − 5 − 3⎥⎦ ⎢⎣ 0 −2 1 3⎥⎦ ⎡ 1 2 −2 −2 ⎤ ( −1/11) R 3 ⎢ ⎯⎯⎯⎯⎯⎯ → 0 1 −6 −7 ⎥ ⎢ ⎥ 1 1⎦⎥ ⎣⎢0 0 ⎧ x + 2 y − 2 z = −2 y − 6(1) = −7 ⎪ y = −1 y − 6 z = −7 ⎨ ⎪ z=1 ⎩
2⎤ ⎡ 1 2 −2 ⎡ 1 2 −2 −2 ⎤ 2 R2 + R3 ⎢ −1R 2 ⎢ ⎥ 1 −6 −7 ⎯⎯⎯⎯⎯ ⎯⎯⎯⎯ → 0 → 0 1 − 6 −7 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣0 −2 ⎢⎣0 0 −11 −11⎥⎦ 1 3⎥⎦
x + 2(−1) − 2(1) = −2 x=2
The solution is (2, −1, 1).
Copyright © Houghton Mifflin Company. All rights reserved.
Section 10.1
14.
693
−2 R1 + R 2 ⎡ 1 −3 1 8⎤ 8⎤ 8⎤ 1 8⎤ ⎡ 1 −3 1 ⎡1 − 3 ⎡ 1 −3 1 (1/ 35) R3 ⎢ −7 R 2 + R3 ⎢ −1R1 + R3 ⎢ ⎥ ⎢ ⎥ ⎥ ⎥ 1 −5 −14 ⎥ 1 −5 −14 ⎯⎯⎯⎯⎯→ 0 1 −5 −14 ⎯⎯⎯⎯⎯→ ⎢0 ⎢2 − 5 − 3 2⎥ ⎯⎯⎯⎯⎯→ ⎢0 ⎢ ⎥ ⎥ ⎢⎣ 1 ⎢⎣0 0 1 13 5 ⎥⎦ 4 1 1⎥⎦ 91⎦⎥ ⎣⎢0 7 0 −7 ⎦⎥ ⎣⎢0 0 35 ⎧ ⎪ x − 37 + z = 8 ⎪ ⎨ y − 5 z = −14 ⎪ z = 13 5 ⎩⎪
The solution is
15.
(
12 , 5
− 1,
13 5
).
⎡ 3 7 −7 −4 ⎤ ⎢ 1 2 −3 0 ⎥ R ←⎯→ R 2 ⎢ ⎥ 1 ⎢⎣5 6 1 −8⎥⎦
0⎤ ⎡ 1 2 −3 0⎤ ⎡ 1 2 −3 0⎤ −3R1 + R 2 ⎡ 1 2 −3 4 R 2 + R3 −5R1 + R 3 ⎢ ⎥ ⎢3 7 −7 −4 ⎥ ⎯⎯⎯⎯⎯⎯ ⎢ 1 2 −4 ⎯⎯⎯⎯⎯ → 0 → 0 1 2 −4 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 −4 16 −8⎥⎦ ⎢⎣0 0 24 −24 ⎥⎦ ⎢⎣5 6 1 −8⎥⎦
1R ⎡ 1 2 −3 0 ⎤ 3 24 ⎯⎯⎯⎯ → ⎢ 0 1 2 −4 ⎥ ⎢ ⎥ ⎢⎣0 0 1 −1⎥⎦
y + 2(−1) = −4 ⎧ x + 2 y − 3z = 0 ⎪ y = −2 2 4 y + z = − ⎨ ⎪ z = −1 ⎩
x + 2( −2) − 3(−1) = 0 x =1
The solution is (1, −2, −1). 16.
13⎤ ⎡ 2 −3 2 13⎤ ⎡ 2 −3 2 − R1 + R 2 ⎢ 3 −4 −3 1⎥ ⎯⎯⎯⎯⎯ → ⎢ 1 −1 −5 −12 ⎥ R1 ←⎯→ R 2 ⎢ ⎥ ⎢ ⎥ ⎢⎣ 3 1 −1 1 −1 2 ⎥⎦ 2 ⎦⎥ ⎣⎢ 3
−2 R1 + R 2 ⎡ 1 −1 −5 −12⎤ −3R1 + R3 ⎢ 2 −3 2 13⎥ ⎯⎯⎯⎯⎯ → ⎢ ⎥ 2⎥⎦ ⎣⎢ 3 1 −1
⎡ 1 −1 −5 −12 ⎤ ⎢0 −1 12 37 ⎥ ⎢ ⎥ 38⎥⎦ ⎣⎢0 4 14
1R ⎡ 1 −1 −5 −12 ⎤ ⎡ 1 −1 −5 −12 ⎤ ⎡ 1 −1 −5 −12⎤ −4 R 2 + R 3 −1R 2 ⎢ 0 1 −12 −37 ⎥ ⎯⎯⎯⎯ 62 3 → ⎢ 0 1 −12 −37⎥ ⎯⎯⎯⎯ → ⎢ 0 1 −12 −37 ⎥ ⎯⎯⎯⎯⎯→ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 14 38⎥⎦ 1 3⎦⎥ ⎣⎢ 0 4 ⎣⎢ 0 0 62 186⎦⎥ ⎣⎢ 0 0 ⎧ x − y − 5 z = −12 ⎪ y − 12 z = −37 ⎨ ⎪ z=3 ⎩
y − 12(3) = −37
x − (−1) − 5(3) = −12
y = −1
x=2
The solution is (2, −1, 3). 17.
−5R1 + R 2 ⎡ 1 2 −2 3⎤ ⎡ 1 2 − 2 3⎤ ⎡ 1 2 −2 ⎡ 1 2 −2 3⎤ −3R1 + R3 ⎢ 2 R2 + R3 ⎢ (1/ 2) R 2 ⎢ ⎥ ⎢ ⎥ ⎯⎯⎯⎯⎯ 1 ⎥ ⎯⎯⎯⎯⎯ 0 2 4 1 0 1 −2 0 1 2 → ⎯⎯⎯⎯⎯ → − − → − − 5 8 6 14 2⎥ ⎥ ⎢ ⎢ ⎢ ⎢ ⎥ ⎢⎣3 4 − 2 8⎥⎦ ⎢ ⎥ ⎢ ⎢⎣ 0 −2 ⎥ 4 −1⎦ 0 4 −1⎦ ⎣0 0 ⎣ 0 −2
⎧⎪ x + 2 y − 2 z = 3 ⎨ y − 2z = 1 ⎪⎩ 2 1 y = 2z + 2 1 ⎛ x + 2⎜ 2 z + ⎞⎟ − 2 z = 3 2⎠ ⎝ x = 2 − 2z
(
)
Let z be any real number c. The solution is 2 − 2c, 2c + 1 , c . 2
Copyright © Houghton Mifflin Company. All rights reserved.
3⎤ ⎥ 2⎥ 0⎥⎦
1
694
18.
Chapter 10: Matrices
⎡ 3 −5 2 4 ⎤ ⎢ 1 −3 2 4 ⎥ R ←⎯→ R 2 ⎢ ⎥ 1 ⎣⎢5 −11 6 12 ⎦⎥
−3R1 + R 2 ⎡ 1 −3 2 4 ⎤ −5R1 + R3 ⎢3 −5 2 4 ⎥ ⎯⎯⎯⎯⎯ → ⎢ ⎥ ⎣⎢5 −11 6 12 ⎦⎥
1R ⎡ 1 −3 2 4 ⎤ ⎢0 4 −4 −8⎥ ⎯⎯⎯ 4 2→ ⎢ ⎥ ⎣⎢0 4 −4 −8⎥⎦
4⎤ ⎡ 1 −3 2 ⎢0 1 − 1 −2 ⎥ ⎢ ⎥ ⎣⎢0 4 −4 −8⎥⎦
4⎤ ⎡ 1 −3 2 −4 R 2 + R3 ⎢0 ⎯⎯⎯⎯⎯→ 1 −1 −2 ⎥ ⎢ ⎥ 0⎥⎦ ⎣⎢ 0 0 0
⎧x − 3 y + 2z = 4 y = z−2 ⎨ y − z = −2 ⎩
x − 3( z − 2) + 2 z = 4
x = z−2 Let z be any real number c. The solution is (c – 2, c – 2, c).
19.
−2 R1 + R 2 ⎡ 3 2 −1 1⎤ ⎡ 1 −1 2 3⎤ −3R1 + R3 ⎢ 2 3 −1 1⎥ R ←⎯→ R ⎢ 2 3 −1 1⎥ ⎯⎯⎯⎯⎯ → 3 ⎢ ⎢ ⎥ 1 ⎥ ⎣⎢ 1 −1 2 3⎥⎦ ⎣⎢ 3 2 −1 1⎦⎥
3⎤ 1R ⎡ 1 −1 2 ⎢ 0 5 −5 −5⎥ ⎯⎯⎯ 5 2→ ⎢ ⎥ ⎣⎢ 0 5 −7 −8⎥⎦
3⎤ ⎡ 1 −1 2 ⎢0 1 −1 −1⎥ ⎢ ⎥ ⎣⎢0 5 −7 −8⎥⎦
3⎤ ⎡ 1 −1 2 3⎤ ⎡ 1 −1 2 − 1 R3 −5R 2 + R3 2 → ⎢ 0 1 −1 −1⎥ ⎯⎯⎯⎯⎯ → ⎢0 1 −1 −1⎥ ⎯⎯⎯⎯ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 0 1 3 2 ⎥⎦ ⎣⎢0 0 −2 −3⎦⎥
⎧ ⎪x − y + 2z = 3 ⎪ y − z = −1 ⎨ ⎪ z=3 ⎪⎩ 2
y−
3 = −1 2 1 y= 2
x−
1 ⎛3⎞ + 2⎜ ⎟ = 3 2 ⎝2⎠ 1 x= 2
⎛1 1 3⎞ The solution is ⎜ , , ⎟ . ⎝2 2 2⎠ 20.
⎡ 2 5 2 −1⎤ ⎢ 1 2 −3 5⎥ R ←⎯→ R 2 ⎢ ⎥ 1 ⎢⎣ 5 12 1 10⎥⎦
⎡ 1 2 −3 5⎤ −1R1 + R 2 −5R1 + R3 ⎢ 2 5 2 −1⎥ ⎯⎯⎯⎯⎯ → ⎢ ⎥ ⎢⎣ 5 12 1 10⎥⎦
5⎤ 5⎤ ⎡ 1 2 −3 ⎡ 1 2 −3 −2 R 2 + R3 ⎢ 0 1 8 −11⎥ ⎢0 1 8 −11⎥ ⎯⎯⎯⎯⎯→ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 0 0 ⎢⎣0 2 16 −15⎥⎦ 7 ⎥⎦
⎧ x + 2 y − 3z = 5 ⎪ y + 8 z = −11 ⎨ ⎪ 0z = 7 ⎩
The system of equations has no solution because the equation 0z = 7 has no solution. 21.
22.
−2 R1 + R 2 ⎡ 1 −3 2 0⎤ 2 0⎤ ⎡1 − 3 ⎡ 1 −3 2 0 ⎤ −4 R1 + R3 ⎢ −1R2 + R3 ⎢ ⎥ ⎢ ⎥ 1 −6 0 ⎯⎯⎯⎯⎯→ 0 1 −6 0⎥ ⎢ 2 − 5 − 2 0⎥ ⎯⎯⎯⎯⎯→ ⎢0 ⎥ ⎢ ⎥ ⎢⎣ 4 − 11 ⎢⎣0 ⎢⎣ 0 0 0 0⎥⎦ 1 −6 0⎥⎦ 2 0⎥⎦ ⎧x − 3 y + 2z = 0 y = 6z x − 3( 6z ) + 2z = 0 ⎨ y − 6 z = 0 x = 16 z ⎩ Let z be any real number c. The solution is (16c, 6c, c). ⎡ 1 1 − 2 0⎤ −3R1 + R 2 ⎡ 1 1 −2 0⎤ ⎡ 1 1 −2 0⎤ −1R 2 + R3 ⎢ ⎥ −5R1 + R3 ⎢ ⎢ ⎥ ⎥ ⎢3 4 − 1 0⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 1 5 0⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 1 5 0⎥ ⎢⎣5 6 − 5 0⎥⎦ ⎢⎣ 0 1 5 0⎦⎥ ⎣⎢ 0 0 0 0⎦⎥ ⎧x + y − 2z = 0 y = −5 z x + ( −5 z ) − 2 z = 0 ⎨ y + 5 z = 0 x = 7z ⎩ Let z be any real number c. The solution is (7c, –5c, c).
Copyright © Houghton Mifflin Company. All rights reserved.
Section 10.1
23.
695
− R1 + R 2 −2 R1 + R 2 ⎡ 2 1 −3 4 ⎤ ⎡ 2 1 −3 4 ⎤ ⎡ 1 1 4 −2 ⎤ → ⎢ ⎢3 2 ⎥ ⎯⎯⎯⎯⎯ ⎥ R1 ←⎯→ R 2 ⎢ 2 1 −3 4 ⎥ ⎯⎯⎯⎯⎯→ 1 2 1 1 4 2 − ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ − 1 1 4 2 −1R 2 ⎡ ⎤ ⎯⎯⎯⎯ → ⎢ ⎥ ⎣0 1 11 −8⎦
4 −2 ⎤ ⎡1 1 ⎢0 −1 −11 8⎥ ⎣ ⎦
⎧ x + y + 4 z = −2 y = −11z − 8 x + (−11z − 8) + 4 z = −2 ⎨ + = − y 11 z 8 x = 7z + 6 ⎩ Let z be any real number c. The solution is (7c + 6, −11c – 8, c).
24.
−2 R1 + R 2 ⎡ 3 −6 2 2 ⎤ ⎡ 1 −11 5 0⎤ −1R 2 + R1 → ⎢ ⎢2 ⎥ ⎯⎯⎯⎯⎯⎯ ⎥ ⎯⎯⎯⎯⎯→ − − 5 3 2 2 5 3 2 ⎣ ⎦ ⎣ ⎦
⎧⎪ x − 11 y + 5 z = 0 ⎨ y − 13 z = 2 ⎪⎩ 27 27
Let z be any real number c. The solution is 25.
(
( 27
)
2
0⎤ ⎥ 27 ⎦
x − 11 13 z + 2 + 5 z = 0
y = 13 z + 2 27
1R 5 5 0⎤ ⎡ 1 −11 27 2 → ⎡ 1 −11 ⎯⎯⎯⎯ ⎢0 ⎢0 27 −13 2⎥ 13 1 − 27 ⎣ ⎦ ⎣
27
27
x = 8 z + 22 27
)
27
8 c + 22 , 13 c + 2 , c . 27 27 27 27
⎡ 2 2 − 4 4⎤ ⎡ 1 1 −2 2 ⎤ −2 R1 + R 2 ⎡ 1 1 −2 2 ⎤ ⎡ 1 1 −2 2 ⎤ −4 R1 + R3 ⎢ −1R 2 + R3 ⎢ (1/ 2) R1 ⎢ ⎥ ⎢ ⎥ ⎥ → 2 3 −5 4 ⎯⎯⎯⎯⎯→ 0 1 −1 0 ⎯⎯⎯⎯⎯→ 0 1 −1 0⎥ ⎢ 2 3 − 5 4⎥ ⎯⎯⎯⎯⎯ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 4 5 − 9 8⎥⎦ ⎢⎣ 4 5 −9 8⎥⎦ ⎢⎣ 0 1 −1 0⎥⎦ ⎢⎣0 0 0 0⎥⎦ y=z ⎧x + y − 2z = 2 x + z − 2z = 2 ⎨ y−z =0 x = z+2 ⎩
Let z be any real number c. The solution is (c + 2, c, c). 26.
−3R1 + R 2 ⎡ 1 −4 ⎡ 1 −4 1 13⎤ 1 13⎤ ⎡3 − 10 2 34⎤ ⎡ 1 −4 1 13⎤ −5R1 + R3 ⎢ (1/ 2) R 2 ⎢ ⎢ ⎥ ⎥ ⎯⎯⎯⎯⎯ ⎢ ⎥ 1 −5 ⎥ 0 2 → − − 3 10 2 34 1 4 1 13 R ←⎯ → R − ⎯⎯⎯⎯⎯ → − − 0 2 1 5 2⎢ 2⎥ ⎢ 2 ⎢ ⎥ 1 ⎥ ⎢ ⎥ ⎢0 18 ⎢⎣5 − 2 7 31⎥⎦ ⎢⎣5 −2 7 31⎥⎦ ⎢⎣ 0 18 2 −34 ⎥⎦ 2 −34⎥⎦ ⎣
⎡ 1 −4 ⎡ 1 −4 1 13⎤ 1 13⎤ −18R 2 + R3 ⎢ ⎥ (1/11) R3 ⎢ ⎥ ⎯⎯⎯⎯⎯⎯ → ⎢0 1 − 12 − 5 2 ⎥ ⎯⎯⎯⎯⎯ 1 − 12 − 5 2 ⎥ → ⎢0 ⎢0 0 11 ⎢0 0 11⎥⎦ 1 1⎥⎦ ⎣ ⎣ ⎧ x − 4 y + z = 13 ⎪⎪ 1 5 ⎨ y− 2z = −2 ⎪ z =1 ⎪⎩
1 2
y − (1) = −
5 2
y = −2
x − 4(−2) + 1 = 13 x=4
The solution is (4, − 2, 1). 27.
−2 R1 + R 2 1 3 4 11⎤ 3 4 11⎤ ⎡1 3 4 11⎤ ⎡ 1 3 4 11⎤ −4 R1 + R3 ⎡ ⎡1 ⎢0 ⎢0 −3 −6 −15⎥ ⎢2 ⎥ 1 2 5⎥ 3R 2 + R 3 ⎢0 − −3R1 + R 4 ⎢ R ( 1/ 3) 1 2 5⎥ 3 2 7 2 ⎥ ⎯⎯⎯⎯⎯→ ⎢ ⎥ ⎯⎯⎯⎯⎯ →⎢ ⎢ 4 9 10 20⎥ ⎯⎯⎯⎯⎯→ ⎥ − 0 0 0 9 ⎢0 −3 −6 −24 ⎥ ⎢0 −3 −6 −24 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 3 2 1 1 0 11 11 32 − − − − ⎣ ⎦ ⎣ ⎦ ⎣0 −11 −11 −32 ⎦ ⎣0 −11 −11 −32 ⎦ ⎡1 3 4 11⎤ ⎡1 ⎢ ⎢0 ⎥ 1 2 5 ( −1/11) R3 ⎢ 0 ⎢ ⎥ R 3 ←⎯→ R 4 ⎯⎯⎯⎯⎯⎯ → ⎢0 ⎢ 0 −11 −11 −32 ⎥ ⎢ ⎢ ⎥ 0 0 −9 ⎦ ⎢⎣ 0 ⎣0 ⎧ x + 3 y + 4 z = 11 ⎪⎪ y + 2z = 5 ⎨ 23 z = 11 ⎪ 0 z = −9 ⎩⎪
11⎤ ⎡1 ⎢0 5⎥ − R + R ⎥ ⎯⎯⎯⎯⎯ 2 3→ ⎢ ⎥ ⎢0 1 1 32 11 ⎥ ⎢ 0 0 −9 ⎥⎦ ⎢⎣0 3 4 1 2
11⎤ ⎡1 ⎢ 5⎥ −1R 3 ⎢0 ⎥ ⎯⎯⎯⎯ → 23 ⎥ ⎢0 0 −1 − 11 ⎥ ⎢ 0 0 −9 ⎥⎦ ⎢⎣0 3 1
4 2
Because 0 z = −9 has no solutions, the system of equations has no solution.
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11⎤ 5⎥ ⎥ 23 ⎥ 0 1 11 ⎥ 0 0 −9 ⎥⎦ 3 4 1 2
696
28.
Chapter 10: Matrices
3 4⎤ 3 4⎤ 3 4⎤ 3 4⎤ −3R1 + R 2 ⎡ 1 −4 ⎡ 1 −4 ⎡ 1 −4 ⎡ 1 −4 ⎥ −5R1 + R3 ⎢ ⎢ ⎥ ⎢0 ⎥ ⎢0 0 2 6 8 − − − − − − 1 3 4 1 3 4⎥ 3 10 3 4 − R R − + 2 2 R R R − + (1/ 2) 1 3→ ⎢ 2→ ⎢ 1 4→ ⎢ ⎥ ⎯⎯⎯⎯⎯ ⎢ ⎥ ⎥ ⎯⎯⎯⎯⎯ ⎥ ⎯⎯⎯⎯⎯ ⎢ 5 − 18 ⎢0 ⎢0 2 −6 −10⎥ ⎢ 0 2 −6 −10⎥ 0 0 −2 ⎥ 9 10⎥ ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 2 − 3 − 11⎦⎥ ⎣0 10 −9 −19 ⎦ ⎣0 10 −9 −19 ⎦ ⎣ 0 10 −9 −19 ⎦ ⎣⎢ 2 3 4⎤ 3 4⎤ 3 4⎤ ⎡ 1 −4 ⎡ 1 −4 ⎡ 1 −4 ⎢0 1 −3 −4 ⎥ −10 R 2 + R3 ⎢0 1 −3 −4 ⎥ (1/ 21) R 3 ⎢0 1 −3 − 4 ⎥ ⎥ ⎯⎯⎯⎯⎯⎯ ⎥ ⎯⎯⎯⎯⎯ ⎥ R 3 ←⎯→ R 4 ⎢ →⎢ →⎢ ⎢0 ⎢0 ⎢0 10 −9 −19 ⎥ 0 21 21⎥ 0 1 1⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 −2 ⎦ 0 0 −3⎦ 0 0 −2 ⎦ ⎣0 ⎣0 ⎣0 ⎧ x − 4 y + 3z = 4 ⎪ y − 3 z = −4 ⎪ ⎨ z =1 ⎪ ⎪⎩ 0 z = −2
Because 0 z = −2 has no solution, the system of equations has no solution. 29.
30.
−3R1 + R 2 1 2 − 3 1 −7 ⎤ 1 −7 ⎤ −7 ⎤ ⎡ 1 2 −3 −2 R1 + R3 ⎡ ⎥ ⎢ 0 −1 1 2 13⎥ ⎢0 1 −1 −2 −13⎥ 5 − 8 5 − 8⎥ −4 R1 + R 4 ⎢ −1R 2 ⎢ ⎥ ⎯⎯⎯⎯ ⎥ ⎯⎯⎯⎯⎯ → → ⎢ 0 −1 −1 1 3⎥ ⎢ 0 −1 − 1 1 3⎥ 3 − 7 3 − 11⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 3 18⎦ 8 − 10 7 − 10⎥⎦ ⎣ 0 0 2 3 18⎦ ⎣0 0 2 1 −7 ⎤ 1 −7 ⎤ 1 −7 ⎤ ⎡ 1 2 −3 ⎡ 1 2 −3 ⎡ 1 2 −3 ⎢0 1 −1 −2 −13⎥ ⎢0 1 −1 −2 −13⎥ ⎢0 1 −1 −2 −13⎥ −2 R3 + R 4 ⎢ R2 + R3 ⎢ ( −1/ 2) R3 ⎢ ⎥ ⎯⎯⎯⎯⎯→ ⎥ ⎥ ⎯⎯⎯⎯⎯ ⎯⎯⎯⎯ → → ⎢0 0 ⎢0 0 1 12 5⎥ 1 12 5⎥ ⎢0 0 −2 −1 −10⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 3 18⎦ 3 18⎦⎥ 8⎥⎦ ⎣0 0 2 ⎣⎢0 0 2 ⎣⎢0 0 0 2 1 ⎧t + 2u − 3v + w = − 7 u − 3 − 2(4) = −13 t + 2(−2) − 3(3) + 4 = −7 v+ (4)=5 ⎪ 2 u v w 2 13 − − = − ⎪⎪ u = − 2 t= 2 v=3 ⎨ 1 v+ w= 5 ⎪ 2 ⎪ ⎪⎩ 2w = 8 The solution is (2, − 2, 3, 4). ⎡1 ⎢ ⎢3 ⎢2 ⎢ ⎢⎣ 4
2
−3 1
−2 R1 + R 2 1 ⎡ 1 4 2 −3 11⎤ 4 R1 + R3 ⎡ − ⎥ ⎢ ⎢ −1R1 + R 4 ⎢0 ⎢ 2 10 3 − 5 17 ⎥ ⎯⎯⎯⎯⎯ → ⎢ 4 16 7 − 9 34⎥ ⎢0 ⎥ ⎢ ⎢ − 1 4 1 1 4 ⎥⎦ ⎣0 ⎣⎢ ⎡1 ⎢ R3 + R 4 ⎢0 ⎯⎯⎯⎯ → ⎢0 ⎢ ⎣⎢0
4
( 2,
2 −3
4 1 0 0
11⎤ ⎥ − 25 ⎥ 1 −3 10⎥ ⎥ −1 2 −7 ⎦⎥
− 12
1 2
11⎤ ⎥ − 25 ⎥ 1 −3 10⎥ ⎥ 0 −1 3⎦⎥
⎧t + 4u + 2v − 3w = 11 ⎪ u− 1v+ 1 w= − 5 ⎪ 2 2 2 ⎨ v − 3w = 10 ⎪ ⎪ −w= 3 ⎩
The solution is
⎡1 (1/ 2) R 2 ⎢ 0 −1R3 ⎯⎯⎯⎯⎯ →⎢ ⎢0 ⎢ ⎣⎢0
2 −3
1 − 12 0 0
4 2 −3 11⎤ 2 −1 1 −5⎥ ⎥ 0 −1 3 −10⎥ ⎥ 0 −1 2 −7 ⎦
1 2
v − 3(3) = 10 v= 1
u − 1 (1) + 1 (−3) = − 5 2
2
2 u =− 1 2
t + 4(− 1 ) + 2(1) − 3(−3) = 11
)
− 1 , 1, − 3 . 2
Copyright © Houghton Mifflin Company. All rights reserved.
2
t= 2
Section 10.1
31.
697
3 −1R1 + R 2 ⎡ 1 − 1 3 1 1⎤ ⎡1 − 1 1 1⎤ 3 2 2⎤ ⎡ 2 −1 2 2 2 2 −3R1 + R3 ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ 1 1 0 1⎥ 2 1 2⎥ −2 R1 + R4 ⎢0 − 2 (1/ 2) R1 ⎢ 1 −1 2 1 2⎥ 2 ⎢ 1 −1 ⎯⎯⎯⎯⎯ → ⎯⎯⎯⎯⎯→ ⎢ ⎥ ⎢3 ⎢ 3 0 − 2 − 3 13⎥ 3 − 13 −6 10 0 −2 −3 13⎥ ⎢0 ⎥ 2 2 ⎢ ⎥ ⎥ ⎢ ⎢ ⎥ 0 − 2 6⎥⎦ ⎢⎣ 2 2 2 0 −2 6⎦⎥ ⎢⎣ 2 − − 0 3 3 4 4 ⎣ ⎦ 3 3 ⎡ 1 − 12 1 1⎤ ⎡1 − 1 1 1⎤ 2 2 2 ( −3/ 2) R 2 + R3 ⎢ ⎢ ⎥ ⎥ 1 −1 0 −2 ⎥ −2 R 2 ⎢ 0 − 3R 2 + R 4 ⎢ 0 1 −1 0 −2 ⎥ ⎯⎯⎯⎯ →⎢ ⎯⎯⎯⎯⎯⎯⎯ → ⎢0 3 − 13 −6 10⎥ 0 −5 −6 13⎥ ⎢0 ⎥ 2 2 ⎢ ⎥ ⎢ 0 0 −4 10⎥⎦ ⎢⎣0 3 −3 −4 4 ⎦⎥ ⎣0 3 1 3 ⎡ 1 − 12 1⎤ ⎡ 1 − 12 1 1⎤ 2 2 ⎢ ⎥ ⎢ ⎥ 1 −1 0 −2 ⎥ 1 −1 0 −2 ⎥ (1/5) R3 ⎢0 ( −1/ 4) R 4 ⎢ 0 ⎯⎯⎯⎯⎯ →⎢ ⎥ ⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 0 1 6 − 13 0 1 6 − 13 ⎢ 5⎥ ⎢0 5 5⎥ 5 ⎢ ⎥ ⎢0 0 0 1 − 25 ⎥⎦ 0 0 −4 10⎥⎦ ⎣ ⎣⎢ 0 ⎧t − 1 u + 3 v + w = 1 v + 6 ( − 5 ) = − 13 u − 2 = −2 t − 1 (− 8 ) + 3 ( 2 ) − 5 = 1 2 ⎪ 2 5 5 2 5 2 5 2 5 2 ⎪ u− v = −2 8 2 ⎪ v= u=− t = 21 5 5 ⎨ 10 6 w = − 13 v + ⎪ 5 5 ⎪ w= − 5 ⎪ 2 ⎩
8 2 5⎞ ⎛ 21 The solution is ⎜ , − , , − ⎟ . 5 5 2⎠ ⎝ 10
32.
3 ⎡ 4 7 −10 ⎢ 3 5 7 2 − ⎢ ⎢1 2 − 3 1 ⎢ 2 −4 ⎢⎣ 2 − 1 ⎡1 − R2 ⎢ − R 2 + R 3 ⎢0 ⎯⎯⎯⎯⎯→ ⎢0 ⎢ ⎣0
⎡1 ⎢ − (1/ 2) R 3 ⎢ 0 ⎯⎯⎯⎯⎯⎯ → ⎢0 ⎢ ⎣⎢ 0
−3R1 + R 2 1 −29⎤ 1 −9 ⎤ −4 R + R ⎡ 1 2 −3 ⎡ 1 2 −3 1 3 ⎥ ⎢ 3 5 −7 2 −20⎥ −2 R1 + R 4 ⎢ 0 −1 2 −1 − 20⎥ ⎥ ⎯⎯⎯⎯⎯⎯ →⎢ R1 ←⎯→ R 3 ⎢ ⎢ 0 −1 2 −1 ⎢ 4 7 −10 3 −29 ⎥ − 9⎥ ⎥ ⎢ ⎢ ⎥ 2 1 2 4 15 − − 15⎥⎦ ⎣ 0 −5 8 −6 ⎣ ⎦ 2 −3 1 −9 ⎤ 1 −9 ⎤ ⎡ 1 2 −3 ⎡1 2 ⎢0 ⎢0 1 1 −2 1 −7 ⎥ 1 −2 1 −7 ⎥ + R R 5 2 3→ ⎢ ⎥ R 3 ←⎯→ R 4 ⎢ ⎥ ⎯⎯⎯⎯⎯⎯ ⎢0 −5 8 −6 33⎥ ⎢0 0 0 0 0 0⎥ ⎥ ⎢ ⎥ ⎢ −5 8 −6 33⎦ 0 0 0 0 0 ⎣ ⎦ ⎣0 0
−9 ⎤ 7⎥ ⎥ 7⎥ ⎥ 33⎦ − 3 1 −9 ⎤ −2 1 −7 ⎥ ⎥ −2 −1 −2 ⎥ ⎥ 0 0 0⎦
2 −3 1 −9 ⎤ 1 −2 1 −7 ⎥ ⎥ 0 1 12 1⎥ ⎥ 0 0 0 0⎦⎥
⎧ ⎪t + 2u − 3v + w = − 9 ⎪ u − 2v + w = − 7 ⎨ ⎪ v+ 1w= 1 ⎪⎩ 2
v = − 1 w +1 2
u − 2(− 1 w + 1) + w = −7
t + 2(2 w − 5) − 3( − 1 w + 1) + w = −9
u = − 2w − 5
t = 3 w+ 4
2
2
2
1 ⎛3 ⎞ Let w be any real number c. The solution is ⎜ c + 4, − 2c − 5, − c + 1, c ⎟ . 2 2 ⎝ ⎠
Copyright © Houghton Mifflin Company. All rights reserved.
698
33.
Chapter 10: Matrices
⎡ 3 10 7 −6 ⎢ 2 8 6 −5 ⎢ 1 4 2 −3 ⎢ ⎣ 4 14 9 −8
⎡1 4 2 7⎤ ⎢2 8 6 5⎥ R1 ←⎯→ R 3 ⎢ ⎥ 2⎥ ⎢ 3 10 7 ⎢ 8⎦ ⎣ 4 14 9
−3 −5 −6 −8
−2 R + R 2 ⎤ −3R 1 + R 2 ⎡ 1 4 2 −3 2 ⎤ 1 3 5⎥ −4 R1 + R 4 ⎢0 0 2 1 1⎥ ⎥ ⎯⎯⎯⎯⎯⎯ ⎥ →⎢ 7⎥ ⎢0 −2 1 3 1⎥ ⎥ ⎢0 −2 1 4 0⎥ 8⎦ ⎣ ⎦
2 −3 2⎤ 2 −3 2⎤ ⎡1 4 ⎡1 4 ⎢ ⎢ 1 − 3 − 1⎥ 1 − 3 − 1⎥ 0 1 0 1 − − ( −1/ 2) R 2 ⎢ 2R 2 + R 4 ⎢ 2 2 2 ⎥ ⎯⎯⎯⎯⎯→ 2 2 2⎥ ⎯⎯⎯⎯⎯→ ⎢0 0 ⎢0 0 2 1 1⎥ 2 1 1⎥ ⎢ ⎥ ⎢ ⎥ 1 4 0⎥⎦ 0 1 −1⎦⎥ ⎣⎢ 0 0 ⎣⎢0 −2 ⎧t + 4u + 2v − 3w = 2 u − 12 (1) − 32 (−1) = − 12 2v + (−1) = 1 ⎪⎪ u − 1 v − 3 w = − 1 v =1 u = − 32 2 2 2 ⎨ 2v + w = 1 ⎪ w = −1 ⎪⎩
⎡ 1 4 2 −3 2 ⎤ ⎢0 −2 1 3 1⎥ ⎥ R 2 ←⎯→ R 3 ⎢ ⎢0 0 2 1 1⎥ ⎢ ⎥ ⎣0 −2 1 4 0⎦
t + 4(− 32 ) + 2(1) − 3(−1) = 2 t =3
The solution is ⎛⎜ 3, − 3 , 1, − 1⎞⎟ . 2 ⎝ ⎠ 34.
−3R1 + R 2 4 13⎤ ⎡ 1 −3 2 R2 + R3 2 4 13⎤ −2 R1 + R 3 ⎡ 1 −3 2 4 13⎤ ⎢0 ⎢ ⎥ ⎥ 1 2 1 −4 ⎥ − − − 0 1 2 1 4 −1R 2 + R 4 ⎢ −4 R1 + R 4 4 13 35 ⎥ ⎥ ⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯⎯ →⎢ → ⎥ 8 5 28⎥ ⎢ 0 0 2 −2 − 2 ⎥ ⎢0 −1 4 −3 2 ⎥ ⎢ ⎥ 6 17 56 ⎦ ⎢0 1 −2 1 4 ⎥⎦ 0 0 8⎦ ⎣ ⎣0 0 ⎧t − 3u + 2v + 4w = 13 ⎪ u − 2v + w = − 4 ⎨ v − w = −1 ⎪ 0w = 8 ⎩ ⎡ 1 −3 ⎢ 3 −8 ⎢ 2 −7 ⎢ ⎣ 4 −11
Because equation 0w = 8 has no solution, the system of equations has no solution. 35.
−4 R1 + R 2 ⎡ 1 −1 2 −3 9 ⎤ ⎡ 1 −1 2 −3 9 ⎤ ⎡ 1 −1 2 −3 9 ⎤ −3R1 + R 3 ⎢ ⎥ ⎢ 4 0 11 −10 46 ⎥ ⎯⎯⎯⎯⎯⎯ → 0 4 3 2 10 R 2 ←⎯→ R 3 ⎢0 2 2 3 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 3 −1 8 −6 27 ⎥ ⎢⎣0 2 2 ⎣ ⎦ ⎢⎣0 4 3 2 10⎥⎦ 3 0⎥⎦ ⎡ 1 − 1 2 −3 9 ⎤ −2 R 2 + R3 3 0⎥ ⎯⎯⎯⎯⎯→ ⎢0 2 2 ⎢ ⎥ ⎢⎣0 0 −1 −4 10⎥⎦ ⎧t − u + 2v − 3w = 9 ⎪ ⎨ 2u + 2v + 3w = 0 ⎪ − v − 4 w = 10 ⎩
v = −4w − 10
2u + 2( −4 w − 10) + 3w = 0
Let w be any real number c. The solution is ⎛⎜ 27 c + 39, 5 c + 10, − 4c − 10, 2 ⎝ 2 36.
t − ( 5 w + 10) + 2(−4 w − 10) − 3w = 9 2
u = 5 w + 10 2
t=
c ⎞⎟ . ⎠
−2 R1 + R 2 1 −1 3 −5 10⎤ 1 −1 3 −5 10⎤ ⎡ 1 −1 3 −5 10 ⎤ −3R1 + R 3 ⎡⎢ −1R 2 ⎡⎢ ⎢ 2 −3 4 → 0 1 1 7 ⎥ ⎯⎯⎯⎯⎯⎯ 2 −11 13⎥ → 0 −1 −2 11 −13⎥ ⎯⎯⎯⎯ ⎢ 3 1 −2 −2 6 ⎥ ⎢0 4 −11 13 −24 ⎥ ⎢0 4 −11 13 −24 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 1 −1 3 −5 10⎤ 1 −1 3 −5 10⎤ −4 R 2 + R 3 ⎡⎢ ( −1/19) R 3 ⎡⎢ ⎯⎯⎯⎯⎯⎯ → 0 1 → 0 1 2 −11 13⎥ 2 −11 13⎥ ⎯⎯⎯⎯⎯⎯ ⎢0 0 −19 57 −76⎥ ⎢ 0 0 1 −3 4 ⎥ ⎣ ⎦ ⎣ ⎦ ⎧t − u + 3v − 5w = 10 ⎪ ⎨ u + 2v − 11w = 13 ⎪ v − 3w = 4 ⎩
v = 3w + 4
u + 2(3w + 4) − 11w = 13 u = 5w + 5
t − (5w + 5) + 3(3w + 4) − 5w = 10 t = w+3
Let w be any real number c. The solution is (c + 3, 5c + 5, 3c + 4, c ). Copyright © Houghton Mifflin Company. All rights reserved.
27 w + 39 2
Section 10.1
37.
699
1 0 1 0 2⎤ ⎡3 −4 (1/ 3) R1 ⎡1 − 43 3 →⎢ ⎥ ⎯⎯⎯⎯⎯ ⎢ 1 −2 3 ⎣ 1 1 − 2 3 1⎦ ⎣⎢1
2 ⎧t − 4 u + 1 v = ⎪ 3 3 3 ⎨ 9 1 ⎪ u− v+ = 7 7 ⎩
4 − R1 + R 2 ⎡1 − 3 →⎢ ⎥ ⎯⎯⎯⎯⎯ 7 ⎢⎣1 1⎦⎥ 3
u = v− 9 w+ 1 7
1 3 7 −3
2⎤ 3
7
0 3
2⎤ 3 ⎥ 1⎥ 3⎦
4 1 0 (3/ 7) R 2 ⎡ 1 − 3 3 ⎯⎯⎯⎯⎯ →⎢ ⎢⎣ 0 1 −1 97
2⎤ 3 ⎥ 1⎥ 7⎦
t − 4 (v − 9 w + 1 ) + 1 v = 2 3
7 7 6 12 t = v− w+ 7 7
3
3
(
)
Let v be any real number c1 and w be any real number c2 . The solution is c1 − 12 c + 76 , c1 − 79 c2 + 71 , c1 , c2 . 7 2 38.
3 −2 3 −2 1⎤ 1⎤ 1R1 + R 2 ⎡ 1 0 3 −4 2⎤ (1/ 2) R1 ⎡1 0 ⎡2 0 2 2 →⎢ ⎥ ⎥ ⎯⎯⎯⎯⎯→ ⎢ ⎢ ⎥ ⎯⎯⎯⎯⎯ 11 1 − 3⎦ 3 −4 ⎥⎦ 1 −3⎥⎦ ⎢⎣ 0 2 − 2 ⎢⎣1 2 −4 ⎣1 2 − 4 3 −2 1⎤ (1/ 2) R 2 ⎡ 1 0 2 ⎥ ⎯⎯⎯⎯⎯ →⎢ 3 −2 ⎥ 11 2 ⎣⎢ 0 1 − 4 ⎦ ⎧ t + 3 v − 2w = 1 ⎪ 2 ⎨ ⎪u − 11 v + 3 w = −2 4 2 ⎩
t = − 3 v + 2w + 1 2
u = 11 v − 3 w − 2 4
2
(
Let v be any real number c1 and w be any real number c2 . The solution is − 3 c1 + 2c2 + 1, 39.
2
11 c 4 1
)
− 23 c2 − 2, c1 , c2 .
Because there are two points, the degree of the interpolating polynomial is at most 1. The form of the polynomial is p ( x) = a1 x + a0 Use this polynomial and the given points to find the system of equations. p (−2) = a1 (−2) + a0 = −7
p (1) = a1 (1) + a0 = −1 ⎧−2a1 + a0 = −7 ⎡ −2 1 −7 ⎤ The system of equations and the associated augmented matrix are ⎨ ⎢ ⎥ ⎣ 1 1 − 1⎦ ⎩ a1 + a0 = −1 The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. ⎧⎪a 1 − 1 a 0 = 7 ⎡ 1 −1 / 2 7 / 2 ⎤ 2 2 The augmented matrix in echelon form and resulting system of equations are ⎢ ⎨ ⎥ 1 −3⎦ a 0 = −3 ⎣0 ⎪⎩ Solving by back substitution yields a0 = −3 and a1 = 2 . The interpolating polynomial is p(x) = 2x – 3. 40.
Because there are two points, the degree of the interpolating polynomial is at most 1. The form of the polynomial is p( x) = a1x + a0 . Use this polynomial and the given points to find the system of equations. p(−3) = a1 (−3) + a0 = −8 p(1) = a1 (1) + a0 = 4 ⎧−3a1 + a0 = −8 ⎡−3 1 −8⎤ The system of equations and the associated augmented matrix are ⎨ ⎢ ⎥ ⎣ 1 1 4⎦ ⎩ a1 + a0 = 4 The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. ⎧a − 1 a = 8 ⎡ 1 − 1 / 3 8 / 3⎤ ⎪ 1 3 0 3 The augmented matrix in echelon form and resulting system of equations are ⎨ ⎥ ⎢ 0 1 1 ⎦ ⎣ ⎪⎩ a0 = 1 Solving by back substitution yields a0 = 1 and a1 = 3 . The interpolating polynomial is p(x) = 3x + 1.
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700
41.
Chapter 10: Matrices
Because there are three points, the degree of the interpolating polynomial is at most 2. The form of the polynomial is p ( x) = a2 x 2 + a1x + a0 . Use this polynomial and the given points to find the system of equations. p(−1) = a2 ( −1) 2 + a1 (−1) + a0 = 6 p (1) = a2 (1) 2 + a1 (1) + a0 = 2 p (2) = a2 ( 2) 2 + a1 (2) + a0 = 3 ⎧ a2 − a1 + a0 = 6 ⎡ 1 − 1 1 6⎤ ⎪ ⎥ ⎢ The system of equations and the associated augmented matrix are ⎨ a2 + a1 + a0 = 2 ⎢ 1 1 1 2⎥ ⎪4a + 2 a + a = 3 ⎢⎣ 4 2 1 3⎥⎦ 1 0 ⎩ 2 The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. ⎧a + 1 a + 1 a = 3 ⎪ 2 2 1 4 0 4 1 1/ 4 3 / 4⎤ ⎡ 1/ 2 ⎪⎪ 1 7 ⎢ ⎥ a1 − a0 = − The augmented matrix in echelon form and resulting system of equations are ⎢0 1 − 1 / 2 − 7 / 2⎥ ⎨ 2 2 ⎪ ⎢⎣0 0 1 3⎥⎦ a0 = 3 ⎪ ⎪⎩ Solving by back substitution yields a0 = 3, a1 = −2, and a2 = 1 .
The interpolating polynomial is p( x) = x 2 − 2 x + 3 . 42.
Because there are three points, the degree of the interpolating polynomial is at most 2. The form of the polynomial is p ( x) = a2 x 2 + a1x + a0 . Use this polynomial and the given points to find the system of equations. p (−2) = a2 ( −2) 2 + a1 ( −2) + a0 = −3 p (0) = a2 (0) 2 + a1 (0) + a0 = −1 p(3) = a2 (3) 2 + a1 (3) + a0 = 17 ⎧4a2 − 2a1 + a0 = −3 ⎡ 4 − 2 1 − 3⎤ ⎪ ⎥ ⎢ a0 = −1 The system of equations and the associated augmented matrix are ⎨ 0 1 − 1⎥ ⎢0 ⎪ 9a + 3a + a = 17 ⎢⎣ 9 3 1 17 ⎥⎦ 1 0 ⎩ 2 The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. ⎧a + 1 a + 1 a = 17 ⎪ 2 3 1 9 0 9 1 / 9 17 / 9⎤ ⎡ 1 1/ 3 ⎪⎪ 1 19 ⎢ ⎥ a1 − a0 = The augmented matrix in echelon form and resulting system of equations are ⎢0 1 − 1 / 6 19 / 6⎥ ⎨ 6 6 ⎪ ⎢⎣0 0 1 − 1⎥⎦ a0 = −1 ⎪ ⎪⎩ Solving by back substitution yields a0 = −1, a1 = 3, and a2 = 1 .
The interpolating polynomial is p( x ) = x 2 + 3 x − 1 .
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Section 10.1
43.
701
Because there are four points, the degree of the interpolating polynomial is at most 3. The form of the polynomial is p( x) = a3 x 3 + a2 x 2 + a1x + a0 . Use this polynomial and the given points to find the system of equations. p (−2) = a3 (−2) 3 + a2 ( −2) 2 + a1 (−2) + a0 = −12 p(0) = a3 (0)3 + a2 (0) 2 + a1 (0) + a0 = 2 p (1) = a3 (1) 3 + a2 (1) 2 + a1 (1) + a0 = 0 p(3) = a3 (3)3 + a2 (3) 2 + a1 (3) + a0 = 8
⎧−8a3 + 4a2 − 2a1 + a0 = −12 ⎡−8 4 −2 1 −12⎤ ⎪ ⎢ ⎥ a0 = 2 2⎥ ⎪ ⎢ 0 0 0 1 The system of equations and the associated augmented matrix are ⎨ ⎢ 1 1 1 1 0⎥ ⎪ a3 + a2 + a1 + a0 = 0 ⎢ ⎥ ⎪ 27 a3 + 9a2 + 3a1 + a0 = 8 27 9 3 1 8⎦⎥ ⎣⎢ ⎩ The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. The augmented matrix in echelon form and resulting system of equations are ⎧a + 1 a + 1 a + 1 a = 8 ⎪ 3 3 2 9 1 27 0 27 1 / 9 1 / 27 8 / 27 ⎤ ⎡ 1 1/ 3 ⎪ 1 7 13 ⎢ ⎥ 1 − 1 / 6 7 / 36 − 13 / 9⎥ a2 − a1 + a0 = − ⎪⎪ ⎢0 6 36 9 ⎨ ⎢0 0 1 5/6 2 / 3⎥ ⎪ 5 2 ⎢ ⎥ a1 + a0 = ⎪ 0 0 1 2⎦⎥ 6 3 ⎣⎢0 ⎪ a0 = 2 ⎪⎩ Solving by back substitution yields a0 = 2, a1 = −1, a2 = −2 and a3 = 1 . The interpolating polynomial is p( x) = x 3 − 2 x 2 − x + 2 . 44.
Because there are four points, the degree of the interpolating polynomial is at most 3. The form of the polynomial is p( x) = a3 x 3 + a2 x 2 + a1x + a0 . Use this polynomial and the given points to find the system of equations. p (−1) = a3 (−1) 3 + a2 (−1) 2 + a1 (−1) + a0 = −5 p (0) = a3 (0) 3 + a2 (0) 2 + a1 (0) + a0 = 0 p (1) = a3 (1)3 + a2 (1) 2 + a1 (1) + a0 = 1 p( 2) = a3 (2) 3 + a2 (2) 2 + a1 (2) + a0 = 4 ⎧ − a3 + a2 − a1 + a0 ⎪ a0 ⎪ The system of equations and the associated augmented matrix are ⎨ + + + a a a a 2 1 0 ⎪ 3 ⎪⎩8a3 + 4a2 + 2a1 + a0
= −5
1 −5⎤ 1 0⎥ ⎥ =1 1 1⎥ ⎥ 1 4⎦ =4 The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. The augmented matrix in echelon form and resulting system of equations are ⎧a + 1 a + 1 a + 1 a = 1 ⎪ 3 2 2 4 1 8 0 2 1 / 4 1 / 8 1 / 2⎤ ⎡ 1 1/ 2 ⎪ 1 3 ⎢ ⎥ ⎪⎪ 1 − 1 / 2 3 / 4 − 3⎥ a2 − a1 + a0 = −3 ⎢0 2 4 ⎨ ⎢0 0 1 1/ 2 2⎥ ⎪ 1 ⎢ ⎥ a1 + a0 = 2 ⎪ 0 0 1 0⎥⎦ ⎢⎣0 2 ⎪ a0 = 0 ⎪⎩ Solving by back substitution yields a0 = 0, a1 = 2, a2 = −2 and a3 = 1 . =0
The interpolating polynomial is p( x ) = x 3 − 2 x 2 + 2 x .
Copyright © Houghton Mifflin Company. All rights reserved.
⎡ −1 ⎢ 0 ⎢ ⎢ 1 ⎢ ⎣ 8
1 −1 0 0 1 1 4 2
702
45.
Chapter 10: Matrices
Because there are three points, the degree of the interpolating polynomial is at most 2. The form of the polynomial is p ( x) = a2 x 2 + a1x + a0 . Use this polynomial and the given points to find the system of equations. p(−1) = a2 (−1) 2 + a1 (−1) + a0 = 3 p (1) = a2 (1) 2 + a1 (1) + a0 = 7 p (2) = a2 (2) 2 + a1 (2) + a0 = 9 ⎧ a2 − a1 + a0 = 3 ⎡ 1 − 1 1 3⎤ ⎪ ⎢ ⎥ The system of equations and the associated augmented matrix are ⎨ a2 + a1 + a0 = 7 ⎢ 1 1 1 7⎥ ⎪4a + 2 a + a = 9 ⎢⎣ 4 2 1 9⎥⎦ 1 0 ⎩ 2 The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. ⎧a + 1 a + 1 a = 9 ⎪ 2 2 1 4 0 4 1 / 2 1 / 4 9 / 4 1 ⎤ ⎡ ⎪⎪ 1 1 ⎥ ⎢ a1 − a0 = − The augmented matrix in echelon form and resulting system of equations are ⎢0 1 − 1 / 2 − 1 / 2⎥ ⎨ 2 2 ⎪ ⎢⎣0 0 1 5⎥⎦ a0 = 5 ⎪ ⎪⎩ Solving by back substitution yields a0 = 5, a1 = 2, and a2 = 0 .
The interpolating polynomial is p( x) = 2 x + 5 . 46.
Because there are three points, the degree of the interpolating polynomial is at most 2. The form of the polynomial is p ( x) = a2 x 2 + a1x + a0 . Use this polynomial and the given points to find the system of equations. p(−2) = a2 (−2) 2 + a1 (−2) + a0 = 7 p (1) = a2 (1) 2 + a1 (1) + a0 = −2 p (2) = a2 (2) 2 + a1 (2) + a0 = −5
7⎤ ⎧4a2 − 2a1 + a0 = 7 ⎡4 − 2 1 ⎪ ⎢ ⎥ 1 1 1 2⎥ − The system of equations and the associated augmented matrix are ⎨ a2 + a1 + a0 = −2 ⎢ ⎪4a + 2a + a = −5 ⎢⎣4 2 1 − 5⎥⎦ 1 0 ⎩ 2 The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. ⎧a − 1 a + 1 a = 7 ⎪ 2 2 1 4 0 4 ⎡ 1 −1 / 2 1 / 4 7 / 4⎤ ⎪ ⎥ = −3 a1 The augmented matrix in echelon form and resulting system of equations are ⎢⎢0 1 0 − 3⎥ ⎨ ⎪ ⎢⎣0 ⎥ 0 1 1⎦ a0 = 1 ⎪ ⎩ Solving by back substitution yields a0 = 1, a1 = −3, and a2 = 0 . The interpolating polynomial is p ( x ) = −3 x + 1 .
Copyright © Houghton Mifflin Company. All rights reserved.
Section 10.1
47.
703
The form of the polynomial is: z = ax + by + c. Use this polynomial and the given points to find the system of equations. −4 = a( −1) + b(0) + c 5 = a(2) + b(1) + c −1= a( −1) + b(1) + c
⎧ −a + c = −4 ⎡ −1 0 1 −4 ⎤ ⎪ ⎢ 2 1 1 5⎥ The system of equations and the associated augmented matrix are ⎨ 2a + b + c = 5 ⎢ ⎥ ⎪−a + b + c = −1 ⎩ ⎣⎢ −1 1 1 −1⎦⎥ The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. ⎧a + 1 b + 1 c = 5 ⎡ 1 1/ 2 1/ 2 5/ 2 ⎤ ⎪⎪ 2 2 2 b + c =1 The augmented matrix in echelon form and resulting system of equations are ⎢ 0 1 1 1⎥ ⎨ ⎢ ⎥ ⎪ 0 1 −2 ⎥⎦ c = −2 ⎢⎣ 0 ⎪⎩ Solving by back substitution yields a = 2, b = 3, and c = −2 . The interpolating polynomial is z = 2x + 3y – 2. 48.
The form of the polynomial is: z = ax + by + c. Use this polynomial and the given points to find the system of equations. −3 = a (1) + b(2) + c −7 = a (−2) + b(0) + c −4 = a (0) + b(1) + c
⎧ a + 2b + c = −3 ⎡ 1 2 1 −3⎤ ⎪ ⎢ −2 0 1 −7 ⎥ + c = −7 The system of equations and the associated augmented matrix are ⎨−2a ⎢ ⎥ ⎪ b + c = −4 ⎩ ⎣⎢ 0 1 1 −4⎦⎥ The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. ⎧a − 1 c = 7 2 2 7 / 2⎤ ⎡ 1 0 −1/ 2 ⎪ ⎪ 3 13 ⎢ ⎥ The augmented matrix in echelon form and resulting system of equations are 0 1 3/ 4 −13 / 4 ⎨ b+ 4 c = − 4 ⎢ ⎥ ⎪ 1 −3⎦⎥ ⎣⎢0 0 c = −2 ⎪ ⎩ Solving by back substitution yields a = 2, b = −1, and c = −3 . The interpolating polynomial is z = 2x –y – 3. 49.
The form of the polynomial is: x 2 + y 2 + ax + by = c Use this polynomial and the given points to find the system of equations. (2) 2 + (6)2 + a (2) + b(6) = c (−4)2 + (−2)2 + a (−4) + b(−2) = c (3)2 + (−1) 2 + a(3) + b( −1) = c
⎧ 2a + 6b − c = −40 ⎡ 2 6 −1 −40⎤ ⎪ ⎢ −4 −2 −1 −20⎥ The system of equations and the associated augmented matrix are ⎨−4a − 2b − c = −20 ⎢ ⎥ ⎪ 3a − b − c = −10 ⎩ ⎣⎢ 3 −1 −1 −10 ⎦⎥ The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. ⎧a − 1 b + 1 c = 5 1/ 4 5⎤ ⎡ 1 1/ 2 ⎪ 2 4 ⎪ ⎢ ⎥ b − 3 c = −10 The augmented matrix in echelon form and resulting system of equations are 0 1 −3/10 −10 ⎨ ⎢ ⎥ 10 ⎪ 0 1 20 ⎦⎥ ⎣⎢ 0 c = 20 ⎪ ⎩ Solving by back substitution yields a = 2, b = −4, and c = 20 . The interpolating polynomial is x 2 + y 2 + 2 x − 4 y = 20 . Copyright © Houghton Mifflin Company. All rights reserved.
704
50.
Chapter 10: Matrices
The form of the polynomial is: x 2 + y 2 + ax + by = c Use this polynomial and the given points to find the system of equations. (2)2 + (1) 2 + a(2) + b(1) = c (0) + ( −7) 2 + a(0) + b(−7) = c (5)2 + (−2)2 + a(5) + b(−2) = c 2
⎧ 2a + b − c = −5 ⎪ The system of equations and the associated augmented matrix are ⎨ −7b − c = −49 ⎩⎪5a − 2b − c = −29
1 −1 −5⎤ ⎡2 ⎢ 0 −7 −1 −49 ⎥ ⎢ ⎥ ⎣ 5 −2 −1 −29⎦
The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. ⎧a − 2 b − 1 c = − 29 5 ⎡ 1 −2 / 5 −1/ 5 −29 / 5 ⎤ ⎪ 5 5 ⎪ The augmented matrix in echelon form and resulting system of equations are ⎢ 0 b− 1 c =7 1 1/ 7 7⎥ ⎨ 7 ⎢0 ⎪ 0 1 7 ⎥⎦ ⎣ c =7 ⎪⎩ Solving by back substitution yields a = –2, b = 6, and c = 7. The interpolating polynomial is x2 + y2 – 2x + 6y = 7.. 51.
Using a calculator, ⎡ 1 2 −1 2 ⎢ 1 −1 2 − 1 ⎢ 2 1 −1 2 ⎢ ⎢ 3 2 −1 1 ⎣⎢ 2 1 −1 −2 ⎡1 ⎢0 ⎢ ⎯⎯ → ⎢0 ⎢ ⎢0 ⎢0 ⎣
2 −1 1 −1 0 0 0
52. 3 11⎤ 2 0⎥ −1 4 ⎥ ⎥ −2 2 ⎥ 1 4 ⎥⎦
−3 2 5⎤ ⎡ 1 −2 2 ⎢0 1 3 −5 3 9⎥ ⎢ 45 69 ⎥ 24 ⎯⎯ → ⎢0 0 1 − 29 29 29 ⎥ ⎢0 0 0 1 − 32 − 52 ⎥ ⎢ ⎥ 0 1 1⎦⎥ ⎣⎢0 0 0
3 11⎤ 11 ⎥ 3⎥ 1 − 12 3 72 ⎥ ⎥ 14 0 1 11 6 3⎥ 0 0 1 2 ⎥⎦ 2 1
1 3
⎧ x − 2 x + 2 x − 3x + 2 x 2 3 4 5 ⎪ 1 x2 + 3 x3 − 5x4 + 3 x5 ⎪ ⎪ 45 x + 24 x x3 − 29 ⎨ 4 29 5 ⎪ 3 x 4 − 2 x5 ⎪ ⎪⎩ x5 The solution is (2, 1, 0, –1, 1).
⎧ x1 + 2 x2 − x3 + 2 x4 + 3x5 = 11 ⎪ x2 − x3 + x4 + 13 x5 = 11 3 ⎪ ⎪ x3 − 12 x4 + 3x5 = 72 ⎨ ⎪ x4 + 11 x = 14 ⎪ 6 5 3 x5 = 2 ⎩⎪ The solution is (1, 0, –2, 1, 2).
53.
Using a calculator, ⎡ 1 −2 2 −3 2 5⎤ ⎢ 1 −3 −1 2 −1 −4 ⎥ ⎢ 3 1 −2 1 3 9⎥ ⎢ ⎥ ⎢ 2 −1 3 −1 −2 2 ⎥ ⎣⎢ −1 2 −2 3 −1 −4 ⎥⎦
= = =
2 −10 ⎤ ⎡ 1 2 −3 −1 −10 ⎤ ⎢0 1 2 0 −1 6⎥ 4⎥ ⎢ ⎥ ⎥ 4 2 1 3 −3 2⎥ → ⎢0 0 −20 ⎯⎯ ⎥ −16 ⎥ ⎢0 0 0 1 3 −7 ⎥ ⎢0 0 0 0 −12 ⎥⎦ 0 0⎥⎦ ⎣ x4 = −3x5 − 7 x4 + 2 x5 = −10 x3 + 43 (−3 x5 − 7) − 23 x5 = 2 − x = 6
⎡ 1 2 −3 −1 2 ⎢ −1 −3 1 1 −1 ⎢ 2 3 −5 2 3 ⎢ ⎢ 3 4 −7 3 −2 ⎢⎣ 2 1 −6 4 −3
x3 = 14 x + 34 3 5 3 x2 + 2
( 143 x5 + 343 ) − x5 = 6
x2 = − 25 x − 50 3 5 3
(
) ( 143 x5 + 343 ) −( −3x5 − 7 )+ 2 x5 = −10
x1 + 2 − 25 x − 50 − 3 3 5 3
(
x1 = 77 x + 151 3 5 3
)
Let x5 be any real number c. The solution is 77c +151 , −25c −50 , 14c+34 , − 3c − 7, c . 3
3
3
Copyright © Houghton Mifflin Company. All rights reserved.
69 29 − 52
= = 1
Using a calculator,
⎧ x1 + 2 x2 − 3 x3 − ⎪⎪ x2 + 2 x3 5 ⎨ x3 + 43 x4 − 23 x5 = 2 ⎪ ⎪⎩ x4 + 3 x5 = − 7
5 9
Section 10.1
54.
705
Using a calculator, 1 5⎤ ⎡ 1 −2 2 −3 −2 2 − 3 1 5⎤ ⎢0 1 0 1 −3 3⎥ −3 4 −5 −1 13⎥ ⎢ ⎥ → ⎢0 0 1 − 12 − 52 25 1 −2 2 2 −11⎥ ⎯⎯ 4 ⎥ ⎥ −2 2 − 2 − 2 7⎥ ⎢0 0 0 1 −3 5⎥ ⎢0 0 0 −4 4 −5 −1 12 ⎥⎦ 0 0 0 ⎥⎦ ⎣ x4 = 3 x5 + 5 ⎧ x1 − 2 x2 + 2 x3 − 3 x4 + x5 = 5 ⎪ x2 + x4 − 3 x5 = 3 x3 − 1 (3 x5 + 5) − 5 x5 = 25 ⎪ 2 2 4 ⎨ 35 x3 − 1 x4 − 5 x5 = 25 = 4 + x x ⎪ 4 3 5 4 2 2 ⎪ x4 − 3 x5 = 5 ⎩ x2 + (3x5 + 5) − 3 x5 = 3 ⎡1 ⎢2 ⎢1 ⎢ ⎢3 ⎢⎣ 4
x2 = −2
(
)
x1 − 2(− 2) + 2 4 x5 + 35 − 3(3x5 + 5) + x5 = 5 4
x1 = − 3 2
Let x5 be any real number c. The solution is ⎛⎜ − 3 , − 2, 4c + 35 , 3c + 5, c ⎞⎟ . 4 ⎝ 2 ⎠
....................................................... 55.
⎡1 3 − a2 ⎢ 2 ⎢3 4 ⎢2 3 a ⎢⎣
−3R1 + R2 ⎡ 1 3 a2 ⎤ ⎢ ⎥ − 2R + R 3 → ⎢0 − 5 3 ⎥ ⎯⎯ ⎯1⎯ ⎯ ⎯ ⎢0 − 3 2 ⎥⎥ ⎦ ⎣⎢
⎡1 3 − a2 R2 + R3 ⎢ ⎯⎯ ⎯ ⎯ ⎯→ ⎢0 15 − 9a 2 −6 ⎢0 0 2 + − a a 5 6 ⎣⎢
− a2 2
3a + 2 2a 2 + a
Connecting Concepts ⎤ −3R2 ⎡ 1 3 ⎥ 5R3 ⎢ − 3a 2 + 3 ⎥ ⎯⎯ ⎯ ⎯→ ⎢0 15 ⎢0 − 15 − 2a 2 + 2 ⎥⎥ ⎦ ⎣⎢ a2
− a2 2
− 9a − 6 10a 2 + 5a
⎤ ⎥ 9a − 9 ⎥ − 10a 2 + 10 ⎥⎥ ⎦ a2 2
⎤ ⎥ 9a − 9 ⎥ − a 2 + 1 ⎥⎥ ⎦ a2 2
⎧ x + 3y − a2z = a2 ⎪ ⎪ 2 2 ⎨15 y − (9a + 6) z = 9a − 9 ⎪ 2 2 ⎪⎩ a + 5a − 6 z = − a + 1
(
)
For the system of equations to have a unique solution, a 2 + 5a − 6 cannot be zero. Thus a 2 + 5a − 6 ≠ 0, or a ≠ 1 and a ≠ −6. The system of equations has a unique solution for all values of a except 1 and −6. 56.
See the solution to exercise 55. For the system to have an infinite number of solutions, a 2 + 5a − 6 and − a 2 + 1 both must be zero. Thus a 2 + 5a − 6 = 0
and
− a2 +1 = 0
a2 = 1 a = −6 or a = 1 a = 1 or a = −1 Since both equations are zero when a = 1, the system of equations has an infinite number of solutions when a = 1. (a + 6)(a − 1) = 0
57.
See the solution to exercise 55. For the system of equations to have no solution, a 2 + 5a − 6 must be zero and − a 2 + 1 must not equal zero. Thus a 2 + 5a − 6 = 0
and
− a2 +1 ≠ 0
a2 ≠ 1 a = −6 or a = 1 a ≠ 1 or a ≠ −1 The system of equations will have no solution when a = −6. (a + 6)(a − 1) = 0
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706
Chapter 10: Matrices
.......................................................
Prepare for Section 10.2
PS1. 0
PS2. –c
PS3. 1
PS4. 1 c
PS5. 3 × 1
PS6. ⎡1 −3 4 ⎤ ⎢ 0 7 −5⎥ ⎢ ⎥ ⎣⎢ 0 −7 13 ⎦⎥
Section 10.2 1.
a. b. c. d.
2.
a. b. c. d.
3.
a.
⎡ 0 −2 ⎤ ⎡5 −1⎤ ⎡5 −3⎤ + = A+ B = ⎢ 3⎥⎦ ⎢⎣3 0 ⎥⎦ ⎢⎣5 3⎥⎦ ⎣2 ⎡ 0 −2 ⎤ ⎡5 −1⎤ ⎡ −5 −1⎤ − = A− B = ⎢ 3⎥⎦ ⎢⎣3 0 ⎥⎦ ⎢⎣ −1 3⎥⎦ ⎣2 ⎡5 −1⎤ ⎡10 −2 ⎤ = 2B = 2 ⎢ 0 ⎥⎦ ⎣3 0 ⎥⎦ ⎢⎣ 6 ⎡ 0 −2 ⎤ ⎡5 −1⎤ ⎡ 0 −4 ⎤ ⎡15 −3 ⎤ ⎡ −15 −1⎤ −3 = − = 2 A − 3B = 2 ⎢ 3⎥⎦ ⎢⎣3 0 ⎥⎦ ⎢⎣ 4 6 ⎥⎦ ⎢⎣ 9 0 ⎥⎦ ⎢⎣ − 5 6⎥⎦ ⎣2 ⎡0 A+ B = ⎢ ⎣1 ⎡0 A− B = ⎢ ⎣1 ⎡ −3 2B = 2 ⎢ ⎣ 2
−1 0
d.
⎡0 2 A − 3B = 2 ⎢ ⎣1
a.
⎡ 2 −2 A+ B = ⎢ ⎣ 0 −3
4 ⎤ ⎡ 1 −5 6 ⎤ ⎡ 3 − 7 + = −4 ⎥⎦ ⎢⎣ 4 −2 −3⎥⎦ ⎢⎣ 4 −5
c.
b. c. d.
−1 0
3 ⎤ ⎡ −3 1 2 ⎤ ⎡ −3 0 5⎤ + = −2 ⎥⎦ ⎢⎣ 2 5 −3 ⎥⎦ ⎢⎣ 3 5 −5⎥⎦ 3⎤ ⎡ −3 1 2 ⎤ ⎡ 3 −2 − = −2 ⎥⎦ ⎢⎣ 2 5 −3 ⎥⎦ ⎢⎣ −1 −5 2 ⎤ ⎡ −6 2 4⎤ = −3 ⎥⎦ ⎢⎣ 4 10 −6 ⎥⎦ 2⎤ ⎡0 −1 3 ⎤ ⎡ −3 1 = −3 ⎢ 0 −2 ⎥⎦ ⎣ 2 5 −3 ⎥⎦ ⎢⎣ 2
b.
4.
−1⎤ ⎡ −1 3⎤ ⎡1 2⎤ + = 3⎥⎦ ⎢⎣ 2 1⎥⎦ ⎢⎣5 4⎥⎦ −1⎤ ⎡ −1 3⎤ ⎡3 −4 ⎤ − = 3⎥⎦ ⎢⎣ 2 1⎥⎦ ⎢⎣1 2⎥⎦ 3⎤ ⎡ −2 6 ⎤ = 1⎥⎦ ⎢⎣ 4 2 ⎥⎦ ⎡ 2 −1⎤ ⎡ −1 3⎤ ⎡ 4 −2 ⎤ ⎡ −3 9⎤ ⎡ 7 −11⎤ −3 = − = 2 A − 3B = 2 ⎢ 6 ⎥⎦ ⎢⎣ 6 3⎥⎦ ⎢⎣ 0 3 ⎥⎦ ⎣ 3 3⎥⎦ ⎢⎣ 2 1⎥⎦ ⎢⎣ 6
⎡2 A+ B = ⎢ ⎣3 ⎡2 A− B = ⎢ ⎣3 ⎡ −1 2B = 2 ⎢ ⎣ 2
1 5
4 ⎤ ⎡1 −5 6 ⎤ ⎡ 1 ⎡2 − 2 A− B = ⎢ ⎥−⎢ ⎥=⎢ ⎣ 0 − 3 −4 ⎦ ⎣ 4 −2 −3 ⎦ ⎣ −4 ⎡ 1 −5 6 ⎤ ⎡ 2 −10 12 ⎤ = 2B = 2− ⎢ ⎣ 4 −2 −3⎥⎦ ⎢⎣ 8 − 4 − 6 ⎥⎦ 4 ⎤ ⎡ 1 −5 6 ⎤ ⎡ 4 ⎡2 − 2 −3 = 2 A − 3B = 2 ⎢ ⎣ 0 − 3 −4 ⎥⎦ ⎢⎣ 4 −2 −3⎥⎦ ⎢⎣ 0
1⎤ 1⎥⎦
−2 6 ⎤ ⎡ −9 3 6 ⎤ ⎡ 9 − = 0 −4 ⎥⎦ ⎢⎣ 6 15 −9 ⎥⎦ ⎢⎣ −4
− 5 0⎤ −15 5⎥⎦
10 ⎤ − 7 ⎥⎦ 3 −2 ⎤ −1 −1⎥⎦
−4 −6
8⎤ ⎡ 3 − − 8⎥⎦ ⎢⎣12
−15 −6
18⎤ ⎡ 1 11 −10 ⎤ = 1 ⎥⎦ − 9 ⎥⎦ ⎢⎣ −12 0
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Section 10.2
5.
a.
b.
6.
4 ⎤ ⎡4 − 3⎥ + ⎢ 1 0 ⎥⎦ ⎢⎣ 3
1 ⎤ ⎡1 − 2⎥ = ⎢ 3 − 4 ⎥⎦ ⎢⎣ 2
4 ⎤ ⎡4 − 3⎥ − ⎢ 1 0 ⎥⎦ ⎢⎣ 3
1 ⎤ ⎡ −7 − 2⎥ = ⎢ 1 − 4 ⎥⎦ ⎢⎣ −4
⎡4 2 B = 2 ⎢1 ⎢3 ⎣
d.
⎡ −3 2 A − 3B = 2 ⎢ 2 ⎢ −1 ⎣
a.
c.
d.
a.
b.
c.
d.
8.
⎡ −3 A+ B = ⎢ 2 ⎢ −1 ⎣ ⎡ −3 A− B = ⎢ 2 ⎢ −1 ⎣
c.
b.
7.
707
a.
b.
c.
d.
1 ⎤ ⎡8 − 2⎥ = ⎢ 2 − 4 ⎥⎦ ⎢⎣ 6
5⎤ −5⎥ − 4 ⎥⎦ 3⎤ −1 ⎥ 4 ⎥⎦
2⎤ −4⎥ −8 ⎥⎦
4 ⎤ ⎡4 − 3⎥ − 3 ⎢ 1 0 ⎥⎦ ⎢⎣ 3
1 ⎤ ⎡ −6 − 2⎥ = ⎢ 4 − 4 ⎥⎦ ⎢⎣ −2
8⎤ ⎡12 −6⎥ − ⎢ 3 0 ⎥⎦ ⎢⎣ 9
8 ⎤ ⎡ 1 6⎤ ⎡ 2 − 2 ⎤ ⎡ −1 A+ B = ⎢3 4 ⎥ + ⎢ 2 − 2⎥ = ⎢ 5 2⎥ ⎢1 0 ⎥⎦ ⎢⎣ −4 3 ⎥⎦ ⎢⎣ −3 3⎥⎦ ⎣ 8 ⎤ ⎡3 −10 ⎤ ⎡ 2 − 2 ⎤ ⎡ −1 A− B = ⎢3 4 ⎥ − ⎢ 2 − 2 ⎥ = ⎢1 6⎥ ⎢1 ⎥ ⎢ −4 ⎥ ⎢5 − 0 3 3⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 8 ⎤ ⎡ −2 16 ⎤ ⎡ −1 2 B = 2 ⎢ 2 − 2 ⎥ = ⎢ 4 −4 ⎥ ⎢ −4 3 ⎥⎦ ⎢⎣ −8 6 ⎥⎦ ⎣ 8 ⎤ ⎡ 4 −4 ⎤ ⎡ − 3 ⎡ 2 − 2 ⎤ ⎡ −1 2 A − 3B = 2 ⎢ 3 4 ⎥ − 3 ⎢ 2 − 2⎥ = ⎢6 8⎥ − ⎢ 6 ⎢1 0 ⎥⎦ ⎢⎣ −4 3 ⎥⎦ ⎢⎣ 2 0⎥⎦ ⎢⎣ −12 ⎣ 0 ⎤ ⎡ −1 1 ⎡ −2 3 −1⎤ ⎡1 −2 A + B = ⎢ 0 −1 2 ⎥ + ⎢ 2 3 −1⎥ = ⎢ 2 2 ⎢ −4 3 3 ⎥ ⎢ 3 −1 2⎥ ⎢ −1 2 ⎣ ⎦ ⎣ ⎦ ⎣ 0 ⎤ ⎡ −3 5 ⎡ −2 3 −1⎤ ⎡ 1 −2 A − B = ⎢ 0 −1 2 ⎥ − ⎢ 2 3 −1⎥ = ⎢ −2 −4 ⎢ −4 3 3 ⎥ ⎢ 3 −1 2 ⎥ ⎢ −7 4 ⎣ ⎦ ⎣ ⎦ ⎣ 0 ⎤ ⎡ 2 −4 0⎤ ⎡1 −2 2 B = 2 ⎢ 2 3 −1⎥ = ⎢ 4 6 −2 ⎥ ⎢ 3 −1 2 ⎥ ⎢ 6 −2 4 ⎥⎦ ⎣ ⎦ ⎣ 0 ⎤ ⎡ −4 ⎡ −2 3 −1⎤ ⎡ 1 −2 2 A − 3B = 2 ⎢ 0 −1 2 ⎥ − 3 ⎢ 2 3 −1⎥ = ⎢ 0 ⎢ −4 3 3 ⎥ ⎢ 3 −1 2 ⎥ ⎢ −8 ⎣ ⎦ ⎣ ⎦ ⎣ ⎡0 2 A + B = ⎢1 −3 ⎢5 4 ⎣ ⎡0 2 A − B = ⎢1 −3 ⎢5 4 ⎣
0 ⎤ ⎡ −1 3⎥ + ⎢ 3 −2 ⎥⎦ ⎢⎣ −4 0 ⎤ ⎡ −1 3⎥ − ⎢ 3 −2 ⎥⎦ ⎢⎣ −4
2 3 4 2 3 4
3 ⎤ ⎡ −18 5 ⎤ − 6 ⎥ = ⎢ 1 0⎥ −12 ⎥⎦ ⎢⎣ −11 12 ⎥⎦
24 ⎤ ⎡ 7 −6 ⎥ = ⎢ 0 9 ⎥⎦ ⎢⎣14
−28 ⎤ 14 ⎥ − 9 ⎥⎦
−1⎤ 1⎥ 5⎥⎦ −1⎤ 3⎥ 1⎥⎦
6 −2 ⎤ ⎡ 3 −6 0⎤ ⎡ − 7 12 −2 ⎤ −2 4⎥ − ⎢6 9 −3⎥ = ⎢ − 6 −11 7 ⎥ 6 6 ⎥⎦ ⎢⎣9 −3 6⎥⎦ ⎢⎣ −17 9 0 ⎥⎦
4 ⎤ ⎡ −1 4 4⎤ −2 ⎥ = ⎢ 4 0 1 ⎥ 3⎥⎦ ⎢⎣ 1 8 1 ⎥⎦ 4 ⎤ ⎡ 1 0 −4⎤ −2 ⎥ = ⎢ −2 −6 5⎥ 3⎥⎦ ⎢⎣ 9 0 −5⎥⎦
4 ⎤ ⎡ −2 4 8 ⎤ ⎡ −1 2 2 B = 2 ⎢ 3 3 −2 ⎥ = ⎢ 6 6 −4 ⎥ ⎢ −4 4 3⎥⎦ ⎢⎣ −8 8 6 ⎥⎦ ⎣ 0 ⎤ ⎡ −1 2 4⎤ ⎡ 0 4 0⎤ ⎡ − 3 6 ⎡0 2 2 A − 3B = 2 ⎢1 −3 3 ⎥ − 3 ⎢ 3 3 −2 ⎥ = ⎢ 2 − 6 6⎥ − ⎢ 9 9 ⎢5 ⎥ ⎢ −4 4 ⎥ ⎢10 ⎥ ⎢ −12 12 − − 4 2 3 8 4 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣
12 ⎤ ⎡ 3 −6 ⎥ = ⎢ − 7 9 ⎥⎦ ⎢⎣ 22
−2 −15 −4
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−12 ⎤ 12⎥ −13 ⎥⎦
708
9.
10.
11.
12.
13.
14.
15.
Chapter 10: Matrices
4 ⎤ ⎡ (2)(−2) + (−3)(2) ⎡ 2 −3⎤ ⎡ −2 AB = ⎢ = ⎣ 1 4 ⎥⎦ ⎢⎣ 2 −3⎥⎦ ⎢⎣ (1)( − 2) + (4)(2) ⎡ −2 4 ⎤ ⎡ 2 −3⎤ ⎡ (−2)(2) + (4)(1) = BA = ⎢ ⎣ 2 −3⎥⎦ ⎢⎣ 1 4 ⎥⎦ ⎢⎣ (2)(2) + ( − 3)(1)
(−2)( − 3) + (4)(4) ⎤ ⎡0 = (2)( −3) + (−3)(4) ⎥⎦ ⎢⎣ 1
⎡ 3 −2 ⎤ ⎡ −1 −1⎤ ⎡ (3)(−1) + (−2)(0) = AB = ⎢ 1⎥⎦ ⎢⎣ 0 4 ⎥⎦ ⎢⎣ (4)( −1) + (1)(0) ⎣4 ⎡ −1 −1⎤ ⎡ 3 −2 ⎤ ⎡(−1)(3) + (−1)(4) = BA = ⎢ 4 ⎥⎦ ⎢⎣ 4 1⎥⎦ ⎢⎣ (0)(3) + (4)(4) ⎣ 0
(3)( −1) + ( − 2)(4) ⎤ ⎡ −3 −11⎤ = (4)(−1) + (1)(4) ⎥⎦ ⎢⎣ −4 0⎥⎦ (−1)( − 2) + ( −1)(1) ⎤ ⎡ −7 1⎤ = (0)(−2) + (4)(1) ⎥⎦ ⎢⎣ 16 4 ⎥⎦
1⎤ ⎡ (3)(4) + (−1)(2) ⎡ 3 −1⎤ ⎡ 4 = AB = ⎢ ⎣ 2 3⎥⎦ ⎢⎣ 2 −3⎥⎦ ⎢⎣ (2)(4) + (3)(2) 1⎤ ⎡ 3 −1⎤ ⎡ (4)(3) + (1)(2) ⎡4 = BA = ⎢ ⎣ 2 −3⎥⎦ ⎢⎣ 2 −3⎥⎦ ⎢⎣ (2)(3) + ( − 3)(2) ⎡ −3 2 ⎤ ⎡ 0 AB = ⎢ ⎣ 2 −2 ⎥⎦ ⎢⎣ −2 ⎡ 0 2 ⎤ ⎡ −3 BA = ⎢ ⎣ −2 4 ⎥⎦ ⎢⎣ 2
(2)(4) + ( − 3)( − 3) ⎤ ⎡ −10 = (1)(4) + (4)(−3) ⎥⎦ ⎢⎣ 6
17 ⎤ − 8⎥⎦ 22 ⎤ −18⎥⎦
(3)(1) + ( −1)( − 3) ⎤ ⎡10 6⎤ = (2)(1) + (3)( −3) ⎥⎦ ⎢⎣14 −7 ⎥⎦ (4)( −1) + (1)(3) ⎤ ⎡14 = (2)(−1) + (−3)(3) ⎥⎦ ⎢⎣ 0
2 ⎤ ⎡ (−3)(0) + (2)(−2) = 4 ⎥⎦ ⎢⎣ (2)(0) + ( − 2)( − 2) 2 ⎤ ⎡ (0)(−3) + (2)(2) = −2 ⎥⎦ ⎢⎣ ( −2)(−3) + (4)(2)
−1⎤ −11⎥⎦
(−3)(2) + (2)(4) ⎤ ⎡ −4 2⎤ = (2)(2) + (−2)(4) ⎥⎦ ⎢⎣ 4 −4⎥⎦ (0)(2) + (2)( − 2) ⎤ ⎡ 4 − 4 ⎤ = ( −2)(2) + (4)(−2) ⎥⎦ ⎢⎣ 14 −12 ⎥⎦
(2)( − 2) + ( −1)(0) (2)(3) + (−1)(1) ⎤ ⎡ 0 −4 5⎤ −1⎤ ⎡ (2)(1) + (−1)(2) ⎡ 1 −2 3⎤ ⎢ = (0)(1) + (3)(2) 3⎥ ⎢ (0)(−2) + (3)(0) (0)(3) + (3)(1) ⎥ = ⎢ 6 0 3⎥ ⎥ 2 0 1⎦ ⎢ ⎥ ⎢ −3 −2 1⎥ −2 ⎥⎦ ⎣ + − − + − + − (1)(1) ( 2)(2) (1)( 2) ( 2)(0) (1)(3) ( 2)(1) ⎦ ⎣ ⎦ ⎣ ⎡ 2 −1⎤ (1)( −1) + ( − 2)(3) + ( − 2) ⎤ ⎡5 −13⎤ ⎡ 1 −2 3⎤ ⎢ ⎡ (1)(2) + (−2)(0) + (3)(1) = BA = ⎢ 0 3⎥ = ⎣ 2 0 1⎥⎦ ⎢ 1 −2 ⎥ ⎢⎣ (2)(2) + (0)(0) + (1)(1) (2)(−1) + (0)(3) + (1)(−2) ⎥⎦ ⎢⎣5 − 4 ⎥⎦ ⎣ ⎦ ⎡2 AB = ⎢ 0 ⎢ 1 ⎣
(−1)( −1) + (3)(2) (−1)(2) + (3)(−4) ⎤ ⎡ 3 7 −14 ⎤ ⎡ −1 3⎤ ⎡ (−1)(0) + (3)(1) ⎡0 −1 2 ⎤ ⎢ = AB = ⎢ 2 0 0⎥ 1⎥ ⎢ (2)(0) + (1)(1) (2)(−1) + (1)(2) (2)(2) + (1)(−4) ⎥ = ⎢ 1 ⎥ ⎢ −3 −2 ⎥ ⎣ 1 2 −4 ⎦ ⎢ ( − 3)(0) + ( − 2)(1) (−3)(−1) + (−2)(2) (−3)(2) + (−2)(−4) ⎥ ⎢ −2 −1 ⎥ 2 ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎡ −1 3⎤ (0)(3) + ( −1)(1) + (2)( − 2) ⎤ ⎡ −8 − 5⎤ ⎡0 −1 2 ⎤ ⎢ ⎡ (0)(−1) + (−1)(2) + (2)(−3) =⎢ BA = ⎢ 2 1⎥ = ⎢ ⎥ ⎥ − − + + − − + + − − 1 2 4 (1)( 1) (2)(2) ( 4)( 3) (1)(3) (2)(1) ( 4)( 2) 15 13⎥⎦ ⎢ ⎥ ⎣ ⎦ −3 −2 ⎣ ⎦ ⎣ ⎣ ⎦ −1 3⎤ ⎡ 2 0 0⎤ 2 −1⎥ ⎢ 1 −1 0⎥ 0 2 ⎥⎦ ⎢⎣ 2 −1 −2 ⎥⎦ (2)(0) + (−1)(0)(3)(−2)⎤ ⎡9 − 2 − 6⎤ ⎡ (2)(2) + (−1)(1) + (3)(2) (2)(0) + (−1)(−1) + (3)(−1) ⎢ ⎥ ⎢ ⎥ = ⎢ (0)(2) + (2)(1) + (−1)(2) (0)(0) + (2)(−1) + (−1)(−1) (0)(0) + (2)(0) + (−1)(−2) ⎥ = ⎢0 − 1 2⎥ ⎢⎣ (0)(2) + (0)(1) + (2)(2) (0)(0) + (0)(−1) + (2)(−1) (0)(0) + (0)(0) + (2)(−2)⎥⎦ ⎢⎣4 − 2 − 4⎥⎦ ⎡ 2 0 0 ⎤ ⎡ 2 −1 3⎤ BA = ⎢ 1 −1 0 ⎥ ⎢ 0 2 −1⎥ ⎢ 2 −1 −2 ⎥ ⎢ 0 0 2 ⎥⎦ ⎣ ⎦ ⎣ (2)(−1) + (0)(2) + (0)(0) (2)(3) + (0)(−1) + (0)( 2) ⎤ ⎡4 − 2 6⎤ ⎡ (2)(2) + (0)(0) + (0)(0) ⎥ ⎢ ⎥ ⎢ = ⎢ (1)(2) + (−1)(0) + (0)(0) (1)(−1) + (−1)(2)(−1) + (0)(0) (1)(3) + (−1)(−1) + (0)(2) ⎥ = ⎢2 − 3 4⎥ ⎢⎣(2)(2) + (−1)(0) + (−2)(0) (2)(−1) + (−1)(2) + ( −2)(0) (2)(3) + (−1)(−1) + (−2)(2)⎥⎦ ⎢⎣4 − 4 3⎥⎦ ⎡2 AB = ⎢ 0 ⎢0 ⎣
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Section 10.2
16.
709
⎡ −1 2 0 ⎤ ⎡ 2 −1 0⎤ AB = ⎢ 2 −1 1⎥ ⎢ 1 5 −1⎥ ⎢ −2 2 −1⎥⎦ ⎢⎣ 0 −1 3⎥⎦ ⎣ (−1)(0) + (2)(−1) + (0)(3)⎤ ⎡ 0 11 − 2⎤ ⎡ (−1)(2) + (2)(1) + (0)(0) (−1)(−1) + (2)(5) + (0)(−1) ⎢ ⎥ ⎢ ⎥ = ⎢ (2)(2) + (−1)(1) + (1)(0) (2)(−1) + (−1)(5) + (1)(−1) (2)(0) + (−1)(−1) + (1)(3)⎥ = ⎢ 3 − 8 4⎥ ⎢⎣ (−2)(2) + (2)(1) + (−1)(0) (−2)(−1) + (2)(5) + (−1)(−1) (−2)(0) + (2)(−1) + (−1)(3)⎥⎦ ⎢⎣− 2 13 − 5⎥⎦ ⎡ 2 −1 0 ⎤ ⎡ −1 2 0 ⎤ BA = ⎢ 1 5 −1⎥ ⎢ 2 −1 1⎥ ⎢ 0 −1 3⎥ ⎢ −2 2 −1⎥⎦ ⎣ ⎦ ⎣ 5 − 1⎤ ⎡ (2)(−1) + (−1)(2) + (0)(−2) ( 2)(2) + ( −1)(−1) + (0)(2) ( 2)(0) + (−1)(1) + (0)(−1)⎤ ⎡ − 4 ⎥ ⎢ ⎥ ⎢ 6⎥ = ⎢ (1)(−1) + (5)(2) + (−1)(−2) (1)(2) + (5)(−1) + ( −1)(2) (1)(0) + (5)(1) + ( −1)( −1) ⎥ = ⎢ 11 − 5 ⎢⎣ (0)(−1) + ( −1)(2) + (3)(−2) 7 − 4⎥⎦ (0)(2) + (−1)(−1) + (3)(2) (0)(0) + (−1)(1) + (3)(−1)⎥⎦ ⎢⎣ − 8
17.
⎡ 1 0⎤ AB = [1 −2 3] ⎢ 2 −1⎥ = [ (1)(1) + (−2)(2) + (3)(1) ⎢ 1 2⎥ ⎣ ⎦
18.
⎡ −2 AB = ⎢ 1 ⎢ ⎣ 0
19.
The number of columns of the first matrix is not equal to the number of rows of the second matrix. The product is not possible.
20.
The number of columns of the first matrix is not equal to the number of rows of the second matrix. The product is not possible.
21.
3⎤ ⎡ 3 6 ⎤ ⎡ (2)(3) + (3)( −2) (2)(6) + (3)(−4) ⎤ ⎡ 0 0 ⎤ ⎡ 2 AB = ⎢ = = ⎣ −4 −6 ⎥⎦ ⎢⎣ −2 −4 ⎥⎦ ⎢⎣( −4)(3) + ( −6)( −2) ( −4)(6) + (−6)(−4) ⎥⎦ ⎢⎣ 0 0 ⎥⎦
22.
23.
24.
(1)(0) + ( − 2)( −1) + (3)(2)] = [ 0 8]
3⎤ ⎡( −2)(3)) + (3)(−2) ⎤ ⎡ −12 ⎤ ⎡ 3⎤ −2 ⎥ ⎢ ⎥ = ⎢ (1)(3) + (−2)(−2) ⎥ = ⎢ 7 ⎥ −2 2 ⎦⎥ ⎣ ⎦ ⎣⎢ (0)(3) + (2)(−2) ⎦⎥ ⎣⎢ −4 ⎦⎥
⎡1 3 ⎡ 2 −1 3⎤ ⎢ AB = ⎢ 2 −1 ⎣ −1 2 1⎥⎦ ⎢ 3 1 ⎣ ⎡(2)(1) + (−1)( 2) + (3)(3) =⎢ ⎣ (−1)(1) + (2)(2) + (1)(3)
2⎤ 0⎥ 2 ⎥⎦ ( 2)(3) + (−1)( −1) + (3)(1) (2)(2) + (−1)(0) + (3)(2)⎤ ⎡9 10 10⎤ ⎥=⎢ ⎥ (−1)(3) + (2)(−1) + (1)(1) (−1)( 2) + (2)(0) + (1)(2)⎦ ⎣6 − 4 0⎦
The number of columns of the first matrix is not equal to the number of rows of the second matrix. The product is not possible. 4⎤ ⎡ 2 1 −3 0 ⎤ ⎡ 2 −2 AB = ⎢ 1 0 −1⎥ ⎢ 0 −2 1 −2 ⎥ ⎢2 1 3⎥⎦ ⎢⎣ 1 −1 0 2 ⎥⎦ ⎣ 2 − 8 12⎤ ⎡ 2(2) + (−2)(0) + 4(1) 2(1) + (−2)(−2) + 4(1) 2(−3) + (−2)(1) + 4(0) 2(0) + (−2)(−2) + 4(2)⎤ ⎡ 8 ⎥ ⎥ ⎢ ⎢ = ⎢ 1(2) + 0(0) + (−1)(1) 1(1) + 0(−2) + ( −1)(−1) 1( −3) + 0(1) + (−1)(0) 1(0) + 0(−2) + (−1)(2)⎥ = ⎢ 1 2 − 3 − 2⎥ ⎢⎣ 2(2) + 1(0) + 3(1) 2(1) + 1(−2) + 3(−1) 2(−3) + 1(1) + 3(0) 2(0) + 1(−2) + 3(2)⎥⎦ ⎢⎣7 − 3 − 5 4⎥⎦
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710
25.
Chapter 10: Matrices
3X + A = B 3⎤ ⎡ 0 −1⎥ = ⎢ 1 ⎥ ⎢ 1⎦⎥ ⎣⎢ 4 ⎡0 3X = ⎢ 1 ⎢ ⎢⎣ 4
⎡ −1 3X + ⎢ 2 ⎢ ⎣⎢ 3
⎡ 1 3 X = ⎢ −1 ⎢ ⎢⎣ 1 ⎡ 1 ⎢ 3 X = ⎢ − 13 ⎢ ⎢ 1 ⎣ 3
27.
29.
30.
31.
2X ⎡ −1 2X −⎢ 2 ⎢ 3 ⎣ ⎡ −1 X −⎢ 2 ⎢ 3 ⎣
26. ⎡ −1 2⎢ 2 ⎢ 3 ⎣ ⎡ −2 ⎢ 4 ⎢ 6 ⎣
− 2⎤ 3⎥ ⎥ − 3⎦⎥ −2 ⎤ 3⎥ − ⎥ −3⎥⎦
⎡ −1 3 ⎤ ⎢ 2 −1⎥ ⎢ ⎥ ⎢⎣ 3 1⎥⎦
−5 ⎤ 4⎥ ⎥ −4 ⎥⎦ − 53 ⎤ ⎥ 4⎥ 3⎥ − 34 ⎥ ⎦
−A= X +B 3⎤ ⎡0 − 2⎤ −1⎥ = X + ⎢ 1 3⎥ ⎢ 4 − 3⎥ 1⎥⎦ ⎣ ⎦ 3⎤ ⎡ 0 − 2 ⎤ −1⎥ = ⎢ 1 3⎥ 1⎥⎦ ⎢⎣ 4 − 3⎥⎦ ⎡ 0 − 2 ⎤ ⎡ −1 X = ⎢1 3⎥ + ⎢ 2 ⎢ 4 − 3⎥ ⎢ 3 ⎣ ⎦ ⎣ 1⎤ ⎡ −1 2⎥ X= ⎢ 3 ⎢ 7 −2 ⎥ ⎣ ⎦
28.
3⎤ −1⎥ 1⎥⎦
2 A − 3 X = 5B 3⎤ ⎡0 −1⎥ − 3 X = 5 ⎢ 1 ⎢4 1⎥⎦ ⎣ 6⎤ ⎡ 0 − 2⎥ − 3 X = ⎢ 5 ⎢ 20 2 ⎥⎦ ⎣ ⎡ 0 −3 X = ⎢ 5 ⎢ 20 ⎣ ⎡ 2 −3 X = ⎢ 1 ⎢ 14 ⎣ ⎡ − 23 X = ⎢⎢ − 1 3 ⎢⎣ − 14 3
3X ⎡0 3X + 2 ⎢ 1 ⎢4 ⎣ ⎡0 3 X + ⎢2 ⎢8 ⎣ ⎡0 2 X + ⎢2 ⎢8 ⎣
− 2⎤ 3⎥ − 3⎥⎦ −10 ⎤ 15⎥ −15⎥⎦ −10 ⎤ ⎡ −2 15⎥ − ⎢ 4 −15 ⎥⎦ ⎢⎣ 6 −16 ⎤ 17 ⎥ −17 ⎥⎦ 16 ⎤ 3 −17 3 ⎥⎥ 17 ⎥ 3⎦
+ 2B = X − 2 A −2 ⎤ 3⎤ ⎡ −1 3⎥ = X − 2 ⎢ 2 −1⎥ ⎢ 3 −3⎥⎦ 1⎥⎦ ⎣ −4 ⎤ 6⎤ ⎡ −2 6 ⎥ = X − ⎢ 4 −2 ⎥ ⎢ 6 −6 ⎥⎦ 2 ⎥⎦ ⎣ −4 ⎤ ⎡ − 2 6⎤ 6 ⎥ = − ⎢ 4 −2 ⎥ −6 ⎥⎦ ⎢⎣ 6 2 ⎥⎦ ⎡ 0 −4 ⎤ ⎡ − 2 2 X = − ⎢2 6⎥ − ⎢ 4 ⎢ 8 −6 ⎥ ⎢ 6 ⎣ ⎦ ⎣ ⎡ 1 −1⎤ X = ⎢ −3 − 2 ⎥ ⎢ −7 2 ⎥⎦ ⎣
⎡ 2 −3⎤ ⎡ 2 −3⎤ ⎡ 2(2) + (−3)(1) 2(−3) + (−3)(−1) ⎤ ⎡1 −3⎤ = = A 2 = A⋅ A = ⎢ ⎣ 1 −1⎥⎦ ⎢⎣ 1 −1⎥⎦ ⎢⎣ 1(2) + (−1)(1) 1( −3) + ( −1)( −1) ⎥⎦ ⎢⎣1 −2 ⎥⎦ ⎡ 2 −3⎤ ⎡ 2 −3⎤ ⎡ 2 −3⎤ ⎡ 2(2) + (−3)(1) 2(−3) + (−3)(−1) ⎤ ⎡ 2 −3⎤ = A3 = A⋅ A⋅ A = ⎢ ⎣ 1 −1⎥⎦ ⎢⎣ 1 −1⎥⎦ ⎢⎣ 1 −1⎥⎦ ⎢⎣ 1(2) + (−1)(1) 1(−3) + (−1)( −1) ⎥⎦ ⎢⎣ 1 −1⎥⎦ ⎡1 −3⎤ ⎡ 2 −3⎤ ⎡1(2) + (−3)(1) 1(−3) + (−3)(−1) ⎤ ⎡ −1 0 ⎤ = = =⎢ ⎣1 −2 ⎥⎦ ⎢⎣ 1 −1⎥⎦ ⎢⎣1(2) + (−2)(1) 1(−3) + ( −2)( −1) ⎥⎦ ⎢⎣ 0 −1⎥⎦
⎡ 3 −1 0 ⎤ ⎡ 3 −1 0 ⎤ B 2 = B ⋅ B = ⎢ 2 −2 −1⎥ ⎢ 2 −2 −1⎥ ⎢1 0 2 ⎥⎦ ⎢⎣ 1 0 2 ⎥⎦ ⎣ 3(−1) + (−1)(−2) + 0(0) 3(0) + (−1)(−1) + 0(2) ⎤ ⎡7 −1 1⎤ ⎡ 3(3) + ( −1)(2) + 0(1) = ⎢ 2(3) + (−2)(2) + (−1)(1) 2(−1) + (−2)( − 2) + ( −1)(0) 2(0) + (−2)(−1) + (−1)(2) ⎥ = ⎢ 1 2 0⎥ ⎢ 1(3) + 0(2) + 2(1) 1(−1) + 0(−2) + 2(0) 1(0) + 0(−1) + 2(2) ⎥⎦ ⎢⎣ 5 −1 4⎥⎦ ⎣
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6⎤ − 2⎥ 2 ⎥⎦
6 ⎤ ⎡ 2 −2 ⎤ − 2 ⎥ = ⎢ − 6 −4 ⎥ 4 ⎥⎦ 2 ⎥⎦ ⎢⎣ −14
Section 10.2
32.
711
⎡ 3 −1 0⎤ ⎡ 3 −1 0⎤ ⎡ 3 −1 0⎤ B 3 = B ⋅ B ⋅ B = ⎢ 2 −2 −1⎥ ⎢ 2 −2 −1⎥ ⎢ 2 −2 −1⎥ ⎢ 1 0 2⎥ ⎢ 1 0 2⎥⎦ ⎢⎣ 1 0 2⎥⎦ ⎣ ⎦ ⎣ 3(−1) + (−1)(−2) + 0(0) 3(0) + (−1)(−1) + 0( 2) ⎤ ⎡ 3 − 1 0⎤ ⎡ 3(3) + (−1)(2) + 0(1) ⎢ ⎥⎢ ⎥ = ⎢2(3) + (−2)(2) + (−1)(1) 2(−1) + (−2)(−2) + (−1)(0) 2(0) + (−2)(−1) + ( −1)(2)⎥ ⎢2 − 2 − 1⎥ ⎢⎣ 1(3) + 0(2) + 2(1) 1(−1) + 0(−2) + 2(0) 1(0) + 0(−1) + 2(2) ⎥⎦ ⎢⎣ 1 0 2⎥⎦ 3⎤ ⎡7 −1 1⎤ ⎡ 3 −1 0 ⎤ ⎡ 7(3) + (−1(2) +1(1) 7(−1) + (−1)(−2) +1(0) 7(0) + (−1)(−1) +1(2) ⎤ ⎡ 20 −5 = ⎢ 1 2 0 ⎥ ⎢ 2 −2 −1⎥ = ⎢ 1(3) + 2(2) + 0(1) 1(−1) +−2( − 2) + (0)(0) 1(0) + 2(−1) + 0(2) ⎥ = ⎢ 7 −5 −2⎥ ⎢ 5 −1 4 ⎥ ⎢ 1 0 2 ⎥⎦ ⎢⎣5(3) + (−1)(2) + 4(1) 5( −1) + (−1)(−2) + 4(0) 5(0) + (−1)(−1) + 4(2) ⎥⎦ ⎢⎣ 17 −3 9⎥⎦ ⎣ ⎦ ⎣
33.
⎡ 3 −8⎤ ⎡ x ⎤ ⎡11⎤ ⎢ ⎥⎢ ⎥=⎢ ⎥ 3⎦ ⎣ y ⎦ ⎣ 1⎦ ⎣4 ⎧ 3 x − 8 y = 11 ⎨ ⎩4 x + 3 y = 1
34.
7 ⎤ ⎡ x ⎤ ⎡ 1⎤ ⎡2 ⎢ ⎥⎢ ⎥=⎢ ⎥ 3 4⎦ ⎣ y ⎦ ⎣16⎦ − ⎣ ⎧2 x + 7 y = 1 ⎨ ⎩ 3 x − 4 y = 16
35.
⎡ 1 −3 −2 ⎤ ⎡ x ⎤ ⎢3 1 0⎥ ⎢ y ⎥ = ⎢ 2 −4 5⎥⎦ ⎢⎣ z ⎥⎦ ⎣ ⎧ x − 3 y − 2z = 6 ⎪ =2 ⎨3x + y ⎪⎩ 2x − 4 y + 5 z =1
36.
0 5⎤ ⎡ x ⎤ ⎡ 9 ⎤ ⎡2 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ 3 − 5 1⎥ ⎢ y ⎥ = ⎢ 7 ⎥ ⎢⎣ 4 − 7 6⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣14⎥⎦ 5z = 9 ⎧ 2x + ⎪ − + x y z = 7 3 5 ⎨ ⎪ 4x − 7 y + 6 z = 14 ⎩
37.
−1 0 2 ⎤ ⎡ x1 ⎤ ⎡ 5⎤ 1 2 −3⎥ ⎢ x2 ⎥ ⎢ 6 ⎥ = 0 1 −2 ⎥ ⎢⎢ x3 ⎥⎥ ⎢ 10 ⎥ ⎥ ⎢ ⎥ 2 −1 −4 ⎦ ⎣⎢ x4 ⎥⎦ ⎣ 8⎦ + 2 x4 = 5 ⎧ 2 x1 − x2 ⎪ ⎪ 4 x1 + x2 + 2 x3 − 3 x4 = 6 ⎨ + x3 − 2 x4 = 10 ⎪6 x1 ⎪ 5 x1 + 2 x2 − x3 − 4 x4 = 8 ⎩
38.
⎡ 5 −1 ⎢4 0 ⎢ 2 −2 ⎢ 1 ⎣3
39.
a. b. c.
3 × 4. There are three different fish in four different samples. Fish A was caught in sample number 4. Fish B. There are more 1’s in this row than in any other row.
40.
a. b. c. d.
4 × 4. There are four animals in this system. Rabbits do not prey on hawks. The coyote is not preyed on by any other animal in this system. The rabbit does not prey on any animal in this system.
41.
⎡ 2. 0 ⎢ 0.98⎢ 0.8 ⎢⎣3.6
1.4
3.0
1.1
2.0
1. 2
4. 5
⎡2 ⎢4 ⎢6 ⎢ ⎣5
1.4⎤ ⎡0.98(2.0) 0.98(1.4) ⎥ ⎢ 0.9⎥ = ⎢0.98(0.8) 0.98(1.1) 1.5⎥⎦ ⎢⎣0.98(3.6) 0.98(1.2)
⎡ 6⎤ ⎢ 2⎥ ⎢ 1⎥ ⎣ ⎦
2 −3⎤ ⎡ x1 ⎤ ⎡ −2 ⎤ 2 0 ⎥ ⎢ x2 ⎥ ⎢ 2 ⎥ = 5 −4 ⎥ ⎢⎢ x3 ⎥⎥ ⎢ −1⎥ ⎥ ⎢ ⎥ −3 4 ⎦ ⎢⎣ x4 ⎥⎦ ⎣ 2 ⎦ ⎧5 x1 − x2 + 2 x3 − 3 x4 = −2 ⎪ + 2 x3 = 2 ⎪ 4 x1 ⎨ x x x x 2 2 5 4 − + − 2 3 4 = −1 ⎪ 1 ⎪ 3 x1 + x2 − 3 x3 + 4 x4 = 2 ⎩
0.98(1.4)⎤ ⎡1.96 1.37 ⎥ ⎢ 0.98( 2.0) 0.98(0.9)⎥ = ⎢0.78 1.08 0.98( 4.5) 0.98(1.5)⎥⎦ ⎢⎣3.53 1.18 0.98(3.0)
2.94 1.37 ⎤ ⎥ 1.96 0.88⎥ 4.41 1.47 ⎥⎦
42.
⎡ 28.0 28.9 30.0 31.5⎤ ⎡1.06(28.0) 1.06(28.9) 1.06(30.0) 1.06(31.5) ⎤ ⎡ 29.7 30.6 31.8 33.4 ⎤ 1.06 ⎢ 29.0 30.3 32.5 34.5⎥ = ⎢1.06(29.0) 1.06(30.3) 1.06(32.5) 1.06(34.5) ⎥ = ⎢ 30.7 32.1 34.5 36.6⎥ ⎢ 31.8 33.3 36.0 39.2 ⎥ ⎢ 30.0 31.4 34.0 37.0⎥ ⎢1.06(30.0) 1.06(31.4) 1.06(34.0) 1.06(37.0) ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
43.
a.
b.
⎡14 3⎤ ⎡12 5⎤ ⎡ 26 8⎤ H + A = ⎢14 3⎥ + ⎢ 7 10⎥ = ⎢ 21 13⎥ ⎢10 7 ⎥ ⎢ 8 9⎥ ⎢ 18 16⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ The matrix represents the total number of wins and losses for each team for the season. ⎡14 3⎤ ⎡12 5⎤ ⎡ 2 −2⎤ H − A = ⎢14 3⎥ − ⎢ 7 10⎥ = ⎢ 7 −7⎥ ⎢10 7 ⎥ ⎢ 8 9⎥ ⎢ 2 −2⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ The matrix represents the difference between performances at home and performances away.
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712
44.
Chapter 10: Matrices
a. b.
45.
47.
⎡ 23 35 49⎤ ⎡19 28 36 ⎤ ⎡ 42 63 85⎤ A+ B = ⎢ ⎥+⎢ ⎥=⎢ ⎥ ⎣32 41 24⎦ ⎣25 38 26⎦ ⎣57 79 50⎦ The sum of A and B represents the total number of each of the three television sets available from both stores.
⎡ 530 ⎢ 190 A= ⎢ ⎢ 485 ⎣ 150
650 385 600 210
815⎤ ⎡ 480 500 675⎤ ⎢ 175 215 345⎥ 715⎥ , B= ⎢ ⎥ 610 ⎥⎥ ⎢ 400 350 480⎥ 305⎦ ⎣ 70 95 280 ⎦ ⎡50 150 140 ⎤ ⎢ 15 170 370 ⎥ A− B = ⎢ ⎥ ⎢ 85 250 130 ⎥ ⎣80 115 25⎦ A − B is number sold of each item during the week. ⎡0.04 C = ⎢0.04 ⎢ ⎢⎣ 0.03 ⎡0.04 CS = ⎢0.04 ⎢ ⎣⎢ 0.03
0.06 0.05 ⎤ ⎡ 500 600 ⎤ 0.04 0.04 ⎥ , S = ⎢ 250 450 ⎥ ⎥ ⎢ ⎥ 0.07 0.06 ⎥⎦ ⎢⎣ 600 750 ⎥⎦ 0.06 0.05 ⎤ ⎡500 600 ⎤ 0.04 0.04 ⎥ ⎢ 250 450 ⎥ ⎥⎢ ⎥ 0.07 0.06 ⎦⎥ ⎣⎢ 600 750 ⎦⎥
46.
48.
= ⎡ 65 88.5⎤ ⎢ 54 72 ⎥ ⎢ ⎥ ⎢⎣68.5 94.5⎥⎦ To minimize commissions costs, customer S1 should use company T2.
⎡315 200 415⎤ ⎡ 200 175 350 ⎤ A = ⎢ 285 175 300 ⎥ , B = ⎢ 150 90 180⎥ ⎢ 275 195 250 ⎥ ⎢ 105 50 175⎥ ⎣ ⎦ ⎣ ⎦ ⎡ 515 375 765 ⎤ A + B = ⎢ 435 265 480 ⎥ ⎢380 245 425⎥ ⎣ ⎦ A + B gives the number of employees in each area for both branches of the company. ⎡.25 .50 ⎤ ⎢.30 .75⎥ ⎢.15 .45⎥ ⎢ ⎥ ⎣.10 .50 ⎦ Total cost and total revenue can be determined from the product of S and P. ⎡.25 .50 ⎤ ⎡ 52 50 75 20 ⎤ ⎢ .30 .75⎥⎥ SP = ⎢⎢ 45 48 80 20 ⎥⎥ ⎢ ⎢.15 .45⎥ ⎢⎣ 62 70 78 25⎥⎦ ⎢ ⎥ ⎣ 10 .50 ⎦ ⎡ 41.25 107.25⎤ = ⎢⎢ 39.65 104.50 ⎥⎥ ⎣⎢ 50.70 131.10 ⎦⎥ ⎡52 50 75 20 ⎤ S = ⎢ 45 48 80 20 ⎥ , P = ⎢ 62 70 78 25⎥ ⎣ ⎦
49.
⎡ 2 −3⎤ ⎡ 1 0 0⎤ ⎡ 2 −3⎤ ⎡ 2 −3⎤ P '(2, − 5) Rx ⋅ ⎢ 5 6⎥ = ⎢0 −1 0⎥ ⎢ 5 6⎥ = ⎢ −5 −6⎥ ⇒ Q '( −3, − 6) ⎢ 1 1⎥ ⎢0 0 1⎥ ⎢ 1 1⎥ ⎢ 1 1⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
50.
⎡ −1 2 ⎤ ⎡ −1 0 0 ⎤ ⎡ −1 2 ⎤ ⎡ 1 −2 ⎤ P '(1, 2) R y ⋅ ⎢ 2 −4 ⎥ = ⎢ 0 1 0⎥ ⎢ 2 −4⎥ = ⎢ 2 −4⎥ ⇒ Q '( −2, − 4) ⎢ 1 1⎥ ⎢ 0 0 1⎥ ⎢ 1 1⎥ ⎢ 1 1⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
51.
⎡ −3 −5⎤ ⎡0 1 0⎤ ⎡ −3 −5⎤ ⎡ 1 3⎤ P '(1, − 3) Rxy ⋅ ⎢ 1 3⎥ = ⎢ 1 0 0⎥ ⎢ 1 3⎥ = ⎢ −3 −5⎥ ⇒ Q '(3, − 5) ⎢ 1 1⎥ ⎢0 0 1⎥ ⎢ 1 1⎥ ⎢ 1 1⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
52.
⎡ 2 −1⎤ ⎡ −1 0 0⎤ ⎡0 −1 0⎤ ⎡ 2 −1⎤ ⎡ −1 0 0⎤ ⎡ −4 −3⎤ ⎡ 4 3⎤ P '(4, 2) R y ⋅ R90 ⋅ ⎢ 4 3⎥ = ⎢ 0 1 0⎥ ⎢ 1 0 0⎥ ⎢ 4 3⎥ = ⎢ 0 1 0⎥ ⎢ 2 −1⎥ = ⎢ 2 −1⎥ ⇒ '(3, −1) Q ⎢ 1 1⎥ ⎢ 0 0 1⎥ ⎢0 0 1⎥ ⎢ 1 1⎥ ⎢ 0 0 1⎥ ⎢ 1 1⎥ ⎢ 1 1⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
53.
⎡ −1 1 3⎤ ⎡0 1 0⎤ ⎡ 1 0 3⎤ Rxy ⋅T3,−1 ⋅ ⎢ 5 −2 4 ⎥ = ⎢ 1 0 0⎥ ⎢0 1 −1⎥ ⎢ 1 1 1⎥ ⎢0 0 1⎥ ⎢0 0 1⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
54.
A '( −4, 1) ⎡ −4 −2 2 ⎤ ⎡ −1 0 0⎤ ⎡ −1 0 0⎤ ⎡ −4 −2 2⎤ ⎡ − 1 0 0 ⎤ ⎡ 4 2 −2 ⎤ ⎡ −4 −2 2 ⎤ R180 ⋅ R y ⋅ ⎢ −1 1 −5⎥ = ⎢ 0 −1 0⎥ ⎢ 0 1 0⎥ ⎢ −1 1 −5⎥ = ⎢ 0 −1 0⎥ ⎢ −1 1 −5⎥ = ⎢ 1 −1 5⎥ ⇒ B '( −2, −1) ⎢ 1 1 1⎥ ⎢ 0 0 1⎥ ⎢ 0 0 1⎥ ⎢ 1 1 1⎥ ⎢ 0 0 1⎥ ⎢ 1 1 1⎥ ⎢ 1 1 1⎥ C '(2, 5) ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
A '(4, 2) ⎡ −1 1 3⎤ ⎡ 0 1 0⎤ ⎡ 2 4 6 ⎤ ⎡ 4 −3 3⎤ ⎢ 5 −2 4⎥ = ⎢ 1 0 0⎥ ⎢ 4 −3 3⎥ = ⎢ 2 4 6⎥ ⇒ B '( −3, 4) ⎢ 1 1 1⎥ ⎢ 0 0 1⎥ ⎢ 1 1 1⎥ ⎢ 1 1 1⎥ C '(3, 6) ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
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Section 10.2
55.
56.
57.
713
⎡ −1 T1,−6 ⋅ ⎢ 2 ⎢ 1 ⎣ ⎡ 0 R180 ⋅ ⎢ −4 ⎢ 1 ⎣
−1 4 4 ⎤ 1⎤ ⎡1 0 6 6 2 ⎥ = ⎢ 0 1 −6 ⎥ ⎢0 0 1 1 1⎥⎦ 1⎥⎦ ⎣ 0 5 5⎤ ⎡ −1 0 0⎤ 0 0 −4 ⎥ = ⎢ 0 −1 0 ⎥ ⎢ 0 0 1⎥ 1 1 1⎥⎦ ⎣ ⎦ ⎡ 0 0 −5 −5⎤ ⎡ 1 0 −1⎤ T−1, 6 ⋅ ⎢ 4 0 0 4 ⎥ = ⎢ 0 1 6⎥ ⎢ 1 1 1 1⎥ ⎢ 0 0 1⎥ ⎣ ⎦ ⎣ ⎦
⎡ −1 −1 4 4 ⎤ ⎡ 0 ⎢ 2 6 6 2 ⎥ = ⎢ −4 ⎢ 1 1 1 1⎥ ⎢ 1 ⎣ ⎦ ⎣ 0 0 5 5 ⎡ ⎤ ⎡0 ⎢ −4 0 0 − 4 ⎥ = ⎢ 4 ⎢ 1 1 1 1⎥ ⎢1 ⎣ ⎦ ⎣
⎡ 0 0 −5 −5⎤ ⎡ −1 − 1 − 6 − 6 ⎤ ⎢ 4 0 0 4 ⎥ = ⎢ 10 6 6 10⎥ ⇒ A '( −1, 10) C '( −6, 6) B '( −1, 6) D '( −6, 10) ⎢ 1 1 1 1⎥ ⎢ 1 1 1 1⎥ ⎣ ⎦ ⎣ ⎦
⎡ −3 −1 6 4 ⎤ ⎡ 1 0 −6 ⎤ ⎡ − 3 T−6, 0 ⋅ ⎢ −1 4 0 −5⎥ = ⎢0 1 0⎥ ⎢ −1 ⎢ 1 1 1 1⎥ ⎢0 0 1⎥⎦ ⎢⎣ 1 ⎣ ⎦ ⎣ ⎡ −9 −7 0 −2 ⎤ ⎡ 0 1 0 ⎤ ⎡ −9 R270 ⋅ ⎢ −1 4 0 −5⎥ = ⎢ −1 0 0⎥ ⎢ −1 ⎢ 1 1 1 1⎥ ⎢ 0 0 1⎥ ⎢ 1 ⎣ ⎦ ⎣ ⎦ ⎣ ⎡ −1 4 0 −5⎤ ⎡ 1 0 6 ⎤ ⎡ −1 4 T6, 0 ⋅ ⎢ 9 7 0 2 ⎥ = ⎢0 1 0⎥ ⎢ 9 7 ⎢ 1 1 1 1⎥ ⎢0 0 1⎥ ⎢ 1 1 ⎣ ⎦ ⎣ ⎦ ⎣ a.
−1 6 4 ⎤ ⎡ −9 − 7 0 − 2 ⎤ 4 0 −5⎥ = ⎢ −1 4 0 −5⎥ ⎢ 1 1 1 1⎥ 1 1 1⎥⎦ ⎣ ⎦ −7 0 − 2 ⎤ ⎡ −1 4 0 −5⎤ 4 0 −5⎥ = ⎢ 9 7 0 2⎥ ⎢ 1 1 1 1⎥ 1 1 1⎥⎦ ⎣ ⎦ 0 −5⎤ ⎡ 5 10 6 1⎤ A '(5, 9) C '(6, 0) 0 2⎥ = ⎢9 7 0 2 ⎥ ⇒ B '(10, 7) D '(1, 2) ⎥ ⎢ ⎥ 1 1⎦ ⎣ 1 1 1 1⎦
For 1 year from now, n = 2.
b. 2
⎡ 0.989 0.011⎤ 0.45] ⎢ ⎥ = [ 0.5443 0.4557 ] ⎣ 0.007 0.993⎦ 1 year from now 45.6% of the customers will be drinking diet soda.
[0.55
0 5 5⎤ 0 0 −4 ⎥ 1 1 1⎥⎦ 0 −5 −5⎤ 0 0 4⎥ 1 1 1⎥⎦
For 3 years from now, n = 6. 6
⎡ 0.989 0.011⎤ 0.45] ⎢ ⎥ = [0.5334 0.4666] ⎣ 0.007 0.993⎦ 3 years from now 46.7% of the customers will be drinking diet soda.
[0.55
58.
When n = 21, d > r. After 10 years, the number of diet soda drinkers out numbers the number of regular soda drinkers.
59.
a.
For 12 months from now, n = 12.
b.
12
⎡ 0.975 0.025⎤ 0.85] ⎢ ⎥ = [ 0.2293 0.7707] ⎣ 0.014 0.986⎦ 12 months from now 22.9% of the customers will be renting DVD movies online.
[0.15
60.
For 5 months from now, n = 5. ⎡ 0.98 0.02 ⎤ 0.75] ⎢ ⎥ = [ 0.3913 0.6087 ] ⎣ 0.05 0.95⎦ 5 months from now, 39.1% of the customers will stop at Store A.
[0.25
62.
When n = 11, a > 0.5053.
⎡ 0 0.65⎤ 375,000] ⎢ 0 ⎥⎦ ⎣1.25 After 4 years,
n
4
⎡ 0 0.65⎤ 375,000] ⎢ = [313,574 247,558] 0 ⎥⎦ ⎣1.25 313,574 + 247,558 = 561,132 There will be approximately 561,000 plants in the reserve after 4 years.
[ 475,000
11
⎡0.98 0.02 ⎤ 0.75] ⎢ ⎥ = [ 0.5053 0.4947 ] ⎣0.05 0.95 ⎦ After 11 months, Store A will have 50% of the town’s customers.
[0.25
In n years,
[ 475,000
24
⎡ 0.975 0.025⎤ 0.85] ⎢ ⎥ = [ 0.2785 0.7215] ⎣ 0.014 0.986⎦ 24 months from now 27.9% of the customers will be renting DVD movies online.
[0.15
61.
5
For 24 months from now, n = 24.
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714
Chapter 10: Matrices
63.
Using A and B as given and a calculator, 21 −12 32 0⎤ ⎡ 24 ⎢ −7 − 8 3 21 20 ⎥ AB = ⎢ 32 10 −32 1 5⎥ ⎢ ⎥ ⎢ 19 −15 −17 30 20 ⎥ 9 −28 13 −6 ⎦⎥ ⎣⎢ 29
64.
Using A and B as given and a calculator, 3 13 28⎤ ⎡ 30 − 27 ⎢ 7 −13 −3 −5 21⎥ BA = ⎢ 17 33⎥ − 7 −3 12 ⎢ ⎥ −8 23 2 −17 ⎥ ⎢ 34 14 −7 14 −8⎦⎥ ⎣⎢ 12
65.
Using A as given and a calculator, −100 36 273 93⎤ ⎡ 46 ⎢ 82 − 93 19 27 97 ⎥ A3 = ⎢ 73 −10 −23 109 83⎥ ⎢ ⎥ −189 52 37 156 ⎥ ⎢ 212 − 22 54 221 58⎦⎥ ⎣⎢ 68
66.
Using B as given and a calculator, ⎡ 55 − 65 65 291 −154 ⎤ ⎢ −60 − 72 69 87 − 26 ⎥ B3 = ⎢ 98 −94 −33 128 −124 ⎥ ⎢ ⎥ − 93⎥ ⎢ 149 213 −49 114 55 63 −121⎦⎥ ⎣⎢ 44 −57
67.
Using A and B as given and a calculator, 6⎤ ⎡76 − 8 −25 30 ⎢14 16 −10 14 2⎥ A2 + B 2 = ⎢39 0 −45 22 27 ⎥ ⎢ ⎥ 23 83 −16 ⎥ ⎢ 0 −4 7 5⎦⎥ ⎣⎢56 −20 −22
68.
Using A and B as given and a calculator, 48 −15 19 −28⎤ ⎡ −6 ⎢ −14 −1⎥ 5 6 26 AB − BA = ⎢ 15 17 −29 −11 −28⎥ ⎢ ⎥ 28 37 ⎥ − 7 −40 ⎢ −15 2 ⎦⎥ − 5 −21 −1 ⎣⎢ 17
....................................................... 69.
⎡ 2 + 3i 1 − 2i ⎤ ⎡6 + 9i 3A = 3 ⎢ = ⎣ 1 + i 2 − i ⎥⎦ ⎢⎣ 3 + 3i
71.
⎡ 1 − i 2 + 3i ⎤ ⎡ 2 + 2i −6 + 4i ⎤ = 2iB = 2i ⎢ 2 + 8i ⎥⎦ ⎣3 + 2i 4 − i ⎥⎦ ⎢⎣ −4 + 6i
73.
⎡ 2 + 3i 1 − 2i ⎤ ⎡ 1 − i 2 + 3i ⎤ ⎡3 + 2i A+ B = ⎢ + = ⎣ 1 + i 2 − i ⎥⎦ ⎢⎣3 + 2i 4 − i ⎥⎦ ⎢⎣ 4 + 3i
74.
⎡ 2 + 3i 1 − 2i ⎤ ⎡ 1 − i 2 + 3i ⎤ ⎡ 1 + 4i −1 − 5i ⎤ A− B = ⎢ − = − 2 ⎥⎦ ⎣ 1 + i 2 − i ⎥⎦ ⎢⎣3 + 2i 4 − i ⎥⎦ ⎢⎣ −2 − i
75.
⎡ 2 + 3i 1 − 2i ⎤ ⎡ 1 − i 2 + 3i ⎤ ⎡ (2 + 3i )(1 − i ) + (1 − 2i )(3 + 2i ) (2 + 3i )(2 + 3i ) + (1 − 2i )(4 − i) ⎤ ⎡12 − 3i −3 + 3i ⎤ = AB = ⎢ ⎥ ⎢ ⎥=⎢ (1 + i )(2 + 3i ) + (2 − i )(4 − i ) ⎥⎦ ⎢⎣ 10 + i 6 − i ⎥⎦ ⎣ 1 + i 2 − i ⎦ ⎣3 + 2i 4 − i ⎦ ⎣ (1 + i )(1 − i ) + (2 − i )(3 + 2i )
76.
⎡ 1 − i 2 + 3i ⎤ ⎡ 2 + 3i 1 − 2i ⎤ ⎡ (1 − i)(2 + 3i ) + (2 + 3i )(1 + i ) BA = ⎢ ⎥ ⎢ ⎥ = ⎢ ⎣3 + 2i 4 − i ⎦ ⎣ 1 + i 2 − i ⎦ ⎣ (3 + 2i )(2 + 3i ) + (4 − i )(1 + i )
77.
3 − 6i ⎤ 6 − 3i ⎥⎦
Connecting Concepts
⎡ 2 + 3i 1 − 2i ⎤ ⎡ 2 + 3i 1 − 2i ⎤ A2 = A ⋅ A = ⎢ ⎣ 1 + i 2 − i ⎥⎦ ⎢⎣ 1 + i 2 − i ⎥⎦ ⎡ (2 + 3i)(2 + 3i ) + (1 − 2i )(1 + i ) =⎢ ⎣ (1 + i )(2 + 3i ) + (2 − i )(1 + i ) ⎡ −2 + 11i 8 − 6i ⎤ =⎢ ⎣ 2 + 6i 6 − 5i ⎥⎦
70.
⎡ 1 − i 2 + 3i ⎤ ⎡ −2 + 2i −2 B = −2 ⎢ = ⎣3 + 2i 4 − i ⎥⎦ ⎢⎣ −6 − 4i
72.
⎡ 2 + 3i 1 − 2i ⎤ ⎡ −9 + 6i 3iA = 3i ⎢ = ⎣ 1 + i 2 − i ⎥⎦ ⎢⎣ −3 + 3i
−4 − 6i ⎤ − 8 + 2i ⎥⎦ 6 + 3i ⎤ 3 + 6i ⎥⎦
3 + i⎤ 6 − 2i ⎥⎦
(1 − i )(1 − 2i) + (2 + 3i )(2 − i ) ⎤ ⎡ 4 + 6i = (3 + 2i )(1 − 2i ) + (4 − i )(2 − i ) ⎥⎦ ⎢⎣5 + 16i
(2 + 3i)(1 − 2i ) + (1 − 2i)(2 − i) ⎤ (1 + i )(1 − 2i ) + (2 − i)(2 − i ) ⎥⎦
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6 + i⎤ 14 − 10i ⎥⎦
Section 10.3
78.
79.
715
⎡ 1 − i 2 + 3i ⎤ ⎡ 1 − i 2 + 3i ⎤ B2 = B ⋅ B = ⎢ ⎣3 + 2i 4 − i ⎥⎦ ⎢⎣3 + 2i 4 − i ⎥⎦ (1 − i)(2 + 3i ) + (2 + 3i )(4 − i ) ⎤ ⎡(1 − i )(1 − i ) + (2 + 3i )(3 + 2i ) =⎢ ⎣ (3 + 2i )(1 − i ) + (4 − i )(3 + 2i ) (3 + 2i )(2 + 3i ) + (4 − i )(4 − i ) ⎥⎦ ⎡ +11i 16 + 11i ⎤ =⎢ ⎣19 + 4i 15 + 5i ⎥⎦ 2
⎡ 0 −1 0⎤ ⎡ 0 −1 0⎤ ⎡ 0 −1 0⎤ ⎡ −1 0 0 ⎤ ( R90 ) = ⎢ 1 0 0⎥ = ⎢ 1 0 0⎥ ⎢ 1 0 0⎥ = ⎢ 0 −1 0⎥ = R180 ⎢ 0 0 1⎥ ⎢ 0 0 1⎥ ⎢ 0 0 1⎥ ⎢ 0 0 1⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Rotating 90° twice around the origin is the same as rotating 180° once around the origin. 2
3
⎡ 0 −1 0 ⎤ ⎡0 −1 0⎤ ⎡0 −1 0⎤ ⎡0 −1 0⎤ ⎡ 0 −1 0 ⎤ ⎡ − 1 0 0 ⎤ ( R90 ) = ⎢ 1 0 0⎥ = ⎢ 1 0 0⎥ ⎢ 1 0 0⎥ ⎢ 1 0 0⎥ = ⎢ 1 0 0⎥ ⎢ 0 −1 0⎥ = ⎢ 0 0 1⎥ ⎢0 0 1⎥ ⎢0 0 1⎥ ⎢0 0 1⎥ ⎢ 0 0 1⎥ ⎢ 0 0 1⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Rotating 90° three times around the origin is the same as rotating 270° once around the origin. 3
80.
⎡ 1 0 0⎤ ⎡ −1 R x ⋅ R y = ⎢ 0 −1 0 ⎥ ⎢ 0 ⎢ 0 0 1⎥ ⎢ 0 ⎣ ⎦ ⎣ ⎡ −1 0 0 ⎤ ⎡ 1 R y ⋅ Rx = ⎢ 0 1 0⎥ ⎢ 0 ⎢ 0 0 1⎥ ⎢0 ⎣ ⎦ ⎣ Thus, Rx ⋅ R y = R y ⋅ Rx .
0 0⎤ ⎡ −1 0 1 0 ⎥ = ⎢ 0 −1 ⎢ 0 0 0 1⎥⎦ ⎣ 0 0⎤ ⎡ −1 0 −1 0⎥ = ⎢ 0 −1 ⎢ 0 0 0 1⎥⎦ ⎣
⎡ 0 1 0⎤ ⎢ −1 0 0⎥ = R270 ⎢ 0 0 1⎥ ⎣ ⎦
0⎤ 0⎥ 1⎥⎦ 0⎤ 0⎥ 1⎥⎦
Reflecting across the x-axis and then the y-axis is the same as reflecting across the y-axis and then the x-axis.
....................................................... 3 PS1. − 2
PS2. ⎡1 0 0 ⎤ ⎢0 1 0⎥ ⎢ ⎥ ⎣⎢ 0 0 1 ⎥⎦
PS3. 1. Interchange any two rows. 2. Multiply all elements in a row by the same nonzero number. 3. Replace a row by the sum of that row and a nonzero multiple of any other row.
AX = B
PS5. −1
Prepare for Section 10.3
−2 R1 + R2 ⎡1 −2 3 ⎤ PS4. ⎡ 1 −2 3⎤ 3R1 + R3 ⎢ ⎢ 2 −1 4 ⎥ ⎯⎯⎯⎯⎯ → 0 3 −2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ −3 2 2 ⎦⎥ ⎣⎢0 −4 11 ⎥⎦
PS6. ⎧2 x + 3 y = 9 ⎨ ⎩4 x − 5 y = 7
−1
A AX = A B X = A−1B
Section 10.3 1.
2 R1 + R 2 ⎡ 1 −3 1 0⎤ −1R 2 ⎡ 1 −3 1 0⎤ 3R 2 + R1 ⎡ 1 0 −5 −3⎤ ⎡ 1 −3 1 0⎤ ⎯⎯⎯⎯⎯ →⎢ ⎯⎯⎯⎯ →⎢ ⎯⎯⎯⎯⎯ →⎢ ⎢⎣ −2 5 0 1⎥⎦ ⎥ 0 1 2 1 0 1 2 − − −1⎥⎦ ⎣ ⎦ ⎣ ⎣0 1 −2 −1⎥⎦ ⎡ −5 −3⎤ The inverse matrix is ⎢ ⎥. ⎣ − 2 − 1⎦
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716
Chapter 10: Matrices
2.
2 1 0⎤ 2 R1 + R 2 ⎡ 1 2 1 0⎤ −2 R 2 + R1 ⎡ 1 0 −3 −2⎤ ⎡ 1 →⎢ ⎢ ⎥ ⎯⎯⎯⎯⎯ ⎥ ⎯⎯⎯⎯⎯→ ⎢0 1 2 −1⎥ − 2 − 3 0 1⎦ ⎣0 1 2 1⎦ ⎣ ⎦ ⎣ ⎡ −3 −2⎤ The inverse matrix is ⎢ ⎥. 1⎦ ⎣ 2
3.
1 0⎤ −4 R 2 + R1 ⎡ 1 0 5 −2 ⎤ ⎡ 1 4 1 0⎤ −2 R1 + R2 ⎡ 1 4 1 0⎤ (1/ 2) R 2 ⎡ 1 4 →⎢ 1 ⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 1 −1 1⎥ ⎢ ⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 2 −2 1⎥ ⎯⎯⎯⎯⎯ 0 1 1 − ⎣ ⎦ 2⎦ 2⎦ ⎣ 2 10 0 1⎦ ⎣ ⎣ ⎡ 5 −2⎤ The inverse matrix is ⎢ ⎥. −1 1 ⎥ 2⎦ ⎣⎢
4.
3 1 0⎤ ( −1/ 2) R1 ⎡ 1 − 3 − 1 0⎤ 6 R1 + R 2 ⎡ 1 − 3 − 1 0⎤ ⎡ −2 2 2 2 2 →⎢ ⎢ ⎥ ⎯⎯⎯⎯⎯→ ⎢ ⎥ ⎯⎯⎯⎯⎯ ⎥ − 6 − 8 0 1⎦ 0 1⎦ ⎣ ⎣ −6 −8 ⎣ 0 −17 −3 1⎦
( −1/17) R 2 ⎡ 1 − 23 − 12 ⎯⎯⎯⎯⎯⎯ →⎢ 3 1 17 ⎣⎢0 ⎡− 4 The inverse matrix is ⎢ 17 ⎢ 3 ⎣ 17
5.
6.
0⎤
1⎥ − 17 ⎦⎥
4 (3/ 2) R 2 + R1 ⎡ 1 0 − 17 ⎯⎯⎯⎯⎯⎯ →⎢ 3 17 ⎣⎢0 1
3⎤ − 34 1⎥ − 17 ⎦⎥
− 3⎤ 34 ⎥. − 1⎥ 17 ⎦
−2 R1 + R2 1 2 −1 1 0 ⎡ 1 2 −1 1 0 0 ⎤ −3R1 + R3 ⎡ ⎢2 5 1 0 1 0 ⎥ ⎯⎯⎯⎯⎯ → ⎢0 1 3 −2 1 ⎢ 3 6 −2 0 0 1⎥ ⎢0 0 1 −3 0 ⎣ ⎦ ⎣ 7 R3 + R1 1 0 −7 5 −2 0⎤ 1 −3R3 + R 2 ⎡ −2 R 2 + R1 ⎡ 1 0⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 ⎯⎯⎯⎯⎯→ ⎢0 1 3 −2 ⎢0 0 ⎢0 1 −3 0 1⎥⎦ ⎣ ⎣ ⎡ −16 −2 7 ⎤ The inverse matrix is ⎢ 7 1 −3⎥ . ⎢ −3 0 1⎥⎦ ⎣
0⎤ 0⎥ 1⎥⎦ 0 0 −16 −2 7 ⎤ 1 0 7 1 −3⎥ 0 1 −3 0 1⎥⎦
R1 + R 2 1 0 0⎤ ⎡ 1 3 −2 1 3 −2 1 0 0⎤ ⎡ 1 3 −2 1 0 0 ⎤ − 2 ( −1/ 2) R 2 ⎢ R1 + R3 ⎡ ⎢ −1 −5 6 0 1 0 ⎥ ⎯⎯⎯⎯⎯ → 0 1 −2 − 12 − 12 0⎥ → ⎢0 −2 4 1 1 0⎥ ⎯⎯⎯⎯⎯ ⎢ ⎥ ⎢ 2 6 −3 0 0 1⎥ ⎢0 0 1 −2 0 1⎥⎦ 1 −2 0 1⎥⎦ ⎢⎣0 0 ⎣ ⎣ ⎦ 3 −4 ⎤ 2 R3 + R 2 ⎡ 1 3 0 −3 ⎡ 1 0 0 21 0 2⎤ 2 2 ⎥ 2 R3 + R1 ⎢ −3R2 + R1 ⎢ 9 9 1 1 ⎥ 2⎥ ⎯⎯⎯⎯⎯ → 0 1 0 − 2 − 2 2 ⎯⎯⎯⎯⎯→ ⎢0 1 0 − 2 − 2 ⎢ ⎥ ⎢ 0 0 1 −2 0 1⎥⎦ 0 1⎥ ⎢⎣ 0 0 1 −2 ⎣ ⎦
3 −4 ⎤ ⎡ 21 2 ⎢ 29 ⎥ The inverse matrix is ⎢ − 2 − 12 2⎥ . ⎢ −2 0 1⎥ ⎢⎣ ⎥⎦
Copyright © Houghton Mifflin Company. All rights reserved.
Section 10.3
7.
717
−2 R1 + R 2 ⎡ 1 2 − 1 1 0 0⎤ ⎡ 1 2 −1 1 0 0 ⎤ 1 2 −1 1 0 0⎤ −3R1 + R3 ⎡ (1/ 2) R 2 ⎢ ⎥ ⎢ ⎢ ⎥ ⎯⎯⎯⎯⎯ → − 0 2 3 2 1 0 ⎯⎯⎯⎯⎯ → 0 1 23 −1 21 0⎥ 1 0 1 0⎥ ⎢2 6 ⎢ ⎥ ⎢0 0 −1 −3 0 1⎥ ⎢⎣ 3 6 − 4 0 0 1⎥⎦ ⎣ ⎦ ⎣⎢0 0 −1 −3 0 1⎦⎥ 4 R3 + R1 ⎡ 1 2 −1 1 0 0 ⎤ ⎡ 1 0 −4 3 −1 0⎤ ⎡ 1 0 0 15 −1 −4⎤ −1R3 ⎢ − ( 3/ 2) R3 + R 2 ⎢ R R − + 2 3 3 3⎥ 1 1 1 2 1 → ⎢0 1 ⎥ ⎯⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯ → 0 1 2 −1 2 − → 0⎥ ⎯⎯⎯⎯⎯ 1 0 0 1 0 − 11 2 2 2 2 2⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1 3 0 −1⎦⎥ 3 0 −1⎥⎦ ⎣⎢0 0 1 3 0 −1⎦⎥ ⎣⎢0 0 ⎣⎢0 0 1 ⎡ 15 − 1 − 4⎤ ⎢ 3 ⎥. The inverse matrix is ⎢− 11 1 2 2 2⎥ ⎢⎣ 3 0 − 1⎥⎦
8.
1 0 0⎤ ⎡ 1 12 − 12 12 0 0⎤ −6 R1 + R 2 ⎡ 1 12 − 21 ⎡ 2 1 − 1 1 0 0⎤ 2 ⎥ (1/ 2) R1 ⎢ ⎢ ⎥ −4 R1 + R3 ⎢ ⎥ → ⎢ 6 4 −1 0 1 0⎥ ⎯⎯⎯⎯⎯→ ⎢0 1 2 −3 1 0 ⎥ ⎢ 6 4 − 1 0 1 0⎥ ⎯⎯⎯⎯⎯ ⎢⎣ 4 2 − 3 0 0 1⎥⎦ ⎢⎣ 4 2 −3 0 0 1⎥⎦ ⎢⎣0 0 −1 −2 0 1⎥⎦ 1 0 ⎡1 0 − 3 ⎡ 1 12 − 12 0⎤ 2 − 12 0⎤ ( −3/ 2) R3 + R1 ⎡ 1 0 0 5 − 12 − 23 ⎤ 2 2 −1R3 ⎢ −2 R3 + R 2 ⎥ ⎢ ⎥ ⎥ ( −1/ 2) R 2 + R1 ⎢ ⎯⎯⎯⎯ → ⎢0 1 → ⎢0 1 → ⎢0 1 0 −7 2 −3 1 0⎥ ⎯⎯⎯⎯⎯⎯⎯ 2 −3 1 0⎥ ⎯⎯⎯⎯⎯⎯⎯ 1 2⎥ 1 2 0 −1⎥ 1 2 0 −1⎥ 0 −1⎥ ⎢⎣0 0 ⎢⎣0 0 1 2 ⎢⎣0 0 ⎦ ⎦ ⎦
⎡ 5 − 1 − 3⎤ 2 2⎥ ⎢ 1 2⎥. The inverse matrix is ⎢− 7 ⎢ 2 0 1⎥⎥ − ⎢⎣ ⎦ 9.
1 0 0⎤ ⎡ 1 2 −2 12 0 0⎤ −1R1 + R 2 ⎡ 1 2 −2 ⎡ 2 4 − 4 1 0 0⎤ 2 −2 R1 + R3 ⎢ ⎥ (1/ 2) R1 ⎢ ⎢ ⎥ 1 1 0⎥ ⎯⎯⎯⎯⎯ → − ⎯⎯⎯⎯⎯ → − − 1 3 4 0 1 0 0 1 2 1 3 − 4 0 1 0 ⎢ ⎥ 2 ⎥ ⎢ ⎢ ⎥ ⎢0 0 ⎢⎣ 2 4 − 3 0 0 1⎥⎦ 1 −1 0 1⎥ ⎢⎣ 2 4 −3 0 0 1⎥⎦ ⎣ ⎦ 7 −2 −2 ⎤ 3 −2 0 ⎤ −2 R3 + R1 ⎡ 1 0 0 ⎡1 0 2 2 2 ⎢ ⎥ R R ⎥ + 2 −2 R 2 + R1 ⎢ 5 3 2 ⎯⎯⎯⎯⎯→ ⎢0 1 −2 − 12 1 0⎥ ⎯⎯⎯⎯⎯→ 1 2⎥ ⎢0 1 0 − 2 ⎢ 0 0 1 −1 0 ⎢0 0 1 −1 0 1⎥ 1⎥ ⎣ ⎦ ⎣ ⎦ 7 ⎡ − 2 − 2⎤ ⎢ 2 ⎥ 5 1 2⎥. The inverse matrix is ⎢− ⎢ 2 ⎥ 0 1⎥ ⎢ −1 ⎣ ⎦
10.
−2 R1 + R 2 ⎡ 1 − 2 2 1 0 0⎤ 1 −2 2 1 0 0⎤ −3R1 + R3 ⎡ ⎥ ⎢ ⎢0 ⎯⎯⎯⎯⎯ → − − 1 3 2 1 0⎥ 2 − 3 1 0 1 0 ⎥ ⎢ ⎢0 0 0 −3 0 1⎥ ⎢⎣ 3 − 6 6 0 0 1⎥⎦ ⎣ ⎦ The matrix does not have an inverse.
Copyright © Houghton Mifflin Company. All rights reserved.
718
11.
Chapter 10: Matrices
2 1 ⎡ 1 −1 ⎢ − 2 1 5 1 ⎢ ⎢ 3 −3 7 5 ⎢ 3 − 4 −1 ⎣⎢ − 2
2 R1 + R 2 1 0 0 0⎤ −3R1 + R3 ⎡ 1 −1 ⎥ 0 1 0 0⎥ 2 R1 + R 4 ⎢0 1 ⎯⎯⎯⎯⎯ →⎢ 0 0 0 0 1 0⎥ ⎢ ⎥ 0 1 ⎣ 0 0 0 1⎦⎥
⎡ 1 −1 2 1 1 0 −1R2 + R 4 ⎢0 1 1 −1 −2 1 ⎯⎯⎯⎯⎯⎯ →⎢ 0 0 1 2 −3 0 ⎢ ⎣0 0 −1 2 4 −1 ⎡ 1 −1 (1/4)R4 ⎢0 1 ⎯⎯⎯⎯⎯ → ⎢0 0 ⎢ ⎣⎢0 0
0 0 1 0
2 1 1 0 1 −1 −2 1 1 2 −3 0 0 1 2 0
0⎤ ⎡ 1 −1 0⎥ 1R3 + R 4 ⎢ 0 1 ⎯⎯⎯⎯⎯ →⎢ 0⎥ 0 0 ⎥ ⎢ 1⎦ ⎣0 0
2 1 1 0 0 0⎤ ⎡1 1 −1 −2 1 0 0⎥ R 2 + R1 ⎢0 1 2 −3 0 1 0⎥ ⎯⎯⎯⎯→ ⎢0 ⎥ ⎢ 1 0 1 4 − 14 41 14 ⎦⎥ ⎣⎢0
12.
0 1 0 0
0 0 0 1 0 0 0 1 0 0 0 1
2 ⎡ 1 −1 ⎢ − 3 2 1 ⎢ ⎢2 2 − 1 ⎢ ⎣⎢ 4 4 − 4
−3R1 + R 2 1 1 0 0 0⎤ 1 −2 R1 + R3 ⎡ 1 1 −1 2 ⎥ 5 0 1 0 0⎥ −4R1 + R 4 ⎢0 −1 2 −1 −3 ⎯⎯⎯⎯⎯→ ⎢ 0 0 1 1 −2 5 0 0 1 0⎥ ⎢ ⎥ 0 0 0 −1 −4 ⎣ 7 0 0 0 1⎦⎥
−1R 2 ⎡ 1 −1R 4 ⎢0 ⎯⎯⎯⎯ →⎢ 0 ⎢ ⎣0
1 −1 1 −2 0 1 0 0
− R3 + R1 ⎡ 1 2 R3 + R 2 ⎢0 ⎯⎯⎯⎯⎯ →⎢ 0 ⎢ ⎣0
0 1 0 0
0 0⎤ ⎡1 0 − R 2 + R1 ⎢0 1 0 0⎥ ⎯⎯⎯⎯⎯ →⎢ 0 0 1 0⎥ ⎥ ⎢ 0 −1⎦ ⎣0 0 1 −1 0⎤ −3R 4 + R 2 ⎡ 1 ⎢0 −1 2 0 ⎥ − R 4 + R 3 ⎯⎯⎯⎯⎯→ ⎢ 0 1 0⎥ 0 ⎢ ⎥ 0 0 −1⎦ ⎣0
2 1 0 1 3 −1 1 −2 0 1 4 0 0 0 1 0
0 0 3 −1 1 −2 1 4
2 1 1 0 1 −1 −2 1 1 2 −3 0 0 4 1 −1
0 0 1 1
0⎤ 0⎥ 0⎥ ⎥ 1⎦
3 0 −1 1 0 0⎤ 1 −1 −2 1 0 0⎥ 1 2 −3 0 1 0⎥ ⎥ 1 0 1 4 − 14 14 41 ⎦⎥
0 1 0 0
6R 4 + R1 ⎡ 1 0 −6 8 1 −3 0⎤ 3R 4 + R 2 ⎢ 0 −3 1 1 − 1 0 ⎥ −2 R 4 + R 3 ⎢ 0 1 2 −3 0 1 0⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 ⎥ ⎢ 1 1 1 1 0 1 4 −4 ⎥ 4 4⎦ ⎣⎢ 0 3⎤ ⎡ 19 − 1 − 3 2 2 2⎥ ⎢ 2 3⎥ 1 −1 ⎢ 7 4 4 4 ⎥. The inverse matrix is ⎢ 4 1 1 − 1⎥ ⎢− 7 2 2 2⎥ ⎢ 2 1 1⎥ ⎢ 1 −1 4 4 4⎦ ⎣ 4 −3R3 + R1 ⎡ 1 ⎢0 − R3 + R 2 ⎯⎯⎯⎯⎯ → ⎢0 ⎢ ⎣⎢0
0⎤ 0⎥ 0⎥ ⎥ 1⎦
0 0 1 0
1 −2 1 0 0 1 0 0
0 1 0 0
0 0 1 0
19 2 7 4 − 72 1 4
1 4 1 2 − 14
− 23 − 41
1 2 1 4
3⎤ 2 1⎥ 4⎥ − 21 ⎥ ⎥ 1 ⎥ 4⎦
0⎤ 0⎥ 0⎥ ⎥ 1⎦
1 −2 1 1 3 −1 1 −2 0 1 4 0 0 0 1 0
− 12
0 0⎤ 0 0⎥ 1 0⎥ ⎥ 0 −1⎦
0 0 1 −1 0 ⎤ 0 −13 −1 2 3⎥ 0 −6 0 1 1⎥ ⎥ 1 4 0 0 −1⎦
1 −1 0⎤ ⎡ 0 ⎥ ⎢ 13 1 2 3⎥ − − The inverse matrix is ⎢ . ⎢ − 6 0 1 1⎥ ⎥ ⎢ ⎣⎢ 4 0 0 − 1⎦⎥
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Section 10.3
13.
719
⎡ 1 −1 ⎢ 2 −1 ⎢ 1 1 ⎢ ⎣ −1 5
1 3 1 4 8 0 6 10 0 5 4 0
0 1 0 0
0 0 1 0
−2R 2 + R3 ⎡ 1 −1 1 −4 R 2 + R 4 ⎢0 1 2 ⎯⎯⎯⎯⎯→ ⎢ 0 0 1 ⎢ ⎣0 0 −2
−2R1 + R 2 1 0 ⎤ −1R1 + R3 ⎡ 1 −1 1 3 1R1 + R4 ⎢0 1 2 2 −2 0⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 2 5 7 −1 0 ⎥⎥ ⎢ 1 1⎦ ⎣0 4 6 7 3 1 0 0 0⎤ ⎡1 2 −2 1 0 0⎥ 2R3 + R 4 ⎢0 ⎯⎯⎯⎯⎯ →⎢ 3 3 −2 1 0⎥ 0 ⎥ ⎢ −1 9 −4 0 1⎦ ⎣0
1 0 0 0⎤ ⎡ 1 −1 1 3 ⎡1 0 3 1 0 0⎥ R 2 + R1 ⎢0 1 2 (1/5) R 4 ⎢0 1 2 2 −2 ⎯⎯⎯⎯⎯ → ⎢0 0 1 3 3 −2 1 0⎥ ⎯⎯⎯⎯→ ⎢0 0 1 ⎢ ⎥ ⎢ 8 2 1 ⎣⎢0 0 0 1 3 − 5 5 5 ⎥⎦ ⎣⎢0 0 0 4 R 4 + R1 ⎡ 1 7 −3 0⎤ −3R3 + R1 ⎡ 1 0 0 −4 −10 4 R4 + R2 ⎢ 5 −2 0⎥ −3R 4 + R3 ⎢0 −2 R3 + R 2 ⎢0 1 0 −4 −8 ⎯⎯⎯⎯⎯→ ⎢0 0 1 3 3 −2 1 0⎥ ⎯⎯⎯⎯⎯→ ⎢0 ⎢ ⎥ ⎢ 8 2 1 1 3 −5 ⎥ 5 5⎦ ⎣⎢0 0 0 ⎢⎣0 3 ⎡ 2 5 ⎢ ⎢ 4 −7 5 The inverse matrix is ⎢ ⎢ −6 14 5 ⎢ ⎢ 3 −8 5 ⎣
14.
15.
−7
5 −2 5 −1 5 2 5
0 1 0 0
0 0 1 0
−1 1 0 0
1 2 1 0
0⎤ 0⎥ 0⎥ ⎥ 1⎦ 3 1 0 2 −2 1 3 3 −2 5 15 −8
5 −1 1 0 0⎤ 2 −2 1 0 0⎥ 3 3 −2 1 0⎥ ⎥ 1 3 − 85 25 15 ⎦⎥ 0 0 0
2
1 0 0
4
0 1 0 −6 0 0 1
−1 1 2 6 6 −1 −1 12 12 −1 −14 −10
1 0 0 0
0 1 0 0
0 0 1 0
3
3 5 − 75 14 5 − 85
− 75
− 25
− 15 2 5
4⎤ 5 4⎥ 5⎥ − 53 ⎥ ⎥ 1⎥ 5⎦
4⎤ 5⎥ 4⎥ 5⎥ . − 3⎥ 5⎥ 1⎥ 5⎦
−2R1 + R 2 1 2 1 0 0 ⎤ −3R1 + R3 ⎡ 1 −1 2 R1 + R 4 ⎢0 1 4 2 −2 1 0⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 2 9 6 −3 0 0 ⎥⎥ ⎢ − − −6 2 0 1⎦ 0 3 12 ⎣ 1 0 0 0⎤ − R 2 + R3 ⎡ 1 −1 1 2 3R 2 + R 4 ⎢0 1 4 2 −2 1 0 0⎥ ⎯⎯⎯⎯⎯ →⎢ 0 0 1 2 1 −2 1 0 ⎥ ⎢ ⎥ 3 0 1⎦ ⎣ 0 0 0 0 −4 The matrix does not have an inverse. ⎡ 1 ⎢ 2 ⎢ 3 ⎢ ⎣ −2
0⎤ 0⎥ 0⎥ ⎥ 1⎦
0 0 1 2
0 0 1 0
0⎤ 0⎥ 1⎥ ⎥ 1⎦
⎡1 4⎤ ⎡ x ⎤ ⎡ 6 ⎤ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ (1) ⎣2 7 ⎦ ⎣ y ⎦ ⎣11⎦ ⎡1 4 ⎤ Find the inverse of ⎢ ⎥. ⎣2 7 ⎦ 1 0⎤ − 1R2 ⎡1 4 1 0 ⎤ − 4 R2 + R1 ⎡1 0 −7 4 ⎤ ⎡1 4 1 0⎤ − 2 R1 + R2 ⎡1 4 ⎯→ ⎢ ⎯→ ⎢ ⎯→⎢ ⎢ ⎥ ⎯⎯ ⎯ ⎯ ⎯ ⎥ ⎯⎯ ⎯ ⎥ ⎯⎯ ⎯ ⎯ ⎯ ⎥ 0 1 2 1⎦ 2 7 0 1 − − ⎣ ⎦ ⎣ ⎣0 1 2 − 1⎦ ⎣0 1 2 − 1⎦ ⎡1 4⎤ ⎡−7 4⎤ The inverse of ⎢ ⎥ is ⎢ ⎥. Multiply each side of Eq. (1) by the inverse matrix. ⎣ 2 7 ⎦ ⎣ 2 − 1⎦ ⎡−7 4⎤ ⎡1 4⎤ ⎡ x ⎤ ⎡−7 4⎤ ⎡ 6 ⎤ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎣ 2 − 1⎦ ⎣2 7 ⎦ ⎣ y ⎦ ⎣ 2 − 1⎦ ⎣11⎦
⎡ x ⎤ ⎡2⎤ ⎢ ⎥=⎢ ⎥ ⎣ y ⎦ ⎣1 ⎦ The solution is (2, 1).
Copyright © Houghton Mifflin Company. All rights reserved.
720
16.
Chapter 10: Matrices
⎡ 2 3⎤ ⎡ x ⎤ ⎡5⎤ ⎢⎣ 1 2 ⎥⎦ ⎢⎣ y ⎥⎦ = ⎢⎣ 4⎥⎦ (1) ⎡ 2 3⎤ . Find the inverse of ⎢ ⎣ 1 2 ⎥⎦
0⎤ − R1 + R 2 ⎡1 ⎥ ⎯⎯⎯⎯⎯→ ⎢1 0 1⎦ ⎣⎢
1R 3 ⎡ 2 3 1 0⎤ ⎯⎯⎯ 2 1 → ⎡1 2 ⎢ ⎢⎣ 1 2 0 1⎦⎥ ⎣1 2
1 2
3 2 1 2
1 2 − 12
1 0⎤ − 23 R 2 + R1 ⎡ 1 0 2 −3⎤ 0⎤ 2 R 2 ⎡ 1 3 2 2 ⎯⎯⎯⎯⎯⎯ →⎢ ⎥ ⎯⎯⎯→ ⎢ ⎥ 1⎦⎥ ⎣0 1 −1 2 ⎦⎥ ⎣ 0 1 −1 2 ⎦
⎡ 2 −3⎤ . Multiply each side of Eq. (1) by the inverse matrix. The inverse matrix is ⎢ ⎣ −1 2 ⎥⎦
⎡ 2 −3⎤ ⎡ 2 3⎤ ⎡ x ⎤ ⎡ 2 −3⎤ ⎡ 5 ⎤ ⎢⎣ −1 2 ⎥⎦ ⎢⎣ 1 2 ⎥⎦ ⎢⎣ y ⎥⎦ = ⎢⎣ −1 2⎥⎦ ⎢⎣ 4⎥⎦ ⎡ x ⎤ ⎡ −2 ⎤ ⎢⎣ y ⎥⎦ = ⎢⎣ 3⎥⎦ The solution is (−2, 3). 17.
⎡ 1 −2⎤ ⎡ x ⎤ ⎡ 8⎤ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ (1) ⎣3 2⎦ ⎣ y ⎦ ⎣− 1⎦ ⎡ 1 −2⎤ Find the inverse matrix of ⎢ ⎥. ⎣3 2⎦ 1 1 0⎤ 2 R 2 + R1 ⎡ 1 0 −3R1 + R 2 ⎡ 1 −2 1 0⎤ (1/ 8) R 2 ⎡ 1 −2 ⎡ 1 −2 1 0⎤ ⎯⎯⎯⎯⎯⎯ 4 → ⎯⎯⎯⎯⎯ → ⎯⎯⎯⎯⎯ → ⎢ 3 1 ⎢ ⎥ 3 ⎢⎣3 2 0 1⎥⎦ ⎢⎣ 0 1 − 8 8⎦ 8 −3 1⎥⎦ ⎣0 ⎣⎢0 1 − 8
⎡ 1 The inverse matrix is ⎢ 43 ⎢− ⎣ 8
⎡ 1 ⎢ 4 ⎢− 3 ⎣ 8
1⎤ 4 1⎥ ⎥ 8⎦
1⎤ 4 ⎥. Multiply each side of Eq. (1) by the inverse matrix. 1⎥ 8⎦
1⎤ ⎡ 1 4 ⎥ ⎡ 1 − 2⎤ ⎡ x ⎤ = ⎢ 4 ⎥ ⎢ ⎥ ⎢ 1⎥ 3 2⎦ ⎣ y ⎦ ⎢− 3 8⎦ ⎣ ⎣ 8
1⎤ 4 ⎥ ⎡ 8⎤ 1 ⎥ ⎢− 1⎥ 8⎦ ⎣ ⎦
7 ⎡ x ⎤ ⎡⎢ 4 ⎤⎥ = ⎢ ⎥ ⎢ 25 ⎥ ⎣ y ⎦ ⎣− 8 ⎦
⎛ 7 25 ⎞ The solution is ⎜ ,− ⎟. ⎝4 8 ⎠ 18.
⎡ 3 −5 ⎤ ⎡ x ⎤ ⎡ −18⎤ ⎢ ⎥⎢ ⎥=⎢ ⎥ (1) ⎣ 2 − 3⎦ ⎣ y ⎦ ⎣− 11⎦ ⎡ 3 −5 ⎤ Find the inverse matrix of ⎢ ⎥. ⎣ 2 − 3⎦ (1/ 3) R1 5 −2 R1 + R 2 ⎡1 − 3 3 5 1 0 − ⎡ ⎤ ⎯⎯⎯⎯⎯⎯ → ⎢ 1 ⎢⎣ 2 −3 0 1⎥⎦ 3 ⎣⎢1
1 3 − 23
1 0⎤ 0 ⎤ 3R 2 ⎡ 1 − 5 (5/ 3) R 2 + R1 ⎡ 1 0 −3 5⎤ 3 3 →⎢ ⎥ ⎯⎯⎯→ ⎢ ⎥ ⎯⎯⎯⎯⎯⎯⎯ 1⎦⎥ 0 1 2 3 − ⎣0 1 −2 3⎥⎦ ⎣ ⎦
⎡ −3 5⎤ The inverse matrix is ⎢ ⎥. Multiply each side of Eq. (1) by the inverse matrix. ⎣ − 2 3⎦ ⎡ −3 5⎤ ⎡ 3 −5⎤ ⎡ x ⎤ ⎡ −3 5⎤ ⎡ −18⎤ ⎢ −2 3⎥ ⎢ 2 −3⎥ ⎢ y ⎥ = ⎢ −2 3⎥ ⎢ −11⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ x ⎤ ⎡ −1⎤ ⎢ y ⎥ = ⎢ 3⎥ ⎣ ⎦ ⎣ ⎦ The solution is (–1, 3).
Copyright © Houghton Mifflin Company. All rights reserved.
Section 10.3
19.
721
⎡ 1 1 2⎤ ⎡ x ⎤ ⎡ 4⎤ ⎢ 2 3 3⎥ ⎢ y ⎥ = ⎢ 5 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ 3 3 7 ⎦⎥ ⎢⎣ z ⎦⎥ ⎣⎢14 ⎥⎦
(1)
⎡1 1 2 ⎤ ⎥ ⎢ Find the inverse matrix of ⎢ 2 3 3⎥. ⎢⎣ 3 3 7 ⎥⎦ 2 R1 + R 2 1 0 0⎤ ⎡ 1 1 2 1 0 0⎤ ⎡1 1 2 ⎡ 1 0 3 3 −1 0⎤ −3R1 + R 3 ⎢ − R 2 + R1 ⎢ ⎢ 2 3 3 0 1 0⎥ ⎯⎯⎯⎯⎯⎯ → 0 1 −1 −2 1 0⎥ ⎯⎯⎯⎯⎯ → 0 1 −1 −2 1 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 3 3 7 0 0 1⎦ ⎣ 0 0 1 −3 0 1⎦ ⎣0 0 1 −3 0 1⎦ 3R 2 + R1 ⎡ 1 0 0 12 −1 −3⎤ R3 + R2 ⎯⎯⎯⎯⎯ → ⎢0 1 0 −5 1 1⎥ ⎢ ⎥ 1⎦ ⎣0 0 1 −3 0
⎡ 12 −1 −3⎤ The inverse matrix is ⎢ −5 1 1⎥ . Multiply each side of Eq. (1) by the inverse matrix. ⎢ ⎥ 1⎥⎦ ⎣⎢ −3 0 ⎡ 12 −1 −3⎤ ⎡ 1 1 2 ⎤ ⎡ x ⎤ ⎡ 12 −1 −3⎤ ⎡ 4⎤ ⎢ −5 1 1⎥ ⎢ 2 3 3⎥ ⎢ y ⎥ = ⎢ −5 1 1⎥ ⎢ 5⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 1⎥⎦ ⎢⎣ 3 3 7 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ −3 0 1⎥⎦ ⎢⎣14⎥⎦ ⎢⎣ −3 0 ⎡ x ⎤ ⎡ 1⎤ ⎢ y ⎥ = ⎢ −1⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ z ⎦⎥ ⎣⎢ 2 ⎦⎥ The solution is (1, −1, 2). 20.
⎡ 1 2 −1⎤ ⎢ 2 3 −1⎥ ⎢ 3 6 −2 ⎥ ⎣ ⎦
⎡ x ⎤ ⎡ 5⎤ ⎢ y ⎥ = ⎢ 8⎥ ⎢ z ⎥ ⎢14⎥ ⎣ ⎦ ⎣ ⎦
(1)
⎡1 2 − 1 ⎤ ⎥ ⎢ Find the inverse matrix of ⎢ 2 3 − 1 ⎥. ⎢⎣ 3 6 − 2⎥⎦ −2 R1 + R 2 ⎡ 1 2 −1 1 0 0⎤ −3R + R ⎡ 1 2 −1 1 0 0⎤ −1R ⎡ 1 2 −1 1 0 0 ⎤ 1 3 → ⎢ 0 −1 2 → ⎢ 0 1 −1 ⎢ 2 3 −1 0 1 0⎥ ⎯⎯⎯⎯⎯⎯ 1 −2 1 0⎥ ⎯⎯⎯⎯ 2 −1 0 ⎥ ⎢ 3 6 −2 0 0 1⎥ ⎢0 0 1 −3 0 1⎥ ⎢0 0 1 −3 0 1⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ − R 3 + R1 1 0 1 −3 2 0⎤ ⎡ 1 0 0 0 2 −1⎤ R3 + R2 −2 R 2 + R1 ⎡⎢ ⎯⎯⎯⎯⎯⎯ → 0 1 −1 2 −1 0⎥ ⎯⎯⎯⎯⎯ → ⎢0 1 0 −1 −1 1⎥ ⎢ ⎥ ⎢0 0 1 −3 0 1⎥ ⎣ ⎦ ⎣0 0 1 −3 0 1⎦ ⎡ 0 2 −1⎤ The inverse matrix is ⎢ −1 −1 1⎥ . Multiply each side of Eq. (1) by the inverse matrix. ⎢ ⎥ ⎣ −3 0 1⎦
⎡ 0 2 −1⎤ ⎡ 1 2 −1⎤ ⎢ −1 −1 1⎥ ⎢ 2 3 −1⎥ ⎢ ⎥ ⎢ ⎥ ⎣ −3 0 1⎦ ⎣ 3 6 −2 ⎦
⎡ x ⎤ ⎡ 0 2 −1⎤ ⎡ 5⎤ ⎢ y ⎥ = ⎢ −1 −1 1⎥ ⎢ 8⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ z ⎦ ⎣ −3 0 1⎦ ⎣14⎦ ⎡ x ⎤ ⎡ 2⎤ ⎢ y ⎥ = ⎢ 1⎥ ⎢ ⎥ ⎢ ⎥ ⎣ z ⎦ ⎣ −1⎦
The solution is (2, 1, −1).
Copyright © Houghton Mifflin Company. All rights reserved.
722
21.
Chapter 10: Matrices
⎡ 1 2 2 ⎤ ⎡ x ⎤ ⎡ 5⎤ ⎢ −2 −5 −2 ⎥ ⎢ y ⎥ = ⎢ 8⎥ ⎢⎣ 2 4 7 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣19⎥⎦
(1)
⎡ 1 2 2⎤ Find the inverse matrix of ⎢ −2 −5 −2 ⎥ . ⎢⎣ 2 4 7 ⎥⎦ 2 R1 + R 2 1 0 0⎤ 1 2 2 1 0 0⎤ ⎡ 1 2 2 1 0 0 ⎤ −2 R ⎡1 2 2 −1R 2 ⎡⎢ 1 + R 3 → ⎢ 0 −1 2 ⎢ −2 −5 −2 0 1 0⎥ ⎯⎯⎯⎯⎯⎯ 2 1 0⎥ ⎯⎯⎯⎯ → 0 1 −2 −2 −1 0⎥ ⎢ 2 4 7 0 0 1⎥ ⎢ 0 0 3 −2 0 1⎥ ⎢0 0 3 −2 0 1⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 2R 3 + R 2 ⎡1 0 0 ⎡1 0 ⎡1 2 9 2 −2 ⎤ 6 5 2 0⎤ 2 1 0 0⎤ ⎥ (1/ 3) R 3 ⎢ −6 R 3 + R1 ⎢ −2 R 2 + R1 ⎢ ⎥ ⎥ 10 → ⎢ 0 1 −2 −2 −1 0⎥ ⎯⎯⎯⎯⎯⎯ → ⎢0 1 0 − 3 −1 23 ⎥ ⎯⎯⎯⎯⎯ → ⎢0 1 −2 −2 −1 0⎥ ⎯⎯⎯⎯⎯⎯ 1⎥ ⎢0 0 1 − 2 1 − 23 0 13 ⎥⎦ 1 − 23 1 13 ⎥⎦ ⎢⎣0 0 ⎢⎣ 0 0 0 3 3⎦ ⎣
⎡ 9 2 −2 ⎤ ⎢ ⎥ The inverse matrix is ⎢ − 10 −1 32 ⎥ . Multiply each side of Eq. (1) by the inverse matrix. 3 ⎢ 2 1⎥ ⎢⎣ − 3 0 3 ⎥⎦ ⎡ 9 2 −2 ⎤ ⎡ 9 2 −2 ⎤ ⎢ 10 ⎥ ⎡ 1 2 2 ⎤ ⎡ x ⎤ ⎢ 10 ⎥ ⎡ 5⎤ 2 ⎢ − 3 −1 3 ⎥ ⎢ −2 −5 −2 ⎥ ⎢ y ⎥ = ⎢ − 3 −1 32 ⎥ ⎢ 8⎥ ⎢ ⎢ ⎥ ⎢ 2 4 7 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢ − 2 0 1⎥ ⎣ 2 1 ⎥ ⎣19 ⎦ 3 ⎥⎦ 3 ⎦⎥ ⎣⎢ − 3 0 ⎣⎢ 3 ⎡ x ⎤ ⎡ 23⎤ ⎢ y ⎥ = ⎢ −12 ⎥ ⎢ z ⎥ ⎢ 3⎥ ⎣ ⎦ ⎣ ⎦
The solution is (23, −12, 3). 22.
⎡ 1 −1 3⎤ ⎡ x ⎤ ⎡ 5⎤ ⎢ 3 −1 10 ⎥ ⎢ y ⎥ = ⎢16 ⎥ ⎢⎣ 2 −2 5⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ 9 ⎥⎦
(1)
⎡ 1 −1 3⎤ Find the inverse matrix of ⎢ 3 −1 10 ⎥ . ⎢⎣ 2 −2 5⎥⎦ −3R1 + R 2 1 −1 3 1 ⎡ 1 −1 3 1 0 0 ⎤ −2 R1 + R 3 ⎡⎢ ⎢ 3 −1 10 0 1 0 ⎥ ⎯⎯⎯⎯⎯⎯ → 0 2 1 −3 ⎢0 0 −1 −2 ⎢⎣ 2 −2 5 0 0 1⎥⎦ ⎣ 1 0 0⎤ ⎡ 1 −1 ⎡ 1 −1 3 −1R 3 ⎢ (1/ 2) R 2 ⎢ → 0 1 ⎯⎯⎯⎯⎯ → 0 1 12 − 23 12 0⎥ ⎯⎯⎯⎯ ⎢ ⎢ ⎥ ⎣⎢0 0 ⎣⎢0 0 −1 −2 0 1⎦⎥ ⎡ 1 0 72 − 12 R 2 + R1 ⎢ ⎯⎯⎯⎯⎯ → ⎢ 0 1 21 − 23 ⎢0 0 1 2 ⎣
0 0⎤ 1 0⎥ 0 1⎥⎦ 3 1 0 0⎤ 1 −3 1 0⎥ 2 2 2 ⎥ 1 2 0 −1⎦⎥
0⎤ ( −7 / 2) R 3 + R1 ⎡ 1 0 0 − 15 2 ⎥ ( −1/ 2) R 3 + R 2 ⎢ ⎯ → ⎢0 1 0 − 25 0⎥ ⎯⎯⎯⎯⎯⎯⎯ ⎢0 0 1 0 −1⎥ 2 ⎦ ⎣ 1 2 1 2
1 2 1 2
7⎤ 2 1⎥ 2⎥
0 −1⎥ ⎦
7⎤ ⎡ − 15 1 2⎥ ⎢ 25 12 1 . Multiply each side of Eq. (1) by the inverse matrix. The inverse matrix is ⎢ − 2 2 2⎥ ⎢ 2 0 −1⎥ ⎢⎣ ⎥⎦
⎡ − 15 ⎢ 25 ⎢ −2 ⎢ 2 ⎣⎢
⎡ − 15 ⎡ 1 −1 3⎤ ⎡ x ⎤ ⎢ 2 ⎢ 3 −1 10 ⎥ ⎢ y ⎥ = ⎢ − 5 2 ⎢ ⎥ ⎢ ⎥ ⎥ 0 −1 ⎣ 2 −2 5⎦ ⎣ z ⎦ ⎢ 2 ⎦⎥ ⎣⎢ ⎡ x ⎤ ⎡ 2⎤ ⎢ y ⎥ = ⎢ 0⎥ ⎢ z ⎥ ⎢ 1⎥ ⎣ ⎦ ⎣ ⎦ 1 2 1 2
7⎤ 2⎥ 1 2⎥
1 2 1 2
7⎤ 2⎥ 1 2⎥
⎡ 5⎤ ⎢16⎥ ⎢ ⎥ 0 −1⎥ ⎣ 9 ⎦ ⎦⎥
The solution is (2, 0, 1).
Copyright © Houghton Mifflin Company. All rights reserved.
Section 10.3
23.
⎡1 ⎢2 ⎢ ⎢2 ⎢ ⎢⎣ 3
723
2 0 1⎤ ⎡ w⎤ ⎡ 6 ⎤ 5 1 2 ⎥⎥ ⎢⎢ x ⎥⎥ ⎢⎢10 ⎥⎥ = 4 1 1⎥ ⎢ y ⎥ ⎢ 8⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 6 0 4 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣16 ⎥⎦
(1)
⎡1 ⎢ 2 Find the inverse matrix of ⎢ ⎢2 ⎢ ⎢⎣ 3 ⎡1 ⎢ ⎢2 ⎢2 ⎢ ⎣⎢ 3
1 1 0 0 0⎤ ⎥ 5 1 2 0 1 0 0⎥ 4 1 1 0 0 1 0⎥ ⎥ 6 0 4 0 0 0 1⎦⎥ 2 0
1⎤ ⎥ 5 1 2⎥ . 4 1 1⎥ ⎥ 6 0 4⎥⎦ −2 R1 + R 2 −2 R1 + R 3 ⎡ 1 −3R1 + R 4 ⎢0 ⎯⎯⎯⎯⎯⎯ →⎢ ⎢0 ⎣⎢0 2 0
2 1 0 0
0 1 1 1 0 −2 1 −1 −2 0 1 −3
0 1 0 0
0 0 1 0
0⎤ ⎡1 0⎥ −2 R 2 + R1 ⎢0 →⎢ ⎥ ⎯⎯⎯⎯⎯⎯ 0⎥ ⎢0 ⎢⎣0 1⎥⎦
0 −2 1 5 −2 0 0 ⎤ 1 1 −0 −2 1 0 0⎥ ⎥ 0 1 −1 −2 0 1 0⎥ 0 0 1 −3 0 0 1⎥⎦
R 4 + R1 0 −1 1 −2 2 0⎤ − R 4 + R 2 ⎡ 1 0 0 0 −2 −2 2 1⎤ ⎢0 1 0 0 0 1 0 1 −1 0 ⎥ 3 1 −1 −1⎥ R4 + R3 ⎥ ⎯⎯⎯⎯⎯→ ⎢ ⎥ 1 −1 −2 0 1 0⎥ ⎢0 0 1 0 −5 0 1 1⎥ ⎢⎣0 0 0 1 −3 0 0 1⎥⎦ 0 1 −3 0 0 1⎥⎦ 1⎤ ⎡ −2 −2 2 ⎥ ⎢ 3 1 1 1⎥ − − . Multiply each side of Eq. (1) by the inverse matrix. The inverse matrix is ⎢ ⎢− 5 0 1 1⎥ ⎥ ⎢ 1⎥⎦ ⎢⎣ − 3 0 0 2 R 3 + R1 ⎡1 ⎢0 −1R 3 + R 2 ⎯⎯⎯⎯⎯⎯ →+⎢ ⎢0 ⎢⎣0
0 1 0 0
⎡ −2 −2 2 1⎤ ⎡ 1 2 ⎢ −3 1 −1 −1⎥ ⎢ 2 5 ⎢ ⎥ ⎢ ⎢ −5 0 1 1⎥ ⎢ 2 4 ⎢ −3 0 0 1⎥ ⎢ 3 6 ⎣ ⎦ ⎣
0 1 1 0
1⎤ 2⎥ ⎥ 1⎥ 4 ⎥⎦
⎡ w⎤ ⎡ −2 −2 2 1⎤ ⎡ 6⎤ ⎢ x⎥ ⎢ 3 1 −1 −1⎥ ⎢10⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ y ⎥ ⎢ −5 0 1 1⎥ ⎢ 8⎥ ⎢ z ⎥ ⎢ −3 0 0 1⎥ ⎢16⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ w⎤ ⎡ 0⎤ ⎢ x ⎥ ⎢ 4⎥ ⎢ ⎥=⎢ ⎥ ⎢ y ⎥ ⎢ −6 ⎥ ⎢ z ⎥ ⎢ −2 ⎥ ⎣ ⎦ ⎣ ⎦
The solution is (0, 4, −6, −2).
Copyright © Houghton Mifflin Company. All rights reserved.
724
24.
Chapter 10: Matrices
⎡1 ⎢2 ⎢ ⎢3 ⎢ ⎢⎣ 1
−1 2 −1 6 −2 9
0⎤ 2 ⎥⎥ 4⎥ ⎥ −2 0 −1⎥⎦
⎡ w⎤ ⎡ 5⎤ ⎢ x ⎥ ⎢ 16 ⎥ ⎢ ⎥=⎢ ⎥ ⎢ y ⎥ ⎢ 28⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ z ⎥⎦ ⎢⎣ 2 ⎥⎦
(1)
⎡ 1 −1 ⎢ 2 −1 Find the inverse matrix of ⎢ ⎢3 − 2 ⎢ ⎢⎣ 1 − 2 ⎡ 1 −1 ⎢ ⎢2 − 1 ⎢3 − 2 ⎢ ⎣⎢ 1 − 2
1 0 0 0⎤ ⎥ 6 2 0 1 0 0⎥ 9 4 0 0 1 0⎥ ⎥ 0 − 1 0 0 0 1⎦⎥ 2
0
0⎤ ⎥ 2⎥ . 4⎥ ⎥ 0 − 1⎥⎦ −2 R1 + R 2 1 0 −3R1 + R 3 ⎡ 1 −1 2 0 −1R1 + R 4 ⎢0 1 2 2 −2 1 ⎯⎯⎯⎯⎯⎯ →⎢ ⎢0 1 3 4 −3 0 ⎣⎢0 −1 −2 −1 −1 0 2
6 9
1 0 −1R 2 + R 3 ⎡ 1 −1 2 0 1R 2 + R 4 ⎢0 1 2 2 −2 1 ⎯⎯⎯⎯⎯⎯ →⎢ ⎢0 0 1 2 −1 −1 ⎢⎣0 0 0 1 −3 1 − 4 R 3 + R1 ⎡ 1 − 2R 3 + R 2 ⎢0 ⎯⎯⎯⎯⎯⎯→ ⎢ ⎢0 ⎢⎣0
0 1 0 0
0 0 1 0
0 −6 3 5 0 −2 0 3 1 2 −1 −1 0 1 −3 1
0⎤ ⎡1 0 4 0⎥ R 2 + R1 ⎢ 0 1 2 →⎢ ⎥ ⎯⎯⎯⎯⎯ 0⎥ ⎢0 0 1 ⎢⎣0 0 0 1⎥⎦ 6 R 4 + R1 −4 0⎤ 2 R 4 + R 2 ⎡1 −2 0⎥ − 2R 4 + R 3 ⎢0 ⎥ ⎯⎯⎯⎯⎯⎯→ ⎢ 1 0⎥ ⎢0 ⎢⎣0 0 1⎥⎦
0 0 1 0
0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦
2 2 2 1
−1 1 −2 1 −1 −1 −3 1
0 1 0 0
0 0 1 0
0 0 1 0
0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦
0 −15 11 −4 6⎤ 0 −6 5 −2 2 ⎥ ⎥ 0 5 −3 1 −2 ⎥ 1 −3 1 0 1⎥⎦
6⎤ ⎡ −15 11 −4 ⎥ ⎢ 6 5 2 2 − − ⎥ . Multiply each side of Eq. (1) by the inverse matrix. The inverse matrix is ⎢ ⎢ 5 −3 1 − 2⎥ ⎥ ⎢ 1 0 1⎥⎦ ⎢⎣ − 3
⎡ −15 11 −4 6 ⎤ ⎡ 1 ⎢ −6 5 −2 2 ⎥ ⎢ 2 ⎢ ⎥ ⎢ 1 −2 ⎥ ⎢ 3 ⎢ 5 −3 ⎢ −3 1 0 1⎥⎦ ⎢⎣ 1 ⎣
−1 −1 −2 −2
2 0⎤ 6 2⎥ ⎥ 9 4⎥ 0 −1⎥⎦
⎡ w⎤ ⎡ −15 11 −4 6⎤ ⎡ 5⎤ ⎢ x ⎥ ⎢ −6 5 −2 2⎥ ⎢ 16⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ 1 −2⎥ ⎢ 28⎥ ⎢ y ⎥ ⎢ 5 −3 ⎢ z ⎥ ⎢ −3 1 0 1⎥⎦ ⎢⎣ 2⎥⎦ ⎣ ⎦ ⎣ ⎡ w⎤ ⎡ 1⎤ ⎢ x ⎥ ⎢ −2 ⎥ ⎢ ⎥=⎢ ⎥ ⎢ y ⎥ ⎢ 1⎥ ⎢ z ⎥ ⎢ 3⎥ ⎣ ⎦ ⎣ ⎦
The solution is (1, −2, 1, 3). 25.
The average temperature for the two points, 35 + 50 + x2 + 60 145 + x2 = x1 = or 4 x1 − x2 =145 4 4 x + 50 + 55 + 60 165 + x1 = x2 = 1 or − x1 + 4 x2 =165 4 4 The system of equations and associated matrix equation are ⎧ 4 x1 − x2 =145 ⎡ 4 −1⎤ ⎡ x1 ⎤ ⎡145⎤ ⎨ ⎢ −1 4 ⎥ ⎢ x ⎥ = ⎢165⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦ ⎩− x1 + 4 x2 =165 ⎡ x ⎤ ⎡ 49.7 ⎤ Solving the matrix equation by using an inverse matrix gives ⎢ 1 ⎥ = ⎢ ⎥ ⎣ x2 ⎦ ⎣53.7 ⎦ The temperatures are x1 = 49.7°F, x2 = 53.7°F.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 10.3
26.
725
The average temperature for the two points, 40 + 25 + x2 + 40 105 + x2 x1 = or 4 x1 − x2 =105 = 4 4 x + 25 + 60 + 40 125 + x1 or − x1 + 4 x2 =125 x2 = 1 = 4 4 The system of equations and associated matrix equation are ⎧ 4 x1 − x2 =105 ⎡ 4 −1⎤ ⎡ x1 ⎤ ⎡105⎤ ⎨ ⎢ −1 4 ⎥ ⎢ x ⎥ = ⎢125⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦ ⎩− x1 + 4 x2 =125 ⎡ x ⎤ ⎡36.3⎤ Solving the matrix equation by using an inverse matrix gives ⎢ 1 ⎥ = ⎢ ⎥ ⎣ x2 ⎦ ⎣ 40.3⎦ The temperatures are x1 = 36.3°F, x2 = 40.3°F.
27.
The average temperature for the two points, 50 + 60 + x2 + x3 110 + x2 + x3 or 4 x1 − x2 − x3 =110 = x1 = 4 4 x + 60 + 60 + x4 120 + x1 + x4 = x2 = 1 or − x1 + 4 x2 − x4 =120 4 4 50 + x1 + x4 + 50 100 + x1 + x4 or − x1 + 4 x3 − x4 =100 = x3 = 4 4 x + x + 60 + 50 110 + x2 + x3 = x4 = 3 2 or − x2 − x3 + 4 x4 =110 4 4 The system of equations and associated matrix equation are ⎧ 4 x1 − x2 − x3 =110 ⎡ 4 −1 −1 0 ⎤ ⎡ x1 ⎤ ⎡110 ⎤ ⎪ ⎢ −1 4 0 −1⎥ ⎢ x ⎥ ⎢120 ⎥ ⎪ − x1 + 4 x2 − x4 =120 ⎢ ⎥⎢ 2⎥=⎢ ⎥ ⎨ ⎢ −1 0 4 −1⎥ ⎢ x3 ⎥ ⎢100 ⎥ ⎪ − x1 + 4 x3 − x4 =100 ⎢ 0 −1 −1 4 ⎥ ⎢ x ⎥ ⎢110⎥ ⎪⎩− x2 − x3 + 4 x4 =110 ⎣ ⎦⎣ 4⎦ ⎣ ⎦ ⎡ x1 ⎤ ⎡ 55 ⎤ ⎢ x ⎥ ⎢57.5⎥ ⎥ Solving the matrix equation by using an inverse matrix gives ⎢ 2 ⎥ = ⎢ ⎢ x3 ⎥ ⎢52.5⎥ ⎢ x ⎥ ⎢ 55 ⎥ ⎦ ⎣ 4⎦ ⎣
The temperatures are x1 = 55°F, x2 = 57.5°F, x3 = 52.5°F, x4 = 55°F. 28.
The average temperature for the two points, 55 + 70 + x2 + x3 125 + x2 + x3 or 4 x1 − x2 − x3 =125 = x1 = 4 4 x + 70 + 65 + x4 135 + x1 + x4 = x2 = 1 or − x1 + 4 x2 − x4 =135 4 4 55 + x1 + x4 + 40 95 + x1 + x4 or − x1 + 4 x3 − x4 = 95 = x3 = 4 4 x + x + 65 + 40 105 + x2 + x3 = x4 = 3 2 or − x2 − x3 + 4 x4 =105 4 4 The system of equations and associated matrix equation are ⎧ 4 x1 − x2 − x3 =125 ⎡ 4 −1 −1 0 ⎤ ⎡ x1 ⎤ ⎡125⎤ ⎪ ⎢ −1 4 0 −1⎥ ⎢ x ⎥ ⎢135⎥ ⎪ − x1 + 4 x2 − x4 =135 ⎢ ⎥⎢ 2⎥=⎢ ⎥ ⎨ ⎢ −1 0 4 −1⎥ ⎢ x3 ⎥ ⎢ 95 ⎥ ⎪ − x1 + 4 x3 − x4 = 95 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0 −1 −1 4 ⎦ ⎣ x4 ⎦ ⎣105⎦ ⎩⎪− x2 − x3 + 4 x4 =105 ⎡ x1 ⎤ ⎡ 60 ⎤ ⎢ x ⎥ ⎢ 62.5⎥ ⎥ Solving the matrix equation by using an inverse matrix gives ⎢ 2 ⎥ = ⎢ ⎢ x3 ⎥ ⎢52.5⎥ ⎢ x ⎥ ⎢ 55 ⎥ ⎦ ⎣ 4⎦ ⎣
The temperatures are x1 = 60°F, x2 = 62.5°F, x3 = 52.5°F, x4 = 55°F. Copyright © Houghton Mifflin Company. All rights reserved.
726
29.
Chapter 10: Matrices
A = number of adult tickets C = number of child tickets Saturday
A + C =100 20 A +15C =1900 ⎡ 1 1⎤ ⎡ A ⎤ ⎡ 100⎤ ⎢ 20 15⎥ ⎢C ⎥ = ⎢1900⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
1⎤ 1⎤ ⎡ −3 1 ⎤ ⎡ A⎤ ⎡ −3 5⎥ ⎡ 1 5 ⎥ ⎡ 100 ⎤ ⎢ ⎢ = ⎢ 4 − 1 ⎥ ⎢⎣ 20 15⎥⎦ ⎢⎣C ⎥⎦ ⎢ 4 − 1 ⎥ ⎢⎣1900⎥⎦ 5⎦ 5⎦ ⎣ ⎣ ⎡ A ⎤ ⎡ 80 ⎤ ⎢C ⎥ = ⎢ 20 ⎥ ⎣ ⎦ ⎣ ⎦ On Saturday, 80 adults and 20 children took the tour.
Sunday
A + C =120 20 A +15C = 2275 ⎡ 1 1⎤ ⎡ A ⎤ ⎡ 120 ⎤ ⎢ 20 15⎥ ⎢C ⎥ = ⎢ 2275⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
1⎤ ⎡ −3 5⎥ ⎢ ⎢ 4 − 1⎥ 5⎦ ⎣
1⎤ ⎡ A⎤ ⎡⎢ −3 5 ⎥ ⎡ 120 ⎤ ⎢C ⎥ = ⎢ ⎢ ⎥ 1 ⎣ ⎦ ⎣ 4 − 5 ⎥⎦ ⎣ 2275⎦ ⎡ A ⎤ ⎡ 95⎤ ⎢C ⎥ = ⎢ 25⎥ ⎣ ⎦ ⎣ ⎦ On Sunday, 95 adults and 25 children took the tour.
30.
⎡1 1⎤ ⎢ 20 15⎥ ⎣ ⎦
S = number of standard models D = number of deluxe models January:
S + D = 90 45S + 60 D = 4650 ⎡ 1 1⎤ ⎡ S ⎤ ⎡ 90 ⎤ ⎢ 45 60 ⎥ ⎢ D ⎥ = ⎢ 4650 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ 4 − 1 ⎤ ⎡ 1 1 ⎤ ⎡ S ⎤ ⎡ 4 − 1 ⎤ ⎡ 90⎤ 15 ⎥ 15 ⎥ ⎢ =⎢ 1 ⎥ ⎢ 45 60 ⎥ ⎢ D ⎥ ⎢ −3 1 ⎥ ⎢ 4650 ⎥ ⎢ −3 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 15 ⎦ 15 ⎦ ⎣ ⎣ ⎡ S ⎤ ⎡ 50 ⎤ ⎢ D ⎥ = ⎢ 40 ⎥ ⎣ ⎦ ⎣ ⎦
In January, 50 standard models and 40 deluxe models were manufactured. February
S + D =100 45S + 60 D = 5250 ⎡ 1 1⎤ ⎡ S ⎤ ⎡ 100 ⎤ ⎢ 45 60 ⎥ ⎢ D ⎥ = ⎢5250⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ 4 − 1 ⎤ ⎡ 1 1 ⎤ ⎡ S⎤ ⎡ 4 − 1 ⎤ 15 ⎥ 15 ⎥ ⎢ =⎢ 1 ⎥ ⎢ 45 60 ⎥ ⎢ D ⎥ ⎢ −3 1⎥ ⎢ −3 ⎣ ⎦ ⎣ ⎦ 15 ⎦ 15 ⎦ ⎣ ⎣ ⎡ S ⎤ ⎡50 ⎤ ⎢ D ⎥ = ⎢50 ⎥ ⎣ ⎦ ⎣ ⎦
⎡ 100 ⎤ ⎢5250 ⎥ ⎣ ⎦
In February, 50 standard models and 50 deluxe models were manufactured.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 10.3
31.
727
x1 = number of 100-gram portions of additive 1 x2 = number of 100-gram portions of additive 2 x3 = number of 100-gram portions of additive 3
30 x1 + 40 x2 + 50 x3 = 380 10 x1 + 15 x2 + 5 x3 = 95
Sample 1:
10 x1 + 10 x2 + 5 x3 = 85 ⎡30 40 50⎤ ⎡ x1 ⎤ ⎡380⎤ ⎢10 15 5⎥ ⎢ x ⎥ = ⎢ 95⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎣⎢10 10 5⎦⎥ ⎣⎢ x3 ⎦⎥ ⎣⎢ 85⎦⎥
⎡− 1 − 6 35 ⎢ 70 1 ⎢ 0 5 ⎢ ⎢ 1 −2 35 ⎣ 35
11 ⎤ 35 ⎥ − 15 ⎥ ⎥ 1⎥ − 35 ⎦
11 ⎤ ⎡− 1 − 6 35 35 ⎥ ⎡380 ⎤ ⎡30 40 50⎤ ⎡ x1 ⎤ ⎢ 70 ⎢10 15 5⎥ ⎢ x ⎥ = ⎢ 0 1 1 ⎥ ⎢ 95⎥ − 2 5 5⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢⎣10 10 5⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢ 1 − 2 − 1 ⎥ ⎢⎣ 85⎥⎦ 35 35 ⎦ ⎣ 35 ⎡ S ⎤ ⎡50⎤ ⎢ D ⎥ = ⎢50⎥ ⎣ ⎦ ⎣ ⎦ For Sample 1, 500 g of additive 1, 200 g of additive 2, and 300 g of additive 3 are required.
30 x1 + 40 x 2 + 50 x 3 = 380 10 x1 + 15 x 2 + 5 x 3 = 110 10 x1 + 10 x 2 + 5 x 3 = 90
Sample 2:
⎡30 40 50⎤ ⎡ x1 ⎤ ⎢10 15 5⎥ ⎢ x ⎥ = ⎢ ⎥ ⎢ 2⎥ ⎢⎣10 10 5⎥⎦ ⎢⎣ x3 ⎥⎦ ⎡− 1 − 6 35 ⎢ 70 1 ⎢ 0 5 ⎢ ⎢ 1 −2 35 ⎣ 35
11 ⎤ 35 ⎥ − 15 ⎥ ⎥ 1⎥ − 35 ⎦
⎡380⎤ ⎢100⎥ ⎢ ⎥ ⎢⎣ 90⎥⎦
⎡− 1 − 6 35 ⎡30 40 50⎤ ⎡ x1 ⎤ ⎢ 70 ⎢10 15 5⎥ ⎢ x ⎥ = ⎢ 0 1 5 ⎢ ⎥ ⎢ 2⎥ ⎢ ⎣⎢10 10 5⎦⎥ ⎣⎢ x3 ⎦⎥ ⎢ 1 − 2 35 ⎣ 35 ⎡ x1 ⎤ ⎡ 4⎤ ⎢ x ⎥ = ⎢ 4⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ x3 ⎥⎦ ⎢⎣ 2⎥⎦
11 ⎤ 35 ⎥ − 15 ⎥ ⎥ 1 ⎥ − 35 ⎦
⎡380⎤ ⎢110⎥ ⎢ ⎥ ⎣⎢ 90⎦⎥
For Sample 2, 400 g of additive 1, 400 g of additive 2, and 200 g of additive 3 are required.
Copyright © Houghton Mifflin Company. All rights reserved.
728
32.
Chapter 10: Matrices
x1 = number of 100-gram portions of Food Type I x2 = number of 100-gram portions of Food Type II x3 = number of 100-gram portions of Food Type III
First diet: 13x1 + 4 x2 + 10 x1 + 4 x2
x3 = 23 = 18
13x1 + 4 x2 + 10x3 = 39
⎡13 4 1⎤ ⎢10 4 0⎥ ⎢ ⎥ ⎣13 3 10⎦ ⎡ 20 49 ⎢ 50 − ⎢ 49 ⎢ 11 ⎣⎢ − 49
− 37 98 117 98 13 98
2 ⎤ − 49 ⎥ 5 49 ⎥ 6 ⎥ 49 ⎦⎥
⎡13 4 1⎤ ⎢10 4 0⎥ ⎢ ⎥ ⎣13 3 10⎦
⎡ x1 ⎤ ⎢x ⎥= ⎢ 2⎥ ⎣ x3 ⎦
⎡ 23⎤ ⎢ 18⎥ ⎢ ⎥ ⎣39⎦
⎡ 20 ⎡ x1 ⎤ ⎢ 49 ⎢ x ⎥ = ⎢ − 50 49 ⎢ 2⎥ ⎣ x3 ⎦ ⎢⎢ − 11 ⎣ 49 ⎡ x1 ⎤ ⎡ 1⎤ ⎢ x ⎥ = ⎢2⎥ ⎢ 2⎥ ⎢ ⎥ ⎣ x3 ⎦ ⎣ 2 ⎦
− 37 98 117 98 13 98
2⎤ − 49 ⎥ 5 49 ⎥ 6⎥ 49 ⎦⎥
⎡ 23⎤ ⎢ 18⎥ ⎢ ⎥ ⎣39⎦
For the first diet, 100g of Food Type I, 200g of Food Type II, and 200g of Food Type III are required. 13x1 + 4 x2 + x3 = 35 Second diet: 10 x1 + 4 x2
= 28
13x1 + 4 x2 + 10x3 = 42
⎡13 4 1⎤ ⎢10 4 0⎥ ⎢ ⎥ ⎣13 3 10⎦ ⎡ 20 49 ⎢ 50 ⎢ − 49 ⎢ 11 ⎣⎢ − 49
− 37 98 117 98 13 98
2 ⎤ − 49 ⎥ 5 49 ⎥ 6 ⎥ 49 ⎦⎥
⎡13 4 1⎤ ⎢10 4 0⎥ ⎢ ⎥ ⎣13 3 10⎦
⎡ x1 ⎤ ⎢x ⎥= ⎢ 2⎥ ⎣ x3 ⎦
⎡ 35⎤ ⎢ 28⎥ ⎢ ⎥ ⎣42⎦
⎡ 20 ⎡ x1 ⎤ ⎢ 49 ⎢ x ⎥ = ⎢ − 50 ⎢ 2 ⎥ ⎢ 49 ⎣ x3 ⎦ ⎢ − 11 ⎣ 49 ⎡ x1 ⎤ ⎡ 2⎤ ⎢ x ⎥ = ⎢ 2⎥ ⎢ 2⎥ ⎢ ⎥ ⎣ x3 ⎦ ⎣1 ⎦
− 37 98 117 98 13 98
2⎤ − 49 ⎥ 5 49 ⎥ 6⎥ 49 ⎦⎥
⎡ 35⎤ ⎢ 28⎥ ⎢ ⎥ ⎣ 42⎦
For the second diet, 200g of Food Type I, 200g of Food Type II, and 100g of Food Type III are required. 33.
Using a calculator,
⎡ 2 −2 3 ⎢ 5 2 −2 ⎢ 6 −1 2 ⎢ 3 −1 ⎣2 34.
−1
5 ⎡ −5.667 −3.667 ⎢ −27.667 −18.667 24 ≈⎢ −19.333 −13.333 17 ⎢ −13 10 ⎣ 15
0.333 ⎤ 2.333 ⎥ 1.667 ⎥ ⎥ −1 ⎦
Using a calculator,
⎡ 3 −1 0 ⎢ 2 −2 −3 ⎢ −1 −3 5 ⎢ ⎣ 5 3 −2 35.
1⎤ 3⎥ 3⎥ ⎥ 5⎦
−1
1⎤ 0⎥ 3⎥ ⎥ 1⎦
⎡ 0.1 0.143 −0.057 ⎢ −0.5 0.071 0.071 ≈⎢ −0.4 0.286 0.086 ⎢ ⎣ 0.2 −0.357 0.243
0.071⎤ 0.286⎥ 0.143⎥ ⎥ 0.071⎦
Using a calculator,
⎡− 2 ⎢ 7 ⎢ ⎢ −2 ⎢ 3 ⎢⎣
− 1⎤ 6⎥ ⎥ −3⎥ 2 3 − 5⎥ ⎥⎦ 4
−1
⎡ −0.150 −0.217 0.302⎤ ≈ ⎢⎢ 0.248 −0.024 0.013⎥⎥ ⎢⎣ 0.217 −0.200 −0.195⎥⎦
Copyright © Houghton Mifflin Company. All rights reserved.
Section 10.3
36.
Using a calculator,
⎡ 6 π ⎢ ⎢ −5 7 ⎢ 5 − 3 ⎢ ⎣ 6 37.
729
− 4⎤ 7⎥ 2⎥ ⎥ 10 ⎥ ⎦
−1
⎡0.097 −0.073 0.064⎤ ≈ ⎢ 0.143 0.159 −0.075⎥ ⎢ 0.053 0.106 0.258⎥ ⎣ ⎦
X = ( I − A) −1 D, where X is consumer demand, I is the identity matrix, A is the input-output matrix, and D is the final demand. Thus −1
⎛ ⎡ 1 0 0⎤ ⎡0.20 0.15 0.10⎤ ⎞ ⎡120 ⎤ X = ⎜ ⎢0 1 0⎥ − ⎢0.10 0.30 0.25⎥ ⎟ ⎢ 60 ⎥ ⎜⎢ ⎥ ⎢0.20 0.10 0.10⎥ ⎟ ⎢ 55⎥ ⎣ ⎦⎠ ⎣ ⎦ ⎝ ⎣0 0 1⎦ −1 ⎡ 0.80 −0.15 −0.10⎤ ⎡120⎤ ⎢ 60⎥ = ⎢ −0.10 0.70 −0.25⎥ ⎢ −0.20 −0.10 0.90 ⎥ ⎢ 55⎥ ⎣ ⎦ ⎣ ⎦ ⎡194.67 ⎤ ≈ ⎢157.03⎥ ⎢121.82⎥ ⎣ ⎦ $194.67 million worth of manufacturing, $157.03 million worth of transportation, $121.82 million worth of services. 38.
X = ( I − A) −1 D, where X is consumer demand, I the identity matrix, A is the input-output, and D is the final demand. Thus −1
⎛ ⎡1 0 0 0 ⎤ ⎡0.10 0.05 0.20 0.15⎤ ⎞ ⎡ 80⎤ ⎜ ⎢0 1 0 0⎥ ⎢0.20 0.10 0.30 0.10 ⎥ ⎟ ⎢100⎥ − ⎢ X =⎜ ⎢ ⎟ ⎥ 0.05 0.30 0.20 0.40 ⎥ ⎟ ⎢ 50⎥ ⎜ ⎢0 0 1 0⎥ ⎢ ⎥ ⎜ 0 0 0 1 ⎟ ⎢ ⎥ ⎦ ⎣0.10 0.20 0.15 0.20 ⎦ ⎠ ⎣ 80⎦ ⎝⎣ −1 ⎡ 0.90 −0.05 −0.20 −0.15⎤ ⎡ 80⎤ ⎢ −0.20 0.90 −0.30 −0.10 ⎥ ⎢100⎥ =⎢ −0.05 −0.30 0.80 −0.40 ⎥ ⎢ 50⎥ ⎢ ⎥ ⎢ ⎥ − ⎣ 0.10 −0.20 −0.15 0.80⎦ ⎣ 80⎦ −1 ⎡1.30 0.40 0.58 0.58⎤ ⎡ 80 ⎤ ⎢ 0.48 1.58 0.84 0.71⎥ ⎢100 ⎥ ≈⎢ (A computer program was used to calculate the inverse matrix.) 0.44 0.92 1.92 1.16 ⎥ ⎢ 50 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣0.36 0.62 0.64 1.72 ⎦ ⎣ 80 ⎦ $219.0 million worth of manufacturing, $294.3 million worth of agriculture, $316.7 million worth of service, $260.3 million worth of transportation. 39.
The input-output matrix, A, is given by ⎡ 0.05 0.20 0.15⎤ A = ⎢0.02 0.03 0.25⎥ ⎢0.10 0.12 0.05⎥ ⎣ ⎦ Consumer demand is given by X = ( I − A) −1 D −1
⎛ ⎡ 1 0 0⎤ ⎡ 0.05 0.20 0.15⎤ ⎞ ⎡ 30 ⎤ ⎜ ⎟ X = ⎜ ⎢0 1 0 ⎥ − ⎢0.02 0.03 0.25⎥ ⎟ ⎢ 5⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎜ ⎢0 0 1⎥ ⎟ ⎢⎣0.10 0.12 0.05⎥⎦ ⎠ ⎢⎣ 25⎥⎦ ⎦ ⎝⎣ ⎡ 0.95 −0.20 −0.15⎤ = ⎢ −0.02 0.97 −0.25⎥ ⎢ ⎥ ⎣⎢ −0.10 −0.12 0.95⎦⎥
−1
⎡30 ⎤ ⎢ 5⎥ ⎢ ⎥ ⎣⎢ 25⎦⎥
⎡39.69 ⎤ ≈ ⎢14.30 ⎥ ⎢ ⎥ ⎣⎢32.30 ⎦⎥ $39.69 million worth of coal, $14.30 million worth of iron, $32.30 million worth of steel.
Copyright © Houghton Mifflin Company. All rights reserved.
730
40.
Chapter 10: Matrices
The input-output matrix, A, is given by ⎡ 0.01 0.08 0.20 ⎤ A = ⎢ 0.03 0.05 0.20 ⎥ ⎢ 0.10 0.15 0.10 ⎥ ⎣ ⎦ Consumer demand is given by X = ( I − A) −1 D ⎛ ⎡ 1 0 0 ⎤ ⎡ 0.01 0.08 0.20 ⎤ ⎞ X = ⎜ ⎢ 0 1 0 ⎥ − ⎢ 0.03 0.05 0.20 ⎥ ⎟ ⎜⎢ ⎥ ⎢ ⎥⎟ ⎝ ⎣ 0 0 1⎦ ⎣ 0.10 0.15 0.10 ⎦ ⎠ −1 ⎡ 0.99 −0.08 −0.20 ⎤ ⎡100 ⎤ ⎢ ⎥ = −0.03 0.95 −0.20 ⎢ 75⎥ ⎢ −0.10 −0.15 0.90 ⎥ ⎢150 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ 152.63⎤ = ⎢ 126.88⎥ ⎢ 204.77 ⎥ ⎣ ⎦
−1
⎡100 ⎤ ⎢ 75⎥ ⎢150 ⎥ ⎣ ⎦
$152.63 million from the plastics division, $126.88 million from the semiconductor division, $204.77 million from the computer division.
....................................................... 41.
42.
Connecting Concepts
⎡ 2 −3⎤ ⎡ −3 15⎤ ⎡2(−3) + (−3)(−2) 2(15) + (−3)(10) ⎤ ⎡0 0⎤ AB = ⎢ ⎥=⎢ ⎥=O ⎥=⎢ ⎥⎢ ⎣− 6 9⎦ ⎣− 2 10⎦ ⎣ − 6(−3) + 9(−2) − 6(15) + 9(10) ⎦ ⎣0 0⎦
Let A be a matrix with an inverse matrix A −1 . If AB=O, then A−1 ( AB) = A−1O ( A−1 A) B = O IB = O B=O
43.
⎡ 2 −1⎤ ⎡3 4 ⎤ AB = ⎢ ⎥ ⎢ ⎥ ⎣ −4 2 ⎦ ⎣ 1 5⎦ 3⎤ ⎡ 2(3) + (−1)(1) 2(4) + (−1)(5) ⎤ ⎡ 5 =⎢ ⎥ = ⎢ −10 −6 ⎥ − + − + 4(3) 2(1) 4(4) 2(5) ⎣ ⎦ ⎣ ⎦
44.
Let A be a matrix with an inverse matrix A −1 . If AB=AC, then A −1 ( AB) = A −1 ( AC ) ( A −1 A) B = ( A −1 A)C
⎡ 2 −1⎤ ⎡ 4 7 ⎤ AC = ⎢ ⎥ ⎢ ⎥ ⎣ −4 2 ⎦ ⎣ 3 11⎦ ⎡ 2(4) + (−1)(3) 2(7) + ( −1)(11) ⎤ ⎡ 5 3⎤ =⎢ = −4(7) + 2(11) ⎥⎦ ⎢⎣ −10 −6 ⎥⎦ ⎣ −4(4) + 2(3) 45.
1 ⎡ a b 1 0 ⎤ a R1 ⎡1 ba ⎯⎯⎯ →⎢ ⎢ c d 0 1⎥ ⎢⎣c d ⎣ ⎦
⎡ 0 ⎤ ( −c) R1 + R2 ⎢1 ⎥ ⎯⎯⎯⎯⎯⎯ → ⎢0 0 1 ⎥⎦ ⎣⎢
d − b R2 + R1 ⎡1 0 ad −bc a ⎯⎯⎯⎯⎯ ⎯ →⎢ ⎢0 1 − c ⎢⎣ ad −bc
−
d ⎡ A−1 = ⎢ ad − bc ⎢− c ⎣ ad − bc
−
1 a
IB = IC B=C
b a ad −bc a
1 a −c a
⎡1 b a R 0⎤ a ⎥ ⎯⎯⎯⎯⎯ ad −bc 2 → ⎢ ⎥ ⎢ 1⎥ 0 1 ⎦ ⎣⎢
b ⎤ ad −bc ⎥ ⎥ a ad −bc ⎥⎦
b ⎤ ad − bc ⎥ = 1 ⎡ d a ⎥ ad − bc ⎢ − c ⎣ ad − bc ⎦
− b⎤ ⎥ a⎦
Copyright © Houghton Mifflin Company. All rights reserved.
1 a − c ad −bc
⎤ ⎥ a ⎥ ad −bc ⎦⎥ 0
Section 10.3
46.
731
d ⎡ A−1 = ⎢ ad − bc ⎢− c ⎣ ad − bc
−
b ⎤ ad − bc ⎥ a ⎥ ad − bc ⎦
A −1 exists if and only if the denominator ad − bc ≠ 0. 47.
a.
a = 2, b = −3, c = 4, d = −5
b.
−b ⎤ ⎡ −5 3 ⎤ 1 ⎡ −5 3⎤ ⎡ − 5 2 1 = = = a ⎥⎦ 2(−5) − (−3)(4) ⎢⎣ −4 2 ⎥⎦ 2 ⎢⎣ −4 2⎥⎦ ⎢⎣ −2 a = 5, b = 6, c = 3, d = 4
c.
−b ⎤ ⎡ 4 −6 ⎤ 1 ⎡ 4 −6 ⎤ ⎡ 2 −3⎤ 1 = = = a ⎥⎦ 5(4) − (6)(3) ⎢⎣ −3 5⎥⎦ 2 ⎢⎣ −3 5⎥⎦ ⎢⎣ − 3 2 5 2 ⎥⎦ a = 0, b = −1, c = 4, d = 4
3 ⎤ 2⎥
1 ⎡ d ad −bc ⎢⎣ −c
1⎦
1 ⎡ d ad −bc ⎢⎣ −c 1 ⎡ d ad −bc ⎢⎣ −c
48.
−b ⎤ ⎡ 4 1⎤ 1 ⎡ 4 1⎤ ⎡ 1 1 = = = a ⎥⎦ 0(4) − (−1)(4) ⎢⎣ −4 0 ⎥⎦ 4 ⎢⎣ −4 0⎥⎦ ⎢⎣ −1
A−1 =
1 ⎡ d ad −bc ⎢⎣ −c
−b ⎤ ⎡ 1 2⎤ 1 ⎡ 1 2⎤ 1 = = a ⎥⎦ 3(1) − (−2)(1) ⎢⎣ −1 3⎥⎦ 5 ⎢⎣ −1 3⎥⎦
B −1 =
1 ⎡ d ad −bc ⎣⎢ −c
−b ⎤ ⎡ 3 1⎤ 1 ⎡ 3 1⎤ 1 = = a ⎦⎥ 2(3) − (−1)(2) ⎣⎢ −2 2 ⎦⎥ 8 ⎣⎢ −2 2⎦⎥
⎛ ⎡3 −2 ⎤ ⎡ 2 −1⎤ ⎞ ( AB )−1 = ⎜ ⎢ ⎟ ⎝ ⎣ 1 1⎥⎦ ⎢⎣ 2 3⎥⎦ ⎠ ⎡ 3 B −1 ⋅ A−1 = ⎢ 8 ⎢⎣ − 1 4 49.
1 ⎤ 8⎥ 1 ⎥ 4⎦
−1
⎡ 1 ⎢ 5 ⎢ 1 ⎣− 5
⎡ 1 ⎡ 2 −9⎤ ⎢ 20 =⎢ = 1 ⎣ 4 2⎥⎦ ⎢ − 10 ⎣ 2 ⎤ ⎡ 1 5⎥ = ⎢ 20 3 ⎥ ⎢− 1 5 ⎦ ⎣ 10
9 1
4⎤ 0 ⎥⎦
1
9 ⎤ 40 ⎥ 1 ⎥ 20 ⎦
⎤
40 ⎥
20 ⎥⎦
( AB )( AB) −1 = I
A−1 ( AB)( AB) −1 = A−1I ( A−1 A) B( AB) −1 = A−1 IB( AB) −1 = A−1 B( AB) −1 = A−1 −1 B B( AB) −1 = B −1 A−1 I ( AB) −1 = B −1 A−1 ( AB) −1 = B −1 A−1
.......................................................
Prepare for Section 10.4 PS2. (−1)i+ j
PS1. 2
(−1) 2+6 = (−1)8 =1
PS3. (−1)1+1 (−3) + (−1)1+2 (−2) + (−1)1+3 (5) = (−1) 2 (−3) + (−1)3 (−2) + (−1) 4 (5) = −3 + ( −1)(−2) + 5 = −3 + 2 + 5 =4
PS4. a23 = 1
1⎤ ⎡ −6 3 ⎤ PS5. 3 ⎡ −2 ⎢ 3 −5⎥ = ⎢ 9 −15⎥ ⎣ ⎦ ⎣ ⎦
2 R1+ R 2 PS6. ⎡ 1 3 −2 ⎤ 1 3 −2 ⎤ −4 R1+ R 3 ⎡ ⎢ −2 −1 1 ⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 5 −3⎥ ⎢4 0 1⎥ ⎢ 0 −12 9 ⎥ ⎣ ⎦ ⎣ ⎦ Copyright © Houghton Mifflin Company. All rights reserved.
732
Chapter 10: Matrices
Section 10.4 1.
2 −1 = 2(5) − (−1)(3) =10 − (−3) =13 3 5
2.
2 9 = 2(2) − (−6)(9) = 4 − (−54) = 58 −6 2
3.
5 0 = 5(−3) − (2)(0) = −15 − 0 = −15 2 −3
4.
0 −8 = 0(4) − (3)(−8) = 0 − (−24) = 24 3 4
5.
4 6 = 4(3) − (2)(6) =12 −12 = 0 2 3
6.
−3 6 = −3(−8) − (4)(6) = 24 − 24 = 0 4 −8
7.
0 9 = 0(−2) − (0)(9) = 0 − 0 = 0 0 −2
8.
−3 9 = −3(0) − (0)(9) = 0 − 0 = 0 0 0
9.
M11 =
10.
M 21 =
4 −1 = 4(6) − (−5)( −1) =19 −5 6
C11 = (−1)1+1 M 11 = M 11 = 19 11.
M 32 =
C21 = (−1) 2 +1 M 21 = − M 21 = −(−27) = 27
5 −3 = 5( −1) − 2(−3) =1 2 −1
12.
C32 = (−1) 3+ 2 M 32 = − M 32 = −1 13.
M 22 =
3 3 = 3(3) − 6(3) = −9 6 3
M 31 =
14.
−2 3 = −2(0) − 3(3) = −9 3 0
M 33 =
5 −2 = 5(4) − 2(−2) = 24 2 4
C33 = ( −1) 3+ 3 M 33 = M 33 = 24
C22 = (−1) 2 + 2 M 22 = M 22 = −9 15.
−2 −3 = −2(6) − ( −5)(−3) = −27 −5 6
M13 =
1 3 =1( −2) − 6(3) = −20 6 −2
C13 = (−1)1+ 3 M 13 = M 13 = −20 16.
C31 = (−1)3+1 M 31 = M 31 = −9
M 23 =
3 −2 = 3(−2) − 6(−2) = 6 6 −2
C23 = (−1) 2 + 3 M 23 = − M 23 = −6
17.
2 −3 1 −3 1 2 1 2 −3 +0 −2 2 0 2 = −2 −2 4 3 4 3 −2 3 −2 4 = −2(−10) + 0 − 2(5) = 20 −10 =10
18.
3 1 −2 2 −5 −5 4 2 4 2 −5 4 = 3 −1 + (−2) 2 1 3 1 3 2 3 2 1 = 3(−13) −1( −10) − 2(19) = −39 +10 − 38 = −67
19.
−2 3 2 2 −3 1 −3 1 2 −3 +2 1 2 −3 = −2 −2 −4 1 −4 −2 1 −4 −2 1 = −2(−4) − 3(−11) + 2(6) = 8 + 33+12 = 53
20.
3 −2 0 2 2 2 −3 −3 2 2 −3 2 = 3 − (−2) +0 8 5 8 −2 −2 5 8 −2 5
2 −3 10 2 −3 −3 10 −3 10 0 2 −3 = 2 −0 +0 0 5 0 5 2 −3 0 0 5 = 2(10) − 0 + 0 = 20
22.
21.
= 3(−11) + 2(−6) + 0 = −33 −12 = −45 6 0 0 −3 0 2 0 2 −3 −0 +0 2 −3 0 = 6 −8 2 7 2 7 −8 7 −8 2 = 6(−6) − 0 + 0 = −36
Copyright © Houghton Mifflin Company. All rights reserved.
Section 10.4
733
23.
0 −2 4 0 −7 1 −7 1 0 − (−2) +4 1 0 −7 = 0 −6 0 5 0 5 −6 5 −6 0 = 0 + 2(35) + 4(−6) = 70 − 24 = 46
24.
5 −8 0 0 −7 2 −7 2 0 − (−8) +0 2 0 −7 = 5 −2 −1 0 −1 0 −2 0 −2 −1 = 5(−14) + 8(−2) + 0 = −70 −16 = −86
25.
4 −3 3 1 −4 2 −4 2 1 − ( −3) +3 2 1 −4 = 4 −2 −1 6 −1 6 −2 6 −2 −1 = 4(−9) + 3(22) + 3(−10) = −36 + 66 − 30 =0
26.
3 9 −2 3 9 3 9 −2 −6 4 −2 −6 = −2 −4 +0 −8 −24 −8 −24 −2 −6 0 −8 −24 = −2(0) − 4(0) + 0 =0
27.
Row 2 consists entirely of zeros. Therefore, the determinant is zero.
28.
Column 3 consists entirely of zeros. Therefore, the determinant is zero.
29.
2 was factored from row 2.
30.
−3 was factored from column 2.
31.
Row 1 was multiplied by −2 and added to row 2.
32.
Row 1 was multiplied by −1 and added to row 3.
33.
2 was factored from column 1.
34.
Row 3 is a constant multiple of row 1. −2 R1 = R3 . Therefore, the determinant is zero.
35.
The matrix is in triangular form. The product of the elements on the main diagonal is −12. Therefore, the value of the determinant is −12.
36.
The matrix is in triangular form. The product of the elements on the main diagonal is −15. Therefore, the value of the determinant is −15.
37.
Row 1 and row 3 were interchanged. Therefore, the sign of the determinant was changed.
38.
Column 1 and column 2 were interchanged. Therefore, the sign of the determinant was changed.
39.
Each row of the first determinant was multiplied by a to produce the second determinant.
40.
Columns 1, 2, and 3 are identical. Therefore, the determinant is zero.
41.
2 4 1 1 2 −1 1 2 −1 = − 2 4 −1 R1 ↔ R2 1 2 2 1 2 2 1 2 −1 −2 R1 + R2 =− 0 0 1 − R1 + R3 0 0 3 = −(1)(0)(3) = 0
42.
3 −2 −1 −2 1 2 4 =− 2 2 −2 3 −2 1 = 2 −1 1 1 = 2 −1 1
3 1 2 3 1 2 0 4 −1
−1 4 C1 ↔ C2 3 −1 4 Factor 2 from C1 3 0 −3C1 + C2 3 C1 + C3 4
1 0 = 2 −1 4 1 −1 = 2(1)(4)
43.
1 2 −1 1 2 −1 −2 R1 + R2 2 3 1 = 0 −1 3 −3R1 + R3 3 4 3 0 −2 6 1 2 −1 = 0 −1 3 − 2 R2 + R3 0 0 0 = (−1)(0) = 0
44.
0 0 − 34 C2 + C3
19 4
( ) = 38 19 4
1 2 5 1 2 5 R +R −1 1 −2 = 0 3 3 1 2 −3R1 + R3 3 1 10 0 −5 −5 1 2 5 = 0 3 3 5 R2 + R3 0 0 0 3 =1(0)(3) = 0
Copyright © Houghton Mifflin Company. All rights reserved.
734
45.
Chapter 10: Matrices
0 −1 1 1 0 −2 1 0 −2 = − 0 −1 1 R1 ↔ R2 2 2 0 2 2 0
46.
1 0 −2 = − 0 −1 1 − 2 R1 + R3 0 2 4
2 −1 3 1 1 1 1 1 = − 2 −1 3 −4 5 3 −4 1 1 = − 0 −3 0 −7
1 1 1 1 − 7 R2 + R3 = − 0 −3 3 0 0 − 13
1 0 −2 = − 0 −1 1 2 R2 + R3 0 0 6
⎛ ⎞ = −(1)(−3) ⎜ − 1 ⎟ = −1 ⎝ 3⎠
= −(1)(−1)(6) = 6
47.
1 2 −1 1 −2 0 3 0 1 −2 −4 1
2 1 3 0 = 5 0 6 0 1 0 = 0 0 1 0 = 0 0
=1(−4) 49.
2 3 −1 1 5 9 8 6 =3 4 12 −1 2 2 6 −1 1 1 0 =3 0 0 1 0 =3 0 0
1 6 2 1
−1 2 −1R1 + R2 1 1 −3R1 + R3 4 −1 2 R1 + R4 −1 10 2 −1 1 1 3 5 − 5 − R2 + R3 2 2 2 −1 10 2 −1 1 12 5 − 5 R3 + R4 2 2 5 0 9
2 −4 −6 0 2 −4 0 0 2 −4 0 0
48.
1 −1 −1 2 1 −1 −1 0 2 4 6 0 2 4 = 1 1 4 12 0 2 5 1 −1 0 8 0 0 2 1 −1 −1 0 2 4 = 0 0 1 0 0 1 1 −1 −1 0 2 4 = 0 0 1 0 0 0 =1(2)(1)(2) = 4
50.
1 −1 2 −2
( ) (9) = −90
2 5 4 2 2 −7 0 0 2 −7 0 0
1 3 R1 ↔ R2 5 1 −2 R1 + R2 1 −3R1 + R3 2
5 2
1 −1 3 8 Factor 3 from C3 4 −1 2 −1 1 −1 −6 R1 + R2 −3 14 −2 R1 + R3 2 1 −1R1 + R4 1 0 1 −1 −3 14 1 2 1 − 2 R3 + R4 0 − 12
( )
= 3(1)(−7)(2) − 1 = 21
2 1 1 5
0 −2 1 2 0 −2 1R1 + R2 3 5 0 3 3 3 = −2 R1 + R3 4 0 0 −3 4 4 2 R1 + R4 2 6 0 9 2 2 1 2 0 −2 0 3 3 3 R2 + R3 = 0 0 7 7 −3R2 + R4 0 0 −7 −7 1 2 0 −2 0 3 3 3 = R +R 0 0 7 7 3 4 0 0 0 0 =1(3)(7)(0) = 0
2
51.
Using a calculator, 2 −2 3 1 5 2 −2 3 =3 6 −1 2 3 2 3 −1 5
52.
Using a calculator, 3 −1 0 1 2 −2 3 0 =140 −1 −3 5 3 5 3 −2 1
53.
Using a calculator,
54.
Using a calculator,
−2 7
4
−1 6
−2
2
−3 ≈ −38.933
3
3 − 5
2 6 −1R1 + R3 10 −1R1 + R4 6 2 6 −1R2 + R3 4 6 2 6 −R +R 4 3 4 2
6
π
−5
7
5 6
− 3
−4 7
2 ≈ 122.204 10
Copyright © Houghton Mifflin Company. All rights reserved.
Section 10.4
735
....................................................... 55.
57.
56.
2 3 1 1 −1 0 1 = 1 3C + 0C + 8C [ 12 22 32 ] 2 4 8 1 2 ⎡ −1 1 2 1⎤ = 1 ⎢ −3 + 0 −8 −1 1 ⎦⎥ 4 1 2⎣ = 1 [ −3(5) −8(3)] = 1 (15 − 24) = 1 (−9) = − 9 2 2 2 2 −9 =9 2 2 1 The area of the triangle is 4 square units. 2 1 2
4 9 1 8 2 1= −3 −2 1
1 2
[ 4C11 + 8C21 + (−3)C31]
=
1 2
[ 4M11 − 8M 21 − 3M 31]
Connecting Concepts −3 4 1 1 1 5 1 = 1 −3C +1C + 5C [ 11 21 31 ] 2 5 −2 1 2 = 1 [ −3M11 −1M 21 + 5M 31 ] 2
⎡ = 1 ⎢ −3 2⎣
5 1 4 1 4 1⎤ −1 +5 −2 1 −2 1 5 1 ⎥⎦ 1 = [ −3(7) −1(6) + 5(−1)] 2
= 1 [ −21− 6 − 5] = 1 (−32) = −16 2
2
−16 =16 The area of the triangle is 16 square units.
2 1 9 1 9 1⎤ 1⎡ = ⎢4 −8 −3 2 2 1 ⎥⎦ −2 1 ⎣ −2 1 1 1 1 93 = [ 4(4) − 8(11) − 3(7)] = [16 − 88 − 21] = (−93) = − −
93 2
=
2 93 2
2
2
2
1 2
The area of the triangle is 46 square units. 58.
0 4 1 1 −5 7 1 = 1 (0)C + (−5)C + 2C [ 11 21 31 ] 2 2 2 9 1 = 1 [ 0 M11 + 5M 21 + 2 M 31 ] 2 4 1 4 1⎤ ⎡ 1 = ⎢0 + 5 +2 9 1 7 1 ⎥⎦ 2⎣ = 1 [5(−5) + 2(−3) ] = 1 [ −25 − 6] = 1 (−31) = − 31 2 2 2 2 − 31 = 31 2
2
1 2
The area of the triangle is 15 square units. 59.
a
b
c
60.
a b
c
a
d
na nb nc = a b
c
b
e 0 = 0C13 + 0C 23 + 0C33
f
c
f
d
e
f
d
e
0 = 0+0+0 = 0
0
Since two rows are identical, the determinant is 0. 61.
x x1 x2
y 1 y1 1 = xC11 + yC12 +1C13 = 0 y2 0 = xM11 − yM12 +1M13 = 0 = x( y1 − y2 ) − y ( x1 − x2 ) + ( x1 y2 − x2 y1 ) = 0
Since x1, x2 , y1 and y2 are constants, x( y1 − y 2 ) − y ( x1 − x2 ) + ( x1 y 2 − x2 y1 ) = 0 is a line in the form ax + by + c = 0, and (x1, y1) and ( x2 , y2 ) satisfy this equation.
Copyright © Houghton Mifflin Company. All rights reserved.
736
62.
Chapter 10: Matrices
x y 1 2 3 1 = xC11 + yC12 +1C13 = 0 −1 4 1 = xM11 − yM12 +1M13 = x
3 1 2 1 2 3 −y +1 =0 4 1 −1 1 −1 4
= x(−1) − y (3) +1(11) = 0 = − x − 3 y +11= 0 x + 3 y = 11 is the equation of the line passing through the points (2,3) and (−1,4). 63.
x y 1 −3 4 1 = xC11 + yC12 +1C13 = 0 2 −3 1 = xM11 − yM12 +1M13 = 0 = x(7) − y (−5) +1(1) = 0 = 7 x + 5 y +1= 0 7 x + 5 y = −1 is the equation of the line passing through the points (−3,4) and (2,−3).
64.
a1 b1 = a1b2 − a2b1 a2 b2 a1 Ka1 + a2
b1 = a1b1K + a1b2 − (a1b1K + a2b1 ) = a1b2 − a2b1 Kb1 + b2
Adding a multiple of a row to another row does not change the value of the determinant. 65.
25 15 15 17 17 0 0 8⎞ 1 ⎛ 8 25 A= ⎜ + + + + ⎟ 2 −4 5 5 9 9 20 20 10 10 −4 ⎠ ⎝ = =
[
1 140 + 150 + 147 + 170 + (−80) 2 1 (527) = 263.5 square units 2
]
....................................................... PS1.
Prepare for Section 10.5 PS2.
−5 2 = −5(1) − 3(2) = −5 − 6 = −11 3 1
3 −1 6 9 0 2 0 2 9 2 9 0 =3 +1 +6 1 −2 −2 3 1 3 1 −2 3 = 3(27) +1(6) + 6(−13) = 81+ 6 − 78 =9
PS3. ⎡ 2 −7 ⎤ ⎢3 5 ⎥ ⎣ ⎦
PS5.
3 −1 = −9 + 2 = −7 2 −3
PS4.
1 4 = 5 + 8 =13 −2 5
3 2 1 −2
−1 −3 =− 7 4 13 5
1 −2 1 1 −2 −1 −2 −1 1 −1 1 −2 =1 +2 +1 3 −1 2 −1 2 3 2 3 −1 = (5) + 2(5) + (−5) = 5 +10 − 5 =10
PS6. No
Copyright © Houghton Mifflin Company. All rights reserved.
Section 10.5
737
Section 10.5 1.
8
2.
4
1 − 5 − 44 44 = = 3 4 − 31 31 1 −5
x1 =
3
8
9 8 x1 = 2 5
1
5.
5 7 23 = = − 23 5 −11 11 7
5 8 2 5
=
0
4
3 −7
x2 =
9.
3
− 21 x2 = = = −7 7 2 3 2 1
x2 =
4
2 −1
3 0
5 −3
2.1
x2 =
10.
0.3
− 1.6 − 1.4 − 2.46 x1 = = = 1.28125 1. 2 0.3 − 1.92 0. 8 − 1. 4 1.2
2. 1
0.8 − 1.6
− 3. 6 x2 = = = 1.875 1.2 0.3 − 1.92 0.8 − 1.4
3 −8 5
4
0 −1 5
1
4 −2 4
−3 x1 =
2 0 0 = =0 3 − 7 26 2 4
− 14 1 = − 42 3
1 −8
0
0 =0 26
=
− 2 5 − 11 11 = =− x1 = 3 −8 47 47 4 5
4
2
5 4 3 −6
6.
8. =
−1 4 5 −6
5 −1 3 5 28 2 = =− x2 = 5 4 3 −42 3 −6
2 −3
0 −7 x1 =
9
0 2
7
− 29 29 = − 11 11
5 7
7.
x1 =
−3 1 6 x1 = = =2 7 2 3 2 1
2 9 x2 =
3.
−3
2 − 3 − 21 21 x2 = = =− 1 −3 2 2 2 −4
1 − 29 29 = = 4 − 31 31 4 −5
4 x2 = 3
4.
9
− 3 − 4 − 45 45 = =− x1 = 1 −3 2 2 2 −4
=
3 3 =− − 13 13
2 5
0 6 6 = =− 4 − 13 13 2 −1
1.1 −4.2 −3.4 3.2 −10.76 x1 = = ≈ −0.82 3.2 −4.2 13.18 0.7 3.2 3.2 1.1 0.7 −3.4 −11.65 x2 = = ≈ −0.88 3.2 −4.2 13.18 0.7 3.2
Copyright © Houghton Mifflin Company. All rights reserved.
=
− 10 10 =− 47 47
738
11.
14.
Chapter 10: Matrices
3 −4 2 D = 1 −1 2 = −17 2 2 3 1 −4 2 D1 = −2 −1 2 = −21 −3 2 3 3 1 2 D2 = 1 −2 2 = 3 2 −3 3 3 −4 1 D3 = 1 −1 −2 = 29 2 2 −3 D1 −21 21 x1 = = = D −17 17 D x2 = 2 = 3 = − 3 D −17 17 D x3 = 3 = 29 = − 29 17 D −17
12.
4 −1 2 D = 1 3 −1 = −18 2 3 −2
15.
4 −5 1 D= 3 1 0 = 53 1 −1 3 −2 D1 = 4 0 4 D2 = 3 1 4 D3 = 3 1
−5 1 −1 −2 4 0 −5 1 −1
D1 50 = D 53 D x2 = 2 = 62 D 53 D x3 = 3 = 4 D 53 x1 =
1 0 = 50 3 1 0 = 62 3 −2 4 =4 0
13.
0 2 −3 D = 3 −5 1 = −64 4 0 2 1 D1 = 0 −3 0 D2 = 3 4 0 D3 = 3 4
2 −5 0 1 0 −3 2 −5 0
16.
−3 1 = 29 2 −3 1 = 25 2 1 0 = 38 −3
5 D1 = −2 0 3 D2 = 1 2 3 D3 = 1 2
−1 0 2 5 −2 0 −1 0 2
1 3 = −24 −5 1 3 = 89 −5 5 −2 = 26 0
D1 −24 8 = = D −27 9 D x2 = 2 = 89 = − 89 D −27 27 D x3 = 3 = 26 = − 26 D −27 27 x1 =
5 0 −1 1 −2 4 5 0 −1
0 −3 = −43 2 0 −3 = 8 2 1 −2 = −45 4
D x1 = 1 = −43 = 43 D −46 46 D2 = 8 =− 4 x2 = D −46 23 D3 −45 45 = = x3 = D −46 46
D1 29 = = − 29 D −64 64 D2 25 = = − 25 x2 = D −64 64 D3 38 19 = =− x3 = D −64 32
3 −1 1 D = 1 0 3 = −27 2 2 −5
2 5 0 D = 1 0 −3 = −46 2 −1 2 1 D1 = −2 4 2 D2 = 1 2 2 D3 = 1 2
x1 =
18.
1 4 −2 D = 3 −2 3 = 49 2 1 −3 0 4 −2 3 = 32 D1 = 4 −2 −1 1 −3 1 0 −2 D2 = 3 4 3 =13 2 −1 −3 1 4 0 D3 = 3 −2 4 = 42 2 1 −1 D1 32 x1 = = D 49 D2 13 x2 = = D 49 D x3 = 3 = 42 = 6 D 49 7
3 −2 −2 1 −2 =1 D1 = 3 3 −1 −2 5 −2 3 D2 = 3 3 −2 = 39 1 −1 3 5 −2 −2 D3 = 3 1 3 = 27 1 −2 −1 D x1 = 1 = 1 = − 1 4 D −4 D2 39 = = − 39 x2 = D −4 4 D x3 = 3 = 27 = − 27 D −4 4
6 −1 2 D1 = −1 3 −1 = −47 5 3 −2 4 6 2 D2 = 1 −1 −1 = 42 2 5 −2 4 −1 6 D3 = 1 3 −1 = 61 2 3 5 D1 −47 47 x1 = = = D −18 18 D x2 = 2 = 42 = − 7 D −18 3 D3 61 61 x3 = = =− D −18 18
17.
5 −2 3 D= 3 1 −2 = −4 1 −2 3
19.
2 2 −3 D = 1 −3 2 = −37 4 −1 3 0 D1 = 0 0 2 D2 = 1 4 2 D3 = 1 4
2 −3 −1 0 0 0 2 −3 −1
−3 2 =0 3 −3 2 =0 3 0 0 =0 0
D1 = 0 =0 D −37 D x2 = 2 = 0 = 0 D −37 D x3 = 3 = 0 = 0 D −37 x1 =
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Section 10.5
20.
739
21.
1 3 0 D = 2 −3 1 = 25 4 5 −2 −2 1 0 1 D2 = 2 4 1 D3 = 2 4 D1 =
3 −3 5 −2 1 0 3 −3 5
−3 4
2 1 D= 3
0 1 =4 −2 0 1 = −18 −2 −2 1 = −37 0
1 −3 2
x2 =
3
1 4
−1 2
3
2 0 −2
1
3 2
−2
1
3
2 −3 2 D= 1 1 −2
−1
1 −1 0
2 D2 =
22.
−1
2 0 2 = −38 1 0 −2
2
= 70
−1 x4 =
1
2 −3
D4 =
D2 70 35 = =− D − 38 19
3 −2
0 3
0 = −3 2
−2
4
2 −2
1
1 −2
3
2
0
4
3
= 51
D4 51 = = −17 D −3
D1 4 = D 25 D x2 = 2 = −18 = − 18 D 25 25 D3 −37 37 = =− x3 = D 25 25 x1 =
23.
26.
24.
1 −3 2 4 3 5 −6 2 = −1310 D= 2 −1 9 8 1 1 1 −8
2 1 D= 4 3
0 −3 2 4 −2 5 −6 2 = 1210 D1 = 0 −1 9 8 −3 1 1 −8
2 1 D3 = 4 3
D 1210 121 x1 = 1 = =− D − 1310 131
x3 =
4 1 0 −3 5 2 −2 1 = −254 , D1 = D= 1 −3 2 −2 0 0 3 4
5 −5 −3 7 8 −1 = 168 0 1 1 2 −1 0
25.
5 −3 −3 7 4 −1 = 157 0 3 1 2 0 0
0 3 −1 2 5 1 3 −1 = 120 D= 1 −2 0 9 2 0 2 0 0 3 −1 1 5 1 3 −4 = 160 D4 = 1 −2 0 5 2 0 2 3
D3 157 = D 168
x4 =
D4 160 4 = = D 120 3
4 1 0 −3 7 2 −2 1 = −77 −6 −3 2 −2 3 4 −7 0
D 77 −77 x1 = 1 = = D − 254 254
....................................................... 27.
2 −3
28.
1
a11 x1 + a12 x 2 = b1 a 21 x1 + a 22 x 2 = b2
D= 1 1 −2 = 0 4 −1 − 3
In order for us to use Cramer’s Rule, the determinant of the coefficient matrix cannot be zero. The system of equations has infinitely many solutions.
− a21a11 x1 − a21a12 x2 = − a21b1 a11a21 x1 + a11a22 x2 = a11b2
(a11a22 − a21a12 ) x2 = a11b2 − a21b1 a11 a12 a b x = 11 1 a21 a22 2 a21 b2
Connecting Concepts 29.
D=
k k
3 = −5k −2
For the system of equations to have a unique solution, the determinant of the coefficient matrix cannot be zero. −5k = 0 k =0 The system of equations has a unique solution for all values of k except k = 0.
a11 b1 a21 b2 x2 = a11 a12 a21 a22
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740
30.
Chapter 10: Matrices
D=
31.
4 k = − k 2 − 36 9 −k
For the system of equations to have a unique solution, the determinant of the coefficient matrix cannot be zero.
For the system of equations to have a unique solution, the determinant of the coefficient matrix cannot be zero.
− k 2 − 36 = 0 k 2 = −36 k = ± −36
4k − 8 = 0 4( k − 2) = 0 k −2=0 k=2
− 36 is not a real number. The system of equations has a unique solution for all real values of k. 32.
k
1
33.
0 2
D = 0 1 −4 = k −4 1 0 k
For the system of equations to have a unique solution, the determinant of the coefficient matrix cannot be zero. k2 −4 = 0 k2 = 4 k = ±2
The system of equations has a unique solution for all values of k except k = 2 and k = −2.
1 2 −3 D = 2 k −4 = 4k − 8 1 −2 1
The system of equations has a unique solution for all values of k except k = 2.
ru + sv = w (2 + 3i )r + (4 − 2i ) s = −6 +15i 2r + 3ri + 4 s − 2si = −6 +15i (2r + 4 s ) + (3r − 2 s )i = −6 +15i 2r + 4 s = −6 3r − 2 s = 15 2 4 = −16 D= 3 −2 −6 4 Dr = = −48 15 −2 2 −6 Ds = = 48 3 15 D r = r = −48 = 3 D −16 Ds s= = 48 = −3 D −16
.......................................................
34.
ru + sv = w (3 − 4i )r + (1+ 2i ) s = 4 − 22i 3r − 4ri + s + 2si = 4 − 22i (3r + s ) + (−4r + 2 s )i = 4 − 22i 3r + s = 4 − 4r + 2s = −22 3 1 = 10 D= −4 2 4 1 = 30 Dr = −22 2 3 4 Ds = = −50 −4 −22 D r = r = 30 = 3 D 10 Ds −50 s= = = −5 10 D
Exploring Concepts with Technology
Stochastic Matrices XT = [0.428 0.572] XT 2 ≈ [0.45236 0.54764] XT 3 ≈ [0.47355 0.52645]
XT 20 ≈ [0.60209 0.39791] XT 40 ≈ [0.61456 0.38544] XT 60 ≈ [0.61533 0.38467] XT 100 ≈ [0.61538 0.38462] It appears that as the number of weeks increases, Super A will get slightly more than 61.5% of the neighborhood and Super B will get slightly less than 38.5% of the neighborhood. Changing the market share does not affect the result (to 6 decimal places) after 100 weeks. Three Department Stores: After 100 weeks, Super A will have 23.87% of the market share, Super B will have 33.65% of the market share, and Super C will have 42.48% of the market share.
....................................................... 1.
See the Chapter Summary under 10.1.
Assessing Concepts 2.
See the Chapter Summary under 10.1.
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Chapter Review
741
3.
Yes.
4.
They must be of the same order.
5.
The number of columns of the first matrix must equal the number of rows of the second matrix.
6.
A matrix does not have a multiplicative inverse.
7.
A square matrix with 1’s along the main diagonal and zeros elsewhere.
8.
No. Nonsquare and singular matrices do not have multiplicative inverses.
9.
It is the determinant of the matrix obtained by deleting the ith row and the jth column of A.
10.
No. If the determinant of the coefficient matrix is zero, Cramer’s Rule cannot be used.
....................................................... 1.
3.
9⎤ ⎡ 2 −1 3⎤ ⎡6 −3 3A = 3⎢ ⎥ [10.2] ⎥=⎢ − − 3 2 1 9 6 3⎦ ⎦ ⎣ ⎣ ⎡ −3 4 − A + D = − ⎡ 2 −1 3⎤ + ⎢ ⎢⎣ 3 2 −1⎥⎦ ⎣ 4 −2 1 −3⎤ ⎡ −3 4 ⎡ −2 =⎢ + 1⎥⎦ ⎢⎣ 4 −2 ⎣ −3 −2 ⎡ −5 5 −1⎤ =⎢ ⎣ 1 −4 6 ⎥⎦
2⎤ [10.2] 5⎥⎦
Chapter Review 2.
4.
2⎤ 5⎥⎦
⎡ 0 −2 ⎤ ⎡ 0 4 ⎤ −2 B = −2 ⎢ 4 2 ⎥ = ⎢ −8 −4 ⎥ [10.2] ⎢ 1 −3⎥ ⎢ −2 6 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ 2 −1 3⎤ ⎡ −3 4 2 ⎤ 2 A − 3D = 2 ⎢ − 3⎢ [10.2] ⎣ 3 2 −1⎥⎦ ⎣ 4 −2 5⎥⎦ ⎡ 4 −2 6 ⎤ ⎡ −9 12 6 ⎤ =⎢ − ⎣ 6 4 −2 ⎥⎦ ⎢⎣ 12 −6 15⎥⎦ 0⎤ ⎡ 13 −14 =⎢ ⎣ −6 10 −17 ⎥⎦
5.
⎡ 0 −2 ⎤ ⎡ 2 −1 3⎤ ⎢ ⎡ −1 −5⎤ AB = ⎢ [10.2] 4 2⎥ = ⎥ 1⎥⎦ ⎣ 3 2 −1⎦ ⎢ 1 −3⎥ ⎢⎣ 7 ⎣ ⎦
6.
⎡ 0 −2 ⎤ 8⎤ ⎡ −3 4 2 ⎤ ⎢ ⎡ 18 DB = ⎢ 4 2⎥ = [10.2] ⎥ ⎣ 4 −2 5⎦ ⎢ 1 −3⎥ ⎢⎣ −3 −27 ⎥⎦ ⎣ ⎦
7.
⎡ 0 −2 ⎤ ⎡ −6 −4 2 ⎤ ⎡ 2 −1 3⎤ ⎢ = BA = ⎢ 4 2 ⎥ ⎢ 14 0 10 ⎥ [10.2] ⎢ ⎥ ⎣ 3 2 −1⎥⎦ ⎢ −7 −7 6 ⎥ ⎣ 1 −3⎦ ⎣ ⎦
8.
⎡ 0 −2 ⎤ ⎡ −8 4 −10 ⎤ ⎡ −3 4 2 ⎤ ⎢ BD = ⎢ 4 2 ⎥ ⎢ = −4 12 18⎥ [10.2] ⎢ 1 −3⎥ ⎣ 4 −2 5⎥⎦ ⎢ −15 10 −13⎥ ⎣ ⎦ ⎣ ⎦
9.
⎡2 C = C ⋅ C = ⎢⎢ 1 ⎢⎣ 2 ⎡12 = ⎢⎢ 2 ⎢⎣ 6 2
6 1⎤ ⎡ 2 6 1⎤ 2 −1⎥⎥ ⎢⎢ 1 2 −1⎥⎥ [10.2] 4 −1⎥⎦ ⎢⎣ 2 4 −1⎥⎦ 28 −5⎤ 6 0 ⎥⎥ 16 −1⎥⎦
10.
⎡2 C3 = C ⋅ C ⋅ C = ⎢ 1 ⎢2 ⎣ ⎡12 =⎢ 2 ⎢ 6 ⎣ ⎡ 42 = ⎢ 10 ⎢ 26 ⎣
6 1⎤ ⎡ 2 6 2 −1⎥ ⎢ 1 2 4 −1⎥⎦ ⎢⎣ 2 4 28 −5⎤ ⎡ 2 6 0⎥ ⎢ 1 16 −1⎥⎦ ⎢⎣ 2 108 −11⎤ 24 −4 ⎥ 64 −9 ⎥⎦
11.
⎡ 0 −2 ⎤ ⎡ 2 6 1⎤ ⎡ −6 −4 2 ⎤ ⎡ 2 6 1⎤ ⎡ −12 −36 −4 ⎤ ⎡ 2 −1 3⎤ ⎢ ⎥ = ⎢ 14 0 10 ⎥ ⎢ 1 2 −1⎥ = ⎢ 48 124 4 ⎥ [10.2] − BAC = ⎢ 4 2 ⎥ ⎢ 1 2 1 ⎢ 1 −3⎥ ⎣ 3 2 −1⎥⎦ ⎢ 2 4 −1⎥ ⎢ −7 −7 6 ⎥ ⎢ 2 4 −1⎥ ⎢ −9 −32 −6 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
12.
Not possible since A is of order 2 × 3 and D is of order 2 × 3. [10.2]
13.
⎡ −6 −4 2 ⎤ ⎡ 0 −2 ⎤ ⎡ 0 −2 ⎤ ⎡ −1 −15⎤ ⎡ 2 −1 3⎤ ⎢ ⎡ 2 −1 3⎤ AB − BA = ⎢ 4 2⎥ − ⎢ 4 2⎥ ⎢ = ⎢ − ⎢ 14 0 10 ⎥ ⎥ ⎥ ⎥ 1⎦ ⎢ ⎥ ⎣ 3 2 −1⎦ ⎢ −7 −7 6 ⎥ ⎣ 3 2 −1⎦ ⎢ 1 −3⎥ ⎣ 7 ⎣ ⎦ ⎣ 1 −3⎦ ⎣ ⎦ Not possible since AB is of order 2 × 2 and BA is of order 3 × 3. [10.2]
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1⎤ ⎡ 2 6 1⎤ −1⎥ ⎢ 1 2 −1⎥ [10.2] −1⎥⎦ ⎢⎣ 2 4 −1⎥⎦ 6 1⎤ 2 −1⎥ 4 −1⎥⎦
742
14.
Chapter 10: Matrices
⎡ 0 −2 ⎤ ⎡ 0 −2 ⎤ ⎡ −8 4 −10 ⎤ 8⎤ ⎡ −3 4 2 ⎤ ⎢ ⎥ − ⎢ 4 2 ⎥ ⎡ −3 4 2 ⎤ = ⎡ 18 ⎢ −4 12 18⎥ DB − BD = ⎢ 4 2 − ⎢⎣ −3 −27 ⎥⎦ ⎢ 1 −3⎥ ⎢⎣ 4 −2 5⎥⎦ ⎢ −15 10 −13⎥ ⎣ 4 −2 5⎥⎦ ⎢ 1 −3⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Not possible since DB is of order 2 × 2 and BD is of order 3 × 3. [10.2]
15.
⎡ 2 6 1⎤ ⎡ 2 6 1⎤ 1⎤ ⎢ ⎛ ⎡ 2 −1 3⎤ ⎡ 7 24 9 ⎤ ⎡ −3 4 2 ⎤ ⎞ ⎢ ⎥ = ⎡ 5 −5 1 2 −1⎥ = ⎢ ( A − D )C = ⎜ ⎢ 1 2 1 [10.2] − − ⎟ ⎢ ⎥ ⎥ ⎢ ⎥ 1 4 6 3 2 1 4 2 5 − − − − ⎢ ⎥ ⎢ ⎥ ⎦ ⎣ ⎦ ⎠ 2 4 −1 ⎣ ⎦ 2 4 −1 ⎣ −10 −22 1⎥⎦ ⎝⎣ ⎣ ⎦ ⎣ ⎦
16.
⎡ 2 6 1⎤ ⎡ 2 6 1⎤ ⎡ 2 −1 3⎤ ⎢ ⎥ − ⎡ −3 4 2 ⎤ ⎢ 1 2 −1⎥ = ⎡9 22 0 ⎤ ⎡ 2 −2 −9 ⎤ = ⎡ 7 24 9 ⎤ [10.2] − AC − DC = ⎢ 1 2 1 ⎢⎣ 4 −2 5⎥⎦ ⎢ ⎢⎣6 18 2 ⎥⎦ ⎢⎣16 40 ⎢⎣ −10 −22 1⎥⎦ 1⎥⎦ ⎥ ⎣ 3 2 −1⎥⎦ ⎢ 2 4 −1⎥ ⎣ ⎦ ⎣ 2 4 −1⎦
17.
⎡ 2 6 1 1 0 0⎤ 1R 2 1→ ⎢ 1 2 −1 0 1 0⎥ ⎯⎯⎯ ⎢ 2 4 −1 0 0 1⎥ ⎣ ⎦
−1R1 + R 2 ⎡ 1 3 12 12 0 0⎤ −2 R 1 + R 3 ⎢ ⎥ ⎢ 1 2 −1 0 1 0⎥ ⎯⎯⎯⎯⎯→ ⎢⎣ 2 4 −1 0 0 1⎥⎦
−1R 2 −3R 2 + R1
4 R 3 + R1 ⎡ 1 0 −4 −1 3 0⎤ 2R 2 +R3 ( − 3/ 2) R 3 + R 2 3 1 ⎯⎯⎯⎯⎯→ ⎢ 0 1 2 −1 0⎥ ⎯⎯⎯⎯⎯⎯⎯ → 2 ⎢ ⎥ − 0 0 1 0 2 1 ⎥⎦ ⎣⎢ C
18.
−1
4⎤ ⎡ 1 0 0 −1 − 5 3⎥ ⎢0 1 0 1 − 2 2 2⎥ ⎢ 1⎥⎦ ⎣⎢0 0 1 0 −2
4⎤ ⎡ −1 −5 ⎢ 3 1 =⎢ 2 − ⎥⎥ [10.3] 2 2 ⎢⎣ 0 −2 1⎥⎦
⎡ 2 6 1⎤ Determinant of C = ⎢ 1 2 −1⎥ = 2C11 +1C21 + 2C31 ⎢ ⎥ ⎢⎣ 2 4 −1⎥⎦ = 2M11 −1M 21 + 2M 31 =2
19.
1 1 0 0⎤ ⎡1 3 2 2 ⎢ ⎥ 3 1 ⎢0 −1 − 2 − 2 1 0⎥ ⎢ 0 −2 −2 −1 0 1⎥ ⎣ ⎦
[10.4]
2 −1 6 1 6 1 −1 +2 = 2(2) −1(−10) + 2(−8) = 4 +10 −16 = −2 4 −1 4 −1 2 −1
R1 + R 2 −2 R1 + R 2 ⎡ 2 −3 7 ⎤ ⎡ 2 −3 7 ⎤ ⎡ 1 −1 3⎤ ⎡ 1 −1 3⎤ −1R 1 + R 2 → ⎢ ⎢ 3 −4 10⎥ ⎯⎯⎯⎯⎯⎯ ⎥ ←⎯⎯⎯→ ⎢ 2 −3 7⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 −1 1⎥ 1 1 3 − ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎧x − y = 3 ⎨ y = −1 ⎩
x − ( −1) = 3 x=2
The solution is (2, −1). [10.1] 20.
− R 2 + R1 −2 R 1 + R 2 ⎡ 3 4 −9 ⎤ ⎡ 1 1 −2 ⎤ ⎡ 1 1 −2 ⎤ → ⎢ ⎢ 2 3 −7 ⎥ ⎯⎯⎯⎯⎯ ⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 1 −3⎥ 2 3 7 − ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎧ x + y = −2 ⎨ y = −3 ⎩
x + (−3) = −2 x =1
The solution is (1, −3). [10.1]
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−1R 2 ⎡ 1 −1 3⎤ ⎯⎯⎯⎯ → ⎢ ⎥ ⎣0 1 −1⎦
Chapter Review
21.
743
−1R 2 + R1 ⎡ 4 −5 12 ⎤ ⎢ 3 1 9 ⎥ ⎯⎯⎯⎯⎯→ ⎣ ⎦ ⎧x − 6 y = 3 ⎨ y=0 ⎩
−3R 1 + R 2 ⎡ 1 −6 3⎤ ⎢3 ⎥ ⎯⎯⎯⎯⎯→ 1 9 ⎣ ⎦
1R ⎡ 1 −6 3⎤ 19 2 → ⎢ 0 19 0⎥ ⎯⎯⎯⎯ ⎣ ⎦
⎡ 1 −6 3⎤ ⎢0 1 0⎥⎦ ⎣
x − 6(0) = 3 x=3
The solution is (3, 0). [10.1] 22.
−2 R1 + R 2 ⎡ 2 −5 10⎤ ⎡ 2 −5 10⎤ ⎡ 1 12 −16⎤ ⎡ 1 12 −16⎤ −2 R 1 + R 2 → ⎢ ⎢ 5 2 4 ⎥ ⎯⎯⎯⎯⎯→ ⎢ 1 12 −16⎥ R 1 ←⎯→ R 2 ⎢ 2 −5 10⎥ ⎯⎯⎯⎯⎯⎯ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣0 −29 42⎦ 1R − 29 ⎡ 1 12 −16⎤ 2 ⎯⎯⎯⎯ → ⎢ 42 ⎥ ⎣0 1 − 29 ⎦
( 29 )
x + 12 − 42 = −16
⎧⎪ x + 12 y = −16 ⎨ y = − 42 ⎪⎩ 29
x − 504 = −16 29
x = 40 29
( 29
)
The solution is 40 , − 42 . [10.1] 23.
29
⎡ 1 2 3 5⎤ −3R1 + R 2 ⎡ 1 2 3 5⎤ ⎡ 1 2 3 5⎤ ⎡ 1 2 3 5⎤ ⎡ 1 2 3 5⎤ −2 R1 + R 3 ⎢ −2 R 2 + R 3 ⎢ −1R 3 ⎢ (1/ 2) R 2 ⎢ ⎢ 3 8 11 17 ⎥ ⎯⎯⎯⎯⎯ ⎥ ⎥ ⎥ → 0 2 2 2 ⎯⎯⎯⎯⎯ → 0 1 1 1 ⎯⎯⎯⎯⎯→ 0 1 1 1 ⎯⎯⎯⎯ → 0 1 1 1⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 2 6 7 12 ⎥⎦ ⎢⎣ 0 2 1 2 ⎥⎦ ⎢⎣ 0 2 1 2⎥⎦ ⎢⎣0 0 −1 0⎥⎦ ⎢⎣0 0 1 0⎥⎦ ⎧ x + 2 y + 3z = 5 ⎪ y + z =1 ⎨ ⎪ z=0 ⎩
y +0 =1 y =1
x + 2(1) + 3(0) = 5 x=3
The solution is (3, 1, 0). [10.1] 24.
⎡ 1 −1 3 10⎤ −2 R1 + R 2 ⎡ 1 −1 3 10⎤ ⎡ 1 −1 3 10⎤ −3R1 + R 3 ⎢ 3R 2 + R 3 ⎢ ⎢ 2 −1 7 24 ⎥ ⎯⎯⎯⎯⎯→ → 0 1 1 4⎥ 0 1 1 4 ⎥ ⎯⎯⎯⎯⎯ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣0 −3 −2 −9 ⎦⎥ ⎣⎢ 3 −6 7 21⎥⎦ ⎣⎢ 0 0 1 3⎦⎥ ⎧ x − y + 3 z = 10 ⎪ y+ z=4 ⎨ ⎪ z=3 ⎩
y+3= 4 y =1
x − 1 + 3(3) = 10 x=2
The solution is (2, 1, 3). [10.1] 25.
−2 R1 + R 2 5⎤ ⎡ 1 −2 −2 5⎤ ⎡ 1 −2 − 2 1R −3R1 + R 3 ⎢ 2 −1 −1 4 ⎥ ⎯⎯⎯⎯⎯→ ⎢0 3 2→ 3 3 −6⎥ ⎯⎯⎯ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 3 −2 4 ⎦⎥ ⎣⎢ 3 −3 −8 19 ⎦⎥
⎡ 2 −1 −1 4 ⎤ ⎢ 1 −2 −2 5⎥ R ←⎯→ R 2 ⎢ ⎥ 1 ⎣⎢ 3 −3 −8 19 ⎦⎥
5⎤ 5⎤ ⎡ 1 −2 −2 ⎡ 1 −2 − 2 − 15 R 3 −3R 2 + R 3 ⎢0 1 1 −2 ⎥ ⎯⎯⎯⎯ 1 1 −2 ⎥ ⎯⎯⎯⎯⎯→ → ⎢0 ⎢ ⎥ ⎢ ⎥ ⎢⎣0 ⎢⎣ 0 0 −5 10⎥⎦ 0 1 −2 ⎥⎦ ⎧x − 2 y − 2z = 5 ⎪ y + z = −2 ⎨ ⎪ z = −2 ⎩
y + (−2) = −2 y=0
x − 2(0) − 2(−2) = 5 x =1
The solution is (1, 0, −2). [10.1]
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5⎤ ⎡ 1 −2 −2 ⎢0 1 1 −2 ⎥ ⎢ ⎥ 3 −2 4 ⎥⎦ ⎣⎢0
744
26.
Chapter 10: Matrices
⎡ 3 −7 8 10⎤ ⎢ 1 −3 2 0⎥ R ←⎯→ R 2 ⎢ ⎥ 1 ⎣⎢ 2 −8 7 5⎥⎦
− 3R 1 + R 2 ⎡ 1 −3 2 0 ⎤ −2 R 1 + R 3 ⎢ 3 −7 8 10⎥ ⎯⎯⎯⎯⎯ → ⎢ ⎥ ⎣⎢ 2 −8 7 5⎥⎦
⎡ 1 −3 2 0 ⎤ 1 R 2R 2 +R3 5 3→ ⎯⎯⎯⎯⎯ → ⎢0 1 1 5⎥ ⎯⎯⎯ ⎢ ⎥ ⎣⎢ 0 0 5 15⎥⎦ ⎧x − 3 y + 2z = 0 ⎪ y+ z =5 ⎨ ⎪ z =3 ⎩
y+3= 5 y=2
⎡ 1 −3 2 0 ⎤ ⎡ 1 −3 2 0 ⎤ 1 R ⎢ 0 2 2 10⎥ ⎯⎯⎯ 2 2 → ⎢0 1 1 5⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 −2 3 5⎦⎥ ⎣⎢ 0 −2 3 5⎥⎦
⎡ 1 −3 2 0 ⎤ ⎢0 1 1 5⎥ ⎢ ⎥ ⎣⎢ 0 0 1 3⎥⎦
x − 3(2) + 2(3) = 0 x=0
The solution is (0, 2, 3). [10.1] 27.
⎡ 4 −9 6 54 ⎤ ⎢ 3 −8 8 49 ⎥ R ←⎯→ R 3 ⎢ ⎥ 1 ⎣⎢ 1 −3 2 17 ⎥⎦
−3R1 + R 2 ⎡ 1 −3 2 17 ⎤ −4 R1 + R 3 ⎢ 3 −8 8 49 ⎥ ⎯⎯⎯⎯⎯ → ⎢ ⎥ ⎣⎢ 4 −9 6 54 ⎥⎦
17 ⎤ ⎡ 1 −3 2 17 ⎤ ⎡ 1 −3 2 −3R 2 + R 3 ⎢0 ⎢0 1 2 −2 ⎥ 1 2 −2 ⎥ ⎯⎯⎯⎯⎯→ ⎢ ⎥ ⎢ ⎥ ⎢⎣0 0 −8 −8⎦⎥ ⎣⎢0 3 −2 −14 ⎦⎥
⎡ 1 −3 2 17 ⎤ ( −1/8) R 3 ⎯⎯⎯⎯⎯ → ⎢0 1 2 −2 ⎥ ⎢ ⎥ ⎢⎣ 0 0 1 1⎥⎦ ⎧ x − 3 y + 2 z = 17 ⎪ y + 2 z = −2 ⎨ ⎪ z= 1 ⎩
y + 2(1) = −2 y = −4
x − 3(−4) + 2(1) = 17 x=3
The solution is (3, −4, 1). [10.1] 28.
⎡ 3 8 −5 6⎤ ⎢ 2 9 −1 −8⎥ R ←⎯→ R 3 ⎢ ⎥ 1 ⎢⎣ 1 −4 −2 16⎥⎦
−2 R 1 + R 2 16⎤ ⎡ 1 −4 −2 16⎤ ⎡ 1 −4 − 2 1R −3R1 + R 3 ⎢ 2 9 −1 −8⎥ ⎯⎯⎯⎯⎯→ ⎢0 17 17 2⎯ → 3 −40⎥ ⎯⎯⎯ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 3 8 −5 6⎥⎦ ⎢⎣0 20 1 −42⎥⎦
⎡ 1 −4 −2 16⎤ 16⎤ ⎡ 1 −4 −2 ⎢ ⎥ − 17 R3 −20 R 2 + R 3 ⎢ 3 40 43 3 40 ⎥ 1 1 17 − 17 ⎯⎯⎯⎯⎯⎯ → ⎢0 − 17 ⎥ ⎯⎯⎯⎯→ ⎢0 ⎥ 17 ⎢ ⎥ ⎢0 0 43 86 1 −2 ⎥⎦ ⎢⎣0 0 − 17 ⎥ ⎣ 17 ⎦ ⎧ x − 4 y − 2 z = 16 ⎪⎪ y + 3 z = − 40 ⎨ 17 17 ⎪ z = −2 ⎪⎩
y + 3 (−2) = − 40 17
17
x − 4(−2) − 2(−2) = 16 x=4
y = −2
The solution is (4, −2, −2). [10.1] 29.
−2 R1 + R 2 ⎡ 1 1 2 −5⎤ ⎡ 1 1 2 −5⎤ ⎡ 1 1 2 −5⎤ −2 R1 + R 3 −3R 2 + R 3 ⎢ 2 3 5 −13⎥ ⎯⎯⎯⎯⎯→ ⎢ 0 1 1 −3⎥ ⎯⎯⎯⎯⎯→ ⎢0 1 1 −3⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 2 5 7 −19 ⎥⎦ ⎢⎣ 0 3 3 −9 ⎥⎦ ⎢⎣0 0 0 0⎥⎦ ⎧ x + y + 2 z = −5 ⎨ ⎩ y + z = −3
y = −z − 3
x + (− z − 3) + 2 z = −5 x = −z − 2
Let z be any real number c. The solution is (−c −2, −c −3, c). [10.1]
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16⎤ ⎡ 1 −4 −2 ⎢ 3 40 ⎥ 1 17 − 17 ⎢0 ⎥ ⎢0 20 ⎥ − 1 42 ⎣ ⎦
Chapter Review
30.
745
− 3R 1 + R 2 ⎡ 1 −2 3 9 ⎤ −1R1 + R 3 ⎢ 3 −5 8 25⎥ ⎯⎯⎯⎯⎯ → ⎢ ⎥ ⎣⎢ 1 0 −1 5⎥⎦ ⎧ x − 2 y + 3z = 9 ⎪ y − z = −2 ⎨ ⎪ z= 0 ⎩
3 9⎤ 3 9⎤ ⎡ 1 −2 ⎡ 1 −2 −2 R 2 + R 3 ( −1/ 2) R 3 ⎢ ⎢0 ⎥ → 1 −1 −2 ⎥ ⎯⎯⎯⎯⎯ 1 −1 −2 ⎯⎯⎯⎯⎯→ 0 ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 0 −2 0⎦⎥ ⎣⎢ 0 2 −4 −4 ⎥⎦
y − 0 = −2 y = −2
⎡ 1 −2 3 9 ⎤ ⎢0 1 −1 −2 ⎥ ⎢ ⎥ 0 1 0⎥⎦ ⎣⎢0
x − 2(−2) + 3(0) = 9 x=5
The solution is (5, –2, 0). [10.1] 31.
−1 2 1⎤ 1 4 1⎥ ⎥ 7 3 2 0⎥ ⎥ 3 −2 5 6⎦ ⎡1 −3R 2 + R 3 ⎢0 −1R 2 + R 4 ⎯⎯⎯⎯⎯→ ⎢ ⎢0 ⎢ ⎣0
⎡1 ⎢3 ⎢ ⎢2 ⎢ ⎣1
2 8
−3R1 + R 2 −2 R1 + R 3 −1R1 + R 4 ⎯⎯⎯⎯⎯ →
2 −1 2 1 2 −1 0 −1 0 −3
⎧w + 2 x − y + 2 z = 1 ⎪ x + 2 y − z = −1 ⎪ ⎨ y − z = −1 ⎪ ⎪⎩ z= 3
1 4
1⎤ ⎡ 1 −2 − 1 2 ⎡ 1 2 −1 ⎢ 0 2 4 −2 −2 ⎥ ⎢0 1 2 R ( 1/ 2) − 2→ ⎢ ⎢ ⎥ ⎯⎯⎯⎯⎯ ⎢0 3 5 ⎢0 3 5 −2 −2 ⎥ ⎢ ⎢ ⎥ 0 1 1 3 5 − ⎣ ⎦ ⎣ 0 1 −1 1⎤ 1⎤ ⎡ 1 2 −1 2 ⎢ 0 1 2 −1 −1⎥ −1⎥ 1 R − − 3R 3 + R 4 3→ ⎢ ⎥ ⎯⎯⎯⎯ ⎥ ⎯⎯⎯⎯⎯→ ⎢0 0 1⎥ 1 −1 −1⎥ ⎥ ⎢ ⎥ − 6⎦ 0 0 3 4 6⎦ ⎣
y − 3 = −1 y=2
x + 2(2) − 3 = −1
w + 2(−2) − 2 + 2(3) = 1
x = −2
w =1
2 1⎤ −1 −1⎥ ⎥ −2 −2 ⎥ ⎥ 3 5⎦ 1⎤ ⎡ 1 2 −1 2 ⎢ 0 1 2 −1 −1⎥ ⎢ ⎥ ⎢ 0 0 1 −1 −1⎥ ⎢ ⎥ ⎣ 0 0 0 1 3⎦
The solution is (1, –2, 2, 3). [10.1] 32.
⎡1 ⎢2 ⎢ ⎢3 ⎢ ⎣2
−3 −2 −5 0
1 3
−7 3 0 −3 −5 −2
⎡1 ⎢0 13R 3 + R 4 ⎯⎯⎯⎯⎯ → ⎢ ⎢0 ⎢ ⎣0
−1⎤ 1⎥ ⎥ −18⎥ ⎥ −8 ⎦ −3 1 0 0
⎧w − 3x − 2 y + z = − 1 ⎪ x + 4y + z = 3 ⎪ ⎨ y − 5 z = −21 ⎪ ⎪⎩ z= 4
−2 R1 + R 2 −3R1 + R 3 −2 R1 + R 4 ⎯⎯⎯⎯⎯→
−2 1 4 1 1 −5 0 −72
1 −1⎤ 1 −3 −2 1 −1⎤ ⎡ 1 −3 − 2 −2 R 2 + R 3 ⎡ ⎢0 ⎥ ⎢ 1 4 1 3 1 4 1 3⎥ −3R 2 + R 4 ⎢0 ⎢ ⎥ ⎯⎯⎯⎯⎯→ ⎥ ⎢0 2 ⎢0 0 9 −3 −15⎥ 1 −5 −21⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 3 −1 −4 − 6 ⎦ ⎣0 0 −13 −7 −15⎦ −1⎤ 1 −1⎤ ⎡ 1 − 3 −2 ⎥ ⎢ 3 0 1 4 1 3⎥ ( −1/ 72) R 4 ⎥ ⎯⎯⎯⎯⎯⎯ ⎥ → ⎢ ⎢0 0 −21⎥ 1 −5 −21⎥ ⎥ ⎢ ⎥ −288⎦ 0 1 4⎦ ⎣0 0
y − 5(4) = −21 y = −1
x + 4(−1) + 4 = 3 x=3
w − 3(3) − 2( −1) + 4 = −1 w=2
The solution is (2, 3, –1, 4). [10.1]
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746
33.
Chapter 10: Matrices
⎡1 ⎢1 ⎢ ⎢2 ⎢ ⎣2
−1R1 + R 2 3⎤ 1 −4 3⎤ ⎡1 3 ⎡1 −2 R 1 + R 3 −2 R 2 + R 3 ⎥ ⎢ 0 1 2 −2 ⎥ ⎢0 5 2 − + R R + 2 R R 1 2 4 1 4 ⎥ ⎯⎯⎯⎯⎯→ ⎢ ⎥ ⎯⎯⎯⎯⎯→ ⎢ ⎢0 2 ⎢0 8 7 −5 11⎥ 5 3 5⎥ ⎥ ⎢ ⎥ ⎢ 5 0 −6 4 ⎦ 2 −2 ⎦ ⎣ 0 −1 − 2 ⎣0
3 1 −4 4 3 −6
⎧w + 3x + y − 4 z = 3 ⎪ x + 2 y − 2z = 2 ⎨ ⎪ y + 7z = 1 ⎩
y = −7 z + 1
x + 2(−7 z + 1) − 2 z = 2
3 1 −4 3⎤ 1 2 −2 2 ⎥ ⎥ 0 1 7 1⎥ ⎥ 0 0 0 0⎦
w + 3(16 z ) + (−7 z + 1) − 4 z = 3
x = 16 z
w = −37 z + 2
Let z be any real number c. The solution is ( −37c + 2, 16c, − 7c + 1, c ). [10.1] 34.
−2 R + R ⎡ 1 4 −2 3 6⎤ −1R 1+ R 2 ⎡ 1 1 3 ⎢ 2 9 −1 5 13⎥ ⎢ −3R1 + R 4 ⎢ 0 ⎢ ⎥ ⎯⎯⎯⎯⎯→ ⎢1 7 ⎢0 6 5 9⎥ ⎢ ⎥ ⎢ 0 7 20⎦ ⎣ 3 14 ⎣0 ⎧w + 4 x − 2 y + 3z = 6 ⎪ x + 3y − z = 1 ⎨ ⎪ y − 5z = 0 ⎩
4 −2 3 1 3 −1 3 8 2 2 6 −2
6⎤ 1 −3R 2 + R 3 ⎡ ⎢0 1⎥ − + 2 R R 2 4 ⎥ ⎯⎯⎯⎯⎯→ ⎢ ⎢0 3⎥ ⎢ ⎥ 2⎦ ⎣0
x + 3(5 z ) − z = 1
y = 5z
4 −2 3 1 3 −1 0 −1 5 0 0 0
6⎤ ⎡1 1⎥ −1R 3 ⎢0 ⎥ ⎯⎯⎯⎯ →⎢ ⎢0 0⎥ ⎥ ⎢ 0⎦ ⎣0
4 −2 3 1 3 −1 0 1 −5 0 0 0
6⎤ 1⎥ ⎥ 0⎥ ⎥ 0⎦
w + 4(−14 z + 1) − 2(5 z ) + 3z = 6
x = −14 z + 1
w = 63z + 2
Let z be any real number c. The solution is (63c + 2, − 14c + 1, 5c, c ). [10.1] 35.
Because there are three points, the degree of the interpolating polynomial is at most 2. The form of the polynomial is p ( x) = a2 x 2 + a1x + a0 . Use this polynomial and the given points to find the system of equations.
p(−1) = a2 (−1) 2 + a1 (−1) + a0 = −4 p (2) = a2 (2) 2 + a1 (2) + a0 = 8 p (3) = a2 (3) 2 + a1 (3) + a0 = 16 ⎧ a2 − a1 + a0 = −4 ⎪ The system of equations and the associated augmented matrix are ⎨4a2 + 2a1 + a0 = 8 ⎪ 9a + 3a + a = 16 1 0 ⎩ 2
⎡ 1 − 1 1 − 4⎤ ⎢ ⎥ 8⎥ ⎢4 2 1 ⎢⎣9 3 1 16⎥⎦ The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. ⎧a + 1 a + 1 a = 16 ⎪ 2 3 1 9 0 9 1 / 9 16 / 9⎤ ⎡ 1 1/ 3 ⎪⎪ 13 2 ⎢ ⎥ a1 − a0 = The augmented matrix in echelon form and resulting system of equations are ⎢0 1 − 2 / 3 13 / 3⎥ ⎨ 2 3 ⎪ ⎢⎣0 0 1 − 2⎥⎦ a0 = −2 ⎪ ⎪⎩ Solving by back substitution yields a0 = −2, a1 = 3, and a2 = 1 .
The interpolating polynomial is p( x) = x 2 + 3 x − 2 . [10.1]
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Chapter Review
36.
747
Because there are three points, the degree of the interpolating polynomial is at most 2. The form of the polynomial is p ( x) = a2 x 2 + a1x + a0 . Use this polynomial and the given points to find the system of equations. p (−1) = a2 (−1) 2 + a1 (−1) + a0 = 4 p(1) = a2 (1)2 + a1 (1) + a0 = 0
p(2) = a2 (2) 2 + a1 (2) + a0 = −5 ⎧ a2 − a1 + a0 = 4 ⎡ 1 −1 1 4 ⎤ ⎪ ⎢ 1 1 1 0⎥ The system of equations and the associated augmented matrix are ⎨ a2 + a1 + a0 = 0 ⎢ 4 2 1 −5⎥ ⎪⎩4a2 + 2a1 + a0 = −5 ⎣ ⎦ The ref (row echelon form) feature of a graphing calculator can be used to rewrite the augmented matrix in echelon form. Consider using the function of your calculator that converts a decimal to a fraction. ⎧a + 1 a + 1 a = − 5 1 / 4 − 5 / 4⎤ ⎡ 1 1/ 2 4 ⎪ 2 2 1 4 0 ⎪ 1 7 ⎢ ⎥ a1 − a0 = − The augmented matrix in echelon form and resulting system of equations are ⎢0 1 − 1 / 2 − 7 / 2⎥ ⎨ 2 2 ⎪ ⎢⎣0 0 1 3⎥⎦ a0 = 3 ⎪⎩ Solving by back substitution yields a0 = 3, a1 = −2, and a2 = −1 . The interpolating polynomial is p(x) = –x2 – 2x + 3. [10.1]
37.
38.
1 0⎤ ⎡ 2 −2 1 0⎤ (1/ 2) R 1 ⎡ 1 −1 12 0⎤ −3R 1 + R 2 ⎡ 1 −1 R 2 + R 1 ⎡ 1 0 −1 1⎤ 2 ⎥ ⎯⎯⎯⎯⎯ →⎢ →⎢ →⎢ ⎥ ⎥ ⎯⎯⎯⎯⎯⎯ ⎢ 3 −2 0 1⎥ ⎯⎯⎯⎯⎯ 3 3 ⎣ ⎦ ⎣⎢3 −2 0 1⎦⎥ ⎣⎢ 0 1 − 2 1⎦⎥ ⎣⎢0 1 − 2 1⎦⎥ ⎡ −1 1⎤ The inverse matrix is ⎢ 3 ⎥. [10.3] ⎢⎣− 2 1⎥⎦
⎡ 3 4 1 0⎤ 13 R 1 ⎡ 1 43 13 0⎤ −2 R 1 + R 2 ⎡ 1 →⎢ ⎥ ⎯⎯⎯⎯⎯⎯ ⎢ 2 3 0 1⎥ ⎯⎯⎯→ ⎢ ⎢⎣ 0 ⎣ ⎦ ⎣⎢ 2 3 0 1⎥⎦
4 3 1 3
1 3 2 −3
1 0⎤ 0 ⎤ 3R 2 ⎡ 1 4 − 43 R 2 + R 1 ⎡ 1 0 3 −4 ⎤ 3 3 →⎢ ⎥ ⎯⎯⎯→ ⎢ ⎥ ⎯⎯⎯⎯⎯⎯ 0 1 2 3⎥⎦ − 1⎦⎥ 0 1 2 3 − ⎣ ⎥⎦ ⎣⎢
⎡ 3 −4⎤ The inverse matrix is ⎢ ⎥. [10.3] 3⎦ ⎣− 2
39.
1R ⎡ 1 − 23 − 12 ⎡ −2 3 1 0⎤ − 21 R 1 ⎡ 1 − 23 − 12 0⎤ −2 R 1 + R 2 ⎡ 1 − 23 − 12 0⎤ 7 2 ⎯⎯⎯→ ⎯⎯⎯⎯ → ⎯⎯⎯⎯⎯⎯ → ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ 2 4 0 1⎥ 1 1 4 0 1⎥⎦ 7 1 1⎥⎦ ⎢⎣ 0 ⎣ ⎦ ⎢⎣ 2 ⎢⎣0 7
3R +R ⎡ 1 0 − 72 1 2 2 ⎯⎯⎯⎯⎯→ ⎢ 1 ⎢⎣0 1 7
⎡− 2 The inverse matrix is ⎢ 7 ⎢ 1 ⎣ 7
40.
0⎤ ⎥ 1 7⎥ ⎦
3⎤ 14 ⎥ 1 7⎥ ⎦
3⎤ 14 ⎥. [10.3] 1⎥ 7⎦
4 ⎡5 −4 1 0⎤ (1/ 5) R 1 ⎡ 1 − 45 15 0⎤ −3R 1 + R 2 ⎡ 1 − 5 ⎢ ⎯⎯⎯⎯⎯ → ⎯⎯⎯⎯⎯⎯ → ⎢ ⎥ ⎢ ⎥ ⎢⎣0 22 2 0 1⎥⎦ ⎣ 3 2 0 1⎦ ⎢⎣3 5
1 5 − 53
1 ⎡ 1 − 45 0⎤ (5/ 22) R 5 2 ⎥ ⎯⎯⎯⎯⎯→ ⎢ 3 ⎢⎣0 1⎥⎦ 1 − 22
1 2⎤ (4 / 5) R 2 + R 1 ⎡ 1 0 11 11 ⎥ ⎯⎯⎯⎯⎯⎯⎯ →⎢ 3 5 ⎥ ⎢⎣0 1 − 22 22 ⎦ 2⎤ ⎡ 1 11 ⎥. [10.3] The inverse matrix is ⎢ 11 5 ⎥ ⎢− 3 ⎣ 22 22 ⎦
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0⎤ ⎥
5 ⎥ 22 ⎦
748
41.
Chapter 10: Matrices
−2 R 1 + R 2 ⎡ 1 2 1 1 0 0⎤ ⎡ 1 2 1 1 0 0⎤ ⎡ 1 2 1 1 0 0⎤ −3R 1 + R 3 ⎢ (1/ 2) R 2 ⎢ ⎢ 2 6 4 0 1 0⎥ ⎯⎯⎯⎯⎯⎯ → 0 2 2 −2 1 0⎥ ⎯⎯⎯⎯⎯ → 0 1 1 −1 12 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣0 2 3 −3 0 1⎥⎦ ⎣ 3 8 6 0 0 1⎦ ⎣ 0 2 3 −3 0 1⎦ ⎡ 1 0 0 2 −2 1⎤ ⎡ 1 2 1 1 0 0⎤ ⎡ 1 0 −1 3 − 1 0 ⎤ R 3 + R 1 −2 R 2 + R 3 ⎢ −R 3 + R 2 ⎢ −2 R 2 + R 1 ⎢ ⎥ 3 −1⎥ 1 ⎯⎯⎯⎯⎯⎯ → 0 1 1 −1 2 0 ⎯⎯⎯⎯⎯⎯ → 0 1 1 −1 12 0⎥ ⎯⎯⎯⎯⎯→ 0 1 0 0 2 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢0 0 1 −1 −1 1⎦⎥ ⎣⎢ 0 0 1 −1 −1 1⎦⎥ ⎣⎢0 0 1 −1 −1 1⎦⎥ ⎡ 2 −2 1⎤ The inverse matrix is ⎢ 0 32 −1⎥ . [10.3] ⎢ ⎥ ⎢⎣ −1 −1 1⎥⎦
42.
−3R 1 + R 2 1 0 0⎤ 1 ⎡ 1 −3 2 1 0 0⎤ ⎡ 1 −3 2 ⎡ 1 −3 2 −2 R 1 + R 3 ⎢ −3 R 2 + R 3 ⎢ ⎢ 3 −8 7 0 1 0⎥ ⎯⎯⎯⎯⎯⎯ 1 1 −3 1 0⎥ ⎯⎯⎯⎯⎯⎯ 1 1 −3 → 0 → 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎣ 2 −3 6 0 0 1⎦ ⎣ 0 3 2 −2 0 1⎦ ⎣ 0 0 −1 7 −5R 3 + R 1 1 0 0⎤ ⎡ 1 −3 2 ⎡ 1 0 5 −8 3 0⎤ ⎡1 0 −1R 3 ⎢ −R 3 + R 2 3R 2 + R 1 ⎢ 1 1 −3 1 0⎥ ⎯⎯⎯⎯⎯ ⎯⎯⎯⎯ → 0 → 0 1 1 −3 1 0⎥ ⎯⎯⎯⎯⎯⎯ → ⎢0 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎣ 0 0 1 −7 3 −1⎦ ⎣0 0 1 −7 3 −1⎦ ⎣0 0
0 0⎤ 1 0⎥ ⎥ −3 1⎦ 0 27 −12 5⎤ 0 4 −2 1⎥ ⎥ 1 −7 3 −1⎦
⎡ 27 −12 5⎤ The inverse matrix is ⎢ 4 −2 1⎥ . [10.3] ⎢ −7 3 −1⎥⎦ ⎣ 43.
⎡ 1 − 23 73 13 0 0⎤ −2 R 1 + R 2 ⎡ 1 − 23 ⎡ 3 −2 7 1 0 0⎤ ⎢ ⎥ −3R 1 + R 3 ⎢ (1/ 3) R 1→ 2 1 ⎢ 2 −1 5 0 1 0⎥ ⎯⎯⎯⎯⎯ −1 5 0 1 0⎥ ⎯⎯⎯⎯⎯⎯ → ⎢0 ⎢ 3 ⎢ ⎥ ⎢ 0 10 0 0 1⎥ ⎢3 0 2 ⎣ 3 0 10 0 0 1⎦ ⎣ ⎦ ⎣ ⎡ 1 − 23 73 3R 2 ⎢ 1 1 ⎯⎯⎯ → ⎢0 ⎢0 2 3 ⎢⎣ −3R 3 + R 1 ⎡1 0 −1R 3 + R 2 ⎢ ⎯⎯⎯⎯⎯⎯ → 0 1 ⎢ ⎢⎣0 0
7 3 1 3
3
0 0⎤ ⎥ 1 0⎥ −1 0 1⎥ ⎦
1 3 − 23
1 ⎡ 1 − 23 73 0 0⎤ 0 0⎤ ⎡ 1 0 3 −1 2 0 ⎤ 3 ⎥ ⎢ ⎥ −2 R 2 + R 3 (2 / 3) R 2 + R 1 ⎢ 1 1 −2 3 0⎥ ⎯⎯⎯⎯⎯⎯⎯ −2 3 0⎥ ⎯⎯⎯⎯⎯⎯ → ⎢0 3 0⎥ → 0 1 1 −2 ⎢ ⎥ ⎢0 ⎥ ⎢⎣0 0 1 3 −6 1⎥⎦ 0 1 3 6 1 −1 0 1⎥⎥ − ⎢⎣ ⎦ ⎦⎥ 1 3
0 −10 20 −3⎤ 0 −5 9 −1⎥ ⎥ 1 3 −6 1⎥⎦ ⎡ −10 20 −3⎤ The inverse matrix is ⎢ −5 9 −1⎥ . [10.3] ⎢ 3 −6 1⎥⎦ ⎣ 44.
−3R 1 + R 2 ⎡ 1 1 0 0⎤ ⎡ 1 94 − 11 ⎡ 4 9 −11 1 0 0⎤ 4 4 −2 R 1 + R 3 ⎢ R (1/ 4) ⎢ ⎥ 1→ 3 7 ⎢ 3 7 −8 0 1 0⎥ ⎯⎯⎯⎯⎯ −8 0 1 0⎥ ⎯⎯⎯⎯⎯⎯ → ⎢0 ⎢ ⎢ ⎥ ⎢ ⎢ 2 6 −3 0 0 1⎥ ⎣ 2 6 −3 0 0 1⎦ ⎣ ⎦ ⎣0 1 0 0⎤ ⎡ 1 94 − 11 ⎡ 1 94 − 11 4 4 4 4R 2 ⎢ ⎥ ( −3/ 2) R 2 + R 3 ⎢ 1 −3 4 0⎥ ⎯⎯⎯⎯⎯⎯⎯⎯ 1 ⎯⎯⎯→ ⎢0 1 → ⎢0 1 5 − 1 0 1⎥ 1 ⎢0 3 ⎢0 0 2 2 2 ⎣ ⎦ ⎣ 5R 3 + R 1 ⎡ 1 0 −5 7 −9 0⎤ ⎡1 0 −1R 3 + R 2 ⎢ ( −9 / 4) R 2 + R 1 ⎢ → 0 1 ⎯⎯⎯⎯⎯⎯⎯⎯ → 0 1 1 −3 4 0⎥ ⎯⎯⎯⎯⎯⎯ ⎢ ⎥ ⎢ 1 4 −6 1⎦ ⎣0 0 ⎣0 0 ⎡ 27 −39 5⎤ The inverse matrix is ⎢ −7 10 −1⎥ . [10.3] ⎢ 4 −6 1⎥ ⎣ ⎦
9 4 1 4 3 2
− 11 4 1 4 5 2
1 4 − 43 − 21
0 0⎤ ⎥ −3 4 0⎥ 4 −6 1⎥ ⎦ 1 4
0 27 −39 5⎤ 0 −7 10 −1⎥ ⎥ 1 4 −6 1⎦
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0 0⎤ ⎥ 1 0⎥ ⎥ 0 1⎦
Chapter Review
45.
⎡1 ⎢2 ⎢ ⎢3 ⎣⎢ 2
−1 −1 −1 −2
749
2 6 9 4
3 5 6 7
1 0 0 0
⎡ 1 −1 −1R 3 ⎢ 0 1 ⎯⎯⎯⎯ →⎢ ⎢0 0 ⎢⎣ 0 0 −4 R 3 + R 1 ⎡ 1 −2 R 3 + R 2 ⎢ 0 ⎯⎯⎯⎯⎯⎯ →⎢ ⎢0 ⎣⎢ 0
0 1 0 0
0 0 1 0
−2 R 1 + R 2 0⎤ −3R 1 + R 3 ⎡ 1 −1 0⎥ −2 R 1 + R 4 ⎢0 1 →⎢ ⎥ ⎯⎯⎯⎯⎯⎯ 0⎥ ⎢0 2 1⎦⎥ ⎣⎢0 0
2 3 1 0 0 2 −1 −2 1 0 1 1 −1 2 −1 0 1 −2 0 0 0 1 0 0
2 3 1 0 2 −1 − 2 1 3 −3 −3 0 0 1 −2 0
0⎤ ⎡1 0⎥ R 2 + R 1 ⎢ 0 →⎢ ⎥ ⎯⎯⎯⎯⎯ 0⎥ ⎢0 1⎥⎦ ⎢⎣0
0 1 0 0
4 2 −1 2 − 1 −2 1 1 −1 0 1 −2
2R 4 + R 1 0 ⎤ 3R 4 + R 2 ⎡1 0⎥ −1R 4 + R 3 ⎢0 →⎢ ⎥ ⎯⎯⎯⎯⎯⎯ 0⎥ ⎢0 1⎦⎥ ⎣⎢0
0 −2 3 −7 4 0 −3 0 −3 2 1 1 −1 2 −1 0 1 −2 0 0
0⎤ 1 0 ⎡ 1 −1 2 3 0 ⎥ − 2 R 2 + R 3 ⎢ 0 1 2 −1 − 2 1 →⎢ ⎥ ⎯⎯⎯⎯⎯⎯ 0⎥ 0 0 1 1 1 2 − − − ⎢ 1⎦⎥ 0 ⎣⎢0 0 0 1 −2
0 0 1 0
0 1 0 0
0 0 1 0
1 0 1 0 2 −1 0 0
0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦
0 −1 −7 4 2 ⎤ 0 −6 −3 2 3⎥ ⎥ 0 1 2 −1 −1⎥ 1 −2 0 0 1⎦⎥
4 2⎤ ⎡ −1 −7 ⎥ ⎢ 3⎥ −6 −3 2 The inverse matrix is ⎢ . [10.3] ⎢ 1 2 − 1 − 1⎥ ⎥ ⎢ 0 0 1⎦⎥ ⎣⎢ − 2
46.
⎡1 ⎢ ⎢3 ⎢2 ⎢ ⎣⎢ 1
−3R 1 + R 2 1 1 0 0 0⎤ −2 R + R 1 1 ⎡ 1 2 −2 1 3 ⎥ 7 − 3 1 0 1 0 0⎥ −1R 1 + R 4 ⎢0 1 3 −2 −3 ⎯⎯⎯⎯⎯⎯ →⎢ 1 −2 7 4 3 0 0 1 0⎥ ⎢0 3 8 ⎥ 4 3 −1 ⎣⎢0 2 4 2 4 0 0 0 1⎦⎥ 2
−2
−3R 2 + R 3 ⎡ 1 −2 R 2 + R 4 ⎢0 ⎯⎯⎯⎯⎯⎯ →⎢ ⎢0 ⎣⎢0 ⎡1 2 R 3 + R 4 ⎢0 ⎯⎯⎯⎯⎯→ ⎢ ⎢0 ⎢⎣ 0 8R 3 + R 1 ⎡1 −3R 3 + R 2 ⎢0 ⎯⎯⎯⎯⎯⎯ →⎢ ⎢0 ⎢⎣0
7R 4 + R 3 −19 R 4 R 2 ⎡⎢ 1 51R 4 + R 1 ⎢0 ⎯⎯⎯⎯⎯⎯ →⎢ 0 ⎢ ⎢⎣0
2 −2 1 1 0 1 3 −2 −3 1 0 −1 7 7 −3 0 −2 7 5 −2 2 −2 1 1 1 3 −2 −3 0 1 −7 −7 0 0 −7 −9 0 1 0 0
0 0 1 0
0 0 1 0 3 −1 4 −2
0⎤ ⎡1 2 0⎥ −1R 3 ⎢ 0 1 →⎢ ⎥ ⎯⎯⎯⎯ 0⎥ ⎢0 0 1⎥⎦ ⎢⎣ 0 0 0⎤ ⎡1 0⎥ −2 R 2 + R 1 ⎢0 →⎢ ⎥ ⎯⎯⎯⎯⎯⎯ 0⎥ ⎢0 1⎥⎦ ⎢⎣0
0 −51 −49 22 −8 0 19 18 −8 3 1 −7 −7 3 −1 0 −7 −9 4 −2
0 0 0 1 0 0 0 1 0 0 0 1
116 7 − 45 7
2 9 7
− 50 7 20 7
−1 − 74
0 1 0 0
0 0 1 0
0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦
1 1 0 0 −2 3 −2 −3 1 0 1 −7 −7 3 −1 −2 7 5 −2 0
0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦
0 −8 5 7 −2 0 1 3 −2 −3 1 0 0 1 −7 −7 3 −1 0 0 −7 −9 4 −2
⎡1 0 0⎤ ⎥ 0 ( −1/ 7) R 4 ⎢⎢0 1 ⎥ ⎯⎯⎯⎯⎯→ 0⎥ ⎢0 0 ⎢0 0 1⎥⎦ ⎣
0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦
0 −51 −49 22 −8 0⎤ 0 19 18 −8 3 0⎥ ⎥ 1 −7 −7 3 −1 0⎥ 9 −4 2 − 1⎥ 0 1 7 7 7 7⎦
⎤ − 51 7 ⎥ 19 7⎥ 1 −1⎥ ⎥ 2 − 71 ⎥⎦ 7
46 7 − 17 7
46 − 51 ⎤ ⎡ 116 − 50 7 7 7⎥ ⎢ 7 20 − 17 19 ⎥ ⎢ − 45 7 7 7 ⎥. [10.3] The inverse matrix is ⎢ 7 1 −1 − 1⎥ ⎢ 2 ⎢ 9 2 −4 − 1 ⎥⎥ 7 7 7⎦ ⎣⎢ 7
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0 0 1 0
0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦
750
47.
Chapter 10: Matrices
⎡3 ⎢2 ⎢ ⎢3 ⎣⎢ 2
7 −1 5 0 6 −4 4 −4
8 5 8 4
1 0 0 0
0 1 0 0
0 0 1 0
⎡ 1 73 − 13 83 13 0⎤ ⎥ 0 (1/ 3) R 1 ⎢⎢ 2 5 0 5 0 → ⎥ ⎯⎯⎯⎯⎯ 0⎥ ⎢ 3 6 −4 8 0 ⎢ 1⎦⎥ ⎣ 2 4 −4 4 0
7 8 1 ⎡1 − 13 3 3 3 ⎢ 1 2 −1 −2 3R 2 ⎢0 ⎯⎯⎯→ 0 −1 ⎢0 −1 −3 ⎢0 − 2 − 10 − 4 − 2 3 3 3 3 ⎣
0 3 0 0
0 0 1 0
8 1 ⎡ 1 73 − 13 0 0 3 3 −1R 3 ⎢0 1 2 −1 −2 3 0 ⎯⎯⎯⎯ →⎢ 1 1 3 −3 −1 ⎢0 0 ⎢0 0 −2 −2 −2 2 0 ⎣
0 1 0 0
0 0 1 0
−2 R 1 + R 2 7 8 1 − 13 0⎤ −3R 1 + R 3 ⎡ 1 3 3 3 ⎢ ⎥ −2 R + R 1 2 −1 −2 0⎥ ⎯⎯⎯⎯⎯⎯ 1 4 → ⎢0 3 3 3 3 ⎢0 −1 −3 0 −1 0⎥ ⎢ 10 2 4 2 1⎦⎥ ⎣⎢0 − 3 − 3 − 3 − 3
8 1 0 0⎤ 1R 2 + R 3 ⎡ 1 73 − 13 3 3 ⎥ ⎢ 0⎥ (2 / 3) R 2 + R 4 2 −1 −2 3 ⎯⎯⎯⎯⎯⎯⎯→ ⎢0 1 0⎥ ⎢0 0 −1 −1 −3 3 ⎢ 0 0 −2 −2 − 2 2 1⎥⎦ ⎣
0⎤ ⎥ 0⎥ 0⎥ 1⎥⎦
0 0 1 0
8 1 ⎡ 1 73 − 13 0⎤ 0 0 3 3 ⎢ ⎥ 2R + R 0⎥ ⎯⎯⎯⎯⎯→ 2 −1 −2 3 0 3 4 ⎢0 1 ⎢0 0 0⎥ 1 1 3 −3 −1 ⎢0 0 1⎥⎦ 0 0 4 −4 −2 ⎣
0 0 0⎤ ⎥ 1 0 0⎥ 0 1 0⎥ ⎥ 0 0 1⎥⎦
0⎤ ⎥ 0⎥ 0⎥ 1⎥⎦
The matrix does not have an inverse. [10.3]
48.
⎡3 ⎢2 ⎢ ⎢3 ⎢⎣ 4
1 1 0 1
5 −5 1 0 4 −3 0 1 4 −3 0 0 8 1 0 0
0 0 1 0
⎡ 1 13 53 − 53 13 0 0⎤ ⎢ ⎥ 0 (1/ 3) R 1 ⎢ 2 1 4 −3 0 1 → ⎥ ⎯⎯⎯⎯⎯ 0⎥ ⎢ 3 0 4 −3 0 0 ⎢4 1 8 1⎦⎥ 1 0 0 ⎣
3R 2 1 ⎡ 1 13 53 − 53 1R 2 + R 3 3 (1/ 3) R 2 + R 4 ⎢0 1 2 1 2 − ⎯⎯⎯⎯⎯⎯⎯ →⎢ 3 −3 ⎢0 0 1 ⎢0 0 2 8 −2 ⎣
0 3 3 1
0 0 1 0
0 0 1 0
−2 R 1 + R 2 5 −5 1 1 0⎤ −3R 1 + R 3 ⎡ 1 3 3 3 3 ⎢ ⎥ −4 R + R 1 2 1 −2 0⎥ ⎯⎯⎯⎯⎯⎯ 1 4 → ⎢0 3 3 3 3 ⎢0 −1 −1 2 −1 0⎥ ⎢ 23 − 4 1 4 1⎥⎦ 3 3 3 ⎣⎢0 − 3
1 ⎡ 1 13 53 − 53 0 0 0⎤ 3 ⎢ ⎥ − + 2 R R 1 −2 3 0 0⎥ ⎯⎯⎯⎯⎯⎯⎯ 3 4 → ⎢0 1 2 ⎢0 0 1 3 −3 3 1 0⎥ ⎢0 0 0 2 4 −5 −2 1⎥⎦ ⎣
1 ⎡1 1 5 − 5 0 0 0⎤ ⎡1 3 3 3 3 ⎢ ⎥ 1 −2 3 0 0⎥ (1/2)R 4 ⎢ 0 1 2 ( −1/3)R 2 + R 1 ⎢0 ⎯⎯⎯⎯⎯⎯ →⎢ ⎯⎯⎯⎯⎯⎯⎯⎯→ ⎢0 0 0 1 3 −3 3 1 0⎥ ⎢ ⎢ ⎥ ⎢0 5 1 ⎢0 0 0 ⎥ ⎣ 1 2 − 2 −1 2 ⎦ ⎣ 5R 4 + R 1 ⎡1 0 0 −2R 3 + R 2 ⎡ 1 0 0 −5 4 −4 −1 0⎤ 5R 4 + R 2 ⎢ ⎢ ⎥ − − − 0 1 0 5 4 3 2 0 −3 R 4 + R 3 ⎢ 0 1 0 −1R 3 + R 1 ⎥ ⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯⎯ → ⎢ 0 0 1 3 −3 → ⎢ 3 1 0⎥ ⎢ ⎢0 0 1 5 1 ⎢0 0 0 ⎥ − − 1 2 1 ⎢0 0 0 2 2⎦ ⎣ ⎣
⎡ 14 − 33 2 ⎢ ⎢ 14 − 31 2 The inverse matrix is ⎢ 21 ⎢− 9 2 ⎢ ⎢ 2 −5 2 ⎣
0 1 0 0
1 −2 1 −1 0 0⎤ 2 1 −2 3 0 0⎥ ⎥ 1 3 −3 3 1 0⎥ 0 1 2 − 25 −1 12 ⎥⎦
0 14 − 33 2
−6
−7 0 14 − 31 2 0 −9 1
0⎤ ⎥ 0⎥ 0⎥ 1⎥⎦
2
21 2 − 25
5⎤ 2⎥ 5⎥ −7 2 ⎥. [10.3] 4 − 3⎥ 2⎥ 1⎥ −1 2⎦
−6
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4 −1
5⎤ 2 ⎥ 5 2⎥ ⎥ − 23 ⎥ 1⎥ 2⎦
0 0 0⎤ ⎥ 1 0 0⎥ 0 1 0⎥ ⎥ 0 0 1⎥⎦
Chapter Review
49.
a.
751
⎡3 ⎢2 ⎣ ⎡ 3 −4 ⎤ ⎡ 3 ⎢ −2 3⎥⎦ ⎢⎣ 2 ⎣
50.
4⎤ ⎡ x ⎤ ⎡ 2⎤ = 3 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ −3⎥⎦ 4⎤ ⎡ x ⎤ ⎡ 3 −4 ⎤ ⎡ 2⎤ = 3⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ −2 3⎥⎦ ⎢⎣ −3⎥⎦ ⎡ x ⎤ ⎡ 18⎤ ⎢ y ⎥ = ⎢ −13⎥ ⎣ ⎦ ⎣ ⎦
a.
The solution is (18, −13). b.
⎡ 3 −4 ⎤ ⎢ −2 3⎥⎦ ⎣
⎡3 ⎢2 ⎣ ⎡3 ⎢2 ⎣
b.
4 ⎤ ⎡ x ⎤ ⎡ −2⎤ = 3 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ 4⎥⎦ 4⎤ ⎡ x ⎤ ⎡ 3 −4 ⎤ ⎡ −2⎤ = 3⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ −2 3⎥⎦ ⎢⎣ 4⎥⎦ ⎡ x ⎤ ⎡ −22⎤ ⎢ y ⎥ = ⎢ 16⎥ ⎣ ⎦ ⎣ ⎦
⎡ −7 5⎤ ⎢ −3 2 ⎥ ⎣ ⎦
a. ⎡ 1 −1 14 ⎢ 2 ⎢ −1 7 ⎢ 1 ⎢ 0 7 ⎣
b.
⎡ 2 −5 ⎤ ⎡ x ⎤ ⎡ −3⎤ ⎢ 3 −7 ⎥ ⎢ y ⎥ = ⎢ 4⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 5⎤ ⎡ 1 −1 − 14 14 ⎥ ⎡ 2 1 −1⎤ ⎡ x ⎤ ⎢ ⎥ ⎢ y ⎥ = ⎢ −1 3 ⎥ ⎢4 4 2 1 7⎥ ⎢ 7 ⎥ ⎢ ⎥ ⎢ 1 − 72 ⎥⎦ ⎣⎢ 2 2 −3⎦⎥ ⎣⎢ z ⎦⎥ ⎢⎣ 0 7 ⎡ − 18 ⎤ ⎡x⎤ ⎢ 7 ⎥ ⎢ y ⎥ = ⎢ 23 ⎥ ⎢ ⎥ ⎢ 7⎥ ⎣⎢ z ⎦⎥ ⎢ − 6 ⎥ ⎣ 7⎦
⎡2 ⎢3 ⎣ ⎡2 ⎢3 ⎣
−5 ⎤ ⎡ x ⎤ ⎡ 2⎤ = −7 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ −5⎥⎦ −5 ⎤ ⎡ x ⎤ ⎡ −7 5⎤ ⎡ 2⎤ = −7 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ −3 2 ⎥⎦ ⎢⎣ −5⎥⎦ ⎡ x ⎤ ⎡ −39⎤ ⎢ y ⎥ = ⎢ −16⎥ ⎣ ⎦ ⎣ ⎦
The solution is (−39, −16). [10.3]
The solution is (−22, 16). [10.3] 51.
−5 ⎤ ⎡ x ⎤ ⎡ −3⎤ = −7 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ 4⎥⎦ −5 ⎤ ⎡ x ⎤ ⎡ −7 5⎤ ⎡ −3⎤ = −7 ⎥⎦ ⎢⎣ y ⎥⎦ ⎢⎣ −3 2 ⎥⎦ ⎢⎣ 4⎥⎦ ⎡ x ⎤ ⎡ 41⎤ ⎢ y ⎥ = ⎢17 ⎥ ⎣ ⎦ ⎣ ⎦ The solution is (41, 17). ⎡2 ⎢3 ⎣ ⎡ −7 5⎤ ⎡ 2 ⎢ −3 2 ⎥ ⎢ 3 ⎣ ⎦ ⎣
5⎤ − 14 ⎥ 3⎥ 7⎥ − 72 ⎥⎦
⎡ −1⎤ ⎢ 2⎥ ⎢ ⎥ ⎣⎢ 4 ⎦⎥
5⎤ − 14 ⎥ 3⎥ 7⎥ − 72 ⎥⎦
⎡ −2⎤ ⎢ 3⎥ ⎢ ⎥ ⎣⎢ 0⎦⎥
⎛ 18 23 6 ⎞ The solution is ⎜ − , , − ⎟. 7⎠ ⎝ 7 7 ⎡ 2 1 −1⎤ ⎡ x ⎤ ⎡ −2⎤ ⎢4 4 1⎥ ⎢ y ⎥ = ⎢ 3⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 2 2 −3⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ 0⎥⎦ ⎡ 1 −1 14 ⎢ 2 ⎢ −1 7 ⎢ 1 ⎢ 0 7 ⎣
5⎤ − 14 ⎥ 3⎥ 7⎥ − 72 ⎥⎦
⎡ 2 1 −1⎤ ⎢4 4 1⎥ ⎢ ⎥ ⎣⎢ 2 2 −3⎦⎥
⎡ 1 −1 14 ⎡x⎤ ⎢ ⎢ y ⎥ = ⎢ −1 2 7 ⎢ ⎥ ⎢ 1 ⎣⎢ z ⎦⎥ ⎢ 0 7 ⎣ ⎡ − 31 ⎤ ⎡ x ⎤ ⎢ 14 ⎥ ⎢ y ⎥ = ⎢ 20 ⎥ ⎢ ⎥ ⎢ 7⎥ ⎣⎢ z ⎦⎥ ⎢ 3 ⎥ ⎣ 7⎦
⎛ 31 20 3 ⎞ The solution is ⎜ − , , ⎟. [10.3] ⎝ 14 7 7 ⎠
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752
52.
Chapter 10: Matrices
a.
5⎤ ⎡ − 11 2 3⎥ ⎢ 3 3 ⎢ −5 1 2 ⎥ ⎢ 2 0 −1⎥ ⎢⎣ ⎥⎦
b.
⎡ 3 −2 1⎤ ⎡ x ⎤ ⎡ 0⎤ ⎢ 3 −1 3⎥ ⎢ y ⎥ = ⎢ 3⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣6 −4 1⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ −2⎥⎦ 5⎤ 11 2 ⎡ 3 −2 1⎤ ⎡ x ⎤ ⎡⎢ − 3 3 3 ⎢ 3 −1 3⎥ ⎢ y ⎥ = ⎢ −5 1 2 ⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣6 −4 1⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎢ 2 0 −1⎥⎥ ⎣ ⎦ 4⎤ ⎡ − ⎡ x⎤ ⎢ 3 ⎥ ⎢ y ⎥ = ⎢ −1⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ z ⎥⎦ ⎢ 2 ⎥ ⎣ ⎦
⎛ 4 ⎞ The solution is ⎜ − , − 1, 2 ⎟. 3 ⎝ ⎠ ⎡ 3 −2 1⎤ ⎢ 3 −1 3⎥ ⎢ ⎥ ⎢⎣6 −4 1⎥⎦ 5⎤ ⎡− 1 2 3⎥ ⎢ 13 3 ⎢ −5 1 2 ⎥ ⎢ 2 0 −1⎥ ⎢⎣ ⎥⎦
⎡ 3 −2 1⎤ ⎢ 3 −1 3⎥ ⎢ ⎥ ⎢⎣6 −4 1⎥⎦
⎡ x ⎤ ⎡ 1⎤ ⎢ y ⎥ = ⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ z ⎥⎦ ⎢⎣ −4 ⎥⎦ 5⎤ 1 2 ⎡ x ⎤ ⎡⎢ − 13 3 3 ⎢ y ⎥ = ⎢ −5 1 2 ⎥⎥ ⎢ ⎥ ⎢⎣ z ⎥⎦ ⎢⎢ 2 0 −1⎥⎥ ⎣ ⎦ ⎡ x ⎤ ⎡ −9 ⎤ ⎢ y ⎥ = ⎢ −11⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ z ⎦⎥ ⎣⎢ 6 ⎦⎥
⎡ 0⎤ ⎢ 3⎥ ⎢ ⎥ ⎢⎣ −2⎥⎦
⎡ 1⎤ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ −4⎥⎦
The solution is (−9, − 11, 6). [10.3]
53.
54.
⎡ −5 4 ⎤ ⎡1 T3,−1 ⋅ ⎢ 3 −2 ⎥ = ⎢0 ⎢ 1 1⎥ ⎢0 ⎣ ⎦ ⎣ ⎡ −2 7 ⎤ ⎡0 Rxy ⋅ ⎢ 2 −3⎥ = ⎢ 1 ⎢ 1 1⎥ ⎢0 ⎣ ⎦ ⎣
0 3⎤ ⎡ −5 4⎤ 1 −1⎥ ⎢ 3 −2⎥ 0 1⎥⎦ ⎢⎣ 1 1⎥⎦ 1 0⎤ ⎡ −2 7⎤ 0 0⎥ ⎢ 2 −3⎥ = 0 1⎥⎦ ⎢⎣ 1 1⎥⎦
⎡ −3 1 0⎤ ⎡1 T−2,−3 ⋅ ⎢ 4 −1 5⎥ = ⎢0 ⎢ 1 1 1⎥ ⎢0 ⎣ ⎦ ⎣ ⎡ −5 −1 −2 ⎤ ⎡0 R90 ⋅ ⎢ 1 −4 2 ⎥ = ⎢ 1 ⎢ 1 1 1⎥ ⎢0 ⎣ ⎦ ⎣ 1 4 2 − − ⎡ ⎤ ⎡1 T2, 3 ⋅ ⎢ −5 −1 −2 ⎥ = ⎢ 0 ⎢ 1 1 1⎥ ⎢0 ⎣ ⎦ ⎣
⎡ −2 7 ⎤ = ⎢ 2 −3⎥ ⎢ 1 1⎥ ⎣ ⎦ ⎡ 2 −3⎤ ⎢ −2 7⎥ ⇒ A '(2, − 2) [10.2] B '( −3, 7) ⎢ 1 1⎥ ⎣ ⎦
0 −2⎤ ⎡ −3 1 0⎤ ⎡ −5 1 −3⎥ ⎢ 4 −1 5⎥ = ⎢ 1 ⎢ 1 0 1⎥⎦ ⎢⎣ 1 1 1⎥⎦ ⎣ −1 0⎤ ⎡ −5 −1 −2⎤ ⎡ −1 0 0⎥ ⎢ 1 −4 2⎥ = ⎢ −5 ⎢ 1 0 1⎥⎦ ⎢⎣ 1 1 1⎥⎦ ⎣ 0 2 ⎤ ⎡ −1 4 −2⎤ ⎡ 1 1 3⎥ ⎢ −5 −1 −2⎥ = ⎢ −2 ⎢ 1 0 1⎥⎦ ⎢⎣ 1 1 1⎥⎦ ⎣
−1 − 2 ⎤ −4 2 ⎥ 1 1⎥⎦ 4 −2 ⎤ −1 −2⎥ 1 1⎥⎦ 6 0⎤ A '(1, − 2) 2 1⎥ ⇒ B '(6, 2) [10.2] 1 1⎥⎦ C '(0, 1)
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Chapter Review
55.
57.
753
2 6 4 1 3 2 Factor 2 from row 1 1 2 1 ⎯⎯⎯⎯⎯⎯⎯→ 2 1 2 1 3 8 6 3 8 6 −1R 1 + R 2 1 3 2 − 3R 1 + R 3 ⎯⎯⎯⎯⎯⎯ → 2 0 −1 −1 0 −1 0 1 3 2 −R 2 + R 3 2 0 −1 −1 ⎯⎯⎯⎯⎯→ 0 0 1 = 2(1)( −1)(1) = −2 [10.4] ( −2 / 3) R 1 + R 2 3 −8 7 3 −8 7 ( −1/ 3) R 1 + R 3 7 4 2 −3 6 ⎯⎯⎯⎯⎯⎯⎯⎯ → 0 3 3 1 −3 2 0 − 13 − 13 3 −8 (1/ 7) R 2 + R 3 7 ⎯⎯⎯⎯⎯⎯⎯ → 0 3 0
0
56.
3 0 10 3 0 0 (−10 / 3)C1 + C3 → 3 −2 −3 3 −2 7 ⎯⎯⎯⎯⎯⎯⎯ 2 −1 5 2 −1 − 53 3 0 (−3/ 2)C2 + C3 ⎯⎯⎯⎯⎯⎯⎯ → 3 −2 2
−1 − 16
( )
= 3(−2) − 16 =1 [10.4]
58.
7 4 3 − 17
( −1/ 2) R 1 + R 2 4 9 −11 4 9 −11 ( −3/ 4) R 1 + R 3 5 2 6 −3 ⎯⎯⎯⎯⎯⎯⎯⎯ → 0 23 2 3 7 −8 1 0 14 4 4 9 −11 ( −1/ 6) R 2 + R 3 5 ⎯⎯⎯⎯⎯⎯⎯⎯ → 0 23 2
7⎛ 1⎞ = 3 ⎜ − ⎟ = −1 [10.4] 3⎝ 7⎠
59.
0 0
1 −1 2 1 1 −1 2 1 2 −1 6 3 Factor 3 from row 4 2 −1 6 3 ⎯⎯⎯⎯⎯⎯⎯⎯⎯→ 3 3 −1 8 7 3 −1 8 7 3 0 9 9 1 0 3 3 −R1 + R 2 −3R 1 + R 3 1 −1 2 1 0 1 2 1 −1R 1 + R 4 ⎯⎯⎯⎯⎯⎯ → 3 0 2 2 4 0 1 1 2 −2 R 2 + R 3 1 −1 2 1 0 1 2 1 −1R 2 + R 4 ⎯⎯⎯⎯⎯⎯ → 3 0 0 −2 2 0 0 −1 1 1 −1 2 1 Factor − 2 from row 3 0 1 2 1 R3 + R4 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ − 6 0 0 1 −1 0 0 0 0 = −6(1)(1)(1)(0) = 0 [10.4]
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0
0
−1 6
⎛ 3⎞ ⎛ 1⎞ = 4 ⎜ ⎟ ⎜ − ⎟ = −1 [10.4] ⎝ 2⎠ ⎝ 6⎠
754
60.
Chapter 10: Matrices
1 3 2 2
−3R 1 + R 2 2 −2 3 −2 R 1 + R 3 7 −3 11 −2 R 1 + R 4 ⎯⎯⎯⎯⎯⎯ → 3 −5 11 6 1 8 R2 + R3 −2 R 2 + R 4 ⎯⎯⎯⎯⎯⎯ →
1 0 0 0 1 0 0 0 1 0 (1/ 2) R 3 + R 4 ⎯⎯⎯⎯⎯⎯⎯ → 0 0
2 1 −1 2 2 1 0 0 2 1 0 0
−2 3 −1 5 −2 3 2 −1 −2 3 2 0
3 2 5 2 3 2 7 −2 3 2 7
61.
1 2 2 3
1 3 −2 3 11 −4 2 9 −8 3 12 −10
63.
2 −3 2 5
x1 =
3 2
2 −3 3 5
0 Factor − 2 from column 3 1 3 1 0 4 3 11 2 2 Factor 2 from column 4 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ → −4 2 2 9 4 1 2 3 12 5 1 −2 R 4 + R 2 1 3 1 0 −3 −13 −8 0 −1R 4 + R 3 ⎯⎯⎯⎯⎯⎯ → −4 −1 −3 −1 0 3 12 5 1 R 3 + R1 0 0 0 0 5 11 0 0 −8 R 3 + R 2 ⎯⎯⎯⎯⎯⎯ → −4 −1 −3 −1 0 3 12 5 1 = −4(0)(11)( −1)(1) = 0 [10.4]
64.
=
16 19
x2 =
3 2 2 −3 3
5
−3 x1 =
4
2 −2 3
4
=
1 −2 = − 26 13
5 −2 3 −3
2 2 =
−2 2 =− [10.5] 19 19
0 0 0 1
− 73 − 26 0 0 (1/ 6) R 3 + R 2 3 7 13 (1/ 3) R 3 + R 1 0 0 ⎯⎯⎯⎯⎯⎯⎯ → −6 − 3 −1 −8 −6 0 3 8 −4 1 0 0 0 0 − 76 − 13 0 0 −2 R 2 + R 1 3 ⎯⎯⎯⎯⎯⎯ → −1 −8 −6 0 3 8 −4 1 13 ⎛ ⎞ = 0 ⎜ − ⎟ ( −6)( −1) = 0 [10.4] ⎝ 3⎠
⎛ 3⎞ = 1(1)(2) ⎜ ⎟ = 3 [10.4] ⎝2⎠
62.
−R 4 + R1 −2 −6 2 1 −R 4 + R 2 −1 − 3 1 −R 4 + R 3 1 ⎯⎯⎯⎯⎯⎯ → −1 −8 −6 1 1 3 8 −4
2 −2 5 −3 0 −10 8 −4
5 x2 = 3
2 21 21 = =− [10.5] 4 − 26 26 5 −2
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Chapter Review
65.
2
755
1 −3
66.
67.
−1
2
0
2 5
D = 2 − 5 1 = 138 [10.5] 4 3 0
0 2 −1 D1 = 3 3 − 2 = 34 −1 −1 − 5
2 2 5 D1 = 4 − 5 1 = 108 2 3 0
2 2 −3 D2 = 3 1 1 = 13 1 −2 4
3 0 −1 D2 = 1 3 − 2 = −38 4 −1 − 5
0 2 5 D2 = 2 4 1 = −52 4 2 0
2 1 2 D3 = 3 2 1 = −17 1 −3 4
3 2 0 D3 = 1 3 3 = 26 4 −1 −1
0 2 2 D3 = 2 − 5 4 = 76 4 3 2
D1 13 = D 44 D 2 11 1 x2 = = = D 44 4 D 17 − 17 x3 = 3 = =− D 44 44
D 34 17 x1 = 1 = =− D − 44 22 D2 − 38 19 x2 = = = − 44 22 D D 26 13 x3 = 3 = =− D − 44 22
D 108 54 18 x1 = 1 = = = D 138 69 23 26 D −52 x2 = 2 = =− 138 69 D 76 38 D x3 = 3 = = D 138 69
D1 =
2 1 −3 1 2 1 = 13 −2 −3 4
x1 =
68.
3
D = 1 3 − 2 = −44 [10.5] 4 −1 − 5
2 D= 3 1 −3
1 = 44 [10.5] 4
2
−3 −4
D = 1 −2 2 7
2 = −83 [10.5] −1
2 −3 −4 2 = −21 D1 = − 1 − 2 2 7 −1 2 2 −4 2 = −12 D2 = 1 − 1 2 2 −1 2 −3 2 D3 = 1 − 2 − 1 = 40 2 7 2
69.
1 −3 D=
2 −1 3
1
2
7 −3
1
4
2
−3
70. = −252 [10.5]
D=
1 −1 − 2
1 −3 3 2 2 7 2 1 D3 = = −230 −1 4 −1 −3 3 1 0 −2 x3 =
2
D3 −230 115 = = D −252 126
1 3
−2
1
−1 − 3
2
3
3 − 4 −1
5 −5
−1
= −230 [10.5]
2
2 −2 −2 1 1 2 −3 2 D2 = = 289 3 4 −4 −1 5 7 −1 2 x2 =
D −21 21 = x1 = 1 = D − 83 83 D − 12 12 = x2 = 2 = − 83 83 D D 40 40 x3 = 3 = =− D − 83 83
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289 289 D2 = =− 230 D −230
756
71.
Chapter 10: Matrices
The input-output matrix A is given by ⎡ 0.05 0.06 0.08 ⎤ A = ⎢ 0.02 0.04 0.04 ⎥ ⎢ 0.03 0.03 0.05 ⎥ ⎣ ⎦
72.
Consumer demand X is given by X = ( I − A) −1 D.
The input-output matrix A is given by ⎡ 0.07 0.04 0.07 ⎤ A = ⎢ 0.03 0.07 0.04 ⎥ ⎢ 0.03 0.03 0.02 ⎥ ⎣ ⎦ The consumer demand X is given by X = ( I − A) −1 D. −1
−1
⎛ ⎡1 0 0 ⎤ ⎡ 0.05 0.06 0.08⎤ ⎞ ⎡30 ⎤ X = ⎜ ⎢ 0 1 0 ⎥ − ⎢0.02 0.04 0.04 ⎥ ⎟ ⎢12 ⎥ ⎜⎢ ⎥ ⎢ 0.03 0.03 0.05⎥ ⎟ ⎢ 21⎥ ⎣ ⎦⎠ ⎣ ⎦ ⎝ ⎣0 0 1 ⎦ −1 0.95 0.06 0.08 30 − − ⎡ ⎤ ⎡ ⎤ = ⎢ −0.02 0.96 −0.04 ⎥ ⎢12 ⎥ ⎢ −0.03 −0.03 0.95⎥ ⎢ 21⎥ ⎣ ⎦ ⎣ ⎦ ⎡34.47 ⎤ ≈ ⎢14.20 ⎥ ⎢ 23.64 ⎥ ⎣ ⎦ $34.47 million computer division, $14.20 million monitor division, $23.64 million disk drive division. [10.3]
⎛ ⎡1 0 0 ⎤ ⎡0.07 0.04 0.07 ⎤ ⎞ ⎡ 27 ⎤ X = ⎜ ⎢0 1 0 ⎥ − ⎢ 0.03 0.07 0.04 ⎥ ⎟ ⎢18 ⎥ ⎜⎢ ⎥ ⎢ ⎥⎟ ⎢ ⎥ ⎝ ⎣0 0 1 ⎦ ⎣ 0.03 0.03 0.02 ⎦ ⎠ ⎣10 ⎦ −1 ⎡ 0.93 −0.04 −0.07 ⎤ ⎡ 27 ⎤ = ⎢ −0.03 0.93 −0.04 ⎥ ⎢18 ⎥ ⎢ −0.03 −0.03 0.98⎥ ⎢10 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ 30.82 ⎤ = ⎢ 20.86 ⎥ ⎢11.79 ⎥ ⎣ ⎦ $30.82 million lumber division, $20.86 million paper division, $11.79 million prefabricated wall division. [10.3]
....................................................... QR1. a.
There are 4 vertices, so the adjacency matrix A is a 4 × 4 matrix. ⎡ 0 1 0 1⎤ ⎢ 1 0 2 1⎥ A= ⎢ 0 2 0 0⎥ ⎢ ⎥ ⎣ 1 1 0 0⎦
Quantitative Reasoning b.
For a walk of length 3, find A3 ⎡0 3 ⎢1 A =⎢ 0 ⎢ ⎣1
1 0 2 1
0 2 0 0
3
1⎤ ⎡ 2 7 2 3⎤ ⎢ 7 2 12 7 ⎥ 1⎥ =⎢ 0⎥ 2 12 0 2 ⎥ ⎥ ⎢ ⎥ 0⎦ ⎣ 3 7 2 2⎦
....................................................... 1.
3.
Chapter Test 2.
3 −3 4 ⎤ ⎡ 2 3 −3⎤ ⎡ 4 ⎤ ⎡2 ⎢ 3 0 2 −1⎥ , ⎢ 3 0 2 ⎥ , ⎢ −1⎥ [10.1] ⎢ 4 −4 2 3⎥ ⎢ 4 −4 2 ⎥ ⎢ 3⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎧3x − 2 y + 5 z − w = 9 ⎪ ⎨2 x + 3 y − z + 4 w = 8 [10.1] + 3z + 2 w = −1 ⎩⎪ x
−2 R1 + R 2 ⎡ 1 −2 3 10⎤ 3 10⎤ ⎡ 1 −2 3 10⎤ ⎡ 1 −2 3 10⎤ ⎡ 1 −2 R1 + R 3 ⎢ −2 R 2 + R 3 ⎢ −R3 ⎢ ⎥ ⎢ 2 −3 8 23⎥ ⎯⎯⎯⎯⎯→ ⎥ 1 2 3 ⎯⎯⎯→ 0 1 2 3⎥ 0 1 2 3 ⎯⎯⎯⎯⎯→ 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 ⎢⎣ 0 ⎢⎣ −1 3 −2 −9 ⎥⎦ ⎢⎣0 0 −1 −2 ⎥⎦ 0 1 2 ⎥⎦ 1 1 1⎥⎦
⎧ x − 2 y + 3 z = 10 ⎪ y + 2z = 3 ⎨ ⎪⎩ z=2
y + 2(2) = 3 y = −1
x − 2(−1) + 3(2) = 10 x=2
The solution is (2, –1, 2). [10.1] 4.
⎡ 1 3 −1 1⎤ ⎡ 2 6 −1 1⎤ ⎡ 1 3 −1 1⎤ −2 R1 + R 2 ⎡ 1 3 −1 1⎤ −3R1 + R 3 ⎢ ⎢ 1 3 −1 1⎥ R ↔ R ⎢ 2 6 −1 1⎥ ⎯⎯⎯⎯⎯→ ⎥ R ↔ R ⎢0 1 1 −2 ⎥ − 0 0 1 1 2⎢ 3⎢ ⎢ ⎥ 2 ⎥ ⎢ ⎥ 1 ⎥ ⎢⎣0 1 1 −2 ⎦⎥ ⎣⎢ 3 10 −2 1⎥⎦ ⎣⎢ 3 10 −2 1⎦⎥ ⎣⎢0 0 1 −1⎦⎥ ⎧ x + 3y − z = 1 ⎪ y + z = −2 ⎨ ⎪ z = −1 ⎩
y + ( −1) = −2 y = −1
x + 3(−1) − (−1) = 1 x=3
The solution is (3, –1, –1). [10.1] Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Test
5.
757
−2 R1 + R 2 ⎡ 1 2 −3 2 11⎤ 11⎤ ⎡ 1 2 −3 2 2 R1 + R 3 ⎢ ⎢ 2 ⎥ − ⎯⎯⎯⎯⎯→ 5 8 5 28 0 1 −2 1 6 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 0 1 3 4 ⎦⎥ ⎣⎢ −2 −4 7 −1 −18⎦⎥ ⎧ w + 2 x − 3 y + 2 z = 11 ⎪ x − 2y + z = 6 ⎨ ⎪ y + 3z = 4 ⎩
y = 4 − 3z
x − 2(4 − 3z ) + z = 6
w + 2(−7 z + 14) − 3(4 − 3 z ) + 2 z = 11
x = −7 z + 14
w = 3z − 5
Let z be any real number c. The solution is (3c − 5, − 7c + 14, 4 − 3c, c) [10.1]
⎡ 2 −1 3⎤ ⎡ −1 3 2 ⎤ ⎢ A+ B = ⎢ + 4 −2 −1⎥ ⎣ 1 4 −1⎦⎥ ⎢ 3 2 2 ⎥ ⎣ ⎦ A + B is not defined because the matrices do not have the same order. [10.2]
6.
⎡ −1 3 2⎤ ⎡ 3 −9 −6⎤ − 3A = −3⎢ ⎥=⎢ ⎥ [10.2] 3⎦ ⎣ 1 4 − 1⎦ ⎣− 3 − 12
8.
3⎤ 3⎤ ⎡ 6 −3 9 ⎤ ⎡ 2 −4 6 ⎤ ⎡ 4 1 ⎡ 2 −1 3⎤ ⎡ 1 −2 3B − 2C = 3 ⎢ 4 −2 −1⎥ − 2 ⎢ 2 −3 8⎥ = ⎢12 −6 −3⎥ − ⎢ 4 −6 16 ⎥ = ⎢ 8 0 −19 ⎥ [10.2] ⎢ 3 2 2⎥ ⎢ −1 3 −2 ⎥ ⎢ 9 6 6 ⎥ ⎢ −2 6 −4 ⎥ ⎢11 0 10 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
9.
⎡ 2 −1 3⎤ −1 3 2 ⎤ ⎢ ⎡16 −1 −2 ⎤ ⎡ (−1)(2) + 3(4) + 2(3) (−1)(−1) + 3(−2) + 2(2) (−1)3 + 3(−1) + 2(2) ⎤ 4 −2 −1⎥ = ⎢ AB = ⎢⎡ = ⎢ [10.2] ⎥ ⎥ 1 4 1 + + − − + − + − + − + − − 1(2) 4(4) ( 1)(3) 1( 1) 4( 2) ( 1)(2) 1(3) 4( 1) ( 1)(2) ⎦ ⎣15 −11 −3⎥⎦ ⎣ ⎦ ⎢ 3 2 2⎥ ⎣ ⎣ ⎦
10.
⎡16 −1 −2 ⎤ ⎡ −1 3 2 ⎤ ⎡17 −4 −4 ⎤ Use AB from Problem 9. AB − A= ⎢ [10.2] − = ⎣15 −11 −3⎥⎦ ⎢⎣ 1 4 −1⎥⎦ ⎢⎣14 −15 −2⎥⎦
11.
12.
13.
7.
3⎤ ⎡ 1 −2 ⎡ −1 3 2 ⎤ [10.2] CA = ⎢ 2 −3 8⎥ ⎢ ⎢ −1 3 −2 ⎥ ⎣ 1 4 −1⎥⎦ ⎣ ⎦ The number of columns of the first matrix is not equal to the number of rows of the second matrix. The product is not possible. 3⎤ ⎡ 1 −2 3⎤ ⎡ 2 −1 3⎤ ⎡ 1 −2 BC − CB = ⎢ 4 −2 −1⎥ ⎢ 2 −3 8⎥ − ⎢ 2 −3 8⎥ ⎢ 3 2 2 ⎥ ⎢ −1 3 −2⎥ ⎢ −1 3 −2⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 11⎤ ⎡ −3 8 −8⎤ ⎡ 3 9 = ⎢ 1 −5 −2 ⎥ − ⎢16 20 25⎥ ⎢ 5 −6 21⎥ ⎢ 4 −9 −10 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ −6 −1 −19 ⎤ = ⎢ −15 −25 −27 ⎥ ⎢ 1 3 31⎥⎦ ⎣ ⎡−1 3 2⎤ ⎡ −1 3 2⎤ A2 = ⎢ ⎥⎢ ⎥ ⎣ 1 4 − 1⎦ ⎣ 1 4 − 1⎦ The number of columns of the first matrix is not equal to the number of rows of the second matrix. The product is not possible. [10.2]
⎡ 2 −1 3⎤ ⎢ 4 −2 −1⎥ ⎢ 3 2 2⎥ ⎣ ⎦
[10.2]
14.
6 13⎤ ⎡ 2 − 1 3⎤ ⎡2 − 1 3⎤ ⎡ 9 ⎢ ⎥⎢ ⎥ ⎢ ⎥ B 2 = ⎢ 4 − 2 − 1⎥ ⎢4 − 2 − 1⎥ = ⎢ − 3 − 2 12⎥ [10.2] ⎢⎣ 3 2 2⎥⎦ ⎢⎣ 3 2 2⎥⎦ ⎢⎣ 20 − 3 11⎥⎦
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758
15.
Chapter 10: Matrices
3 1 0 0⎤ −2 R1 + R 2 ⎡ 1 −2 3 1 ⎡ 1 −2 R1 + R 3 ⎢ ⎢ 2 −3 8 0 1 0⎥ ⎯⎯⎯⎯⎯→ 0 1 2 −2 ⎢ ⎥ ⎢ 1 1 1 ⎣⎢ −1 3 −2 0 0 1⎥⎦ ⎣⎢0 7 R 3 + R1 ⎡ 1 0 7 −3 2 0⎤ 2 R 3 + R 2 ⎡ 1 −R3 2 R 2 + R1 ⎢ ⎯⎯⎯⎯⎯ → 0 1 2 −2 1 0⎥ ⎯⎯⎯⎯⎯ → ⎢0 ⎢ ⎥ ⎢ ⎣⎢0 0 −1 3 −1 1⎦⎥ ⎣⎢0
1 0 0⎤ 0 0⎤ ⎡ 1 −2 3 −2 R 2 + R 3 ⎢ ⎥ ⎯⎯⎯⎯⎯→ − 0 1 2 2 1 0⎥ 1 0 ⎢ ⎥ ⎥ ⎢⎣ 0 0 −1 3 −1 1⎥⎦ 0 1⎦⎥
0 0 18 −5 7 ⎤ 1 0 4 −1 2 ⎥ ⎥ 0 1 −3 1 −1⎦⎥
⎡ 18 −5 7 ⎤ The inverse matrix is ⎢ 4 −1 2 ⎥ . [10.3] ⎢ −3 1 −1⎥⎦ ⎣
16.
M 21 =
−1 3 = −2 − 6 = −8 2 2
2 −1 3 B = 4 −2 −1 = 3C31 + 2C32 + 2C33 [10.4] 3 2 2 −1 3 2 3 2 −1 = 3( −1)3+1 + 2( −1)3+2 + 2( −1)3+3 −2 −1 4 −1 4 −2 = 3(1+ 6) − 2( −2 −12) + 2( −4 + 4) = 21+ 28 + 0 = 49
17.
[10.4]
C21 = (−1)2+1 M 21 = −(−8) = 8
18.
1 −2
3
2 −3 8= 3 −2 −1 3
−2 R1 + R2
1 −2 3
0 R1 + R3 0
2 −1
19.
D = 2 −3 5 6
20.
X = ( I − A) −1 D [10.3]
2 = −82 3
1 −2
1 2 = − R2 + R3 0 1 1 0 3
3 1 2 = (1) (1) (–1) = –1 [10.4] 0 −1
2 12
D3 = 2 − 3 − 1 = 280 5 6 4
⎛ ⎡1 0 0 ⎤ ⎡ 0.15 0.23 0.11⎤ ⎞ ⎜ ⎢0 1 0 ⎥ − ⎢ 0.08 0.10 0.05 ⎥ ⎟ ⎜⎢ ⎥ ⎢ 0.16 0.11 0.07 ⎥ ⎟ ⎣ ⎦⎠ ⎝ ⎣0 0 1 ⎦
−1
z=
D3 280 140 [10.5] = =− 41 D − 82
⎡50 ⎤ ⎢32 ⎥ ⎢ 8⎥ ⎣ ⎦
....................................................... 1.
y − 5 = − 1 ( x − ( −4)) [2.3] y −5= y=
2 −1 2 −1 2
x−2 x+3
2.
2 x 2 + x − 10 x − 3 2 x3 − 5 x 2 − 13 x + 30 2 x3 − 6 x 2 x 2 − 13x x 2 − 3x − 10 x + 30 −10 x + 30 0
Cumulative Review 3.
[3.1]
h[k (0)] = h[3 0 ] [4.2] = h[1] = e −1 ≈ 0.3679
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Cumulative Review
4.
759
5.
[8.1] 2 x 2 − 4 x + 3 y −1= 0 2 2 x − 4 x = −3 y +1 2( x 2 − 2 x) = −3 y +1 2( x −1) 2 = −3 y + 3 2( x −1) 2 = −3( y −1) Vertex = (1, 1)
{
3x − 4 y = 4 (1) 2 x − 3 y =1 (2)
Solve (2) for x and substitute into (1). 2 x − 3 y =1 2 x = 3 y +1 3 y +1 x= 2 ⎛ 3 y +1 ⎞ 3⎜ ⎟−4 y = 4 ⎝ 2 ⎠ 9 y + 3−8 y = 8 y =5 3(5) +1 =8 2 The solution is (8, 5). [9.1] x=
6.
Domain: {x | –3 < x < 3} [2.2]
7.
x 2 + 4 x − 5 = ( x + 5)( x − 1) [3.5] x+5=0 x −1 = 0 x = −5 x =1 Vertical asymptotes: x = –5, x = 1
8.
[2.5] 9.
x+2 >0 x +1
10.
2a =12 a=6 a 2 = b2 + c2 (6) 2 = b 2 + (4)2 36 −16 = b 2 20 = b 2
The quotient x + 2 is positive. x +1 The critical values are –2 and –1. x+2 x +1
( x − 3)2 ( y + 4) 2 + =1 [8.2] 36 20
(−∞, −2) ∪ (−1, ∞) [1.5] 11.
f ( x + h) − f ( x) ( x + h) 2 − 3( x + h) + 2 − ( x 2 − 3x + 2) = [2.6] h h 2 2 2 = x + 2 xh + h − 3x − 3h + 2 − x + 3 x − 2 h 2 2 3 h xh h + − = h = 2 x − 3+ h
12.
125 x = 1 [4.5] 25 53 x = 5−2 3 x = −2 x=−2 3
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760
13.
Chapter 10: Matrices
x−2 x−2 = x − 5 x − 6 ( x + 1)( x − 6) x−2 = A + B ( x + 1)( x − 6) x + 1 x − 6
14.
2
Let u =10 x
x − 2 = A( x − 6) + B( x + 1) x − 2 = Ax − 6 A + Bx + B x − 2 = ( A + B ) x + ( −6 A + B )
u 2 − 2u −1= 0 2 ± 4 − 4(1)(−1) u= 2 = 2± 8 2 =1± 2 10 x =1+ 2 log10 x = log(1+ 2) x = log(1+ 2) ≈ 0.3828
{
1= A+ B (1) −2 = −6 A + B (2)
−1 = − A − B −2 = −6 A + B −3 = −7 A 3 =A 7
[4.5] 10 x −10− x = 2 −x x x x 10 (10 −10 ) = 2(10 ) (10 x ) 2 − 2(10 x ) −1= 0
− 1 times (1) (2)
1= 3 + B 7 4=B 7 4 3 x−2 [9.4] = + x 2 − 5 x − 6 7( x + 1) 7( x − 6)
15.
cos30o sin 30o + sec 45o tan 60o = cos30o sin 30o + 1 o tan 60o sin 45 3 1 2 = ⋅ + ⋅ 3 2 2 2 = 3+2 6 4 2 3 4 6 + = 4
[5.2]
16.
o
sin 292 1 = sin 585 2 2
o
o = − 1 − cos585 2 o = − 1 − cos 225 2
=−
1 − ⎛⎜ − ⎝
2⎞ ⎟ 2 ⎠
2
=− 2+ 2 4 = −1 2+ 2 2 17.
K = 1 bc sin A 2 = 1 (11)(4)sin 65o 2 = 22sin 65o ≈ 20 m 2
[7.2]
18.
[6.3]
cosθ (1 − sin θ ) tan θ + cos θ = tan θ + 1 − sin θ (1 + sin θ )(1 − sin θ ) cosθ (1 − sin θ ) = tan θ + 1 − sin 2 θ cosθ (1 − sin θ ) = tan θ + cos 2 θ − sin 1 sin θ θ = + cos θ cosθ = sin θ + 1 − sin θ cos θ 1 = cos θ = secθ [5.4]
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Cumulative Review
19.
761
z = 2 − 2i 3 2
[7.4] 2
r = 2 + ( −2 3) = 4 + 12 = 16 = 4
α = tan −1 b = tan −1 −2 3 = tan −1 − 3 = 60o a
2
z is in the fourth quadrant 270o < θ < 360o ,
θ = 360o − 60o = 300o z = rcisθ = 4(cos300o + i sin 300o )
20.
sin x + cos x = 1 [6.6] 2 1 − cos x + cos x = 1 2 1 − cos x = 1 − cos x 2 1 − cos x = 1 − 2cos x + cos 2 x 2 1 − cos x = 2 − 4cos x + 2cos 2 x 0 = 2cos 2 x − 3cos x + 1 0 = (2cos x − 1)(cos x − 1) 2cos x − 1 = 0 cos x = 1 2
x = π , 5π 3 3
Copyright © Houghton Mifflin Company. All rights reserved.
cos x = 1 x=0
Chapter 11
Sequences, Series, and Probability Section 11.1 1.
a1 = 1(1 − 1) = 1 ⋅ 0 = 0
2.
a2 = 2(2 − 1) = 2 ⋅ 1 = 2
a3 = 2 ⋅ 3 = 6
a8 = 8(8 − 1) = 8 ⋅ 7 = 56
a8 = 2 ⋅ 8 = 16
1+1 2 = =2 1 1 2 +1 3 = = 2 2 3 +1 4 = = 3 3 8 +1 9 = = 8 8
a1 = a2 a3 a8
5.
a1 = a2 = a3 = a8 =
7.
10.
a1 =
( −1)2⋅1−1 (−1)2−1 1 = =− 3 ⋅1 3 3
a2 =
3.
a2 = 2 ⋅ 2 = 4
a3 = 3(3 − 1) = 3 ⋅ 2 = 6
4.
a1 = 2 ⋅ 1 = 2
(−1)1+1 12 (−1)2+1 2
2
(−1)3+1 32 (−1)8+1 82
a1 =
(−1)1+1 (−1)2 1 = = 1(1 + 1) 1⋅ 2 2
(−1)3 1 =− 4 4
a2 =
(−1)2+1 (−1)3 1 = =− 2(2 + 1) 2⋅3 6
=
(−1)4 1 = 9 9
a3 =
(−1)3+1 (−1)4 1 = = 3(3 + 1) 3⋅ 4 12
=
(−1)9 1 =− 64 64
a8 =
(−1)8+1 (−1)9 1 = =− 8(8 + 1) 8⋅9 72
=
(−1)2 =1 1
=
a1 =
−1 (−1)1 = = −1 2 ⋅1 − 1 2 − 1
( −1)2⋅2−1 (−1)4−1 1 = =− 3⋅ 2 6 6
a2 =
(−1)2 1 = 2 ⋅ 2 −1 3
a3 =
( −1)2⋅3−1 (−1)6−1 1 = =− 3⋅3 9 9
a3 =
(−1)3 1 =− 2 ⋅ 3 −1 5
a8 =
( −1)2⋅8−1 (−1)16−1 1 = =− 3⋅8 24 24
a8 =
(−1)8 1 = 2 ⋅ 8 − 1 15
8.
1
1 ⎛ −1 ⎞ a1 = ⎜ ⎟ =− 2 ⎝ 2 ⎠
1 a1 = 1 − = 0 1 1 1 a2 = 1 − = 2 2 1 2 a3 = 1 − = 3 3 1 7 a8 = 1 − = 8 8
11.
6.
9.
1
2 ⎛ 2⎞ a1 = ⎜ ⎟ = 3 3 ⎝ ⎠ 2
4 ⎛2⎞ a2 = ⎜ ⎟ = 3 9 ⎝ ⎠ 3
8 ⎛2⎞ a3 = ⎜ ⎟ = 27 ⎝3⎠ 8
256 ⎛2⎞ a8 = ⎜ ⎟ = 3 6561 ⎝ ⎠
a1 = 1 + (−1)1 = 1 + (−1) = 0
2
a2 = 1 + (−1) 2 = 1 + 1 = 2
3
a8 = 1 + (−1)8 = 1 + 1 = 2
1 ⎛ −1 ⎞ a2 = ⎜ ⎟ = 4 ⎝ 2 ⎠
a3 = 1 + (−1)3 = 1 + (−1) = 0
1 ⎛ −1 ⎞ a3 = ⎜ ⎟ = − 2 8 ⎝ ⎠ 8
1 ⎛ −1 ⎞ a8 = ⎜ ⎟ = 256 ⎝ 2 ⎠
12.
a1 = 1 + (−0.1)1 = 1 + (−0.1) = 0.9
13.
a1 = (1.1)1 = 1.1
a2 = 1 + (−0.1)2 = 1 + 0.01 = 1.01
a2 = (1.1)2 = 1.21
a3 = 1 + (−0.1)3 = 1 + (−0.001) = 0.999
a3 = (1.1)3 = 1.331
a8 = 1 + (−0.1)8 = 1 + 0.00000001 = 1.00000001
a8 = (1.1)8 = 2.14358881
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Section 11.1
14.
16.
763
1 = 1 =1 1 +1 1+1 2 a2 = 22 = 2 = 2 2 +1 4 +1 5 a3 = 23 = 3 = 3 3 + 1 9 + 1 10 a8 = 28 = 8 = 8 8 + 1 64 + 1 65 a1 =
15.
2
1−1 0 a1 = 3 1 = 3 = 1 2 2 2
17.
a1 =
1! = 1 = 1 = 1 (1 − 1)! 0! 1
19.
a1 = ln 1 = 0
21.
a8 = ln 8 ≈ 2.0794
26.
29.
a3 =
(−1)3+1 (−1)4 1 3 = = = 3 3 3 3
a8 =
(−1)8+1 (−1)9 1 2 = =− =− 4 8 2 2 2 2
a1 = 1!= 1
a1 = log1 = 0
a8 = log 8 ≈ 0.9031
a3 = ln 3 ≈ 1.0986
a1 = 3
(−1) 2+1 ( −1)3 1 2 = =− =− 2 2 2 2
a3 = log 3 ≈ 0.4771
a2 = ln 2 ≈ 0.6931
23.
a2 =
a2 = log 2 ≈ 0.3010
2! = 2 ⋅ 1 = 2 = 2 (2 − 1)! 1! 1 a3 = 3! = 3 ⋅ 2 ⋅ 1 = 3 ⋅ 2 ⋅ 1 = 3 (3 − 1)! 2! 2 ⋅1 a8 = 8! (8 − 1)! 8 = ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 7! ⋅ ⋅ ⋅ ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = 8 8 7 6 5 = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
a2 =
20.
(−1)1+1 ( −1)2 = =1 1 1
a2 = 2! = 2 ⋅ 1 = 2 a3 = 3! = 3 ⋅ 2 ⋅ 1 = 6 a8 = 8! = 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 40,320
2 −1 1 a2 = 3 2 = 3 = 3 4 4 2 3−1 2 3 3 =9 a3 = 3 = 8 8 2 8 −1 7 a8 = 3 8 = 3 = 2187 256 256 2
18.
a1 =
24.
a1 = 1
1 = 0.142857 7
22.
a1 = 0
a2 = 4
a2 = 7
a3 = 2
a3 = 6
a8 = 4
a8 = 7
a1 = −2
25.
1 = 0.076923 13
a1 = 5
a2 = 3
a2 = −2
a2 = 2 ⋅ a1 = 2 ⋅ 5 = 10
a3 = 3
a3 = −2
a3 = 2 ⋅ a2 = 2 ⋅ 10 = 20
a8 = 3
a8 = −2
a1 = 2
27.
a1 = 2
28.
a2 = 3 ⋅ a1 = 3 ⋅ 2 = 6
a2 = 2 ⋅ a1 = 2 ⋅ 2 = 4
a3 = 3 ⋅ a2 = 3 ⋅ 6 = 18
a3 = 3 ⋅ a2 = 3 ⋅ 4 = 12
a1 = 2
a2 = (a1 )2 = (2 )2 = 4 a3 = (a2 )2 = (4)2 = 16
30.
a1 = 4
a2 = 22 ⋅ a1 = 4 ⋅ 1 = 4 a3 = 32 ⋅ a2 = 9 ⋅ 4 = 36 31.
a2 = 1 = 1 a1 4 a3 = 1 = 11 = 4 a2
a1 = 1
a1 = 2
a2 = 2 ⋅ 2 ⋅ a1 = 4 ⋅ 2 = 8 a3 = 2 ⋅ 3 ⋅ a2 = 6 ⋅ 8 = 48
4
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764
32.
Chapter 11: Sequences, Series, and Probability
a1 = 2
33.
a2 = (−3) ⋅ 2 ⋅ a1 = −6 ⋅ 2 = −12
a2 = ( a1 )1 2 = (3)1 2 = 3
a3 = (−3) ⋅ 3 ⋅ a2 = −9 ⋅ (−12) = 108
34.
a1 = 2 2
a1 = 3 a3 = ( a2 )1 3 = ( 31 2 )
13
= 31 6 = 6 3
35.
a1 = 1 a2 = 3 a3 = 1 ( a2 + a1 ) = 1 (3 + 1) = 1 (4) = 2 2 2 2 a 4 = 1 ( a3 + a 2 ) = 1 ( 2 + 3 ) = 1 ( 5 ) = 5 2 2 2 2 1 1 5 1 9 ⎛ ⎞ ⎛ ⎞ a5 = ( a4 + a3 ) = ⎜ + 2 ⎟ = ⎜ ⎟ = 9 2 2⎝2 ⎠ 2⎝2⎠ 4
38.
668 + 866 1534 Sorting 1534 gives 1345
2
a2 = ( a1) = (2) = 4 a3 = ( a2 )3 = (4)3 = 64
36.
a1 = 1 a2 = 4 a3 = a2 ⋅ a1 = 4 ⋅ 1 = 4 a4 = a3 ⋅ a2 = 4 ⋅ 4 = 16 a5 = a4 ⋅ a3 = 16 ⋅ 4 = 64
37.
an a3 a4 a5
39.
7!− 6! = 7 ⋅ 6!− 6! = 6!( 7 − 1) = 6!⋅ 6 = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ⋅ 6 = 4320
40.
(4!)2 = (4 ⋅ 3 ⋅ 2 ⋅ 1)2 = (24)2 = 576
41.
9! = 9 ⋅ 8 ⋅ 7! = 72 7! 7!
42.
10! = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5! = 30, 240 5! 5!
43.
8! = 8 ⋅ 7 ⋅ 6 ⋅ 5! = 56 3!5! 3 ⋅ 2 ⋅ 1 ⋅ 5!
44.
12! = 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8! = 495 4!8! 4 ⋅ 3 ⋅ 2 ⋅ 1 ⋅ 8!
45.
100! = 100 ⋅ 99! = 100 99! 99!
46.
100! = 100 ⋅ 99 ⋅ 98! = 4950 98!2! 98!⋅ 2 ⋅ 1
47.
∑i = 1 + 2 + 3 + 4 + 5 = 15
48.
= an −1 + an − 2 = 3+1 = 4 =4+3=7 = 7 + 4 = 11
4
∑i
2
5
i =1
= 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30
i =1 5
49.
∑i (i − 1) = 1(1 − 1) + 2 ( 2 − 1) + 3(3 − 1) + 4 ( 4 − 1) + 5(5 − 1) i =1
= 1⋅ 0 + 2 ⋅1 + 3 ⋅ 2 + 4 ⋅ 3 + 5 ⋅ 4 = 0 + 2 + 6 + 12 + 20 = 40 7
50.
∑i ( 2i + 1) = ( 2 ⋅1 + 1) + ( 2 ⋅ 2 + 1) + 2 ( 2 ⋅ 3 + 1) + ( 2 ⋅ 4 + 11) + ( 2 ⋅ 5 + 1) + ( 2 ⋅ 6 + 1) + ( 2 ⋅ 7 + 1) i =1
= ( 2 + 1) + ( 4 + 1) + ( 6 + 1) + ( 8 − 1) + (10 + 1) + (12 + 1) + (14 + 1) = 3 + 5 + 7 + 9 + 11 + 13 + 15 = 63
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1345 + 5431 6776 Sorting 6776 gives 6677
Section 11.1
765
4
51.
∑ 1k = 11 + 12 + 13 + 14 = 1212 + 126 + 124 + 123 = 1225 k =1 6
52.
∑ k ( k1+ 1) = 1(11+ 1) + 2 ( 21+ 1) + 3(31+ 1) + 4 ( 41+ 1) + 5(51+ 1) + 6 ( 61+ 1) k =1
1 1 1 1 1 1 + + + + + 1⋅ 2 2 ⋅ 3 3 ⋅ 4 4 ⋅ 5 5 ⋅ 6 6 ⋅ 7 210 70 35 21 14 10 360 6 = + + + + + = = 420 420 420 420 420 420 420 7 =
8
53.
∑
8
∑ j =2 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 2 (36) = 72
2j =2
j =1
6
54.
∑
j =1
( 2i + 1)( 2i − 1) =
i =1
6
∑(
6
6
i =1
i =1
) ∑ i2 − ∑ (1) = 4 (12 + 22 + 32 + 42 + 52 + 62 ) − (1 + 1 + 1 + 1 + 1 + 1)
4i 2 − 1 = 4
i =1
= 4(1 + 4 + 9 + 16 + 25 + 36 ) − 6 = 4(91) − 6 = 358 5
55.
( −1)i 2i = (−1)3 23 + (−1) 4 24 + (−1)5 25 = (−1)8 + (1)16 + (−1)32 = −8 + 16 − 32 = −24 ∑ i=3 5
56.
i
(−1) ∑ i i=3 2
=
(−1)3 3
2
+
( −1)4 2
4
+
(−1)5 5
2
=
−1 1 − 1 − 4 2 − 1 3 + + = + + =− 8 16 32 32 32 32 32
7
57.
⎛ n +1⎞ ⎛1+1⎞ ⎛ 2 +1⎞ ⎛ 3 +1⎞ ⎛ 4 + 1⎞ ⎛ 5 +1⎞ ⎛ 6 +1⎞ ⎛ 7 +1⎞ log ⎜ ⎟ = log ⎜ ⎟ + log ⎜ ⎟ + log ⎜ ⎟ + log ⎜ ⎟ + log ⎜ ⎟ + log ⎜ ⎟ + log ⎜ ⎟ ∑ ⎝ n ⎠ ⎝ 1 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 4 ⎠ ⎝ 5 ⎠ ⎝ 6 ⎠ ⎝ 7 ⎠ n =1 ⎛ 3 4 5 6 7 8⎞ = log ⎜ 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⎟ = log8 = 3log 2 ≈ 0.9031 ⎝ 2 3 4 5 6 7⎠ 8
58.
2 n 2 3 5 6 7 8 ⎛2 3 4 5 6 7 8⎞ ln = ln + ln + ln + ln + ln + ln = ln⎜ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⎟ = ln ≈ −1.5041 ∑ 3 4 6 7 8 9 n +1 3 4 5 6 7 8 9⎠ 9 ⎝ n=2 8
59.
∑0 k !(88!− k )! = 0!(88!− 0)! + 1!(88!− 1)! + 2!(88!− 2)! + 3!(88!− 3)! + 4!(88!− 4)! + 5!(88!− 5)! + 6!(88!− 6)! + 7!(88!− 7)! + 8!(88!− 8)! k=
8! 8! 8! 8! 8! 8! 8! 8! 8! + + + + + + + + 0!8! 1!7! 2!6! 3!5! 4!4! 5!3! 6!2! 7!1! 8!0! 8⋅7 8⋅7 ⋅6 8⋅7⋅6⋅5 8⋅7⋅6 8⋅7 =1+ 8 + + + + + + 8 +1 4⋅3⋅ 2 3 ⋅ 2 ⋅1 2 2 3⋅ 2 = 1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 256 =
Copyright © Houghton Mifflin Company. All rights reserved.
766
Chapter 11: Sequences, Series, and Probability 7
60.
1 ∑0 k1! = 0!1 + 1!1 + 2!1 + 3!1 + 4!1 + 5!1 + 6!1 + 7!1 = 11 + 11 + 12 + 16 + 241 + 1201 + 7201 + 5040 k=
5040 5040 2520 840 210 42 7 1 + + + + + + + 5040 5040 5040 5040 5040 5040 5040 5040 13700 1370 685 = = = 5040 504 252 =
61.
62.
1 1 1 1 1 1 1 1 1 1 1 1 + + + + + = + + + + + = 1 4 9 16 25 36 12 22 32 42 52 62
6
∑1 i12 i=
2 + 4 + 6 + 8 + 10 + 12 + 14 = 2(1) + 2(2) + 2(3) + 2(4 ) + 2(5) + 2(6 ) + 2(7 ) =
7
∑1 2i i=
63.
2 − 4 + 8 −16 + 32 − 64 +128 = 21( −1)1+1 + 22 (−1)2+1 + 23 (−1)3+1 + 24 (−1) 4+1 + 25 (−1)5+1 + 26 (−1)6+1 + 27 (−1)7 +1 =
7
∑1 2 (−1) i
i +1
i=
64.
1 − 8 + 27 − 64 + 125 = 13 − 23 + 33 − 43 + 53 = 13 (−1)1+1 + 23 (−1)2 +1 + 33 (−1)3+1 + 43 (−1)4 +1 +53 (−1)5+1 =
5
∑1 i3(−1)
i +1
i=
65.
7 + 10 + 13 + 16 + 19 = 7 + (7 + 3) + (7 + 3 ⋅ 2 ) + (7 + 3 ⋅ 3) + (7 + 3 ⋅ 4 ) =
4
∑0 (7 + 3i) i=
66.
30 + 26 + 22 + 18 + 14 + 10 = 30 + ( 30 − 4 ) + ( 30 − 4 ⋅ 2 ) + ( 30 − 4 ⋅ 3) + ( 30 − 4 ⋅ 4 ) + ( 30 − 4 ⋅ 5) =
5
∑(30 − 4i ) i =0
67.
68.
1 1 1 1 1 1 1 1 + + + = + + + = 2 4 8 16 2 22 23 24
1−
4
∑1 21i i=
5 2 3 4 5 i 5 ⎛ 2 ⎞i 2 4 8 16 32 2 ⎛2⎞ ⎛2⎞ ⎛2⎞ ⎛2⎞ ⎛ 2⎞ + − + − = 1 − + ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ − ⎜ ⎟ = ∑ ⎜ ⎟ (− 1)i = ⎜− ⎟ 3 9 27 81 243 3 ⎝3⎠ ⎝3⎠ ⎝3⎠ ⎝3⎠ ⎝ 3⎠ i = 0⎝ 3 ⎠ i =0
∑
....................................................... 69.
Let N = 7. a1 = 7 = 3.5 2 1 a2 = ⎛⎜ 3.5 + 7 ⎞⎟ = 2.75 2⎝ 3.5 ⎠
Connecting Concepts 70.
Let N = 10. a1 = 10 = 5 2 1 a2 = ⎛⎜ 5 + 10 ⎞⎟ = 3.5 2⎝ 5⎠
a3 = 1 ⎛⎜ 2.75 + 7 ⎞⎟ ≈ 2.6477273 2⎝ 2.75 ⎠
a3 = 1 ⎛⎜ 3.5 + 10 ⎞⎟ ≈ 3.1785714 2⎝ 3.5 ⎠
7 ⎞ ≈ 2.6457520 a4 ≈ 1 ⎛⎜ 2.6477273 + ⎟ 2⎝ 2.6477273 ⎠
⎞ ≈ 3.1623194 10 a4 = 1 ⎛⎜ 3.1785714 + ⎟ 2⎝ 3.1785714 ⎠ ⎞ ≈ 3.1622777 10 a5 = 1 ⎛⎜ 3.1623194 + ⎟ 2⎝ 3.1623194 ⎠
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Section 11.1
71.
767
a20 ≈ 1.0000037
72.
a1 = i, a2 = −1, a3 = −i, a4 = 1 a5 = i, a6 = −1, a7 = −i, a8 = 1
a100 ≈ 1
Notice that the sequence repeats itself in groups of 4. To find a237 divide 237 by 4. a237 = i r where r is the remainder after
the division. Thus a237 = i1 = i. 73.
1+1 = 2 1+1+ 2 = 4 1+1+ 2 + 3 = 7 The sum of the first n terms of a Fibonacci sequence equals the (n + 2) term – 1. Therefore the sum of the first ten terms is (10+2) term – 1, i.e. 12th term – 1. 144 1 = 143 10
75.
⎛1+ 5 ⎞ ⎜ ⎟ 2 ⎠ ⎝ F10 =
10
⎛ ⎞ − ⎜1− 5 ⎟ ⎝ 2 ⎠ 5
15
74.
76.
= 55
⎛n⎞ 2π n ⎜ ⎟ ⎝e⎠
n
10
⎛ 10 ⎞ When n = 10 we have 2π (10) ⎜ ⎟ ⎝ e ⎠
15
⎛1+ 5 ⎞ ⎛1− 5 ⎞ ⎜ ⎟ −⎜ ⎟ 2 2 ⎠ ⎝ ⎠ ⎝ F15 = 5
81 = 5 + 21 + 55
= 610
⎛ 20 ⎞ When n = 20 we have 2π (20) ⎜ ⎟ ⎝ e ⎠ ⎛ 30 ⎞ When n = 30 we have 2π (30) ⎜ ⎟ ⎝ e ⎠
≈ 3.5986956 × 106.
20
30
≈ 2.4227869 × 1018. ≈ 2.6451710 × 1032.
n
77.
cai = ca1 + ca2 + ca3 + ⋅ ⋅ ⋅ + can ∑ i =1
= c ( a1 + a2 + a3 + ⋅ ⋅ ⋅ + an ) n
=c
∑a
i
i =1
....................................................... PS1.
−3 = 25 + (15 − 1)d
Prepare for Section 11.2 PS2. 13 = 3 + (5 − 1)d 13 = 3 + 4d 10 = 4d 5 =d 2
−3 = 25 + 14d −28 = 14d −2 = d 50 ⎡ 2(2) + (50 − 1) 5 ⎤ 50 ⎡ 4 + 245 ⎤ 4 ⎦⎥ = ⎣⎢ 4 ⎦⎥ ⎣⎢ 2 2 50 ⎡ 261 ⎤ ⎢ ⎥ = ⎣ 4 ⎦ = 6525 2 4
PS3. S =
PS4. a5 = 5 + (5 − 1)4 = 21
PS5. a20 = 52 + (20 − 1)( −3) = −5
PS6. 5 – 2 = 3 8–5=3 Yes
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768
Chapter 11: Sequences, Series, and Probability
Section 11.2 1.
d = 10 − 6 = 4
2.
an = 6 + ( n − 1)4 = 6 + 4n − 4 = 4n + 2 a9 = 4 ⋅ 9 + 2 = 36 + 2 = 38 a24 = 4 ⋅ 24 + 2 = 98 3.
d = 4 − 6 = −2
a n = 7 + ( n − 1)5 = 5n + 2 a 9 = 5 ⋅ 9 + 2 = 45 + 2 = 47 a 24 = 5 ⋅ 24 + 2 = 120 + 2 = 122 4.
a n = 6 + ( n − 1)( −2) = 6 − 2n + 2 = 8 − 2n a 9 = 8 − 2 ⋅ 9 = 8 − 18 = −10 a 24 = 8 − 2 ⋅ 24 = 8 − 48 = −40
5.
d = −5 − (−8) = 3
d = 12 − 7 = 5
d = 4 − 11 = −7 a n = 11 + ( n − 1)( −7) = 11 − 7n + 7 = 18 − 7n a 9 = 18 − 7 ⋅ 9 = 18 − 63 = −45 a 24 = 18 − 7 ⋅ 24 = 18 − 168 = −150
6.
a n = −8 + ( n − 1)3 = −8 + 3n − 3 = 3n − 11 a 9 = 3 ⋅ 9 − 11 = 27 − 11 = 16 a 24 = 3 ⋅ 24 − 11 = 72 − 11 = 61
d = −9 − (−15) = 6 a n = −15 + ( n − 1)6 = −15 + 6n − 6 = 6n − 21 a 9 = 6 ⋅ 9 − 21 = 33 a 24 = 6 ⋅ 24 − 21 = 123
7.
d = 4 −1 = 3 a n = 1 + ( n − 1)3 = 1 + 3n − 3 = 3n − 2 a 9 = 3 ⋅ 9 − 2 = 27 − 2 = 25 a 24 = 3 ⋅ 24 − 2 = 72 − 2 = 70
8.
d = 1 − (−4 ) = 5 a n = −4 + ( n − 1)5 = −4 + 5n − 5 = 5n − 9 a 9 = 5 ⋅ 9 − 9 = 45 − 9 = 36 a 24 = 5 ⋅ 24 − 9 = 120 − 9 = 111
9.
d = (a + 2 ) − a = 2
10.
d = (a + 1) − (a − 3) = a + 1 − a + 3 = 4 a n = a − 3 + ( n − 1)4 = a − 3 + 4n − 4 = a + 4n − 7 a 9 = a + 4 ⋅ 9 − 7 = a + 36 − 7 = a + 29 a 24 = a + 4 ⋅ 24 − 7 = a + 96 − 7 = a + 89
a n = a + ( n − 1)2 = a + 2n − 2 a 9 = a + 2 ⋅ 9 − 2 = a + 18 − 2 = a + 16 a 24 = a + 2 ⋅ 24 − 2 = a + 48 − 2 = a + 46
11.
13.
d = log14 − log 7 = log 14 = log 2 7 a n = log 7 + ( n − 1) log 2 a 9 = log 7 + 8log 2 a 24 = log 7 + 23log 2
2
d = log a 2 − log a = log a = log a a a n = log a + ( n − 1) log a = (1 + n − 1)log a = n log a a 9 = 9log a a 24 = 24log a
12.
d = ln16 − ln 4 = ln 16 = ln 4 4
a n = ln 4 + ( n − 1) ln 4 = ln 4 + ln 4 ( n −1) = ln ( 4 ⋅ 4 ( n − 1) ) = ln 4 n = n ln 4 a 9 = 9ln 4 a 24 = 24ln 4 14.
d = log 2 5a − log 2 5 = log 2 5a = log 2 a 5 a n = log 2 5 + ( n − 1) log 2 a a 9 = log 2 5 + 8log 2 a a 24 = log 2 5 + 23log 2 a
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Section 11.2
15.
769
d = 15 − 13 = 2 a 4 = a1 + ( 4 − 1) 2 = 13 a1 + 6 = 13 a1 = 7 a 20 = 7 + ( 20 − 1) 2 = 7 + (19 ) 2 = 7 + 38 = 45
16.
a 6 = −14 = a1 + ( 6 − 1) d
a8 = a1 + ( 8 − 1) d = −20 a1 + 7d 14 − 5d + 7d −14 + 2d 2d d
−14 = a1 + 5d −14 − 5d = a1
= −20 = −20 = −20 = −6 = −3
a1 = −14 − 5( −3) = −14 + 15 = 1 a15 = 1 + (15 − 1)( −3) = 1 + 14( −3) = 1 − 42 = −41
17.
a5 = −19 = a1 + ( 5 − 1) d
a 7 = −29 = a1 + ( 7 − 1) d
18.
a1 = 3 (1) + 2 = 5
3d = 12 d =4
22 = a1 + 3d a1 = 22 − 3d a1 = 22 − 3 ⋅ 4 = 10 a 23 = 10 + ( 23 − 1) 4 = 10 + ( 22 ) 4 = 10 + 88 = 98
a10 = 3 (10 ) + 2 = 32
a1 = 4(1) − 3 = 1 a12 = 4(12) − 3 = 45
S10 = 10 ( a1 + a10 ) = 5(5 + 32) = 185
S12 = 12 ( a1 + a12 ) = 6(1 + 45) = 276
20.
2
2
21.
23.
a1 = 3 − 5 (1) = −2
22.
a 20 = 1 − 2 ( 20 ) = −39
S15 = 15 ( a1 + a15 ) = 15 ( −2 + ( −72 ) ) = 15 ( −74 ) = −555 2 2 2
S 20 = 20 ( a1 + a 20 ) = 10 ( −1 + ( −39 ) ) = 10 ( −40 ) = −400 2
a1 = 6 (1) = 6 a12 = 6 (12 ) = 72
24.
a1 = 1 + 8 = 9 a25 = 25 + 8 = 33
S14 = 14 ( a1 + a14 ) = 7 ( 7 + 98 ) = 735 2
26.
a1 = −1 a30 = −30
28.
a1 = 1 + x a12 = 12 + x
a1 = 4 − 1 = 3 a40 = 4 − 40 = −36
S40 = 40 ( a1 + a40 ) = 20 ( 3 + ( −36 ) ) = 20 ( −33) = −660 2 30.
S12 = 12 ( a1 + a12 ) = 6 (1 + x + 12 + x ) = 78 + 12 x 2
a1 = 1 − 4 = −3 a 25 = 25 − 4 = 21 S25 = 25 ( a1 + a25 ) = 25 ( −3 + 21) = 25 (18 ) = 225 2 2 2
S30 = 30 ( a1 + a30 ) = 15 ( −1 + ( −30 ) ) = 15 ( −31) = −465 2
29.
a1 = 7 (1) = 7
a14 = 7 (14 ) = 98
S25 = 25 ( a1 + a25 ) = 25 ( 9 + 33) = 25 ( 42 ) = 525 2 2 2
27.
a1 = 1 − 2 (1) = −1
a15 = 3 − 5 (15) = −72
S12 = 12 ( a1 + a12 ) = 6 ( 6 + 72 ) = 6 ( 78 ) = 468 2
25.
34 = a1 + 6d 34 = 22 − 3d + 6d 34 = 22 + 3d
a1 = −4( −5) − 19 = 1 a17 = 1 + (17 − 1)( −5) = 1 + 16( −5) = 1 − 80 = −79
19.
a 4 = 22 = a1 + ( 4 − 1) d a 7 = 34 = a1 + ( 7 − 1) d
−29 = a1 + 6d −29 = ( −19 − 4d ) + 6d −29 = −19 + 2d −10 = 2d d = −5
−19 = a1 + 4d −19 − 4d = a1
a1 = 2 (1) − x = 2 − x a15 = 2 (15) − x = 30 − x S15 = 15 ( a1 + a15 ) = 15 ( 2 − x + 30 − x ) = 15 ( 32 − 2 x ) 2 2 2 = 240 − 15 x
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770
31.
Chapter 11: Sequences, Series, and Probability
a1 = (1) x = x a20 = ( 20 ) x = 20 x
32.
S20 = 20 ( a1 + a20 ) = 10 ( x + 20 x ) = 210 x 2
33.
a1 = − (1) x = − x
a14 = − (14 ) x = −14 x
− 1, c2 , c3 , c4 , c5 , c6 , 23
S14 = 14 ( a1 + a14 ) = 7 ⎣⎡ − x + ( −14 x )⎦⎤ = 7 ( −15 x ) = −105 x 2
34.
7, c2 , c3 , c4 , c5 , c6 , 19
a1 = 7 a7 = a1 + ( n − 1) d 19 = 7 + ( 7 − 1) d 12 = 6d d =2 c2 = 7 + ( 2 − 1) 2 = 7 + 2 = 9 c3 = 7 + ( 3 − 1) 2 = 7 + ( 2 ) 2 = 11 c4 = 7 + ( 4 − 1) 2 = 7 + ( 3) 2 = 13 c5 = 7 + ( 5 − 1) 2 = 7 + ( 4 ) 2 = 15 c6 = 7 + ( 6 − 1) 2 = 7 + ( 5) 2 = 17
a1 = −1 a7 = a1 + ( n − 1) d 23 = −1 + ( 7 − 1) d 23 = −1 + 6d 24 = 6d d =4 c2 = −1 + ( 2 − 1) 4 = −1 + 4 = 3 c3 = −1 + ( 3 − 1) 4 = −1 + ( 2 ) 4 = 7 c4 = −1 + ( 4 − 1) 4 = −1 + ( 3) 4 = 11 c5 = −1 + ( 5 − 1) 4 = −1 + ( 4 ) 4 = 15 c6 = −1 + ( 6 − 1) 4 = −1 + ( 5) 4 = 19 35.
3, c2 , c3 , c4 , c5 , 1 2 a1 = 3 a6 = a1 + ( n − 1)d 1 = 3 + (6 − 1)d 2 − 5 = 5d 2 d =−1 2 c2 = 3 + (2 − 1) ⎛⎜ − 1 ⎞⎟ = 3 − 1 = 5 2 2 ⎝ 2⎠ 1 ⎛ ⎞ c3 = 3 + (3 − 1) ⎜ − ⎟ = 3 − 1 = 2 ⎝ 2⎠ c4 = 3 + (4 − 1) ⎛⎜ − 1 ⎞⎟ = 3 − 3 = 3 2 2 ⎝ 2⎠ 1 ⎛ ⎞ c5 = 3 + (5 − 1) ⎜ − ⎟ = 3 − 2 = 1 ⎝ 2⎠
36.
11 , c , c , c , c , 6 2 3 4 5 3 a1 = 11 3 a6 = a1 + ( n − 1)d 6 = 11 + (6 − 1)d 3 7 = 5d 3 d= 7 15 c2 = 11 + (2 − 1) 7 = 11 + 7 = 62 3 15 3 15 15 c3 = 11 + (3 − 1) 7 = 11 + 14 = 69 3 15 3 15 15 11 c4 = + (4 − 1) 7 = 11 + 21 = 76 3 15 3 15 15 11 c5 = + (5 − 1) 7 = 11 + 28 = 83 3 15 3 15 15
37.
a1 = 1, d = 2
38.
a1 = 2 an = 2 n
n[2(1) + (n − 1)2] n Sn = = [2n] = n 2 2 2
39.
a1 = 25, d = –1 a6 = 25 + (6 − 1)(−1) = 25 − 5 = 20 S6 = 6 (25 + 20) = 3(45) = 135 2 20 logs stacked on sixth row, 135 logs in the six rows
Sn = n ( an + a1 ) = n [ 2n + 2] = n 2 + n 2 2
40.
a1 = 27, d = 2 a10 = 27 + (10 − 1)(2) = 27 + 18 = 45 S10 = 10 (27 + 45) = 5(72) = 360 2 45 seats in the tenth row, 360 seats in the ten rows
Copyright © Houghton Mifflin Company. All rights reserved.
Section 11.2
41.
771
a1 = 5000, d = −250
42.
a15 = 5000 + (15 − 1)(−250) = 5000 − 3500 = 1500
The fifteenth prize is $1500. 15 15 ( 5000 + 1500 ) = ( 6500 ) = 48,750 2 2 The total amount of money distributed is $48,750. S15 =
43.
a1 = 16, d = 32
a1 = 15, d = 5 an = a1 + ( n − 1) d 60 = 15 + ( n − 1) 5 60 = 15 + 5n − 5 50 = 5n n = 10 In 10 weeks, a person will be walking 60 minutes a day.
44.
S7 = 7 ⎡⎣ 2 (16 ) + ( 7 − 1) 32 ⎤⎦ = 7 [32 + 192] = 7 [ 224] = 784 2 2 2
an = 2n – 1 a10 = 2 (10 ) − 1 = 20 − 1 = 19 a1 = 2 (1) − 1 = 1 S10 = 10 (1 + 19 ) = 100 2
The total distance the object falls is 784 ft.
The distance the ball rolls in the tenth second is 19 ft. The total distance is 100 ft. 45.
a n = a1 + ( n − 1)d 46.9 = 3.5 + ( n − 1)1.4 43.4 = ( n − 1)1.4 31 = n − 1 32 = n
46.
S n = n ( a1 + a n ) 2 S 32 = 32 ( 3.5 + 46.9 ) 2 S 32 = 806.4 mm
....................................................... 47.
If f(x) is a linear function, then f(x) = mx + b. To show that f(n), where n is a positive integer, is an arithmetic sequence, we must show that f ( n + 1) − f ( n ) is a constant. We have f (n + 1) − f (n) = (m(n + 1) + b) − (m(n) + b) = mn + m + b − mn − b =m
Thus, the difference between any two successive terms is m, the slope of the linear function. 49.
Connecting Concepts 48.
a1 = 3, an = an–1 + 5 Rewriting an = an −1 + 5 as an − an −1 = 5, the difference between successive terms is the same constant 5. Thus the sequence is an arithmetic sequence with a1 = 3 and d = 5. an = a1 + (n – 1)d Substituting, an + 3 + (n – 1)5 = 5n – 2
a1 = 4, an = an −1 − 3
Rewriting an = an −1 − 3 as an − an −1 = −3, we find that the difference between successive terms is the same constant –3. Thus the sequence is an arithmetic sequence with a1 = 4 and d = −3. an = a1 + ( n − 1)d Substituting, an = 4 + ( n − 1)( −3) = 4 − 3n + 3 = 7 − 3n
Copyright © Houghton Mifflin Company. All rights reserved.
772
50.
Chapter 11: Sequences, Series, and Probability
a1 = 4, an = bn −1 + 5; b1 = 2, bn = an −1 + 1
To show that an is an arithmetic sequence, we must show that the difference between successive terms is some constant d. We begin by finding a relationship between an and an − 2 . an = bn −1 + 5 = an −1−1 + 1 + 5 = an − 2 + 6
a1 = 4 a2 = b1 + 5 = 2 + 5 = 7 a3 = a1 + 6 a4 = a2 + 6 a5 = a3 + 6 = ( a1 + 6 ) + 6 = a1 + 2 ( 6 )
a6 = a4 + 6 = ( a2 + 6 ) + 6 = a2 + 2 ( 6 ) a7 = a5 + 6 = ( a3 + 6 ) + 6 = a1 + 3 ( 6 )
a8 = a6 + 6 = ( a4 + 6 ) + 6 = a2 + 3 ( 6 ) Thus we have an = an − 2 + 6. This establishes a relationship between alternate successive terms. We now examine the terms of an . Now consider two cases. First, n is an even integer, n = 2 k . From the list of terms, we have a2 k = a2 + ( k − 1) 6
k≥2
Now consider the case when n is an odd integer, n = 2 k − 1. a2 k −1 = a1 + ( k − 1) 6
k≥2
Thus a2 k − a2 k −1 = ( a2 + ( k − 1) 6 ) − ( a1 + ( k − 1) 6 ) = a2 − a1 = 7 − 4 = 3 Therefore, the difference between successive terms is the constant 3. To find a100 , use an = a1 + ( n − 1) d . a100 = 4 + (100 − 1)( 3 ) = 4 + ( 99 )( 3 ) = 301
To show that bn is also an arithmetic sequence, we have
bn − bn −1 = ( an −1 + 1) − ( an − 2 + 1) = an −1 − an − 2
Because an is an arithmetic sequence, an −1 − an − 2 is a constant. Thus bn is an arithmetic sequence.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 11.2
51.
773
a1 = 1, an = bn −1 + 7; b1 = −2, bn = an −1 + 1 To show that an is an arithmetic sequence, we must show that an − an −1 = d , where d is a constant. We begin by finding a relationship between an and an − 2 . an = bn−1 + 7 = an−2 + 1 + 7
an = an−2 + 8
This establishes a relationship between alternate successive terms. We now examine some terms of an . a1 = 1 a2 = b1 + 7 = −2 + 7 = 5 a3 = a1 + 8 ( an = a n − 2 + 8 ) a4 = a2 + 8 a5 = a3 + 8 = ( a1 + 8 ) + 8 = a1 + 2 (8 ) a6 = a4 + 8 = ( a2 + 8 ) + 8 = a2 + 2 (8 )
a7 = a5 + 8 = ( a1 + 2 (8 ) ) + 8 = a1 + 3 (8 ) a8 = a6 + 8 = ( a2 + 2 (8 ) ) = a2 + 3 ( 8 )
Now consider two cases. First, n is an even integer, n = 2 k . a2 k = a2 + ( k − 1) 8 k≥2 When n is an odd integer, n = 2 k − 1. a2 k −1 = a1 + ( k − 1) 8
k≥2
Thus a2 k − a2 k −1 = ( a2 + ( k − 1) 8 ) − ( a1 + ( k − 1) 8 ) = a2 − a1 = 5 − 1 = 4 Therefore, the difference between successive terms is the constant. To find a50 , use an = a1 + ( n − 1) d . a50 = 1 + ( 49 )( 4 ) = 197 To show that bn is an arithmetic sequence, we have
bn − bn −1 = ( an −1 + 1) − ( an − 2 + 1) = an −1 − an − 2 Because an is an arithmetic sequence, an −1 − an − 2 is a constant. Thus bn is an arithmetic sequence.
....................................................... PS1.
4 = 2, 8 = 2 2 4 The ratio is 2.
5 PS3. S = 3(1 − ( −2) ) = 33 1 − ( −2)
Prepare for Section 11.3 PS2.
4
∑ 2n1−1 = 210 + 211 + 212 + 213 = 158
n =1
PS4.
S − rS = a − ar 2 S (1− r ) = a (1− r 2 ) a (1+ r )(1− r ) S= (1− r ) S = a (1+ r )
1
PS5.
⎛ ⎞ a1 = 3 ⎜ − 1 ⎟ = − 3 2 ⎝ 2⎠ 2
⎛ ⎞ a2 = 3 ⎜ − 1 ⎟ = 3 ⎝ 2⎠ 4 3 ⎛ ⎞ a3 = 3 ⎜ − 1 ⎟ = − 3 ⎝ 2⎠ 8
PS6. S1 = 2
S2 = 2 + 4 = 6 S3 = 2 + 4 + 8 = 14
Copyright © Houghton Mifflin Company. All rights reserved.
774
Chapter 11: Sequences, Series, and Probability
Section 11.3 1.
3.
5.
4i +1 = 4, geometric, r = 4 4i 1 i + 1 = i , not geometric 1 i +1 i 2(i +1) x 2ix
=
2ix + x 2ix
7.
3(2(i +1) −1)
9.
x 2(i +1)
3(2i −1)
x 11.
2.
2i
=
=
x
(− 1)(i +1)−1e(i +1)x = (− 1)i eix + x = (− 1)e x (− 1)i −1eix (− 1)i −1eix geometric, r = (− 1)e x
6.
= 2, geometric, r = 2
8.
5(− 2 )(i +1)−1
10.
3(i + x )xi +1
2i −1
2i
1 i 2i +1 = 2 = 1 , geometric, r = 1 i + 1 2 2 1 2 i 2
= 2 x , geometric, r = 2 x
2i
x 2i + 2
4.
6i +1 = 6, geometric, r = 6 6i
= x 2 , geometric, r = x 2
13.
r = 8 = 4, 2 2 n −1 an = 2 ⋅ 4n −1 = 2 ⋅ 2 ( ) = 22 n −1
15.
r = 12 = −3, an = −4 ( −3) −4
18.
r = 6 = 3 , an = 8 ⎛⎜ 3 ⎞⎟ 8 4 ⎝4⎠
21.
r = −3 = − 1 , an = 9 ⎛⎜ − 1 ⎞⎟ 9 3 ⎝ 3⎠
23.
n −1 n −1 r = − x = − x, an = 1( − x ) = ( − x ) 1
n −1
( )
n −1
n −1
12.
= ⎛⎜ − 1 ⎞⎟ ⎝ 3⎠
r=
=
2i −1
n −1 r = 6 = −2, an = −3 ( −2 ) −3
19.
r = − 5 , an = −6 ⎛⎜ − 5 ⎞⎟ 6 ⎝ 6⎠
n −1
n −3
(− 2)i = −2, geometric, r = −2 (− 2)i −1
(i + 1)x , not geometric
=
i
2i log x i −1
2
log x
=
2i i −1
22
17.
r = 4 = 2 , an = 6 ⎛⎜ 2 ⎞⎟ 6 3 ⎝3⎠
20.
r=−
−4 3
= −1 , an = 8 ⎛⎜ − 1 ⎞⎟ 8 6 ⎝ 6⎠
= − 4 = − 2 , an = −2 ⎛⎜ − 2 ⎞⎟ −2 6 3 ⎝ 3⎠
n −1
r=
24.
n −1 r = 2a = a, an = 2 ( a ) = 2a n −1 2
26.
n −1 4 n −1 n r = x 2 = − x 2 , an = − x 2 ( − x 2 ) = −1 ⋅ x 2 ⎡ ( −1) ⋅ x 2 n − 2 ⎤ = ( −1) x 2 n ⎣ ⎦ −x
27.
r=
29.
n −1 n r = 0.05 = 0.1, an = 0.5 ( 0.1) = 5 ( 0.1) 0.5
n −1
4 3
22.
5 r = c 2 = c 3 , an = c 2 c 3 c
= 2, geometric, r = 2
5 = 5, an = 1(5)n −1 = 5n −1 1
16.
25.
3 10,000 3 100
i
log x 2(i +1)−1 log x
14.
n −1
5(− 2 )i −1
3ix
ln 5(i + 1) ⎛ ln 5 ⎞⎛ ln(i + 1) ⎞ =⎜ ⎟⎜ ⎟, not geometric ln 5i ⎝ ln 5i ⎠⎝ ln 5i ⎠
=
= c 2c3n −3 = c3n −1
= 1 , an = 3 ⎛⎜ 1 ⎞⎟ 100 100 ⎝ 100 ⎠
n −1
= 3 ⎛⎜ 1 ⎞⎟ ⎝ 100 ⎠
n
7 10,000 7 10
=
1 ,a = 7 ⎛ 1 ⎞ ⎜ ⎟ n 1000 100 ⎝ 1000 ⎠
n −1
= 7 ⎛⎜ 1 ⎞⎟ ⎝ 10 ⎠
3n − 2
28.
r=
30.
n −1 2 n −1 r = 0.004 = 0.01, an = 0.4 ( 0.01) = 4 ( 0.1) 0.4
Copyright © Houghton Mifflin Company. All rights reserved.
n −1
Section 11.3
775
31.
n −1 n r = 0.0045 = 0.01, an = 0.45 ( 0.01) = 45 ( 0.01) 0.45
33.
a1 = 2,
a5 = 162,
an = a1r n −1
34.
162 = 2r 5−1
32.
a3 = 1,
a8 =
a3 = 2(3)3−1 = 2 ⋅ 9 = 18
a4 = 4 ⎜⎛ 1 ⎟⎞ ⎝2⎠
a4 = 8 9 64 243
8 , 9
=
a7 =
64 243
37.
a1r 4 −1
2 a1 = , 3
a1 = 3,
S5 =
7 −1
a1r 27 = 1 8 r3 r3 = 8 27 r=2 3 4 −1 8 = a ⎛2⎞ 1⎜ ⎟ 9 ⎝3⎠ 8 a1 = ⋅ 27 = 3 9 8 5 −1 a5 = 3 ⎛⎜ 2 ⎞⎟ = 16 27 ⎝3⎠
39.
1 32
4 2 9 r= = 2 3 3
4 a2 = , 9
41.
a3 =
4 −1
4 3 − 32 81
a2 = 9,
( )
40.
38.
a1 = 4 , a2 = 16 , r = 3 9 14 ⎤ ⎡ 4 1− ⎛ 4 ⎞ ⎢ ⎥ ⎜ ⎟ 3 ⎝3⎠
a6 = −
32 81
3−1
=
a1 ( r )
6 −1
a1 = 2,
S7 =
16 9 4 3
(
a2 = 4,
)
=4 3
1 1 a1 = 1, a2 = − , r = − 3 3
⎡ ⎛ 1 ⎞8 ⎤ 1⎢1 − ⎜ − ⎟ ⎥ 6560 ⎢ ⎝ 3⎠ ⎥ ⎦ = 6561 = 1640 S8 = ⎣ 4 ⎛ 1⎞ 2187 1− ⎜− ⎟ 3 ⎝ 3⎠
Copyright © Houghton Mifflin Company. All rights reserved.
r=
4 =2 2
2 1 − 27 2(− 127 ) = = 254 1− 2 −1
4 ⎛ −263,652,487 ⎞ ⎢ ⎥⎦ 3 ⎝⎜ 4,782,969 ⎠⎟ S14 = ⎣ = −1 1− 4 3 3 1,054,609,948 = ≈ 220.49 4,782,969 42.
9⎤ ⎡ 1 ⎢1 − ⎛⎜ − 2 ⎞⎟ ⎥ 1,953,637 ⎢ ⎝ 5 ⎠ ⎦⎥ 1,953,125 279,091 = = S9 = ⎣ 7 390,625 1 − ⎛⎜ − 2 ⎞⎟ 5 ⎝ 5⎠
9 =3 3
3 1 − 35 3(− 242) = = 363 1− 3 −2
3
2 2 a1 = 1, a2 = − , r = − 5 5
r=
4 , 3
a1 ( r ) −27 = 1 8 r3 3 r =− 8 27 r = −2 3 3−1 4 = a ⎛− 2⎞ ⎟ 1⎜ 3 ⎝ 3⎠ ⎛ 4 a1 = ⎜ 9 ⎞⎟ = 3 3⎝ 4⎠ a2 = 3 ⎛⎜ −2 ⎞⎟ = −2 ⎝ 3 ⎠
= 4 ⎜⎛ 1 ⎟⎞ = 1 ⎝8⎠ 2
⎡ ⎛ 2 6 ⎞⎤ 2 ⎢1 − ⎜ ⎟ ⎥ 2 ⎛ 665 ⎞ 3 ⎢ ⎜ 3 ⎟⎥ ⎟ ⎜ ⎝ ⎠ ⎦ 3 ⎝ 729 ⎠ 1330 S6 = ⎣ = = 2 1 729 1− 3
35.
2 1 = a1r 1 7 a1r 32 32 = 15 r 5 r = 1 32 r=1 2 2 1 = a1 ⎛⎜ 1 ⎞⎟ ⎝2⎠ a1 = 4
r 4 = 81 r =3
36.
n −1 n r = 0.000234 = 0.001, an = 0.234 ( 0.001) = 234 ( 0.001) 0.234
776
43.
Chapter 11: Sequences, Series, and Probability
a1 = 1, a2 = −2, r = −2
S7 =
46.
1[1 − (−2) ] = −341 1 − ( −2) 47.
2[1 − (−4)11] = 1,677,722 1 − (−4)
a1 =
2[1 − 5 ] = 195,312 1− 5
S10 =
1 1 1 a1 = , a2 = , r = 3 3 9
a1 =
48.
50.
2 3
3 5
()
a1 =
51.
53.
−
a1 = 0.1, r = 0.1
a1 = 0.4, r = 0.5
54.
0.1 0.1 1 S= = = 1 − 0.1 0.9 9
7 7 7 10 10 S= = = 7 3 3 1− 10 10
S=
55.
a1 = 1, r = −0.4 1 S= = 1 =5 1 − ( −0.4 ) 1.4 7
56.
a1 = 1, r = −0.8 1 = 1 =5 S= 1 − ( −0.8 ) 1.8 9
57.
0.3 = 3 + 3 + 3 + ⋅ ⋅ ⋅ 10 100 1000
58.
0.5 = 5 + 5 + 5 + ⋅ ⋅ ⋅ 10 100 1000
3
a1 =
0.3 =
59.
=
=1 3
0.5 =
45 0.45 = 45 + 45 + + ⋅⋅⋅ 100 10,000 1,000,000 a1 = 0.45 =
61.
a1 =
10
3 10 9 10
60.
45 10,000 = 1 45 100 100 100 45 45 100 = 100 = 5 99 11 1− 1 100 100 45
,r =
123 1000 1− 1 1000
= 123 = 41 999 333
= 1 , r = 100 5 10 10 5
5 10 1− 1 10
0.63 =
62.
10 5 = 10 9 10
=5 9
63 0.63 = 63 + 63 + + ⋅⋅⋅ 100 10,000 1,000,000 a1 =
123 + + ⋅⋅⋅ 0.123 = 123 + 123 100 1,000,000 1,000,000,000 a1 = 123 , r = 1 1000 1000 0.123 =
0.5 0.5 = =1 1 − 0.5 0.5
5
= 1 , r = 100 3 10 10 3
3 10 1− 1 10
9 9 ,r = 100 100
9 9 9 100 100 S= = = 9 91 91 1− 100 100
3 3 5 S= = =− 8 3⎞ 8 ⎛ 1− ⎜− ⎟ 5 ⎝ 5⎠
7 7 ,r = 10 10
3 9 3 , a2 = , r = 4 4 16
3 S= = 4 =3 1 3 1− 4 4
3 3 a1 = − , r = − 5 5 −
5[1 − 310 ] = 147,620 1− 3
3 4
1 1 S= = 3 = 2 1 2 1− 3 3
2 2 S= =− 3 =− 5 2⎞ 5 ⎛ 1− ⎜− ⎟ 3 ⎝ 3⎠
52.
a1 = 5, a2 = 15, r = 3
45.
8
1 3
2 2 a1 = − , r = − 3 3 −
a1 = 2, a2 = 10, r = 5
S8 =
a1 = 2, a2 = −8, r = −4
S11 =
49.
44.
10
63 100
,r =
63 100 1− 1 100
63 10,000 63 100
= 1 100
= 63 = 7 99 11
⎡ ⎤ 95 0.395 = 3 + ⎢ 95 + 95 + + ⋅ ⋅ ⋅⎥ 10 ⎣1000 100,000 10,000,000 ⎦ 1 95 a1 = ,r = 100 1000 95
0.395 = 3 + 10001 = 3 + 95 = 392 = 196 10 1 − 10 990 990 495 100
Copyright © Houghton Mifflin Company. All rights reserved.
Section 11.3
63.
422 + 422 + ⋅⋅⋅ 0.422 = 422 + 1000 1,000,000 1,000,000,000 a1 = 422 , r = 1 1000 1000 0.422 =
65.
777
422 1000 1− 1 1000
64.
= 422 999
0.355 =
⎤ 25 ⎡ 4 4 4 +⎢ + + + ⋅ ⋅ ⋅⎥ 100 ⎣1,000 10,000 100,000 ⎦ 1 4 ,r = a1 = 10 1000
0.25 4 =
66.
⎡ ⎤ 84 84 1.2084 = 1 + 2 + ⎢ 84 + + + ⋅ ⋅ ⋅⎥ 10 ⎣10,000 1,000,000 100,000,000 ⎦ a1 = 84 , r = 1 10, 000 100
68.
⎡ ⎤ 90 90 + ⋅⋅⋅ 02.2590 = 225 + ⎢ 90 + 100 ⎣10,000 1,000,000 100,000,000 ⎥⎦ a1 = 90 , r = 1 10, 000 100 90
10,000 = 9 + 1 = 497 2.2590 = 225 + 100 1 − 1 4 110 220 100
100
P = 100
⎡ ⎤ 2 0.372 = 37 + ⎢ 2 + 2 + ⋅⋅⋅ 100 ⎣1,000 10,000 100,000 ⎥⎦ a1 = 2 , r = 1 10 1000
10
84
A = 100, i = 0.09, n = 2, r = 8, r =
= 355 999
2
10,000 12 = + 7 = 1994 = 997 1.2084 = 1 + 2 + 10 1− 1 10 825 1650 825
69.
355 1000 1− 1 1000
0.372 = 37 + 10001 = 37 + 2 = 335 = 67 100 1 − 100 900 900 180
4 25 1000 25 4 229 0.25 4 = + = + = 100 1− 1 100 900 900 10
67.
355 355 + + ⋅⋅⋅ 0.355 = 355 + 1000 1,000,000 1,000,000,000 a1 = 355 , r = 1 1000 1000
i 0.09 = = 0.049, m = nt = 2 ⋅ 8 = 16 n 2
[(1+0.045) −1] ≈ 2271.93367 16
0.045
The future value is $2271.93.
70.
A = 250, i = 0.08, n = 12, t = 4, r = 8 = 2 , m = 12 ⋅ 4 = 48 12 3 P = 250
71.
⎡⎛ 0.08 ⎞48 ⎤ ⎢⎜1+ ⎟ −1⎥ ⎣⎢⎝ 12 ⎠ ⎦⎥ 0.08 12
≈ $14,087.48
When a name was removed from the top of the list, the letter had been sent to
72.
A = 0.5, n = 4, k = −0.876, t = 4
74.
( 12 ) S12 = 5 1 − 5 = 305,175,780 1− 5 For a population of 127,000,000, the entire population would receive the letter on the 12th level.
5 ( 55 ) = 15,625 people who sent 10 cents each for a total of 0.10(15,625) = $1562.50 for each recipient..
73.
A + Aekt + Ae2 kt Ae3kt = 0.5 + 0.5e −0.867(4) + 0.5e2( −0.867)(4) + 0.5e3( −0.867)(4) ≈ 0.52 mg
( 11 ) S11 = 5 1 − 5 = 61,035,155 1− 5
A + Aekt + L + Ae( n −1) kt + L = 2 A (1 + ekt + L + e( n −1) kt + L) = 2 A =2 1 − ekt
Or A = 0.5, r = ekt , n = 4, k = −0.876, t = 4
( −.867(4)(3) ) ≈ 0.52 mg S4 = 0.5 1 − e−.867(4) 1− e
Copyright © Houghton Mifflin Company. All rights reserved.
A = 2 (1 − ekt ) A = 2 (1 − e −0.25(12) ) A ≈ 1.90 mg
778
75.
Chapter 11: Sequences, Series, and Probability
D (1 + g ) i−g 1.87(1 + 0.15) = 0.20 − 0.15 = $43.01
Stock value =
76. D = 1.87, g = 0.15, i = 0.20
D (1 + g ) i−g 1.32(1 + g ) 67 = Solve for g . 0.20 − g 67(0.20 − g ) = 1.32(1 + g ) 13.4 − 67 g = 1.32 + 1.32 g 12.08 = 68.32 g 0.1768 = g The dividend growth rather is 17.68%. Stock value =
77.
Stock value (no dividend growth) = D = 2.94 = $19.60 i 0.15
78.
Stock value (no dividend growth) = D i 3.24 16 = i 3.24 i= 16 i = 0.2025 The expected rate of return is 20.25%.
79.
If g was not less than i in the Gordon model of stock valuation, the common ratio of the geometric sequence would be greater than 1 and the sum of the infinite geometric series would not be defined.
80.
Using the multiplier effect, 50 = 500 1 − 0.90 The net effect of $50 million is $500 million.
81.
Using the multiplier effect, 25 = 100 1 − 0.75 The net effect of $25 million is $100 million.
82.
Using the multiplier effect, 500,000 ≈ $833,000 1 − 0.40 About $833,000 is used before it is removed.
....................................................... 83.
Connecting Concepts
Let an be a geometric sequence. Thus
an = a1r n −1, a1 ≠ 0,
r≠0
and log an = log a1r n −1 = log a1 + log r n −1
= log a1 + (n − 1)log r Since r is a constant, log r is a constant. Conjecture: The sequence log an is an arithmetic sequence. To prove this conjecture, we must show that log an − log an −1 is a constant.
log an − log an −1 = ( log a1 + ( n − 1) log r ) − ( log a1 + ( n − 2 ) log r )
= log r Since log r is a constant, the sequence log an is an arithmetic sequence.
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Section 11.3
84.
779
Let an be a geometric sequence. Thus
an = a1 + (n − 1)d where d is a constant.
( )
n −1 2an = 2a1 + (n −1)d = 2a1 ⋅ 2(n −1)d = 2a1 ⋅ 2d
Conjecture : The sequence 2an is a geometric sequence. To prove this conjecture let bn = 2an and show that bn + 1 = r where r is a constant. bn
( )n ( )
2a1 ⋅ 2d bn +1 = = 2d . bn a1 d n −1 2 ⋅2 Because d is a constant (see above), 2d is a constant. Thus 2an is a geometric sequence. 85.
Yes. Because x ≠ 0, the first term is 1 and the common ratio is x. If x < 1, the geometric series converges to
1 . If x ≥ 1, the 1− x
geometric series does not converge. 86.
To find the area of the n th inscribed square, begin by finding the area of the first few squares. The area of the first square is 12 = 1. Thus a1 = 1 The length of a side of the second square can be found by using the Pythagorean Theorem. 2
2
S 2 = ⎛⎜ 1 ⎞⎟ + ⎛⎜ 1 ⎞⎟ = 1 + 1 = 1 4 4 2 ⎝2⎠ ⎝2⎠ Notice here that S 2 is the area of the second square. Thus a2 = 1 . 2 The length of the side of the third square is 2
2
⎛ ⎞ ⎛ ⎞ S2 = ⎜ 1 ⎟ + ⎜ 1 ⎟ = 1 + 1 = 2 = 1 8 8 8 4 ⎝2 2⎠ ⎝2 2⎠ Again S 2 is the area of the square. Thus a3 = 1 . 4 From here we conjecture that the area of the n th square is an = n1−1 . 2
87.
If an is a geometric sequence, then an = a1r n −1
Since an = ar n −1 , then a1 = a. Pn = a1 ⋅ a2 ⋅ a3 ⋅ ⋅ ⋅ ⋅ ⋅ an = a ⋅ a2 ⋅ a3 ⋅ ⋅ ⋅ ⋅ ⋅ an
= a ⋅ ar ⋅ ar 2 ⋅ ⋅ ⋅ ⋅ ⋅ ar n −1 = a n r [( n −1) n ]/ 2
The exponent on r is found by using the sum of an arithmetic series formula. Note that
a ⋅ ar ⋅ ar 2 ⋅ ar 3 ⋅ ⋅ ⋅ ⋅ ⋅ ar n −1 = a n ⋅ r 1 + 2 + 3 + ⋅⋅⋅+ n − 1 and 1 + 2 + 3 + ⋅ ⋅ ⋅ + n − 1 = 88.
( n − 1) n .
Let an = f (n ) = ab n . Thus an +1 ab n +1 = =b an ab n Since b is a constant, an is a geometric sequence.
Copyright © Houghton Mifflin Company. All rights reserved.
2
780
89.
Chapter 11: Sequences, Series, and Probability
The distance the ball travels each bounce is given by a1 = 5 a2 = 2 ( 0.8 ) 5 Multiply by 2 for the distance up and down. a3 = 2 ( 0.8 )( 0.8 )( 5 ) = 2 ( 0.8 ) 5 2
a4 = 2 ( 0.8 )( 0.8 ) 5 = 2 ( 0.8 ) 5 2
M
3
n −1
an = 2 ( 0.8 ) ⋅ 5 This is a geometric sequence (after a1 ). The sum of this sequence is the total distance travelled by the ball. n
2 ( 0.8 ) ( 5 ) an +1 = = 0.8 n +1 an 2 ( 0.8 ) ( 5 ) Because 0.8 < 1, the geometric series converges. S = 5 + 8 = 5 + 8 = 5 + 40 = 45 0.2 1 − 0.8 The distance traveled is 45 feet. Note from our calculation that the geometric series begins with a2 . The first term (a1 = 5) is added to the series. 90.
The distance of each swing of the bob is a term of a sequence 91. a1 = 30 a2 = ( 0.9 )( 30 ) 2
a3 = ( 0.9 )( 0.9 )( 30 ) = ( 0.9 ) ( 30 ) M an = ( 0.9 )
n −1
( 30 ) a ( 0.9 )n ( 30 ) = 0.9, a constant, the sequence is Because n +1 = an ( 0.9 )n −1 ( 30 ) a geometric sequence. Since 0.9 < 1, the infinite geometric series converges. The sum of this series is the total distance traveled by the bob. 30 30 S= = = 300 1 − 0.9 0.1 The bob traveled 300 inches.
The n th generation has an = 2n grandparents. Since this is a geometric sequence, the sum can be found by a formula. Sn =
a1 1 − r n
(
)
1− r 2 1 − 210
)
(
2 (1 − 1024 ) = = 2046 1− 2 −1 When n = 1, an = 2 and there are no grandparents. Therefore there are 2046 − 2 = 2044 grandparents by the 10 th generation. S10 =
....................................................... PS1.
4
∑ i(i 1+ 1) = 1(11+ 1) + 2(21+ 1) + 3(31+ 1) + 4(41+ 1)
Prepare for Section 11.4 PS2. k ( k + 1)(2k + 1) + 6( k + 1)2 = ( k + 1)[k (2k + 1) + 6( k + 1)] = ( k + 1)[2k 2 + k + 6k + 6]
i =1
=1+1+ 1 + 1 =4= 4 2 6 12 20 5 4 + 1
PS3.
k + 1 1 = k +2⋅ k + k + 1 ( k + 1)( k + 2) k + 2 k + 1 ( k + 1)( k + 2) 2 ( k + 1)( k + 1) k + 1 = k + 2k + 1 = = ( k + 1)( k + 2) ( k + 1)( k + 2) k + 2
PS5. Sn + an +1 = n( n + 1) + n + 1 2 2 2 = n + n + 2n + 2 = n + 3n + 2 2 2 2 ( n + 1)( n + 2) = 2
= ( k + 1)[2k 2 + 7k + 6] = ( k + 1)( k + 2)(2k + 3)
PS4. 12 > 2(1) + 1 = 3
22 > 2(2) + 1 = 5 32 > 2(3) + 1 = 7 3 PS6. Sn + an +1 = 2n +1 − 2 + 2n +1
= 2n + 2 − 2 = 2(2n +1 − 1)
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Section 11.4
781
Section 11.4 1.
1. 2.
1(3 ⋅ 1 − 1) 2 Assume the statement is true for some positive integer k. k (3k − 1) (Induction Hypothesis) Sk = 1 + 4 + 7 + ⋅ ⋅ ⋅ + 3k − 2 = 2 Verify that the statement is true when n = k + 1 (k + 1)(3k + 2) . Sk +1 = 2 a k = 3k − 2, a k +1 = 3k + 1
Let n = 1. S1 = 3 ⋅ − 2 = 1 =
Sk +1 = Sk + ak +1 =
3k 2 − k 6k + 2 k (3k − 1) + 3k + 1 = + 2 2 2
3k 2 + 5k + 2 (k + 1)(3k + 2) = 2 2 By the Principle of Mathematical Induction, the statement is true for all positive integers. =
2.
1.
Let n =1. S1 = 2 ⋅ 1 = 2 = 1(1 + 1)
2.
Assume Sk = 2 + 4 + 6 + ⋅ ⋅ ⋅ + 2k = k (k + 1) is true for some positive integer k (Induction Hypothesis). Verify that Sk +1 = ( k + 1)( k + 2) is true when n = k + 1 ak = 2k ,
ak +1 = 2(k + 1)
Sk +1 = Sk + ak +1 = k (k + 1) + 2(k + 1) = (k + 1)(k + 2) By the Principle of Mathematical Induction, the statement is true for all positive integers.
3.
12 (1 + 1)
2
1.
Let n = 1. S1 = 13 = 1 =
2.
Assume Sk = 1 + 8 + 27 + ⋅ ⋅ ⋅ + k 3 = Verify that Sk +1 =
4 k 2 (k + 1)2 is true for some positive integer k.(Induction Hypothesis). 4
( k + 1)2 (k + 2)2 . 4
ak = k 3 , ak +1 = (k + 1)3 Sk +1 = Sk + ak +1 =
2 2 3 k 2 (k + 1)2 3 k ( k + 1) + 4( k + 1) + ( k + 1) = 4 4
( k + 1) 2 (k 2 + 4k + 4) (k + 1)2 (k + 2)2 = 4 4 By the Principle of Mathematical Induction, the statement is true for all positive integers. =
4.
1.
Let n = 1 S1 = 21 = 2 = 2(21−1)
2.
Assume that Sk = 2 + 4 + 8 + ⋅ ⋅ ⋅ + 2k = 2(2k − 1) is true for some positive integer k (Induction Hypothesis).
)
(
Verify that S k +1 = 2 2 k +1 − 1 . k
a k = 2 , a k +1 = 2
k +1
(
)
S k +1 = S k + a k +1 = 2 2 k − 1 + 2 k +1 = 2 k +1 − 2 + 2 k +1
(
)
2 ⋅ 2 k +1 − 2 = 2 2 k +1 − 1 By the Principle of Mathematical Induction, the statement is true for all positive integers. Copyright © Houghton Mifflin Company. All rights reserved.
782
5.
Chapter 11: Sequences, Series, and Probability
1.
Let n = 1. S1 = 4 ⋅ 1 − 1 = 3 = 1(2 ⋅ 1 + 1)
2.
Assume that Sk = 3 + 7 + 11 + ⋅ ⋅ ⋅ + 4k − 1 = k (2k + 1) is true for some positive integer k (Induction Hypothesis). Verify that Sk +1 + (k + 1)(2k + 3) . a k = 4k − 1,
a k +1 = 4k + 3
Sk +1 = Sk + ak +1 = k (2k + 1) + 4k + 3 = 2k 2 + 5k + 3 = (k + 1)(2k + 3) By the Principle of Mathematical Induction, the statement is true for all positive integers.
6.
3(31 − 1) =3 2
1.
Let n = 1. S1 = 31 =
2.
Assume that S k = 3 + 9 + 27 + ⋅ ⋅ ⋅ + 3 k = Verify that S k +1 = a k = 3k ,
3(3k − 1) is true for some positive integer k (Induction Hypothesis). 2
3(3k + 1 − 1) . 2
a k +1 = 3 k +1
Sk +1 = Sk + ak +1 =
3(3k − 1) + 3k +1 2
3k +1 − 3 + 2 ⋅ 3k +1 3 ⋅ 3k +1 − 3 3(3k +1 − 1) = = 2 2 2 By the Principle of Mathematical Induction, the statement is true for all positive integers. =
7.
1.
Let n = 1. S1 = (2 ⋅ 1 − 1)3 = 1 = 12 (2 ⋅ 12 − 1)
2.
Assume that Sk = 1 + 27 + 125 + ⋅ ⋅ ⋅ + (2k − 1)3 = k 2 (2k 2 − 1) is true for some positive integer k (Induction Hypothesis). Verify that Sk +1 = (k + 1)2 (2k 2 + 4k + 1) . ak = (2k − 1)3 ,
ak +1 = (2k + 1)3
Sk +1 = Sk + ak +1 = k 2 (2k 2 − 1) + (2k + 1)3 = 2k 4 − k 2 + 8k 3 + 12k 2 + 6k + 1 = 2k 4 + 8k 3 + 11k 2 + 6k + 1 = (k + 1)(2k 3 + 6k 2 + 5k + 1) = (k + 1)2 (2k 2 + 4k + 1) By the Principle of Mathematical Induction, the statement is true for all positive integers. 8.
1(1 + 1)(1 + 2) 3
1.
Let n = 1. S1 = 1(1 + 1) = 2 =
2.
Assume that Sk = 2 + 6 + 12 + ⋅ ⋅ ⋅ + k ( k + 1) =
k (k + 1)(k + 2) is true for some positive integer k (Induction Hypothesis). 3
(k + 1)(k + 2)(k + 3) . 3 ak +1 = (k + 1)(k + 2)
Verify that Sk +1 = ak = k (k + 1),
k (k + 1)(k + 2) k (k + 1)(k + 2) + 3(k + 1)(k + 2) + (k + 1)(k + 2) = 3 3 (k + 1)( k + 2)( k + 3) = 3 By the Principle of Mathematical Induction, the statement is true for all positive integers. Sk +1 =
Copyright © Houghton Mifflin Company. All rights reserved.
Section 11.4
9.
783
1 1 1 = = (2 ⋅ 1 − 1)(1 ⋅ 1 + 1) 3 2 ⋅ 1 + 1
1.
Let n = 1. S1 =
2.
Assume that Sk =
1 1 1 1 k + + + ⋅⋅⋅ + = for some positive integer k (Induction Hypothesis). 1⋅ 3 3 ⋅ 5 5 ⋅ 7 (2k − 1)(2k + 1) 2k + 1
Verify that S k +1 = ak =
k +1 . 2k + 3
1 , (2k − 1)(2k + 1)
ak +1 =
1 (2k + 1)(2k + 3)
k 1 2k 2 + 3k + 1 (2k + 1)(k + 1) k +1 + = = = 2k + 1 (2k + 1)(2k + 3) (2k + 1)(2k + 3) (2k + 1)(2k + 3) 2k + 3 By the Principle of Mathematical Induction, the statement is true for all positive integers. Sk +1 =
10.
1 1 1 = = 2 ⋅ 1(2 ⋅ 1 + 2) 8 4(1 + 1)
1.
Let n = 1. S1 =
2.
Assume that Sk =
1 1 1 1 k + + + ⋅⋅⋅ + = for some positive integer k (Induction Hypothesis). 2⋅4 4⋅6 6⋅7 2k (2k + 2) 4(k + 1)
Verify that Sk +1 =
k +1 . 4(k + 2)
1 1 , ak +1 = 2k (2k + 2) (2k + 2)(2k + 4) 2 k ( k + 2) + 1 1 1 k k S k +1 = + = + = = k + 2k + 1 4(k + 1) (2k + 2)(2k + 4) 4(k + 1) 4(k + 1)(k + 2) 4(k + 1)(k + 2) 4(k + 1)(k + 2) (k + 1) 2 = = k +1 4(k + 1)(k + 2) 4(k + 2) By the Principle of Mathematical Induction, the statement is true for all positive integers. ak =
11.
1(1 + 1)(2 ⋅ 1 + 1)(3 ⋅ 12 + 3 ⋅ 1 − 1) 2 ⋅ 3 ⋅ 5 = =1 30 30
1.
Let n = 1. S1 = 14 = 1 =
2.
Assume that Sk = 1 + 16 + 81 + ⋅ ⋅ ⋅ + k 4 = Verify that Sk +1 = ak = k 4 ,
k (k + 1)(2k + 1)(3k 2 + 3k − 1) for some positive integer k (Induction Hypothesis). 30
( k + 1)(k + 2)(2k + 3)(3k 2 + 9k + 5) . 30
ak +1 = (k + 14 )
k (k + 1)(2k + 1)(3k 2 + 3k − 1) + (k + 1) 4 30 (k + 1)[k (2k + 1)(3k 2 + 3k − 1) + 30( k + 1)3 ] = 30 (k + 1)[6k 4 + 39k 3 + 91k 2 + 89k + 30] (k + 1)(k + 2)(6k 3 + 27 k 2 + 37 k + 15) = = 30 30 2 (k + 1)(k + 2)(2k + 3)(3k + 9k + 5) = 30 By the Principle of Mathematical Induction, the statement is true for all positive integers. S k +1 =
Copyright © Houghton Mifflin Company. All rights reserved.
784
12.
Chapter 11: Sequences, Series, and Probability
1. 2.
1 ⎞ 1 1 ⎛ Let n = 1. P1 = ⎜1 − ⎟= = 1 1 2 1 + 1 + ⎠ ⎝ 1⎞ ⎛ Assume Pk = ⎜1 − ⎟ 2⎠ ⎝ 1 Verify Pk +1 = . k+2
⎛ 1⎞ ⎜1 − ⎟ 3⎠ ⎝
1⎞ ⎛ 1 ⎞ 1 ⎛ is true for some positive integer k (Induction Hypothesis). ⎟= ⎜1 − ⎟ ⋅ ⋅ ⋅ ⎜ 1 − +1 4 1 + k k ⎠ ⎠ ⎝ ⎝
Pk +1 = ⎛⎜1 − 1 ⎞⎟ ⎛⎜ 1 − 1 ⎞⎟ ⎛⎜ 1 − 1 ⎞⎟ ⋅ ⋅ ⋅ ⎛⎜ 1 − 1 ⎞⎟ ⎛⎜1 − 1 ⎞⎟ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 4 ⎠ ⎝ k +1⎠ ⎝ k + 2 ⎠ = Pk ⎛⎜1 − 1 ⎞⎟ ⎝ k +2⎠ ⎛1 − 1 ⎞ 1 = ⎜ ⎟ k +1 ⎝ k + 2 ⎠ k +1 1 = 1 − = k + 2 −1 = = 1 k + 1 (k + 1)(k + 1) (k + 1)(k + 2) ( k + 1)( k + 2) k + 2 By the Principle of mathematical Induction, the statement is true for all positive integers.
13.
4
1.
1 81 ⎛3⎞ Let n = 4. Then ⎜ ⎟ = = 5 ;4 +1= 5 16 16 ⎝2⎠ n
⎛3⎞ Thus, ⎜ ⎟ > n + 1 for n = 4. ⎝2⎠ k
2.
⎛3⎞ Assume ⎜ ⎟ > k + 1 is true for some positive integer k ≥ 4 (Induction Hypothesis). ⎝2⎠ ⎛3⎞ Verify that ⎜ ⎟ ⎝2⎠ ⎛3⎞ ⎜ ⎟ ⎝2⎠
k +1
> k + 2.
k
1 1 ⎛ 3⎞ ⎛3⎞ ⎛ 3⎞ 1 = ⎜ ⎟ ⎜ ⎟ > ( k + 1) ⎜ ⎟ = ( 3k + 3) = ( 2k + k + 3) > ( 2k + 1 + 3) = k + 2 2 2 ⎝ 2⎠ ⎝2⎠ ⎝ 2⎠ 2
⎛3⎞ Thus ⎜ ⎟ ⎝2⎠ 14.
k +1
k +1
n
⎛3⎞ > k + 2. By the Principle of Mathematical Induction, ⎜ ⎟ > n + 1 for all n ≥ 4. ⎝2⎠ 7
1.
16384 1075 ⎛4⎞ Let n = 7. ⎜ ⎟ = =7 >7 3 2187 2187 ⎝ ⎠ n
⎛4⎞ Thus, ⎜ ⎟ > n for n = 7. ⎝3⎠ k
2.
⎛4⎞ Assume ⎜ ⎟ > k is true for some positive integer k ≥ 7 (Induction Hypothesis). ⎝3⎠ ⎛4⎞ Verify that ⎜ ⎟ ⎝3⎠ ⎛4⎞ ⎜ ⎟ ⎝3⎠
k +1
k +1
> k + 1.
k
1 1 ⎛4⎞ ⎛ 4⎞ ⎛ 4⎞ 1 = ⎜ ⎟ ⎜ ⎟ > k ⎜ ⎟ = ( 4k ) = ( 3k + k ) > ( 3k + 3) = k + 1 3 3 3 3 3 3 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛4⎞ Thus ⎜ ⎟ ⎝3⎠
k +1
n
⎛ 4⎞ > k + 1. By the Principle of Mathematical Induction, ⎜ ⎟ > n for all n ≥ 7. ⎝ 3⎠
Copyright © Houghton Mifflin Company. All rights reserved.
Section 11.4
15.
1.
785
Let n = 1. 0 < a <1 0 < a ⋅ a < a ⋅1 a1+1 = a 2 < 1 Thus, if 0 < a < 1, then a1+1 < a n for n = 1.
2.
Assume a k +1 < a k is true for some positive integer k, if 0 < a < 1 (Induction Hypothesis). Verify a k + 2 < a k +1. 0 < a <1 0 < a ⋅ a k +1 < 1 ⋅ a k +1 a k + 2 < a k +1 By the Principle of Mathematical Induction, if 0 < a < 1, then a n +1 < a n for all positive integers.
16.
1.
Let n = 1. a >1 a ⋅ a1 > 1 ⋅ a1 a1+1 > a1 Thus, if a > 1, then a n +1 > a n for n = 1.
2.
Assume that if a > 1, then a k +1 > a k for some positive integer k (Induction Hypothesis). Verify a k + 2 > a k +1. a >1 a ⋅ a k +1 > 1 ⋅ a k +1 a k + 2 > a k +1 By the Principle of Mathematical Induction, if a > 1, then a n + 2 > a n +1 for all positive integers.
17.
1.
Let n = 4. 1 ⋅ 2 ⋅ 3 ⋅ 4 = 24, 24 = 16 Thus, 1 ⋅ 2 ⋅ 3 ⋅ 4 > 2n for n = 4.
2.
Assume 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ ⋅ ⋅ ⋅ ⋅ k > 2k is true for some positive integer k ≥ 4 (Induction Hypothesis). Verify 1 ⋅ 2 ⋅ 3 ⋅ ⋅ ⋅ ⋅ ⋅ k ⋅ (k + 1) > 2k +1. 1 ⋅ 2 ⋅ 3 ⋅ ⋅ ⋅ ⋅ ⋅ k ⋅ (k + 1) > 2k ( k + 1) > 2k ⋅ 2 = 2k +1
Thus, 1 ⋅ 2 ⋅ 3 ⋅ ⋅ ⋅ ⋅ ⋅ k ⋅ (k + 1) > 2k +1. By the Principle of Mathematical Induction, 1 ⋅ 2 ⋅ 3 ⋅ ⋅ ⋅ ⋅ ⋅ n > 2n for all n ≥ 4. 18.
1.
1
Let n = 1. Thus
1 1
+ 1
Assume
1 1
=1= 1
2
+ 1
+
2
1
1
2
+
1 3
+ ⋅⋅⋅ + 1
+
3
1
1 n
+ ⋅⋅⋅ +
3
+ ⋅⋅⋅ +
1
≥ n for n = 1. 1 k ≥
≥ k is true for some positive integer k (Induction Hypothesis). 1
> k + 1. n k +1 1 1 1 1 1 1 + + + ⋅⋅⋅ + + ≥ k+ 1 2 3 k k +1 k +1
Verify
+
1 1
k k +1 1 k k 1 + > + k +1 k +1 k +1 k +1 k +1 = = k +1 k +1 1 + 1 + 1 + ⋅ ⋅ ⋅ + 1 + 1 > k + 1. By the Principle of Mathematical Induction, 1 + 1 + 1 + ⋅ ⋅ ⋅ + 1 ≥ n for Thus, 1 2 3 k k +1 1 2 3 n all positive integers n. =
Copyright © Houghton Mifflin Company. All rights reserved.
786
19.
Chapter 11: Sequences, Series, and Probability
1.
Let n = 1 and a > 0.
(1 + a )1 = 1 + a = 1 + 1 ⋅ a
Thus (1 + a )n ≥ 1 + na for n = 1. 2.
Assume (1 + a) k ≥ 1 + ka is true for some positive integer k (Induction Hypotheses). Verify (1 + a )k +1 > 1 + (k + 1)a. (1 + a )k +1 = (1 + a )k (1 + a) ≥ (1 + ka)(1 + a) = 1 + (k + 1) a + ka 2 > 1 + ( k + 1) a
Thus (1 + a )k +1 > 1 + (k + 1)a. By the Principle of Mathematical Induction, (1 + a) n > 1 + na for all positive integers n. 20.
1.
Let n = 1. log10 1 = 0 < 1
2.
Assume log10 k < k is true for some positive integer k (Induction Hypothesis).
Thus log10 n < n for n = 1. Verify log10 (k + 1) < k + 1.
log10 (k + 1) ≤ log10 (k + k ) = log10 2k = log10 2 + log10 k < 1 + k because log10 2 < 1 and log10 k < k .
Thus log10 (k + 1) < k + 1. By the Principle of Mathematical Induction, log10 n < n for all positive integers n. 21.
1.
Let n = 1. 12 + 1 = 2,
2 = 2 ⋅1
Thus 2 is a factor of n 2 + n for n = 1. 2.
Assume 2 is a factor of k 2 + k for some positive integer k (Induction Hypothesis). Verify 2 is a factor of (k + 1)2 + k + 1.
(k + 1)2 + k + 1 = (k + 1)(k + 1 + 1) = (k + 1)(k + 2) Since k 2 + k = k (k + 1) , 2 is a factor of k or k + 1. If 2 is a factor of k + 1, then 2 is a factor of (k + 1)(k + 2). If 2 is a factor of k, then 2 is a factor of k + 2. Thus, 2 is a factor of (k + 1)2 + k + 1. By the Principle of Mathematical Induction, 2 is a factor of n 2 + n for all positive integers. 22.
1.
Let n = 1. 13 − 1 = 0,
0 = 0⋅3
Thus, 3 is a factor of n 3 − n for n = 1. 2.
Assume 3 is a factor of k 3 − k for some positive integer k (Induction Hypothesis). Verify 3 is a factor of (k + 1)3 − (k + 1) when n = k + 1. (k + 1)3 − ( k + 1) = (k + 1)[(k + 1)2 − 1] = (k + 1)(k 2 + 2k ) = (k + 1)(k + 2) k = k (k + 1)(k + 2) Since k 3 − k = k (k + 1)( k − 1) , then 3 is a factor of k, k + 1, or k − 1 . If 3 is a factor of k, then 3 is a factor of k (k + 1)(k + 2). If 3 is a factor of k + 1, then 3 is a factor of k (k + 1)(k + 2). If 3 is a factor of k − 1, then 3 is a factor of k + 2. Since k + 2 = k − 1 + 3, the sum of two multiples of 3 is a multiple of 3, so 3 is a factor of k (k + 1)(k + 2).
Thus, 3 is a factor of (k + 1)3 − (k + 1). By the Principle of Mathematical Induction, 3 is a factor of n 3 − n for all positive integers. 23.
1.
Let n = 1. 51 − 1 = 4,
4 = 4 ⋅1
Thus, 4 is a factor of 5 n − 1 for n = 1. 2.
Assume 4 is a factor of 5 k − 1 for some positive integer k (Induction Hypothesis). Verify 4 is a factor of 5 k +1 − 1. Now 5k +1 − 1 = 5 ⋅ 5k − 5 + 4 = 5 (5k − 1) + 4 which is the sum of two multiples of 4.
Thus, 4 is a factor of 5 k +1 − 1. By the Principle of Mathematical Induction, 4 is a factor of 5 n − 1 for all positive integers. Copyright © Houghton Mifflin Company. All rights reserved.
Section 11.4
24.
1.
787
Let n = 1. 61 − 1 = 5,
5 = 5 ⋅1 n
Thus, 5 is a factor of 6 − 1 for n = 1. 2.
Assume 5 is a factor of 6 k − 1 for some positive integer k (Induction Hypothesis). Verify 5 is a factor of 6 k +1 − 1. Now 6k +1 − 1 = 6 ⋅ 6k − 6 + 5 = 6(6k − 1) + 5 which is the sum of two multiples of 5. Thus, 5 is a factor of 6 k +1 − 1. By the Principle of Mathematical Induction, 5 is a factor of 6 n − 1 for all positive integers.
25.
1.
Let n = 1. ( xy )1 = xy = x1 y1 Thus, ( xy )n = x n y n for n = 1.
2.
Assume ( xy )k = x k y k is true for some positive integer k (Induction Hypothesis). Verify ( xy )k +1 = x k +1 y k +1. ( xy )k +1 = ( xy )k ( xy )1 = x k y k ⋅ xy = x k +1 y k +1
Thus ( xy )k +1 = x k +1 y k +1. By the Principle of Mathematical Induction, ( xy )n = x n y n for all positive integers. 1
26.
1.
1 ⎛ ⎞ Let n = 1. ⎜ x ⎟ = x = x1 . y y ⎝ y⎠ n
n ⎛ ⎞ Thus, ⎜ x ⎟ = x n for n = 1. y y ⎝ ⎠ k
2.
k ⎛ ⎞ Assume ⎜ x ⎟ = x k is true for some positive integer k (Induction Hypothesis). y ⎝ y⎠
Verify ⎛⎜ x ⎞⎟ ⎝ y⎠ ⎛x⎞ ⎜ y⎟ ⎝ ⎠
k +1
1.
k +1 = x k +1 . y
k
1
k k +1 = ⎛⎜ x ⎞⎟ ⎛⎜ x ⎞⎟ = x k ⋅ x = x k +1 y y y ⎝ y⎠ ⎝ y⎠
Thus, ⎛⎜ x ⎞⎟ ⎝ y⎠ 27.
k +1
k +1
n
k +1 n = x k +1 . By the Principle of Mathematical Induction, ⎛⎜ x ⎞⎟ = x n for all positive integers. y y ⎝ y⎠
Let n = 1. a1 − b1 = a − b Thus a − b is a factor of a n − b n for n = 1.
2.
Assume a − b is a factor of a k − b k for some positive integer k (Induction Hypothesis). Verify a − b is a factor of a k +1 − b k +1. a k +1 − b k +1 = (a ⋅ a k − ab k ) + (ab k − b ⋅ b k ) = a(a k − b k ) + b k (a − b) The sum of two multiples of a − b is a multiple of a − b. Thus, a − b is a factor of a k +1 − b k +1 . By the Principle of Mathematical Induction, a − b is a factor of a n − b n for all positive integers.
28.
1.
Let n = 1. a 2⋅1+1 + b 2⋅1+1 = a3 + b3 = (a + b)(a 2 − ab + b 2 ) Thus, a + b is a factor of a 2n+1 + b 2n+1 for n = 1.
2.
Assume a + b is a factor of a 2k +1 + b 2k +1 for some positive integer k (Induction Hypothesis). Verify a + b is a factor of a 2k +3 + b 2k +3 . a 2k + 3 + b 2k + 3 = ( a 2k + 2 + b 2k + 2 )(a + b) − ab(a 2k +1 + b 2k +1 ) The sum of two multiples of a + b is a multiple of a + b. Thus, a + b is a factor of a 2 k + 3 + b 2 k + 3. By the Principle of Mathematical Induction, a + b is a factor of a 2 n +1 + b 2 n +1 for all positive integers.
Copyright © Houghton Mifflin Company. All rights reserved.
788
29.
Chapter 11: Sequences, Series, and Probability
1.
a(1 − r1) 1− r Thus, the statement is true for n = 1. Let n = 1. ar1−1 = a ⋅ 1 = a = j
2.
Assume
−
= j +1
Verify
∑ k 1
ar k −1 =
=
j +1
∑
j
a (1 − r ) is true for some positive integer j. ar k 1 = ∑ 1− r k 1 a(1 − r j +1) is true for n = j + 1. 1− r
j
ar k −1 =
k =1
∑ (ar k =1
k +1
− 1) =
a (1 − r j ) + ar j 1− r
a (1 − r j ) + ar j (1 − r ) a[1 − r j + r j − r j +1 ] = = 1− r 1− r j +1 a (1 − r ) = 1− r n
By the Principle of Mathematical Induction,
ar k ∑ k
−1
=
=1
a (1 − r n ) 1− r
1
30.
1[(1 + 1)a + 2b] = a+b ( ak + b ) = a + b and ∑ 2 k =1
1.
Let n = 1.
2.
Therefore, the statement is true for n = 1. Assume the statement is true for n = i. i
i[(i + 1)a + 2b] ( ak + b ) and ∑ 2 k =1 i +1
Prove the statement is true for n = i +1. That is, prove
(i + 1)[(i + 2)a + 2b] ( ak + b ) = ∑ 2 k =1
i +1
i
( ak + b ) = ∑ ( ak + b ) + a ( i + 1) + b ∑ n k =1
=1
i[(i + 1)a + 2b] i[(i + 1)a + 2b] + [2a(i + 1) + 2b] + a (i + 1) + b = 2 2 i (i + 1)a + 2bi + 2a (i + 1) + 2b i (i + 1)a + 2b(i + 1) + 2a (i + 1) = = 2 2 (i + 1)[ai + 2a + 2b] (i + 1)[a (i + 2) + 2b] = = 2 2 Therefore, the statement is true for all positive integers n. =
....................................................... 31.
1. 2.
If N = 25, then log25! ≈ 25.19 > 25. Assume logk! > k for k > 25 (Induction Hypothesis). Prove log(k + 1)! > k + 1. log(k + 1)! = log[(k + 1)k!] = log(k + 1) + logk! > log(k + 1) + k Because k > 25, log(k + 1) > 1. Thus, log(k + 1)! > k + 1. Therefore, log n! > n for all n > 25.
Copyright © Houghton Mifflin Company. All rights reserved.
Connecting Concepts
Section 11.4
32.
1.
789
Let
a n+1 < r for n ≥ N . We are to prove that a N + k < a N r k for each positive integer k . an
a When k = 1, we have N +1 < r. Thus a N +1 < a N r which is true by the condition of the sequence. aN Thus, the statement is true for k = 1. 2.
Assume that for some integer k, a N + k < a N r k (Induction Hypothesis). Prove a N + k +1 < a N r k +1. a N + k +1 < r by the Induction Hypothesis, we have a N =k
Since
a N + k +1 < ra N + k < r (a N r k ) = a N r k +1 Thus, the statement is true for all positive integers n. 33.
1.
When n = 1, we have ( x m )1 = x m and x m ⋅ 1 = x m . Therefore, the statement is true for n = 1.
2.
Assume the statement is true for n = k. That is, assume ( x m ) k = x mk (Induction Hypothesis). Prove the statement is true for n = k + 1. x m( k +1) = x mk + m = x mk ⋅ x m = ( x m ) k ⋅ x m = ( x m )k +1 Thus, the statement is true for all positive integers n and m.
34.
1.
When n = 1, 1
1 1 1 = + =1+1= 2 ∑ i ! 0 ! 1! i =0
1 3 − = 3 −1= 2 1 Thus, the statement is true for n = 1. k
2.
Assume the statement is true for n = k. That is, assume
1 1 ≤ 3 − (Induction Hypothesis). ∑ i! k i =0
Now prove the statement is true for n = k + 1. k +1
∑ i =0
1 = i!
Because
k
1 1 1 1 + ≤ 3− + ∑ i ! (k + 1)! k (k + 1)! i =0
1 1 1 1 1 1 1 ≤ ,3 − + ≤ 3− + = 3− . (k + 1)! k (k + 1) k (k + 1)! k k (k + 1) (k + 1)
k +1
Thus,
1 1 ≤3− . ∑ i! k +1 i =0
The statement is true for all positive integers n.
Copyright © Houghton Mifflin Company. All rights reserved.
790
35.
Chapter 11: Sequences, Series, and Probability 3
1.
3
64 ⎛ 3 +1⎞ ⎛ 4 ⎞ When n = 3, we have ⎜ < 3. ⎟ =⎜ ⎟ = 27 ⎝ 3 ⎠ ⎝3⎠ Thus the statement is true for n = 3. k
2.
⎛ k +1⎞ Assume the statement is true for n = k. That is, assume ⎜ ⎟ < k (Induction Hypothesis). ⎝ k ⎠ ⎛k +2⎞ Prove the statement is true for n = k + 1. That is, prove ⎜ ⎟ ⎝ k +1 ⎠
k +1
< k + 1.
⎛ k + 2 ⎞ k +1 . Therefore We begin by noting that ⎜ ⎟< k ⎝ k +1 ⎠ ⎛k +2⎞ ⎜ ⎟ ⎝ k +1 ⎠
k +1
⎛ k +1⎞ <⎜ ⎟ ⎝ k ⎠
k +1
k
⎛ k +1⎞ ⎛ k +1⎞ =⎜ ⎟ ⎜ ⎟ ⎝ k ⎠ ⎝ k ⎠ k
⎛ k +1⎞ By the Induction Hypothesis, ⎜ ⎟ < k ; thus ⎝ k ⎠ k
⎛ k +1⎞ ⎛ k +1⎞ ⎛ k +1⎞ ⎜ ⎟ ⎜ ⎟ < k⎜ ⎟ = k +1 k k ⎝ ⎠ ⎝ ⎠ ⎝ k ⎠
⎛k +2⎞ We now have ⎜ ⎟ ⎝ k +1 ⎠
k +1
< k + 1. The induction is complete.
n
⎛ n +1⎞ Thus ⎜ ⎟ < n is true for all n ≥ 3. ⎝ n ⎠
....................................................... PS2. 5! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120
PS1. ( a + b)3 = ( a + b)( a + b)( a + b) 3
2
2
= a + 3a b + 3ab + b PS4.
Prepare for Section 11.5 PS3. 0! = 1
3
6! = 720 = 15 2!(6 − 2)! 2(24)
PS5.
7! = 5040 = 35 3!(7 − 3)! 6(24)
PS6.
3,628,800 10! = =1 10!(10 − 10)! 3,628,000(1)
Section 11.5 1.
⎛7⎞ 7! 7 ⋅ 6 ⋅ 5 ⋅ 4! ⎜⎜ ⎟⎟ = = = 35 4 ⎝ ⎠ 4!(7 − 4)! 4!(3 ⋅ 2 ⋅ 1)
2.
⎛8 ⎞ 8! 8 ⋅ 7 ⋅ 6! ⎜⎜ ⎟⎟ = = = 28 6 ⎝ ⎠ 6!(8 − 6)! 6!⋅2 ⋅ 1
3.
⎛9⎞ 9! 9 ⋅ 8 ⋅ 7! ⎜⎜ ⎟⎟ = = = 36 ⎝ 2 ⎠ 2!(9 − 2)! 2 ⋅ 1 ⋅ 7!
4.
⎛10 ⎞ 10! 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5! ⎜⎜ ⎟⎟ = = = 252 5!⋅5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ⎝ 5 ⎠ 5!(10 − 5)!
5.
⎛12 ⎞ 12! 12 ⋅ 11 ⋅ 10 ⋅ 9! ⎜⎜ ⎟⎟ = = = 220 9 − 9 ! ( 12 9 )! 9!⋅3 ⋅ 2 ⋅ 1 ⎝ ⎠
6.
⎛ 6⎞ 6! 6 ⋅ 5! ⎜⎜ ⎟⎟ = = =6 5 − 5 ! ( 6 5 )! 5!⋅1 ⎝ ⎠
7.
⎛11⎞ 11! 11! ⎜⎜ ⎟⎟ = = =1 0 − ⋅ 11! 0 ! ( 11 0 )! 1 ⎝ ⎠
8.
⎛14 ⎞ 14! 14! ⎜⎜ ⎟⎟ = = =1 14 − 14 ! ( 14 14 )! 14 !⋅1 ⎝ ⎠
9.
⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞ ( x − y ) 6 = ⎜⎜ ⎟⎟ x 6 + ⎜⎜ ⎟⎟ x 5 ( − y ) + ⎜⎜ ⎟⎟ x 4 ( − y ) 2 + ⎜⎜ ⎟⎟ x 3 ( − y ) 3 + ⎜⎜ ⎟⎟ x 2 ( − y ) 4 + ⎜⎜ ⎟⎟ x ( − y ) 5 + ⎜⎜ ⎟⎟ ( − y ) 6 ⎝0⎠ ⎝6⎠ ⎝5⎠ ⎝4⎠ ⎝3⎠ ⎝2⎠ ⎝1 ⎠
= x6 − 6 x5 y + 15 x 4 y 2 − 20 x3 y 3 + 15 x 2 y 4 − 6 xy5 + y 6
Copyright © Houghton Mifflin Company. All rights reserved.
Section 11.5
10.
11.
12.
791
⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ (a − b )5 = ⎜ ⎟ a 5 + ⎜ ⎟ a 4 (− b )+ ⎜ ⎟ a 3 (− b )2 + ⎜ ⎟ a 2 (− b )3 + ⎜ ⎟ a (− b )4 + ⎜ ⎟(− b )5 ⎝0⎠ ⎝1 ⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ = a 5 − 5 a 4 b +1 0 a 3b 2 −1 0 a 2 b 3 + 5 a b 4 − b 5
⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ( x + 3 ) 5 = ⎜ ⎟ x 5 + ⎜ ⎟ x 4 ⋅3 + ⎜ ⎟ x 3 ⋅3 2 + ⎜ ⎟ x 2 ⋅3 3 + ⎜ ⎟ x ⋅ 3 4 + ⎜ ⎟ 3 5 ⎝0⎠ ⎝1 ⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ = x 5 +1 5 x 4 + 9 0 x 3 + 2 7 0 x 2 + 4 0 5 x + 2 4 3
⎛ 4⎞ ⎛ 4⎞ ⎛ 4⎞ ⎛ 4⎞ ⎛ 4⎞ ( x − 5) 4 = ⎜ ⎟ x 4 + ⎜ ⎟ x3 (−5) + ⎜ ⎟ x 2 (−5) 2 + ⎜ ⎟ x(−5)3 + ⎜ ⎟ (−5) 4 0 1 2 3 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 4⎠ 4 3 2 = x − 20 x + 150 x − 500 x + 625
13.
( 2 x − 1)7 = ⎛⎜ 70 ⎞⎟ ( 2 x )7 + ⎛⎜17 ⎞⎟ ( 2 x )6 ( −1) + ⎛⎜ 72 ⎞⎟ ( 2 x )5 ( −1)2 + ( −1)2 + ⎛⎜ 37 ⎞⎟ ( 2 x )4 ( −1)3 + ⎛⎜ 74 ⎞⎟ ( 2 x )3 ( −1)4 + ⎛⎜ 57 ⎞⎟ ( 2 x )2 ( −1)5
14.
( 2 x + y )6 = ⎛⎜ 60 ⎞⎟ ( 2 x )6 + ⎛⎜16 ⎞⎟ ( 2 x )5 y + ⎛⎜ 62 ⎞⎟ ( 2 x )4 y 2 + ⎛⎜ 36 ⎞⎟ ( 2 x )3 y3 + ⎛⎜ 64 ⎞⎟ ( 2 x )2 y 4
15.
( x + 3 y )6 = ⎛⎜ 60 ⎞⎟ x6 + ⎛⎜16 ⎞⎟ x5 ( 3 y ) + ⎛⎜ 62 ⎞⎟ x 4 ( 3 y )2 + ⎛⎜ 36 ⎞⎟ x3 ( 3 y )3 + ⎛⎜ 64 ⎞⎟ x 2 ( 3 y )4 + ⎛⎜ 56 ⎞⎟ x ( 3 y )5 + ⎛⎜ 66 ⎞⎟ ( 3 y )6
16.
( x − 4 y )5 = ⎛⎜ 50 ⎞⎟ x5 + ⎛⎜15 ⎞⎟ x 4 ( −4 y ) + ⎛⎜ 52 ⎞⎟ x3 ( −4 y )2 + ⎛⎜ 53 ⎞⎟ x 2 ( −4 y )3 + ⎛⎜ 54 ⎞⎟ x ( −4 y )4 + ⎛⎜ 55 ⎞⎟ ( −4 y )5
17.
( 2 x − 5 y )4 = ⎛⎜ 04 ⎞⎟ ( 2 x )4 + ⎛⎜14 ⎞⎟ ( 2 x )3 ( −5 y ) + ⎛⎜ 42 ⎞⎟ ( 2 x )2 ( −5 y )2 + ⎛⎜ 34 ⎞⎟ ( 2 x )( −5 y )3 + ⎛⎜ 44 ⎞⎟ ( −5 y )4
18.
( 3x + 2 y )4 = ⎛⎜ 04 ⎞⎟ ( 3x )4 + ⎛⎜14 ⎞⎟ ( 3x )2 ( 2 y ) + ⎛⎜ 42 ⎞⎟ ( 3x )2 ( 2 y )2 + ⎛⎜ 34 ⎞⎟ ( 2 x )( 2 y )3 + ⎛⎜ 44 ⎞⎟ ( 2 y )4
19.
20.
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 6 ⎛7⎞ 7 ⎛7⎞ + ⎜ ⎟ ( 2 x )( −1) + ⎜ ⎟ ( −1) ⎝6⎠ ⎝7⎠ = 128 x7 − 448 x6 + 672 x5 − 560 x 4 + 280 x3 − 84 x 2 + 14 x − 1
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = 64 x6 + 192 x5 y + 240 x 4 y 2 + 160 x3 y 3 + 60 x 2 y 4 + 12 xy 5 + y 6
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = x5 − 20 x 4 y + 160 x3 y 2 − 640 x 2 y 3 + 1280 xy 4 − 1024 y 5
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = 16 x 4 − 160 x3 y + 600 x 2 y 2 − 1000 xy 3 + 625 y 4
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = 81x 4 + 216 x3 y + 216 x 2 y 2 + 96 xy 3 + 16 y 4
6
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠
2
⎝ ⎠
⎛6⎞ ⎛ 6⎞ + ⎜ ⎟ ( 2 x ) y5 + ⎜ ⎟ y 6 5 ⎝ ⎠ ⎝ 6⎠
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = x6 + 18 x5 y + 135 x 4 y 2 + 540 x3 y 3 + 1215 x 2 y 4 + 1458 xy 5 + 729 y 6
⎝ ⎠
⎛ x + 1 ⎞ = ⎛ 6 ⎞ x 6 + ⎛ 6 ⎞ x5 ⎛ 1 ⎞ + ⎛ 6 ⎞ x 4 ⎛ 1 ⎞ + ⎛ 6 ⎞ x3 ⎛ 1 ⎞ + ⎛ 6 ⎞ x 2 ⎛ 1 ⎞ + ⎛ 6 ⎞ x ⎛ 1 ⎞ + ⎛ 6 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜1 ⎟ ⎜ ⎟ ⎜ 2⎟ ⎜ ⎟ ⎜3⎟ ⎜ ⎟ ⎜ 4⎟ ⎜ ⎟ ⎜ 5⎟ ⎜ ⎟ ⎜ 6⎟⎜ ⎟ x ⎠ ⎝0⎠ ⎝ ⎝ ⎠ ⎝x⎠ ⎝ ⎠ ⎝ x⎠ ⎝ ⎠ ⎝ x⎠ ⎝ ⎠ ⎝ x⎠ ⎝ ⎠ ⎝ x ⎠ ⎝ ⎠⎝ x ⎠ 6 4 2 15 6 1 = x + 6 x + 15 x + 20 + 2 + 4 + 6 x x x
6
(2x − y )
2 ⎛7⎞ + ⎜ ⎟(2 x) − y ⎝5⎠
7
(
)
3
(
)
4
(
5
)
(
2 3 7 6 5 4 3 ⎛7⎞ ⎛7⎞ ⎛7⎞ ⎛ 7⎞ ⎛7⎞ = ⎜ ⎟( 2 x ) + ⎜ ⎟( 2 x ) − y + ⎜ ⎟( 2 x ) − y + ⎜ ⎟( 2 x ) − y + ⎜ ⎟( 2 x ) − y ⎝0⎠ ⎝1 ⎠ ⎝ 2⎠ ⎝3⎠ ⎝ 4⎠ 6 7 ⎛7⎞ ⎛7⎞ + ⎜ ⎟( 2 x) − y + ⎜ ⎟ − y ⎝6⎠ ⎝7⎠ = 128 x 7 − 448 x 6 y + 672 x5 y − 560 x 4 y y + 280 x3 y 2 − 84 x 2 y 2 y + 14 xy 3 − y 3 y
(
21.
⎝ ⎠
)
(
)
)
( x2 −4) =⎛⎜⎝ 70 ⎞⎟⎠( x2 ) + ⎛⎜⎝17 ⎞⎟⎠(2 x)6 (−4) + ⎛⎜⎝ 72⎞⎟⎠( x2 ) (−4)2 + ⎛⎜⎝ 37⎞⎟⎠( x2 ) (−4)3 + ⎛⎜⎝ 74⎞⎟⎠( x2 ) (−4)4 + ⎛⎜⎝ 57 ⎞⎟⎠( x2 ) (−4)5 6 ⎛7⎞ 7 ⎛7⎞ + ⎜ ⎟( x 2 )( −4) + ⎜ ⎟(−4) 6 7 ⎝ ⎠ ⎝ ⎠ 7
7
5
4
3
= x14 − 28 x12 +336 x10 − 2240 x8 +8960 x 6 − 21504 x 4 + 28672 x 2 −16384
Copyright © Houghton Mifflin Company. All rights reserved.
2
4
(
)
5
792
22.
Chapter 11: Sequences, Series, and Probability
( x − y3 )
6
( ) + ⎛⎜⎝ 62 ⎞⎟⎠ x4 ( − y3 )
⎛ 6⎞ ⎛ 6⎞ = ⎜ ⎟ x 6 + ⎜ ⎟ x5 − y 3 0 ⎝ ⎠ ⎝1 ⎠
2
( )
( )
( )
( )
3 4 5 6 ⎛ 6⎞ ⎛ 6⎞ ⎛ 6⎞ ⎛ 6⎞ + ⎜ ⎟ x3 − y 3 + ⎜ ⎟ x 2 − y 3 + ⎜ ⎟ x − y 3 + ⎜ ⎟ − y 3 3 4 5 6 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= x6 − 6 x5 y3 + 15 x 4 y 6 − 20 x3 y9 + 15 x 2 y12 − 6 xy15 + y18
23.
( 2 x2 + y3 )
( )
5
( ) ( ) + ⎛⎜⎝ 52 ⎞⎟⎠ ( 2x2 ) ( y3 )
5 4 3 ⎛5⎞ ⎛ 5⎞ = ⎜ ⎟ 2 x2 + ⎜ ⎟ 2 x2 y 0 1 ⎝ ⎠ ⎝ ⎠
3
2
( )( )
( )( y3 )
2 3 3 ⎛ 5⎞ ⎛5⎞ + ⎜ ⎟ 2 x2 y + ⎜ ⎟ 2 x2 3 ⎝ ⎠ ⎝ 4⎠
4
( )
5 ⎛ 5⎞ + ⎜ ⎟ y3 5 ⎝ ⎠
= 32 x10 + 80 x8 y3 + 80 x6 y 6 + 40 x 4 y9 + 10 x 2 y12 + y15
24.
( 2 x − y3 )
6
2 ⎛ 6⎞ 3 ⎛6⎞ 4 ⎛ 6⎞ 5 ⎛6⎞ 6 ⎛6⎞ ⎛ 6⎞ ⎛6⎞ = ⎜ ⎟ (2 x)6 + ⎜ ⎟ (2 x)5 (− y3 ) + ⎜ ⎟ (2 x)4 (− y3 ) + ⎜ ⎟ (2 x)3 (− y3 ) + ⎜ ⎟ (2 x) 2 (− y 3 ) + ⎜ ⎟ (2 x)( − y 3 ) + ⎜ ⎟ (− y 3 ) 0 1 2 3 4 5 6 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= 64 x6 − 192 x5 y3 + 240 x 4 y 6 − 160 x3 y9 + 60 x 2 y12 − 12 xy15 + y18
25.
4 4 3 2 2 3 4 ⎛ 2 − x ⎞ = ⎛ 4 ⎞ ⎛ 2 ⎞ + ⎛ 4 ⎞ ⎛ 2 ⎞ ⎛ − x ⎞ + ⎛ 4 ⎞ ⎛ 2 ⎞ ⎛ − x ⎞ + ⎛ 4 ⎞ ⎛ 2 ⎞ ⎛ − x ⎞ + ⎛ 4 ⎞ ⎛ − x ⎞ = 16 − 16 + 6 − x 2 + x 4 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ 16 ⎝ x 2⎠ x4 x2 ⎝ 0 ⎠ ⎝ x ⎠ ⎝1 ⎠ ⎝ x ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ x ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ x ⎠ ⎝ 2 ⎠ ⎝ 4 ⎠ ⎝ 2 ⎠
26.
3 3 2 2 3 a3 3a 3b b3 ⎛ a b ⎞ ⎛ 3 ⎞ ⎛ a ⎞ ⎛ 3 ⎞ ⎛ a ⎞ ⎛ b ⎞ ⎛ 3 ⎞ ⎛ a ⎞⎛ b ⎞ ⎛ 3 ⎞ ⎛ b ⎞ + + ⎜ + ⎟ = ⎜ 0 ⎟ ⎜ ⎟ + ⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ + ⎜ 2 ⎟ ⎜ ⎟⎜ ⎟ + ⎜ 3 ⎟ ⎜ ⎟ = 3 + b a a3 ⎝ b a ⎠ ⎝ ⎠ ⎝ b ⎠ ⎝ ⎠ ⎝ b ⎠ ⎝ a ⎠ ⎝ ⎠ ⎝ b ⎠⎝ a ⎠ ⎝ ⎠ ⎝ a ⎠ b
27.
( s−2 + s2 )
6
( )
( )( )
( )( )
( )( )
( )( )
( )( )
( )
6 ⎛ 6⎞ 5 4 2 2 ⎛ 6 ⎞ −2 3 2 3 ⎛ 6 ⎞ −2 2 2 4 ⎛ 6 ⎞ −2 5 ⎛ 6⎞ 6 ⎛ 6⎞ ⎛ 6⎞ s s s s2 + ⎜ ⎟ s2 = ⎜ ⎟ s −2 + ⎜ ⎟ s − 2 s 2 + ⎜ ⎟ s − 2 +⎜ ⎟ s +⎜ ⎟ s +⎜ ⎟ s ⎝ 0⎠ ⎝1 ⎠ ⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠ ⎝5⎠ ⎝ 6⎠
= s −12 + 6s −8 + 15s −4 + 20 + 15s 4 + 6 s8 + s12 28.
( 2r −1 + s−1 )
5
( )
( ) ( ) + ⎛⎜⎝ 52 ⎞⎟⎠ ( 2r −1 ) ( s−1 )
5 4 ⎛5⎞ ⎛ 5⎞ = ⎜ ⎟ 2r −1 + ⎜ ⎟ 2r −1 s −1 ⎝ 0⎠ ⎝1 ⎠
3
2
( )( )
( )( )
= 32r −5 + 80r −4 s −1 + 80r −3s −2 + 40r −2 s −3 + 10r −1s −4 + s −5 29.
⎛10 ⎞ 3 7 eighth term is ⎜ ⎟ ( 3 x ) ( − y ) = −3240 x3 y 7 7 ⎝ ⎠
30.
⎛12 ⎞ 3 fourth term is ⎜ ⎟ x9 ( 2 y ) = 1760 x9 y3 3 ⎝ ⎠
31.
⎛12 ⎞ 2 third term is ⎜ ⎟ x10 ( 4 y ) = 1056 x10 y 2 ⎝2 ⎠
32.
⎛14 ⎞ 2 12 thirteenth term is ⎜ ⎟ ( 2 x ) ( −1) = 364 x 2 ⎝12 ⎠
33.
⎛9⎞ fifth term is ⎜ ⎟ ⎝ 4⎠
34.
1 5 1 5 ⎛10 ⎞ sixth term is ⎜ ⎟ x − 2 x 2 = 252 5 ⎝ ⎠
35.
⎛11⎞ ⎛ a ⎞ ⎛ b ⎞ 165b5 ninth term is ⎜ ⎟⎜ ⎟ ⎜ ⎟ = a5 ⎝8 ⎠⎝ b ⎠ ⎝ a ⎠
36.
⎛13 ⎞ ⎛ 3 ⎞ ⎛ x ⎞ 5148 seventh term is ⎜ ⎟⎜ ⎟ ⎜ − ⎟ = 6 x ⎝ ⎠⎝ x ⎠ ⎝ 3 ⎠
38.
⎛9⎞ 2 7 eighth term is ⎜ ⎟ ( 3r ) ( 2 s ) = 41, 472r 2 s 7 7 ⎝ ⎠
40.
( b3 )
37.
39.
( x ) (− y ) 5
3
4
= 126 x 2 y 2 x
8
⎛ n ⎞ n −( i −1) i −1 b , if bi-1 = b8 , then i = 9. ⎜ ⎟a 1 i − ⎝ ⎠ ⎛10 ⎞ 2 8 ninth term is ⎜ ⎟ ( 2a ) ( −b ) = 180a 2b8 8 ⎝ ⎠
( y2 )
i −1
= y8 , if 2i − 2 = 8, then 2i = 10 or i = 5.
( )
4 ⎛6⎞ 2 fifth term is ⎜ ⎟ ( 2 x ) y 2 = 60 x 2 y8 4 ⎝ ⎠
( )
2 3 4 5 ⎛ 5⎞ ⎛5⎞ ⎛ 5⎞ + ⎜ ⎟ 2r −1 s −1 + ⎜ ⎟ 2r −1 s −1 + ⎜ ⎟ s −1 ⎝ 3⎠ ⎝ 4⎠ ⎝ 5⎠
( )( ) 7
i −1
6
= b9 , if 3i − 3 = 9, then 3i = 12 or i = 4
( )
3 ⎛8⎞ fourth term is ⎜ ⎟ a5 −b3 = −56a5b9 3 ⎝ ⎠
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Section 11.5
41.
43.
793
⎛10 ⎞ 5 5 sixth term is ⎜ ⎟ ( 3a ) ( −b ) = −61, 236a5b5 ⎝5 ⎠
( )
4 ⎛8 ⎞ 4 fifth term is ⎜ ⎟ ( a ) b 2 = 70a 4b8 4 ⎝ ⎠
42.
( )
5 ⎛9⎞ 4 fifth term is ⎜ ⎟ s −1 ( s ) = 126s −1 ⎝ 4⎠ 4 ⎛9⎞ 5 sixth term is ⎜ ⎟ s −1 ( s ) = 126s 5 ⎝ ⎠
( )( )
3 1 3 ⎛7⎞ 1 4 fourth term is ⎜ ⎟ x 2 − y 2 = −35 x 2 y 2 ⎝3⎠ 3 1 4 ⎛7⎞ 1 3 fifth term is ⎜ ⎟ x 2 − y 2 = 35 x 2 y 2 4 ⎝ ⎠
44.
( )
( )( )
45.
( 2 − i )4 = ⎛⎜ 04 ⎞⎟ ( 24 ) + ⎛⎜14 ⎞⎟ ( 2 )3 ( −i )1 + ⎛⎜ 24 ⎞⎟ ( 2 )2 ( −i )2 + ⎛⎜ 34 ⎞⎟ 2 ( −i )3 + ⎛⎜ 44 ⎞⎟ ( −i )4
46.
( 3 + 2i )3 = ⎛⎜ 30 ⎞⎟ ( 3)3 + ⎛⎜13 ⎞⎟ ( 3)2 ( 2i )1 + ⎛⎜ 32 ⎞⎟ ( 3)( 2i )2 + ⎛⎜ 33 ⎞⎟ ( 2i )3
47.
(1 + 2i )5 = ⎛⎜ 50 ⎞⎟ (1)5 + ⎛⎜15 ⎞⎟ (1)4 ( 2i )1 + ⎛⎜ 52 ⎞⎟ (1)2 ( 2i )2 + ⎛⎜ 53 ⎞⎟ (1)2 ( 2i )3 + ⎛⎜ 54 ⎞⎟ (1)( 2i )4 + ⎛⎜ 55 ⎞⎟ ( 2i )5
48.
(1 − 3i )5 = ⎛⎜ 0 ⎞⎟ (1)5 + ⎛⎜1 ⎞⎟ (1)4 ( −3i )1 + ⎛⎜ 2 ⎞⎟ (1)3 ( −3i )2 + ⎛⎜ 3 ⎞⎟ (1)2 ( −3i )3 + ⎛⎜ 4 ⎞⎟1( −3i )4 + ⎛⎜ 5 ⎞⎟ ( −3i )5
49.
⎛ 2 8 2⎞ ⎛ 2⎞ 1 1 ⎜ 2 + i 2 ⎟ = ⎜ 2 ⎟ (1 + i ) + 16 (1 + 8i − 28 − 56i + 70 + 56i − 28 − 8i + 1) = 16 (16 ) = 1 ⎝ ⎠ ⎝ ⎠
50.
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 3 = 16 + 32 ( −i ) + 24 ( −1) + 8 ( −i ) + 1 = 16 − 32i − 24 + 8i + 1 = −7 − 24i
⎝ ⎠ ⎝ ⎠ = 27 + 54i − 36 − 8i = −9 + 46i
⎝ ⎠
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = 1 + 10i − 40 − 80i + 80 + 32i = 41 − 38i
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠
5 5 5 5 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = 1 − 15i − 90 + 270i + 405 − 243i = 316 + 12i 8
⎝ ⎠
5 ⎝ ⎠
5 ⎝ ⎠
8
6
6 ⎛1 3 ⎞ = ⎛ 1 ⎞ 1+ i 3 ⎜ +i ⎟ ⎜ ⎟ 2 ⎠ ⎝2⎠ ⎝2
(
)
6
(
)
= 1 1 + 6i 3 − 45 − 60i 3 + 135 + 54i 3 − 27 = 1 ( 64 ) = 1 64 64
....................................................... 51.
52.
Connecting Concepts
n ( n − 1) x n − 2 h 2 n ( n − 1)( n − 2 ) x n −3h3 + + ⋅ ⋅ ⋅ + hn − xn ( x + h) − x = 2 6 h h n−2 n n x h n n n − 2 ) x n − 3h 2 1 1 − − ( )( ) ( )( = nx x −1 + + + ⋅ ⋅ ⋅ + h n −1 2 6 n
n
x n + nx n −1h +
⎛n⎞ n! ⎜ ⎟= k k n − k )! ! ( ⎝ ⎠
⎛ n ⎞ n! n! = ⎜ ⎟= n k − n k n n k n k ) !k ! − − − − ! ! ( ) ( ) ( ( ) ⎝ ⎠
⎛n⎞ ⎛ n ⎞ Thus ⎜ ⎟ = ⎜ ⎟ ⎝k⎠ ⎝n − k⎠
53.
( x + y )n =
n
∑ k =0
⎛ n ⎞ n−k k n y . With x = 1 and y = 1, we have (1 + 1) = ⎜ ⎟x ⎝k⎠ n
2 =
n
∑ ⎛⎜⎝ nk ⎞⎟⎠(1)
n−k
(1)k
k =0 n
∑ ⎛⎜⎝ nk ⎞⎟⎠ k =0
Copyright © Houghton Mifflin Company. All rights reserved.
794
54.
56.
Chapter 11: Sequences, Series, and Probability
⎛n⎞ ⎛ n ⎞ n! n! + ⎜ ⎟+⎜ ⎟= ⎝ k ⎠ ⎝ k + 1⎠ k !(n − k )! (k + 1)!(n − (k + 1))! n!(k + 1) + n!(n − k ) = ( k + 1)!(n − k )! = n!k + n!+ n !n − n!k (k + 1)!(n − k )! n!(n + 1) = (k + 1)!(n − k )! (n + 1)! ⎛ n + 1⎞ = =⎜ ⎟ (k + 1)!(n + 1 − (k + 1))! ⎝ k + 1⎠
55.
Let x = 1 and y = −1 in the expansion of ( x + y )n . n
(x + y ) n =
∑⎛⎜⎝ in ⎞⎟⎠ x i =0
n −i i
y
n
(1 + (−1)) n = 0n = 0 =
∑ i =0
n
Thus
⎛ n ⎞ n −i i ⎜ i ⎟1 (−1) = ⎝ ⎠
n
∑ (−1) ⎛⎜⎝ in ⎞⎟⎠ i
i =0
(−1)i ⎜ ⎟ = 0 ∑ ⎝i ⎠ i ⎛n⎞
=0
(0.98)8 = (1 − 0.02)8
⎛8 ⎞ ⎛8⎞ ⎛8 ⎞ The sum of the first three terms is ⎜ ⎟18 + ⎜ ⎟17 (−0.02)1 + ⎜ ⎟16 (−0.02)2 = 18 + 8(−0.02) + 28(−0.02)2 ⎝0⎠ ⎝1 ⎠ ⎝ 2⎠ (0.98)8 ≈ 1 − 0.16 + 0.0112 = 0.8512 57.
8
8
(1.02) = (1 + 0.02) =
8
∑i ⎜⎝ i ⎟⎠ (1) ⎛8⎞
8−1
(0.02)i
58.
8−
10! = 2520 2!3!5!
The sum of the first three terms is ⎛8 ⎞ 0 ⎛8⎞ 1 ⎛8 ⎞ 2 ⎜ 0 ⎟(0.02) +⎜1 ⎟(0.02) +⎜ 2 ⎟(0.02) =1+ 0.16+ 0.0112 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ (1.02)8 ≈1.1712 59.
9! = 756 5!2!2!
60.
9! = 126 4!5!0!
61.
8! = 56 3!5!0!
.......................................................
Prepare for Section 11.6
PS1. 7! = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 5040
PS2. (7 − 3)! = 4! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24
PS4. ⎛ 8 ⎞ 8! = 8 ⋅ 7 ⋅ 6 ⋅ 5! = 56 ⎜ ⎟= ⎝ 5 ⎠ 5!(8 − 5)! 5!(3 ⋅ 2 ⋅ 1)
PS5.
PS3. ⎛ 7 ⎞ 7! = 7 ⋅ 6! = 7 ⎜ ⎟= ⎝ 1 ⎠ 1!(7 − 1)! 1(6!)
10! = 10 ⋅ 9 ⋅ 8! = 90 (10 − 2)! 8!
PS6.
6! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 720 (6 − 6)! 0!
Section 11.6 1.
P (6, 2) =
6! = 6 ⋅ 5 ⋅ 4! = 30 (6 − 2)! 4!
2.
P (8,7) =
8! = 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 40,320 (8 − 7)! 1
3.
C (8, 4) =
8! = 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4! = 70 4!(8 − 4)! 4!⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1
4.
C (9, 2) =
9! = 9 ⋅ 8 ⋅ 7! = 36 2!(9 − 2)! 2 ⋅ 1 ⋅ 7!
5.
P (8,0) =
8! = 8! = 1 (8 − 0)! 8!
6.
P (9,9) =
9! = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 362,880 (9 − 9)! 1
7.
C (7,7) =
7! = 7! = 1 7!(7 − 7)! 7! ⋅ 1
8.
C (6,0) =
6! = 6! = 1 0!(6 − 0)! 1 ⋅ 6!
9.
C (10, 4) =
10.
P (10, 4) =
10! = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6! = 210 4!(10 − 4)! 4 ⋅ 3 ⋅ 2 ⋅ 1 ⋅ 6!
10! = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6! = 5040 (10 − 4)! 6!
Copyright © Houghton Mifflin Company. All rights reserved.
Section 11.6
795
11.
Use the counting principle. 3 ⋅ 2 ⋅ 2 = 12 There are 12 different possible computer systems.
12.
Use the counting principle. 4 ⋅ 4 ⋅ 4 ⋅ 4 = 256 256 possible colors could be formed, assuming that each palate must be used each time.
13.
Use the counting principle. 2 ⋅ 2 ⋅ 2 ⋅ 2 = 16 There are 16 possible light switch configurations.
14.
Use the counting principle. 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2=27 = 128 There are 128 different positive integers that can be stored.
15.
P (6,6) = 6! = 720
18.
C (9,3) = 84
19.
Use the combination formula with n = 25, r = 5. C (25,5) = 25! = 53,130 5!(20)!
20.
Use the combination formula with n = 26, r = 2. 26! C (26, 2) = = 325 2!23!
21.
There are 676 ways to arrange 26 letters taken two at a time (26 ⋅ 26 = 676). Now if there are more than 676 employees, then at least two employees have the same first and last initials.
22.
C (4, 2) ⋅ P(3,3) ⋅ P(3,3) = 216
23.
C (6,3) ⋅ C (8,3) = 1120
24.
a. b.
25.
210 = 1024
26.
420
27.
C (40,6) = 3,838,380
28.
C (10,8) = 45
30.
a.
The number of ways 10 finalists can be selected from 15 semifinalists is the combination of 15 students selected 10 at a time. C (15,10) = 3003 There are 3003 ways the finalists can be chosen.
b.
The number of ways the 10 finalists can contain three seniors is the product of the combination of 7 seniors selected 3 at a time and the combination of 8 remaining students selected 7 at a time. C (7,3)C (8,7) = 35 ⋅ 8 = 280 There are 280 ways the finalists can contain 3 seniors.
c.
At least five seniors means 5 or 6 or 7 seniors are finalists (there are only 7 seniors). Since the events are related by “or,” sum the number of ways each event can occur. C (7,3) ⋅ C (8,5) + C (7,6) ⋅ C (8,3) = 21 ⋅ 56 + 7 ⋅ 70 + 56 = 1176 + 490 + 56 = 1722 There are 1722 ways the finalists can contain at least 5 seniors.
16.
P (12,3) = 1320
17.
29.
32.
31.
3 ⋅ 12 ⋅ 5 ⋅ 107 = 1.8 × 109
33.
C (10,3) − C (8,1) = 120 − 8 = 112
34.
a. b.
36.
7 ⋅ 6 = 42
37.
39.
16 ⋅ 14 = 112 2
40.
C (12,5) = 792 C (10,3) = 120
a. b. c.
a. b. c. d.
5 ⋅ 5 ⋅ 5 = 125 ways
3!⋅ 3! = 6 ⋅ 6 = 36 6 ⋅ 3 ⋅ 2 ⋅ 2 ⋅ 1 ⋅ 1 = 72
C (7,5) = 21 C (7, 4) ⋅ C (3,1) = 35 ⋅ 3 = 105 C (7, 2) ⋅ C (3,3) = 21 ⋅ 1 = 21
C (13,5) = 1287 4 ⋅ C (13,5) = 5148 C (48, 2) ⋅ C (4,3) = 4512 C (4,4)⋅C (48,1) +C (4,3)⋅C (48,2)+C (4,2)⋅C (48,3) =108,336 35.
P (5,5) = 120
C (7, 2) = 21
38.
C (12, 2) = 66
C (12,3) = 220
41.
C (20,10) = 184,756
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796
Chapter 11: Sequences, Series, and Probability
42.
C (15,8) ⋅ P(8,3) = 6, 435 ⋅ 336 = 2,162,160
44.
A total of 14 moves are required, 7 to the right and 7 down. 14! C (14, 7) = = 3432 7!7!
45.
A triple-decker cone could have all one flavor ice cream, or two different flavors with one scoop of the first flavor and two scoops of the second flavor (such as one scoop of vanilla and two scoops of chocolate), or two different flavors with two scoops of the first flavor and one scoop of the second flavor (such as two scoops of vanilla and one scoop of chocolate), or three different flavors. 31! 31! 31! 31! C (31, 1) + C (31, 2) + C (31, 2) + C (31, 3) = + + + 1!30! 2!29! 2!29! 3!28! = 31 + 465 + 465 + 4495 = 5456
46.
( 216 )1024⋅768 = 216⋅1024⋅768
43.
47.
C (20,12) ⋅ C (12, 4) = 125,970 ⋅ 495 = 62,355,150
19!
= 212,582,912
.......................................................
Connecting Concepts
48.
There are n ways to choose the first point and n – 1 ways to choose the second point. Thus there are n(n – 1) ways to choose both points. Since the direction of the line is not important, the number of lines is ⎛n⎞ n(n − 1) or ⎜ ⎟ 2 ⎝2⎠
49.
a. b.
51.
12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 = 3,991,680
50.
C (4, 1) + C (4, 2) + C (4, 3) + C (4, 4) = 15 different sums
7
12 − 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 = 31,840,128
Let a1 , a2 ,..., a5 be the long pieces and b1 , b2 ,..., b5 be the short pieces. The pairs must have one a with one b. For a1 there are 5 b’s, for a2 there are 4 b’s, … . Thus there are 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120 pairs consisting of a long piece and a short piece.
52.
We use the following: (Number of numbers with no repetitions) plus (Number of numbers with two or more repetitions) equals 104 . Number of numbers with no repetitions = 10 ⋅ 9 ⋅ 8 ⋅ 7. Thus the number of numbers with two or more repetitions = 104 − 10 ⋅ 9 ⋅ 8 ⋅ 7 = 4960.
53.
To return to the original spot, the tourist must toss an equal number of heads and tails. This is C (10, 5) = 252. There are 252 different toss combinations that return the tourist to the origin.
....................................................... PS1. See 11.6.
PS3. P (7, 2) =
Prepare for Section 11.7 PS2. Use the counting principle. 4 ⋅ 3 = 12 There are 12 different possible two-digit numbers.
7! 7 ⋅ 6 ⋅ 5! = = 42 (7 − 2)! 5!
5 8− 5 5 3 PS5. ⎛ 8 ⎞ 8! ⎛ 1⎞ ⎛ 3⎞ ⎛ 1 ⎞ ⎛ 3 ⎞ = 189 = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ 5!(8 − 5)! ⎝ 4 ⎠ ⎝ 4 ⎠ 8192 ⎝ 5⎠ ⎝ 4 ⎠ ⎝ 4 ⎠
PS4. C (7, 2) =
7! 7 ⋅ 6 ⋅ 5! = = 21 2!(7 − 2)! 2!⋅ 5!
PS6. Use the counting principle. 2 ⋅ 2 ⋅ 2 ⋅ 2 = 16 There are 16 possible light switch configurations.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 11.7
797
Section 11.7 1.
Label senators S1, S2 and representatives R1, R2 , R3.
The sample space is S = {S1R1, S1R2 , S1R3 , S2 R1, S2 R2 , S2 R3 , R1R2 , R1R3 , R2 R3 , S1S2 } 2.
S = {T, e, n, s}
4.
S = {HHHH, THHH, HTHH, HHHT, TTHH, THHT, HTTH, HHTT, THTH, HTHT, HHTH, HTTT, TTTH, THTT, TTHT, TTTT}
5.
Let the three cans be represented by A, B, and C and (x, y) represent the cans that balls 1 and 2 are placed in; e.g., (A, B) means ball 1 is in can A and ball 2 is in can B.
3.
Label coin H, T and integers 1, 2, 3, 4. S = {H 1, H 2, H 3, H 4, T 1, T 2, T 3, T 4}
S = {( A, A), (A, B ), (A, C ), ( B, A), (B, B ), (B, C ), (C , A), (C , B ), (C , C )} 6.
S = { RD, RI , DI }
8.
Let ( L1, L2 , L3 ) denote letter L1 in envelope A, letter L2 in envelope B, letter L3 in envelope C. Then the sample space is
7.
S = {HSC , HSD, HCD, SCD}
{(ABC ),( ACB),( BAC ),( BCA),(CAB),(CBA)}
9.
S = {ae,ai,ao,au,ei,eo,eu,io,iu,ou}
10.
S = {DN1N 2 , DN1N3 , DN 2 N3 , N1N 2 N3}
11.
E = {HHHH }
12.
E = {TTHH , THHT , HTTH , HHTT , THTH , HTHT }
13.
E = {TTTT , HTTT , THTT , TTHT , TTTH , TTHH , THTH , HTHT , THHT , HTTH , HHTT }
14.
E = {HHHH , HHHT , HHTH , HTHH , THHH }
15.
E =∅
The sample space S for the events in Exercises 16—20 is S = {(1,1), (1, 2), (1,3), (1, 4), (1,5), (1,6), (2,1), (2, 2), (2,3), (2, 4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} 16.
E = {(1,6),(2,5),(3, 4),(4,3),(5, 2),(6,1)}
18.
E = {(2,1),(3,1),(3, 2),(4,1),(4, 2),(4,3),(5,1),(5, 2),(5,3),(5, 4),(6,1),(6, 2),(6,3),(6, 4),(6,5)}
19.
E = {(1, 4),(2, 4),(3, 4),(4, 4),(5, 4),(6, 4)}
20.
E = S = {(1,1), (1, 2), (1,3), (1, 4), (1,5), (1,6), (2,1), (2, 2), (2,3), (2, 4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
21.
a. b.
4 1 = 52 13 13 1 P (spade) = = 52 4
P (king) =
17.
22.
E = {(1,1),(2, 2),(3,3),(4, 4),(5,5),(6,6)}
P (even) + P (÷3) − P (even and ÷ 3) =
23.
P (increase GNP) + P (increase inflation) − P (increase GNP and inflation) = 0.64 + 0.55 − 0.22 = 0.97
24.
P (greater than 3000) = 2 ⋅ 4 ⋅44 ⋅ 4 = 128 = 1 156 2 4
25.
P (1st) + P(2nd) − P (1st and 2nd) =
Copyright © Houghton Mifflin Company. All rights reserved.
3 2 1 5 1 2 + − = − = 6 6 6 6 6 3
1 1 1 6 3 + − = = 2 5 10 10 5
798
26.
Chapter 11: Sequences, Series, and Probability
a.
There are C (10, 2) = 45 ways in which 2 calculators can be randomly chosen from 10 calculators. There is only one way in which 2 defective calculators can be chosen from 2 defective calculators. Therefore, the
27.
a 0 is 9 . Therefore, the probability of not selecting a 0 10
on five trials is
probability is 1 . 45 b.
Because sampling is with replacement, the events are independent. On one trial, the probability of not selecting
5
⎛9⎞ ⎜ ⎟ = 0.59. ⎝ 10 ⎠
At least 1 defective calculator means 1 defective calculator or 2 defective calculators. We first calculate the probability of 1 defective calculator in a sample of 2 defective calculators. The number of ways of choosing 1 defective calculator from 2 is C (2, 1) = 2. Because there are two calculators in the sample and 1 is defective, the other calculator is not defective. There are C (8, 1) ways of selecting 1 calculator that is not defective. The probability of choosing 1 defective calculator is C (2,1)C (8,1) 2 ⋅ 8 16 = = C (10,2) 45 45 From part (a), the probability of 2 defective calculators is 1 . Therefore, the probability of at least 1 de45
fective calculator is 16 + 1 = 17 45 45 45 28.
The number of ways in which 6 people can be arranged in a line is P(6, 6) = 720. To determine how many ways 2 people A and B can be together, consider the number of arrangements of (AB), C, D, E, F. (We have grouped A and B together and considered arranging 5 items.) Since there is the same number of arrangements with BA (instead of AB), the number of arrangements is 2 ⋅ P (5,5) = 240. Therefore, the probability of any 2 people being together is 240 = 1 . 720 3
29.
To receive at least $50, an envelope with $50 in cash or $100 in cash must be selected. The probability is 75 50 125 1 + = = = 0.25 . 500 500 500 4
30.
The number of possible juries that can be chosen is C (30,12) = 86, 493, 225. It is possible to choose 6 women from 15 in C (15,6) = 5005 ways. It is possible to choose 6 men from 15 in C (15, 6) = 5005 ways. Therefore, the probability of 6 men and 6 women is C (15,6)C (15,6) 5005 ⋅ 5005 = ≈ 0.2896. C (30,12) 86493225
31.
There are P(6, 6) = 720 seating arrangements for the 6 children. There are 2 ⋅ 3! ⋅ 3! ways to have boys and girls alternate. Therefore the probability of boys and girls alternating is 2 ⋅ 3! ⋅ 3! 72 1 = = = 0.1. 720 720 10
32.
Four committee members can be selected from 8 people in C(8, 4) = 70 ways. There are C(3, 2) = 3 ways to choose 2 accountants from 3 and C(5, 2) = 10 ways to choose 2 actuaries from 5. The probability of 2 accountants and 2 C (3,2)C (5,2) 3 ⋅ 10 3 = = actuaries is C (8,4) 70 7
33.
The subject can select C(5, 2) = 10 different sets of 2 cards. The magician must name the set the subject has drawn. Therefore, the probability that a magician can guess the
Yes. The probability of drawing an ace from a regular deck
35.
34.
of playing cards is 4 = 1 . Because the card is replaced 52
13
before the second draw, the probability of drawing an ace on the second draw is also 1 . Therefore, the probability of 13
drawing two aces with replacement is 1 ⋅ 1 = 1 . 13 13 169
answers is 1 or 0.1. 10
The probability of choosing Monday is of choosing 8:00 A.M. is
1. 8
The probability
Therefore the probability of
choosing Monday at 8:00 A.M. is 1 ⋅ 1 = 1 or 0.025. 5 8 40
Copyright © Houghton Mifflin Company. All rights reserved.
1. 5
Section 11.7
36.
38.
799
Calculate the probability that the first radar system detects the missile but the second radar system does not plus the probability that the second but not the first detects the missile plus the probability that both detect the missile. Probability = 0.95(0.05) + 0.05(0.95) + 0.95(0.95) = 0.9975
37.
The probability that a single CD-ROM is not defective is
39.
999 1000
1 = 1− . 1000
=1 − [ (0.10)(0.10)(0.10)(0.10)] = 1 − 0.0001 = 0.9999
The probability that three are not defective is
and q =
41.
3 . 4
The probability of at least one defective equals 1 minus the probability of no defectives. The probability of no defectives is (0.95)5 . Therefore the probability of at
⎛ 8 ⎞ ⎛ 1 ⎞3 ⎛ 3 ⎞5 Probability = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ≈ 0.2076 ⎝ 3⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ 42.
The probability of all four
1 ⎛1⎞ companies choosing program A is ⎜ ⎟ = . 16 ⎝2⎠
This is a binomial probability with n = 8, k = 3., p=
1 . 2
4
⎛ 999 ⎞ ⎜ ⎟ ≈ 0.997. ⎝ 1000 ⎠
1 , 4
Assuming there is no preference, then the probability of choosing program A is
3
40.
The probability of at least one unprofitable = 1 − probability of all profitable
least one defective is 1 − (0.95)5 ≈ 0.2262
The probability of winning a prize is 1 minus the probability of not winning a prize. The probability of not winning a prize is C (998,10) 2.58 × 1023 ≈ ≈ 0.98 . C (1000,10) 2.63 × 1023 Therefore the probability of winning a prize is 1 – 0.98, or 0.02.
43.
3 4
1 4
This is a binomial probability with p = , q = , n = 25, and k = 21, 22, 23, 24, and 25. 21 4 22 3 23 2 24 1 25 ⎛ 25 ⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 25 ⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 25 ⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 25 ⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 25 ⎞ ⎛ 3 ⎞ P =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ +⎜ ⎟ ⎜ ⎟ ⎝ 21 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 22 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 23 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 24 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 25 ⎠ ⎝ 4 ⎠ The probability is approximately 0.2137.
44.
In a two-engine plane, safe flight occurs when one or two engines operate. The probability of this event is equal to 1 – probability that both engines fail = 1 − [ (0.03)(0.03)] = 1 − 0.0009 = 0.9991. For a four-engine plane, safe flight occurs when two, three, or four engines operate. The probability of this event is equal to ⎛ 4⎞ 2 2 ⎛ 4⎞ 3 ⎛ 4⎞ 4 ⎜ ⎟ (0.03) (0.97) + ⎜ ⎟ (0.03)(0.97) + ⎜ ⎟ (0.97) ≈ 0.9999 2 1 0 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ This probability is higher than that of two engines. Therefore, a plane that has four engines is safer.
....................................................... 45.
Let the originator tell a member of the club. (Rumor told once.) This person repeats it to any of 7 remaining members with probability
7 . 8
(Rumor told twice.) That person
repeats it to any of 7 members with probability ability of both events is
(). 7 8
2
7 . 8
Prob-
Connecting Concepts 46.
After the cards are shuffled, number the cards from 1 to 10, Let c1, c2, c3, c4, c5 be the numbers of the cards the sub-
{
}
ject said were white. For example, {1, 5, 7, 9, 10} means that the subject named those cards white. The sample space is all possible five-element subsets. Thus, N(S) = C(10, 5). The event E is that the subject named exactly 8 correctly. Then N ( E ) = C ( 5, 4 ) ⋅ C ( 5,1) . P(E) =
5 ⋅ 5 25 = 252 252
Copyright © Houghton Mifflin Company. All rights reserved.
800
47.
Chapter 11: Sequences, Series, and Probability
Let a, b, c, d be the first number chosen. Then event E is choosing a second number such that the first digit is not a, the second digit is not b, the third digit is not c, and the fourth digit matches the third digit. Since there are 9 digits available (digits 1 through 9), the probability that the first
48.
9! = 3780 1!4!2!2! Arrangements in which T occurs first = 8! = 430 4!2!2! 420 1 Probability T occurs first = = 3780 9
50.
Recall from Section 11.3 the Sum of a Geometric Series a S= 1 1− r 5 36 S= a1 = 5 , r = 11 36 36 1 − 11 36 S = 0.2 The probability is 0.2.
digit is not a is 8 , the probability that the second digit is
9 8 not b is , the probability that the third digit is not c is 8 , 9 9
Arrangements of Tennessee =
and the probability that the fourth digit matches the third is 1 . 9
Thus, P(E) = 8 ⋅ 8 ⋅ 8 ⋅ 1 = 512 ≈ 0.078. 9 9 9 9 6561 49.
Recall from Section 11.3 the Sum of a Geometric Series a S= 1 1− r S=
p2 1 − 2 p (1 − p )
a1 = p 2 , r = 2 p (1 − p )
(0.55)2 1 − 2(0.55)(1 − 0.45) S ≈ 0.599 The probability is 0.599.
S=
.......................................................
Exploring Concepts with Technology
Mathematical Expectation
Casino 1
Casino 2
Mark
Catch
Win
Expectation
Mark
Catch
Win
Expectation
6
4
$8
$.23
6
4
$6
$.17
6
5
$176
$.54
6
5
$160
$.50
6
6
$2960
$.38
6
6
$3900
$.50 $1.17
$1.15 Casino 3
Casino 4
Mark
Catch
Win
Expectation
Mark
Catch
Win
Expectation
6
4
$8
$.23
6
4
$6
$.17
6
5
$180
$.56
6
5
$176
$.54
6
6
$3000
$.39
6
6
$3000
$.39
$1.18
$1.10
Each mathematical expectation was determined using the formula C (20, c) ⋅ C (60, 6 − c) Mathematical expectation = W C (80, 6) where W is the number of dollars you win for c catches. Thus, casino 3 offers the greatest mathematical expectation. For each $2 bet at casino 3, the gambler has a mathematical expectation of winning $1.18.
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Review
801
.......................................................
Assessing Concepts
1.
n ! = n ⋅ ( n − 1) ⋅ ( n − 2) ⋅ L ⋅ 2 ⋅ 1
2.
An arithmetic sequence is one in which the difference between any two successive terms is the same constant.
3.
A geometric sequence is one in which the ratio of any two successive terms is the same constant.
4.
A sequence is an ordered list of numbers. A series is the sum of the terms of a sequence.
5.
The nth partial sum of a sequence is the sum of the first n terms of the sequence.
6.
A permutation takes into consideration the order of the elements of a list. A combination does not.
7.
No
8.
See the Chapter 11 Summary under Section 11.4.
9.
The Binomial Theorem for Positive Integers is used
10.
0
12.
P ( A U B ) = P( A) + P ( B ) − P ( A I B )
to expand ( a + b)n , where n is a positive integer. 11.
0 ≤ P ≤1
....................................................... 1.
an = n 2
2.
[11.1]
2
a3 = 3 = 9
a7 = 7 2 = 49
4.
an = 1 − 2n
[11.1]
5.
1 [11.1] n! 1 1 a3 = = 3! 6 1 1 a7 = = 7! 5040
10.
⎛ 4⎞ an = ⎜ − ⎟ ⎝ 3⎠
8.
11.
n
[11.1]
3
64 ⎛ 4⎞ a3 = ⎜ − ⎟ = − 27 ⎝ 3⎠ 7
16,384 ⎛ 4⎞ a7 = ⎜ − ⎟ = − 2187 ⎝ 3⎠
an = 3n + 2 [11.1]
a3 = 3! = 6
a3 = 11
a7 = 7! = 5040
a7 = 23
an = 2− n
6.
[11.1]
a3 = 2 = 1 8 a7 = 2−7 = 1 128
a7 = 1 − 2(7) = 1 − 14 = −13
an =
3.
[11.1]
−3
a3 = 1 − 2(3) = 1 − 6 = −5
7.
an = n!
Chapter Review
a3 = 3 = 27 a7 = 37 = 2187
1 [11.1] n 1 a3 = 3 1 a7 = 7
a1 = 2, an = 3an −1 a3 = 3a2 = 3 ⋅ 6 = 18 • a4 = 3a3 = 3 ⋅ 18 = 54 a5 = 3a4 = 3 ⋅ 54 = 162 a6 = 3a5 = 3 ⋅ 162 = 486 a7 = 3a6 = 3 ⋅ 486 = 1458 •
[11.1]
3
an =
a2 = 3a1 = 3 ⋅ 2 = 6
an = 3n
[11.1]
n
9.
[11.1] an = ⎛⎜ 2 ⎞⎟ ⎝3⎠ 3 a3 = ⎛⎜ 2 ⎞⎟ = 8 27 ⎝3⎠ 7 ⎛ ⎞ 2 a7 = ⎜ ⎟ = 128 2187 ⎝3⎠
12.
a1 = −1, an = 2an −1 [11.1] a2 = 2a1 = 2 ( −1) = −2 a3 = 2a2 = 2 ( −2 ) = −4 • a4 = 2a3 = 2 ( −4 ) = −8 a5 = 2a4 = 2 ( −8 ) = −16 a6 = 2a5 = 2 ( −16 ) = −32 a7 = 2a6 = 2 ( −32 ) = −64 •
Copyright © Houghton Mifflin Company. All rights reserved.
802
13.
Chapter 11: Sequences, Series, and Probability
a1 = 1, an = − nan −1
14.
[11.1]
a2 = −2a1 = −2 ⋅ 1 = −2
2
a3 = 32 a2 = 9 ⋅ 8 = 72 •
a4 = 42 a3 = 16 ⋅ 72 = 1152
a4 = −4a3 = −4 ⋅ 6 = −24
a5 = −5a4 = −5 ⋅ ( −24 ) = 120
a5 = 52 a4 = 25 ⋅ 1152 = 28,800
a6 = 62 a5 = 36 ⋅ 28,800 = 1,036,800
a6 = −6a5 = −6 ⋅ 120 = −720
a7 = 7 2 a6 = 49 ⋅ 1,036,800 = 50,803, 200 •
a7 = −7 a6 = −7 ⋅ ( −720 ) = 5040 •
15.
a1 = 4, an = an −1 + 2 [11.1] a2 = a1 + 2 = 4 + 2 = 6 a3 = a2 + 2 = 6 + 2 = 8 • a4 = a3 + 2 = 8 + 2 = 10 a5 = a4 + 2 = 10 + 2 = 12 a6 = a5 + 2 = 12 + 2 = 14 a7 = a6 + 2 = 14 + 2 = 16 •
16.
17.
a1 = 1, a2 = 2, an = an −1an − 2 [11.1]
18.
a3 = a2 ⋅ a1 = 2 ⋅ 1 = 2 •
a7 = a6 ⋅ a5 = 32 ⋅ 8 = 256 •
a1 = −1, an = 3n an −1 [11.1] a2 = 3(2)a1 = 3(2)(−1) = −6 a3 = 3(3) a2 = 3(3)(−6) = −54 • a4 = 3(4)a3 = 3(4)(−54) = −648 a5 = 3(5)a4 = 3(5)(−648) = −9720 a6 = 3(6) a5 = 3(6)(−9720) = −174,960 a7 = 3(7)a6 = 3(7)(−174,960) = −3,674,160 •
20.
an +1 − an = (n + 1) 2 − n 2 = 2n + 1 ≠ constant.
22.
an +1 − an = ( n + 1)!− n ! = ( n + 1) n!− n! = n ⋅ n! ≠ constant.
an +1 (n + 1) ⎛ 1⎞ = = ⎜1 + ⎟ ≠ constant. 2 an ⎝ n⎠ n Thus not a geometric sequence. Neither an arithmetic nor a geometric sequence. [11.1]
[11.2]
Thus not a geometric sequence. Neither an arithmetic nor a geometric sequence. [11.1] 24.
Arithmetic sequence. 25.
a1 = 2, an = −2nan −1 [11.1]
Thus not an arithmetic sequence. an +1 (n + 1)! ( n − 1)n! = = = n + 1 ≠ constant. an n! n!
2
an +1 − an = 3( n + 1) + 2 − (3n + 2) = 3 = constant.
= 2=2 • 1 = 2 =1 2 1 = 2 1/ = 2=1 1 2 1/ 2 = =1 • 1/ 2
a2 = −2(2)a1 = −2(2)(2) = −8 a3 = −2(3)a2 = −2(3)(−8) = 48 • a4 = −2(4)a3 = −2(4)(48) = −384 a5 = −2(5)a4 = −2(5)(−384) = 3840 a6 = −2(6)a5 = −2(6)(3840) = −46,080 a7 = −2(7)a6 = −2(7)( −46,080) = 645,120 •
Thus not an arithmetic sequence.
23.
a −1 a1 = 1, a2 = 2, an = n [11.1] an − 2 a2 a1 a a4 = 3 a2 a a5 = 4 a3 a a6 = 5 a4 a a7 = 6 a5
a5 = a4 ⋅ a3 = 4 ⋅ 2 = 8 a6 = a5 ⋅ a4 = 8 ⋅ 4 = 32
2
[11.1] a1 = 3, an = an −1 − 3 a2 = a1 − 3 = 3 − 3 = 0 a3 = a2 − 3 = 0 − 3 = −3 • a4 = a3 − 3 = −3 − 3 = −6 a5 = a4 − 3 = −6 − 3 = −9 a6 = a5 − 3 = −9 − 3 = −12 a7 = a6 − 3 = −12 − 3 = −15 •
a3 =
a4 = a3 ⋅ a2 = 2 ⋅ 2 = 4
21.
[11.1]
2
a2 = 2 a1 = 2 ⋅ 2 = 8
a3 = −3a2 = −3 ⋅ ( −2 ) = 6 •
19.
a1 = 2, an = n 2 an −1
an +1 − an = (1 − 2(n + 1)) − (1 − 2n) = −2 = constant. Arithmetic sequence.
an +1 2−( n +1) 2− n ⋅ 2−1 1 = = = 2−1 = = constant. [11.3] −n −n an 2 2 2 Geometric sequence.
26.
an +1 3n +1 3n ⋅ 3 = = = 3 = constant. an 3n 3n Geometric sequence.
Copyright © Houghton Mifflin Company. All rights reserved.
[11.3]
[11.2]
Chapter Review
27.
803
1 −1 (n + 1)! n! 1 = − 1 = 1 ⎛⎜ 1 − 1⎞⎟ ≠ constant (n + 1)n! n! n! ⎝ n + 1 ⎠ 1 1 an +1 (n + 1)! n! n! = = = = ≠ constant 1 an ( n + 1)! (n + 1)n! n + 1 n! Neither an arithmetic nor a geometric sequence. [11.1] an +1 − an =
28.
29.
1 1 1 − =− ≠ constant n +1 n n( n + 1)
1 an +1 n + 1 n = = ≠ constant 1 an n +1 n Neither an arithmetic nor a geometric sequence. [11.1]
n +1
⎛2⎞ an +1 ⎜⎝ 3 ⎟⎠ 2 = = = constant. n 3 an ⎛2⎞ ⎜ ⎟ ⎝3⎠ Geometric sequence. [11.3]
an +1 − an =
n +1
30.
⎛ 4⎞ − an +1 ⎜⎝ 3 ⎟⎠ 4 = = − = constant. n 3 an ⎛ 4⎞ ⎜− ⎟ ⎝ 3⎠ Geometric sequence [11.3]
31.
Since each successive term is 3 times the previous term, the sequence has a common ratio of 3. Geometric sequence. [11.1]
32.
Since each successive term is 2 times the previous term, the sequence has a common ratio of 2. Geometric sequence. [11.1]
33.
Examining the terms of the sequence reveals that there is no common difference and no common ratio. Neither an arithmetic nor a geometric sequence. [11.1]
34.
Examining the terms of the sequence reveals that there is no common difference and no common ratio. Neither an arithmetic nor a geometric sequence. [11.1]
35.
Since each successive term is 2 more than the preceding term, the sequence has a common difference of 2. Arithmetic sequence. [11.2]
36.
Since each successive term is 3 less than the preceding term, the sequence has a common difference of −3 . Arithmetic sequence. [11.2]
37.
Examining the terms of the sequence reveals that there is no common difference and no common ratio. Neither an arithmetic nor a geometric sequence. [11.1]
38.
Examining the terms of the sequence reveals that there is no common difference and no common ratio. Neither an arithmetic nor a geometric sequence. [11.1]
39.
Examining the terms of the sequence reveals that there is no common difference and no common ratio. Neither an arithmetic nor a geometric sequence. [11.1]
40.
Examining the terms of the sequence reveals that there is no common difference and no common ratio. Neither an arithmetic nor a geometric sequence. [11.1] 9
41.
(2n − 3) is an arithmetic series with common difference 2. a1 = −1, a9 = 15, n = 9 ∑ n =1
[11.2]
Sn = n ( a1 + an ) 2 9 S9 = ( −1 + 15 ) = 9 (14 ) = 63 2 2
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804
Chapter 11: Sequences, Series, and Probability 11
42.
(1 − 3i ) is an arithmetic series with common difference − 3. a1 = −2, a11 = −32, n = 11 [11.2] ∑ i =1 S n = n ( a1 + an ) 2 S11 = 11 ( −2 + ( −32 ) ) = 11 ( −34 ) = −187 2 2 8
43.
(4k + 1) is an arithmetic series with common difference 4. a1 = 5, a8 = 33, n = 8 ∑ k =1
[11.2]
S n = n ( a1 + an ) 2 S8 = 8 ( 5 + 33) = 4 ( 38 ) = 152 2 10
44.
(i 2 + 3) is neither an arithmetic nor a geometric series. [11.1] ∑ i =1 10
∑ (i
2
+ 3) = (1 + 3) + ( 4 + 3) + ( 9 + 3) + (16 + 3) + ( 25 + 3) + ( 36 + 3) + ( 49 + 3) + ( 64 + 3) + ( 81 + 3) + (100 + 3) = 415
i =1
6
45.
3 ⋅ 2n is a geometric series with common ratio 2. ∑ n =1
a1 = 6, n = 6, r = 2 [11.3]
a1 (1 − r n ) 1− r 6(1 − 26 ) 6(−63) S6 = = = 378 1− 2 −1 Sn =
5
46.
2 ⋅ 4i −1 is a geometric series with common ratio 4. ∑ i =1
a1 = 2, n = 5, r = 4 [11.3]
a (1 − r n ) Sn = 1 1− r S5 =
2(1 − 45 ) 2(−1023) = = 682 1− 4 −3
9
47.
(−1)k (3k ) is a geometric series with common ratio − 3. ∑ k =1
a1 = −3, n = 9, r = −3 [11.3]
a (1 − r n ) Sn = 1 1− r S9 =
−3(1 − ( −3)9 ) −3(19,684) = = −14,763 1 − ( −3) 4
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Review
805
8
48.
(−1)i +1 2i is a geometric series with common ratio − 2. ∑ i =1
a1 = 2, n = 8, r = −2 [11.3]
a (1 − r n ) Sn = 1 1− r S8 = 10
49.
∑ i =1
2(1 − (−2)8 ) 2(−255) = = −170 1 − (−2) 3 i
2 2 2 ⎛2⎞ [11.3] ⎜ ⎟ is a geometric series with common ratio . a1 = , n = 10, r = 3 3 3 3 ⎝ ⎠
a1 (1− r n ) 1− r 10 10 2 ⎛⎜1− ⎛ 2 ⎞ ⎞⎟ 2 ⎛⎜1− ⎛ 2 ⎞ ⎞⎟ ⎜ ⎟ ⎜ ⎟ 3 ⎜ ⎝ 3 ⎠ ⎠⎟ 3 ⎝⎜ ⎝ 3 ⎠ ⎠⎟ ⎛ ⎛ 2 ⎞10 ⎞ ⎛ 1,024 ⎞ ⎛ 58,025 ⎞ 116,050 = = 2⎜1− ⎜ ⎟ ⎟ = 2⎜1− =2 = ≈1.9653 S10 = ⎝ ⎜ ⎝ 3 ⎠ ⎟ ⎝ 59,049 ⎠⎟ ⎝⎜ 59,049 ⎠⎟ 59,049 1 1− 2 ⎝ ⎠ 3 3 Sn =
11
50.
i
3 ⎛3⎞ ⎜ ⎟ is a geometric series with common ratio . ∑ 2 ⎝2⎠ i =1
a1 =
3 3 [11.3] , n = 11, r = 2 2
a1 (1 − r n ) 1− r 11 3 ⎜⎛ 1 − ⎛ 3 ⎞ ⎟⎞ 3 ⎜ ⎟ ⎟ ⎜ 2 ⎝ 2 ⎠ ⎠ 2 ( −85.4976 ) ≈ ≈ 256.4927 S11 = ⎝ −1 1− 3 2 2 Sn =
9
51.
n +1
(−1) ∑ n2 n =1 9
n +1
(−1) ∑ n2 n =1 5
52.
∑ k =1 ∞
53.
∑ n =1
=1−
1 1 1 1 1 1 1 1 + − + − + − + ≈ 0.8280 4 9 16 25 26 49 64 81
k +1
(−1) ∑ k! k =1 5
is neither a geometric nor an arithmetic series. [11.1]
is neither an arithmetic nor a geometric series. [11.1]
( −1)k +1 1 1 1 1 19 = 1− + − + = ≈ 0.63 2 6 24 120 30 k! n
1 1 1 ⎛1⎞ [11.3] ⎜ ⎟ is an infinite geometric series with common ratio . a1 = , r = 4 4 4 ⎝4⎠
a1 1− r 1 1 1 4 S= = 4 = 1 3 3 1− 4 4
S=
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806
Chapter 11: Sequences, Series, and Probability ∞
54.
i
∑ i =1
5 5 5 ⎛ 5⎞ [11.3] ⎜ − ⎟ is an infinite geometric series with common ratio − . a1 = − , r = − 6 6 6 ⎝ 6⎠
S=
a1 1− r
5 5 S= = 6 =− ⎛ 5 ⎞ 11 11 1− ⎜− ⎟ 6 ⎝ 6⎠ 5 6
−
∞
55.
−
k
∑ k =1
4 4 4 ⎛ 4⎞ ⎜ − ⎟ is an infinite geometric series with common ratio − . a1 = − , r = − [11.3] 5 5 5 ⎝ 5⎠
S=
a1 1− r
4 4 5 S= = =− 9 ⎛ 4⎞ 9 1− ⎜− ⎟ 5 ⎝ 5⎠ 4 5
−
∞
56.
∑ j =0
−
j
⎛1⎞ 1 1 [11.3] ⎜ ⎟ is an infinite geometric series with common ratio . a1 = 1, r = 5 5 ⎝5⎠
a1 1− r S= 1 =1 =5 1− 1 4 4 5 5 S=
n
57.
n(5n + 7) (5i + 1) = ∑ 2 i =1
[11.4] 1
∑ (5i + 1) = 6 and 1(5 2+ 7) = 6. Therefore that statement is true for n = 1.
1.
For n = 1, we have
2.
Assume the statement is true for n = k .
i =1
k
∑ (5i + 1) = k (5k2+ 7) Induction Hypothesis i =1
Prove the statement is true for n = k + 1. That is, prove k +1
∑ (5i + 1) = (k + 1)(52k + 12) i =1 k +1
∑
k
(5i + 1) =
i =1
∑ i =1
k
(5i + 1) + 5( k + 1) + 1 =
∑ (5i + 1) + 5k + 6 i =1
k (5k + 7) + 5k + 6 Using the Induction Hypothesis 2 k (5k + 7) + 10k + 12 5k 2 + 7 k + 10k + 12 = = 2 2 2 (k + 1)(5k + 12) = 5k + 17 k + 12 = 2 2 Thus the statement is true for n = k + 1. By the Induction Axiom, the statement is true for all positive integers. =
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Chapter Review
807
n
58.
∑ (3 − 4i ) = n(1 − 2n)
[11.4]
i =1
1
∑ (3 − 4i ) = −1 and 1[1 − 2(1)] = −1.
1.
For n = 1, we have
2.
Therefore the statement is true for n = 1. Assume the statement is true for n = k .
i =1
k
∑ ( 3 − 4i ) =k (1 − 2k ) Induction Hypothesis i =1
Prove the statement is true for n = k + 1. That is, prove k +1
∑ (3 − 4i ) = (k + 1)( −1 − 2k ) = −(k + 1)(2k + 1)
i =1 k +1
k
k
i =1
i =1
∑ (3 − 4i ) = ∑ (3 − 4i ) + 3 − 4(k + 1) = ∑ (3 − 4i ) − 1 − 4k i =1
= k (1 − 2k ) − 1 − 4k Using the Induction Hypothesis = k − 2k 2 − 1 − 4k = −2k 2 − 3k − 1 = −(2k 2 + 3k + 1) = −( k + 1)(2k + 1)
Thus the statement is true for n = k + 1. By the Induction Axiom, the statement is true for all positive integers.
59.
( )
⎛ 2 ⎜1 − − 1 2 ⎛− 1⎞ = ⎝ ∑ ⎜⎝ 2 ⎟⎠ 3 i =0 i
n
1.
n +1 ⎞
⎟ ⎠ [11.4]
This induction begins with n = 0.
( )
1⎞ ⎛ 2 ⎜1 − − 1 ⎟ 2 ∑ ⎛⎜⎝ − 12 ⎞⎟⎠ = ⎛⎜⎝ − 12 ⎞⎟⎠ = 1 and ⎝ 3 ⎠ = 1 i =0 0
2.
i
0
Thus the statement is true when n = 0. Assume the statement is true for n = k .
( )
⎛ 2 ⎜1 − − 1 2 ∑ ⎛⎜⎝ − 12 ⎞⎟⎠ = ⎝ 3 i =0 k
i
k +1 ⎞
⎟ ⎠ Induction Hypothesis
Prove the statement is true for n = k + 1. That is, prove
( )
⎛ 2 ⎜1 − − 1 2 ∑ ⎛⎜⎝ − 12 ⎞⎟⎠ = ⎝ 3 i =0
k +1
i
k +1
i
k
∑ ⎛⎜⎝ − 12 ⎞⎟⎠ = ∑ ⎛⎜⎝ − 12 ⎞⎟⎠
i =0
i =0
i
⎟ 2−2 −1 ⎠= 2 3
+ ⎛⎜ − 1 ⎞⎟ ⎝ 2⎠
⎛ 2 ⎜1 − − 1 2 = ⎝ 3
( )
k +1 ⎞
⎛ 2 ⎜1 − − 1 2 = ⎝
( )
k +1 ⎞
( )
( )
k +2 ⎞
k +2
( )( )
2−2 −1 −1 2 2 = 3
k +1
( )
2+ −1 2 = 3
k +1
k +1
⎟ k +1 ⎠ + ⎛− 1⎞ ⎜ ⎟ ⎝ 2⎠
( )
1 ⎟+3 − 2 ⎠
3
k +1
( )
2−2 −1 2 =
k +1
( )
+3 −1 2
k +1
3
k +1
2+ −1 2 3 Thus the statement is true for all integers n ≥ 0. =
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808
Chapter 11: Sequences, Series, and Probability n
60.
∑ ( −1)
i
i =0
1.
( )n +1 = 1 − −1 [11.4] 2
This induction begins with n = 0. 0
∑ ( −1)i = ( −1)0 = 1 and
i =0
2.
1 − ( −1) 1 + 1 = =1 2 2
Thus the statement is true for n = 0. Assume the statement is true for n = k . k
∑ ( −1)i =
i =0
1 − ( −1)k +1 2
Prove the statement is true for n = k + 1. That is, prove k +1
∑ ( −1)i =
i =0 k +1
1 − ( −1) k + 2 1 − ( −1)( −1) k +1 1 + ( −1)k +1 = = 2 2 2 k
∑ ( −1)i = ∑ ( −1)i + (−1)k +1
i =0
i =0
1 − ( −1)
k +1
k +1
+ ( −1) Using the Induction Hypothesis 2 1 − ( −1) k +1 + 2( −1)k +1 = 2 1 + ( −1)k +1 = 2 Thus the statement is true for all integers n ≥ 0. =
61.
n n ≥ n! [11.4] 1. 2.
When n = 1, 11 = 1 and n! = 1. The statement is true for n = 1. Assume the statement is true for n = k .
k k ≥ k! Induction Hypothesis Prove the statement is true for n = k + 1. That is, prove (k + 1)k +1 ≥ (k + 1)! (k + 1)k +1 = (k + 1)(k + 1)k > (k + 1)k k By Induction Hypothesis k k ≥ k !. We have (k + 1)k k ≥ (k + 1)k ! = (k + 1)! Therefore (k + 1)k +1 ≥ (k + 1)! Therefore the statement is true for all integers n ≥ 1. 62.
1. 2.
For n = 9, 9! = 362,880, 49 = 262,144 [11.4] Since 362,880 > 262,144, the statement is true for n = 9. Assume the statement is true for n = k .
k ! > 4k Induction Hypothesis Prove the statement is true for n = k + 1. That is, prove (k + 1)! > 4k +1 (k + 1)! = (k + 1) k ! > (k + 1)4k By Induction Hypothesis Since k ≥ 9, k + 1 ≥ 4. Thus (k + 1)4k > 4 ⋅ 4k = 4k +1 Thus (k + 1)! > 4k +1 The statement is true for all integers n ≥ 9.
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Chapter Review
63.
1. 2.
809
When n = 1, we have 13 + 2(1) = 3. Since 3 is a factor of 3, the statement is true for n = 1. [11.4] Assume the statement is true for n = k .
3 is a factor of k 3 + 2k Induction Hypothesis Prove the statement is true for n = k + 1. That is, prove 3 is a factor of (k + 1)3 + 2(k + 1).
(k + 1)3 + 2(k + 1) = k 3 + 3k 2 + 3k + 1 + 2k + 2 = ( k 3 + 2k ) + 3(k 2 + k + 1) By Induction Hypothesis, 3 is a factor of k 3 + 2k . Three is also a factor of 3(k 2 + k + 1). Thus 3 is a factor of (k + 1)3 + 2(k + 1). The statement is true for all positive integers n. 64.
1.
When n = 1, a1 = 2 < 2. The statement is true for n = 1. [11.4]
2.
Assume the statement is true for some integer k . ak < 2 Induction Hypothesis Prove the statement is true for n = k + 1. That is, prove ak +1 < 2. By the Induction Hypothesis, ak < 2. Thus ( 2) ak < ( 2)2 = 2 But ( 2)ak = ak +1. Thus ak +1 < 2 The statement is true for all positive integers n. 5
65.
(4a − b)5 =
∑ ⎛⎜⎝ 5i ⎞⎟⎠ (4a)
5−i
( −b )i
[11.5]
i =0
⎛5⎞ ⎛ 5⎞ ⎛5⎞ ⎛ 5⎞ ⎛5⎞ ⎛ 5⎞ = ⎜ ⎟ (4a )5 + ⎜ ⎟ (4a )4 (−b)1 + ⎜ ⎟ (4a )3 (−b)2 + ⎜ ⎟ (4a ) 2 (−b)3 + ⎜ ⎟ (4a )(−b)4 + ⎜ ⎟ (−b)5 ⎝0⎠ ⎝1 ⎠ ⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠ ⎝ 5⎠ = 1(1024a 5 ) + 5(−256a 4b) + 10(64a3b 2 ) + 10(−16a 2b3 ) + 5(4ab 4 ) + (−b5 ) = 1024a 5 − 1280a 4b + 640a3b 2 − 160a 2b3 + 20ab 4 − b5
66.
6
( x + 3 y )6 = ∑ ⎛⎜ i6 ⎞⎟ x6−i ( 3 y )i i=0 ⎝
[11.5]
⎠
⎛ 6⎞ ⎛ 6⎞ ⎛6⎞ ⎛ 6⎞ ⎛ 6⎞ ⎛ 6⎞ ⎛ 6⎞ = ⎜ ⎟ x6 + ⎜ ⎟ x5 (3 y ) + ⎜ ⎟ x 4 (3 y ) 2 + ⎜ ⎟ x3 (3 y )3 + ⎜ ⎟ x 2 (3 y ) 4 + ⎜ ⎟ x(3 y )5 + ⎜ ⎟ (3 y )6 ⎝ 0⎠ ⎝1 ⎠ ⎝ 2⎠ ⎝3⎠ ⎝ 4⎠ ⎝5⎠ ⎝ 6⎠ = x6 + 6 x5 ( 3 y ) +15 x 4 (9 y 2 ) + 20 x3 (27 y 3 ) +15 x 2 (81 y 4 ) + 6 x(243 y 5 ) +1(729 y 6 ) = x6 +18 x5 y +135 x 4 y 2 + 540 x3 y 3 +1215 x 2 y 4 +1458 xy 5 + 729 y 6
67.
8
( a + 2 b ) = ∑ ⎛⎜⎜ 8i ⎞⎟⎟( a )8−i (2 b )i ⎝ ⎠ 8
[11.5]
i =0
⎛8 ⎞ ⎛8⎞ ⎛8 ⎞ ⎛8⎞ ⎛8 ⎞ ⎛8⎞ = ⎜ ⎟ ( a )8 + ⎜ ⎟ ( a )7 (2 b ) + ⎜ ⎟ ( a )6 (2 b )2 + ⎜ ⎟ ( a )5 (2 b )3 + ⎜ ⎟ ( a )4 (2 b )4 + ⎜ ⎟ ( a )3 (2 b )5 ⎝0⎠ ⎝1 ⎠ ⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠ ⎝ 5⎠ ⎛8 ⎞ ⎛8 ⎞ ⎛8⎞ + ⎜ ⎟ ( a )2 (2 b )6 + ⎜ ⎟ ( a )(2 b )7 + ⎜ ⎟ (2 b )8 ⎝ 6⎠ ⎝7⎠ ⎝8⎠ = 1(a 4 ) + 8a 7 / 2 (2b1/ 2 ) + 28a3 (4b) + 56a5 / 2 (8b3 / 2 ) + 70a 2 (16b 2 ) + 56a3 / 2 (32b5 / 2 ) + 28a(64b3 ) + 8a1/ 2 (128b7 / 2 ) + 1(256b 4 ) = a 4 + 16a 7 / 2b1/ 2 + 112a3b + 448a5 / 2b3 / 2 + 1120a 2b 2 + 1792a3 / 2b5 / 2 + 1792ab3 + 1024a1/ 2b7 / 2 + 256b 4
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810
68.
Chapter 11: Sequences, Series, and Probability 7
1 ⎞ ⎛ ⎟ = ⎜ 2x − 2x ⎠ ⎝
7
∑ i =0
i
⎛7⎞ ⎛ 1 ⎞ ⎜⎜ ⎟⎟(2 x )7 −i ⎜ − ⎟ [11.5] i ⎝ 2x ⎠ ⎝ ⎠ 2
3
4
⎛7⎞ ⎛7⎞ ⎛7⎞ ⎛7⎞ ⎛7⎞ ⎛7⎞ = ⎜ ⎟ (2 x)7 + ⎜ ⎟ (2 x)6 ⎛⎜ − 1 ⎞⎟ + ⎜ ⎟ (2 x)5 ⎛⎜ − 1 ⎞⎟ + ⎜ ⎟ (2 x) 4 ⎛⎜ − 1 ⎞⎟ + ⎜ ⎟ (2 x)3 ⎛⎜ − 1 ⎞⎟ + ⎜ ⎟ (2 x) 2 ⎛⎜ − 1 ⎞⎟ ⎝ 2x ⎠ ⎝ 2 ⎠ ⎝ 2x ⎠ ⎝ 3 ⎠ ⎝ 2x ⎠ ⎝ 4⎠ ⎝ 2x ⎠ ⎝ 5 ⎠ ⎝ 2x ⎠ ⎝0⎠ ⎝1 ⎠ 6 7 7 7 ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ (2 x) ⎛⎜ − 1 ⎞⎟ + ⎜ ⎟ ⎛⎜ − 1 ⎞⎟ ⎝ 2x ⎠ ⎝ 7 ⎠ ⎝ 2x ⎠ ⎝6⎠ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = 1(128 x7 ) − 7(64 x6 ) ⎛⎜ 1 ⎞⎟ + 21(32 x5 ) ⎜ 1 2 ⎟ − 35(16 x 4 ) ⎜ 13 ⎟ + 35(8 x3 ) ⎜ 1 4 ⎟ − 21(4 x 2 ) ⎜ 1 5 ⎟ ⎝ 2x ⎠ ⎝ 4x ⎠ ⎝ 8x ⎠ ⎝ 16 x ⎠ ⎝ 32 x ⎠ ⎛ 1 ⎞ ⎛ 1 ⎞ + 7(2 x) ⎜ ⎟ − 1⎜ ⎟ ⎝ 64 x6 ⎠ ⎝ 128 x7 ⎠ 7 5 3 = 128 x − 224 x + 168 x − 70 x + 35 − 213 + 7 5 − 1 7 2x 8x 32 x 128 x 69.
The fifth term of (3 x − 4 y )7 is
⎛7⎞ 3 4 ⎜⎜ 4 ⎟⎟ (3 x) ( −4 y ) ⎝ ⎠
70.
The eighth term of (1 − 3x)9 is
⎛9⎞ 2 7 ⎜⎜ ⎟⎟ (1) (−3 x ) ⎝7⎠
71.
There are 26 choices for each letter. By the Fundamental Counting Principle, there are 268 possible passwords. [11.6]
72.
Using the Fundamental Counting Principle, we have 106 ⋅ 26 possible serial numbers. [11.6]
73.
This is a permutation with n = 15 and r = 3. [11.6] 15! 15! P (15,3) = = = 2730 (15 − 3)! 12!
74.
There are ⎛⎜
4⎞ ⎟ ⎝1 ⎠
⎛ 4⎞ ⎜⎜ ⎟⎟ ⎝1 ⎠
75.
⎛12 ⎞ ⎜⎜ ⎟⎟ ⎝3 ⎠
5
= 35(27 x3 )(256 y 4 ) = 241,920 x3 y 4 . [11.5]
= 36 ⋅ 1 ⋅ (−2187 x7 ) = −78,732 x7 . [11.5]
ways to choose a supervisor and
⎛12 ⎞ ⎜ ⎟ ⎝3 ⎠
ways to choose 3 regular employees. Thus, there are ⎛⎜
4 ⎞ ⎛12 ⎞ ⎟ ⎜ ⎟ ways ⎝1 ⎠ ⎝ 3 ⎠
to do both.
= 4 ⋅ 220 = 880 shifts have 1 supervisor. [11.6]
This problem is solved in stages. First, there are ⎛⎜
10 ⎞ ⎟ ⎝5 ⎠
together. Second, there are ⎛⎜
10 ⎞⎛ 2 ⎞ ⎟⎜ ⎟ ⎝ 4 ⎠⎝1 ⎠
Altogether there are
⎛10 ⎞ ⎛10 ⎞⎛ 2 ⎞ ⎜ ⎟ + ⎜ ⎟⎜ ⎟ ⎝ 5 ⎠ ⎝ 4 ⎠⎝1 ⎠
ways to choose a committee excluding both people who refuse to serve
ways to choose a committee that includes one person but not the other.
ways to choose the committee.
⎛10 ⎞ ⎛10 ⎞⎛ 2 ⎞ ⎜ ⎟ + ⎜ ⎟⎜ ⎟ = 252 + 210 ( 2 ) = 672 possible committees [11.6] ⎝ 5 ⎠ ⎝ 4 ⎠⎝1 ⎠
76.
⎛10 ⎞ ⎟⎟ ⎝4 ⎠
There are ⎜⎜
= 210 ways of choosing 4 calculators from 10. If the inspector is to choose 1 defective calculator, then 3 nondefective ⎛ 2⎞ ⎛ 8⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⎝1 ⎠ ⎝ 3 ⎠
calculators must also be chosen. There are ⎜⎜ 112 210
77.
=
8 . 15
= 2 ( 56 ) = 112 ways to accomplish that. Therefore, the probability of the event is
[11.7]
The probability is 1 ⋅ 1 ⋅ 1 = 1 . 2 2 2 8 The probability of one tail and therefore two heads is 3 ⎜⎛ 1 ⋅ 1 ⋅ 1 ⎟⎞ = 3 . [11.4] ⎝2 2 2⎠ 8
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Chapter Review
78.
811 ⎛10 ⎞ ⎟⎟ ⎝4 ⎠
There are ⎜⎜
⎛5⎞⎛5⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⎝ 2⎠ ⎝ 2⎠
ways to draw 4 cards from 10. There are ⎜⎜
ways to draw 2 red and 2 black cards The probability of drawing 2 red
⎛ 5 ⎞⎛ 5 ⎞ ⎜ ⎟⎜ ⎟ 2 2 10 ⋅ 10 10 = . [11.7] and 2 black cards is ⎝ ⎠⎝ ⎠ = 210 21 ⎛10 ⎞ ⎜ ⎟ 4 ⎝ ⎠ 79.
We look at the possibility for each case. If the middle digit is zero, there are If the middle digit is one, there is If the middle digit is two, there are If the middle digit is three, there are If the middle digit is four, there are If the middle digit is five, there are If the middle digit is six, there are If the middle digit is seven, there are If the middle digit is eight, there are If the middle digit is nine, there are Total The probability is
80.
0 numbers 1 number 4 numbers 9 numbers 16 numbers 25 numbers 36 numbers 49 numbers 64 numbers 81 numbers 285 numbers
285 = 0.285. [11.7] 1000
The probability that the sum of two numbers is 9 when the numbers are selected with replacement from 1, 2, 3, 4, 5, 6 is The probability that the sum is 7 is
6 36
= 1 . Therefore the probability that it is not 7 and not 9 is 6
1 − ⎜⎛ 1 + 1 ⎟⎞ = 1 − 10 = 13 . 36 18 ⎝9 6⎠ First selection, sum is 9. Second selection, probability of 9 is 1 .
9 13 ⋅ 1 . 18 9 probability is 13 ⋅ 13 ⋅ 1 . 18 18 9
Third selection, probability is Fourth selection,
M
The total probability is the sum of this infinite process. 2
1 + 13 ⋅ 1 + ⎛ 13 ⎞ ⎛ 1 ⎞ + ⋅ ⋅ ⋅ ⎜ ⎟ ⎜ ⎟ 9 18 9 ⎝ 18 ⎠ ⎝ 9 ⎠ This is a geometric series with a1 = 1 , r = 13 . S=
1 9
1 − 13 18
=
1 9 5 18
9
=2 5
The probability is 2 . 5
81.
18
[11.7]
The probability of drawing an ace and a 10 card from one regular deck of playing cards is ⎛ 4 ⎞ ⎛ 16 ⎞ ⎜1 ⎟ ⎜1 ⎟ ⎝ ⎠ ⎝ ⎠ = 4 ⋅ 16 ≈ 0.0483. 52 ⋅ 51 ⎛ 52 ⎞ ⎜2 ⎟ 2 ⎝ ⎠ The probability of drawing an ace and a 10 card from two regular decks of playing cards is ⎛ 8 ⎞ ⎛ 32 ⎞ ⎜1 ⎟ ⎜1 ⎟ ⎝ ⎠ ⎝ ⎠ = 8 ⋅ 32 ≈ 0.0478. 104 ⋅ 103 ⎛ 104 ⎞ ⎜2 ⎟ 2 ⎝ ⎠ Drawing an ace and a 10 card from one deck has the greater probability. [11.7]
Copyright © Houghton Mifflin Company. All rights reserved.
4 36
= 1. 9
812
82.
83.
84.
Chapter 11: Sequences, Series, and Probability
Probability = (probability of 2)(probability of 1) [11.7] + (probability of 3)(probability of 1 or 2) + (probability of 4)(probability of 1 or 2 or 3) + (probability of 5) 1 = ⋅1 +1⋅2+1⋅3+1⋅4 5 4 5 4 5 4 5 4 = 1 + 2 + 3 + 4 = 10 = 1 20 20 20 20 20 2 ⎛12 ⎞ ⎛11⎞ There are ⎜ ⎟ ways of choosing 3 people from 12. There are ⎜ ⎟ ⋅ 1 ways of choosing 2 people and the person with [11.7] 3 ⎝ ⎠ ⎝2 ⎠ badge number 6. ⎛11⎞ 11 ⋅ 10 ⎜⎜ ⎟⎟ ⋅ 1 2⎠ 1 ⎝ 2 Probability = = = 12 ⋅ 11 ⋅ 10 4 ⎛12 ⎞ ⎜⎜ ⎟⎟ 3⋅ 2 ⎝3 ⎠ Stock value =
D (1 + g ) 1.27(1 + 0.03) = ≈ $14.53 [11.3] 0.12 − 0.03 i−g
85.
Using the multiplier effect, [11.3] 15 = 75 1 − 0.80 The net effect of $15 million is $75 million.
....................................................... i ⎤, i = r QR1. PMT = L ⎡ ⎢ 1 − (1 + i ) − nt ⎥ n ⎣ ⎦ 0.09 ⎡ ⎤ ⎢ ⎥ 12 = 12,000 ⎢ −12(5) ⎥ 0.09 ⎢1 − 1 + ⎥ 12 ⎣ ⎦ ≈ $249.10
(
)
Quantitative Reasoning i ⎤, i = r QR2. PMT = L ⎡ ⎢ 1 − (1 + i ) − nt ⎥ n ⎣ ⎦ 0.06 ⎡ ⎤ ⎢ ⎥ 12 = 18,000 ⎢ −12(4) ⎥ 0.06 ⎢1 − 1 + ⎥ 12 ⎣ ⎦ ≈ $422.73
(
)
i ⎤, i = r QR3. PMT = L ⎡ ⎢ 1 − (1 + i ) − nt ⎥ n ⎣ ⎦ 0.085 ⎡ ⎤ ⎢ ⎥ 12 = 15,000 ⎢ −12(5) ⎥ 0.085 ⎢1 − 1 + ⎥ 12 ⎣ ⎦ ≈ $307.75 Total payments made over 12 years: 307.75(12)(5) = 18,465 Total interest paid 18,465 − 15,000 = $3465 .
(
)
....................................................... 1.
3 a3 = 2 = 8 = 4 3! 6 3 5 a5 = 2 = 32 = 4 5! 120 15
[11.1]
Chapter Test 2.
a2 = 2 ⋅ a1 = 2 ⋅ 3 = 6 [11.1]
a3 = 2 ⋅ a2 = 12 • a4 = 2 ⋅ a3 = 24 a5 = 2 ⋅ a4 = 48 • Copyright © Houghton Mifflin Company. All rights reserved.
Chapter Test
3.
813
an +1 − an = [−2(n + 1) + 3] − (−2n + 3) [11.3]
an +1 − an = 2(n + 1)2 − 2n 2
4.
= −2n − 2 + 3 + 2n − 3 = −2 = constant
= 4n + 2 ≠ constant an +1 2(n + 1)2 = an 2n 2 2 1 = 1+ + ≠ constant n n2 neither
arithmetic
5.
( −1)n +1−1 n +1 an +1 −1 = 3 = = constant [11.3] an 3 ( −1)n −1 3n
geometric 6.
6
120 + 60 + 40 + 30 + 24 + 20 ∑ 1i = 1 + 12 + 13 + 14 + 15 + 16 = 120 120 120 120 120 120
[11.1]
i =1
= 294 = 49 120 20
7.
8.
()
1 ⎛1 − 1 ⎜⎜ 2 1 = 1 + 1 + 1 + 1 +L 1 = 2 ⎝ ∑ 2 j 2 4 8 16 1024 1− 1 j =1 2 1 1023 =1− = 1024 1024 10
10 ⎞
⎟⎟ 10 ⎠ =1− ⎛ 1 ⎞ ⎜ ⎟ ⎝ 2 ⎠ [11.3]
20
(1 + 58) = 10 ( 59 ) ∑ ( 3k − 2 ) = 1 + 4 + 7 + 10 + L + 58 = 20 2
k =1
[11.2]
= 590
9.
a3 = a1 + ( 3 − 1) d = 7,
a1 + 2d = 7
a8 = a1 + ( 8 − 1) d = 22
a1 + 7d = 22 − 5d = −15 d =3
10.
3 3 k ⎛ 3 ⎞ = a1 = 8 = 8 = 3 [11.3] ∑⎜ ⎟ 1− r 5 5 k =1 ⎝ 8 ⎠ 1 − ⎛⎜ 3 ⎞⎟ ⎝8⎠ 8
11.
0.15 = 0.15 + 0.0015 + 0.000015 + ... =
a1 = a3 − 2 ( 3) =7−6 =1
a20 = a1 + ( 20 − 1) d [11.2] = 1 + (19 )( 3) = 58
∞
0.15 0.15 15 5 [11.3] = = = 1 − 0.01 0.99 99 33
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[11.1]
814
12.
Chapter 11: Sequences, Series, and Probability
1.
1(1 − 3(1)) = −1 [11.4] 2 Thus the statement is true for n = 1. Let n = 1. 2 − 3(1) = −1 k
2.
Assume
∑( 2 − 3i ) =
k (1 − 3k ) 2
i =1 k +1
Verify
is true for some positive number k .
∑ (2 − 3i) = (k + 1)[1 −2 3(k + 1)] = (k + 1)(12− 3k − 3) = (k + 1)(−23k − 2) = − (k + 1)(32 k + 2) i =1
k (1 − 3k ) k (1 − 3k ) + [2 − 3(k + 1)] = + (−3k − 1) 2 2 =
k − 3k 2 − 6k − 2 2
(3k 2 + 5k + 2) 2 (k + 1)(3k + 2) =− 2 Thus the formula has been established by the extended principle of mathematical induction. =−
13.
1.
Let n = 7 [11.4] 7!= 50,407 37 = 2187
Thus n!> 3n for n = 7. 2.
Assume k!> 3k Verify (k + 1)! > 3k +1 k ! > 3k k +1 > 3 (k + 1)k ! > 3 ⋅ 3k (k + 1)! > 3k +1 Thus the formula has been established by the extended principle of mathematical induction.
14.
( x − 2 y )5 = x5 − 5( x)4 (2 y ) + 10( x)3 (2 y )2 − 10( x)2 (2 y )3 + 5( x)(2 y )4 − (2 y )5 = x5 − 10 x 4 y + 40 x3 y 2 − 80 x 2 y3 + 80 xy 4 − 32 y5
15.
(a + b)6 = a 6 + 6a5b + 15a 4b 2 + 20a3b3 + 15a 2b 4 + 6ab5 + b6 [11.5] 6
2
1⎞ ⎛ 6 5⎛ 1 ⎞ 4⎛1⎞ ⎜ x + ⎟ = x + 6( x) ⎜ ⎟ + 15( x) ⎜ ⎟ x x ⎝ ⎠ ⎝ ⎠ ⎝ x⎠ 15 = x6 + 6 x 4 + 15 x 2 + 20 + + x2 16.
18.
[11.5]
3
4
5
⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ + 20( x)3 ⎜ ⎟ + 15( x)2 ⎜ ⎟ + 6 x ⎜ ⎟ + ⎜ ⎟ x x ⎝ ⎠ ⎝ ⎠ ⎝ x⎠ ⎝ x⎠ 6 1 + x 4 x6
⎛ 8 ⎞ 3 6 −1 [11.5] 6th term of (3 x + 2 y )8 = ⎜ ⎟ (3 x) (2 y ) ⎝ 6 − 1⎠ ⎛8⎞ = ⎜ ⎟ (3x)3 (2 y )5 ⎝ 5⎠ = 56 ⋅ 27 x3 ⋅ 32 y 5 = 48,384 x3 y 5 26 ⋅ 25 ⋅ 24 ⋅ 9 ⋅ 8 ⋅ 23 ⋅ 22 = 568,339,200 [11.6]
6
17.
52 ⋅ 51 ⋅ 50 = 132,600 [11.6]
19.
C (8, 3)C (10, 2) 56 ⋅ 45 5 = = ≈ 0.294118 [11.7] C (18, 5) 8568 17
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Cumulative Review
20.
Stock value =
815
D (1 + g ) 0.86(1 + 0.06) = ≈ $10.13 [11.3] i−g 0.15 − 0.06
....................................................... 1.
Cumulative Review 2.
5+
x = −3 [2.3] 3 x = −8 3 x = −24
y = 1.7x + 3.6 [2.7] 3.
2 x 2 − 3 x = 4 [1.3] 2 x − 3x − 4 = 0
4.
⎛ xy 2 ⎞ logb ⎜ 3 ⎟ = logb x + logb y 2 − logb z 3 [4.4] ⎜ z ⎟ ⎝ ⎠ = logb x + 2logb y − 3logb z
6.
(1) ⎧2 x − 3 y = 8 ⎨ ⎩ x + 4 y = −7 (2)
2
3 ± (−3) 2 − 4(2)(−4) 3 ± 9 + 32 = 2(2) 4 ± 3 41 = 4
x=
5.
16 x 2 + 25 y 2 − 96 x + 100 y − 156 = 0 16( x 2 − 6 x) + 25( y 2 + 4 y ) = 156 16( x − 3) 2 + 25( y + 2) 2 = 156 + 144 + 100 16( x − 3) 2 + 25( y + 2) 2 = 400 16( x − 3) 2 25( y + 2) 2 + =1 400 400 ( x − 3)2 ( y + 2) 2 + =1 25 16
Solve (2) for x and substitute into (1). x = −4 y − 7 2(−4 y − 7) − 3 y = 8 −8 y − 14 − 3 y = 8 −11y = 22 y = −2 x = −4(−2) − 7 = 1 The solution is (1, –2). [9.1]
a 2 = 25, b 2 = 16 c 2 = a 2 − b 2 = 25 − 16 = 9 c=3 e = c = 3 [5.2] a 5
7.
⎡ −1 3 A − 2 B = 3 ⎢⎢ 5 ⎢⎣ 0 ⎡ −3 = ⎢⎢15 ⎣⎢ 0
9.
y = 0 [3.5]
2⎤ ⎡7 −3⎤ ⎥ 3 ⎥ − 2 ⎢⎢ 6 5 ⎥⎥ [10.2] ⎢⎣1 −2 ⎥⎦ 3 ⎥⎦ 6 ⎤ ⎡14 −6 ⎤ ⎡ −17 12 ⎤ 9 ⎥⎥ − ⎢⎢12 10 ⎥⎥ = ⎢⎢ 3 −1⎥⎥ 9 ⎦⎥ ⎣⎢ 2 −4 ⎦⎥ ⎣⎢ −2 13 ⎥⎦
8.
⎛h⎞ h(−3) −3 − 2 = = −5 [2.6] ⎜ ⎟ (−3) = g (−3) (−3)2 − (−3) + 4 16 ⎝g⎠
10.
log1/ 2 64 = −6 [4.3]
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816
11.
Chapter 11: Sequences, Series, and Probability
42 x +1 = 3x − 2 [4.5] ln 42 x +1 = ln 3x − 2 (2 x + 1)ln 4 = ( x − 2)ln 3 2 x ln 4 + ln 4 = x ln 3 − 2ln 3 x ln 42 − x ln 3 = −2ln 3 − ln 4 x (ln16 − ln 3) = −2ln 3 − ln 4 x = −2ln 3 − ln 4 ≈ −2.1 ln16 − ln 3
12.
⎧⎪ x 2 + y 2 + xy = 10 (1) ⎨ x − y = 1 (2) ⎪⎩ Solve (2) for x and substitute into (1). x = y +1 ( y + 1)2 + y 2 + ( y + 1) y = 10 y + 2 y + 1 + y 2 + y 2 + y = 10 3y2 + 3y − 9 = 0 3( y 2 + y − 3) = 0 2
y=
−1 ± 12 − 4(1)( −3) −1 ± 1 + 12 −1 ± 13 = = 2(1) 2 2
x = −1 ± 13 + 1 = 1 ± 13 2 2 ⎛ 1 + 13 −1 + 13 ⎞ The solutions are ⎜ , ⎟ and 2 ⎝ 2 ⎠
⎛ 1 − 13 −1 − 13 ⎞ , ⎜ ⎟ . [9.3] 2 ⎝ 2 ⎠ 13.
⎡3 2⎤ ⎡ 2 9 11 −3⎤ ⎢ −2 1 ⎥ ⎡ 2 3 1 1 ⎤ = ⎢ −6 −6 2 −5⎥ [10.2] ⎢ ⎥ ⎢ −2 0 4 −3⎥ ⎢ ⎥ ⎦ ⎢10 3 −15 13 ⎥ ⎢⎣ 1 −4 ⎥⎦ ⎣ ⎣ ⎦
14.
Let t = 5, [4.5] 5 = − 175 ln ⎛⎜1 − v ⎞⎟ 32 ⎝ 175 ⎠ ⎛ ⎞ ⎛ v ⎞ 32 5⎜ − ⎟ = ln ⎜ 1 − ⎟ ⎝ 175 ⎠ ⎝ 175 ⎠ −32 / 35 v e =1− 175 −175 e −32 / 35 − 1 = v
(
15.
17.
)
v ≈ 105 mph
opp −1 = [5.3] hyp 2 hyp = 2 secθ = 1 = 2 3 = cos θ 3 adj 3 adj = 3 =− 3 cot θ = 1 = tan θ opp −1
16.
sin x + 1 + cos x = sin x(1 − cos x) + 1 + cos x 1 + cos x sin x sin x 1 − cos 2 x sin x(1 − cos x) 1 + cos x = + sin x sin 2 x 1 cos 1 cos − x x = + + sin x sin x sin x = 1 + 1 sin x sin x = 2csc x
C = 180o − 40o − 65o = 75o [7.1] a = c sin A sin C o a = 20sin 40 ≈ 13 cm o sin 75 b = c sin B sin C o b = 20sin 65 ≈ 19 cm o sin 75
18.
A = 9, B = 4, C = 6 [8.4]
sin θ =
cot 2α = A − C = 9 − 6 = 3 B 4 4 2α ≈ 54o
α ≈ 27o
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[6.1]
Cumulative Review
19.
(
sin 1 cos −1 4 2 5 Let
817
)
[6.5]
θ = cos −1 4 cosθ = 4 5
20.
2 v − 3w = 2(2i + 5 j) − 3(3i − 6 j) = 4i + 10 j − 9i + 18 j = −5i + 28 j
5
1− 4 5 sin 1 θ = 1 − cos θ = 2 2 2 = 5−4 = 1 10 10 10 = 10
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[7.3]
Responses to Projects in the Text
819
Chapter P
College Algebra and Trigonometry
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Preliminary Concepts CHAPTER P of College Algebra and Trigonometry P.1 The Real Number System 1. Number Puzzle n is 1 less than a multiple of 6 and also 1 less than a multiple of 5 and also 1 less than a multiple of 4 and also 1 less than a multiple of 3 and also 1 less than a multiple of 2. The smallest value of n is 1 less than the least common multiple of 6, 5, 4, 3, and 2. Thus n is 1 less than 60, or n = 59. 2. Operations on Intervals a. (−4)2 = 16, 02 = 0, 22 = 4 . The square of every number in the interval (−4, 2) is (−4, 2)2 = [0, 16) . b. |4| = 4, |0| = 0, |5| = 5. The absolute value of every number in the interval (−4, 5) is ABS(−4, 5) = [0, 5). c. 0 = 0, 9 = 3 . The square root of every number in the interval (0, 9) is (0,9) = (0,3) d. The reciprocal of every number in the interval (0, 1) is (1,∞ ) . 3.
Factors of a Number Whenever you square a natural number, there is a repeated factor, so in listing the pairs of factors, that one would repeat, therefore the number of factors is an odd number. For example, squaring 6 gives us 36. The factors are 1 × 36, 2 × 18, 3 × 12,4 × 9, and 6 × 6. We do not count the repeated 6 twice, so there are an odd number of factors.
P.2 Integer and Rational Number Exponents ⎡ ⎤ ⎡ ⎤ 1 1 2 ⎢ ⎥ − 1 mv 2 ⎥ − 1 v2 1 1 mc 2 ⎢ − c − 2 2 ⎢ 1 − v2 ⎥ 2 ⎢ 1 − v2 ⎥ 2 c c ⎣ ⎦ ⎣ ⎦ × 100 = × 100 1. Relativity Theory % error = ⎡ ⎤ ⎡ ⎤ 1 1 − 1⎥ − 1⎥ mc 2 ⎢ c2 ⎢ ⎢ 1 − v22 ⎥ ⎢ 1 − v22 ⎥ c c ⎣ ⎦ ⎣ ⎦ a. For v = 30 meters per second, % error = 0.000303176 b. For v = 240 meters per second, % error = 1.75483 × 10−6 c. For v = 3 × 107 meters per second, % error = 0.750628 d. For v = 1.5 × 108 meters per second, % error = 19.1987 e. For v = 2.7 × 108 meters per second, % error = 68.0755 f. The percent errors is very small for everyday speeds. |m −m| m0 × 100 where m = and v = 0.99 c . % change is approximately 609%. g. % change = 0 1− v 2 m0 2 c
h. As the speed of the object approaches the speed of light, the denominator of kinetic energy equation approaches 0, which implies that the kinetic energy is approaching infinity. Thus it would require an infinite amount of energy to move a particle at the speed of light. P.3 Polynomials 1. Odd Numbers An even number is a number that is a multiple of 2. Let m and n be natural numbers. Then 2m and 2n are even natural numbers. The product ( 2m )( 2n ) = 4mn = 2 ( 2mn ) , which is an even number (it is a multiple of 2). Therefore, an even number times an even number is an even number. An odd number is a number that is not a multiple of 2. Let m and n be natural numbers. Then 2m + 1 and 2n + 1 are odd numbers because they are not multiples of 2. The product ( 2m + 1)( 2n + 1) = 4mn + 2 ( m + n ) + 1 is not a multiple of 2 (2 is not a common factor) and is therefore an odd number. Therefore, the product of two odd numbers is an odd number. Consider the product 2m ( 2n + 1) = 4mn + 2m = 2 ( 2mn + m ) . Because the product is a multiple of 2, it is an even number. Thus the product of an odd number and an even number is an even number.
820
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Chapter P 2.
College Algebra and Trigonometry
Responses to Projects
Prime Numbers a. The contrapositive of “If A, then B” is “If not B, then not A.” The contrapositive of “If two triangles are congruent, then they are similar” is “If two triangles are not similar, then they are not congruent.” b. Fermat’s Little Theorem does not say anything about n when n is not a prime number. c. The converse of “If A, then B” is “If B, then A.” The converse of Fermat’s Little Theorem is “If a is any natural number and a n − a is divisible by n, then n is a prime number.” d. No. In fact, 561 is not a prime number. The converse of Fermat’s Little Theorem is false. e. No. See part (d). f. A Carmichael number n is a number for which a n − a is divisible by every natural number a and n is not prime. The first three Carmichael numbers are 561 , 1105 , and 1729 . Any Carmichael number can be used to show that the converse of Fermat’s Little Theorem is false. Carmichael numbers are sometimes called pseudoprimes.
P.4 Factoring 1. Geometry a. I + II + III b. II + III + V c. Because the area of I is the same as the area of V, the sum of the areas of region I, II, and III equals the sum of the areas of regions II, III, and V. 2. Geometry
( x + y) 3.
2
= x 2 + 2 xy + y 2
Geometry x 3 − y 3 = x ( x − y ) + xy ( x − y ) + xy ( x − y ) + y 2 ( x − y ) 2
= ( x − y ) ( x ( x − y ) + 2 xy + y 2 ) = ( x − y ) ( x 2 + xy + y 2 )
P.5 Rational Expressions 1. Continued Fractions 1 1 1 = = = 0.6 a. C2 = 1 + 1+11 1 + 12 1.5 b.
c.
C3 =
C5 =
1 1+
1 1 + 1+11
1 1 + 12
1+
=
1 1 3 = 5 = = 0.6 2 1+ 3 3 5 1
=
1 1 1 1+ 1+1
1
1+ 1+
1
=
1
1+
1 1+
2.
1+
1 1+
d.
1
=
1 1 + 12
1
1+ 1+
1 1 + 23
=
1 1 1+ 1 + 35
=
1 8 = 5 1 + 8 13
8 −1 + 5 ≈ 0.615 ≈ 0.618 , C5 = 13 2
Representation of π An excellent source for π and its history can be found in an article by Dario Castellanos in Mathematics Magazine, vol. 61, no. 2 (April 1988). Here are two results (both attributable to Euler) from that article. 2 ⎛ ⎞ ⎜1 + ⎟ 1⋅ 3 12 + 3 ⎜ ⎟ = and π 2 π = 3+ 3⋅5 ⎟ 32 ⎜ + 4 6+ 2 7 ⎟ ⎜ 4 + 45+⋅L 6 + 65+L ⎝ ⎠ Copyright © Houghton Mifflin Company. All rights reserved.
821
Chapter P
College Algebra and Trigonometry
Responses to Projects
P.6 Complex Numbers 1. – 8.
9.
2 + 5i = 22 + 52 = 4 + 25 = 29
10. 4 − 3i = 42 + (−3)2 = 16 + 9 = 25 =5
11. −2 + 6i = (−2)2 + 62 = 4 + 36 = 40 = 4 ⋅10 = 2 10
12. −3 − 5i = (−3)2 + (−5)2 = 9 + 25 = 34 13.
a + bi = (a )2 + (b)2 = a 2 + b 2
−a − bi = (− a) 2 + (−b)2 = a 2 + b 2 14. A complex number and its additive inverse are the same distance from the origin in the complex plane. The real parts of a complex number and its additive inverse are the same distance from the imaginary axis but on opposite sides of the imaginary axis. The imaginary parts of a complex number and its additive inverse are the same distance from the real axis but on opposite sides of the real axis.
822
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Equations and Inequalities CHAPTER 1 of College Algebra and Trigonometry 1.1 Linear Equations 1. Perfect Games a. The probability that one of the teams will get a perfect game is 0.7 27 . Because every game involves exactly two teams, the probability of a perfect game in any one game is 2 ( 0.7 27 ) . Therefore, the number of perfect games we can expect in x games is p = 2 ( 0.7 27 ) x.
b. A baseball almanac shows that during the years 1962 to 2002, there were 10 perfect games. It is difficult to determine the exact number of games played in this period, but it is about 71,000. c. From part (a), we should expect the number of perfect games during the period 1962 to 2002 to be p = 2 ( 0.7 27 ) 71, 000 ≈ 9.3 . This result is very close to the actual result found in part (b). 1.2 Formulas and Applications 1. A Work Problem and Its Extensions a.
1 . A 1 If a pump can fill a pool in B hours, then the part of the pool it fills every hour is . B Let T be the total time it takes the pumps to fill the pool when they both work together. T ( 1A ) = the part of the pool filled by pump A. T ( B1 ) = the part of the pool filled by pump B. Because
If a pump can fill a pool in A hours, then the part of the pool it fills every hour is
T ⎛⎜ 1 ⎞⎟ + T ⎛⎜ 1 ⎞⎟ = 1 ⎝ A⎠ ⎝B⎠ ⎛ ⎞ ⎛ ⎞ 1 1 T ⎜ ⎟ AB + T ⎜ ⎟ AB = 1 ⋅ AB ⎝ A⎠ ⎝B⎠ TB + TA = AB T ( B + A ) = AB T = AB A+ B
we fill exactly 1 pool, we have
b. Using the procedure in part (a) yields T =
c.
2.
ABC AB + AC + BC
Observe that T is the product of the individual times divided by the sum of the products of the times taken two at a time. A1 A2 A3 L An T= L + L A A A A A A A A ( 2 3 4 n ) ( 1 3 4 n ) + ( A1 A2 A4 L An )L + ( A1 A2 A3 L An −1 )
That is, T is given by the product of the A’s divided by the sum of products of the A’s taken (n – 1) at a time. d. One method is as follows: Use the alignment chart to determine that together the pump that can fill it in 6 hours working with the pump that can fill it in 12 hours would take a total of 4 hours. Think of these two pumps as one 4-hour pump. Now use the alignment chart again, using 4 hours and 6 hours as the individual times for two pumps to produce the final answer of 2.7 hours (nearest 0.1 hour). Resistance of Parallel Circuits In electronics it can be shown that if two resistors (one with resistance R1 ohms and the other with resistance R2 ohms) are placed in parallel, the total resistance R provided by the two resistors is R=
R1R2 R1 + R2
This formula has the same form as the formula give in part 1(a). Although the setting is different, the mathematics needed to solve a parallel resistance problem is exactly the same as that used to solve a combined work problem. The parallel resistance problem can also be extended to consider more than two resistors, in a manner analogous to the work problems in 1(b) and 1(c).
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1.3 Quadratic Equations 1. The Sum and Product of the Roots Theorem The quadratic equation ax 2 + bx + c = 0 has roots r1 =
−b + b 2 − 4ac −b − b 2 − 4ac and r2 = 2a 2a
The sum of the roots is r1 + r2 =
b −b + b 2 − 4ac −b − b 2 − 4ac −2b + = =− 2a 2z 2a a
The product of the roots is ⎛ −b + b 2 − 4ac ⎞ ⎛ −b − b 2 − 4ac ⎞ b 2 − ( b 2 − 4ac ) b 2 − b 2 + 4ac c ⎟⎜ ⎟= = = r1r2 = ⎜ ⎜ ⎟⎜ ⎟ 2a 2a 4a 2 4a 2 a ⎝ ⎠⎝ ⎠ Visual Insight The reasons for the steps in President Garfield’s proof of the Pythagorean Theorem are as follows: i) The area of the large triangle plus the area of the two small triangles is equal to the total area of the region, which is a trapezoid. ii) The height of the trapezoid is a + c. The sum of the bases of the trapezoid is also a + b. iii) Expand the right side and subtract ab from each side of the equation. iv) Multiply each side of the equation by 2. 2
2.
1.4 Other Types of Equations 1. The Reduced Cubic a.
Given the reduced cubic x 3 + mx + n = 0. Let x =
m − z. 3z
3
⎛m ⎞ ⎛m ⎞ ⎜ − z ⎟ + m⎜ − z ⎟ + n = 0 ⎝ 3z ⎠ ⎝ 3z ⎠ 3 2 2 4 6 2 2 m − 9m z + 27 mz − 27 z m − 3mz + +n =0 3 27 z 3z m3 − 9m 2 z 2 + 27mz 4 − 27 z 6 + 9m 2 z 2 − 27 mz 4 + 27 nz 3 =0 27 z 3 m3 − 9m 2 z 2 + 27mz 4 − 27 z 6 + 9m 2 z 2 − 27 mz 4 + 27 nz 3 = 0 −27 z 6 + 27 z 3 n + m3 = 0 z 6 − nz 3 −
m3 =0 27
m3 =0. 27 At this point Francois Vieta knew he could solve the original reduced cubic in part (a), because he knew he could use the quadratic formula, which had been around for centuries, to solve the foregoing quadratic. The work follows in part (c).
b. Let u = z 3 and u 2 = z 6 . Then the last equation in part (a) can be written as u 2 − nu −
n ± n 2 + 427m
3
c.
z =u= 3
2
=
n 1 2 4m 3 n n 2 m3 ± n + = ± + 2 2 27 2 4 27
If we use the positive root in this equation, we can show that z =
824
3
n n 2 m3 + + . 2 4 27
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d. Rewrite the reduced cubic equation x 3 + 3 x = 14 as x 3 + 3 x − 14 = 0. Thus we have a reduced cubic equation with m = 3 and n = −14. Substituting for z in the equation from part (c) gives z=
3
−14 + 2
It is easy to verify that
3
( −14 )
+
33 = 3 −7 + 49 + 1 = 3 −7 + 5 2 = 2 − 1 27
−7 + 5 2 = 2 − 1. Just show that
more difficult to determine that 3
4
2
−7 + 5 2 can be expressed as
3
(
)
3
2 − 1 = −7 + 5 2. It is somewhat
−7 + 5 2 = 2 − 1. The motivating idea is first to suspect that 2 + k for some real constant k and then solve
for k. Hence the real solution of x + 3 x = 14 is 3 m x= −z= − 3z 3 2 −1
(
2+k
)
3
= −7 + 5 2
3
(
2.
)
(
)
2 −1 =
1 2 −1
−
(
)
2 −1 = 2 .
Fermat’s Last Theorem The History of Fermat’s Last Theorem The conjecture that x n + y n = z n is impossible for all integers n > 2 is known as Fermat’s Last Theorem. The actual statement of the conjecture was given by Pierre de Fermat in 1637. Fermat wrote the conjecture as a paragraph in the margin of the text The Arithmetic of Diophantus.
It is impossible to write a cubic as the sum of two cubes, a fourth power as the sum of two fourth powers, and in general any power beyond the second as the sum of two similar powers. For this I have discovered a truly wonderful proof but the margin of this book is too small to contain it. Many mathematicians have tried to find either a proof of Fermat’s Last Theorem or a counterexample to disprove the result. In 1780 Leonhard Euler proved the theorem for n = 3. Other mathematicians proved it true for n = 5, n = 7, and n = 13. Before June of 1993, the theorem had been established for 2 < n < 25, 000 . The Relationship Between Fermat’s Last Theorem and the Pythagorean Theorem If we remove the restriction that n must be larger than 2 and consider the equation x 2 + y 2 = z 2 , then we can find an infinite number of solutions. For example, we know that x = 3, y = 4, and z = 5 are solutions
of this equation, as are 3k, 4k, and 5k for any positive integer k. Because x 2 + y 2 = z 2 is in the form of the Pythagorean Theorem, positive-integer solutions of this equation are called Pythagorean triples. Dr. Andrew Wile’s Proof of Fermat’s Last Theorem In June of 1993, Andrew Wiles of Princeton University announced that he had produced a proof of Fermat’s Last Theorem. At first it appeared that he had, in fact, written a proof of Fermat’s Last Theorem. However, an error was soon discovered. At this time Andrew Wiles was extremely disappointed. He had spent over 7 years working on his proof. He had even given a presentation at Cambridge University in which he outlined his proof to his peers. At first Wiles could not find a way to repair his proof, but eventually, after an additional year of work, and with the assistance of the mathematician Richard Taylor, a valid proof of Fermat’s Last Theorem was achieved. Additional information about Wiles’ proof of Fermat’s Last Theorem can be found on the NOVA ONLINE internet site Solving Fermat: Andrew Wiles at: http://www.pbs.org/wgbh/nova/proof/wiles.html
Andrew Wiles’ proof consists of a 150-page document. It makes use of 20th century mathematics that was not available to Fermat. Thus we are certain that Wiles’ proof is not the same as the proof that Fermat indicated he had produced. It is interesting to note that on the NOVA ONLINE internet site mentioned above, Andrew Wiles states “I don’t believe that Fermat had a proof.”
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1.5 Inequalities 1. Triangles a. x + x + 5 > x + 9, ⇒ x > 4 b. x + x 2 + x > 2 x 2 + x − x2 + x > 0 − x ( x − 1) > 0 − x ( x − 1) | + + + | − − − − − x 0 1 x is between 0 and 1.
c.
1 1 1 + > x + 2 x +1 x 1 1 1 + − >0 x + 2 x +1 x x ( x + 1) + x ( x + 2 ) − ( x + 2 )( x + 1) >0 x ( x + 1)( x + 2 ) x 2 + x + x 2 + 2 x − x 2 − 3x − 2 >0 x ( x + 1)( x + 2 ) x2 − 2 >0 x ( x + 1)( x + 2 ) x2 − 2 > 0 and x > 0 x ( x + 1)( x + 2 )
|− − − − − −| + + x 0 2 The solution of the above inequalities is given by x > 2 .
2.
Fair Coins t − 500 ≤ 2.33 if and only if a. 15.81 t − 500 ≤ 2.33 15.81 −36.8373 ≤ t − 500 ≤ 36.8373 −2.33 ≤
463.1627 ≤
t
≤ 536.8373
Because t must be a nonnegative integer, we have 464 ≤ t ≤ 536. Therefore, according to this definition, a coin will be considered a fair coin if, in 1000 flips of the coin, the number of tails is greater than or equal to 464, but less than or equal to 536. b. Answers will vary.
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1.6 Variation and Applications 1.
A Direct Variation Formula
Given f ( x) = k x , prove f ( x2 ) = f ( x1 )
x1 . x2
Proof: f ( x1 ) = k x1 and f ( x2 ) = k x2 , so f ( x2 ) f ( x1 )
f ( x2 ) f ( x1 )
=
k x2 k x1
=
x2 x1
f ( x2 ) = f ( x1 )
x2 x1
Let f1 = 17 kg , f 2 = 22 kg , and d ( f1 ) = 8.5 centimeters. Then d ( f 2 ) = d ( f1 )
f2 f1
22 17 d ( 22 ) = 11 centimeters d ( 22 ) = 8.5 ⋅
2.
An Inverse Variation Formula x k Given f ( x ) = , prove f ( x2 ) = f ( x1 ) 1 . x x2
Proof: f ( x ) = The ratio
k k k , so f ( x2 ) = and f ( x1 ) = . x x2 x1
f ( x2 ) f ( x1 )
is given by f ( x2 ) f ( x1 )
f ( x2 ) = f ( x1 )
Thus f ( x2 ) = f ( x1 )
k x2 k x1
=
k x2 k x1
= f ( x1 )
k x1 ⋅ x2 k
x1 . x2
Let x1 = 280 , x2 = 330 , and V ( 280 ) = 2.4 . V ( x2 ) = V ( x1 )
x1 x2
V ( 330 ) = V ( 280 )
280 280 = ( 2.4 ) ≈ 2.0 liters 330 330
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827
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Functions and Graphs CHAPTER 2 of College Algebra and Trigonometry 2.1 A Two-Dimensional Coordinate System and Graphs 1. Verify a Geometric Theorem Consider any right triangle ABC. Place the triangle in a coordinate system with the vertex point of its right angle at the origin and its legs on the x-axis and the y-axis as shown on page 177 of the text. B ( 0, b )
By the midpoint formula, the coordinates of point M are ⎛ a+0 0+b⎞ ⎛ a b⎞ , ⎜ ⎟=⎜ , ⎟ 2 ⎠ ⎝ 2 2⎠ ⎝ 2
M C ( 0, 0 )
A ( a, 0 ) 2
2
2
2
2
2
2
2
2
2
a2 + b2 2
2
a 2 + b2 2
d ( A, M ) = ⎛⎜ a − a ⎞⎟ + ⎛⎜ 0 − b ⎞⎟ = ⎛⎜ a ⎞⎟ + ⎛⎜ − b ⎞⎟ = ⎝ ⎝ 2⎠ ⎝ 2⎠ 2⎠ ⎝ 2⎠ d ( B, M ) = ⎛⎜ 0 − a ⎞⎟ + ⎛⎜ b − b ⎞⎟ = ⎛⎜ − a ⎞⎟ + ⎛⎜ b ⎞⎟ = ⎝ ⎝ 2⎠ ⎝2⎠ 2⎠ ⎝ 2⎠
2.
2
2 2 d ( C , M ) = ⎛⎜ 0 − a ⎞⎟ + ⎛⎜ 0 − b ⎞⎟ = ⎛⎜ − a ⎞⎟ + ⎛⎜ − b ⎞⎟ = a + b ⎝ ⎝ 2⎠ ⎝ 2⎠ 2⎠ ⎝ 2⎠ 2 Thus the midpoint of the hypotenuse of any right triangle is equidistant from each of the vertices of the triangle. Solve a Quadratic Equation Geometrically a. See the following figure.
Q
T a
b
E
a
A
a
S
B
P b
C
b. Using the figure from part (a), we see that d ( Q, B ) = a + d ( A, B ) = a + a 2 + b 2
Let x = a + a 2 + b 2 in the equation x 2 = 2ax + b 2 to show that both sides equal the expression 2a 2 + 2a a 2 + b 2 + b 2 . Thus d ( Q, B ) is a solution of x 2 = 2ax + b 2 .
c.
Using the figure from part (a), we see that d ( P, B ) = d ( A, B ) − d ( A, P ) = a 2 + b 2 − a
Let x = a 2 + b 2 − a in the equation x 2 = −2ax + b 2 to show that both sides equal the expression 2a 2 − 2a a 2 + b 2 + b 2 . Thus d ( P, B ) is a solution of x 2 = −2ax + b 2 .
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d. Because d (T , E ) = a 2 − b 2 and d ( E , B ) = a, we know that d (T , B ) = a + a 2 − b 2 . Let x = a + a 2 − b 2 in x 2 = 2ax − b 2 to show that each side simplifies to 2a 2 + 2a a 2 − b 2 − b 2 . Thus d (T , B ) is a solution of x 2 = 2ax − b 2 .
Also note that d ( S , B ) = a − d ( E , T ) = a − a 2 − b 2 . Let x = a − a 2 − b 2 in x 2 = 2ax − b 2 to show that each side simplifies to 2a 2 − 2a a 2 − b 2 − b 2 . Thus d ( S , B ) is a solution of x 2 = 2ax − b 2 . 2.2 Introduction to Functions 1. Day of Week a.
Let m = 10, d = 7, c = 19, and y = 41. Then 13m − 1 y c z = 5 + 4 + 4 + d + y − 2c 13 10 1 19 ⋅ − + 41 + + 7 + 41 − 2 ⋅19 = 5 4 4 = 25 + 10 + 4 + 7 + 41 − 38
= 49 The remainder of 49 divided by 7 is 0. Thus December 7, 1941, was a Sunday.
b. This one is tricky. Because we are finding a date in the month of January, we must use 11 for the month and we must use the previous year, which is 2009. Thus we let m = 11, d = 1, c = 19, and y = 109. Then 13m − 1 y c z= 5 + 4 + 4 + d + y − 2c 13 ⋅11 − 1 109 19 + = 5 4 + 4 + 1 + 109 − 2 ⋅19 = 28 + 27 + 4 + 1 + 109 − 38 = 131 The remainder of 131 divided by 7 is 5. Thus January 1, 2010, will be a Friday. c.
Let m = 5, d = 4, c = 17, and y = 76. Then 13m − 1 y c z = 5 + 4 + 4 + d + y − 2c 13 ⋅ 5 − 1 17 + 76 + + 4 + 76 − 2 ⋅ 17 = 5 4 4 = 12 + 19 + 4 + 4 + 76 − 34 = 81 The remainder of 81 divided by 7 is 4. Thus July 4, 1776 was a Thursday.
d. Answers will vary.
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2.3 Linear Functions 1. Visual Insight y G
x = x1
1 + m2
1 E C ( x1 , mx1 + b )
B
m
F
y = mx + b d
A ( x1 , y1 ) D
x
The distance d between the point A ( x1 , y1 ) and the line y = mx + b is d =
mx1 + b − y1 1 + m2
.
Triangle ABC is similar to triangle EFG. Thus d ( A, B )
d ( A, C )
=
d ( E, F )
• Corresponding sides of similar triangles are proportional.
d ( E, G )
d 1 = mx1 + b − y1 1 + m2
d=
2.
• Substitute for d ( A, B ) , d ( A, C ) , and d ( E , G ) .
mx1 + b − y1
• Solve for d, which is the distance from point A to the line.
1 + m2
Verify Geometric Theorems Place an arbitrary triangle in a coordinate system and label its vertices as shown. y
( b, c ) ( − a, 0 )
( a, 0 )
x
The coordinates of the endpoints of the line segment that connects the midpoints of two of the sides of the triangle are given by ⎛ −a + b 0 + c ⎞ ⎛ a+b 0+c⎞ and , , ⎜ ⎟ ⎜ ⎟ 2 2 2 ⎠ ⎝ ⎠ ⎝ 2 The slope of the line segment that connects the midpoints of two sides of the triangle is c c − 0 0 2 2 = = =0 a + b − a + b 22a a − 2 2 The slope of the third side of the triangle is also 0. Thus the two line segments are parallel.
830
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Chapter 2 3.
College Algebra and Trigonometry
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Verify Geometric Theorems Place an arbitrary square in a coordinate system and label its vertices as shown. y
( a, a )
( 0, a )
( 0, 0 )
x
( a, 0 )
The slope of the diagonal through the origin is a / a = 1. The slope of the other diagonal is a / ( −a ) = −1.
Applying the Parallel and Perpendicular Lines Theorem from Section 2.3 enables us to state that the diagonals are perpendicular.
) = ( , ) . The midpoint ) = ( , ) . Thus the midpoint of each diagonal is M = ( , ) .
The midpoint of the diagonal through the origin is the point of the other diagonal is the point
(
0+a,a+0 2 2
(
a+0,0+a 2 2
a a 2 2
a a 2 2
a a 2 2
Use the distance formula to determine that the distance from the midpoint M to each of the vertices is 2
2
2
2
⎛ a − 0⎞ + ⎛ a − a ⎞ = ⎛ a ⎞ + ⎛ − a ⎞ = a2 = a = 2 a ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ a 2 ⎝2 ⎠ ⎝2 ⎠ ⎝2⎠ ⎝ 2⎠ 2 Thus the diagonals of any square are perpendicular bisectors of each other.
2.4 Quadratic Functions 1. The Cubic Formula In the fifteenth century, the Italian mathematician Luca Pacioli (ca. 1445-1509) expressed his belief that cubic equations of the form x 3 + bx 2 + cx + d = 0 could not be solved, in general, by algebraic procedures involving radicals. His assertion challenged mathematicians to search for a “solution.” The mathematician Scipione del Ferro (1465-1526) was not able to find such a solution, however, he did discover a formula that solved “depressed cubic” equations of the form x 3 + mx = n. At that time del Ferro decided to keep his solution a secret. The reason for this decision was based on the fact that mathematicians of that time period occasionally faced challenges from other mathematicians. If someone challenged him with a set of problems to be solved, then he could challenge the person to solve a set of reduced cubic equations. Even if he did poorly on his problems, he was confident that the challenger would not be able to solve any of the depressed cubic equations. In such a situation, del Ferro would be considered the winner of the challenge. This is in sharp contrast to the present time where a mathematician’s reputation is based on his or her published works. Just prior to his death, del Ferro shared his solution with his student Antonio Fior (ca. 1506-?). In 1535 Fior challenged the scholar Niccolo Fontana (1499-1557), (also known as Tartaglia – the Stammerer) to solve a set of depressed cubic equations. The challenge problems that Fior received from Tartaglia concerned several different mathematical topics. Thus Tartaglia was in a difficult situation. If he could find the solution to depressed cubic equations, then he would be able to solve all of the challenge problems, but if he could not discover the solution to depressed cubics, then he would probably not be able to solve any of the challenge problems. Tartaglia began a desperate assault on the problem of finding the solution for depressed cubics. Many days passed and Tartaglia had not found the solution. However, Tartaglia was a talented mathematician and with continued effort he did discover the solution. With the solution in hand, it was easy for Tartaglia to solve all of the challenge problems. Fior was unable to solve all of the challenge problems he had received from Tartaglia. Thus Tartaglia was the undisputed winner of the challenge. Tartaglia had discovered the solution for the depressed cubic equation, but the solution of the general cubic equations x 3 + bx 2 + cx + d = 0 was still unknown. It was at this point that the Italian mathematician Gerolamo Cardano (1501-1576) contacted Tartaglia to learn about his wonderful solution for depressed cubics. Tartaglia was not eager to share his solution with Cardano, but Cardano was Copyright © Houghton Mifflin Company. All rights reserved.
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persistent and after a period of about 4 years, Tartaglia met with Cardano. Before Tartaglia divulged the solution, he made Cardano take an oath to never publish the solution. For many years Cardano kept Tartaglia’s solution a secret. Cardano even determined how to solve the general cubic equation x 3 + bx 2 + cx + d = 0, but the procedure consisted of using a substitution to write the general cubic equation in the form of a depressed cubic equation. Cardano was stymied. He could not publish his wondrous new solution because of the promise he had made to Tartaglia. Ludovico Ferrari, a protégé of Cardano’s even discovered how to solve general fourth degree polynomial equations, but his solution also depended upon writing the polynomial in the form of the depressed cubic equation and then applying Tartaglia’s solution. Then in 1543, Cardano and Ferrari happened to examine the mathematical papers of Scipione del Ferro. To their surprise, they found that del Ferro had in fact been the first to discover the solution of the depressed cubic equation and that he had left a written copy of his discovery. Cardano felt that he no longer needed to keep his oath to Tartaglia, because the original solution had been done by del Ferro and only rediscovered by Tartaglia. In 1545 Cardano published a mathematical manuscript that included the solution to the general cubic equation x 3 + bx 2 + cx + d = 0. Note that the above account is only a brief discussion of the events surrounding the development of the solution of the general cubic equation. The mathematical techniques used in solving a cubic equation are given in the response to Project 1, of Section 3.4. Simpson’s Rule The equation of the Parabola in the following figure is y = Ax 2 + Bx + C. y
P0 ( − h, y0 )
−h
P1 ( 0, y1 ) P2 ( h, y2 )
h
x
2 y0 = A ( −h ) + B ( − h ) + C = Ah 2 − Bh + C 2 y1 = A ( 0 ) + B ( 0 ) + C = C 2 y2 = A ( h ) + B ( h ) + C = Ah 2 + Bh + C
Thus y0 + 4 y1 + y2 = Ah 2 − Bh + C + 4C + Ah 2 + Bh + C = 2 Ah 2 + 6C. 2.5 Properties of Graphs 1. Dirichlet Function a. The domain of the Dirichlet function is the set of all real numbers. b. The range of the Dirichlet function is {0,1}. c. The Dirichlet function has an x-intercept at every point ( a, 0 ) where a is a rational number. d. The Dirichlet function has a y-intercept of ( 0, 0 ) .
832
e. f.
The Dirichlet function is an even function. Graphing calculators use only rational numbers. Thus a graphing calculator will fail to show any of the points ( b,1) where b is an irrational number.
g.
The Dirichlet function is said to be discontinuous at every point because between any two rational numbers we can find an irrational number, and between any two irrational numbers we can find a rational number. The graph of the Dirichlet function looks like to horizontal lines (one at y = 0 and one at y = 1), except that the lines are “full of holes.” Some mathematicians have called the Dirichlet function a shotgun function because the graph can be thought of as two horizontal lines in which we use a shotgun to blast holes. The holes on the line y = 1 occur at (a, 1) for all rational numbers a. The holes on the line y = 0 occur at (b, 0) for all irrational numbers b. Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 2 2.
4.
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Isolated Point
The point (1,1) is a solution of y =
3.
College Algebra and Trigonometry
( x − 1) ( x − 2 ) + 1 because 1 = ( (1) − 1) ( (1) − 2 ) + 1 . 2
2
Graphing utilities use only a finite number of domain values in the construction of a graph. Your graphing utility may have graphed the function for some values of x < 1 and for some values of x > 1, but not for x = 1. A Line with a Hole Graphing utilities use only a finite number of domain values in the construction of a graph. Thus your graphing utility may fail to show the proper result exactly at x = 2. Finding a Complete Graph To produce the part of the graph to the left of the y-axis, it may be necessary to enter the function as y = −3 x
5/3
−6 x
4/3
+ 2 for x ≤ 0
This procedure may be necessary because some graphing utilities do not evaluate fractional powers of negative numbers. You need y1 as shown in text and y2 as above with a condition at the end:
(
y2 = −3 x
5/3
−6 x
4/3
)
+ 2 ( x ≤ 0 ) (using TI-83)
2.6 The Algebra of Functions 1. A Graphing Utility Project a.
Let x = 5 and y = 9 . Then Maximum ( 5,9 ) =
b. Let x = 201 and y = 80 .
5 + 9 5 − 9 14 4 + = + = 7+2 = 9 . 2 2 2 2
201 + 80 201 − 80 281 121 + = + = 140.5 + 60.5 = 201 . 2 2 2 2 For each real number x ≥ 0, the graph of y3 is a graph whose range value is the maximum of y1 ( x )
Then Maximum ( 201,80 ) = c.
and y2 ( x ) . Thus for each x ≥ 0, y3 can be graphed by plotting points from the graph of y1 if its graph is higher than the graph of y2 , and by plotting points from the graph of y2 if it is the higher graph. d. The domain of y1 :{x x ∈ ℜ} (that is, all real numbers). The domain of y2 :{x x ≥ 0}. The domain of y3 :{x x > 0}. The domain of y3 is the intersection of the domain of y1 and the domain of y2 .
e. 2.
Minimum ( f ( x ) , g ( x ) ) =
f ( x) + g ( x) 2
−
f ( x) − g ( x) 2
The Never-Negative Function a.
b.
6000
0 −1000
3
6000
0
3
−1000
The graph of M + is the graph of M provided that M ≥ 0. For each x such that M < 0, the graph of M + is the point (x, 0). b. The maximum mosquito population is 4850 mosquitoes per acre. This maximum occurs at 1 3 5 t = , t = and t = (the middle of each month). 2 2 2 a.
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Chapter 2 c.
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Assume the months have 30 days. The best time to fish if you wish to minimize your exposure to mosquitoes is during the first 3 or the last 3 days of each month. During this period, the mosquito population will be zero. During the period from the 4th day to the 26th day of the month, you will be exposed to mosquitoes.
2.7 Modeling Data Using Regression The Median-Median Line
834
1.
y ≈ 24.8558 x − 41.9712
2.
y ≈ 3.5429 x + 58.0952
3.
a. y = 2.0000 x + 1.0000 b. y = 2.0000 x + 1.0000 c. If a least-squares regression line provides an exact fit for a set of data, then the median-median line for the data will be exactly the same as the least-squares regression line.
4.
a. median-median line: y ≈ 1.3571x − 0.3333; regression line: y ≈ 1.2636 x + 0.5182 b. median-median line: y ≈ 1.2143 x + 0.3333; regression line: y ≈ 0.6727 x + 3.4727 c. A median for a set of data is generally not changed by increasing one of the larger data values or by decreasing one of the smaller data values. For instance, consider the data 3, 5, 7, 11, 20, which has a median of 7. Now change the 20 to a larger number, say 30. This new set of data, 3, 5, 7, 11, 30, still has a median of 7. In general, a median-median line is less sensitive to a single change in one value than is a least-squares regression line. In fact, any single change in a set of data will cause a change in the least-squares regression line for the data, but often this same change will not change the medianmedian line for the data. Some mathematical texts refer to the median-median line of a set of data as the “resistant line” for the data.
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Chapter 3
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Polynomial and Rational Functions CHAPTER 3 of College Algebra and Trigonometry 3.1 The Remainder Theorem and the Factor Theorem 1. Horner’s Polynomial Form P ( x ) = 3 x5 − 4 x 4 + 5 x 3 − 2 x 2 + 3 x − 8
a.
P (6) = 3(6)5 − 4(6) 4 + 5(6)3 − 2(6) 2 + 3(6) − 8 = 3(7776) − 4(1296) + 5(216) − 2(36) + 3(6) − 8 = 23328 − 5184 + 1080 − 72 + 18 − 8 = 18144 + 1080 − 72 + 18 − 8 = 19224 − 72 + 18 − 8 = 19152 + 18 − 8 = 19170 − 8 = 19162 P ( x ) = 3 x5 − 4 x 4 + 5 x3 − 2 x 2 + 3 x − 8
b.
= (3x 4 − 4 x3 + 5 x 2 − 2 x + 3) x − 8 = ((3 x3 − 4 x 2 + 5 x − 2) x + 3) x − 8 = (((3 x 2 − 4 x + 5) x − 2) x + 3) x − 8 = ((((3x − 4) x + 5) x − 2) x + 3) x − 8 P (6) = ((((3 ⋅ 6 − 4)(6) + 5)(6) − 2)(6) + 3)(6) − 8 = (((18 − 4)(6) + 5)(6) − 2)(6) + 3)(6) − 8 = (((14)(6) + 5)(6) − 2)(6) + 3)(6) − 8 = (((84 + 5)(6) − 2)(6) + 3)(6) − 8 = (((89)(6) − 2)(6) + 3)(6) − 8 = ((534 − 2)(6) + 3)(6) − 8 = ((532)(6) + 3)(6) − 8 = (3192 + 3)(6) − 8 = (3195)(6) − 8 = 19170 − 8 = 19162 The evaluation in part b. involves easier calculations.
3.2
Polynomial Functions of Higher Degree 1.
3.3 1.
The student is correct. The polynomial P (n) = n3 − n can be written in factored form as P (n) = n(n − 1)(n + 1) . In this form it is easy to see that P(n) is the product of three consecutive natural numbers, one of which must be an even number and one of which must be a multiple of three. Thus P (n) must be a multiple of 6 for any natural number n. Zeros of Polynomial Functions Relationships Between Zeros and Coefficients a. r1 = 1, r2 = 2, r3 = 3.
P ( x) = x3 + C1 x 2 + C2 x + C3 = x3 − 6 x 2 + 11x − 6 ⇒ C1 = −6, C2 = 11, C3 = −6 1 + 2 + 3 = 6 = −(−6) ⇒ r1 + r2 + r3 = −C1 1(2) + 1(3) + 2(3) = 2 + 3 + 6 = 11 ⇒ r1r2 + r1r3 + r2 r3 = C2 1(2)(3) = 6 = −(−6) ⇒ r1r2 r3 = −C3
1(2)(3) = 6 = (−1)3 (−6) ⇒ r1r2 r3 = (−1) n Cn Responses will vary.
b. 3.4
The Fundamental Theorem of Algebra 1. Investigate the Roots of Cubic Equation a.
Given x 3 + bx 2 + cx + d = 0. Let x = y −
b and substitute into the equation. Then simplify. 3 x 3 + bx 2 + cx + d = 0
3
⎛ ⎛ ⎛ b⎞ b⎞ b⎞ ⎜ y − ⎟ + b⎜ y − ⎟2 + c⎜ y − ⎟ + d = 0 3⎠ 3⎠ 3⎠ ⎝ ⎝ ⎝ 2 3 2 yb by 2 ⎛ ⎞ − b + b ⎜ y2 − + b ⎟ + c ⎛⎜ y − b ⎞⎟ + d = 0 y 3 − by 2 + 3 27 3 9 ⎠ ⎝ 3⎠ ⎝ 2 2 ⎛ ⎞ ⎛ 3 b b bc 2 − + d ⎟⎞ = 0 y + ⎜c − ⎟ y + ⎜ 3⎠ ⎝ ⎝ 27 3 ⎠ 2 3 Now let m = c − b and n = − ⎛⎜ 2b − bc + d ⎞⎟ . Then y 3 + my = n . 3 ⎝ 27 3 ⎠
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b. To solve the equation x3 − 6 x 2 + 20 x − 33 = 0, let b = −6, c = 20 and d = −33. Then
m = 20 −
( −6 ) 3
2
⎛ 2 ( −6 )3 ( −6 )( 20 ) ⎞ = 8 and n = − ⎜ − − 33 ⎟ = 9 ⎜ 27 ⎟ 3 ⎝ ⎠
A solution is given by y=
3
n n 2 m3 3 n n 2 m3 + + − − + + 2 4 27 2 4 27
=
3
9 92 83 3 9 92 83 + + − − + + 2 4 27 2 4 27
=1 b Substituting y and b into x = y − gives x = 3 is a solution of x3 − 6 x 2 + 20 x − 33 = 0. To find the 3 remaining solutions, use synthetic division to determine the reduced equation, which is 3 + i 35 3 − i 35 x 2 − 3 x + 11 = 0. The solutions of this equation are and . 2 2
3.5
Graphs of Rational Functions and Their Applications 1.
Parabolic Asymptotes a.
Yes. As x → ∞ and x → −∞, the graph of F approaches the graph of the parabola. b. Divide R(x) by S(x) to find the quotient Q(x). The equation y = Q(x) is the equation of the parabolic asymptote. c.
x2 + 2 x2 − 1 x4 + x2 + 2 x4 − x2 2x2 + 2 2x2 − 2 4
Yes d. Answers will vary.
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Exponential and Logarithmic Functions CHAPTER 4 of College Algebra and Trigonometry 4.1 Inverse Functions 1. Intersection Points for the Graphs of f and f–1 i. f ( x) = 2 x − 4 x = 2y − 4 x + 4 = 2y −9.1 1 x+2= y 2 f −1 ( x) = 12 x + 2 f ( x) = − x + 2 x = −y + 2 y = −x + 2
ii.
f −1 ( x) = − x + 2
6
9.1
−6 6
−9.1
9.1
−6
iii.
6
f ( x ) = x3 + 1 x = y3 + 1 x − 1 = y3 3
f
−9.1
x −1 = y
−1
( x) = 3 x − 1
−6
f ( x) = x − 3
iv.
9.1
6
x = y −3 x+3= y f
−1
( x) = x + 3
−9.1
9.1
−6 f ( x ) = −3x + 2
v.
6
x = −3 y + 2 3y = −x + 2 f
−1
( x ) = − 13 x + 32
−9.1
9.1
−6
vi.
f ( x ) = 1x x = 1y y = 1x
a. No
4 −4
f −1 ( x ) = 1x b. They are equal.
4
c. Yes. Consider the function f ( x) = x .
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4.2 Exponential Functions and Their Applications 1. The Saint Louis Gateway Arch a.
b. Use the TRACE feature. Choose a value of x and press ENTER for each successive value of x.
h(0) ≈ 625.1 feet
h(100) ≈ 587.5 feet
h(200) ≈ 433.5 feet
h(299) ≈ 1.6 feet
c.
Width of the catenary at ground level =|−299.2261 – 299.22611| = |−598.45221| ≈ 598.5 feet Maximum height of the catenary = h(0) ≈ 625.1 feet d. 625.1 − 598.5 = 26.6 feet 2.
An Exponential Reward a. From the chart, create a pattern from the total number of grains of wheat on squares 1 to n For square n = 1, 1 = 21 – 1 For square n = 2, 3 = 22 – 1 For square n = 3, 7 = 23 – 1 For square n = 4, 15 = 24 – 1 For square n = 5, 31 = 25 – 1 For square n = 6, 63 = 26 – 1 So for square n = 64, 264 – 1 = 1.8446744 × 1019 grains is the total. b. To find the total weight, multiply the number of grains by the weight for each one (1.8446744 × 1019 )(0.000008) = 1.4757395 × 1014 kilograms c. To find how long it would take, convert the weight to metric tons and divide by the amount per year 1year 1.4757395 × 1014 kg ⋅ 1 metric ton ⋅ ≈ 227 years 1000 kg 6.5 × 108 metric tons
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4.3 Logarithmic Functions and Their Applications 1.
a.
d
1 2 3 4 5 6 7 8 9 b. c.
⎛ 1⎞ P (d ) = log ⎜1 + ⎟ ⎝ d⎠ 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046
(
)
P (6) = log 1 + 16 = log
( 76 ) ≈ 0.067 or 6.7%
( ) P (9) = log (1 + 19 ) = log ( 10 ≈ 0.046 or 4.6% 9) P (1) = log 1 + 11 = log ( 2 ) ≈ 0.301 or 30.1%
P (1) 0.301 = ≈ 6.54 times as many P (9) 0.046
d.
Most high school students are teenagers.
4.4 Logarithms and Logarithmic Scales 1. Logarithmic Scales Mass (g) ln(Mass) a. Animal Rotifer 0.000000006 −8.22 Dwarf Goby 0.30 −0.52 Lobster 15,900 4.20 Leatherback Turtle 851,000 5.93 Giant Squid 1,820,000 6.26 Whale Shark 44,700,000 7.65 Blue Whale 120,000,000 8.08
b.
The logarithmic number line is more helpful when comparing different animals.
c.
101 = 10 times heavier 102 = 100 times heavier
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2. Logarithmic Scales a. b.
Planet
Distance (million km) 58 108 150 228 778 1427 2871 4497 5913
Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto
ln(Distance) 1.76 2.03 2.18 2.36 2.89 3.15 3.46 3.65 3.77
c. Answers will vary. d. 103 = 1000 . One distance is 1000 times greater than the other. 3.
Biological Diversity a.
D = − ( 15 log 2 15 + 15 log 2 15 + 15 log 2 15 + 15 log 2 15 + 15 log 2
b.
D = − ( 18 log 2 18 + 83 log 2 83 + 161 log 2 161 + 18 log 2 18 + 165 log 2 165 )
1 5
log 1
) = − log 2 15 = − log 25 ≈ 2.322
⎛ 1 log 81 + 83 log 83 + 161 log 161 + 18 log 81 + 165 log 165 ⎞ = −⎜ 8 ⎟ ≈ 2.055 log 2 ⎝ ⎠ This system has less diversity than the one given in Table 1. D = − ( 14 log 2 14 + 34 log 2 34 )
c.
⎛ 0.25log 0.25 + 0.75log 0.75 ⎞ = −⎜ ⎟ ≈ 0.811 log 2 ⎝ ⎠ This system has less diversity than the one given in Table 2. d. D = − (1 log1) = 0
This value means that the system has no variety of species. 4.5 Exponential and Logarithmic Equations 1.
Navigating a.
If v = w, then the equation becomes y = 1 − ( 2x ) . The graph of this equation is shown below. The 2
boat reaches the other shore 1 mile from point O. y 1.0 0.8 0.6 0.4 0.2 O
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0.5
1.0
1.5
2.0
x
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b. Now suppose w > v . For instance, let w = 1.1v. Then the equation becomes y = ( 2x )
−0.1
− ( 2x ) . The 2.1
graph of this equation is shown below. In this case, the y-axis is a vertical asymptote and the boat never reaches the shore. y
1.5 1.0 0.5 O
c.
0.5
1.0
1.5
2.0
x
Now suppose w < v . For instance, let w = 0.5v . Then the equation becomes y = ( 2x )
0.5
− ( 2x ) . The 1.5
graph of this equation is shown below. In this case, the boat reaches the shore at point O. y
0.3 0.2 0.1 O
0.5
1.0
1.5
2.0
x
4.6 Exponential Growth and Decay 1.
2.
3.
A Declining Fish Population 1000 1000 = = 1500 fish a. P0 = 1 + (−0.3333) 0.6667 b. As t → ∞, P(t ) → 1000 fish. A Declining Deer Population 1800 1800 = = 2300 deer a. P2 = 1 + (−0.25)(0.869) 0.782 b. As t → ∞, P(t ) → 1800 deer. Modeling World Record Times in the Men’s Mile Race 199.13 = 219.41 s = 3 min, 39.41 s a. WR (107) = 1 + (−0.21726)(0.42536) WR (137) =
b.
199.13 = 214.75 s = 3 min, 34.75 s 1 + (−0.21726)(0.33471)
As t → ∞, WR (t ) → 199.13 s = 3 min, 19.13 s.
4.7 Modeling Data with Exponential and Logarithmic Functions 1.
A Modeling Project a. Answers will vary b. Answers will vary c. Answers will vary d. Answers will vary e. Answers will vary f. Answers will vary
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Trigonometric Functions CHAPTER 5 of College Algebra and Trigonometry 5.1 Angles and Arcs 1. Conversion of Units 14 tollars ⎛ 4 nollars ⎞ ⎛ 3 mollars ⎞ 14 ⋅ 4 ⋅ 3 24 4 a. 14 tollars = mollars = mollars =4 mollars ⎜ ⎟ ⎜ ⎟= 1 7 tollars 5 nollars 7 ⋅ 5 5 5 ⎝ ⎠ ⎝ ⎠ 4 4 4 mollars ⎛ 5 lollars ⎞ b. 4 mollars = 4 mollars + mollars = 4 mollars + ⎜ ⎟ = 4 mollars + 4 lollars 5 5 5 ⎝ 1 mollar ⎠ c. Start with the given quantity. Multiply by unit fractions that eliminate the given units and yield results in terms of the desired units. We knew we wanted to eliminate the tollars unit in the numerator. Thus ⎛ 4 nollars ⎞ we need to multiply by the unit fraction: ⎜ ⎟ ⎝ 7 tollars ⎠ ⎛ π ⎞ d. ⎜ ⎟ ⎝ 180° ⎠ 2.
Space Shuttle Let d be the portion of a revolution that the Galápagos Islands rotates as the shuttle revolves 1 + d revolutions. The time required for the shuttle will be ⎛ (1 + d ) revolutions ⎞ ⎛ 2.231 hours ⎞ t1 = ⎜ ⎟ ⎜ ⎟ = (1+d )( 2.231) hours 1 ⎝ ⎠ ⎝ 1 revolution ⎠ The time required for the earth to rotate d revolutions is ⎛ d revolutions ⎞ ⎛ 23.933 hours ⎞ t2 = ⎜ ⎟ ⎜ ⎟ = 23.933d hours 1 ⎝ ⎠ ⎝ 1 revolution ⎠ Setting t1 = t2 yields 2.231 + 2.231d = 23.933d . Solving this equation gives d ≈ 0.1028 revolution. The time required for the shuttle to complete 1 + d revolutions is 2.231 hours ⎞ (1 + 0.1028) revolutions ⎛⎜ ⎟ = 2.460 hours (to the nearest 0.001 hour) ⎝ 1 revolution ⎠
5.2 Trigonometric Functions of Acute Angles 1. Perimeter of a Regular n-gon a.
P = 2 xn
x
sin 180° n
x
r =1
b.
842
180° x = n r 180° 180° = sin n n 180° P = 2n sin n x = r sin
180° ≈ 6.18034 10 P50 = 2 ( 50 ) sin 180° ≈ 6.27905 50 P100 = 2 (100 ) sin 180° ≈ 6.282152 100 P1000 = 2 (1000 ) sin 180° ≈ 6.283175 1000 P10,000 = 2 (10, 000 ) sin 180° ≈ 6.283185 10, 000 Pn approaches 2π as n increases because the perimeter approaches the circumference of the circle.
P10 = 2 (10 ) sin
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Chapter 5 2.
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Area of a Regular n-gon a. 1 hr ⋅ n 2 n 360° = r ⋅ r sin n 2 n 360° = sin n 2
A= r
h 360° n
r
b.
sin
360° h = n r h = r sin
360° n
r =1
10 360° sin ≈ 2.938926261 2 10 50 360° ≈ 3.133330839 A50 = sin 2 50 100 360° ≈ 3.139525976 sin A100 = 2 100 1000 360° ≈ 3.141571983 sin A1000 = 2 1000 10, 000 360° ≈ 3.141592447 A10,000 = sin 2 10, 000 An approaches π as n increases because the area of the polygon approaches the area of the circle, A20 =
which is π ⋅12 = π . 5.3 Trigonometric Functions of Any Angle 1.
Find Sums or Products a. 0 The sum is 0 because cos n° = − cos (180° − n° ) for all integers n such that 0° ≤ n ≤ 90° . b.
0 The sum is 0 because sin n° = − sin ( 360° − n° ) for all integers n such that 0° ≤ n ≤ 180° .
c.
0 The sum is 0 because cot n° = − cot (180° − n° ) for all integers n such that 1° ≤ n ≤ 89° .
d. e.
0 The product is 0 because cos 90° = 0 and the product of 0 and any real number is 0. 179 This is the sum of 359 numbers. However, the sum can be regrouped so that it consists of 179 pairs of the form ( cos x ) + ( cos 90° − x ) = ( cos x ) + ( sin x ) = 1 . The term that is not paired up 2
2
2
2
with another number in the list is cos 90° = 0 . Thus the sum is 179. 5.4 Trigonometric Functions of Real Numbers 1.
y
Visual Insight E
C
Unit circle OD = cos φ DC = sin φ
a.
φ
F
B
φ O
D
A (1, 0)
x
Consider the triangle ABC. By definition, opp d ( A, B ) d ( A, B ) = = = the length of line segment AB. tan φ = adj d ( O, A ) 1
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b.
Triangle ODC is similar to triangle FEO. Therefore, d ( O, D ) d ( F , E ) d ( F , E ) = = = the length of line segment EF. cot φ = d ( D, C ) d ( E , O ) 1
c.
Consider triangle AOB. By definition, hyp d ( O, B ) d ( O, B ) = = = the length of line segment OB. sec γ = adj d ( O, A ) 1
d.
Triangle OEF is similar to triangle CDO. Therefore, d ( O, C ) d ( F , O ) d ( F , O ) csc γ = = = = the length of line segment OF. d ( C , D ) d ( O, E ) 1
Functions Defined by a Square a. ssin 3.2 = 0.8 because traveling 3.2 units counterclockwise around the square from (1, 0 ) places you
at the point ( −1, 0.8 ) . The y-value of this ordered pair is 0.8. b.
scos 4.4 = −1 because traveling 4.4 units counterclockwise around the square from (1, 0 ) places you at
the point ( −1, −0.4 ) . The x-value of this ordered pair is –1. c. d. e. f.
−1 1 ssin 5.5 = −1 and scos 5.5 = − , thus, stan 5.5 = 1 = 2 . 2 (− 2 )
ssin 11.2 = 0.8 (Note that the function y = ssin x is periodic with a period of 8. Thus ssin 11.2 = ssin 3.2 which equals 0.8 from part (a).) scos − 5.2 = −0.8 . ssin ( −6.5 ) 1 stan − 6.5 = = =2 scos ( −6.5 ) 0.5
5.5 Graphs of the Sine and Cosine Functions 1. Cepheid Variable Stars and the Period-Luminosity Relationship a. A Cepheid is usually a population I giant yellow star, pulsing regularly by expanding and contracting, resulting in a regular oscillation of its luminosity. Named for the prototype of this class found in the constellation Cepheus, classical Cepheids have periods from about 1.5 days to over 50 days and are Population I stars. The longer the period of such a star, the greater its natural brightness; this relationship was discovered in 1912 by the American astronomer Henrietta Leavitt (b. 1868-d. 1921). The relationship between a Cepheid variable's luminosity and variability period is quite precise, and has been used as a standard candle for almost a century. b. Because of this correlation (discovered by Henrietta Leavitt in 1912), a Cepheid variable can be used as a standard candle to determine the distance to its host cluster or galaxy. Since the period-luminosity relation can be calibrated with great precision using the nearest Cepheid stars, the distances found with this method are among the most accurate available. 5.6 Graphs of the Other Trigonometric Functions 1. A Technology Question The function f ( x ) = tan x is undefined at x = 32 π , which is between the domain values 4.7123 and
4.7124. Because y = tan x approaches ∞ as x approaches 32 π from the left and y = tan x approaches −∞ as x approaches 32 π from the right, it is possible to produce large changes in your range values with small changes in your domain values as they change from slightly less than 32 π to slightly greater than 32 π .
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Chapter 5 2.
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Solutions of a Trigonometric Equation The equation tan ( 1x ) = 0 has an infinite number of solutions on the interval −1 ≤ x ≤ 1 . This can be
determined by observing that as x varies from –1 to 1, the fraction 1/x takes on all values less than or equal to –1 and all values greater than or equal to 1. Because the tangent function is periodic with a period of π , the expression tan ( 1x ) will equal 0 whenever 1/x is a multiple of π for – for instance, when x = π1 , 21π , 31π ,K . Thus there are an infinite number of solutions.
5.7 Graphing Techniques 1. Predator-Prey Relationships The graphs are shown below. Because the assumption is that the wolves prey on the rabbits, as the rabbit population increases, there is more food for the wolves, which in turn allows the wolf population to increase. However, as the wolf population increases, the demand for rabbits increases, and the rabbit population starts to decline. This effects a decline in the wolf population. But as the wolf population declines, there is less danger to the rabbits, and their population starts to rise. The process then repeats itself. Rabbit
Population
1000 800 600 400 200
Wolf 4
8
12
16
20
24
t
A possible equation for the graph given in the text of the population model for rabbits is r ( t ) = 503 t + 200sin ( π6t ) + 400 . This equation is based on the assumption that there is a linear increase in the sine function. Other answers may be given.
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5.8 Harmonic Motion—An Application of the Sine and Cosine Functions 1. Three Types of Damped Harmonic Motion The displacement of f ( t ) starts at its maximum when t = 0 and then descends toward its equilibrium
point, getting closer and closer as t increases without bound. See the following figure. 1
15
0
−1
The displacement of g ( t ) starts at its maximum when t = 0 . It next falls to a minimum displacement below the equilibrium position and then rises to approach equilibrium as t increases without bound. See the following figure. 1
15
0
−1
The displacement of h ( t ) starts at its maximum displacement when t = 0 . As t increases, it oscillates about its equilibrium point. As it oscillates, the maximum displacement of each cycle tends to 0. See the following figure. 4.2
15
0
−4.2
2.
Logarithmic Decrement a. γ ≈ 3.51 b.
846
Δ ≈ 1.26; ln γ = Δ
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Chapter 6
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Trigonometric Equations and Identities CHAPTER 6 of College Algebra and Trigonometry 6.1 Verification of Trigonometric Identities 1. Grading a Quiz 1,2, and 4 are correct. 6.2 Sum, Difference, and Cofunction Identities 1. Intersecting Lines a.
In the following figure, we see that if a line intersects the x-axis at an angle of θ , then y −y tan θ = 2 1 = m , which is the slope of the line. x2 − x1 y
P2 ( x2 , y2 )
θ P1 ( x1 , y1 ) x
Let γ be the smallest positive angle from l1 to l 2 , as shown in the following figure. y
l2
l1
γ
α
β x
It is possible to show that β = α + γ . (Use the exterior angle theorem and the theorem on vertical angles.) Thus γ = β − α , and tan γ = tan ( β − α ) =
m − m1 tan β − tan α = 2 . 1 + tan β tan α 1 + m1m2
b. The line l1 given by y = x + 5 has a slope of m1 = 1 . The slope of the line l 2 given by y = 3x − 4 has a slope of m2 = 3 . From part (a), we know that the tangent of the angle γ (the acute angle between the lines) is m − m1 3 −1 2 1 = = = tan γ = 2 1 + m1m2 1 + (1)( 3) 4 2
A graph of y = tan γ and y =
1 (on the interval 0 < γ < 90° ) shows that γ ≈ 26.6° . 2
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Chapter 6
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Let m2 be the slope of the second line. From part (a), m2 − 0.5 tan 60° = 1 + ( 0.5 )( m2 ) 3=
m2 − 0.5 1 + ( 0.5 )( m2 )
3 + 0.5 3m2 = m2 − 0.5 3 + 0.5 = m2 − 0.5 3m2
(
3 + 0.5 = m2 1 − 0.5 3
)
3 + 0.5
= m2 1 − 0.5 3 m2 ≈ 16.66
The second line passes through the point (1,5) with a slope of about 16.66, so y − 5 ≈ 16.66 ( x − 1)
• The equation of line l 2 in slope-intercept form
y ≈ 16.66 x − 11.66
6.3 Double- and Half-Angle Identities 1. Visual Insight The measure of a central angle is equal to the measure of its intercepted arc. The measure of an inscribed angle is one-half the measure of its intercepted arc. Therefore, the measure of the small marked angle at the right must be half the measure of the central angle θ . Thus the measure of the small marked angle is θ / 2. The side opposite θ in the small triangle is sin θ . The side adjacent to θ in the small triangle is cos θ . θ opp sin θ . By definition, tan = = 2 adj 1 + cos θ 6.4 Identities Involving the Sum of Trigonometric Functions 1. Beats a.
Y3
Y3, Y4, Y5 The graph of Y3 is bounded by the graphs of Y4 and Y5. b. Y3 = 2sin 2π 442 + 440 x ⋅ cos 2π 442 − 440 x 2 2 = 2sin ( 2π ⋅ 441x ) cos ( 2π x ) c. 568 − 564 = 4 beats per second d. 2 beats per second
(
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)
(
)
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6.5 Inverse Trigonometric Functions 1. Visual Insight a. Verify the identity tan −1 13 + tan −1 12 = π4 . 1
α
γ
β
1
1
1
1 3
and tanβ = 12 .
Let α = tan −1 13 and β =tan -1 12 , as in the figure. Thus tan α = tan −1 1 + tan −1 1 = α + β 3 2 = tan −1 ⎣⎡ tan (α + β )⎦⎤ ⎛ tan α + tan β ⎞ = tan −1 ⎜ ⎟ ⎝ 1 − tan β tan β ⎠ ⎡ ⎤ = tan −1 ⎢ 1/ 3 + 1/ 2 ⎥ ⎣ 1 − (1/ 3)(1/ 2 ) ⎦
= tan −1 1 = π 4
Hence tan −1 13 + tan −1 12 = π4 . b. The figure in part (a) shows that α = tan −1 13 , β = tan −1 12 , and γ = π4 . Substituting in the result from
part (a) produces α + β = γ . 6.6 Trigonometric Equations 1. The Moons of Saturn a. Titan completes one cycle in about 15.95 days. Thus for Titan,
y 1 ≈ 1.0000sin(0.3963 x − 2.5696) + 0.0026
b. Rhea completes one cycle in about 4.52 days. Thus for Rhea,
y 2 ≈ 0.4002sin( −1.3963x − 2.0944) − 0.0000
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This solution can be found by using the regression formula directly, or using the graphing calculator graph at value x = 40 as shown below. At 11:59PM on May 10, 2000, x = 40, Using the regression formulas from parts a and b, y 1 (40) ≈ 1.0000sin(0.3963(40) − 2.5696) + 0.0026 ≈ 0.65888 y 2 (40) ≈ 0.4002 sin( −1.3963(40) − 2.0944) − 0.0000 ≈ −0.13689 Using the graphing calculator to graph both regression formulas.
They were on opposite sides of Saturn.
850
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Chapter 7
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Applications of Trigonometry CHAPTER 7 of College Algebra and Trigonometry 7.1 The Law of Sines 1. Fermat’s Principle and Snell’s Law Fermat’s Principle: Light traveling from one point to another will follow a path such that, compared with nearby paths, the time required is either a minimum or maximum or will remain unchanged. Snell’s Law, which follows, is derived by using Fermat’s Principle. sin θ Snell’s Law: sin θ12 = n21 , where n21 is a constant called the index of refraction of medium 2 with respect to medium 1. See the diagram below. (Note: The index of refraction depends on the wavelength. As the wavelength increases, the index of refraction decreases.)
θ1 Because a diamond has a higher index of refraction than glass, light entering the diamond is refracted at a greater angle than the same light entering a piece of glass. The result is a narrower “rainbow” as the light leaves the diamond than as it leaves the glass.
θ2
7.2 The Law of Cosines and Area 1. Visual Insight d e
b
c
C
a
a a
Center of circle
bd = ( a + c ) e
• If two chords of a circle intersect, then the product of the lengths of the segments on one chord equals the product of the lengths of the segments on the other chord.
bd = ( a + c )( a − c )
• e = a−c
b ( 2a cos C − b ) = ( a + c )( a − c )
• In the right triangle cos C = ( d + b ) / ( 2a ) . Solving for d gives us d = 2a cos C − b .
2ab cos C − b = a − c 2
2
2
c 2 = a 2 + b 2 − 2ab cos C
• Simplify. • Solve for c 2 .
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7.3 Vectors 1. Same Direction or Opposite Directions
If c > 0 and v = cw , then v and w have the same direction, v⋅w = cos 0° = 1 . and the angle between the vectors is θ = 0° . Thus v w If c < 0 and v = cw , then v and w are vectors that have opposite directions, and the angle between the v⋅w = cos180° = −1 . vectors is θ = 180° . Thus v w
2.
The Law of Cosines and Vectors v−w
2
• Because v ⋅ w = v w cos α , we see that this equation
= ( v − w )( v − w ) = v⋅v − v⋅w − w⋅v + w⋅w 2
is a restatement of the Law of Cosines in vector form.
2
= v + w − 2v ⋅ w
3.
Projection Relationships Let v and w be two nonzero vectors. Let α be the measure of the angle between the vectors. By definition, projw v = v cos α . a.
If projw v = 0 , then v cos α = 0 , and cos α = 0 , which implies α = 90° . Hence v and w are
perpendicular (orthogonal). b. If projw v = v , then v cos α = v , and cos α = 1 . Thus, α = 0° . Hence v and w have the same direction. 7.4 Trigonometric Form of Complex Numbers 1. A Geometrical Interpretation Multiplication of a real number a by i produces the product ai . Note that in an Argand diagram, the numbers a and ai are both placed the same distance from the origin. Also note that the position of ai can be determined by rotating the position of the real umber a 90° counterclockwise about the origin. Multiplication of a complex number a + bi by i also produces a number that is located in an Argand diagram the same distance from the origin as the original number a + bi . Once again, the position of the product −b + ai can be determined by rotating the position of the original number a + bi 90° counterclockwise about the origin. It is also worth noting that multiplicity a number by −1 = i 2 can be thought of geometrically as a 180° counterclockwise rotation of the original number about the origin. 7.5 De Moivre’s Theorem 1. Verify Identities
z = ( cos θ + i sin θ )
z 2 = cos 2θ + i sin 2θ by De Moivre's Theorem. z 2 = cos 2 θ + 2i sin θ cos θ + i 2 sin 2 θ by squaring both sides.
Because z 2 = z 2 , we have ( cos 2θ + i sin 2θ ) = ( cos 2 θ + 2i sin θ cosθ + i 2 sin 2 θ )
( cos 2θ + i sin 2θ ) = ( cos 2 θ + 2i sin θ cosθ − sin 2 θ ) Equating the real parts, we have cos 2θ = cos 2 θ − sin 2 θ
Equating the imaginary parts, we have sin 2θ = 2sin θ cosθ
852
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Chapter 7 2.
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Discover Identities x = ( cos θ + i sin θ ) z 4 = cos 4θ + i sin 4θ by De Moivre's Theorem. z 4 = cos 4 θ + 4i cos3 θ sin θ + 6i 2 cos 2 θ sin 2 θ + 4i 3 cosθ sin 3 θ + i 4 sin θ by taking z to the fourth power.
Because z 4 = z 4 , we have
( cos 4θ + i sin 4θ ) = ( cos 4 θ + 4i cos3 θ sin θ + 6i 2 cos 2 θ sin 2 θ + 4i 3 cosθ sin 3 θ + i 4 sin θ ) = cos 4 θ + 4i cos3 θ sin θ − 6 cos 2 θ sin 2 θ − 4i cosθ sin 3 θ + sin 4 θ
Equating the real parts, we have cos 4θ = cos 4 θ − 6 cos 2 θ sin 2 θ + sin 4 θ = cos 4 θ − 2 cos 2 θ sin 2 θ + sin 4 θ − 4 cos 2 θ sin 2 θ = ( cos 2 θ − sin 2 θ ) − 4 cos 2 θ sin 2 θ 2
= cos 2 2θ − 2sin 2 2θ
Equating the imaginary parts, we have
sin 4θ = 4 cos3 θ sin θ − 4 cosθ sin 3 θ
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Chapter 8
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Topics in Analytic Geometry CHAPTER 8 of College Algebra and Trigonometry 8.1 Parabolas 1.
3-D Optical Illusion Responses will vary. For example, http://www.exploratorium.edu/snacks/parabolas.html gives the following explanation. You are seeing an image formed by two concave mirrors facing one another. The object is placed at the center of the bottom mirror. The curvature of the mirrors is such that the object is at the focal point of the top mirror. When light from a point on the object hits the top mirror, it reflects in parallel rays. These parallel rays hit the bottom mirror and reflect so that they reassemble to form a point located at one focal length from the bottom mirror. The mirrors are placed so that the focal point of the bottom mirror is located at the hole in the top of the device. The end result is that light from every point on the object is assembled into an image in the hole. The ray diagram may help explain this effect. The image produced by this apparatus is known as a real image, because the light that forms it actually passes through the location of the image. However, if you place a piece of wax paper or onionskin paper at the location of the real image, the image will not appear on the paper. The outside regions of the mirrors that do not reflect light to your eyes do reflect light to the paper. The edges of the mirrors have large aberrations and create an image so blurred that it cannot be seen.
8.2 Ellipses 1.
2.
854
Kepler’s Laws Kepler was born Dec. 17, 1571, and died Nov. 15, 1630. He was among the first strong supporters of the heliocentric theory. In 1596 Kepler published Cosmographic Mystery, in which he defended the Copernican theory. Tycho Brahe, mathematician at the court of Emperor Rudolph II, was impressed with the work of Kepler and invited him to Prague as his assistant. When Brahe died the following year (1601), Kepler was appointed to the position held by Brahe. Between 1609 and 1619, Kepler published his three laws of motion: 1. Each planet moves about the sun in an orbit that is an ellipse, with the sun at one focus of the ellipse. 2. The straight line joining a planet and the sun sweeps out equal areas in space in equal intervals of time. 3. The squares of the sidereal periods of the planets are in direct proportion to the cubes of the semimajor axes of their orbits. That is, P 2 = ka 3 . The value of k depends on the units of measurement. If astronomical units are used, then k = 1 . Kepler also made contributions to optics and telescope lenses and gave a physical explanation of nova. His text Introduction to Copernican Astronomy was one of the most widely read treatises on astronomy. a. A planet’s velocity is greatest when it is at perihelion. This follows from Kepler’s second law. b. The period of Mars is 1.87 years. This follows from the third law. Neptune Neptune was discovered as a result of mathematical prediction. The perturbative effects of Jupiter and Saturn on Uranus alone did not allow for observed discrepancies in Uranus’s orbit. Using Newtonian gravitational theory, John Couch Adams produced mathematical evidence that an unknown planet could account for the irregularities in Uranus’s orbit. Adams sent his calculations, along with the region of the sky in which to search for the new planet, to Sir George Airy, the Royal Astronomer, and asked him to begin a search for the planet. But Airy had no faith in Adams’ calculations and did not look for Neptune. Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 8
3.
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In June of 1846, Leverrier, a French mathematician, independently reproduced the work of Adams. His calculations also were sent to Airy. This time the Royal Astronomer suggested that Cambridge University begin a search for the new planet. The work of Challis, director of Cambridge Observatory, was sloppy and negligent. Consequently, he did not find the planet. In September 1846, Leverrier suggested that Galle of the Berlin Observatory search the Aquarius region of the sky for the planet. Galle found the planet during his first observation. The discovery of Neptune was a great achievement for the time and was a major triumph of gravitational theory. Graph the Colosseum a. A graph of the Colosseum produced by using the Maple commands given in the text.
z
200
y
100
200
−300
−100 0 100 x
300
0 −200
b. The following image appears to be constructed with ellipses because the window is not a “square” 9
-4.7
4.7 -4
window. That is, one unit on the x-axis is not the same length as one unit on the y-axis. The graphs of the equations on a “square” window would appear as semicircles, but on this non-square window they are distorted and appear to be elliptical. 8.3 Hyperbolas a. A Hyperbolic Paraboloid The general equation is
y2 b2
− ax 2 = cz . The graph for a = 3, b = 4, and c = 1 is shown below. The 2
graph is symmetric with respect to the planes x = 0 and y = 0 . z Hyperbola
y
x
The section in the plane x = 0 is y 2 =
Parabola b2 c
z , which is a parabola that opens upward and has its vertex at
the origin. 2 The section in the plane y = 0 is x 2 = − ac z , which is a parabola that opens downward and has its vertex at the origin. Copyright © Houghton Mifflin Company. All rights reserved.
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Chapter 8
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If the surface is cut by the plane z = k > 0 , then the section is the hyperbola axis is parallel to the y-axis and whose vertices are on the parabola y 2 = If k < 0 , then the section is the hyperbola
whose vertices lie on the parabola x 2 = −
x2 a2
2
− by2 =
k c
b2 c
y2 b2
− ax 2 = kc , whose focal 2
z.
, whose focal axis is parallel to the x-axis and
a2 z. c
b. A Hyperboloid of One Sheet x2 y 2 z 2 The general equation is 2 + 2 + 2 = 1 . The graph for a = 3, b = 4 , and c = 5 is shown below. The a b c graph is symmetric with respect to each of the coordinate planes. z Ellipse
y
x Hyperbola
The sections cut by the coordinate planes are x = 0:
y2 z2 − =1 b2 c2
y = 0:
x2 z 2 − =1 a2 c2
The plane z = k cuts the surface in an ellipse with equation is on the z-axis, and its vertices fall on the hyperbolas
y2 b2
x2 y 2 + =1 a2 b2
z = 0: x2 a2
=
− cz 2 = 1 and 2
y2 b2
x2 a2
= 1 + kc2 . The center of the ellipse 2
− cz 2 = 1 . 2
Cooling towers are built in the shape of a hyperboloid of one sheet, because such structures are strong, they have a large carrying capacity, and they can be constructed of many straight, narrow boards. 8.4 Rotation of Axes 1.
Use the Invariant Theorems We are given 10 x 2 + 24 xy + 17 y 2 − 26 = 0 . Thus A = 10, B = 24, C = 17, D = 0, E = 0, and F = 26 . We seek the equation of the form A′ ( x′ ) + C ′ ( y ′ ) − F = 0 . 2
2
The invariant theorems in Exercises 41 and 41 of Section 8.4 indicate that 2 A′ + C ′ = A + C and ( B′ ) − 4 A′C ′ = B 2 − 4 AC . A′ + C ′ = A + C implies A′ + C ′ = 10 + 17 = 27. Hence C ′ = 27 − A′.
( B′ )
856
2
− 4 A′C ′ = B 2 − 4 AC = 242 − 4 (10 )(17 ) = −104 . In the x′y ′ -coordinate system B ′ = 0 , so we have
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Chapter 8
College Algebra and Trigonometry
−4 A′C ′ = −104
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• Substitute 27 − A′ for C ′ .
−4 A′ ( 27 − A′ ) = −104 4 ( A′ ) − 108 A′ + 104 = 0 2
( A′ ) − 27 A′ + 26 = 0 ( A′ − 1)( A′ − 26 ) = 0 2
Hence A′ = 1 or A′ = 26. If A′ = 1, then C ′ = 26. If A′ = 26, then C ′ = 1. Thus we conclude that the conic given by 10 x 2 + 24 xy + 17 y 2 − 26 = 0 is also represented by 26 ( x′ ) + 1( y ′ ) − 26 = 0 or 1( x′ ) + 26 ( y ′ ) − 26 = 0 2
2
2
in the x′y ′ -coordinate system. An examination of the following graph shows that we should pick
( x′ )
2
+ 26 ( y ′ ) − 26 = 0 if we plan to obtain the x′ -axis by a 53° counterclockwise rotation of the x-axis 2
and that we should pick 26 ( x′ ) + ( y ′ ) − 26 = 0 if we plan to obtain the x′ -axis by a 143° 2
2
counterclockwise rotation of the x-axis. 5
8
-8
-5
8.5 Introduction to Polar Coordinates 1.
A Polar Distance Formula a. Let P1 , the origin, and P2 form a triangle ΔPOP 1 2. If we let a = the distance from P1 to the origin, b = the distance from the origin to P2 , and c = the
distance from P1 to P1 , the Law of Cosines defines c 2 = a 2 + b 2 − 2ab cos C , where C is the angle formed at the origin. Substituting r1 for the distance to P1 from the origin ( r1 = a ) and r2 for the distance from the origin to P2 ( r2 = b ) , we find that the distance squared between the two points P1 and P2 is now d [ P1 , P2 ] = r12 + r22 − 2r1r2 cos C . 2
Because the angle C is simply the difference between the angles formed by the two points and the polar axis, we can substitute (θ 2 − θ1 ) for C and arrive at the desired result: d ( P1 , P2 ) = r12 + r22 − 2r1r2 cos (θ 2 − θ1 ) .
b.
d ( ( 3, 60° ) , ( 5, 170° ) ) = 32 + 52 − 2 ( 3)( 5 ) cos (170° − 60° ) ≈ 6.65
c.
Because (θ 2 − θ1 ) = − (θ1 − θ 2 ) and because cos x = cos ( − x ) , this distance formula can also be written as d ( P1 , P2 ) = r12 + r22 − 2r1r2 cos (θ1 − θ 2 ) .
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Another Polar Form for a Circle a.
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r = a sin θ + b cos θ
r 2 = ar sin θ + br cos θ x 2 + y 2 = ay + bx x 2 − bx + y 2 − ay = 0 x 2 − bx +
b2 a 2 a 2 b2 + y 2 − ay + = + 4 4 4 4 2
2
b⎞ ⎛ a⎞ a 2 + b2 ⎛ − + − = x y ⎜ ⎟ ⎜ ⎟ 2⎠ ⎝ 2⎠ 4 ⎝ 2 2 b⎞ ⎛ a ⎞ ⎛ a2 + b2 ⎛ ⎜ x − ⎟ + ⎜ y − ⎟ = ⎜⎜ 2⎠ ⎝ 2⎠ ⎝ 2 ⎝
Thus the center of the circle is
( b2 , a2 )
⎞ ⎟ ⎟ ⎠
2
and the radius is r =
a 2 + b2 2
.
8.6 Polar Equations of the Conics 1.
Polar Equation of a Line From the figure k = d sin θ p , h = d cos θ p and the slope of the line segment perpendicular to the line is m = tan θ p . Therefore, the equation of the line is given by
⎛ 1 y − d sin θ p = ⎜ − ⎜ tan θ p ⎝
⎞ ⎟⎟ ( x − d cos θ p ) ⎠
Switching to polar, we have ⎛ 1 ⎞ r sin θ − d sin θ p = ⎜ − r cos θ − d cos θ p ) ⎜ tan θ ⎟⎟ ( p ⎠ ⎝ − r sin θ tan θ p + d sin θ p tan θ p = r cosθ − d cosθ p −r sin θ
sin θ p cos θ p
+ d sin θ p
sin θ p cosθ p
= r cos θ − d cosθ p
−r sin θ sin θ p + d sin 2 θ p = r cosθ cos θ p − d cos 2 θ p d sin 2 θ p + d cos 2 θ p = r cosθ cos θ p + r sin θ sin θ p d ( sin 2 θ p + cos 2 θ p ) = r ( cos θ cosθ p + sin θ sin θ p ) d = r cos (θ − θ p ) r=
858
d
cos (θ − θ p )
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Polar Equation of a Circle That Passes Through the Pole From the figure, we have the center of the circle ( h, k ) with h = a cos θ C , k = a sin θ C , and radius a.
( x − h) + ( y − k ) = a2 2 2 ( x − a cosθ C ) + ( y − a sin θ C ) = a 2 2
2
Changing to polar yields
(r
( r cosθ − a cosθ C )
2
2
+ ( r sin θ − a sin θ C ) = a 2 2
cos 2 θ − 2ra cosθ cos θ C + a 2 cos 2 θ C ) + ( r 2 sin 2 θ − 2ra sin θ sin θ C + a 2 sin 2 θ C1 ) = a 2
r 2 ( cos 2 θ + sin 2 θ ) − 2ra ( cosθ cosθ C + sin θ sin θ C ) + a 2 ( cos 2 θ C + sin 2 θ C ) = a 2 r 2 − 2ra ( cos θ cos θ C + sin θ sin θ C ) + a 2 = a 2 r 2 − 2ra ⎡⎣ cos (θ − θ C ) ⎤⎦ = 0 r − 2a ⎣⎡ cos (θ − θ C ) ⎦⎤ = 0 r = 2a ⎡⎣ cos (θ − θ C ) ⎤⎦
8.7 Parametric Equations 1.
Parametric Equations in an xyz-Coordinate System a. Graph of x = 3 cos t , y = 3 sin t , z = 0.5t , for 0 ≤ t ≤ 12 . z 6
4
y
2 2
-2
2
-2
x
b. Graph of x = 3 cos t , y = 6sin t , z = 0.5t , for 0 ≤ t ≤ 12 . z 6 4
y
2 2
-2 -2
2
x
c. The graph in part (a) is “circular,” whereas the graph in part (b) is “elliptical.” d. Each of the curves is a helix.
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Systems of Equations CHAPTER 9 of College Algebra and Trigonometry 9.1 Systems of Linear Equations in Two Variables 1. Independent and Dependent Conditions a.
⎧ x + y = 30 . Independent. The system of equations is ⎨ ⎩ x − y = 10 The solution of this system of equations is ( 20, 10 ) .
b. Dependent. The equation is 20 x + 30 y = 1000 . Because three are two variables and only one equation, there are an infinite number of possible solutions. The ordered pairs ( 50, 0 ) and ( 35, 10 ) are two possible solutions.
⎧ x + y = 20 The system of equations is ⎨ . The system of equations has no solution. This means it ⎩ 2 x = 10 − 2 y is impossible to satisfy the two conditions of the problem at the same time. d. Answers will vary. The answer should contain a word problem with two independent conditions. e. Answers will vary. The answer should contain a word problem with two dependent conditions. c.
9.2 Nonlinear Systems of Equations 1. Concept of Dimension In Flatland, people are two-dimensional polygons. A person’s station in life is determined by the number of sides of the polygon. People walk by sliding along the plane. The Flatlanders visit the people from the world of one dimension. These people live on a line. When the Flatlanders tell the ruler of the one-dimensional people that they can change position with their neighbors, the one-dimensional people cannot comprehend how such a movement would take place. The Flatlanders are dutifully smug about their ability to move in the plane and think they are superior to the one-dimensional people. Then strange phenomena begin when a three-dimensional person enters the world of the Flatlanders. For example, the Flatlanders cannot understand how the three-dimensional people enter their homes even though all the doors and windows are latched. 2.
Abilities of a Four-Dimensional Human Some of the best accounts have appeared in Scientific American. Descriptions of the capabilities of a fourdimensional person range from turning a tennis ball inside out without tearing it, to reaching into a closed safe and removing the contents.
9.3 Nonlinear Systems of Equations 1. Proving a Geometry Theorem a.
Substitute y = mx into ( x − a ) + y 2 = a 2 and solve for x. 2
( x − a)
2
+ m2 x 2 = a 2 x 2 − 2ax + a 2 + m 2 x 2 = a 2 (1 + m2 ) x 2 − 2ax = 0
(
)
x (1 + m 2 ) x − 2a = 0
Thus x = 0 or (1 + m 2 ) x − 2a = 0 .
2a 1 + m2 If x = 0 , then y = m ( 0 ) = 0 . One intersection is ( 0, 0 ) . x=
If x = 1+2ma 2 , then y = m
860
( )= 2a 1+ m 2
2 am 1+ m2
. A second intersection is
(
2a 1+ m2
)
, 12+am . m2
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b. The slope of the line through P and ( 2a, 0 ) is: Slope =
2.
2am 1+ m 2 2a 1+ m 2
−0 − 2a
=
2 am 1+ m2 −2 am2 1+ m2
=−
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1 . m
c.
The line segment OP is on the line given by y = mx . Thus the slope of OP is m.
d.
OP is perpendicular to PQ because their slopes are negative reciprocals of each other.
e. Since OP is perpendicular to PQ , angle OPQ is a right angle and triangle OPQ is a right triangle. Finding Zeros of a Polynomial Let a, b, and c be zeros of P(x) = x3 + 2x2 + Cx – 6 such that a = b + c . Because these are the zeros of P, the polynomial must factor (by the Factor Theorem) as (x – a)(x – b)(x – c). Now multiply this out and equate the coefficients to those of P. The results enables us to form the system of equations • Equating coefficients of x. a + b + c = −2 abc = −6
• Equating the constant term.
a =b+c
Solving this system yields a = −1, b =
• This is a condition of the problem. −1+ i 23 2
, and c =
−1− i 23 2
.
To find C, solve P ( −1) = 0 = ( −1) + 2 ( −1) + C ( −1) − 6, or C = −9 . 3
9.4 Partial Fractions 1. Using a Computer Algebra System The answer to this question will depend on the rational functions for which the student attempted a partial fraction decomposition using some type of computer algebra system. 9.5 Inequalities in Two Variables and Systems of Inequalities 1.
A Parallelogram Coordinate System Coordinate lines are drawn parallel to the coordinate axes and form a parallelogram. See the accompanying illustration. A point is located in much the same manner as in a rectangular coordinate system. However, displacement is along the edge of a parallelogram rather than of a rectangle.
The graph of 3x + 4 y = 12 is a straight line. In fact, all linear equations in two variables have a graph that is a straight line.
y
x
( 0, 0 ) 3 x + 4 y = 12
Here are some other observations a student may include: Transformation equations between rectangular coordinates and parallelogram coordinates are given by x′ = x and y ′ = y − x , where x′ and y ′ are the coordinates in the parallelogram system. One way to define the distance d between P1 ( x1 , y1 ) and P2 ( x2 , y2 ) is d = x1 − x2 + y1 − y2 .
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9.6 Linear Programming 1. History of Linear Programming Linear programming is a program, or method, of allocating resources. It is used in planning food distribution, building rockets, supplying military units with essential equipment, and farming, as well as in many other applications. George Dantzig developed the “simplex method” of solving linear programming problems in the late 1940s. The simplex method is basically a matrix method, analogous to row reduction. The result is the best method of allocating resources. In the early 1980s, Narendra Karmarkar of AT&T suggested an improvement on the simplex method that greatly reduced the time required for a computer to determine the optimal solution of a linear programming problem. In the mid-1980s, L. G. Khachian introduced a method that was supposed to revolutionize the technique of solving linear programming problems. Although his method has some theoretical importance, its practical applications have not been demonstrated.
862
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Matrices CHAPTER 10 of College Algebra and Trigonometry 10.1 Gaussian Elimination Method 1. Echelon Form by Using a Graphing Calculator 1 4⎤ ⎡ 2 −3 ⎢ a. ⎢ 1 2 −2 −2 ⎥⎥ ⎢⎣ 3 1 −3 4 ⎥⎦ b.
R1 ↔ R2
⎡ 1 2 −2 −2 ⎤ ⎢ 2 −3 1 4 ⎥⎥ ⎢ ⎢⎣ 3 1 −3 4 ⎥⎦
−3R1 + R3 → R3
5R2 + R3 → R3
−2R1 + R2 → R2
⎡ 1 2 −2 −2 ⎤ ⎢ 0 −7 5 8 ⎥⎥ ⎢ ⎢⎣ 0 −5 3 10 ⎥⎦ ⎡ 1 ⎢ ⎢ 0 ⎢⎣ 0
2
−2
1
−
0
−
5 7 4 7
⎛ 1⎞ ⎜ − ⎟ R2 ⎝ 7⎠
−2 ⎤ ⎥ − 87 ⎥ − 307 ⎥⎦
⎛ 7⎞ ⎜ − ⎟ R3 ⎝ 4⎠
⎡ 1 2 −2 −2 ⎤ ⎢ 0 −7 5 8 ⎥⎥ ⎢ ⎢⎣ 3 1 −3 4 ⎥⎦ 2 −2 −2 ⎤ ⎡ 1 ⎢ 0 1 − 75 − 78 ⎥⎥ ⎢ ⎢⎣ 0 −5 3 10 ⎥⎦ ⎡ 1 ⎢ ⎢ 0 ⎢⎣ 0
2 1 0
−2
−2 ⎤ ⎥ − − 87 ⎥ 1 − 152 ⎥⎦ 5 7
10.2 The Algebra of Matrices 1. Transformations ⎡ t ⎤ ⎡ 0 −1 0⎤ ⎡ t ⎤ ⎡ −t − 2⎤ a. R90 ⋅ ⎢ t + 2 ⎥ = ⎢1 0 0⎥ ⎢ t + 2 ⎥ = ⎢ t ⎥ ⇒ ( −t − 2, t ) ⇒ x = −t − 2, y = t ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ Using substitution, x = − y − 2 y = −x − 2 b.
c.
⎡ t ⎤ ⎡ −1 0 0 ⎤ ⎡ t ⎤ ⎡ − t ⎤ R y ⋅ ⎢ 3t − 1⎥ = ⎢ 0 1 0⎥ ⎢ 3t − 1⎥ = ⎢ 3t − 1⎥ ⇒ ( −t , 3t − 1) ⇒ x = − t , y = 3t − 1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ t = −x Using substitution, y = 3( − x ) − 1 y = −3 x − 1 ⎡ t ⎤ ⎡1 0 −1⎤ ⎡ t ⎤ ⎡ t − 1 ⎤ ⎢1⎥ ⎢ ⎢ ⎥ ⎢ ⎥ T−1,−1 ⋅ ⎢ ⎥ = 0 1 −1⎥ ⎢ 1 ⎥ = ⎢ 1 − 1⎥ ⎢ ⎥ ⎢ t ⎥ ⎢0 0 1 ⎥ ⎢ t ⎥ ⎢ t ⎥ ⎣ ⎦ ⎢1⎥ ⎢ 1 ⎥ ⎣⎢ 1 ⎦⎥ ⎣ ⎦ ⎣ ⎦ ⎡ t − 1 ⎤ ⎡ −1 0 0 ⎤ ⎡ t − 1 ⎤ ⎡ − t + 1 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ R180 ⋅ ⎢ 1 − 1⎥ = ⎢ 0 −1 0⎥ ⎢ 1 − 1⎥ = ⎢ − 1 + 1⎥ ⎢ ⎥ ⎢ t ⎥ ⎢ 0 0 1⎥ ⎢ t ⎥ ⎢ t ⎥ ⎣ ⎦⎢ 1 ⎥ ⎢ 1 ⎥ ⎣⎢ 1 ⎦⎥ ⎣ ⎦ ⎣ ⎦
⎡ −t + 1 ⎤ ⎡1 0 1⎤ ⎡ −t + 1 ⎤ ⎡ −t + 2 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ T1, 1 ⋅ ⎢ − 1 + 1⎥ = ⎢ 0 1 1⎥ ⎢ − 1 + 1⎥ = ⎢ − 1 + 2⎥ ⇒ ( −t + 2, − 1 + 2) ⇒ x = −t + 2, y = − 1 + 2 ⎢ ⎥ t t ⎢ t ⎥ ⎢ 0 0 1⎥ ⎢ t ⎥ ⎢ t ⎥ t = −x + 2 ⎣ ⎦ ⎢⎣ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ Using substitution, y = − 1 + 2 = −1 + −2 x + 4 = −2 x + 3 = 2 x − 3 −x + 2 −x + 2 −x + 2 −x + 2 x−2
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⎡ t ⎤ ⎡1 0 2 ⎤ ⎡ t ⎤ ⎡ t + 2 ⎤ d. T2,−1 ⋅ ⎢ t 2 ⎥ = ⎢0 1 −1⎥ ⎢t 2 ⎥ = ⎢ t 2 − 1⎥ ⇒ (t + 2, t 2 − 1) ⇒ x = t + 2, y = t 2 − 1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ t = x−2 ⎣⎢ 1 ⎦⎥ ⎣⎢0 0 1 ⎦⎥ ⎢⎣ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ Using substitution, y = ( x − 2) 2 − 1 y = x2 − 4 x + 3
e.
f.
2 ⎡ t ⎤ ⎡ 0 1 0⎤ ⎡ t ⎤ ⎡t ⎤ ⎢ ⎥ R270 ⋅ ⎢ t 2 ⎥ = ⎢ −1 0 0⎥ ⎢ t 2 ⎥ = ⎢ t ⎥ ⇒ (t 2 , t ) ⇒ x = t 2 , y = t ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣⎢ 1 ⎦⎥ ⎣⎢ 0 0 1⎦⎥ ⎢⎣ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ Using substitution, x = y2
2 ⎡ t ⎤ ⎡ 0 −1 0 ⎤ ⎡ t ⎤ ⎡ − t ⎤ ⎢ ⎥ R90 ⋅ ⎢ t 2 ⎥ = ⎢1 0 0⎥ ⎢ t 2 ⎥ = ⎢ t ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ 0 0 1⎥⎦ ⎢⎣ 1 ⎥⎦ ⎣⎢ 1 ⎦⎥
⎡ − t 2 ⎤ ⎡ 1 0 −2 ⎤ ⎡ − t 2 ⎤ ⎡ − t 2 − 2 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ T−2, 1 ⋅ ⎢ t ⎥ = ⎢0 1 1 ⎥ ⎢ t ⎥ = ⎢ t − 1 ⎥ ⇒ ( −t 2 − 2, t − 1) ⇒ x = − t 2 − 2, y = t − 1 ⎢ ⎥ ⎢⎣ 1 ⎥⎦ ⎣⎢0 0 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ t = y +1 Using substitution, x = −( y + 1) 2 − 2
2.
x = − y2 − 2 y − 3 Translations This project appears on our Internet site at college.hmco.com/info/aufmannCAT.
10.3 The Inverse of a Matrix 1. Multiple Regression Models ⎡ 90,500 ⎤ ⎢ 73,750 ⎥ ⎢ ⎥ B = ⎢117, 200⎥ ⎢ ⎥ ⎢ 59,500 ⎥ ⎢⎣ 74,800 ⎥⎦
⎡ 20 6 1⎤ ⎢16 5 1⎥ ⎢ ⎥ AT = ⎢19 12 1⎥ ⎢ ⎥ ⎢13 4 1⎥ ⎢⎣13 7 1⎥⎦
a.
⎡ 20 16 19 13 13⎤ A = ⎢ 6 5 12 4 7 ⎥ ⎢ ⎥ ⎢⎣ 1 1 1 1 1 ⎥⎦
d.
⎛ ⎡ 20 6 1⎤ ⎞ ⎛ ⎡ 90,500 ⎤ ⎞ ⎜ ⎢16 5 1⎥ ⎟ ⎜ 20 16 19 13 13 ⎢ 73,750 ⎥ ⎟ L 20 16 19 13 13 ⎡ ⎤ ⎜⎡ ⎤ ⎢ ⎤ ⎢ ⎥⎟ ⎜ ⎡ ⎥⎟ ⎢ M ⎥ = ⎜ ⎢ 6 5 12 4 7 ⎥ ⋅ ⎢19 12 1⎥ ⎟ ⋅ ⎜ ⎢ 6 5 12 4 7 ⎥ ⋅ ⎢117, 200⎥ ⎟ ⎢ ⎥ ⎜⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎟ ⎜ ⎢ ⎥⎟ ⎢⎣ N ⎥⎦ ⎜ ⎢⎣ 1 1 1 1 1 ⎥⎦ ⎢13 4 1⎥ ⎟ ⎜ ⎢⎣ 1 1 1 1 1 ⎥⎦ ⎢ 59,500 ⎥ ⎟ ⎜ ⎢⎣13 7 1⎥⎦ ⎟⎠ ⎜⎝ ⎢⎣ 74,800 ⎥⎦ ⎟⎠ ⎝
b.
c.
−1
−1
⎡1355 571 81⎤ ⎡6,962,700⎤ ⎡ 0.03098 −0.01613 −0.39214 ⎤ ⎡6,962,700 ⎤ = ⎢ 571 270 34 ⎥ ⋅ ⎢ 3,079,750 ⎥ ≈ ⎢ −0.01613 0.03417 0.02890 ⎥ ⋅ ⎢ 3,079,750 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 81 34 5 ⎥⎦ ⎢⎣ 415,750 ⎥⎦ ⎢⎣ −0.39214 0.02890 6.35622 ⎥⎦ ⎢⎣ 415,750 ⎥⎦ ⎡ 2974.14 ⎤ ≈ ⎢ 4963.20⎥ ⎢ ⎥ ⎣⎢ 1219.11 ⎦⎥
864
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Chapter 10 e. f. 2.
College Algebra and Trigonometry
Responses to Projects
S = 2974.14 x + 4963.20 y + 1219.11 S (10, 8) = 2974.14(10) + 4963.20(8) + 1219.11
= $70,666.10 Cryptography a. The ASCII (American Standard Code for Information Interchange) is a method by which each letter, punctuation mark, and numeral is given a two-number code. This system is used by all computer systems to exchange information. b. Answers will vary. The students should have an m × n matrix, W, in which m represents the length of a code packet and n represents the number of characters in the message. c. Answers will vary. The student should construct an m × m matrix that has an inverse. We will call this matrix E. d. Answers will vary depending on parts (b) and (c) above. However, the student should exhibit the product EW = M . e. Answers will vary. The student should show that E −1 M = W . That is, the product of the inverse of E and the encoded message should be the original message.
10.4 Determinants 1. Determinants, Matrices, and Area ⎡ 2 1 ⎤ ⎡ 4 4 10 10 ⎤ ⎡ 12 16 28 24 ⎤ a. AM = ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣ 3 2 ⎦ ⎣ 4 8 8 4 ⎦ ⎣ 20 28 46 38⎦ The new figure is a parallelogram. To find the area of the parallelogram, we will use the distance formula to find the length of the base and the formula for the distance between a point and a line to find the height. d=
(12 − 24 )
2
+ ( 20 − 38 ) = 468 = 6 13 2
To find the height, first determine the equation of the line through (12, 20) and (24, 38). 38 − 20 3 m= = 24 − 12 2 3 y − 20 = ( x − 12 ) 2 3 y = x+2 2 3 ( −2 4 13 2 16 ) + 2 − 28 h= = = 2 13 13 4 ( 32 ) + 1 ⎛ 4 13 ⎞ Area of parallelogram = 6 13 ⎜⎜ ⎟⎟ = 24 ⎝ 13 ⎠ Area of rectangle = ( 6 )( 4 ) = 24
(
)
The areas are the same, and det ( A ) = 4 − 3 = 1 .
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865
Chapter 10
b.
College Algebra and Trigonometry
Responses to Projects
⎡3 1⎤ ⎡ 4 4 10 10 ⎤ ⎡ 16 20 38 34⎤ AM = ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣1 1⎦ ⎣ 4 8 8 4 ⎦ ⎣ 8 12 18 14 ⎦ The new figure is a parallelogram. To find the area of the parallelogram, we will use the distance formula to find the length of the base and the formula for the distance between a point and a line to find the height.
d=
(16 − 34 )
2
+ ( 8 − 14 ) = 360 = 6 10 2
To find the height, first determine the equation of the line through (12, 20) and (24, 38). 14 − 8 1 = m= 34 − 16 3 1 ( x − 34 ) 3 1 8 y = x+ 3 3
y − 14 =
h=
( 13 ) 20 + 83 − 12 2 ( − 13 ) + 1
=
− 83 10 9
=
4 10 5
⎛ 4 10 ⎞ Area of parallelogram = 6 10 ⎜⎜ ⎟⎟ = 48 ⎝ 5 ⎠ Area of rectangle = ( 6 )( 4 ) = 24
(
)
The area of the new figure is twice that of the rectangle, and det ( A ) = 3 − 1 = 2 . c.
⎡ 1 2 ⎤ ⎡ 4 4 10 10 ⎤ ⎡ 12 20 26 18⎤ AM = ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣0.5 1 ⎦ ⎣ 4 8 8 4 ⎦ ⎣ 6 10 13 9 ⎦ The new figure is a line segment and therefore has no area. The student should verify this by showing that all the points lie on the same line. The area is 0; det ( A) = 1 − 1 = 0 .
d. The absolute value of the determinant of A times the original area equals the area of the figure produced by AM. e. Some students may not have realized that the absolute value of the determinant is required for their conjecture (in part d).
866
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Chapter 10
College Algebra and Trigonometry
10.5 Cramer’s Rule 1. Cramer’s Rule Prove Cramer’s Rule for a 3 by 3 system of linear equations. and z are variables. ax + by + cz = d ex + fy + gz = h jx + ky + mz = n
Responses to Projects
Consider the following system, where x, y, (I) (II) (III)
Multiply Equation (I) by g, and Equation (II) by (–c), and add. agx + bgy + cgz = dg −cex − cfy − cgz = −ch (ag − ce) x + (bg − cf ) y = dg − ch
(IV)
Multiply Equation (I) by m, and Equation (III) by (–c), and add. amx + bmy + cmz = dm −cjx − cky − cmz = −cn (am − cj ) x + (bm − ck ) y = dm − cn
(V)
Multiply Equation (IV) by ( bm − ck ) , and Equation (V) by ( −bg + cf ) , and add.
( bm − ck )( ag − ce ) x + ( bm − ck )( bg − cf ) y = ( bm − ck )( dg − ch ) ( −bg + cf )( am − cj ) x + ( −bg + cf )( bm − ck ) y = ( −bg + cf )( dm − cn ) ⎡⎣( bm − ck )( ag − ce ) + ( −bg + cf )( am − cj ) ⎤⎦ x = ( bm − ck )( dg − ch ) + ( −bg + cf )( dm − cn ) Thus x= = =
=
The results y =
( bm − ck )( dg − ch ) + ( −bh + cf )( dm − cn ) ( bm − ck )( ag − ce ) + ( −bg + cf )( am − ej ) bdgm − bchm − cdgk + c 2 hk − bdgm + bcgn + cdfm − c 2 fn abgm − bcem − acgk + c 2 ek − abgm + begj + acfm − cefj c {( dfm + bgn + chk ) + ( −cfn − bhm − dgk )} c {( afm + bgj + cek ) + ( −efj − bem − akg )} d b c h f g n k m a b c e f g j k m
a d c e h g j n m a b c e f g j k m
, z=
a b d e f h j k n a b c e f g j k m
can by established by using a similar approach.
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867
Chapter 11
College Algebra and Trigonometry
Responses to Projects
Sequences, Series, and Probability CHAPTER 11 of College Algebra and Trigonometry 11.1 Infinite Sequences and Summation Notation 1. Formulas for Infinite Sequences a. The idea is to write a formula that gives 2n for n = 1, 2, 3, 4 and 43 when n = 5. The formula contains two parts. The first part is zero for n = 1, 2, 3, 4. The second part is 2n. ⎡ n ( n − 1)( n − 2 )( n − 3)( n − 4 ) ⎤ an = ( 43 − 2n ) ⎢ ⎥ + 2n n! ⎣ ⎦ The first five terms of this sequence are 2, 4, 6, 8, 43. b. The idea is to write a formula that gives 2n for n = 1, 2, 3, 4 and x when n = 5. The formula contains two parts. The first part is zero for n = 1, 2, 3, 4. The second part is 2n. ⎡ n ( n − 1)( n − 2 )( n − 3)( n − 4 ) ⎤ an = ( x − 2n ) ⎢ ⎥ + 2n n! ⎣ ⎦ The first five terms of this sequence are 2, 4, 6, 8, x. 11.2 Arithmetic Sequences and Series 1. Angles of a Triangle a. 360° b. 540° c. 720° d. ( n − 3)180° 2.
Prove a Formula We know that S n =
( a1 + an )
n 2
and that an = a1 + ( n − 1) d . Substitute for an in the formula for S n and
simplify. Sn =
n n ⎡⎣ a1 + a1 + ( n − 1) d ⎤⎦ = ⎡⎣ 2a1 + ( n − 1) d ⎤⎦ 2 2
11.3 Geometric Sequences and Series 1. Fractals a.
The perimeter after completing the process n times is Pn = 4 ( 53 )
b. As n → ∞, Pn = 4 (
868
)
5 n −1 3
n −1
.
approaches infinity. Therefore, the perimeter is infinite.
k −1
c.
An = 1 + 4Σ nk =1 532 k
d.
25 An = 1 + 4Σ nk =1 532 k = 1 + 4 ( 19 + 815 + 729 +L) k −1
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Chapter 11
Beginning with
5 81
College Algebra and Trigonometry
, the series is an infinite geometric series with a =
5 81
Responses to Projects
and r = 95 .
25 ⎛1 5 ⎞ ⎛ 1 5 / 81 ⎞ 1+ 4⎜ + + + L⎟ = 1 + 4 ⎜ + ⎟ ⎝ 9 81 729 ⎠ ⎝ 9 1− 5 / 9 ⎠ ⎛1 9 ⎞ = 1+ 4⎜ + ⎟ ⎝ 9 36 ⎠ ⎛1⎞ = 1+ 4⎜ ⎟ = 2 ⎝4⎠
The area is 2 square units. 11.4 Mathematical Induction 1. Steps in a Mathematical Induction Proof a. Assume 2 + 4 + 8 + L + 2 k = 2 k +1 + 1 . Prove the formula is true for n = k + 1 . That is, prove that 2 + 4 + 8 + L + 2k + 2k +1 = 2k + 2 + 1 . S k +1 = 2 + 4 + 8 + L + 2k + 2k +1 = S k + 2k +1 = 2k +1 + 1 + 2k +1 = 2 ⋅ 2k +1 + 1 = 2k + 2 + 1 Thus the statement is true for n = k + 1 . b. Let n = 1 . 21 ≠ 2 2 + 1 = 5 . This statement is not true for n = 1 .
c.
2 + 4 + 8 + L + 2k = 2 (1 + 2 + 4 + 8 + L + 2k −1 )
Let N = 1 + 2 + 4 + 8 + L + 2k −1 . Thus 2 + 4 + 8 + L + 2k = 2 ( N ) = even number 2k+1 + 1 = 2 ⋅ 2k + 1 = even number +1 = odd number Therefore, the left side is always an even number and the right side is always an odd number. Thus the two values can never be equal. d. The Principle of Mathematical Induction requires that we first establish that there is at least one element in the set S. We did not do that, and consequently, we apparently “proved” a statement that is always false. 2.
The Tower of Hanoi a. The proof is by induction. If there is one disk ( n = 1) , then 21 − 1 = 2 − 1 = 1 and the game is
completed in one move. Assume that for k disks, the game can be completed in 2k − 1 moves. Prove that for n = k + 1 disks, the game is completed in 2k +1 − 1 moves. Consider one peg in which k + 1 disks are arranged. By the Induction Hypothesis, 2k − 1 moves are required to move the first k disks to another peg. Now move the k + 1 disk to the unoccupied peg. The total number of moves is now 2k − 1 + 1 . Now move the k disks back to the disk containing the k + 1 disk. This requires 2k − 1 moves (Induction Hypothesis). The total number of moves is 2k − 1 + 1 + 2k − 1 = 2 ⋅ 2k − 1 = 2k +1 − 1 Thus the statement is true for n = k + 1 , and the formula is established. b. From Exercise 2(a), it will take 264 − 1 seconds to complete the transfer. 264 − 1 seconds ≈ 5.85 × 1011 years = 585 billion years Thus the legend predicts that the universe will exist for approximately 580 billion more years.
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869
Chapter 11
College Algebra and Trigonometry
Responses to Projects
11.5 The Binomial Theorem 1. Pascal’s Triangle Two references a student can check are The History of Mathematics: An Introduction by David M. Burton, Dubuque, Iowa: William C. Brown, publishers, 1988 and The History of Mathematics: An Introduction by Victor J. Katz, New York: Harper Collins, 1993. 2.
Some Other Factorial Functions a. b.
For integers, Pochammer ( m, n ) =
( m + n − 2)! ( n −1)!
.
n !! = n ( n − 2 )( n − 4 )K 2 when n is an even integer and n !! = n ( n − 2 )( n − 4 )K1 when n is an odd
integer. 8.6 Permutations and Combinations 1. Explain Permutations and Combinations The student should prepare a lesson to explain permutations and combinations to a classmate. The lesson should contain at least five examples of permutations and five examples of combinations. 2.
Application of Counting a. n = 5, k = 3 ⎛ 3 ⎞ 5 ⎛ 3⎞ ⎛3⎞ 5 5 5 5 ⎜ ⎟ 3 − ⎜ ⎟ ( 3 − 1) + ⎜ ⎟ ( 3 − 2 ) = 3 − 3 ( 2 ) + 3 = 150 0 1 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
b.
n = 10, k = 4 ⎛ 4 ⎞ 10 ⎛ 4 ⎞ ⎛ 4⎞ ⎛ 4⎞ 10 10 10 10 10 10 ⎜ ⎟ 4 − ⎜ ⎟ ( 4 − 1) + ⎜ ⎟ ( 4 − 2 ) − ⎜ ⎟ ( 4 − 3) = 4 − 4 ( 3 ) + 6 ( 2 ) − 4 = 818,520 ⎝0⎠ ⎝1 ⎠ ⎝ 2⎠ ⎝3⎠
11.7 Introduction to Probability 1. Monte Hall Problem If the contestant stays with his or her first choice, then the probability that the contestant will win the grand prize is 1/3. The only other possibility is to go with one of the other curtains when given the opportunity to switch to a different curtain. This event of switching to another curtain is the compliment of staying with the first choice. Thus the probability of winning the grand prize by switching to a different curtain is 1 2 1 − = . This analysis shows that a contestant will double their chance of winning the grand prize by 3 3 switching rather than staying with their first choice. Many Internet sites discuss this famous problem. Here are two recommended sites: http://www.cut-the-knot.com/hall.html http://www.math.rice.edu/~ddonavan/montyurl.html
870
2.
Probability and Automatic Garage Door Openers The probability of at least two having the same garage door opener sequence is 1 – (the probability of none having the same sequence). 0 500 ⎛ 500 ⎞ ⎛ 1 ⎞ ⎛ 63 ⎞ P = 1− ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ≈ 1 − 0.00038 = 0.99962 ⎝ 0 ⎠ ⎝ 64 ⎠ ⎝ 64 ⎠ There is more than a 99.9% chance that at least two people will choose the same code sequence.
3.
Overbooking by Airlines This project appears on our Internet site at http://www.college.hmco.com.
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Additional College Trigonometry Solutions
Additional College Trigonometry, 6e Solutions Section 1.1 1.
2 x + 10 = 40 2 x = 40 − 10 2 x = 30 x = 15
2.
−3 y + 20 = 2 −3 y = 2 − 20 −3 y = −18 y=6
3.
5 x + 2 = 2 x − 10 5 x − 2 x = −10 − 2 3x = −12 x = −4
4.
4 x − 11 = 7 x + 20 4 x − 7 x = 20 + 11 −3x = 31 x = − 31 3
5.
2( x − 3) − 5 = 4( x − 5)
6.
6(5s − 11) − 12(2 s + 5) = 0 30s − 66 − 24 s − 60 = 0 6s − 126 = 0 6s = 126 s = 21
2 x − 6 − 5 = 4 x − 20 2 x − 11 = 4 x − 20 2 x − 4 x = −20 + 11 −2 x = −9 x=
9 2
7.
3x+1 = 2 4 2 3 ⎛ 3 1 12 ⎜ x + ⎞⎟ = 12 ⎛⎜ 2 ⎞⎟ 2⎠ ⎝4 ⎝ 3⎠ 9x + 6 = 8 9x = 8 − 6 9x = 2 x=2 9
8.
x −5= 1 4 2 4 ⎛⎜ x − 5 ⎞⎟ = 4 ⎛⎜ 1 ⎞⎟ ⎝4 ⎠ ⎝2⎠ x − 20 = 2 x = 2 + 20 x = 22
9.
2 x −5= 1 x −3 3 2 ⎛ ⎞ 2 6 ⎜ x − 5 ⎟ = 6 ⎛⎜ 1 x − 3 ⎞⎟ ⎝2 ⎠ ⎝3 ⎠ 4 x − 30 = 3x − 18 4 x − 3x = −18 + 30 x = 12
10.
1 x + 7 − 1 x = 19 2 4 2 4 ⎛⎜ 1 x + 7 − 1 x ⎞⎟ = 4 ⎛⎜ 19 ⎞⎟ 4 ⎠ ⎝2 ⎝ 2⎠ 2 x + 28 − x = 38 x = 38 − 28 x = 10
11.
0.2 x + 0.4 = 3.6 0.2 x = 3.2 x = 16
12.
0.04 x − 0.2 = 0.07 0.04 x = 0.27 x = 6.75
13.
3 ( n + 5) − 3 ( n − 11) = 0 5 4 3 3 ⎡ 20 n + 5) − ( n − 11)⎤ = 20 − 0 ⎢⎣ 5 ( ⎥⎦ 4 12 ( n + 5) − 15 ( n − 11) = 0 12n + 60 − 15n + 165 = 0 −3n = −225 n = 75
15.
3( x + 5)( x − 1) = (3 x + 4)( x − 2) 2
16.
2
18.
5( x + 4)( x − 4) = ( x − 3)(5 x + 4) 2
3x + 12 x − 15 = 3 x − 2 x − 8 14 x = 7 x=
− 5 ( p + 11) + 2 ( 2 p − 5) = 0 7 5 ⎡ 5 2 35 − ( p + 11) + ( 2 p − 5) ⎤ = 35 − 0 ⎢⎣ 7 ⎥⎦ 5 −25 ( p + 11) + 14 ( 2 p − 5) = 0 −25 p − 275 + 28 p − 70 = 0 3 p = 345 p = 115
14.
17.
0.08 x + 0.12(4000 − x) = 432 0.08 x + 480 − 0.12 x = 432 −0.04 x = −48 x = 1200
20.
3x − 5 y = 15 −5 y = −3x + 15 y = 3x−3 5
2
5 x − 80 = 5 x − 11x − 12 11x = 68
1 2
x=
0.075 y + 0.06(10,000 − y ) = 727.50 0.075 y + 600 − 0.06 y = 727.50 0.015 y = 127.5 y = 8500
19.
68 11
x + 2y = 8 2 y = −x + 8 y =−1x+4 2
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874
Additional College Trigonometry Solutions
2 x + 5 y = 10 2 x = −5 y + 10 x = −5 y +5 2
22.
24.
ax + by = c by = − ax + c y = − ax + c b
25.
27.
x 2 − 2 x − 15 = 0 a = 1 b = −2 c = −15
28.
21.
x 2 − 5 x − 24 = 0 a = 1 b = −5 c = −24
ay − by = c y ( a − b) = c y=
26.
29.
c a−b
2y − 3 y −1 x ( y − 1) = 2 y − 3 xy − x = 2 y − 3 xy − 2 y = x − 3 y ( x − 2) = x − 3 y = x−3 x−2 x=
x2 + x − 1 = 0 a = 1 b = 1 c = −1
− (−2) ± ( −2) 2 − 4(1)(−15) 2(1)
x=
− (−5) ± (−5) 2 − 4(1)(−24) 2(1)
x=
x=
2 ± 4 + 60 2
x=
5 ± 25 + 96 2
x=
−1± 1+ 4 2
x=
−1± 5 2
2 ± 64 2 ± 8 = 2 2 2 + 8 10 = = 5 or x= 2 2 2−8 −6 = = −3 x= 2 2
5 ± 121 5 ± 11 = 2 2 5 + 11 16 x= = = 8 or 2 2 5 − 11 − 6 x= = = −3 2 2 x=
x2 + x − 2 = 0 a = 1 b = 1 c = −2 x=
31.
− 1 ± 12 − 4(1)(−2) 2(1)
−1± 1+ 8 1± 9 −1± 3 = = 2 2 2 −1+ 3 2 x= = = 1 or 2 2 −1− 3 − 4 x= = = −2 2 2 x=
33.
y 1− y x (1 − y ) = y x − xy = y x = xy + y x = y ( x + 1) x =y x +1 x=
23.
x=
x=
30.
5 x − 4 y = 10 5 x = 4 y + 10 x= 4 y+2 5
3x 2 − 5 x − 3 = 0 a = 3 b = −5 c = −3
34.
2 x2 + 4 x + 1 = 0 a = 2 b = 4 c =1
32.
− 1 ± 12 − 4(1)(−1) 2
2 x2 + 4 x − 1 = 0 a = 2 b = 4 c = −1
x=
− 4 ± 42 − 4( 2)(1) 2(2)
x=
− 4 ± 42 − 4(2)(−1) 2( 2)
x=
− 4 ± 16 − 8 4
x=
− 4 ± 16 + 8 4
x=
−4± 8 −4±2 2 = 4 4
x=
− 4 ± 24 − 4 ± 2 6 = 4 4
x=
−2± 2 2
x=
−2± 6 2
3x 2 − 5 x − 4 = 0 a = 3 b = −5 c = −4
x=
− (−5) ± (−5) 2 − 4(3)(−3) 2(3)
x=
− (−5) ± (−5) 2 − 4(3)(−4) 2(3)
x=
5 ± 25 + 36 6
x=
5 ± 25 + 48 6
x=
5 ± 61 6
x=
5 ± 73 6
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Additional College Trigonometry Solutions
35.
1 2 3 x + x −1 = 0 2 4 1 3 a= b= c = −1 2 4
x=
x=
−3 ± 4
−3 ± 4
()
875
36.
()
2
3 2 − 4 1 (−1) 4 2 2 1 2
()
2 2 1 x − 5x + = 0 3 2 2 1 a= b = −5 c = 3 2
x=
9 +2 16
x=
1 3 41 x=− ± 4 16
x=
3 41 x=− ± 4 4 −3 ± 41 x= 4
x=
a= 2
( 3 ) ( 12 )
−( −5) ± (−5) − 4 2
(3)
2 2
x=
−3 ± 32 − 4 ⋅ 2 ⋅ 2 2 2
x=
−3 ± 9 − 8 2 2
x=
3
4 3 71 3
5±
b=3 c= 2
−3 + 1 = 2 2 −3 − 1 = x= 2 2
25 − 4
5±
2 x2 + 3x + 2 = 0
37.
−2 2 =− 2 2 2 −4 =− 2 2 2
or
4 3
(5 ± 213 ) ⎛ 3 ⎞ 3 ⎜ ⎟ 4 ⎝3⎠ 3
15 ± 213 x= 4 38.
2 x2 + 5 x − 3 = 0
a=2 b= 5
39.
x 2 − 3x − 5 = 0 a =1 b = −3
c = −3
− 5 ± ( 5 )2 − 4(2)(−3) x= 2⋅2 x=
− 5 ± 5 + 24 4
− 5 ± 29 x= 4
41.
44.
47.
x 2 − 2 x − 15 = 0 ( x + 3)( x − 5) = 0 or x − 5 = 0 x+3= 0 x = −3 x=5
42.
12w2 − 41w + 24 = 0 (4 w − 3)(3w − 8) = 0 4 w − 3 = 0 or 3w − 8 = 0 3 8 w= w= 4 3
45.
( x − 5) 2 − 9 = 0 [( x − 5) − 3][( x − 5) + 3] = 0
48.
x − 8 = 0 or x − 2 = 0 x=8 x=2
x 2 = 3x + 5
40.
− x2 = 7 x − 1 − x2 − 7 x + 1 = 0 a = −1 b = −7
c = −5
c =1
x=
− (−3) ± (−3)2 − 4(1)(5) 2(1)
x=
− (−7) ± (−7)2 − 4(−1)(1) 2( −1)
x=
3 ± 9 + 20 2
x=
7 ± 49 + 4 −2
x=
3 ± 29 2
x=
7 ± 53 −2
x=
− 7 ± 53 2
y 2 + 3 y − 10 = 0 ( y + 5)( y − 2) = 0 y+5 = 0 or y − 2 = 0 y = −5 y=2
43.
3x 2 − 7 x = 0 x (3 x − 7 ) = 0 x = 0 or 3x − 7 = 0 7 x= 3
46.
(3 x + 4)2 − 16 = 0 [(3x + 4) − 4)][(3 x + 4) + 4] = 0 3x = 0 or 3x + 8 = 0
49.
x=0
x=−
8 y 2 + 189 y − 72 = 0 (8 y − 3)( y + 24) = 0
8y − 3 = 0 3 y= 8
or
y + 24 = 0 y = −24
5 x 2 = −8 x 2
5x + 8x = 0 x(5 x + 8) = 0 x = 0 or 5 x + 8 = 0 x = −8 5 2 x + 3 < 11 2x < 8 x<4
8 3
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876
Additional College Trigonometry Solutions
50.
3x − 5 > 16 3 x > 21 x>7
51.
x + 4 > 3x + 16 −2 x > 12 x < −6
52.
5x + 6 < 2x + 1 3x < −5 5 x<− 3
53.
−6 x + 1 ≥ 19 −6 x ≥ 18 x ≤ −3
54.
−5 x + 2 ≤ 37 −5 x ≤ 35 x ≥ −7
55.
−3( x + 2) ≤ 5 x + 7 −3x − 6 ≤ 5 x + 7 − 8 ≤ 13 13 x≥− 8
56.
−4( x − 5) ≥ 2 x + 15 −4 x + 20 ≥ 2 x + 15 − 6 x ≥ −5 5 x≤ 6
57.
−4(3x − 5) > 2( x − 4) −12 x + 20 > 2 x − 8 − 14 x > −28 x<2
58.
3( x + 7) ≤ 5( 2 x − 8) 3x + 21 ≤ 10 x − 40 − 7 x ≤ −61 61 x≥ 7
59.
x2 + 7 x > 0 x( x + 7) > 0
60.
The product is positive. The critical values are 0 and –7.
The product is negative or zero. The critical values are 0 and 5.
x( x − 7)
x( x − 5)
(−∞, − 7) ∪ (0, ∞) 61.
[0, 5]
x 2 + 7 x + 10 < 0
62.
x2 + 5x + 6 < 0
( x + 5)( x + 2) < 0
( x + 3)( x + 2) < 0
The product is negative. The critical values are –5 and –2.
The product is negative. The critical values are –3 and –2.
( x + 5)( x + 2)
( x + 3)( x + 2)
(−3, − 2)
(−5, − 2) 63.
x2 − 5x ≤ 0 x( x − 5) ≤ 0
x 2 − 3x ≥ 28 x − 3x − 28 ≥ 0 ( x + 4)( x − 7) ≥ 0
x 2 + x − 30 < 0 ( x − 5)( x + 6) < 0
The product is positive or zero. The critical values are –4 and 7. ( x + 4)( x − 7)
(−∞, − 4] ∪ [7, ∞)
x 2 < − x + 30
64.
2
The product is negative. The critical values are 5 and –6. ( x − 5)( x + 6)
(−6, 5)
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Additional College Trigonometry Solutions
877
6 x2 − 4 ≤ 5x
65.
12 x 2 + 8 x ≥ 15
66.
6 x2 − 5x − 4 ≤ 0 (3x − 4)(2 x + 1) ≤ 0
12 x 2 + 8 x − 15 ≥ 0 (6 x − 5)(2 x + 3) ≥ 0
The product is negative or zero.
The product is positive or zero.
The critical values are 4 and − 1 . 3 2
The critical values are 5 and − 3 . 6
(6 x − 5)(2 x + 3)
(3 x − 4)(2 x + 1)
⎛ 3⎤ ⎡5 ⎞ ⎜⎜ − ∞,− ⎥ ∪ ⎢ , ∞ ⎟⎟ 2⎦ ⎣6 ⎠ ⎝
⎡ 1 4⎤ ⎢− 2 , 3 ⎥ ⎣ ⎦ 67.
70.
68.
x <4
69.
x >2
x −1 < 9
−4 < x < 4
x < −2 or x > 2
−9 < x − 1 < 9 − 8 < x < 10
(−4, 4)
(−∞, − 2) ∪ (2, ∞)
(−8, 10)
71.
x − 3 < 10
72.
x + 3 > 30 x + 3 < −30 or x + 3 > 30
−10 < x − 3 < 10 − 7 < x < 13
x < −33
74.
2x − 1 > 4 2 x − 1 < −4 or 2 x − 1 > 4 2 x < −3 2x > 5 3 5 x<− x> 2 2
2x − 9 < 7
x+4 < 2 −2 < x + 4 < 2
x > 27
− 6 < x < −2
(−∞, − 33) ∪ (27, ∞)
(−7, 13) 73.
2
(−6, − 2) 75.
−7 < 2 x − 9 < 7
x+3 ≥ 5 x + 3 ≤ −5 or x + 3 ≥ 5
2 < 2 x < 16 1< x < 8
x ≤ −8
x≥2
(−∞, − 8] ∪ [2, ∞)
(1, 8)
3⎞ ⎛ ⎛5 ⎞ ⎜ − ∞, − ⎟ ∪ ⎜ , ∞ ⎟ 2⎠ ⎝ ⎝2 ⎠ 76.
77.
x − 10 ≥ 2
78.
−14 ≤ 3 x − 10 ≤ 14 − 4 ≤ 3 x ≤ 24 − 4 ≤ x≤8 3
x − 10 ≤ −2 or x − 10 ≥ 2 x≤8
3 x − 10 ≤ 14
x ≥ 12
(−∞, 8) ∪ (12, ∞)
2 x − 5 ≤ −1 2x ≤ 4 x≤2
80.
4 − 5 x ≥ 24 4 − 5 x ≤ 24 −5 x ≤ −28 x ≥ 28 5
or
4 − 5 x ≥ 24 −5 x ≥ 20 x ≤ −4
⎡ (−∞, − 4] ∪ ⎢ 28 , ∞ ⎟⎞ ⎠ ⎣5
3 − 2x ≤ 5 −5 ≤ 3 − 2 x ≤ 5 − 8 ≤ −2 x ≤ 2 4 ≥ x ≥ −1
[−1, 4]
or 2 x − 5 ≥ 1 or
2x ≥ 6 x≥3
(−∞, 2] ∪ [3, ∞)
⎡ 4 ⎤ ⎢ − 3 , 8⎥ ⎣ ⎦ 79.
2x − 5 ≥ 1
81.
x −5 ≥ 0
Because an absolute value is always nonnegative, the inequality is always true. The solution set consists of all real numbers. (−∞, ∞)
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878
82.
Additional College Trigonometry Solutions
83.
x−7 ≥ 0
Because an absolute value is always nonnegative, the inequality is always true. The solution set consists of all real numbers.
84.
x−4 ≤ 0
2x + 7 ≤ 0
Because an absolute value is always nonnegative, the inequality x − 4 < 0
2x + 7 = 0 2 x = −7 7 x=− 2
has no solution. Thus the only solution of the inequality x − 4 ≤ 0 is the solution of the equation x – 4 = 0.
(−∞, ∞)
x=4 85.
A = 35
86.
LW = 60
LW = 35 L=
A = 60 A = LW
A = LW 35 W
W =
P = 34 P = 2 L + 2W 2 L + 2W = 34 L + W = 17 60 L + =17 L
P = 27 P = 2 L + 2W 2 L + 2W = 27
⎛ 35 ⎞ 2 ⎜ ⎟ + 2W = 27 ⎝W ⎠ 70 + 2W 2 = 27W
L2 − 17 L + 60 = 0 ( L − 12)( L − 5) = 0
2W 2 − 27W + 70 = 0 (2W − 7)(W − 10) = 0 W =
7 2
L=5
L = 12
or W = 10
W =
7 L 2 70 = 7 L 10 = L 35 =
60 L
35 = LW 35 = 10 L 3.5 = L
60 =5 12
W =
60 = 12 5
The rectangle measures 5 ft by 12 ft.
The rectangle measures 3.5 cm by 10 cm. 87.
A = 1500 = lw P = 600 − 2l + 3w 15000 l= w
88.
2l + 3w = 600 ⎛ 15000 ⎞ 2⎜ ⎟ + 3w = 600 ⎝ w ⎠
P = 4 w + 2l = 400 2 w + l = 200 A = 4800 = lw 4800 l= w
2w +
4800 = 200 w
2 w2 + 4800 = 200w
30,000 + 3w2 = 600w
w2 − 100w + 2400 = 0 ( w − 60)( w − 40) = 0
3w2 − 600 w + 30,000 = 0 3( w2 − 200w + 10,000) = 0 3( w − 100)( w − 100) = 0
w = 60 4800 = 80 l= 60
w = 100 ft 15000 l= = 150 ft 100
The dimensions are 100 feet by 150 feet.
w = 40 4800 = 120 l= 40
There are two solutions: 60 yd × 80 yd or 40 yd × 120 yd.
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Additional College Trigonometry Solutions
89.
91.
879
Plan A: 5 + 0.01x
90.
Company A: 19 + 0.12M
Plan B: 1 + 0.08x
Company B: 12 + 0.21M
5 + 0.01x < 1 + 0.08x 4 < .07x 57.1 < x
19 + 0.12M < 12 + 0.21M 7 < 0.09M 77.7 < M
Plan A is less expensive if you use at least 58 checks.
Company A is less expensive if you drive at least 78 miles.
Plan A: 100 + 8x
92.
Plan A: 15 + 1.49x
93.
Plan B: 250 + 3.5x
Plan B: 1.99x
100 + 8x > 250 + 3.5x 4.5x > 150 x > 33.3
1.99x < 15 + 1.49x 0.50x < 15 x < 30
Plan A pays better if at least 34 sales are made.
If fewer than 30 videos are rented, Plan B is less expensive.
68 ≤ F ≤ 104 9 68 ≤ C + 32 ≤ 104 5 9 36 ≤ C ≤ 72 5 20 ≤ C ≤ 40
....................................................... 94.
a.
l w = w l−w
Connecting Concepts 95.
l (l − w) = w2 l 2 − lw = w2
So 1 + 2 + 3 + L + 21 + 22 = 253 .
0 = w2 + lw − l 2 w=
− l ± l 2 − 4(−l 2 ) 2
Since w > 0, w =
b.
96.
⎛ −1+ 5 ⎞ −l +l 5 ⎟ = l⎜ ⎟ ⎜ 2 2 ⎠ ⎝
⎛ −l + 5 ⎞ ⎟ ≈ 62.4 ft w = 101 ⎜ ⎟ ⎜ 2 ⎠ ⎝ 464 =
a.
n (n − 3) 2
n 2 − 3n − 928 = 0
R = 420 x − 2 x 2
2 x(210 − x) > 0 The product is positive. The critical values are 0 and 210.
The polygon has 32 sides. 12 =
97.
420 x − 2 x 2 > 0
(n − 32)(n + 29) = 0 n = 32
b.
253 = 1 n( n + 1) 2 n 2 + n − 506 = 0 ( n + 23)( n − 22) = 0 n = 22
2 x(210 − x)
n (n − 3) 2
24 = n 2 − 3 0 = n 2 − 3 − 24
(0, 210)
n 2 − 3n − 24 is not factorable over the integers. Thus, the polygon in a. cannot have 12 diagonals.
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880
98.
Additional College Trigonometry Solutions
99.
x −3 < 8
a.
s = −16t 2 + v0t + s0
v0 = 64,
s0 = 0
−16t + 64t > 48
x− j < k
b.
s > 48,
2
− 16t + 64t − 48 > 0 − 16(t 2 − 4t + 3) > 0 − 16(t − 3)(t − 1) > 0 The product is positive. The critical values are t = 3 and t = 1. −16(t − 3)(t − 1)
1 second < t < 3 seconds The ball is higher than 48 ft between 1 and 3 seconds. 100.
v0 = 80,
s0 = 32
101. a.
s − 4.25 ≤ 0.01
b.
s − 4.25 = 0.01,
− 16t 2 + 80t + 32 > 96 − 16t 2 + 80t − 64 > 0
or
s = 4.26
− 16(t 2 − 5t + 4) > 0 − 16(t − 4)(t − 1) > 0
s − 4.25 = −0.01 s = 4.24 critical values
4.24 ≤ s ≤ 4.26
The product is positive. The critical values are t = 4 and t = 1. −16(t − 4)(t − 1)
1 second < t < 4 seconds The ball is higher than 96 ft between 1 and 4 seconds.
.......................................................
Prepare for Section 1.2
PS1.
4 + (−7) −3 = 2 2
PS2.
50 = 25 2 = 5 2
PS3.
y = 3x − 2
PS4.
y = (−3) 2 − 3( −3) − 2 y =9+9−2 y = 16
?
5 = 3(−1) − 2 No, the equation is not true. 5 ≠ −5
PS5.
−3 − ( −1) = −3 + 1 = −2 = 2
PS6.
(−3)2 − 4(−2)(2) = 9 + 16 = 25 = 5
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Additional College Trigonometry Solutions
881
Section 1.2
....................................................... PS1.
y = 3x + 12 0 = 3x + 12 −12 = 3x −4 = x
PS3.
y2 = x a2 =9 a = −3 or 3
Prepare for Section 1.3 PS2.
y = x 2 − 4x + 3 0 = x 2 − 4x + 3 0 = ( x − 1)( x − 3) x −1 = 0 x − 3 = 0 x =1 x=3 1 and 3
PS4. d = (3 − (−4))2 + (−2 −1)2 = 49 + 9 = 58
y2 = x b2 = 9 b = −3 or 3
PS5. –4
PS6. Decrease
Section 1.3
....................................................... PS1.
Prepare for Section 1.4 PS2.
f (3) =
3(3)4 (3) 2 + 1
f ( −3) =
= 243 = 24.3 10
3( −3)4
( −3)2 + 1 f (3) = f ( −3)
= 243 = 24.3 10
The graph of g is one unit above the graph of f. PS3.
f ( −2) = 2( −2)3 − 5( −2) = −16 + 10 = −6
PS4.
− f (2) = −[2(2)3 − 5(2)] = −[16 − 10] = −6 f ( −2) = − f (2)
f ( −2) − g ( −2) = ( −2) 2 − [ −2 + 3] = 4 − 1 = 3 f ( −1) − g ( −1) = ( −1)2 − [ −1 + 3] = 1 − 2 = −1 f (0) − g (0) = (0) 2 − [0 + 3] = 0 − 3 = −3 f (1) − g (1) = (1)2 − [1 + 3] = 1 − 4 = −3 f (2) − g (2) = (2)2 − [2 + 3] = 4 − 5 = −1
PS5.
− a + a = 0, b + b = b 2 2 midpoint is (0, b)
PS6.
− a + a = 0, −b + b = 0 2 2 midpoint is (0, 0)
Section 1.4
....................................................... PS1.
f (3) − g (3) = ( 3 2 + 3(3) + 1) − ( 4(3) + 5) = 19 − 17 =2
Prepare for Section 1.5 PS2.
f ( −2) ⋅ g ( −2) = ( 3( −2) 2 − ( −2) − 4 ) ⋅ ( 2( −2) − 5) = (12 + 2 − 4 ) ⋅ ( −9 ) = 10 ⋅ ( −9) = −90
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882
PS3.
Additional College Trigonometry Solutions
f (3a ) = 2(3a )2 − 5(3a ) + 2
f (2 + h ) = 2(2 + h )2 − 5(2 + h ) + 2
PS4.
= 2h 2 + 8h + 8 − 5h − 10 + 2
= 18a 2 − 15a + 2
= 2h 2 + 3h PS6. 2 x − 8 = 0 x=4 Domain: x > 4 or [4, ∞)
PS5. Domain: all real numbers except x = 1
Section 1.5
.......................................................
Prepare for Section 1.6
y +1 y xy = y + 1 xy − y = 1 y ( x − 1) = 1 y= 1 x −1 x=
PS1. 2 x + 5 y = 15 5 y = −2 x + 15 y =−2x+3 5
PS2.
PS4. (3, 7)
PS5. All real numbers.
PS3.
f (−1) =
2(−1) 2 = 2 = −1 (−1) − 1 −2
PS6. x + 2 ≥ 0 x ≥ −2 { x x ≥ −2}
Section 1.6
.......................................................
Prepare for Section 1.7
See CAT Prepare for Section 2.7 solutions on page 156.
....................................................... 1.
a, c, d, e
2.
Chapter 1 Assessing Concepts
f [ g ( x )] = 3(2 x + 4) + 8 = 6 x + 12 + 8 = 6 x + 20
3.
f (2) = 3 f ( x ) = f ( x + 4) f (2) = f (2 + 4) = f (6) f (6) = f (6 + 4) = f (10) f (10) = f (10 + 4) = f (14) f (14) = f (14 + 4) = f (18) Thus, f (18) = f (2) = 3.
6.
(3, –2)
g [ f ( x )] = 2(3x + 8) + 4 = 6 x + 16 + 4 = 6 x + 20 Thus f [ g ( x )] = 6 x + 20 = g [ f ( x ) ] .
To be inverse functions, f [ g ( x )] = x = g [ f ( x )] . No. They are not inverse functions. 4.
7.
x+2 <3
(7, 3)
5.
It is the slope of the line between ( a, f (a ) ) and ( b, f (b) ) . 8.
(–3, 6)
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Additional College Trigonometry Solutions
9.
883
(3, 4)
10.
Yes. The slope of the regression line is negative.
.......................................................
Chapter 1 Chapter Review
1.
3 − 4 z = 12 −4 z = 9 z =−9 4
2.
4y − 3 = 6y + 5 −2 y = 8 y = −4
3.
2 x − 3(2 − 3x ) = 14 x 2 x − 6 + 9 x = 14 x −6 = 3 x −2 = x
4.
5 − 2(3m + 2) = 3(1 − m ) 5 − 6m − 4 = 3 − 3m 1 − 6m = 3 − 3m −3m = 2 m=−2 3
5.
y 2 − 3 y − 18 = 0 ( y − 6)( y + 3) = 0
6.
2z 2 − 9z + 4 = 0 (2 z − 1)( z − 4) = 0
3v 2 + v = 1 3v + v − 1 = 0
8.
7.
y − 6 = 0 or y + 3 = 0 y = 6 or y = −3
2
v= v=
10.
−1±
2z − 1 = 0 2z = 1 1 z= 2
3s = 4 − 2 s 2
9.
3c − 5 ≤ 5c + 7 −2c ≤ 12 c ≥ −6
12.
2x 2 − x < 1 2x − x − 1 < 0 (2 x + 1)( x − 1) < 0
2 s 2 + 3s − 4 = 0
12 − 4(3)(−1) 2(3)
s=
− 1 ± 13 6
7a > 5 − 2(3a − 4) 7 a > 5 − 6a + 8 13a > 13 a >1
11.
−3±
s=
− 3 ± 9 + 32 4
s=
− 3 ± 41 4
x 2 − x − 12 ≥ 0 ( x − 4)( x + 3) ≥ 0
2
Critical values are 4 and − 3.
Critical values are − −
14.
2x − 5 > 3 2 x − 5 > 3 or 2x > 8 x>4
2x < 2 x <1
(−∞, 1) ∪ (4, ∞) 16.
c2 = a2 + b2 20 2 = 11 2 + b 2 b = 400 − 121 b = 279 b ≈ 16.7
1 − 3x ≤ 4
15.
−4 ≤ 1 − 3x ≤ 4 − 5 ≤ −3 x ≤ 3 5 ≥ x ≥ −1 3
2 x − 5 < −3
17.
z−4= 0 z=4
32 − 4(2)(−4) 2( 2)
(−∞, − 3] ∪ [4, ∞)
13.
or
c2 = a2 + b2 13 2 = a 2 + 12 2 a = 169 − 144 a = 25 a =5
1 < x <1 2
c2 = a 2 + b2 c 2 = 6 2 + 82 c = 36 + 64 c = 100 = 10
18.
c2 = a2 + b2 c 2 = 7 2 + 14 2 c = 49 + 196 c = 245 c ≈ 15.7
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1 and 1. 2
884
Additional College Trigonometry Solutions
(7 − ( −3)) 2 + (11 − 2) 2 100 + 81 181
19.
d= = =
22.
⎛ −4 + 8 7 + (−11) ⎞ , ⎟ = (2, − 2) ⎜ 2 ⎠ ⎝ 2
25.
y = x2 − 7 Replace y with − y.
20.
d = ( −3 − 5) 2 + ( −8 − ( −4)) 2 = 64 + 16 = 80 = 4 5
23.
21.
⎞ ⎛ 2 + ( −3) 8 + 12 ⎞ ⎛ 1 , ⎟ = ⎜ − , 10 ⎟ ⎜ 2 2 2 ⎠ ⎠ ⎝ ⎝
24.
26.
x = y2 + 3 y2 = x − 3
2
−y = x − 7
y = ± x−3
y = − x2 + 7 Thus, y is not symmetric with respect to the x-axis. Replace x with − x. y = (− x)2 − 7 y = x2 − 7 Thus, y is symmetric with respect to the y-axis. Replace x with − x and replace y with − y. − y = (− x)2 − 7 y = − x2 + 7 Thus, y is not symmetric with respect to the origin. 2
Therefore, the graph of y = x − 7 is symmetric with respect to the y-axis.
Replace y with − y. x = (− y )2 + 3 x = y2 + 3 y2 = x − 3 y = ± x−3 Thus, y is symmetric with respect to the x-axis. Replace x with − x. −x = y2 + 3 y2 = − x − 3 y = ± −x − 3 Thus, y is not symmetric with respect to the y-axis. Replace x with − x and replace y with − y.
− x = (− y )2 + 3 −x = y2 + 3 y2 = − x − 3 y = ± −x − 3 Thus, y is not symmetric with respect to the origin. Therefore, the graph of x = y 2 + 3 is symmetric with respect to the x-axis.
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Additional College Trigonometry Solutions
27.
885
y = x3 − 4 x Replace y with − y.
28.
y 2 = x2 + 4 y = ± x2 + 4 Replace y with − y.
3
− y = x − 4x y = − x3 + 4 x Thus, y is not symmetric with respect to the x-axis. Replace x with − x.
(− y )2 = x 2 + 4 y 2 = x2 + 4 y = ± x2 + 4 y is symmetric with respect to the x-axis. Replace x with − x.
y = (− x)3 − 4(− x) y = − x3 + 4 x Thus, y is not symmetric with respect to the y-axis. Replace x with − x and replace y with − y.
y 2 = (− x)2 + 4 y 2 = x2 + 4
− y = (− x)3 − 4(− x)
y = ± x2 + 4 Thus, y is symmetric with respect to the y-axis. Replace x with − x and replace y with − y.
− y = − x3 + 4 x y = x3 − 4 x Thus, y is symmetric with respect to the origin.
(− y ) 2 = (− x)2 + 4
Therefore, the graph of y = x3 − 4 x is symmetric with respect to the origin.
y 2 = x2 + 4 y = ± x2 + 4 Thus, y is symmetric with respect to the origin.
Therefore, the graph of y 2 = x 2 + 4 is symmetric with respect to the x-axis, the y-axis, and the origin. 29.
x2 2
+
y2
30.
=1
2
3 4 Replace y with − y. x2
+
(− y )2
x2
y2
(− y )2 2
=1⇒
+
y2
x2
x2
2
+
y2
=1 3 4 3 42 Thus, y is symmetric with respect to the origin. 2
+
=1⇒
x2
=1 32 42 32 42 Thus, y is symmetric with respect to the y-axis. Replace x with − x and replace y with − y. (− x)2
+
+
y2
=1 32 42 32 42 Thus, y is symmetric with respect to the x-axis. Replace x with − x. (− x)2
=1⇒
y2
+ = 1 is symmetric with 32 42 respect to the x-axis, the y-axis, and the origin.
Therefore, the graph of
xy = 8 8 y= x Replace y with − y. x(− y ) = 8 8 y=− x Thus, y is not symmetric with respect to the x-axis. Replace x with − x. − xy = 8 8 y=− x Thus, y is not symmetric with respect to the y-axis. Replace x with − x and replace y with − y. − x(− y ) = 8 8 x Thus, y is symmetric with respect to the origin. y=
Therefore, the graph of xy = 8 is symmetric with respect to the origin.
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886
31.
Additional College Trigonometry Solutions
32.
y = x
x+ y = 4 Replace y with − y.
Replace y with − y. −y = x ⇒ y = x
x + (− y ) = 4 ⇒ x − y = 4
Thus, y is symmetric with respect to the x-axis. Replace x with − x.
Thus, y is not symmetric with respect to the x-axis. Replace x with − x.
y = −x ⇒ y = x
(− x) + y = 4 ⇒ y − x = 4
Thus, y is symmetric with respect to the y-axis. Replace x with − x and replace y with − y.
Thus, y is not symmetric with respect to the y-axis. Replace x with − x and replace y with − y.
− y = −x ⇒ y = x
( − x ) + ( − y ) = 4 ⇒ −( x + y ) = 4 ⇒ x + y = 4
Thus, y is symmetric with respect to the origin.
33.
Thus, y is symmetric with respect to the origin.
Therefore, the graph of y = x is symmetric with respect to
Therefore, the graph of x + y = 4 is symmetric with respect
the x-axis, the y-axis, and the origin.
to the origin.
center (3, − 4), radius 9
34.
x 2 + 10 x
+ y2 + 4 y
= −20
x 2 + 10 x + 25 + y 2 + 4 y + 4 = 25 + 4 − 20 ( x + 5)2
+ ( y + 2) 2
=9
center ( −5, − 2), radius 3 35.
( x − 2)2 + ( y + 3)2 = 52
37.
a.
f (1) = 3(1)2 + 4(1) − 5 =2
36.
( x + 5) 2 + ( y − 1)2 = 82 , radius = − 5 − (3) = 8
38.
a.
= 55
b.
f (−3) = 27 − 12 − 5 = 10
c.
2
f (t ) = 3t + 4t − 5
d.
f ( x + h) = 3( x + h)2 + 4( x + h) − 5
b.
f.
= 39
3 f (t ) = 9t 2 + 12t − 15
c.
g (8) = 64 − 82 =0
d.
g (− x) = 64 − (− x)2 = 64 − x 2
f (3t ) = 3(3t )2 + 4(3t ) − 5 = 27t 2 + 12t − 5
g (−5) = 64 − (−5)2 = 64 − 25
= 3x 2 + 6 xh + 3h 2 + 4 x + 4h − 5
e.
g (3) = 64 − 32
e. f.
2 g (t ) = 2 64 − t 2 g (2t ) = 64 − (2t )2 = 64 − 4t 2 = 2 16 − t 2
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Additional College Trigonometry Solutions
39.
a.
( f ° g )(3) = f [ g (3)] = f [3 − 8] = f [−5]
887
40.
a.
( f ° g )(−5) = f [ g (−5)] = f ⎡⎣ (−5) − 1 ⎦⎤ = f [6]
= (−5) 2 + 4(−5) =5 b.
= 2(6)2 + 7 = 72 + 7 = 79
( g ° f )(−3) = g[ f (−3)] = g[(−3) 2 + 4(−3)] = g[−3] = [−3 − 8] = −11
c.
b.
= g[2(−5)2 + 7] = g[57] = 57 − 1 = 56
( f ° g )( x) = f [ g ( x)] = f [ x − 8]
c.
= ( x − 8)2 + 4( x − 8) = x 2 − 16 x + 64 + 4 x − 32
( f ° g )( x) = f [ g ( x)] = f ⎡⎣ x − 1 ⎤⎦ =2
2
= x − 12 x + 32 d.
( g ° f )(−5) = g[ f (−5)]
( x − 1 )2 + 7
= 2( x − 1) 2 + 7
( g ° f )( x ) = g[ f ( x)]
= 2( x 2 − 2 x + 1) + 7
2
= g[ x + 4 x ]
= 2 x2 − 4 x + 2 + 7
= [ x 2 + 4 x] − 8
= 2 x2 − 4 x + 9
= x2 + 4 x − 8
d.
( g ° f )( x) = g[ f ( x)] = g[2 x 2 + 7] = 2 x2 + 7 − 1 = 2 x2 + 6 = 2 x2 + 6
41.
f ( x + h) − f ( x) 4( x + h)2 − 3( x + h) − 1 − (4 x 2 − 3 x − 1) = h h =
4 x 2 + 8 xh + 4h 2 − 3x − 3h − 1 − 4 x 2 + 3 x + 1 h
8 xh + 4h 2 − 3h h = 8 x + 4h − 3 =
42.
g ( x + h ) − g ( x ) ( x + h )3 − ( x + h ) − ( x 3 − x ) = h h =
x3 + 3x 2h + 3 xh 2 + h3 − x − h − x3 + x h
=
3 x 2h + 3 xh 2 + h3 − h h
43.
f ( x) = −2 x 2 + 3
Domain: All real numbers
= 3 x 2 + 3 xh + h 2 − 1
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888
44.
Additional College Trigonometry Solutions
f ( x) = 6 − x
45.
f ( x) = 25 − x 2
46.
25 − x 2 ≥ 0 (5 − x)(5 + x) ≥ 0
6− x ≥ 0 − x ≥ −6 x≤6
x − 2 x − 15
( x + 3)( x − 5) = 0 x = −3 x = 5 Domain: {x x ≠ −3 and x ≠ 5}
Domain: {x −5 ≤ x ≤ 5}
47.
3 2
x 2 − 2 x − 15 = 0
Critical values −5 and 5.
Domain: {x x ≤ 6}
f ( x) =
48.
f is increasing on [3, ∞) f is increasing on [0, ∞)
f is decreasing on (−∞, 3]
f is decreasing on (−∞, 0]
49.
50.
f is increasing on [−2, 2] f is constant on (−∞, − 2] ∪ [2, ∞)
51.
f is constant on ...,[−6, − 5), [ −5, − 4), [−4, − 3), [ −3, − 2), [−2, − 1), [ −1, 0), [0, 1),... 52.
f is increasing on (−∞, ∞)
f is increasing on (−∞, ∞) 53.
54.
a.
Domain {x x is a real number}
a.
Range { y y ≤ 4} b.
g is an even function
55.
b.
Domain all real numbers Range all real numbers g is neither even nor odd
56.
a. a.
Domain all real numbers Range { y y ≥ 4}
b.
g is an even function
Domain {x − 4 ≤ x ≤ 4} Range { y 0 ≤ y ≤ 4}
b.
g is an even function
Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
889
57.
58.
a.
Domain {x x is a real number} Range { y y is a real number}
b.
a.
g is an odd function
Range { y y is an even integer} b.
59.
Domain {x x is a real number}
( f + g )( x) = x 2 − 9 + x + 3
60.
g is neither even or odd.
( f + g )( x) = x3 + 8 + x 2 − 2 x + 4
= x2 + x − 6
= x3 + x 2 − 2 x + 12
Domain of ( f + g )( x) is {x x is a real number}.
Domain of (f + g)(x) is the set of all real numbers. ( f − g )( x) = x3 + 8 − ( x 2 − 2 x + 4)
( f − g )( x) = x 2 − 9 − ( x + 3)
= x3 + 8 − x 2 + 2 x − 4
= x 2 − x − 12
= x3 − x 2 + 2 x + 4
Domain of ( f − g )( x) is {x x is a real number}. 2
( fg )( x) = ( x − 9)( x + 3) 3
( fg )( x) = ( x3 + 8)( x 2 − 2 x + 4)
2
= x + 3 x − 9 x − 27
= x5 − 2 x 4 + 4 x3 + 8 x 2 − 16 x + 32
Domain of ( fg )( x) is {x x is a real number}. ⎛f ⎞ ⎜ ⎟ ( x) = ⎝g⎠ =
Domain of ( f − g )( x) is the set of all real numbers.
x2 − 9 x+3
Domain of (fg)(x) is the set of all real numbers. ⎛ f ⎞ x3 + 8 ⎜⎜ ⎟⎟ ( x) = x2 − 2 x + 4 ⎝g⎠
x−3
=
( x + 2)( x 2 − 2 x + 4)
x2 − 2 x + 4 = x+2
⎛f ⎞ Domain of ⎜ ⎟ ( x) is {x x ≠ −3}. ⎝g⎠
⎛ f ⎞ Domain of ⎜⎜ ⎟⎟ ( x) is the set of all real numbers. ⎝g⎠ 61.
⎛ x +5⎞ F [G ( x)] = 2 ⎜ ⎟−5 ⎝ 2 ⎠ = x+5−5 =x
and
(2 x − 5) + 5 2 2x = 2 =x
G[ F ( x)] =
Because F[G(x)] = x and G[F(x)] = x for all real numbers x, F and G are inverses.
Copyright © Houghton Mifflin Company. All rights reserved.
890
62.
Additional College Trigonometry Solutions
h[ k ( x)] = k[h( x)] =
x2 = x
( x )2 = x
Since x ≥ 0
63.
Because h[k(x)] = x for all x in the domain of k and k[h(x)] = x for all x in the domain of h, we have shown that h and k are inverses.
3 +3
l[m( x)] = x −1
3 x −1
⎡ x + 3⎤ m[l ( x)] = m ⎢ ⎥ ⎣ x ⎦
=
3 + 3x − 3 x − 1 , 3 x −1
=
=
3x 3
=
=x
2x − 5 p[q( x)] = x −5 ⎛ 2x ⎞ 2⎜ ⎟ ⎝ x −5⎠
=
65.
2 x −5 x + 25 x −5 4x x −5
− 3 x + 25 x − 5 ⋅ x−5 4x − 3 x + 25 = 4x =
f ( x) = 3x − 4 y = 3x − 4 x = 3y − 4 x + 4 = 3y x+4 =y 3 Thus f −1( x) =
4 x+4 1 = x+ . 3 3 3
Thus, p and q are not inverse functions.
66.
g ( x ) = −2 x + 3 y = −2 x + 3 x = −2 y + 3 x − 3 = −2 y x−3 = y −2 3− x = y 2 Thus, g −1( x) =
67.
3− x 1 3 = − x+ . 2 2 2
3 x + 3− x x
= 3⋅
Thus, l and m are inverse functions.
64.
3 x +3 − 1 x
h( x ) = − 1 x − 2 2 y =−1 x−2 2 x =−1 y−2 2 x+2=−1 y 2 −2 x − 4 = y Thus, h −1( x) = −2 x − 4.
Copyright © Houghton Mifflin Company. All rights reserved.
x =x 3
Additional College Trigonometry Solutions
68.
1 x 1 y= x 1 x= y 1 y= x
k ( x) =
69.
Now y takes on its maximum value when x= 1 . x
−b −50 = 25 = 2a 2(−1)
Thus, the two numbers are 25 and (50 − 25) = 25. That is, both numbers are 25.
Let x = the smaller number. Let x + 10 = the larger number. The sum of their squares y is given by
71.
h(t ) = −16t 2 + 220 and h(t ) = 0 when
− 16t 2 + 220 = 0
y = x 2 + ( x + 10) 2
220 = 16t 2 220 = t2 16
= x 2 + x 2 + 20 x + 100 = 2 x 2 + 20 x + 100
220 =t 16
Now y takes on its minimum value when x=
Let x = one of the numbers and 50 − x = the other number. Their product y is given by y = x(50 − x) = 50 x − x 2 = x 2 + 50 x
Thus k −1 ( x) = k ( x) =
70.
891
−b −20 = = −5 2a 2(2)
2 55 =t 4
Thus, the numbers are −5 and (−5 + 10) = 5. 1 (1050)2 + 25 8820 1,102,500 = + 25 8,820 = 125 + 25 = 150 feet
Thus, t ≈ 3.7 seconds. 1 (2100)2 + 25 8820 4, 410,000 = + 25 8,820 = 500 + 25 = 525 feet
h(2100) =
72.
a.
h(1050) =
73.
a.
Enter the data on your calculator. The technique for a TI-83 is illustrated here.
b.
y = 0.018024687 x + 0.00050045744 b.
Yes. r ≈ 0.999, which is very close to 1.
c.
y = 0.018024687(100) + 0.00050045744 = 1.8024687 + 0.00050045744 ≈ 1.8 seconds Copyright © Houghton Mifflin Company. All rights reserved.
892
74.
Additional College Trigonometry Solutions
a.
Enter the data on your calculator. The technique for a TI-83 is illustrated here.
h = 0.0047952048t 2 − 1.756843157t + 180.4065934 b.
If the can is empty, then height = 0. 0 = 0.0047952048t 2 − 1.756843157t + 180.4065934 t=
−(−1.756843157) ± (−1.756843157) 2 − 4(0.0047952048)(180.4065934) 2(0.0047952048)
1.756843157 ± 3.086497878 − 3.46034625 0.0095904096 1.756843157 ± −0.373848372 = 0.0095904096 This does not represent a real number. Thus, no, according to the model, the can will never empty. =
c.
The regression line is a model of the data and is not based on physical principles.
....................................................... QR1. a. 15 ÷ 4 = 3 remainder 3 15mod 4 ≡ 3
b.
Quantitative Reasoning
37 ÷ 5 = 7 remainder 2 37 mod 5 ≡ 2
QR2. Factor the modulus, M.
52 ÷ 321 = 0 remainder 52 52 mod 321 ≡ 52
c.
QR3. Answers will vary.
....................................................... 1.
4.
4 x − 2(2 − x ) = 5 − 3(2 x + 1) 4x − 4 + 2x = 5 − 6x − 3 6x − 4 = 2 − 6x 12 x = 6 x = 12 3x 2 − x = 2 3x − x − 2 = 0 (3x + 2)( x − 1) = 0 2
2.
Chapter 1 Chapter Test
6 − 3x ≥ 3 − 4(2 − 2 x ) 6 − 3x ≥ 3 − 8 + 8 x −11x ≥ −11 x ≤1
2 x 2 − 3x = 2 2 x − 3x − 2 = 0 (2 x + 1)( x − 2) = 0
3.
2
x = − 1 or x = 2 2
5.
4 − 5x > 6
4 − 5 x < −6 or −5 x < −10
x = − 2 or x = 1
4 − 5x > 6 − 5x > 2 x < −2
x>2
3
( −∞,
−2 5
)
5
∪ (2, ∞)
Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
6.
d = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2
893
7.
d = [4 − ( −2)] 2 + ( −2 − 5) 2 d = 36 + 49 d = 85
4−2 =1 2 −1 + 3 =1 ym = 2
xm =
length = ( x2 − x1)2 + ( y2 − y1)2
midpoint = (1, 1)
= [4 − (−2)]2 + (−1 − 3)2 = 62 + ( −4)2 = 36 + 16 = 52 = 2 13
8.
x = 2 y2 − 4 x=0
9.
2 y2 − 4 = 0 2 y2 = 4 y2 = 2 y=± 2
If y = 0,
x = −4
intercepts (0, − 2), (0,
10.
x 2 − 4x + y 2 + 2 y − 4 = 0 ( x 2 − 4 x + 4) + ( y 2 + 2 y + 1) = 4 + 4 + 1 ( x − 2) 2 + ( y + 1) 2 = 9
center: (2, − 1) 12.
2), (−4, 0)
11.
f ( −3) = − 25 − ( −3) 2 f ( −3) = − 16 f ( −3) = −4
radius: 3
x 2 − 16 ≥ 0 ( x − 4)( x + 4) ≥ 0
f ( x ) = − 25 − x 2
13.
Critical values 4 and −4. Domain {x | x ≥ 4 or x ≤ −4}
f is increasing on (−∞, 2] f is decreasing on [2, ∞ )
Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
894
14.
First shift the graph of f(x) horizontally 2 units to the left. Next, reflect the graph across the x-axis. Finally, shift the graph vertically down 1 unit.
15.
a.
f (− x) = (− x)4 − ( − x) 2
= x4 − x2 = f ( x) The function of f ( x) = x 4 − x 2 is an even function. b.
f ( − x ) = ( − x )3 − ( − x ) = − x3 + x = −( x3 − x ) = − f ( x)
The function f ( x) = x3 − x is an odd function. c.
f (− x) = − x − 1 The function f ( x) = x − 1 is neither an even nor an odd function.
Thus, only b defines an odd function. 16.
( f + g )( x ) = ( x 2 − 1) + ( x − 2) = x2 + x−3 ⎛ f ⎞ x 2 −1, x ≠ 2 ⎜ ⎟ ( x) = g x−2 ⎝ ⎠
17.
18.
( f o g )( x ) = ( x − 2 ) 2 − 2 x − 2 + 1 = x − 2 − 2 x − 2 +1 = x − 2 x − 2 −1
19.
2 2 f ( x + h ) − f ( x ) ⎡⎣ ( x + h ) + 1⎤⎦ − ( x + 1) = h h 2 2 2 = x + 2 xh + h + 1 − x − 1 h 2 = 2 xh + h h = 2x + h
x x +1 Interchange x and y. Then solve for y. y x= y +1 x( y + 1) = y y=
xy + x = y xy − y = − x y ( x − 1) = − x x −x y= = x −1 1− x x −1 f ( x) = 1− x 20.
a.
Enter the data on your calculator. The technique for a TI-83 is illustrated here.
y = −7.98245614 x + 767.122807 b.
y = −7.98245614(89) + 767.122807 ≈ 57 Calories
Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
895
....................................................... 1.
d = (4 − (−3))2 + (1 − 2) 2 = 49 + 1 = 50
[1.1]
Chapter 2 Cumulative Review 2.
c2 = a 2 + b2
[1.1]
2
1 = ⎛⎜ 1 ⎞⎟ + b 2 ⎝2⎠ 3 = b2 4 3 =b 2 2
3.
Intercepts: (–3, 0), (3, 0), (–9, 0) [1.2]
4.
−x = − x = − f ( x ) [1.4] (− x) 2 + 1 x 2 + 1 Odd function
5.
f ( x) =
x [1.5] 2x − 3 y x= 2y − 3 x(2 y − 3) = 2 xy − 3 x = y 2 xy − y = y (2 x − 1) = 3 x y = 3x 2x − 1 f −1 ( x) = 3 x 2x − 1
6.
Domain: (−∞, 4) ∪ (4, ∞) [1.3]
7.
x2 + x − 6 = 0 ( x + 3)( x − 2) = 0
8.
Shift the graph of y = f (x) horizontally 3 units to the right. [1.4]
f (− x) =
x+3=0 x−2=0 x = −3 x=2 The solutions are –3 and 2. [1.1] 9.
Reflect the graph of y = f (x) across the y-axis. [1.4[
10.
⎛ ⎞ 300o = 300o ⎜ π o ⎟ = 5π [2.1] ⎝ 180 ⎠ 3
11.
5π = 5π ⎛ 180o ⎞ = 225o [2.1] ⎜ ⎟ 4 4 ⎝ π ⎠
12.
f ⎜⎛ π ⎟⎞ = sin ⎜⎛ π + π ⎟⎞ = sin ⎜⎛ π ⎟⎞ = 1 [2.3] ⎝2⎠ ⎝3⎠ ⎝3 6⎠
13.
f ⎛⎜ π ⎞⎟ = sin ⎛⎜ π ⎞⎟ + sin ⎛⎜ π ⎞⎟ = 3 + 1 = 3 + 1 2 ⎝3⎠ ⎝3⎠ ⎝6⎠ 2 2
14.
⎞ ⎛ ⎞ ⎛ cos 2 45o + sin 2 60o + = ⎜ 2 ⎟ + ⎜ 3 ⎟ = 2 + 3 = 5 [2.2] 4 4 4 ⎝ 2 ⎠ ⎝ 2 ⎠
16.
θ = 210o [2.3] Since 180o < θ < 270o , θ ′ + 180o = θ θ ′ = 30o
15.
negative [2.3]
[2.2]
2
Copyright © Houghton Mifflin Company. All rights reserved.
2
Additional College Trigonometry Solutions
896
17.
θ = 2π 3
[2.3]
18.
Domain: ( −∞, ∞ ) [2.4]
19.
Range: [–1, 1]
[2.4]
Since π < θ < π , 2 θ +θ′ = π
θ′ = π
3
20.
tan θ =
opp 3 = adj 4
hypotenuse = 32 + 42
sin θ =
= 9 + 16
opp 3 = hyp 5
[2.2]
= 25 =5
.......................................................
Chapter 3 Cumulative Review
1.
−2 x + 1 < 7 −2 x < 6 x > −3
2.
Shift the graph of y = f (x) horizontally 1 unit to the left and up 2 units.
3.
Reflect the graph of y = f (x) across the x-axis.
4.
f (− x) = − x − sin(− x) = − x + sin x = −( x − sin x) = − f ( x) odd function
5.
f ( x) = 5 x x −1 y = 5x x −1 5y x= y −1 x( y − 1) = 5 y xy − 5 y = x y ( x − 5) = x y= x x−5 f −1 ( x) = x x−5
6.
8.
sin π = 3 3 2
9.
⎛ ⎞ 240o = 240o ⎜ π o ⎟ = 4π ⎝ 180 ⎠ 3
csc60o =
[2.1]
1 = 2 =2 3 3 3 sin 60o
7.
10.
5π = 5π ⎛ 180o ⎞ = 300o ⎜ ⎟ 3 3 ⎝ π ⎠
sin θ =
opp 2 = hyp 3
adjacent side = 32 − 22 = 9−4 = 5 opp tan θ = = 2 =2 5 adj 5 5
Copyright © Houghton Mifflin Company. All rights reserved.
[2.1]
Additional College Trigonometry Solutions
11.
cot θ > 0 in quadrant III Positive [2.3]
897
12.
θ = 310o
13.
[2.3] o
θ = 5π
[2.3]
3
o
Since 270 < θ < 360 ,
Since 3π < θ < 2π , 2 θ = θ ′ = 2π
θ = θ ′ = 360o θ ′ = 50o
θ′ = π
3
14.
t=π 3
15.
[2.4]
y = sin t = sin π = 3 3 2 x = cos t = cos π = 1 3 2 The point on the unit circle corresponding to t = π is ⎛⎜ 1 , 3 ⎞⎟ . 3 ⎝2 2 ⎠
17.
cos −1 ( −0.8 ) = 2.498
19.
Range: − π , π 2 2
(
)
16.
y = 0.43cos ⎛⎜ 2 x − π ⎞⎟ [2.7] 6⎠ ⎝ amplitude: 0.43
y = sin −1 sin y =
0 ≤ 2 x − π ≤ 2π
6 π ≤ 2 x ≤ 13π 6 6 π ≤ x ≤ 13π 12 12
y=
1 2
π
1 2 −
[3.5]
π 2
≤y≤
π 2
6
period = π , phase shift = π 12
[3.5]
[3.5]
18.
Domain: [–1, 1].
[3.5]
20.
2cos 2 x − 1 = − sin x 1 − 2sin 2 x = − sin x 0 = 2sin 2 x − sin x − 1 0 = (2sin x + 1)(sin x − 1)
2sin x + 1 = 0 sin x = − 1 2 π 7 , 11π x= 6 6 The solutions are
sin x − 1 = 0 sin x = 1 x=
π 2
π 7π 11π
, , . 2 6 6
.......................................................
Chapter 4 Assessing Concepts
1.
An oblique triangle that does not contain a right angle.
2.
The Law of Cosines
3.
SSA
4.
The variable s represents the semiperimeter of the triangle.
5.
A scalar
6.
A scalar
7.
True
8.
False
9.
True
10.
True
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898
Additional College Trigonometry Solutions
....................................................... 1.
Chapter 4 Test 2.
B = 180° − 70° − 16° B = 94°
b a = sin B sin A
c a = sin C sin A 14 sin 70° a= sin 16° a ≈ 48
⎛ b sin A ⎞ B = sin −1 ⎜ ⎟ [4.1] ⎝ a ⎠ ⎛ 13 sin 140° ⎞ B = sin −1 ⎜ ⎟ 45 ⎝ ⎠ B ≈ 11°
c b [4.1] = sin C sin B 14 sin 94° b= sin16° b ≈ 51
3.
4.
a 2 + c2 − b2 2ac ⎛ 322 + 182 − 242 ⎞ ⎟ [4.2] B = cos −1 ⎜ ⎜ ⎟ 2(32)(18) ⎝ ⎠ B ≈ 48°
c 2 = a 2 + b 2 − 2ab cos C
cos B =
c 2 = 202 + 122 − 2(20)(12) cos 42° [4.2] c ≈ 14
5.
1 ab sin C 2 1 K = (7)(12)(sin110°) 2 K ≈ 39 square units [4.2] K=
7.
A = 180° − 42° − 75° A = 63°
6.
2 K = b sin A sin C 2sin B 2 12 sin 63° sin 75° K= 2sin 42° K ≈ 93 square units [4.2]
8.
v = (−2)2 + (3) 2 = 13
s = 1 (a + b + c) 2 s = 1 (17 + 55 + 42 ) = 57 2 K = s ( s − a )( s − b)( s − c) K = 57(57 − 17)(57 − 55)(57 − 42) K ≈ 260 square units [4.2]
Copyright © Houghton Mifflin Company. All rights reserved.
[4.3]
Additional College Trigonometry Solutions
899
9.
a1 = 12cos 220° ≈ −9.2 [4.3] a2 = 12sin 220° ≈ −7.7 v = a1i + a2 j v = −9.2i − 7.7 j
10.
3u − 5v = 3(2i − 3 j) − 5(5i + 4 j) [4.3] = (6i − 9 j) − (25i + 20 j) = (6 − 25)i + (−9 − 20) j = −19i − 29 j
11.
u ⋅ v = (−2i + 3 j) ⋅ (5i + 3 j) [4.3]
12.
cosθ =
= (−2 ⋅ 5) + (3 ⋅ 3) = −10 + 9 = −1
3,5 ⋅ −6, 2 u⋅v = 2 u v 3 + 52 (−6) 2 + 22
[4.3]
−18 + 10 −8 = 34 40 34 40 θ ≈ 103°
cosθ =
13.
14.
A = 142° − 65° = 77°
A = 82° − 32° = 50°
R 2 = 242 + 182 − 2(24)(18) cos 77° R ≈ 27 miles
C = 180° − 50° − 32° − 72° = 26° a = 12 sin 50° sin 26° a = 12sin 50° sin 26° a ≈ 21 miles [4.3]
[4.3]
15.
1 (112 + 165 + 140 ) = 208.5 2 K = 208.5(208.5 − 112)(208.5 − 165)(208.5 − 140) K ≈ 7743 cost ≈ 8.50(7743) cost ≈ $66,000 [4.2] S=
....................................................... 1.
d = (−3 − 4)2 + (4 − (−1)) 2 = 49 + 25 = 74
3.
( f o g )( x) = f [ g ( x)] [1.5] = f [cos x] = sec(cos x))
[1.2]
Chapter 4 Cumulative Review 2.
f ( x) + g ( x) = sin x + cos x
4.
f ( x) = 1 x − 3 2 y = 1 x−3 2 x = 1 y −3 2 2( x + 3) = y
[1.5]
[1.6]
f −1 ( x ) = 2 x + 6
5.
Shifted 2 units to the right and 3 units up. [1.4]
6.
sin 27o = 15 a a = 15 o ≈ 33 cm sin 27
Copyright © Houghton Mifflin Company. All rights reserved.
[2.2]
Additional College Trigonometry Solutions
900
7.
y = 3sin(2π x)
8.
y = 1 tan(2 x) 4
[2.6]
[2.5]
9.
( )
y = 2sin π x + 1 2
10.
y = sin x + cos x Amplitude:
13.
15.
12.
[2.7]
( )
4
[3.1]
tan ⎛⎜ cos −1 12 ⎞⎟ [3.5] 13 ⎠ ⎝ Let θ = cos −1 12 and find y = tan θ . 13 12 Then cosθ = , and 0 ≤ θ ≤ π . 13 132 − 122 = 25 = 5 Thus tanθ = 5 . 12 y= 5 12
[2.7]
c 2 = a 2 + b 2 − 2ab cos C
[4.2]
c 2 = (10)2 + (12)2 − 2(10)(12) cos50o
2 , period: 2π , phase shift: π
1 − cos x = 1 − cos 2 x cos x cos x 2 x sin = cos x = sin x tan x
)
amplitude = 3, period = 6π , phase shift = 3π 2
[2.7]
11.
(
y = 3sin 1 x − π 3 2 1 π 0 ≤ x − ≤ 2π 3 2 π ≤ 1 x ≤ 5π 2 3 2 3π ≤ x ≤ 15π 2 2
= 244 − 240cos50o ≈ 9.473 c ≈ 9.5 cm
14.
16.
( )
⎛ ⎞ sin −1 sin 2π = sin −1 ⎜ 3 ⎟ = π 3 ⎝ 2 ⎠ 3
sin x tan x − 1 tan x = 0 [3.6] 2 tan x sin x − 1 = 0 2 sin x − 1 = 0 tan x = 0 2 x = 0, π sin x = 1 2
(
[3.5]
)
x = π , 5π 6 6
Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
17.
v = 42 + (−3)2 α = tan −1 v = 16 + 9
α θ θ θ
v =5
901
−3 = tan −1 3 4 4
≈ 36.9° = 360° − α ≈ 360° − 36.9° ≈ 323.1°
19.
α = sin −1
1 mph 1 mph
20.
α
18.
[4.3]
cos θ =
v⋅w v w
[4.3]
1, 2 ⋅ −2, 3 2
1 + (2)2 ( −2)2 + 32
1(−2) + (2)(3) 5 13 4 cos θ = ≈ 0.49613 5 13 θ = 60.3° cos θ =
heading = θ = 270° + α θ ≈ 270° + 19.5° θ ≈ 289.5°
1 1 = sin −1 3 3
α ≈ 19.5°
3 mph
cos θ =
[4.3]
θ
AB = 515(cos36i + sin 36 j) ≈ 416.6i + 302.7 j [4.3] AD = 150[cos(−30°)i + sin(−30°) j] ≈ 129.9i − 75 j AC = AB + AD AC = 416.6i + 302.7 j + 129.9i − 75 j AC ≈ 546.5i + 227.7 j AC = 546.52 + (227.7)2 AC ≈ 592 mph
( 546.5 )
α = 90o − θ = 90o − tan −1 227.7 ≈ 67.4o
Section 5.1 See CAT Section P.6 solutions on page 26 for exercises 1 – 62. 63.
65.
z 1 = ( −2 + i ) + 4 + 3i = 2 + 4i z 2 = (2 + 4i ) + 4 + 3i = 6 + 7i z 3 = (6 + 7i ) + 4 + 3i = 10 + 10i z 4 = (10 + 10i ) + 4 + 3i = 14 + 13i z 5 = (14 + 13i ) + 4 + 3i = 18 + 16i
64.
z 1 = i (1 − i ) = i − i 2 = 1 + i z 2 = i (1 + i ) = i + i 2 = −1 + i z 3 = i ( −1 + i ) = −i + i 2 = −1 − i
66.
z4 z5 z6 z7 z8
= i ( −1 − i ) = −i − i 2 = 1 − i = z1 = z2 = z3 = z4
z 1 = 2i (1 + 3i ) = 2i + 6i 2 = −6 + 2i z 2 = 2i ( −6 + 2i ) = −12i + 4i 2 = −4 − 12i z 3 = 2i ( −4 − 12i ) = −8i − 24i 2 = 24 − 8i
z 4 = 2i (24 − 8i ) = 48i − 16i 2 = 16 + 48i z 5 = 2i (16 + 48i ) = 32i + 96i 2 = −96 + 32i z 1 = (0.5i ) 2 = 0.25i 2 = −0.25 z 2 = ( −0.25) 2 = 0.0625 z 3 = (0.0625) 2 = 0.00390625
Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
902
67.
Use a = 3, b = −3, c = 3.
68.
Use a = 2, b = 4, c = 4. 2 −b + b2 − 4ac = −(4) + (4) − 4(2)(4) 2a 2(2)
2
−b + b − 4ac = −( −3) + ( −3) − 4(3)(3) 2a 2(3) 2
= −4 + 16 − 32 = −4 + −16 4 4 i 4 16 − + i 4 4 − + = = 4 4 i 4 4 − = + = −1 + i 4 4
= 3 + 9 − 36 = 3 + −27 6 6 i i + + 3 27 3 3 3 = = 6 6 = 3+ 3 3i = 1 + 3i 6 6 2 2 69.
Use a = 2 b = 6, c = 6.
70. 2
Use a =2, b = 1, c = 3. 2 −b + b2 − 4ac = −(1) + (1) − 4(2)(3) 2a 2(2)
−b + b − 4ac = −(6) + (6) − 4(2)(6) 2a 2(2) 2
= −1 + 1 − 24 4 = −1 + −23 = −1 + i 23 4 4 23 1 =− + i 4 4
= −6 + 36 − 48 = −6 + −12 4 4 6 2 3 − + i 6 12 − + i = = 4 4 2 3 i 6 3 − = + =− + 3i 4 4 2 2 71.
Use a = 4, b = −4, c = 2.
72. 2
Use a = 3, b = −2, c = 4. 2 −b + b2 − 4ac = −( −2) + ( −2) − 4(3)(4) 2a 2(3)
−b + b − 4ac = −( −4) + ( −4) − 4(4)(2) 2a 2(4) 2
= 2 + 4 − 48 = 2 + −44 6 6 = 2 + i 44 = 2 + 2i 11 6 6 2 11 i 2 1 = + = + 11 i 6 6 3 3
= 4 + 16 − 32 = 4 + −16 8 8 = 4 + i 16 = 4 + 4i 8 8 4 4 1 1 i = + = + i 8 8 2 2
.......................................................
Connecting Concepts
73.
x 2 + 16 = x 2 + 42 = ( x + 4i )( x − 4i )
74.
x 2 + 9 = x 2 + 32 = ( x + 3i )( x − 3i )
75.
z 2 + 25 = z 2 + 52 = ( z + 5i )( z − 5i )
76.
z 2 + 64 = z 2 + 82 = ( z + 8i )( z − 8i )
77.
4 x 2 + 81 = (2 x )2 + 92 = (2 x + 9i )(2 x − 9i )
78.
9 x 2 + 1 = (3x )2 + 12 = (3x + i )(3x − i )
79.
If x = 1 + 2i, then x 2 − 2 x + 5 = (1 + 2i )2 − 2(1 + 2i ) + 5 = 1 + 4i + 4i 2 − 2 − 4i + 5 = 1 + 4i + 4(−1) − 2 − 4i + 5 = 1 + 4i − 4 − 2 − 4i + 5 = (1 − 4 − 2 + 5) + (4i − 4i ) = 0
80.
If x = 1 − 2i, then x 2 − 2 x + 5 = (1 − 2i )2 − 2(1 − 2i ) + 5 = 1 − 4i + 4i 2 − 2 + 4i + 5 = 1 − 4i + 4( −1) − 2 + 4i + 5 = 1 − 4i − 4 − 2 + 4i + 5 = (1 − 4 − 2 + 5) + ( −4i + 4i ) = 0
Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
81.
903
Verify that ( −1 + i 3)3 = 8 . ( −1 + i 3)3 = ( −1 + i 3)( −1 + i 3)2 = ( −1 + i 3)[( −1)2 + 2( −1)(i 3) + (i 3)2 ] = ( −1 + i 3)[1 − 2i 3 + 3i 2 ] = ( −1 + i 3)[1 − 2i 3 + 3( −1)] = ( −1 + i 3)[1 − 2i 3 − 3] = ( −1 + i 3)( −2 − 2i 3) = −1( −2 − 2i 3) + i 3( −2 − 2i 3) = 2 + 2i 3 − 2i 3 − 2i 2 ( 3)2 = 2 + 2i 3 − 2i 3 − 2( −1)(3) = 2 + 2i 3 − 2i 3 + 6 = (2 + 6) + (2i 3 − 2i 3) =8
Verify that ( −1 − i 3)3 = 8 . ( −1 − i 3)3 = ( −1 − i 3)( −1 − i 3)2 = ( −1 − i 3)[( −1)2 + 2( −1)( −i 3) + ( −i 3)2 ] = ( −1 − i 3)[1 + 2i 3 + 3i 2 ] = ( −1 − i 3)[1 + 2i 3 + 3( −1)] = ( −1 − i 3)[1 + 2i 3 − 3] = ( −1 − i 3)( −2 + 2i 3) = −1( −2 + 2i 3) − i 3( −2 + 2i 3) = 2 − 2i 3 + 2i 3 − 2i 2 ( 3)2 = 2 − 2i 3 + 2i 3 − 2( −1)(3) = 2 − 2i 3 + 2i 3 + 6 = (2 + 6) + ( −2i 3 + 2i 3) =8 82.
⎡
⎤
2
Verify that ⎢ 2 (1 + i ) ⎥ = i . ⎣ 2 ⎦ 2
2
⎡ 2 ⎤ 2 2 2 2 1 1 1 ⎢ 2 (1 + i ) ⎥ = 2 (1 + i ) = 4 (1 + 2i + i ) = 2 [1 + 2i + ( −1)] = 2 (1 + 2i − 1) = 2 (2i ) = i 2 ⎣ ⎦ 83.
i + i 2 + i 3 + i 4 + ... + i 28 = 7(i + i 2 + i 3 + i 4 ) = 7(i + ( −1) + ( −i ) + 1) = 7(0) = 0
84.
i + i 2 + i 3 + i 4 + ... + i100 = 25(i + i 2 + i 3 + i 4 ) = 25(i + ( −1) + ( −i ) + 1) = 25(0) = 0
.......................................................
Prepare for Section 5.2
See CAT Prepare for Section 7.4 solutions on page 462.
.......................................................
Exploring Concepts with Technology
The Mandelbrot Iteration Procedure 1.
z 5 ≈ 0.4001878333
z 10 ≈ 0.4353725328
z 100 ≈ 0.4906925008
z 200 ≈ 0.4951947016
Answers will vary. 2.
z0 = i z 1 = −1 + i z 2 = −1 z 3 = −1 + i z 4 = −i The iterates continue to cycles back and forth between –1 + i and –i.
Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
904
.......................................................
Chapter 5 Assessing Concepts
1.
True
True
5.
The four roots are equally spaced around a circle with center 6. (0, 0) and radius 1.
1
7.
5
–3 + 5i
2.
No
8.
True
3.
9.
4.
False
10.
2
....................................................... 1.
4.
3 − −64 = 3 − 8i 3 + 8i
2.
[5.1]
5.
−5 + −27 = −5 + 3i 3 [5.1] −5 − 3i 3
Chapter 5 Review
−4 + 6 = 6 + 2i 6 – 2i
3.
[5.1]
[5.1]
−2 − i 5
(
−4 )( −4 ) = (2i )(2i ) = 4i 2 = −4 [5.1]
6.
9.
(3 + 7i) + (2 − 5i) = 5 + 2i [5.1]
8.
(3 − 4i ) + (−6 + 8i ) = −3 + 4i
10.
(−3 − 5i ) − (2 + 10i ) = −5 − 15i [5.1]
11.
(5 + 3i )(2 − 5i) = 10 − 19i − 15i 2 [5.1] 12. = 25 − 19i
13.
−2i ⋅ 3 + 4i = −6i − 8i 2 = 8 − 6 i 3 − 4i 3 + 4i 9 + 16 25 25
[5.1]
15.
i(2i ) − (1 + i ) 2 = 2i 2 − 1 − 2i − i 2 = −2 − 2i
[5.1]
(−
−27 )( −3 ) = ( −3i 3 )( i 3 ) [5.1] = −9i 2 = 9
7.
[5.1]
−2 + −5 = −2 + i 5
(6 − 8i ) − (9 − 11i ) = −3 + 3i [5.1] (−2 − 3i )(−4 + 7i ) = 8 − 2i − 21i 2 = 29 − 2i
14.
4 + i ⋅ 7 + 2i = 28 + 15i + 2i 2 = 26 + 15 i 7 − 2i 7 + 2i 49 + 4 53 53
16.
(2 − i )3 = (4 − 4i + i 2 )(2 − i) [5.1] = (3 − 4i )(2 − i )
[5.1]
[5.1]
= 6 − 11i + 4i 2 = 2 − 11i 17.
(3 + −4) − (−3 − −16) = 3 + 2i + 3 + 4i = 6 + 6i
19.
(2 − −3)(2 + −3) = 4 − (−3) = 7 [5.1]
[5.1]
18.
(−2 + −9) + ( −3 − −81) = −2 + 3i − 3 − 9i = −5 − 6i
20.
(3 − −5)(2 + −5) = 6 + i 5 − (−5)
[5.1]
[5.1]
= 11 + i 5 21.
25.
22.
i 27 = i 3 = −i [5.1]
( 0 ) 2 + ( −8 ) 2
−8i =
i105 = i 26.
[5.2]
2 − 3i =
=8 28.
−1 − i =
23.
[5.1]
24.
i = 1 = 1 = 1 [5.1] i17 i16 1
( 2 )2 + ( −3)2
[5.2]
27.
−4 + 5i =
= 4 + 9 = 13
( −1)2 + ( −1)2
= 1+1 = 2
[5.2]
29.
r=
( 2 ) 2 + ( −2 ) 2
r=2 2
α = tan −1
( −4 ) 2 + ( 5 ) 2
= 16 + 25 = 41
30.
[5.2]
i 62 = i 2 = −1 [5.1]
r=
(− 3)
2
+ (1)
2
r=2 −2 2
1 3 −1 1 = tan = 30o 3
α = tan −1
= tan −1 1 = 45o
α = 360o − 45o = 315o o
z = 2 2 cis 315
α = 180o − 30o = 150o z = 2 cis 150o
Copyright © Houghton Mifflin Company. All rights reserved.
[5.2]
[5.2]
Additional College Trigonometry Solutions
31.
r=
( −3)2 + ( 2 )2
905
32.
[5.2]
r = 13
( 4 )2 + ( −1)2
[5.2]
r = 17 −1 4 1 = tan −1 = 14.04o 4
2 −3 2 = tan −1 = 33.7o 3
33.
r=
α = tan −1
α = tan −1
α = 180o − 33.7o = 146.3o
α = 360o − 14.04o = 345.96o
z = 13 cis 146.3o
z = 17 cis 345.96o
z = 5(cos 315o + i sin 315o ) [5.2]
z = 2 ( cos 2 + i sin 2 ) [5.2] z ≈ 2 ( −0.4161 + 0.9093i ) z ≈ −0.832 + 1.819i
37.
z1z2 z1z2 z1z2 z1z2 z1z2
= 3(cos 225° + i sin 225°) ⋅ 10(cos 45° + i sin 45°) = 30[cos(225° + 45°) + i sin(225° + 45°)] = 30(cos 270° + i sin 270°) = 30(0 − 1i ) = −30i
39.
z1z2 z1z2 z1z2 z1z2 z1z2
= 3(cos 12° + i sin 12°) ⋅ 4(cos 126° + i sin 126°) = 12[cos(12° + 126°) + i sin(12° + 126°)] = 12(cos 138° + i sin 138°) ≈ 12( −0.74314 + 0.66913i ) ≈ −8.918 + 8.030i
36.
z = 3(cos 115o + i sin 115o ) [5.2] z ≈ 3 ( −0.4226 + 0.9063i ) z ≈ −1.27 + 2.72i
41.
z1z2 z1z2 z1z2 z1z2 z1z2
= 3(cos 1.8 + i sin 1.8) ⋅ 5(cos 2.5 + i sin 2.5) [5.2] = 15[cos(1.8 + 2.5) + i sin(1.8 + 2.5)] = 15(cos 4.3 + i sin 4.3) ≈ 15(−0.4008 − 0.9162i ) ≈ −6.012 − 13.743i
43.
z1 6(cos 50o + i sin 50o ) [5.2] = z2 2(cos 150o + i sin 150o ) z1 = 3 [cos(50o − 150o ) + i sin(50o − 150o )] z2 z1 = 3(cos − 100o + i sin − 100o ) = 3cis (−100o ) z2
[5.2] 38.
[5.2]
)
z = 6 cos 4π + i sin 4π 3 3 ⎛ 1 ⎞ z = 6⎜ − − 3 i ⎟ ⎝ 2 2 ⎠ z = −3 − 3 3i
⎛ ⎞ z = 5⎜ 2 − 2 i ⎟ 2 ⎠ ⎝ 2 5 2 5 z= − 2i 2 2 ≈ 3.536 − 3.536i
35.
(
34.
[5.2]
z1z2 = 5(cos 162° + i sin 162°) ⋅ 2(cos 63° + i sin 63°) [5.2] z1z2 = 10[cos(162° + 63°) + i sin(162° + 63°)] z1z2 = 10(cos 225° + i sin 225°) ⎛ 2 2 ⎞ − z1z2 = 10 ⎜ − i⎟ ⎝ 2 2 ⎠ z1z2 = −5 2 − 5i 2
40.
z1z2 z1z2 z1z2 z1z2 z1z2
= (cos 23° + i sin 23°) ⋅ 4(cos 233° + i sin 233°) [5.2] = 4[cos(23° + 233°) + i sin(23° + 233°)] = 4(cos 256° + i sin 256°) ≈ 4 ( −0.24192 − 0.97030i ) ≈ −0.968 − 3.881i
42.
z1z2 z1z2 z1z2 z1z2 z1z2
= 6(cos 3.1 + i sin 1.8) ⋅ 5(cos 4.3 + i sin 4.3) = 30[cos(3.1 + 4.3) + i sin(3.1 + 4.3)] = 30(cos 7.4 + i sin 7.4) ≈ 30(0.439 + 0.899i ) ≈ 13.2 + 27.0i
44.
z1 30(cos 165o + i sin 165o ) [5.2] = z2 10(cos 55o + i sin 55o ) z1 = 3 [cos(165o − 55o ) + i sin(165o − 55o )] z2 z1 = 3(cos 110o + i sin 110o ) = 3cis (110o ) z2
Copyright © Houghton Mifflin Company. All rights reserved.
[5.2]
906
Additional College Trigonometry Solutions
45.
z1 40(cos 66o + i sin 66o ) [5.2] = z2 8(cos 125o + i sin 125o ) z1 = 5 [cos(66o − 125o ) + i sin(66o − 125o )] z2 z1 = 5(cos − 59o + i sin − 59o ) = 5cis (−59o ) z2
46.
z1 2(cos 150o + i sin 150o ) [5.2] = z2 2(cos 200o + i sin 200o ) z1 = 2 [cos(150o − 200o ) + i sin(150o − 200o )] z2 z1 = 2(cos − 50o + i sin − 50o ) = 2cis (−50o ) z2
47.
z1 10(cos 3.7 + i sin 3.7) = z2 6(cos 1.8 + i sin 1.8)
48.
z1 4(cos 1.2 + i sin 1.2) = z2 8(cos 5.2 + i sin 5.2)
[5.2]
z1 5 = [cos(3.7 − 1.8) + i sin(3.7 − 1.8)] z2 3 z1 5 = (cos 1.9 + i sin 1.9) = 5 cis (1.9) z2 3 3 49.
z1 1 = [cos(1.2 − 5.2) + i sin(1.2 − 5.2)] z2 2 z1 1 = (cos − 4 + i sin − 4) = 1 cis (−4) z2 2 2
[3(cos 45° + i sin 45°)]5 = 35[cos(5 ⋅ 45°) + i sin(5 ⋅ 45°)] [5.3] o
z =1− i 3
[5.3]
r = 12 + ( − 3 )
2
(
52.
z = −2 − 2i [5.3]
8
α = tan −1
r = (−2)2 + (−2)2
1
r=2 2
θ = 300o
)
z = 2 2(cos 225° + i sin 225°) 7
10
(1 − i 3) = [2(cos300° + i sin 300°)]
(−2 − 2i )
= [2 2(cos 225° + i sin 225°)]10
= 128[cos(7 ⋅ 300°) + i sin(7 ⋅ 300°)] = 128(cos 2100° + i sin 2100°)
= 32,768[cos(10 ⋅ 225°) + i sin(10 ⋅ 225°)]
= 128(cos300° + i sin 300°)
= 32,768(cos90° + i sin 90°)
= 64 − 64i 3
= 0 + 32,768i
= 32,768(cos 2250° + i sin 2250°)
≈ 64 − 110.851i
53.
z = 2 −i 2 2
54.
[5.3]
r = ( 2 ) + (− 2 ) r=2
2
α = tan −1 − 2 = 45o 2
o
θ = 315
z = 2(cos315° + i sin 315°) ( 2 − i 2)5 = [2(cos315° + i sin 315°)]5 = 32[cos(5 ⋅ 315°) + i sin(5 ⋅ 315°)] = 32(cos1575° + i sin1575°) = 32(cos135° + i sin135°) = −16 2 + 16i 2
z = 3 − 4i
[5.3]
r = 32 + ( −4)2 r =5
α = tan −1
−4 = 53.13o 3
θ = 306.87o z = 5(cos306.87° + i sin 306.87°) 5
(3 − 4i ) = [5(cos306.87° + i sin 306.87°)]5 = 3125[cos(5 ⋅ 306.87°) + i sin(5 ⋅ 306.87°)] = 3125(cos1534.35° + i sin1534.35°) = 3125(cos94.35° + i sin 94.35°) = −237 + 3116i
≈ −22.627 + 22.627i
Copyright © Houghton Mifflin Company. All rights reserved.
[5.3]
−2 = 45o −2
θ = 225o
z = 2(cos300° + i sin 300°) 7
)
⎡cos 11π + i sin 11π ⎤ = cos 8 ⋅ 11π + i sin 8 ⋅ 11π ⎢⎣ 8 8 ⎥⎦ 8 8 = cos (11π ) + i sin (11π ) = cos π + i sin π = −1 + 0i = −1
α = tan −1 − 3 = 60o
r=2
(
50.
o
= 243[cos 225 + i sin 225 ] ⎛ ⎞ = 243 ⎜ − 2 − 2 i ⎟ ⎝ 2 2 ⎠ = − 243 2 − 243 2 i 2 2 ≈ −171.827 − 171.827i 51.
[5.2]
Additional College Trigonometry Solutions
55.
(
27i = 27 cos90o + i sin 90o
)
907
[5.3]
wk = 271/ 3 ⎛⎜ cos 90° + 360°k + i sin 90° + 360°k ⎞⎟ 3 3 ⎝ ⎠
(
w0 = 3 cos 90° + i sin 90° 3 3 w0 = 3(cos30° + i sin 30°) w0 = 3 cis 30° 56.
(
8i = 8 cos90o + i sin 90o
)
)
w1 = 3 ⎜⎛ cos 90° + 360° + i sin 90° + 360° ⎟⎞ 3 3 ⎝ ⎠ w1 = 3(cos150° + i sin150°) w1 = 3 cis 150°
(
)
w0 = 4 8 cis 22.5°
w1 = 4 8 cis 112.5°
w2 = 4 8 ⎜⎛ cos 90° + 360° ⋅ 2 + i sin 90° + 360° ⋅ 2 ⎟⎞ 4 4 ⎝ ⎠ 4 w2 = 8(cos 202.5° + i sin 202.5°) w2 = 4 8 cis 202.5°
(
k = 0, 1, 2, 3
w1 = 4 8 ⎜⎛ cos 90° + 360° + i sin 90° + 360° ⎟⎞ 4 4 ⎝ ⎠ 4 w1 = 8(cos112.5° + i sin112.5°)
w0 = 4 8 cos 90° + i sin 90° 4 4 4 w0 = 8(cos 22.5° + i sin 22.5°)
256 = 256 cos0o + i sin 0o
w2 = 3 ⎜⎛ cos 90° + 360° ⋅ 2 + i sin 90° + 360° ⋅ 2 ⎟⎞ 3 3 ⎝ ⎠ w2 = 3(cos 270° + i sin 270°) w2 = 3 cis 270°
[5.3]
wk = 81/ 4 ⎛⎜ cos 90° + 360°k + i sin 9° + 360°k ⎞⎟ 4 4 ⎝ ⎠
57.
k = 0, 1, 2
w3 = 4 8 ⎛⎜ cos 90° + 360° ⋅ 3 + i sin 90° + 360° ⋅ 3 ⎞⎟ ⎝ ⎠ 4 4 4 w3 = 8(cos 292.5° + i sin 292.5°) w3 = 4 8 cis 292.5°
)
[5.3]
wk = 811/ 4 ⎛⎜ cos 0° + 360°k + i sin 0° + 360°k ⎞⎟ 4 4 ⎝ ⎠
k = 0, 1, 2, 3
w0 = 3 cis 0°
w1 = 3 ⎜⎛ cos 0° + 360° + i sin 0° + 360° ⎟⎞ 4 4 ⎝ ⎠ w1 = 3(cos90° + i sin 90°) w1 = 3 cis 90°
w2 = 3 ⎜⎛ cos 0° + 360° ⋅ 2 + i sin 0° + 360° ⋅ 2 ⎟⎞ 4 4 ⎝ ⎠ w2 = 3(cos180° + i sin180°) w2 = 3 cis 180°
w3 = 3 ⎛⎜ cos 0° + 360° ⋅ 3 + i sin 0° + 360° ⋅ 3 ⎞⎟ 4 4 ⎝ ⎠ w3 = 3(cos 270° + i sin 270°) w3 = 3 cis 270°
w0 = 3(cos 0° + i sin 0°)
Copyright © Houghton Mifflin Company. All rights reserved.
908
58.
Additional College Trigonometry Solutions
−16 2 − 16i 2 = 32(cos 225° + i sin 225°)
[5.3]
wk = 321/ 5 ⎜⎛ cos 225° + 360°k + i sin 225° + 360°k ⎟⎞ 5 5 ⎝ ⎠
k = 0, 1, 2, 3, 4
w0 = 2 ⎜⎛ cos 225° + i sin 225° ⎟⎞ 5 5 ⎠ ⎝ w0 = 2(cos 45° + i sin 45°) w0 = 2 cis 45°
o o o o⎞ ⎛ w1 = 2 ⎜ cos 225 + 360 + i sin 225 + 360 ⎟ 5 5 ⎝ ⎠ w1 = 2(cos117° + i sin117°) w1 = 2 cis 117°
w2 = 2 ⎛⎜ cos 225° + 360° ⋅ 2 + i sin 225° + 360° ⋅ 2 ⎞⎟ 5 5 ⎝ ⎠ w2 = 2(cos189° + i sin189°) w2 = 2 cis 189°
w3 = 2 ⎜⎛ cos 225° + 360° ⋅ 3 + i sin 225° + 360° ⋅ 3 ⎟⎞ 5 5 ⎝ ⎠ w3 = 2(cos 261° + i sin 261°) w3 = 2 cis 261°
w4 = 2 ⎜⎛ cos 225° + 360° ⋅ 4 + i sin 225° + 360° ⋅ 4 ⎟⎞ 5 5 ⎝ ⎠ w4 = 2(cos333° + i sin 333°) w4 = 2 cis 333° 59.
(
81 = 81 cos 0o + i sin 0o
)
[5.3]
wk = 2561/ 4 ⎜⎛ cos 0° + 360°k + i sin 0° + 360°k ⎟⎞ 4 4 ⎝ ⎠
k = 0, 1, 2, 3
w0 = 4 cis 0°
w1 = 4 ⎛⎜ cos 0° + 360° + i sin 0° + 360° ⎞⎟ 4 4 ⎝ ⎠ w1 = 4(cos90° + i sin 90°) w1 = 4 cis 90°
w2 = 4 ⎜⎛ cos 0° + 360° ⋅ 2 + i sin 0° + 360° ⋅ 2 ⎟⎞ ⎝ 4 4 ⎠ w2 = 4(cos180° + i sin180°) w2 = 4 cis 180°
w3 = 4 ⎛⎜ cos 0° + 360° ⋅ 3 + i sin 0° + 360° ⋅ 3 ⎞⎟ 4 4 ⎝ ⎠ w3 = 4(cos 270° + i sin 270°) w3 = 4 cis 270°
w0 = 4(cos 0° + i sin 0°)
60.
(
−125 = 125 cos180o + i sin180o
)
[5.3]
wk = 1251/ 3 ⎛⎜ cos 180° + 360°k + i sin 180° + 360°k ⎞⎟ 3 3 ⎝ ⎠
(
w0 = 5 cos 180° + i sin 180° 3 3 w0 = 5(cos 60° + i sin 60°) w0 = 5 cis 60°
)
k = 0, 1, 2
w1 = 5 ⎛⎜ cos 180° + 360° + i sin 180° + 360° ⎞⎟ 3 3 ⎝ ⎠ w1 = 5(cos180° + i sin180°) w1 = 5 cis 180° = −5
....................................................... QR1.
w2 = 5 ⎛⎜ cos 180° + 360° ⋅ 2 + i sin 180° + 360° ⋅ 2 ⎞⎟ 3 3 ⎝ ⎠ w2 = 5(cos300° + i sin 300°) w2 = 5 cis 300°
Chapter 5 Quantitative Reasoning QR2. –0.25 + 0.25i, –1 + 0.1i, and 0.1 + 0.2i
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Additional College Trigonometry Solutions
909
QR4. The first iterate of 2i is z 1 = −4 + 2i . 2i is not an element of
QR3. –2 is an element of the Mendelbrot set because all of its iterates equal 2. QR5.
the Mandelbrot set because − 4 + 2i > 3 .
(Z 2 + S ) → Z
QR6. Answers will vary.
....................................................... 1.
6 + −9 = 6 + 3i
2.
4.
(−1 + −25) − (8 − −16) = −1 + 5i − 8 + 4i = −9 + 9i
[5.1]
6.
i 263 = i 3 = −i
9.
(3 − 5i )(−3 + 5i ) = −9 + 30i − 25i 2 [5.1] = 16 + 30i
11.
2 − 7i ⋅ 4 − 3i = 8 − 34i + 21i 2 = − 13 − 34 i 4 + 3i 4 − 3i 16 + 9 25 25
13.
z =
7.
[5.1]
( 3)2 + ( −5)2
14.
[5.2]
−18 = 3i 2
Chapter 5 Test
(
5.
[5.1]
3.
[5.1]
−12 ) ( −3 ) = ( 2i 3 )( i 3 ) [5.1] = 6i 2 = −6
(3 + 7i) − (−2 − 9i ) = 5 + 16i
[5.1]
r=
[5.1]
8.
4 − 5i ⋅ −i = −4i + 5i 2 = −5 − 4i −i i −i 2
12.
6 + 2i ⋅ 1 + i = 6 + 8i + 2i 2 = 4 + 8 i = 2 + 4i 1− i 1+ i 1+1 2 2
( 3)2 + ( −3)2
α = tan −1
(−6 − 9i )(4 + 3i ) = −24 − 54i − 27i 2 = 3 − 54i [5.1]
10.
15.
[5.2]
r =3 2
= 9 + 25 = 34
(3 + −4) + (7 − −9) = 3 + 2i + 7 − 3i = 10 − i [5.1]
−3 3
[5.1]
2
r = ( 0 ) + ( −6 ) r =6
α = tan −1
2
[5.1]
[5.2]
−6 = 90o 0
α = 360o − 90o = 270o
= tan −1 1 = 45o
α = 360o − 45o = 315o
z = 6 cis 270o
z = 3 2 cis 315o 16.
17.
z = 4(cos 120o + i sin 120o ) [5.2] ⎛ ⎞ z = 4⎜ − 1 + 3 i ⎟ ⎝ 2 2 ⎠ z = −2 + 2i 3
18.
z = 5(cos 225o + i sin 225o )
[5.2]
⎛ ⎞ z = 5⎜ − 2 − 2 i ⎟ 2 ⎠ ⎝ 2 z = −5 2 − 5 2 i 2 2
z1z2 = 3(cos 28° + i sin 28°) ⋅ 4(cos 17° + i sin 17°) z1z2 = 12[cos(28° + 17°) + i sin(28° + 17°)] z1z2 = 12(cos 45° + i sin 45°) ⎛ 2 2 ⎞ + z1z2 = 12 ⎜ i⎟ ⎝ 2 2 ⎠ z1z2 = 6 2 + 6i 2
[5.2]
19.
z1z2 z1z2 z1z2 z1z2 z1z2
= 5(cos 115° + i sin 115°) ⋅ 4(cos 10° + i sin 10°) = 20[cos(115° + 10°) + i sin(115° + 10°)] = 20(cos 125° + i sin 125°) ≈ 20(−0.5736 + 0.81915i ) ≈ −11.472 + 16.383i
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[5.2]
910
20.
Additional College Trigonometry Solutions
21.
z1 24(cos 258o + i sin 258o ) = [5.2] z2 6(cos 78o + i sin 78o ) z1 = 4 [cos(258o − 78o ) + i sin(258o − 78o )] z2 z1 = 4(cos 180o + i sin 180o ) = −4 + 0i = −4 z2
22.
z1 18(cos 50o + i sin 50o ) = [5.2] z2 3(cos 140o + i sin 140o ) z1 = 6 [cos(50o − 140o ) + i sin(50o − 140o )] z2 z1 = 6(cos − 90o + i sin − 90o ) = 0 − 6i = −6i z2
z = 2 − 2i 3 [5.3]
r = 22 + ( −2 3 ) r=4
2
α = tan −1 −2 3 = 60o 2
θ = 300o
z = 4(cos300° + i sin 300°) 12
(2 − 3i 3)
23.
= [4(cos300° + i sin 300°)]12 = 16,777, 216[cos(12 ⋅ 300°) + i sin(10 ⋅ 300°)] = 16,777, 216(cos3600° + i sin 3600°) = 16,777, 216(cos 0° + i sin 0°) = 16,777, 216 + 0i
64 = 64(cos 0° + i sin 0°) [5.3] 0° + 360°k 0° + 360°k ⎞ ⎛ + i sin wk = 641/ 6 ⎜ cos ⎟ 6 6 ⎝ ⎠
k = 0, 1, 2, 3, 4, 5
w0 = 2 cis 0°
w1 = 2 ⎛⎜ cos 0 + 360° + i sin 0 + 360° ⎞⎟ 6 6 ⎝ ⎠ w1 = 2(cos 60° + i sin 60°) w1 = 2 cis 60°
w2 = 2 ⎜⎛ cos 0° + 360° ⋅ 2 + i sin 0° + 360° ⋅ 2 ⎟⎞ 6 6 ⎝ ⎠ w2 = 2(cos120° + i sin120°) w2 = 2 cis 120°
w3 = 2 ⎛⎜ cos 0° + 360° ⋅ 3 + i sin 0° + 360° ⋅ 3 ⎞⎟ 6 6 ⎝ ⎠ w3 = 2(cos180° + i sin180°) w3 = 2 cis 180°
w4 = 2 ⎜⎛ cos 0° + 360° ⋅ 4 + i sin 0° + 360° ⋅ 4 ⎟⎞ 6 6 ⎝ ⎠ w4 = 2(cos 240° + i sin 240°) w4 = 2 cis 240°
w5 = 2 ⎛⎜ cos 0° + 360° ⋅ 5 + i sin 0° + 360° ⋅ 5 ⎞⎟ 6 6 ⎝ ⎠ w5 = 2(cos300° + i sin 300°) w5 = 2 cis 300°
w0 = 2(cos 0° + i sin 0°)
24.
−1 + 3i = 2(cos120° + i sin120°) [5.3] wk = 21/ 3 ⎛⎜ cos 120° + 360°k + i sin 120° + 360°k ⎞⎟ 3 3 ⎝ ⎠ w0 = 3 2 ⎛⎜ cos 120° + i sin 120° ⎞⎟ 3 3 ⎠ ⎝ w0 = 3 2(cos40°+ i sin 40°) w0 = 3 2 cis 40°
k = 0, 1, 2
w1 = 3 2 ⎛⎜ cos 120° + 360° + i sin 120°+ 360° ⎞⎟ 3 3 ⎝ ⎠ w1 = 3 2(cos160°+ i sin160°) w1 = 3 2 cis 160°
w2 = 3 2 ⎛⎜ cos 120° + 360°⋅ 2 + i sin 120° + 360°⋅ 2 ⎞⎟ 3 3 ⎝ ⎠ w2 = 3 2(cos280° + i sin 280°) w2 = 3 2 cis 280°
Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
25.
x5 + 32 = 0
911
[5.3]
5
x = −32
Find the five fifth roots of –32. −32 = 32(cos180° + i sin180°) wk = 321/ 5 ⎛⎜ cos 180° + 360°k + i sin 180° + 360°k ⎞⎟ 5 5 ⎝ ⎠
k = 0, 1, 2, 3, 4 o o w2 = 2 cis 180 + 360 ⋅ 2 5 w2 = 2 cis 180°
w0 = 2 cis 180° 5 w0 = 2 cis 36°
w1 = 2 cis 180° + 360° 5 w1 = 2 cis 108°
o o w3 = 2 cis 180 + 360 ⋅ 3 5 w3 = 2 cis 252°
o o w4 = 2 cis 180 + 360 ⋅ 4 5 w4 = 2 cis 324°
....................................................... 1.
Chapter 5 Cumulative Review 2.
x 2 − x − 6 ≤ 0 [1.1 ( x − 3)( x + 2) ≤ 0 The product is negative or zero. The critical values are –2 and 3.
x2 − 4 = 0 ( x − 2)( x + 2) = 0 x = −2, 2 Domain: all real numbers except –2 and 2.
( x − 3)( x + 2) [–2, 3] 3.
f (c ) = 2
[1.3]
4.
( f o g )( x) = f [ g ( x)] ⎡ 2 ⎤ = f ⎢ x − 1⎥ ⎣ 3 ⎦ 2 ⎛ ⎞ = sin ⎜ 3 ⋅ x − 1 ⎟ 3 ⎠ ⎝ = sin( x 2 − 1)
7.
[2.2] cos38o = 20 c c = 20 o ≈ 25.4 cm cos38
2= c c +1 2(c + 1) = c 2c + 2 = c c = −2
6.
3π ⎛ 180o ⎞ = 270o [2.1] ⎜ ⎟ 2 ⎝ π ⎠
9.
y = 3sin π x
[2.5]
10.
[1.5]
y = 1 tan π x 2 4
5.
[1.6] f ( x) = x x −1 y= x x −1 y x= y −1 x( y − 1) = y xy − x = y xy − y = x y ( x − 1) = x y= x x −1 f −1 ( x) = x x −1 f −1 (3) = 3 = 3 3 −1 2
8.
−1 ≤ sin t ≤ 1 [2.4] a = –1, b = 1
[2.6]
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912
Additional College Trigonometry Solutions
11.
cos x = cos x ⋅ 1 − sin x 1 + sin x 1 + sin x 1 − sin x = cos x − sin 2x cos x 1 − sin x x cos x = cos x − sin cos 2 x = 1 − sin x cos x cos x = sec x − tan x
13.
[3.2] sin α = 4 , cos α = 3 5 5 cos β = 12 , sin β = − 5 13 13 cos(α + β ) = cos α cos β − sin α sin β
[3.1]
12.
[3.2] sin 2 x cos3 x − cos 2 x sin 3 x = sin(2 x − 3 x) = sin( − x) or − sin x
14.
⎡ ⎛3⎞ ⎛ 5 ⎞⎤ y = sin ⎢sin −1 ⎜ ⎟ + cos−1 ⎜ − ⎟ ⎥ ⎝5⎠ ⎝ 13 ⎠ ⎦ ⎣
2
Let α = sin −1 3 , sin α = 3 , cosα = 1− ⎛⎜ 3 ⎞⎟ = 4 . 5 5 5 ⎝5⎠
( 13)
⎛ 3 ⎞⎛ 12 ⎞ ⎛ 4 ⎞⎛ 5 ⎞ = ⎜ ⎟⎜ ⎟ − ⎜ ⎟⎜ − ⎟ ⎝ 5 ⎠⎝ 13 ⎠ ⎝ 5 ⎠⎝ 13 ⎠ 48 15 56 =− + = 65 65 65
15.
16.
20.
(
−27 = 27 cos180o + i sin180o
[4.3]
(
)
c 2 = a 2 + b 2 − 2ab cos C
[4.2]
= 36,500 − 36, 400cos 78o ≈ 28,932 c ≈ 170 cm
18.
W = F ⋅s W= F
[4.3] s cos α
W = (100)(15)(cos15°) W ≈ 1449 foot-pounds
19.
r=
( 2 ) 2 + ( 2 )2
[5.2]
r=2 2
α = tan −1
2 2
= tan −1 1 = 45o z = 2 2 cis 45o
)
[5.3]
wk = 271/ 3 ⎛⎜ cos 180° + 360°k + i sin 180° + 360°k ⎞⎟ 3 3 ⎝ ⎠ w0 = 3 cos 180° + i sin 180° 3 3 w0 = 3(cos 60° + i sin 60°) ⎛ ⎞ w0 = 3 ⎜ 1 + 3 i ⎟ = 3 + 3 3 2 ⎝2 2 ⎠ 2
13
= (140)2 + (130)2 − 2(140)(130) cos 78o
2cos x − 3 = 0
cos θ = v ⋅ w v w (3i + 2 j) ⋅ (5i − 3 j) cos θ = 32 + 22 52 + (−3) 2 3(5) + (2)(−3) cos θ = 13 34 9 cos θ = =0 442 o θ = 64.7
⎝ 13 ⎠
13
y = sin(α + β ) = sin α cos β + cos α sin β = 3 ⎛⎜ − 5 ⎞⎟ + 4 ⎛⎜ 12 ⎞⎟ 5 ⎝ 13 ⎠ 5 ⎝ 13 ⎠ = − 15 + 48 = 33 65 65 65
cos x = 3 2 x = π , 11π 6 6 π 11 π The solutions are 0, , π , . 6 6
17.
2
β = cos −1 − 5 , cosβ = − 5 , sinβ = 1− ⎛⎜ − 5 ⎞⎟ = 12 .
sin 2 x = 3 sin x [3.6] 2sin x cos x − 3 sin x = 0 sin x (2cos x − 3) = 0
sin x = 0 x = 0, π
[3.5]
k = 0, 1, 2
w1 = 3⎛⎜ cos 180°+ 360° + i sin 180°+ 360° ⎞⎟ 3 3 ⎝ ⎠ w1 = 3(cos180° + i sin180°) w1 = 3( −1+ 0i ) = −3
w2 = 3⎛⎜ cos 180°+ 360°⋅ 2 + i sin 180° + 360°⋅ 2 ⎞⎟ 3 3 ⎝ ⎠ w2 = 3(cos300° + i sin300°) ⎛ ⎞ w2 = 3⎜ 1 − 3 i ⎟ = 3 − 3 3 i ⎝2 2 ⎠ 2 2
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Additional College Trigonometry Solutions
913
Section 6.7 See CAT Section 8.7 solutions on page 577 for exercises 1 – 18. 19.
y = 3 + 2sin t y − 3 = 2sin t y−3 = sin t 2
x = 2 + 3cos t x − 2 = 3cos t x − 2 = cos t 3
( ) ( x −3 2 ) x−2 3
2
2
2
⎛ y − 3⎞ 2 2 +⎜ ⎟ = cos t + sin t ⎝ 2 ⎠ 2
⎛ y − 3⎞ +⎜ ⎟ =1 ⎝ 2 ⎠
At t = 0 , x = 2 + 3cos0 = 5 At t = π , x = 2 + 3cos π = −1
y = 3 + 2sin 0 = 3 y = 3 + 2sin π = 3
The point traces the top half of the ellipse
( )
2
2 x − 2 + ⎛ y − 3 ⎞ = 1 , as shown in the figure. The point ⎜ ⎟ 3 ⎝ 2 ⎠ starts at (5, 3) and moves counterclockwise along the ellipse until it reaches the point (–1, 3) at time t = π .
20.
x = sin t
y = − cos t
x 2 + y 2 = sin 2 t + cos2 t x2 + y2 = 1
At t = 0 , x = sin 0 = 0 3 At t = π , 2 x = sin 3π = −1 2
y = − cos0 = −1
y = − cos 3π = 0 2
The point traces a portion of the circle x 2 + y 2 = 1 , as shown in the figure. The point starts at (0, –1) and moves counter clockwise along the circle until it reaches the point (–1, 0) at time t = 3π . 2
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Additional College Trigonometry Solutions
914
21.
y = t +1 ⇒ t = y −1 x = 2t − 1 x = 2( y − 1) − 1
22.
y= t y = x − 1 or y 2 = x − 1
x = 2y − 3 x + 3 = 2y y = 1 x+3 2
At t = 0 ,
At t = 0 , x = 2(0) − 1 = −1 At t = 3 , x = 2(3) − 1 = 5
y = 0 +1=1
( )
x = tan π − t 4
y = 3+1 = 4
y2 − x2
( ) = sec ( π − t ) − tan ( π − t ) 4 4
y2 − x2 = 1 At t = 0 ,
y = sec π − t 4
2
At t = π , 2
24.
x = 4 +1= 5
y= 4 =2
x =1− t ⇒ t = 1− x 2
y = (1 − x ) or y = ( x − 1)
( )
y = sec π − 0 = 2 4
x =1− 0 =1 At t = 2 , x = 1 − 2 = −1
y = (0)2 = 0 y = (2)2 = 4
The point traces a portion of the parabola given by
(
)
x = tan π − π = −1 4 2
(
)
y = sec π − π = 2 4 2
The point traces a portion of the top branch of the
y = ( x − 1)2 , as shown in the figure. The point starts at (1, 0) and moves along the parabola until it reaches (–1, 4) at time t = 2.
hyperbola y 2 − x 2 = 1 , as shown in the figure. The point starts at (1,
2
At t = 0 ,
Since 1 + tan 2 θ = sec2 θ
( )
y= 0 =0
y = t2
2
x = tan π − 0 = 1 4
x = 0 +1=1 At t = 4 ,
The point traces a portion of the parabola y 2 = x − 1 , as shown in the figure. The point starts at (1, 0) and moves along the parabola until it reaches the point (5, 2) at time t = 4.
The point traces a line segment, as shown in the figure. The point starts at (–1, 1) and moves along the line segment until it reaches the point (5, 4) at time t = 3.
23.
x = t +1 ⇒ t = x −1
2 ) and moves along the hyperbola until it
reaches the point ( −1,
2 ) at time t = π . 2
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Additional College Trigonometry Solutions
25.
915
26.
C1 : x = 2 + t 2 y = 1 − 2t 2
C1 : x = sec2 t y = tan 2 t
x = 2 + t2 → t2 = x − 2 y = 1 − 2( x − 2) y = −2 x + 5 x ≥ 2, y ≤ 1 C2 : x = 2 + t
y = 1− 2t x = 2+t →t = x−2 y = 1 − 2( x − 2) y = −2 x + 5 x ∈ R, y ∈ R The graph of C1 is a ray beginning at (2, 1) with slope −2. The graph of C 2 is a line passing through (2, 1) with slope –2.
tan 2 t + 1 = sec 2 t y +1 = x y = x −1
Because 0 ≤ t < π , 2 1 ≤ sec2 t < ∞ and 0 ≤ tan 2 t < ∞ 1 ≤ x < ∞ and 0 ≤ y <∞ Thus, x ≥ 1 and y ≥ 0. C2 : x = 1 + t 2 y = t2 x =1+ y y = x −1
Because 0 < t <
π 2
,
2
2
0 ≤ t 2 < π and 1 ≤ t 2 + 1 < 1 + π 4 4 2
2
Thus, 1 ≤ x < 1 + π and 0 ≤ y < π . 4 4
.
2 2⎞ ⎛ C1 is a ray from (1, 0) in the direction of ⎜1 + π , π ⎟ . 4 4 ⎝ ⎠ C 2 is the points on the line y = x − 1 between (1, 0) and
⎛ π2 π2 ⎞ ⎛ π2 π2 ⎞ ⎜1 + 4 , 4 ⎟ . C 2 includes (1, 0) but not ⎜1 + 4 , 4 ⎟ . ⎝ ⎠ ⎝ ⎠ 27.
x = sin t y = csc t 1 sin t 1 y= x
csc t =
Range for graph 1 : 0 < t ≤
π
2 0 < x ≤1 y ≥1
3π 2 −1 ≤ x ≤ 0 y ≤ −1
Range for graph 2 : π ≤ t ≤
28.
C1 : x = cos t 2
y = cos t
0 ≤ t ≤ π , −1 ≤ x ≤ 1 0 ≤ y ≤1
C2 : x = sin t 2
y = sin t
(cos t )2 = cos 2 t
(sin t )2 = sin 2 t
( x)2 = y
( x)2 = y
y = x 2 for − 1 ≤ x ≤ 1 and 0 ≤ y ≤ 1
0 ≤ t ≤ π, 0 ≤ x ≤1 0 ≤ y ≤1
y = x 2 for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1
C1 is the graph of the parabola y = x 2 for − 1 ≤ x ≤ 1, while C2 is the graph of the parabola y = x 2 for 0 ≤ x ≤ 1.
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Additional College Trigonometry Solutions
916
29.
31.
30.
a.
b.
For the Hummer, x=6 y = 60t for t ≥ 0 Using the graphing calculator in SIMUL and PAR mode, the Hummer is the first to reach the intersection.
33.
32.
a.
For the Learjet, x = 300 − 420t y = 200 for t ≥ 0
b.
Using the graphing calculator in SIMUL and PAR mode, the Piper Seneca is the first to reach the intersection point.
34.
Maximum height (to the nearest foot) of 462 feet is attained when t ≈ 5.38 seconds.
Maximum height (to the nearest foot) of 195 feet is attained when t ≈ 3.50 seconds.
The projectile has a range (to the nearest foot) of 1295 feet and hits the ground in about 10.75 seconds.
The projectile has a range (to the nearest foot) of 1117 feet and hits the ground in about 6.99 seconds.
35.
36.
Maximum height: 963 ft at time t ≈ 7.76 sec Range: 3009 ft at time t ≈ 15.51 sec
Maximum height (to the nearest foot) of 694 feet is attained when t ≈ 6.59 seconds. The projectile has a range (to the nearest foot) of 3084 feet and hits the ground in about 13.17 seconds. 37.
38.
39.
40.
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Additional College Trigonometry Solutions
917
....................................................... 41.
Connecting Concepts
Let P1 ( x1 , y1 ) and P2 ( x2 , y 2 ) be two distinct points on a line. 42. If P(x, y) is any other point on the line, then y − y1 y2 − y1 = (Slope is constant along entire line.)This x − x1 x2 − x1 equation can be rewritten as y − y1 x − x1 = Let this value equal t. y2 − y1 x2 − x1
sin 2 t + cos 2 t = 1 2
2 ⎛ x − h ⎞ + ⎛ y − k ⎞ =1 ⎜ ⎟ ⎜ ⎟ ⎝ a ⎠ ⎝ b ⎠
x − x1 y − y1 = t and = t. x2 − x1 y2 − y1 Solving for x and y, respectively, we have x = x1 + t ( x2 − x1) and y = y1 + t ( y 2 − y1 ) Thus,
43.
radius = a, θ = ∠TOR
(x − h )2 + ( y − k )2
= 1, which is the standard equation a2 b2 for an ellipse at (h, k). 44.
The x-coordinate of P(x, y) is given by x = OR + QP. The y-coordinate is given by y = TR − TQ.
a > 0, x ∈ R, 0 ≤ t < 2π b > 0, y ∈ R, 0 ≤ t < 2π x = h + a sin t → sin t = x − h a y−k y = k + b cos t → cos t = b
x = h + a sin t y = k + b cos t
Let α = ∠COP. Because the smaller circle does not slip, b bθ = aα or α = θ . a
From the figure, OR = a cosθ and QP = aθ sin θ . Thus, x = a cosθ + aθ sin θ TR = a sin θ and TQ = aθ cosθ . Thus, y = a sin θ − aθ cosθ The parametric equations are x = a cosθ + aθ sin θ y = a sin θ − aθ cosθ
The coordinates of P(x, y) are given by x = BC + DP y = OB − OD
π⎞ ⎛ a+b Thus, x = (a + b) cosθ + a sin ⎜ θ− ⎟ 2⎠ ⎝ a ⎛a+b ⎞ = (a + b)cosθ − a cos ⎜ θ⎟ ⎝ a ⎠ π⎞ ⎛a+b θ− ⎟ y = (a + b)sin θ − a cos ⎜ 2⎠ ⎝ a ⎛a+b ⎞ = (a + b)sin θ − a cos ⎜ θ⎟ ⎝ a ⎠ The parametric equations are ⎛ a+b ⎞ θ⎟ x = (a + b) cosθ − a cos⎜ ⎠ ⎝ a ⎛ a+b ⎞ θ⎟ y = (a + b) sin θ − a sin ⎜ ⎝ a ⎠
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Additional College Trigonometry Solutions
918
45.
Because the circle moves without slipping, bθ = aα . bθ Therefore, α = . Let P( x, y ) be the coordinates of the moving point. a π ⎛b−a⎞ Angle φ = − ⎜ ⎟θ 2 ⎝ a ⎠ ⎡π ⎛ b − a ⎞ ⎤ Thus, x = (b − a ) cosθ + a sin ⎢ − ⎜ ⎟θ ⎥ ⎣2 ⎝ a ⎠ ⎦ ⎡π ⎛ b − a ⎞ ⎤ y = (b − a )sin θ − a cos ⎢ − ⎜ ⎟θ⎥ ⎣2 ⎝ a ⎠ ⎦
Simplifying, we have ⎛b−a ⎞ θ⎟ x = (b − a) cosθ + a cos⎜ ⎠ ⎝ a ⎛b−a ⎞ θ⎟ y = (b − a ) sin θ − a sin ⎜ ⎝ a ⎠
....................................................... 1.
log5 25 = x [7.2]
2.
x
5.
log3 81 = x [7.2]
Chapter 6 Chapter Review 3.
x
5 = 25
3 = 81
5 x = 52 x=2
3x = 34 x=4
32 x + 7 = 27 [7.4]
6.
x
5 x − 4 = 625 [7.4]
7.
2x + 7 = 3 2 x = −4 x = −2
2x = 1 8
ln eπ = x
[7.2]
π
x
3
e =e x=3
5 x − 4 = 54 x−4=4 x =8
32 x + 7 = 33
4.
ln e3 = x [7.2]
e =e x =π
8.
[7.4]
27 ( 3x ) = 3−1 [7.4] 27 ( 3x ) = 1 3 x 3 = 1 81
2 x = 2 −3 x = −3
3x = 3−4 x = −4
9.
log x 2 = 6 106 = x 2 1,000,000 = x 2 ± 1,000,000 = x ±1000 = x
13.
[7.4]
10.
1 log x = 5 2 log x = 10 1010 = x
[7.4]
11.
10log 2 x = 14 [7.4] 2 x = 14 x=7
x = ±1010
14.
15.
Copyright © Houghton Mifflin Company. All rights reserved.
12.
2
eln x = 64 [7.4] x 2 = 64 x = ±8
Additional College Trigonometry Solutions
919
16.
17.
19.
18.
20.
21.
23.
25.
22.
24.
log 4 64 = 3
[7.2]
log1/ 2 8 = −3 [7.2]
26.
3
4 = 64
29.
53 = 125 [7.2] log5 125 = 3
33.
log b
35.
ln xy 3 = ln x + 3 ln y [7.3]
⎛1⎞ ⎜ ⎟ ⎝2⎠
−3
x2 y3 = 2 log b x + 3 log b y − log b z [7.3] z
log
(
=8
210 = 1024 [7.2] log2 1024 = 10
30.
27.
2
4 = 4 [7.2]
28.
31.
100 = 1 [7.2] log10 1 = 0
34.
logb
36.
ln
ln1 = 0 [7.2] e0 = 1
2) = 4 4
32.
81/ 2 = 2 2 [7.2] log8 2 2 = 1 2
x = 1 log x − 2log y + log z [7.3] ( b b b ) y2 z 2 = 1 logb x − 2logb y − logb z 2 xy
z
4
= 1 ( ln x + ln y ) − 4 ln z [7.3] 2 1 = ln x + 1 ln y − 4 ln z 2 2
37.
2 log x + 1 log ( x + 1) = log ( x 2 3 x + 1 ) [7.3] 3
38.
5log x − 2log( x + 5) = log
39.
1 ln 2 xy − 3ln z = ln 2 xy [7.3] 2 z3
40.
ln x − (ln y − ln z ) = ln
41.
log 5 101 =
log 101 ≈ 2.86754 [7.3] log 5
42.
log 3 40 =
log 40 ≈ 3.35776 [7.3] log 3
43.
log 4 0.85 =
log 0.85 ≈ −0.117233 [7.3] log 4
44.
log 8 0.3 =
log 0.3 ≈ −0.578989 [7.3] log 8
x5 [7.3] ( x + 5)2
x xz = ln [7.3] y z y
Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
920
45.
4 x = 30
46.
[7.4]
x
log 4 = log 30 x log 4 = log 30 log 30 x= log 4
48.
ln ( 3x ) + ln 2 = ln1 [7.4]
49.
ln ( 3x ⋅ 2 ) = 1 ln ( 6 x ) = 1 1
e = 6x e=x 6
5 x +1 = 41 [7.4] ( x + 1) log5 = log 41 log 41 x +1= log5 log 41 x= −1 log5
47.
ln x + 2 e ( ) = 6 [7.4]
50.
3x = 4 x − 4 4=x log 2 x +1) 10 ( = 31 [7.4]
2 x + 1 = 31
x+2=6 x=4
2 x = 30 x = 15
5 x + 5− x = 8 2
52.
5 x ( 5 x + 5− x ) = 16 ( 5x )
4 x ( 4 x + 4− x ) = 2 ( 4 x − 4− x ) 4 x
52 x + 1 = 16 ( 5x )
42 x + 1 = 2 ( 42 x − 1)
52 x − 16 ( 5 x ) + 1 = 0
42 x + 1 = 2 ( 42 x − 1)
Let 5 x = u
42 x − 2 ⋅ 42 x + 3 = 0
u 2 − 16u + 1 = 0
42 x = 3 2 x ln 4 = ln 3 x = ln 3 2 ln 4
[7.4]
ln 3x = ln 4 x −1 3x = 4 x −1 3x = 4 ( x − 1)
( x + 2) = 6
4 x + 4− x = 2 4 x − 4− x
51.
ln ( 3x ) − ln ( x − 1) = ln 4
u=
16 ±
162 − 4 (1)(1)
2 ± 16 252 u= 2 ± 16 6 7 u= 2 u =8±3 7
[7.4]
5x = 8 ± 3 7 x= 53.
log ( log x ) = 3
[7.4]
54.
ln ( ln x ) = 2
=x
log x − 5 = 3
e = ln x
10 = log x 10
55.
2
3
(10 3 )
[7.4]
ln (8 ± 3 7 ) [7.4] ln 5
e
( e2 )
=x
3
10 = x − 5 106 = x − 5 106 + 5 = x x = 1,000,005
101000 = x
Copyright © Houghton Mifflin Company. All rights reserved.
[7.4]
Additional College Trigonometry Solutions
56.
log x + log ( x − 15) = 1
921
57.
log x ( x − 15) = 1
3 =x 81 = x
1 = log5 x 2
[7.4]
5 = x2
15 ± 152 − 4 (1)( −10 )
± 5=x
[7.4]
2
x = 15 ± 265 2 + 15 265 x= 2
[7.4]
log5 x 3 = log5 16 x [7.4]
60.
25 = 16log4 x [7.4] 25 = 4
2
25 = 4log4 x
x = 16 x=4
⎛ I ⎞ m = log ⎜ ⎟ [7.3] ⎝ I0 ⎠ ⎛ 51,782,000 I 0 ⎞ = log ⎜ ⎟ I0 ⎝ ⎠ = log 51,782,000 ≈ 7.7
61.
2 log 4 x
3
x = 16 x
62.
70 = log5 x 2
4
0 = x 2 − 15 x − 10
59.
log7 (log5 x 2 ) = 0
4 = log3 x
10 = x 2 − 15 x
x=
58.
log 4 (log3 x ) = 1
2
25 = x 2 ±5 = x 5= x
M = log A + 3log8t − 2.92
[7.3]
63.
= log18 + 3log8(21) − 2.92 = log18 + 3log168 − 2.92 ≈ 5.0
⎛I ⎞ log ⎜ 1 ⎟ = 7.2 ⎝ I0 ⎠ I1 = 107.2 I0
⎛I ⎞ log ⎜ 2 ⎟ = 3.7 ⎝ I0 ⎠ I2 = 103.7 I0
and
I1 = 107.2 I 0
[7.3]
I 2 = 103.7 I 0
I1 107.2 I 0 103.5 3162 = = ≈ I 2 103.7 I 0 1 1 3162 to 1
64.
I1 = 600 = 10 x [7.3] I2
65.
log 600 = log10 x log 600 = x 2.8 ≈ x
[7.3] pH = − log ⎡⎢ H3O + ⎤⎥ ⎣ ⎦ = − log ⎡⎢6.28 × 10−5 ⎤⎥ ⎣ ⎦ ≈ 4.2
5.4 = − log ⎡⎢ H3O + ⎤⎥ [7.3] ⎣ ⎦
66.
−5.4 = log ⎡⎢ H3O + ⎤⎥ ⎣ ⎦ 10−5.4 = H3O + 0.00000398 ≈ H3O + H3O + ≈ 3.98 × 10−6
67.
P = 16,000, r = 0.08, t = 3 [7.5] a. b.
0.08 ⎞ ⎛ B = 16,000⎜1 + ⎟ 12 ⎠ ⎝
36
68.
≈ $20,323.79
B = 16,000e 0.08(3)
P = 19,000, r = 0.06, t = 5 [7.5] 1825
a.
0.06 ⎞ ⎛ B = 19,000⎜1 + ⎟ 365 ⎠ ⎝
b.
B = 19,000e 0.3 ≈ $25,647.32
B = 16,000e 0.24 ≈ $20,339.99
Copyright © Houghton Mifflin Company. All rights reserved.
≈ $25,646.69
Additional College Trigonometry Solutions
922
69.
S (n ) = P (1 − r )n , P = 12,400, r = 0.29, t = 3 [7.5]
70.
N ( t ) = N 0 e −0.12t
a.
S (n ) = 12,400(1 − 0.29 )3 ≈ $4438.10
N (10 ) = N 0 e N (10 ) N0
[7.5]
−0.12(10 )
= e −1.2 = .301
N (10 ) N0
= 30.1% healed
100% − 30.1% = 69.9% healed b.
N (t ) N0
= 0.5
0.5 = e −0.12t ln 0.5 = −0.12t ln 0.5 = t −0.12 t ≈ 6 days c.
N (t ) N0
= 0.1
0.1 = e −0.12t ln 0.1 = −0.12t ln 0.1 = t −0.12 t ≈ 19 days 71.
N (2) = 5
N ( 0) = 1 k 0 1 = N 0e ( )
72.
5 = e2 k ln 5 = 2k k = ln 5 ≈ 0.8047 2
1 = N0
N (0) = N 0 = 2 and N (3) = N 0 e 3k = 2e 3k = 11 e3k = 11 2 3k 11 e = 2 3k = ln ⎛⎜ 11 ⎞⎟ ⎝2⎠
Thus N ( t ) = e0.8047 t [7.5]
k = 1 ln ⎛⎜ 11 ⎞⎟ 3 ⎝2⎠ ≈ 0.5682
Thus N (t ) = 2e 0.5682t [7.5] 73.
4 = N (1) = N 0e k and thus
4 = e k . Now, we also N0 5
⎛ 4 ⎞ 1024 ⎟ = . have N (5) = 5 = N 0e5k = N 0 ⎜⎜ ⎟ N 04 ⎝ N0 ⎠ N0 = 4
1024 ≈ 3.783 5
74.
1 = N (0 ) = N 0 and 2 = N (− 1) = N 0e− k .
Since N 0 = 1 , we have 2 = 1 ⋅ e− k . ln 2 = − k k ≈ −0.6931 Thus N ( t ) = e−0.6931 t . [7.5]
Thus 4 = 3.783ek . ⎛ 4 ⎞ ln⎜ ⎟=k ⎝ 3.783 ⎠ k ≈ 0.0558 Thus N 0 = 3.783e0.0558 t . [7.5] Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
75.
a.
923
[7.5] N (1) = 25, 200ek (1) = 26,800 k 26,800 e = 25, 200 ⎛ 26,800 ⎞ k ln e = ln ⎜ ⎟ ⎝ 25, 200 ⎠ k ≈ 0.061557893
76.
P(t) = 0.5 t / 5730 = 0.96 [7.5]
t log 0.5 = log 0.96 5730 t log 0.96 = 5730 log 0.5
N (t ) = 25, 200e0.061557893 t
b.
)
(
log 0.5 t / 5730 = log 0.96
⎛ log 0.96 ⎞ t = 5730 ⎜ ⎟ ≈ 340 years ⎝ log 0.5 ⎠
N (7) = 25, 200e0.061557893(7) = 25, 200e0.430905251 ≈ 38,800
77.
Answers will vary.
78.
a.
b. c. 79.
a.
exponential: R ≈ 179.949 ( 0.968094t ) , r ≈ −0.99277 logarithmic: R ≈ 171.19665 − 35.71341ln t , r ≈ −0.98574 The exponential equation provides a better fit to the data.
R ≈ 179.949 ( 0.968094108 ) ≈ 5.4 per 1000 live births [7.6] P(t ) =
mP0 P0 + ( m − P0 )e
P(3) = 360 =
− kt
80.
128 = 128 = 128 = 128 = 21 1 3 6 1 + 5e −0.27(0) 1 + 5e0 1 + 5
a.
P(0) =
b.
As t → ∞, e−0.27t → 0. P(t ) → 128 = 128 = 128 [7.5] 1 + 5(0) 1
1400(210)
210 + (1400 − 210)e − k (3) 294000 360 = 210 + 1190e −3k
360 ( 210 + 1190e −3k ) = 294000
210 + 1190e −3k = 294000 360 −3k 29400 = − 210 1190e 36 e −3k = 29400 / 36 − 210 1190 −3k 29400 / 36 − 210 ⎞ ⎛ ln e = ln ⎜ ⎟ 1190 ⎝ ⎠ −3k = ln ⎛⎜ 29400 / 36 − 210 ⎞⎟ 1190 ⎝ ⎠
b.
k = − 1 ln ⎛⎜ 29400 / 36 − 210 ⎞⎟ 3 ⎝ 1190 ⎠ k ≈ 0.2245763649 294000 1400 P(t ) = = 210 + 1190e −0.22458t 1 + 17 e −0.22458t 3 294000 P(13) = 210 + 1190e −0.22458(13) 294000 = 210 + 1190e −2.919492744 ≈ 1070 coyotes [7.5]
Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
924
....................................................... 1.
a.
logb (5 x − 3) = c [7.2]
Chapter 6 Chapter Test b.
bc = 5 x − 3
2.
logb
z2 y
3
x
3x / 2 = y x 2
log3 y =
= logb z 2 − logb y 3 − logb x1/ 2 [7.3]
3.
3 log10 ( 2 x + 3) − 3log10 ( x − 2 ) = log10 ( 2 x + 3) − log10 ( x − 2 )
= log10 2 x + 33 [7.3] ( x − 2)
1 = 2logb z − 3logb y − logb x 2
4.
log12 [7.3] log 4 ≈ 1.7925
5.
log 4 12 =
45− x = 7 x
8.
5− x
6.
[7.4] x
ln 4 = ln 7 (5 − x ) ln 4 = x ln 7 5ln 4 − x ln 4 = x ln 7 5ln 4 = x ln 7 + x ln 4 5ln 4 = x (ln 7 + ln 4) 5ln 4 = x ln 28
11.
a.
(
) 12(5) = 20,000 (1 + 0.078 ) 12
nt A = P 1+ r n
= 20,000(1.0065) = $29,502.36 b.
60
A = Pert = 20,000e0.078(5) = 20,000e0.39 = $29,539.62
[7.5]
9.
7.
log( x + 99) − log(3x − 2) = 2 [7.4] x + 99 log =2 3x − 2 x + 99 = 102 3x − 2 x + 99 = 100(3x − 2) x + 99 = 300 x − 200 −299 x = −299 x =1
12.
10 − 7 x + x 2 = 37 − x x 2 − 6 x − 27 = 0 ( x − 9)( x + 3) = 0 x = 9 (not in domain) or x = −3 x = −3 [7.4]
(
(
nt
[7.5]
)
0.04 12t 12 0.04 12t
2P = P 1 + 2 = 1+
ln(2 − x) + ln(5 − x) = ln(37 − x) ln(2 − x)(5 − x) = ln(37 − x) (2 − x)(5 − x) = (37 − x)
10.
r⎞ ⎛ A = P ⎜1 + ⎟ ⎝ n⎠
(
12
)
)
0.04 12t 12 0.04 ln 2 = 12t ln 1 + 12
ln 2 = ln 1 +
12t = t=
(
(
ln 2
ln 1 +
0.04 12
)
)
1 ln 2 ⋅ 12 ln 1 + 0.04
(
5 x = 22 [7.4] x log5 = log 22 log 22 x= log5 x ≈ 1.9206
t ≈ 17.36 years
12
)
Copyright © Houghton Mifflin Company. All rights reserved.
Additional College Trigonometry Solutions
13.
a.
925
⎛ I ⎞ [7.3] M = log ⎜ ⎟ ⎝ I0 ⎠ ⎛ 42,304,000 I 0 ⎞ = log ⎜ ⎟ I0 ⎝ ⎠ = log 42,304,000 ≈ 7.6
14.
a.
N (3) = 34600ek (3) = 39800 34600e3k = 39800 39800 e3k = 34600 ⎛ 398 ⎞ ln e3k = ln ⎜ ⎟ ⎝ 346 ⎠ ⎛ 398 ⎞ 3k = ln ⎜ ⎟ ⎝ 346 ⎠ ⎛ 398 ⎞ k = 1 ln ⎜ 3 ⎝ 346 ⎟⎠ k ≈ 0.0466710767 N (t ) = 34600e0.0466710767 t [7.5]
b.
⎛I ⎞ log ⎜ 1 ⎟ = 6.3 ⎝ I0 ⎠ I1 = 106.3 I0
and
⎛I ⎞ log ⎜ 2 ⎟ = 4.5 ⎝ I0 ⎠ I2 = 104.5 I0
I1 = 106.3 I 0
b.
N (10) = 34600e0.0466710767(10) = 34600e0.466710767 ≈ 55,000
I 2 = 104.5 I 0
I1 106.3 I 0 101.8 63 = = ≈ 1 1 I 2 104.5 I 0 Therefore the ratio is 63 to 1.
P (t ) = 0.5 t / 5730 = 0.92 [7.5]
15.
log 0.5 t / 5730 = log 0.92 t log 0.5 = log 0.92 5730 log 0.92 t = 5730 log 0.5 ⎛ log 0.92 ⎞ t = 5730 ⎜ ⎟ ⎝ log 0.5 ⎠ t ≈ 690 years
16.
a.
y = 1.671991998(2.471878247) x
b.
y = 1.671991998(2.471878247)7.8 ≈ 1945
17.
a.
Logarithmic: d ≈ 67.35501 + 2.54015ln t ; Logistic: d ≈
b.
Logarithmic: d ≈ 67.35501 + 2.54015ln(12) ≈ 73.67 m ; 72.03783 Logistic: d ≈ ≈ 72.03 m [7.6] 1 + 0.15279e −0.67752 (12)
72.03783 1 + 0.15279e −0.67752 t
Copyright © Houghton Mifflin Company. All rights reserved.
[7.6]
926
18.
Additional College Trigonometry Solutions
a.
a=
c − P0 1100 − 160 = = 5.875 P0 160
b.
P(t ) =
1100 ≈ 457 raccoons 1 + 5.875e −0.20429(7)
c 1 + ae − bt 1100 P(1) = 1 + 5.875e − b(1) 1100 190 = 1 + 5.875e − b P (t ) =
190(1 + 5.875e − b ) = 1100 190 + 1116.25e − b = 1100 1116.25e − b = 910 e − b = 910 1116.25 −b ln e = ln 910 1116.25 −b = ln 910 1116.25 b = − ln 910 1116.25 b ≈ 0.20429 1100 P(t ) = [7.5] 1 + 5.875e −0.20429 t
....................................................... 1.
x 2 + 4 x − 6 = 0 [1.1]
2.
y = x3 − 4 x [1.4]
Chapter 6 Cumulative Review 3.
( g o f )( x) = g[ f ( x)] = g [sin x ] = 3sin x − 2
6.
opp sin t = − 3 = 2 hyp
2
−4 ± 4 − 4(1)(−6) −4 ± 40 = 2(1) 2 4 2 10 − ± = = −2 ± 10 2 The solutions are 2 ± 10. x=
[1.5]
symmetric with respect to the origin
4.
240o = 240o ⎛⎜ π o ⎞⎟ = 4π ⎝ 180 ⎠ 3
[2.1]
5.
[2.1] v = ωr ⋅ ⋅ ⋅ ⋅ 3 2π 60 60 10 = 12 ⋅ 5280 ≈ 11 mph
[2.4]
adj = 22 − ( −3 ) = 1 = 1 2
tan t =
Copyright © Houghton Mifflin Company. All rights reserved.
opp − 3 = =− 3 adj 1
Additional College Trigonometry Solutions
7.
9.
11.
13.
927
tan 43o = a [2.4] 20 o a = 20 tan 43 a = 19 cm
( )
y = 2 tan π x 3 period: 3
8.
10.
[3.2] sin α = 3 , cos α = − 4 5 5 cos β = − 5 , sin β = 12 13 13 sin(α + β ) = sin α cos β + cos α sin β 5 ⎞ ⎛ 4 ⎞⎛ 12 ⎞ = ⎛⎜ 3 ⎞⎛ ⎟⎜ − ⎟ + ⎜ − ⎟⎜ ⎟ ⎝ 5 ⎠⎝ 13 ⎠ ⎝ 5 ⎠⎝ 13 ⎠ = − 15 − 48 = − 63 65 65 65
12.
( )
a d c b C = 180o − A − B = 180o − 71o − 80o = 29o b = c sin B sin C o b = 155sin 80 ≈ 314.9 ft o sin 29 a = c sin A sin C o a = 155sin 71 ≈ 302.3 ft o sin 29 sin 71o = d 314.9 d = 314.9sin 71o [4.1] ≈ 298 ft
[4.2] 2
2
cos B = a + c − b = 4 + 3.6 − 2.5 = 22.7 2ac 2(4)(3.6) 28.8 B ≈ 38o
C A
b 2 = a 2 + c 2 − 2ac cos B 2
52 − (1) opp = =2 6 hyp 5 5
B
14.
[3.6]
( )
2
[3.5]
⎡ ⎛ 1 ⎞⎤ 2 6 y = sin ⎢cos −1 ⎜ ⎟ ⎥ = 5 ⎝ 5 ⎠⎦ ⎣
( )
2
[3.1]
2
2 2 adj = 13 −12 = 5 hyp 13 13
2
⎡ ⎛ 1 ⎞⎤ y = sin ⎢ cos −1 ⎜ ⎟ ⎥ ⎝ 5 ⎠⎦ ⎣
sin α =
x = cos ⎡sin −1 12 ⎤ = 5 ⎢⎣ 13 ⎥⎦ 13
15.
sin x = sin x ⋅ 1 + cos x 1 − cos x 1 − cos x 1 + cos x sin x (1 + cos x) = 1 − cos 2 x sin x (1 + cos x) = sin 2 x 1 = + cos x sin x sin x = csc x + cot x
adj Let α = cos −1 1 , cosα = 1 = 5 5 hyp
opp Let α = sin −1 12 , sin α = 12 = 13 13 hyp cosα =
[2.5]
period: 6, amplitude: 1 2
[2.6]
cos −1 x = sin −1 12 13 ⎡ x = cos sin −1 12 ⎤ ⎢⎣ 13 ⎥⎦
( )
y = 1 cos π x 2 3
16.
a1 = 30cos145o ≈ −24.6 [4.3] a2 = 30sin145o ≈ 17.2
v = −24.6i + 17.2 j
Copyright © Houghton Mifflin Company. All rights reserved.
928
17.
Additional College Trigonometry Solutions
v ⋅ w = (3i + 2 j) ⋅ (5i − 7 j) [4.3] = 3(5) + (2)( −7) =1≠ 0 No. the vectors are not orthogonal.
18.
z = −2 + 2i 3
[5.2]
2
r = (−2) + (2 3) 2 = 16 = 4
α = tan −1 2 3 = tan −1 3 = π −2
3
θ = π − π = 2π 3
3
( )
z = 4 cis 2π 3
19.
r = 3sin(2θ ) [6.5]
20.
x = 2t − 1 [6.7] x + 1 = 2t t = x +1 2 y = 4t 2 + 1
( )
2 ⎛ 2 ⎞ y = 4 x + 1 + 1 = 4⎜ x + 2x + 1 ⎟ + 1 2 4 ⎝ ⎠ y = x2 + 2 x + 2
....................................................... 1.
( f o g )( x) = f [ g ( x)]
[1.5]
2.
f ( x) = 2 x + 8 y = 2x + 8 x = 2y + 8 x − 8 = 2y x −8 = y 2 f −1 ( x) = 1 x − 4 2
5.
y = 3sin 1 x − π 3 2
2
= f ( x + 1) = cos( x 2 + 1)
4.
[2.2] cos 26o = 15 a a = 15 o ≈ 16.7 cm cos 26
(
Chapter 7 Cumulative Review
)
[1.6]
3.
c = 32 + 42 = 25 = 5 sin θ = 3 , cosθ = 4 , tan θ = 3 5 5 4
[2.5]
6.
y = sin x + cos x
0 ≤ 2 x − π ≤ 2π 2 π ≤ 2 x ≤ 5π 2 2 π ≤ x ≤ 5π 4 4 amplitude = 4, period = π ,
Amplitude:
phase shift = π 4
Copyright © Houghton Mifflin Company. All rights reserved.
[3.4]
2 , period: 2π
[2.2]
Additional College Trigonometry Solutions
7.
929
f ( − x) = sin(− x) = − sin x = − f ( x) odd function [2.4]
8.
1 − sin x = 1 − sin 2 x sin x sin x 2 x cos = sin x = cos x cos x sin x = cos x cot x
9.
[3.1]
( )
tan ⎛⎜ sin −1 12 ⎞⎟ 13 ⎠ ⎝
[3.5]
opp Let α = sin −1 12 , sin α = 12 = 13 13 hyp opp 12 tanα = = = 12 adj 132 −122 5
( )
tan ⎡sin −1 12 ⎤ = 12 ⎢⎣ 13 ⎥⎦ 5
10.
2cos 2 x + sin x − 1 = 0
[3.6]
11.
2
2(1 − sin x) + sin x − 1 = 0 2sin 2 x − sin x − 1 = 0 (2sin x + 1)(sin x − 1) = 0 2sin x + 1 = 0 sin x = − 1 2 x = 7π , 11π 6 6
12.
14.
v = (−3)2 + 42
α = tan −1
v = 9 + 16
α θ θ θ
v =5 sin x − 1 = 0 sin x = 1 x=π 2
cos θ = v ⋅ w v w (2i − 3 j) ⋅ ( −3i + 4 j) cos θ = 22 + (−3)2 (−3) 2 + 42 2( −3) + (−3)(4) cos θ = 13 25 cos θ = −18 = −0.9985 325 θ = 176.8o
13
[4.3]
a = b sin A sin B o sin A = a sin B = 42sin 32 ≈ 0.445 b 50 A ≈ 26o
AB = 400(cos 42i + sin 42 j) ≈ 297.3i + 267.7 j [4.3] AD = 55[cos(−25°)i + sin(−25°) j] ≈ 49.8i − 23.2 j AC = AB + AD AC = 297.3i + 267.7 j + 49.8i − 23.2 j AC ≈ 347.1i + 244.5 j AC = 347.12 + (244.5)2 AC ≈ 425 mph
( 347.1 )
α = 90o − θ = 90o − tan −1 244.5 ≈ 55o 15.
[4.1]
z =1− i
[5.3]
r = 12 + (−1)2 r= 2
α = tan −1 θ = 315o
(1 − i)8 = [ 2(cos315° + i sin 315°)]8 = 16[cos(8 ⋅ 315°) + i sin(8 ⋅ 315°)] = 16(cos 2520° + i sin 2520°) = 16(cos 0° + i sin 0°) = 16 + 0i = 16 i = 1( cos90o + i sin 90o )
[4.3]
≈ 53.1° = 180° − α ≈ 180° − 53.1° ≈ 126.9°
z = 2(cos315° + i sin 315°)
16.
4 = tan −1 4 −3 3
[5.3]
wk = 11/ 2 ⎛⎜ cos 90° + 360°k + i sin 90° + 360°k ⎞⎟ 2 2 ⎝ ⎠
k = 0, 1
w0 = cos 90° + i sin 90° 2 2 w0 = cos 45° + i sin 45°
w1 = cos 90° + 360° + i sin 90° + 360° 2 2 w1 = cos225°+ i sin 225°
w0 = 2 + 2 i 2 2
w1 = − 2 − 2 i 2 2
Copyright © Houghton Mifflin Company. All rights reserved.
−1 = 45o 1
Additional College Trigonometry Solutions
930
17.
y x = tan −1 1 1 o
θ = tan −1
r = x2 + y 2
[6.5]
18.
r = 2 − 2cosθ
[6.5]
()
= 12 + 12 = 2
= 45
The polar coordinates of the point are ( 2, 45o ).
19.
5 x = 10 x
log 5 = log10 x log 5 = 1 x = 1 ≈ 1.43 log 5
[7.4]
20.
N (t ) = N 0ekt N (138) = N 0e138k 0.5 N 0 = N 0e138k 0.5 = e138k ln 0.5 = ln e138k ln 0.5 = 138k ln e ln 0.5 = 138k ln 0.5 =k 138 −0.005023 ≈ k N ( t ) = N 0(0.5)t /138 ≈ N 0e−0.005023t N (100 ) = 3 ( 0.5 )
100 /138
≈ 1.8 mg
Copyright © Houghton Mifflin Company. All rights reserved.
[7.5]
