STUDENT SOLUTIONS MANUAL KEVIN BODDEN
RANDY GALLAHER
LEWIS AND CLARK COMMUNITY COLLEGE
��e6ra &
8 �ri8onometry
MICHAEL SULLIVAN PEARSON
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Prentice Hall
Upper Saddle River, NJ
07458
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Prentice Hall
© 2008 Pearson Education, Inc. Pearson Prentice Hall Pearson Education, Inc.
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Printed in the United States of America
10
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ISBN 13: ISBN 10:
7
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978-0-13-232124-2 0-13-232124-6
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Table of Contents Preface Chapter R
Review
Real Numbers ............................................................................................................................... 1 R.l ... . . . .. . . .. .. . . .. . .. . . 3 R.2 Algebra Essentials . .... R.3 Geometry Essentials ..................................................................................................................... 6 R.4 Polynomials .................................................................................................................................. 8 .. . .. . .. . . . . . . . II R.5 Factoring Polynomials . R.6 Synthetic Division ...................................................................................................................... 14 R.7 Rational expressions .....................: ............................................................................................. 15 R.8 nth Roots; Rational Exponents ................................................................................................... 20 Chapter Review ..................................................................................................................................... 24 Chapter Test .......................................................................................................................................... 28 . .
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Chapter 1 Equations and Inequalities Linear Equations ......................................................................................................................... 30 1.1 Quadratic Equations.................................................................................................................... 39 1.2 Complex Numbers; Quadratic Equations in the Complex Number System ............................... 48 1.3 Radical Equations; Equations Quadratic in Form; Factorable Equations ................................... 50 1.4 Solving Inequalities .................................................................................................................... 59 1.5 Equations and Inequalities Involving Absolute Value ................................................................ 65 1.6 Problem Solving: Interest, Mixture, Uniform Motion, and Constant Rate Job Applications ..... 69 1.7 Chapter Review ..................................................................................................................................... 73 Chapter Test .......................................................................................................................................... 80 Chapter 2
Graphs
The Distance and Midpoint Formulas......................................................................................... 82 2.1 Graphs of Equations in Two Variables; Intercepts; Symmetry .................................................. 87 2.2 Lines 93 2.3 2.4 Circles ....................................................................................................................................... 101 Variation ................................................................................................................................... 106 2.5 Chapter Review ................................................................................................................................... 108 Chapter Test ........................................................................................................................................ 114 Cumulative Review ............................................................................................................................. 116 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 3 Functions and Their Graphs 3.1 Functions................................................................................................................................... 118 The Graph of a Function ........................................................................................................... 125 3.2 Properties of Functions ............................................................................................................. 129 3.3 Library of Functions; Piecewise-defined Functions ................................................................. 136 3.4 3.5 Graphing Techniques: Transformations ................................................................................... 142 Mathematical Models: Building Functions ............................................................................... 150 3.6 Chapter Review................................................................................................................................... 153 Chapter Test ........................................................................................................................................ 159 Cumulative Review ............................................................................................................................. 163
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Chapter 4
Linear and Quadratic Functions
Linear Functions and Their Properties . . . 165 4.1 Building Linear Functions from Data ....................................................................................... 170 4.2 Linear and Quadratic Functions . . . . . . 172 4.3 4.4 Properties of Quadratic Functions ............................................................................................ 182 Inequalities Involving Quadratic Functions . . 185 4.5 Chapter Review . .. . . . . . 195 Chapter Test 201 Cumulative Review ................................................................................................ :............................ 203 ...... ............................. . .. ......................................... ....
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Chapter 5
Polynomial and Rational Functions
Polynomial Functions and Models . . :.................................................................. 205 5.1 Properties of Rational Functions . . . . . . . . .. 215 5.2 The Graph of a Rational Function . . . . . .. . 219 5.3 Polynomial and Rational Inequalities . . . . . . 242 5.4 The Real Zeros of a Polynomial Function . . . . . 249 5.5 Complex Zeros; Fundamental Theorem of Algebra . . . . 270 5.6 Chapter Review . . . . : ................................................................... 273 Chapter Test . . . .. . . . . . . 289 Cumulative Review . . . . . . . . 293 .... ............. ......
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Exponential and Logarithmic Functions
Composite Functions ................................................................................................................ 296 One-to-One Functions; Inverse Functions . . . .. . . . . 303 Exponential Functions . . . . . . . . 313 Logarithmic Functions .............................................................................................................. 321 Properties of Logarithms .......................................................................................................... 329 Logarithmic and Exponential Equations . ... . . 333 Compound Interest . . . . . . . . . 341 Exponential Growth and Decay Models; Newton's Law; Logistic Growth and Decay Models . . . . . . 345 Building Exponential, Logarithmic, and Logistic Models from Data....................................... 348 6.9 Chapter Review . . . . .. 351 Chapter Test . . . . 361 Cumulative Review . . . . . . .. . . . 364 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8
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Chapter 7
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Trigonometric Functions
Angles and Their Measure . . . . . . . 367 7.1 Right Triangle Trigonometry . . . . . 371 7.2 Computing the Values of Trigonometric Functions of Acute Angles . . . 378 7.3 Trigonometric Functions of General Angles ............................................................................ 384 7.4 Unit Circle Approach: Properties of the Trigonometric Functions . . . 391 7.5 7.6 Graphs of the Sine and Cosine Functions ... . . .. . . 394 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions.. . .. . 404 7.7 Phase Shift; Sinusoidal Curve Fitting . . . . . 407 7.8 Chapter Review . . . . 413 Chapter Test . . . . . . . . . . . 421 Cumulative Review . . . . .. . . . .. . . . . . . . 425 .. ........ ......... ...........................................
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© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8
Analytic Trigonometry
The Inverse Sine, Cosine, and Tangent Functions . 8.1 The Inverse Trigonometric Functions (continued) 8.2 Trigonometric Identities 8.3 Sum and Difference Formulas 8.4 Double-Angle and Half-Angle Formulas 8.5 Product-to-Sum and Sum-to-Product Formulas 8.6 Trigonometric Equations I 8.7 Trigonometric Equations 11. 8.8 Chapter Review Chapter Test Cumulative Review
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. 428 433 439 445 455 466 469 476 481 494 . 499
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Chapter 9 Applications of Trigonometric Functions Applications Involving Right Triangles . 9.1 The Law of Sines . . . . . . . 9.2 The Law of Cosines . . .. . . 9.3 9.4 Area of a Triangle .. . .. . Simple Harmonic Motion; Damped Motion; Combining Waves 9.5 Chapter Review . . . . . . .. Chapter Test .. .. . . . Cumulative Review . . .
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Chapter 10 Polar Coordinates; Vectors 10.1 Polar Coordinates . 10.2 Polar Equations and Graphs 10.3 The Complex Plane; DeMoivre's Theorem 10.4 Vectors 10.5 The Dot Product. Chapter Review . . Chapter Test . Cumulative Review
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Chapter 1 1
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535 538 550 555 558 562 569 572
Analytic Geometry
. 11.2 The Parabola 11.3 The Ellipse . . . . .. . . 11.4 The Hyperbola 11.5 Rotation of Axes; General Form of a Conic . . . . . . . 11.6 Polar Equations of Conics .. . . . 11.7 Plane Curves and Parametric Equations Chapter Review . .. . . .. Chapter Test .. . . . . .. . . . Cumulative Review . .
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© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12
12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8
Systems of Equations and Inequalities ... . ...... . .
. . Systems of Linear Equations: Determinants . . . . . . Matrix Algebra. . . . . . Partial Fraction Decomposition . . .. . .. . . Systems of Linear Equations: Matrices
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Systems of Nonlinear Equations Systems of Inequalities Linear Programming
Chapter Review Chapter Test
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Arithmetic Sequences
Mathematical Induction The Binomial Theorem
Chapter Review
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Chapter 14 Counting and Probability .. . . 14.1 Sets and Counting . . 14.2 Permutations and Combinations 14.3 Probability . . . . . .. . . . . Chapter Review . . . . . . . .. . . . Chapter Test .. . . . . Cumulative Review . .. . .
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Graphing Utilities
Section I The Viewing Rectangle
. .. . . 2 Using a Graphing Utility to Graph Equations .. . Section 3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry Section 5 Square Screens Section
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Sequences; Induction; the Binomial Theorem
Sequences
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Chapter 13
Chapter Test
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Cumulative Review
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Systems of Linear Equations: Substitution and Elimination
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© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Preface This solution manual accompanies Algebra & Trigonometry, 8e by Michael Sullivan.
The Instructor Solutions Manual (ISM) contains detailed solutions to all exercises in the text and the chapter projects (both in the text and those posted on the internet). The Student Solutions Manual (SSM) contains detailed solutions to all odd exercises in the text and all solutions to chapter tests. In both manuals, some TI-84 Plus graphing calculator screenshots have been included to demonstrate how technology can be used to solve problems and check solutions. A concerted effort has been made to make this manual as user-friendly and error free as possible. Please feel free to send us any suggestions or corrections. We would like to extend our thanks to Dawn Murrin, Christine Whitlock and Bob Walters from Prentice Hall for all their help with manuscript pages and logistics. Thanks for everything! We would also like to thank our wives (Angie and Karen) and our children (Annie, Ben, Ethan,Logan, Payton, and Shawn) for their patient support and for enduring many late evemngs.
Kevin Bodden and Randy Gallaher Department of Mathematics Lewis and Clark Community College 5800 Godfrey Road Godfrey,IL 62035
[email protected]
[email protected]
Chapter R Review
Section R.t
23. a.
I. rational
b.
3. Distributive
c.
5. True
II. 13.
15. 1 7.
1 9.
{O,I}
{0,1'21 '3I '4I }
d. None
7. False; 6 is the Greatest Common Factor of 1 2 and 1 8 . The Least Common Multiple i s the smallest value that both numbers will divide evenly. The LCM for 1 2 and 1 8 is 36. 9.
{I}
e.
2 5 . a.
AuB = {I, 3, 4,5, 9} u{ 2, 4,6,7,8} = {I, 2,3,4, 5, 6, 7,8, 9} AnB = { l, 3, 4, 5 ;9} n{2, 4,6, 7,8} = {4}
None
b.
None
c.
None
d.
(AuB) nC = ({I, 3, 4,5, 9} u{2,4,6, 7,8})n{I,3,4,6} = {l,2,3,4,5,6, 7,8,9} n{1,3,4,6} = {I, 3,4,6}
{0,1'21 '3I '4I }
e.
{�,Jr,�+1,Jr + �} {�,Jr,�+1,Jr+�}
27. a.
1 8 .953
b.
1 8.952
28 .653
b.
28.653
A = { 0, 2, 6, 7, 8}
29. a. 3I. a.
0.063
b.
0.062
AnB = {I, 3, 4, 5, 9} Il {2, 4, 6, 7, 8} = {4} = {O, 1, 2, 3, 5, 6, 7, 8, 9}
33. a.
9.999
b.
9.998
35. a.
0.429
b.
0.428
37. a.
34.733
b.
34.733
Au B = { 0, 2, 6, 7, 8} u { 0, 1, 3, 5, 9}
39.
3+2 = 5
{2,5}
4I.
x+2 = 3·4
b. {-6,2,5}
43.
3y = 1+ 2
45.
x-2 = 6
47.
�=6 2
49.
9 - 4+2 = 5+2 = 7
5I.
-6+ 4 . 3 = -6+ 1 2 = 6
53.
4+5 - 8 = 9 - 8 = 1
= {O, 1, 2, 3, 5, 6, 7, 8, 9}
2I. a.
c.
{-6,� ,- 1 .333 . .. = - 1 .3,2,5 }
d. {Jr} . e.
{-6,� ,-1 .333 ... = -1 .3,Jr,2,5 } 1
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Chapter R: Review
55
. 4+ .!.3 = 12+13 = .!.i3 6-[3 .5+2.(3-2)J = 6-[15+2 . (1)J = 6-17 = -11 2·(3-5)+8 . 2-1 = 2 . (-2)+16-1 = -4+16-1 = 12-1 = 11 10-[6-2·2 +(8-3)}2 = 10-[6-4+5]· 2 = 10-[2+5] .2 = 10-[7] . 2 = 10-14 = -4 (5-3) � = ( 2) � = 1
81.
57.
83.
59.
85.
61.
87.
89. �1 .
63.
93.
67. 6 9.
73.
75. 77. 79.
3 10 3·2·5 ·2· 2 5'21 = 5·3·7 = tt . t't7 = 7 6 10 2·3·5·2 2· · ·2 4 25 ' 27 = 5·5·3·9 = t t. 5 . tt . 9 = 45
95. 97.
99.
5 9 25+54 =-79 -+-= 6 5 -30 30 5 1 10+3 13 18 + 12 = � = 36 1 7 3-35 32 16 ---=-30 18 90 = --90 = --45 3 2 9-8 ---=--= 20 15 60 60
101.
103.
1 05.
107.
5 C�8 ) = 185 '1127 = 5 . 9 . 3 = 5 . , . 3 = 15 9·2·11 ,. 2 . 11 22 ) U 1 3 7 3 7 -3+7 =-= 10 1 -·_+-=-+-= 2 3 3 2 3 3 6 3 6 . -+2 3 2·-+-=4 8 1 . -+-=-+-=4 8 4 8 42 8 12 3 12+3 = -15 =-+-=-8 8 8 8 6 ( x + 4) = 6x + 24 x(X-4) = X2 -4x 2 (�X4 .!.2 ) = 2 . �x4 2 . .!.2 = 2·2·23x 3.2 t ·3x 2 =-x-1 3 =---t·2 2 2 (x+2)(x+4) = x2 +4x+2x+8 = x2 +6x+8 (x -2)( x + 1 ) = x2 + X -2x -2 =x2 -x-2 ( x -8)( x -2) = x2 -2x -8x + 16 =x2 -10x+16 2x+3x = 2·x+3·x = (2+3)·x = ( 5)· x = 5x 2 ( 3·4) = 2 ( 12) = 24 (2 . 3) . ( 2 . 4) = ( 6) (8) = 48 Subtraction is not commutative; for example: 2 -3 = -1 * 1 = 3 -2 . Division is not commutative; for example: 2 *"23 ' "3 The Symmetric Property implies that if 2 x, then x 2 . 5
10
-
10
10
10
_
10
_
=
=
2
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Section R.2: Algebra Essentials
1 09. There are no real numbers that are both rational
29. Graph on the number line:
1
and irrational, since an irrational number, by definition, is a number that cannot be expressed as the ratio of two integers; that is, not a rational number
1
1 1 3 . Answers will vary.
Section R.2 1 . variable
X
1 . 2345678 103
7 . True
1 -I
-2.5
13.
1 0 -> 2
1 5.
-1> - 2
1 7.
7t> 3.14
19.
1 = 0. 5 '2
21.
3' < 0.67
23.
x>O
25.
x <2
27.
x :::; 1
•
-I
3
�.4 0
-
5 2
1
1--
I
..
x>-l
0
I.
1
�
1
•
37.
d(A,E) = d( -3 , 3 ) = 1 3 - ( -3 ) 1 = 161 =6
39.
x+2y = - 2 +2· 3 = -2+6 = 4
41.
5xy +2 = 5 ( -2)( 3 ) + 2 = -30+2 = - 28
43.
2x 2( - 2 ) = -4 = 4 = -x - y --2- 3 -5 5
45.
3x+ 2y = 3( - 2 ) + 2( 3 ) = -6+ 6 = .2. = 0 2+ 3 5 5 2+y
47.
Ix + y l = 1 3 + ( - 2 ) 1 = 1 1 1 = 1
49.
Ix l+IYI = 1 3 1+1 - 2 1 = 3 +2 = 5
.
53.
55.
2
-
( 1
1
d(D,E) = d( 1 , 3 ) = 1 3 -ti = 1 2 1 =2
51
1
1
1
35.
9. False; a number in scientific notation is
expressed as the product of a number, x , 1 :::; x < 10 or - 1 0 < x :::; - 1 , and a power of 10.
1
x �-2
d(C ,D) = d( 0 , 1 ) = II - 0 1 = 1 1 1 = 1
3. strict
0.25
0
1
33.
1 1 1 . Answers will vary.
11.
-2
1
3 1 . Graph on the number line:
Every real number is either a rational number or an irrational number, since the decimal form of a real number either involves an infinitely repeating pattern of digits or an infinite, non repeating string of digits.
5.
liE 1
L:l J�_Li =1 3 3 x 14x - 5 y I = 14( 3 ) - 5( - 2) I = 1 1 2 + 1 01 = 1 221 = 22 11 4x l - 1 5y II = 114( 3 ) 1 - 1 5( - 2) II = 11 121 - 1 - 1 0 11 = 1 12- 1 0 1 = 121 =2 x
Part (c) must be excluded. The value x = 0 must be excluded from the domain because it causes division by O.
3
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Chapter R: Review
59.
x x2 -9 (x - 3)(x+3) x = -3 x = 3 x2 x2 + X
Part (a) must be excluded. The values and must be excluded from the domain because they cause division by o. 61.
87.
89.
1
None of the given values are excluded. The domain is all real numbers. 63.
x2 +5x-10 x2 +5x-10 x3 -x x(x-l)(x+l) x = 0, x = 1, x = -1
from the domain because they cause division by
67.
x x+4 x = -4
Y
-I
1
-
93 .
must be exluded b ecause it makes the denominator equal O. Domain
-I
Y
Y
O.
4 xx-5= 5
-I
-I
Y
Parts (b), (c), and (d) must be excluded. The values and must be excluded
65.
(x2 y )2 = (x2 )2 .(y )2 =x4 y-2 = 7X4 x2 y34 x2-1 3-4 = x1 y = -X --= xy -8x4iz2 9x/z -8 4-1 2 -3 z2 -1 =-x 9 -8 3 z = -x 9 8x3 z 9y [�)-2 = ( 3y ) 2 = ( 4X )2 42 x2 = 1 6x2 4x 3y 32 i 9i 4y-1
= { xl x 5} "#
95.
--
--
2(2) 2xy-1 = 2x = = -4 y
(-1)
must b e excluded sine it makes the denominator equal O. Domain =
69.
71. 73.
{
xl x -4} "#
= �9 (F - 32) = �9 (32 - 32) = �9 (O) = = �9 (F - 32) = �9 (77 - 32) = �9 (45) = 25°e (_ 4) 2 = (-4)(-4) = 16 C
101.
ooe
C
77.
3-6. 34 = 3
79.
(T2
-6
+4 =
1 05.
1 07.
J;z =
Ixl = 1 21 = 2
x -_ 2-1 -_ -21 = 2, 2x3 - 3x2 + 5x -4 = 2 . 23 -3 . 22 + 5 2 - 4 = 16-12+10-4 = 10 Ifx = 1, 2x3 -3x2 +5x-4 2 13 -3 . 12 +5·1-4 = 2 - 3 +5 - 4 =0 y
If x
.
T2
= 1= .!.. 32
r' = 3(-2)(-1) = 32 = 9
9
=
83.
�( _4)2 = 1 -41 = 4
.
4
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Section R.2: Algebra Essentials
1 09.
1 1 1.
1 13 . 1 1 5. 1 1 7. 1 1 9. 121. 1 23. 1 25. 1 27. 1 29. 131 . 133. 135. 137.
139.
141.
1 43.
(666) 4 = ( 666 = 34 = 81 (222)4 222 J (8. 2)6"" 304,006. 671 (6.1)-3 "" 0.004 (-2. 8)6"" 481. 890 (-8 . 11r4 "" 0.000 454.2 = 4. 542 x 102 0. 0 13 = 1. 3 x 10-2 32,155 = 3. 2 155x104 0.000423 = 4. 23x 10-4 6.15 x 104 = 61,500 l.2 14x10-3 = 0. 001214 1.1 x 108 = 11 0, 000, 000 8 . 1x10-2 = 0. 0 81 A =lw C=Trd A = ,J34 x2 4 r3 V =-Tr 3 V = x3 If x = 1000, C = 4000+2x = 4000 + 2 (1000) = 4000+2000 = $6000 The cost of producing 1000 watches is $6000.
If x = 2000, C = 4000+2x = 4000 + 2(2000) = 4000+4000 = $8000 The cost of producing 2000 watches is $8000. We want the difference between x and 4 to be at least 6 units. Since we don't care whether the value for x is larger or smaller than 4, we take b.
147.
the absolute value of the difference. We want the inequality to be non-strict since we are dealing with an 'at least' situation. Thus, we have
IX -41 6 Ix-110 1 = II08-110 1 = 1 -2 1 =2�5 108 volts is acceptable. I x -11 0 I = 1 104 - 1 1 0 I= 1 -6 1 = 6 5 104 volts is acceptable. Ix -3 1 = 1 2. 999 -3 1 = 1 -0 .001 1 = 0. 001 0.0 1 radius of2. 9 99 centimeters is acceptable. Ix-3 1 = 1 2 . 89-3 1 = 1 -0. 1 1 1 = 0. 11{'0. 0 1 radius of2.89 centimeters is acceptable. �
1 49. a.
b.
>
not
1 5 1 . a.
�
A
b.
not
A
1 53 .
1 55.
1 45. a.
1 57.
The distance from Earth to the Moon is about 4 x 108 = 400,000,000 meters. The wavelength of visible light is about 5xl0-7 = 0 .0000005 meters. The smallest commercial copper wire has a diameter of about 0 . 0005 5 x 10-4 inches. =
5
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Chapter R: Review
1 59.
186,000·60·60·24·365 =(1.86 X 105)( 6 x 101 )2 ( 2.4 x 101)( 3.65 X 102)
1 5.
a= 7, b= 24, c2 = a2 + b2 =72 + 242 =49 + 576 =625 :::> c= 25
1 7.
52 =32 + 42 25 =9 + 1 6 25 = 25
= 586.5696 xl 010 = 5.865696 x 1 012 There are about 5.9 x 1 012 miles in one light
year.
161.
1 63 .
1 65.
1
'3= 0.333333
>
... 0.333 � is larger by approximately 0.0003333 ...
The given triangle is a right triangle. The hypotenuse is 5.
No. For any positive number a, the value � is smaller and therefore closer to O. Answers will vary.
19.
The given triangle is not a right triangle.
Section R.3 1. 3. 5.
7.
9. 11.
13.
21.
right; hypotenuse C
=
True. False; the volume of a sphere of radius is given by V=34 7fr3
23.
r
True. Two corresponding angles are equal.
25.
a=5, b=1 2, c2 = a2 + b2 = 52 + 122 =25 + 144 = 1 69 :::> c= 1 3
27.
29.
a=1 0, b=24, c2 = a2 + b2 = 1 02 + 242
31.
62 =32 + 42 36=9 + 1 6 36= 25 false
The given triangle is not a right triangle. A=I·w =4·2=8 in2
}-
}-
A= b .h= (2)(4)=4in2 A=nr2 =n (5)2 = 25n m2 C= 2nr= 2n (5)= I On m V =1 w h=8 . 4 . 7 = 224 ft3 S = 2/w + 21h + 2wh = 2 (8 )( 4 ) + 2 (8 )( 7 )+ 2 ( 4 )( 7 )
= 64+112+56
= 100+576 :::>
252 = 7 2 + 242 625=49 + 576 625= 625
The given triangle is a right triangle. The hypotenuse is 25.
27fr
= 676
6 2 = 4 2 + 52 36 = 1 6 + 25 36= 4 1 false
c= 26
= 232 ft2 33.
4 3 =-n·4 4 3 =-ncm 256 3 V=-nr 3 3 3 S = 4nr2 = 4n·42 = 64n cm2
6
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Section R.3: Geometry Essentials
35.
37.
39.
= 1t r2 h = 1t(9i (8) = 6481t in3 = 27ir2 + 27irh = 27i ( 9)2 + 27i ( 9) ( 8) = 1627i + 1447i = 3067i in2 The diameter of the circle is 2, so its radius is 1 . A = 1t r2 = 1t(1) 2 = 1t square units
45.
V S
Total Distance
47.
49.
The diameter of the circle is the length of the diagonal of the square .
43.
4C
=
647t '" 20 1 . 1 inches ", 1 6 . 8 feet
=
4( 7t d)
=
47t · 1 6
Area of the border area ofEFGH - area of ABCD = 102 _62 = 100-36 = 64 ft 2 Area of the window area of the rectangle + area of the semicircle. =
=
= (6)(4)+ "21 . 1t . 2 2 = 24+21t '" 30.28 ft2 Perimeter of the window 2 heights + width one-half the circumference. = 2(6)+4+ 21 "1t(4) = 12+4+21t = 16 + 21t '" 22 .2 8 feet =
+
P
51.
We can form similar triangles using the Great Pyramid's height/shadow and Thales' height/shadow: h
Since the triangles are similar, the lengths of corresponding sides are proportional. Therefore, we get
126
240
1 14
2�
This allows us to write
8 x 4 2 8·2 -4 = x 4=x In addition, corresponding angles must have the same angle measure. Therefore, we have A = 90° = 60° , and = 30° . , B
=
A
d 2 = 2 2 +22 = 4+4 =8 d = .J8 = 2.fi r = !!...2 = 2.fi2 = .fi The area of the circle is: A = 1t r2 = 1t ( .fi t = 21t square units
41.
The total distance traveled is 4 times the circumference of the wheel.
3
h =-2 -240 3 h = 2·240 3 = 160
53.
C
Since the triangles are similar, the lengths of corresponding sides are proportional. Therefore, we get
The height of the Great Pyramid is 160 paces. Convert 20 feet to miles, and solve the Pythagorean Theorem to find the distance: 1 mile� = 0. 003788 nules. 20 feet = 20 feet · 5280 eet
d 2 = (3960 + 0 . 003788) 2 -39602 = 30 '" 5 .477 miles
30 x 20 45 30·45 = x -20 135 = x or x = 67. 5 2
d
sq. miles
In addition, corresponding angles must have the same angle measure. Therefore, we have A = 60° , = 95° , and C = 25° . B
7
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Chapter R: Review
55.
Convert 100 feet to miles, and solve the Pythagorean Theorem to find the distance: . 1 milefeet = 0. 0 18939 rrules 100 feet = 100 feet · 5280
d 2 = (3960 + 0. 0 18939)2 -39602 150 miles Convert 150 feet to miles, and solve the Pythagorean Theorem to find the distance: I mile 150 feet = 150 feet · 5280 feet = 0. 028409 miles d 2 = (3960 + 0.028409) 2 -39602 225 d 15 . 0 miles Let 1 length of the rectangle and w width of the rectangle. Notice that (l + W) 2 - (I - w) 2 = [(l + w) + (/-w)][(l + w) -(/-w)] = (2/)(2w) = 4/w = 4A So A = ± [(l+ w) 2 _(l-W) 2 ] Since (I -W) 2 0, the largest area will occur when 1 - w 0 or 1 w; that is, when the rectangle is a square. But 1000 = 21 + 2 w = 2(/ + w) 500 = 1 +w= 21 250 = 1 = w The largest possible area is 2502 = 62500 sq ft. A circular pool with circumference 1000 feet yields the equation: 2 7'Cr = 1 000 r = 500 7'C The area enclosed by the circular pool is: A = 7'C r2 = 7'C ( 5�0 J = 5�2 79577.47 ft2 d
""
�
12.2
sq. miles
�
1 1.
1 3.
=
=
1 7. 19. 21.
�
=
x = 8x- 1 k Not a monomial; when written in the form ax , the variable has a negative exponent. -2xi Monomial; Variable: x,y ; Coefficient: Degree: 3 8x = 8xy-l Not a monomial; when written in the form axn , the exponent on the variable y is negative. !
-2 ;
y
y
sq. miles
�
57.
9.
=
23.
25.
=
�
2 7.
�
29.
Thus, a circular pool will enclose the most area.
Not a monomial; the expression contains more than one term. This expression is a binomial. 3x2 -5 Polynomial; Degree: 2 5 Polynomial; Degree: 0
3x2 x5
--
Not a polynomial; the variable in the denominator results in an exponent that is not a nonnegative integer. 2/ - J2 Polynomial; Degree: 3
x2 +5 Not a polynomial; the polynomial in x3 -1 the denominator has a degree greater than O. (x2 +4x+5)+ (3x-3) = x2 +(4x+3x)+(5-3) = x2 +7x+2 (x3 _2X2 +5x+10)-(2x2 -4x+3) = x3 _2X2 +5x+ 1O-2x2 +4x-3 = x 3 + ( 2X 2 _
Section R.4
_
2X 2 ) + (5X + 4X) + (1 0 3) -
= x 3 - 4x2 + 9x + 7
1.
4; 3
3 1.
5.
True
33.
7.
m
3 Monomial; Variable: x ; 2xCoefficient: 2; Degree: 3
( 6x5 +X3 +X ) + ( 5X4 _X3 +3x2 ) = 6x5 +5x4 +3X2 +x (x2 -3x+1)+2(3x2 +x-4) = x2 -3x + 1 + 6x2 + 2x -8 = 7x2 -x-7
8
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Section R.4: Polynomials
35.
37.
39.
41. 43. 45.
47.
49.
51.
53.
55.
57.
59.
61.
6(x3 +x2 -3)-4(2x3 -3x2) = 6x3 +6x2 -18-8x3 +12x2 = -2x3 + 18x2 -18 ( X2 -X + 2 ) + ( 2X2 -3x + 5 ) - ( x2 + 1 ) = x2 -x+2+2� -3x+5-X2 _1 = 2X2 -4x+6 9 (/ -3y+4 ) -6 ( 1-/ ) =9/ -27y+36-6+6/ =15y2 -27y+30 x(x2 +x-4) = x3 +X2 -4x _2x2 (4x3 + 5) = -8x5 -I Ox2 (x+l)(x2 +2x-4) = x(x2 +2x-4)+I(x2 +2x-4) =x3 +2X2 _4X+X2 +2x-4 = x3 + 3x2 -2x -4 (x+ 2)(x+4) = x2 +4x+ 2x+ 8 = x2 +6x+8 (2x+5)(x+2) = 2X2 +4x+5x+l0 = 2X2 +9x+1O (x-4)(x+2) = x2 + 2x -4x-8 = x2 -2x-8 (x -3)(x -2) = x2 -2x -3x + 6 = x2 -5x+6 (2x+3)(x-2) = 2X2 -4x+3x-6 = 2X2 -x-6 (-2x + 3)(x -4) = _2X2 + 8x + 3x -12 = -2X2 + llx -12 (-x-2)(-2x-4) 2X2 +4x+4x+8 = 2x2 +8x+8 (x-2y)(x+ y) = x2 + xy -2xy -2/ = X2 - xy - 2Y2
63.
(- 2x - 3y)(3x + 2y) -6x2 - 4xy - 9xy - 6/ =
=
65. 67.
73. 75. 77. 79.
83.
85.
87.
89.
-6x2 - 1 3xy - 6/
(x-7)(x+7) = x2 _72 = x2 -49 (2x+3)(2x-3) = (2X)2 _32 = 4x2 -9 (3x+4)(3x-4) = (3X)2 _42 = 9x2 -16 (2x-3)2 = (2X)2 _2(2X)(3)+32 =4x2 -12x+9 (x+y)(x_ Y) = (X)2 _(y)2 =x2 _/ (3x+y)(3x-y) =(3x)2 _(y)2 =9x2 -/ (X_2y)2 = x2 +2(x . (-2Y ))+(2y)2 = x2 -4xy+4y2 (X-2)3 = x3 -3.x2 ·2+3·x . 22 _23 = x3 -6x2 + 12x -8 (2x + 1)3 = (2X)3 + 3(2x)2 (1) + 3(2x)· e + e = 8x3 +12x2 +6x+l 4x 2 - l lx + 23 x + 2 4x 3 - 3x 2 + X + 1
)
4x 3
+
8x 2
- l lx 2
+
X
- l lx 2 - 22x 23x + 1 23x + 46 - 45
Check:
(x + 2)(4x2 -llx+23)+(-45) = 4x3 -llx2 + 23x + 8x2 -22x + 46 -45 = 4x3 -3x2 +x+l The quotient is 4x 2 - l lx 23 ; the remainder is --45.
=
+
9
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Chapter R: Review
91.
4x - 3 X 2 4x 3 - 3x 2 + x +
)
97.
)
2X 2 + x + 1 2x4 - 3x 3 + Ox 2 + x + 1
4x 3
2X4 + x 3 + x 2 - 4x 3 - x 2 + X
-4x3 - 2X2 - 2 x x 2 + 3x + 1
x+ 1
Check:
(x 2 )(4x - 3) + (x + 1) = 4x 3 - 3x 2 + x + 1 The quotient is 4x - 3 ; the remainder is x + 1 . 93.
Check:
( 2X2 + X + 1 ) ( x2 - 2x + !) + � x + ! = 2X4 - 4x 3 + x 2 + X 3 _ 2X 2 + ! x
)
x 2 + 2 5x4 + Ox 3 - 3x 2 + X + 1 5x4
1 1 x2 + - x + 2 2 5 1 -x + 2 2
+ 1 0x 2
+ x 2 _ 2x + .12 + 2. 2 x + .1
2
- 1 3x 2 + x + 1 - 26 x + 27
= 2X4 - 3x 3 + X + 1 The quotient is x 2 - 2x + t ; the remainder is 5
Check:
1
"2 x + "2 .
( X2 + 2 )( 5x2 - 1 3 ) + ( x + 27 ) = 5x4 + 1 0x 2 - 1 3x 2 - 26 + x + 27
99.
= 5x4 - 3x 2 + X + 1 The quotient is 5x 2 - 1 3 ; the remainder is x + 27 .
- 4x 2 - 3x - 3 x - 1 - 4x 3 + x 2 + Ox - 4
)
-4x 3 + 4x 2 - 3x 2 -3x 2 + 3x - 3x - 4 -3x + 3 -7
Check:
Check:
( 2x3 - 1 )( 2x 2 ) + ( _x 2 + X + 1 )
(x - 1)(-4x 2 - 3x - 3) + (- 7)
= - 4x 3 - 3x 2 - 3x + 4x 2 + 3x + 3 - 7
= 4x5 _ 2X 2 _ x 2 + x + 1 = 4x5 - 3x 2 + x + 1 The quotient is 2x2 ; the remainder is _x 2 + x + 1 .
= - 4x 3 + x 2 - 4
The quotient is - 4x 2 - 3x - 3 ; the remainder is
-7.
10
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Section R.S: Factoring Polynomials
101.
1 05.
)
x 2 + x + l x4 + 0x 3 - x 2 + 0x + l x4 + x 3 + x 2
PI
_ x2 + x + 1 _x 2 - x - l 2x + 2
Check:
1 07.
(x 2 + x + l)(x 2 - x - l) + 2x + 2
= x4 + x 3 + X 2 _ x 3 _ x 2 - x - x 2 - x - 1 + 2x + 2 = x4 _ x 2 + 1
The quotient is
1 09.
x2 - x - 1 ;
2x + 2 .
1 03 .
When we multiply polynomials PI ( x ) and P2 ( x ) , each term of PI ( x ) will be multiplied by each term of P2 ( x ) . So when the highest powered term of ( x ) multiplies by the highest powered term of P2 ( x ) , the exponents on the variables in those terms will add according to the basic rules of exponents. Therefore, the highest powered term of the product polynomial will have degree equal to the sum of the degrees of PI ( x ) and P2 ( x ) . When we add two polynomials PI ( x ) and P2 ( x ) , where the degree of PI ( x ) = the degree of P2 ( x ) , the new polynomial will have degree :; the degree of PI ( x ) and P2 ( x ) . Answers will vary.
the remainder is
Section R.S
)
x - a x3 + ox2 + Ox - a 3 x3 _ ax2 ax2 ax2 - a 2 x a2 x _ a 3 a2 x - a 3 o
1. 3.
3x ( x - 2 ) ( x + 2 )
True; x2 + 4 is prime over the set of real numbers.
5.
3x + 6 = 3(x + 2)
7.
ax2 + a = a(x2 + 1)
Check:
(x - a)(x2 + ax + a2 ) + O = x3 + ax2 + a2 x - ax2 _ a2 x a3 = x3 _ a 3 The quotient is x2 + ax + a2 ; the remainder is O.
_
11.
2X2 - 2x = 2x(x - l)
1 3.
3x2 y - 6xy2 + 1 2xy = 3xy(x - 2y + 4)
1 5.
x2 - 1 = x2 _ 12 = (x - 1)(x + 1)
1 7.
4x2 - 1 = (2X)2 _ 1 2 = (2x - 1)(2x + 1)
19.
x2 - 1 6 = x2 _ 42 = (x - 4)(x + 4)
21.
25x2 - 4 = (5x - 2)(5x + 2)
23.
x2 + 2x + l = (x + l)2
11
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Chapter R: Review
67. 3x2 + l Ox - S = (3x - 2)(x + 4)
25. x2 + 4x + 4 = (x + 2)2 27.
x2 - l Ox + 25 = (x _ 5)2
69. x2 - 36 = (x - 6)(x + 6)
29.
4x2 + 4x + 1 = (2x + 1)2
71. 2 - Sx2 = 2(1 - 4x2 ) = 2 1 - 2x 1 + 2x
31.
1 6x2 + Sx + l = (4x + l)2
73.
33.
x3 - 27 = x3 - 33 = (x - 3)(x2 + 3x + 9)
75. x2 - 1 0x + 2 1 = x - 7 x - 3
35. x3 + 27 = x3 + 3 3 = (x + 3)(x2 - 3x + 9)
77.
37.
Sx3 + 27 = (2X) 3 + 33 = (2x + 3)(4x2 - 6x + 9)
79.
39.
x2 + 5x + 6 = (x + 2)(x + 3)
(
)( )
x2 + l 1x + l0 = (x + l)(x + l 0)
( )( ) 4x2 - Sx + 32 = 4 ( x2 - 2x + S )
is prime over the reals because there are no factors of 1 6 whose sum is 4. x2 + 4x + 1 6
81. 1 5 + 2x - x2 = - (x2 - 2x - 1 5) = -(x - 5)(x + 3) 83. 3x2 - 1 2x - 36 = 3(x2 - 4x - 1 2) = 3(x - 6)(x + 2)
4 1 . x2 + 7x + 6 = (x + 6)(x + l)
45. x2 - 1 0x + 1 6 = (x - 2)(x - S)
85. / + l l i + 30 / = / ( / + l ly + 30) = y2 (y + 5)(y + 6)
47. x2 - 7x - S = (x + l)(x - S)
87. 4x2 + 1 2x + 9 = (2x + 3)2
43.
49.
x2 + 7x + l 0 = (x + 2)(x + 5)
89.
x2 + 7x - S = (x + S)(x - l)
51. 2X2 + 4x + 3x + 6 = 2x(x + 2) + 3(x + 2) = (x + 2)(2x + 3)
91.
( ) = 2 ( 3x + 1 ) ( x + 1 ) x 4 - S 1 = ( X2 r - 92 = (x2 - 9)(x2 + 9)
6x2 + Sx + 2 = 2 3x2 + 4x + l
= (x - 3)(x + 3)(X2 + 9)
53. 2X2 - 4x + x - 2 = 2x(x - 2) + I(x - 2) = (x - 2)(2x + 1)
93.
x6 - 2x3 + 1 = (x3 _ 1) 2
[
= (x - l)(x2 + x + l)
55. 6x2 + 9x + 4x + 6 = 3x(2x + 3) + 2(2x + 3) = (2x + 3)(3x + 2)
Y
= (x _ l)2 (x2 + x + l)2
57. 3x2 + 4x + l = (3x + l)(x + l)
95.
x7 _ x5 = X5 (x2 - 1) = x \ x - l)(x + 1)
2z2 + 5z + 3 = (2z + 3)(z + l)
97.
1 6x2 + 24x + 9 = 4x + 3 2
59.
61. 3x2 + 2x - S = ( 3x - 4)(x + 2)
99.
63. 3x2 - 2x - S = (3x + 4)(x - 2) 101.
65. 3x2 + 14x + S = (3x + 2)(x + 4)
(
)
5 + 1 6x - 1 6x2 = -(1 6x2 - 1 6x - 5) = -(4x - 5)(4x + l) 4y2 - 1 6y + 1 5 = (2y - 5)(2y - 3)
12
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Section R.5: Factoring Polynomials
103. 1-8x2 -9x4 = -(9x4 +8x2 -1) = -(9x2 -1)(x2 + 1) = -(3x -1)(3x + 1)(x2 + 1) 105. x(x +3) -6(x + 3) = (x + 3)(x -6) 107. (X+2)2 -5(x+2) = (x+2)[(x+2)-5] = (x+2)(x-3) 109. (3x-2)3 _27 = (3x -2)3 _ 33 = [(3x -2) -3 J[(3x -2)2 + 3(3x -2) + 9] = (3x-5)(9x2 -12x+4+9x-6+9) = (3x -5)(9x2 -3x + 7) 111. 3(X2 +lOx+25 )-4(x+5) =3(x+5)2 -4(x+5) =(x+5)[3(x+5)-4J =(x+5)(3x+15-4) = (x + 5)(3x + 11) 113. x3 +2x2 -x-2 = x2(x+2)-I(x+2) = (x+2)(x2 -1) = (x + 2)(x -1)(x + 1) 115. X4 _x3 +x-l =x\x-l) +I(x-l) = (x -1)(x3 + 1) = (x-l)(x+l)(x2 -x+l) 117. 2(3x+4)2 +(2x+3) . 2(3x+4) . 3 = 2(3x+ 4)((3x + 4) + (2x + 3)·3) = 2(3x+ 4)(3x+4 + 6x+9) = 2(3x+4)(9x+13)
119. 2x(2x+5)+x2 . 2 = 2x((2x+5)+x) = 2x(2x+5 +x) = 2x(3x+5) 121. 2(x+3)(x-2)3 +(X+3)2 . 3(x-2)2 = (x+3)(x-2)2 (2(x-2)+(x+3) . 3) = (x+3)(x -2)2 (2x -4 +3x+9) = (x+3)(x -2)2 (5x +5) = 5(x+3)(x-2)2 (x+ 1) 123. (4x-3)2 +x . 2(4x-3) . 4 = (4x-3)(( 4x-3)+8x) = (4x-3)( 4x-3+8x) = (4x-3)(12x-3) =3(4x-3)(4x-l) 125. 2(3x-5) . 3 (2x+l)3 +(3x-5)2 . 3(2x+l/ . 2 = 6(3x-5)(2x+ 1)2 ((2x+ 1)+(3x-5)) = 6(3x -5)(2x + 1)2 (2x + 1 + 3x -5) = 6(3x-5)(2x+l)2 (5x-4) 127. FactorsSum:of4: 1,5 4 2,4 2 -1,-5-4 -2, -2 None of the sums of the factors is 0, so x2 + 4 is prime. Alternatively, the possibilities are ( x ± 1)( x ± 4) = x2 ± 5x + 4 or (x±2 )(x±2) = x2 ±4x+4 , none of which equals x2 + 4 . 129. Answers will vary. -4
13
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Chapter R: Review
Section R.6 1. 3. 5.
17.
quotient; divisor; remainder True 2)1 -1 2 4
2)4 -3 -8 4 8 10 4 4 5 2 8 Remainder 8 of O. Therefore, x -2 is not a factor of 4x3 -3x2 -8x + 4 . 2)3 -6 0 -5 10 6 0 0 -10 3 0 0 -5 0 Remainder O. Therefore, x -2 is a factor of 3x4 -6x3 -5x+10 . -3)3 0 0 82 0 0 27 - 9 27 -81 -3 9 -27 3 - 9 27 1 -3 9 o Remainder O. Therefore, x + 3 is a factor of 3x6 + 82x3 + 27 . -4)4 0 -64 0 1 0 -15 -16 64 0 0 -4 16 4 -16 0 0 - 4 Remainder 1 O . Therefore, x + 3 is not a factor of 4 x -64x4 +X2 -15 . -i )2 -1 0 2 -1 0 2 002 0 Remainder 0; therefore x -1 is a factor of 2X4 _x3 +2x-1 . -2)1 -2 3 5 -2 8 -22 -4 11 -17 3----17 x -2X2 + 3x + 5 -- = x2 - 4x+ 11 + -x+2 + b+c + = 1- 4 + 11-17 = -9 =
2 2 8 4 12 Quotient: x2 +x+4 Remainder: 12 3)3 2 -1 3 9 33 9 6 3 11 32 99 Quotient: 3x2 +l1x+32 Remainder: 99 0 9. -3)1 0 -4 0 -3 9 -15 45 -138 1 -3 5 -15 46 -138 Quotient: X4 -3x3 + 5x2 -15x + 46 Remainder: -138 1)4 0 -3 0 0 5 2 2 4 4 4 4 2 2 7 Quotient: 4x5 +4X4 +X3 +X2 +2x+2 Remainder: 7 -1.1)0.1 0 0. 2 0 -0. 11 0.121 -0. 3 531 0. 1 -0. 11 0. 3 21 -0. 3 531 Quotient: 0 . lx2 -0.l1x+0. 3 21 Remainder: -0 . 3 531 1)1 0 0 0 0 -1 1 o Quotient: X4 + x3 + x2 + X + 1 Remainder: 0
1 9.
=
7.
21.
=
23.
=
6
11.
2 5.
of
o
=
1 3.
27.
15.
a
d
x+2
14
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Section R. 7: Rational Expressions
Section R.7
1.
1 9.
lowest tenns
3 -4x 2X ( X2 -2 ) 2xx-2 x-2 3x+9 3(x + 3) 3 x2 -9 (x-3)(x+3) x-3 x2 -2x x(x-2) = -x 3x-6 3(x -2) 3 24x2 24x2 4x 12x2 -6x 6x(2x -1) 2x-I y2 -25 (y+5)(y-5) 2y2 -8y-IO 2 ( / -4y-5 ) (y+5 )(y-5) 2(y-5)(y + I) y+5 2(y + 1) x2 +4x-5 (x+5)(x-I) x+5 x2 -2x+I (x-I)(x-I) x-I x2 +5x-I4 _ (x+7)(x-2) 2-x 2-x x + 7)(x-2) - ( -I(x-2) = -(x+7) = -x-7 3x+6 x 3(x+2) x 5x2 x2 -4 5x2 (x-2)(x+2) 3 5x(x -2)
3. True;
5.
7.
9.
1 1.
13.
1 5.
21.
_
-
23.
25.
_
17.
-- . _-
27.
4x2 x3 -64 x2 -16 2x 4x2 (X_4) ( X2 +4x+I6) -4 (x )(x + 4) 2x 2x . 2x(x-4) ( x2 +4x+I6 ) 2x(x -4)(x + 4) 2x ( x2 +4x+I6) x+4 4x-8 12 4(x-2) 12 -3x 12-6x -3x 6(2-x) 4(x-2) 2 -3x (-I)(x-2) 8 3x x2 -3x-1O x2 +4x-21 x2 + 2x -35 x2 + 9x + 14 (x-5)(x+2) ' (x+7)(x-3) (x+7)(x-5) (x+7)(x+2) x-3 x+ 7 6x x23x-9-4 = � . 2x+4 3x-9 2x+ 4 x2 -4 6x 2(x+2) (x-2)(x+2) 3(x-3) 4x (x-2)(x-3) 8x x2�-1 = 8x ' x+I x+ 1 x2 -1 lOx 8x x+I (x -1)( x + 1) lOx 4 5(x -I) -- ' -
-- ' --
15
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Chapter R: Review
29.
31.
33.
35.
37.
39.
41 .
43.
4-x 4-x x2 -16 4+x 4x 4+x 4x x2 -16 4-x ( x+4) ( x-4) 4+x 4x ( 4-x ) ( x-4 ) 4x ( x _4 )2 4x x2 + 7x+12 -x -12 x2 _ 7 x + 12 x2 + 7x + 12 -x2 --x2 + x -12 x2 -7 x + 12 x2 + x -12 x2 -x-12 (x + 3)(x + 4) . (x -4)(x + 3) (x-3)(x-4) (x+4)(x-3) (x +3)2 (X_3) 2 2X2 -x-28 3x2 -x-2 2x2 -x - 28 3x2 + l lx + 6 4x2 +16x+7 3x2 -x-2 4x2 +16x+7 3x2 +llx+6 (2x+7)(x-4) (3x + 2)(x + 3) (3x + 2)(x -1) (2x + 7)(2x + 1) (x -4)(x + 3) (x -1)(2x + 1) -x2 + -25 = x+52 ( x -2 ) x2 -4 ( x + 2 )-'-:x2 4 = -2x-3 --2x-3 2x-3 = -'----:-'2x-3 -"x + 1 2x -3 x + 1 + 2x -3 3x -2 x-3 + x-3 = x-3 x-3 3x+5 2x-4 (3x+5)-(2x-4) 2x -1 2x -1 2x -1 3x+5-2x+4 2x-l x+9 2x-l 4 + x = 4 - x = 4-x x-2 2-x x-2 x-2 x-2
45.
-_ . _-
47.
49.
51.
--
--
--
--
-
-
53.
----
-
--
--
--
--
--
--
or
--
--
4 2 = 4(x+2) 2(x-l)x-l) --x-I x+2 (x-l)(x+2) (x+2)( 4x+8-2x+2 (x + 2)(x -1) 2x+l0 (x+2)(x-l) 2(x+5) (x+2)(x-l) x + -2x-3 = x(x-l) + -'--(2x-3)(x+l) -'--'--'x+l x-I (x+l)(x-l) (x-l)(x+l) x2 -x+2x2 -x-3 (x-l)(x+l) 3X2 -2x-3 (x-l)(x+l) x-3 x+4 (x-3)(x-2) _ (x+4)(x+2) x+2 ---= x-2 (x+2)(x-2) (x-2)(x+2) x2 -5x+6-(x2 +6x+8) (x + 2)(x -2) x2 -5x+ 6-x2 -6x-8 (x+2)(x-2) -(llx + 2) -llx-2 (x+2)(x-2) (x + 2)(x -2) x -1 = x2 +X2 -4) --+ x2 -4 x x ( x2 -4 2X2 -4 X ( X2 -4) ) 2 ( x2 -2 x ( x-2 ) ( x+2 ) x2 -4 = ( x+2 ) ( x-2 ) x2 -x -2 = ( x + 1) (x -2 ) LCM ( x 2 ) ( x - 2) ( x + 1 ) . x3 -x = X ( X2 -1 ) = x ( x+l ) ( x-l ) x2 -x = x ( x -1 ) LCM x ( x+l ) ( x-l ) .
55.
Therefore,
=
Therefore,
=
+
-
16
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Section R. 7: Rational Expressions
57.
4x3 -4x2 +x = x ( 4x2 -4x+I ) = x ( 2x -1)( 2x -1 ) 2x3 -x2 = x2 (2x -1) x3 LCM = x3 ( 2x _1) 2 . x3 -x = x ( x2 -I ) = x (x+I)(x-1) x3 -2X2 +x = x ( x2 -2x+1 ) = x(x-I)2 x3 -1 = (x -1) ( X2 + X + 1 ) LCM x ( x 1)( x_1)2 ( x2 + X + 1 ) . x x 2 x2 - 7x + 6 x -2x -24 x x (x -6)(x -1) (x - 6)(x + 4) x(x+4) x(x-1) (x -6)(x -I)(x + 4) (x -6)(x + 4)(x -1) x2 +4x-x2 +x 5x (x -6)(x + 4)(x -1) (x - 6)(x + 4)(x -1) 4x 2 x2 -4 x2 +x- 6 4x 2 (x-2)(x+2) (x+3)(x-2) 2(x+2) ) = (x -24x)(x(x+3 + 2)(x + 3) (x + 3)(x -2)(x + 2) 4x2 +I2x-2x-4 (x -2)(x + 2)(x + 3) 4x2 +lOx-4 (x -2)(x + 2)(x + 3) 2(2x2 + 5x -2) (x -2)(x + 2)(x + 3) 3 + 2 (x-I)2 (x+I) (x-1)(x+1)2 3(x+I)+2 (x-1) (x_I)2 (x+I)2 3x+3+2x-2 (x_I)2 (x+1)2 5x+I
67.
4)(x
Therefore, 59.
Therefore,
61.
63.
65.
=
x+4 2x+3 2x -x-2 x2 +2x-8 x+4 2x+3 (x -2)(x + 1) (x + 4)(x -2) (2x 3)(x + (x + + (x -2)(x + I)(x + 4) (x + 4)(x -2)(x + 1) 2 +5x+3) _- x2 +8x+I6-(2x -2)(x + l)(x + 4) (x _x2 +3x+13 (x -2)(x + l)(x + 4) 1 2 3 -x x 2 +x + x3 _x2 1 2 3 =-X X ( X + 1) + x2 ( x-I) -2x ( x-I) 3 ( x + 1 ) _- x ( x + 1) ( x-I) x2 ( X + 1)( x-I) x ( x2 -1 ) - 2X2 + 2x + 3x + 3 x2 (x + 1)( x-I) x3 -X-2X2 +5x+3 x2 (x + 1)( x -1 ) x3 -2X2 +4x+3 x2 (x + 1) (x -1)
69.
+
--
+
4)
1)
-
---
+
73.
hx(x+h) -1 x(x+h) (� + �) ( X; I ) x+1 x x+I (� _ � ) = ( X � I ) = -x . x-I = -;=}
17
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Chapter R: Review
75.
77.
79.
x+Ix 2x-x-I -2xx --x 3(x+I) +-x-I 3x+3+x-I x+I x+I x+I x-I x-I . x+I = x 2 (2x + I) _ = 4xx_ + 2 x+I (x-I)(x+I ) 2x(2x + 1) x+4 x-3 x-2x+Ix+I X+4)(X+I) (X-3)(X-2) ) ( (x-2)(x+I) (x+I)(x-2) x+I -SX+6) ) ( X2 +SX+4-(x2 (x -2)(x + 1) x+I lOx-2 1 2(Sx - 1) (x-2)(x+I) x+I (x-2)(x+I)2 x-2 x-I --+-x+2 x+I 2x-3 x-- -x+I x X-2)(X+I) + (X-I)(X+2) ) ( (x+2)(x+I) (x+I)(x+2) ( (x +X2I)(x) (2X-3)(X+I) x(x + 1) J -2 + x2 + X -2 ( 2x2 -4 ( x2 -X(x+2)(x+I) (x+2)(x+I) J J ( X2 _ ��:2+�; _3) ) ( -:�::I; 3 ) 2(x2 -2) . x(x+I) (x+2)(x+I) _(x2 -x-3) 2X(X2 -2) -(x + 2)(x2 -x-3) -2X(X2 -2) (x 2)(x2 -x-3)
2- x+Ix x-I 3+x+I
81 .
1 1- --1 1- --= 1- 1.x x-I = I- � x-I x-I-x x-I -1 x-I 2 ( x-I ) +3 -_ _x-I2_ +3 - _x-I2_ + 3(x-I) x-I 2(x-I) 3 3(x-I rl +2 � x-I +2 _x-I_ + x-I 2+3(x-I) x-I 3+2(x-I) x-I 2+3(x-I) ---;x-I--:x-I 3+2(x-I) 2+3(x-I) 2+3x-3 3+2(x-I) 3+2x-2 3x-I = -2x+I ( 2x + 3) . 3 -( 3x - S ) . 2 --6x + 9 -6x + 10(3X-S)2 (3x-S)2 19 x
-1
83.
85.
87.
_
_
-
--;:--
x . 2x-(x2 +1 ) . 1 2x2 -x2 -I (X2 + 1t (X2 + 1 f x2 -1 (X2 +It (x-I)(x+I) (X2 +I t
+
18
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Section R. 7: Rational Expressions
89.
(3x + l ) . 2x - x2 · 3 6x 2 + 2x - 3x2 (3x + l) 2 (3x + 1) 2 3x 2 + 2x (3x + 1) 2 = x(3x + 2)2 (3x + l) ( x2 + 1 ) . 3 -(3x + 4) 2x 3x 2 + 3 - 6x 2 - 8x ( x2 + 1)2 (X 2 + 1 t -3x2 - 8x + 3 ( X 2 + It - ( 3x2 + 8x -3 ) (X2 + 1t ( 3x -1) ( x + 3 ) ( x2 + 1 )2
95.
.
91.
93.
1 + -1 ) -1 = (n -l) (R, R2 I l. = (n - l) ( RR,2 +· RR, ) I 2 R, · R -I-2 = (n - l) ( R2 + R, ) I I R, · R2 (n - l) (R2 + R, ) R, · R2 1 = (n - 1)(R 2 + R, ) 0.1(0.2) I - (1.5 - 1)(0.2 + 0. 1 ) 0.02 2 meters 0.02 = -= 0.5(0.3) 0.15 = 15
x + l =:> a = 1, b = 1, c = 0 1 + -x1 = -x 1 1 = 1 + -x 1 + --� = 1 + --+l x X;I ( ) 1+ x + l + x = -2x + l --x+l x+l =:> a = 2,b = l,c = 1 x+l 1 + _1 = 1 + ( 2X1+ l = 1 +-) 2x + l 1+ _ l + x l + lx 2x + l + x + l = -3x + 2 2x + l 2x + l =:> a = 3,b = 2,c = 1 1+ = 1 + ( 3X 1+ 2 ) = 1 + 3x2x ++21 1 1+ 1 2x + l 1 + __1 1 + -x 3x + 2 + 2x + l 5x +3 3x + 2 3x +2 =:> a = 5,b = 3,c = 2 If we continue this process, the values of a, b and c produce the following sequences: 1, 2,3,5, 8,l3, 21, .. b : 1,1, 2,3,5,8, l3, 21, .. c : 0, 1, 1, 2,3, 5, 8, l3, 21, .... . In each case we have a Fibonacci Sequence, where the next value in the list is obtained from the sum of the previous 2 values in the list. Answers will vary. a :
. .
. . .
__
97.
19
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Chapter R: Review
Section R.8
1.
33.
9; -9
3. index
root m=W=3 r-s = �(_2)3 = -2
5. cube 7. 9.
35.
37.
39.
41.
�3x2 ..l12x = �36x2 . X = 6xvlx
21.
( .j5 �f = ( .j5 f ( � t
23.
= 5 · W = 5 Wi. = 5 · 3 � = 15 �
43.
45.
+
27.
3J2 4J2 = (3 + 4)J2 = 7J2
29.
-.J18 + 2../8 = -J9:2 + 2#2
47.
= -3J2 + 4J2 = (-3 + 4)J2 = J2
( vIx - 1 f = (vlxf - 2v1x + 1 = x - 2v1x + 1 �16x4 - mx = �8x3 · 2x -mx = 2x mx -mx = (2x - 1)mx �8X3 -3..l50x = �4x 2 . 2x -3..l25 · 2x = 2x£ -15£ = ( 2x -15)£
� - 3x�2xy + 5 �-2xl = �8X3 . 2xy _ 3x �2xy + 5 �r-_--y '---:3 . -2-- x-y = 2x�2xy -3x �2xy - 5 y �2xy = ( 2x - 3x - 5y )�2xy = ( -x - 5 y )�2xy or - (x + 5 y )�2xy 1 1 J2 = J2 J2 = J2 ' J2 7:
.j5 = --J15-J3 -J3 = ' .j5 .j5 .j5 5 J3 = J3 5 + J2 5 -J2 5 - J2 ' 5 + J2 J3 ( 5+J2)
25 - 2 J3(5 +J2) 5J3 + ./6 or 23 23
( J3 + 3)( J3 - 1) = ( J3 t + 3J3 - J3 - 3 = 3 + 2J3 - 3 = 2J3
31.
5ifi -2� = 5ifi - 2 · 3ifi = 5 ifi - 6 ifi = (5 - 6) ifi = -ifi
20
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exist. No portion of thi s material may be reproduced, in any form or by any means , without permission in writing from the publisher.
/
Section R. B: n th Roots; Rational Exponents
49.
2 -.[5 2 - .[5 2 -3.[5 2 + 3v5 2 + 3.[5 . 2 - 3.[5 4 - 2.[5 - 6.[5 + 15 4 - 45 - 8.[5 8-15 41 -41 if,t 5 if,t 5 5 if,t lfi lfi . 2r;
19
51.
53.
69.
19
=
( -VSY
(-27t 3 -V-27 -3
=
=
73.
=
=
=
=
= _
=
=
=
=
2
3
=
Y
=
---'-
-'----,:.
-'---
_
27 27 27 .fi 8 · 2.fi 16J2 16.fi · .fi 27.fi 32 33 2 3 (.fi t 27 . .fi 16.fi .fi
xy
=
=
22 4
9-3/ 2 _311_2 _1_ �3 271 9 ( J9 t 3 =
=
=
75.
61.
1 13
( X2 y (yt3 (xt3 (i t 3 X2 l 3 y2 / 3 X2 / 3 / / 3 X2 / 3 y 4 / 3 X2 / 3 y 2 / 3 X2 / 3+ 2 / 3-2 / 3 y I l 3+4 / 3-2 / 3 X2 l 3 yl x2 / 3 ( 16x2 y-1 /3 )3/ 4 163/ 4 (X2 t 4 ( y-1 /3 )3/4 X l / 4 ( y2 )1 / 4 (� t x3/ 2 y-1 / 4 x1 l4/ 1 2 2 3 X 3/ 2 -1 / 4 y -1 / 4 -1 / 2 8x5 / 4 y -3/ 4 8x5/ 4 /1 4 1 + X )1 / 2 x + 2 (I-+ xt 2 ( x -,-,- + 2 ( 1 + x ) 1 / 2 -(1 + X) 1 / 2 (1 + x) 1 I 2 - x(1+ +2 (Ix)+1I 2x) x + 2 + 2x (1 + X) 1 / 2 3x + 2 (I + X) 1I 2 =
( x + h) - 2)x ( x + h ) + x ( x + h) - x x + h - 2�X2 + xh + x x+h-x 2x + h - 2 �r-X-:-2-+-x-h h
57.
=
6
=
=
.Jx+h - Fx ..Jx+h - Fx ..Jx+h - Fx + h + vxr Jx + h + Fx · ..Jx+h - Fx vx�
8 2 /3
6
l
=
55.
( x 3 y ) 1 / 3 ( x3 ) 1 / 3 ( y )
27 27 8 · 2J2 16.fi 27J2
=
=
----n-
21
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exist. No portion of this material may be reproduced, i n any form or by any means, without permi ssion in wri ting from the publi sher.
Chapter R: Review
77.
79.
2x ( x2 + 1 )1/2 + x2 2"1 ( x 2 + 1 )-1/2 . 2x = 2x ( X2 + 1 )112 + ( 2 x3 1/2 X + 1) 2x ( x2 + 1 )112 . ( x2 + 1 )1/2 + x3 ( X 2 + 1 )1/2 2x ( x2 + l )1/2+1/2 + x3 2x ( x 2 + 1 Y + x3 ( X 2 + 1 )1/2 ( X 2 + 1 )112 2x 3 + 2x + x3 3x3 + 2x ( X2 + 1 )1/2 (X 2 + 1 )1/2 X ( 3 X2 + 2 ) ( X2 + 1 )112
83 .
.
--
�4x + 3 ·
,x > 5 � + �x - 5 . 5v4x � +3 2vx - 5 �x - 5 = �4x + 3 + ---.= 2�x - 5 5 �4x + 3 J4;+3 · 5 · J4;+3 + � · 2 · � 10�x - 5 �4x + 3 3 5 - 1 ( 4x x+ -)5+ 24x( x+- 5 ) O�( ) ( 3 ) 20x + 15 + 2x -10 10�( x - 5 )( 4x + 3 ) 22x + 5 10�( x - 5 )( 4x + 3 )
=
85.
_
81.
( �l + x - x · 2�:+ X ) (Ji+; - 2k l+x) l+x l+x -x ) ( 2Ji+;..,fl+; 2 �I + x l+x 2 ( 1 + x) - x 2( I + x) I /2 l + x 2+x 2 (1 + x) 3/ 2
( x+ 4)112 - 2x ( x + 4 rI l 2
[ [ [
x+4 ( X + 4 Y' 2 - X 2xY' 2 ( +4 x+4 2 ( X + 4 Y' 2 . ( x + 4 Y' 2 X 2x 2 ( x + 4 Y' ( + 4 Y' x+4 X + 4 - 2X ( x + 4 Y' 2 x+4 -x + 4 (X + 4( 2 x + 4 -x + 4 ( X + 4 ) 3/ 2 4-x ( x + 4 ) 3/ 2
J
J
J
2 - ( X2 -1 )112 x ( x2 -1 )1/2 ,x -1 x2 X 2 - ( x2 - 1 )112 . ( x 2 - 1 )112 ( x2 - 1 )112 2 x X2 - ( x2 -1 )1/2 . ( x 2 - 1 )1/2 ( x2 -1 )1/2 X2 x2 - ( x2 -1 ) ( x2 - 1 )1/2 X2 x( 2 _x2)1/2+ 1 2 x2 - 1 x
[
<
]
or
x>1
x 2 ( X2 -lt 2
22
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Section R. B: nth Roots; Rational Exponents
l + x2 - 2x"x -2Fx - -'x > 0 '----' ( 1 + X2 -)2 1 + X 2 - ( 2Fx) ( 2XFx) 2 Fx ( l + x2 t 1 + x2 - ( 2Fx) ( 2xFx) 2Fx r
87.
[
97.
) 99.
3X- 1 / 2 + % XII 2 ,X > 0 = l3/ 2 + 23 x 1 / 2 X ' = 3 · 2 +2X3xII //22 X 1 / 2 = 62X+l3x1 2 = 3(x2Xl+/ 22 ) J2 "'" 1 . 41
�(2)
1 . 4 1 42 1 3 5 6 2
_
101.
if4 "'" 1.59
' � � 4 ) 1 . 58740 1 0 52
89.
(x + l )3 / 2 + x .l2 (x + l) 1 / 2 1 03 .
2 + 1 "", 4 . 89 3 -"S
)
91.
93.
6Xl / 2 ( x2 + x) -8x3 / 2 - 8x l / 2 = 2X I / 2 ( 3(X2 + x) - 4x - 4) = 2X l / 2 ( 3x 2 - X -4 ) = 2X I l 2 (3x -4)(x + l) 3 ( X2 + 4f / 3 + x . 4 ( x 2 + 4t 3 · 2x
2 . 1 45268638
= ( X 2 + 4t 3 [ 3 ( x 2 + 4) + 8x 2 ] = ( X2 + 4t 3 [3x 2 + 12 + 8x 2 ] = ( x 2 + 4t 3 ( l lx2 + 12 )
95.
1 07. a.
b.
4(3x + St3 (2x + 3)3 / 2 + 3 (3x + St 3 ( 2x + 3t 2 = (3x + S) 1 / 3 ( 2x + 3 y / 2 [ 4(2x + 3) + 3(3x + S) ] = ( 3x + St3 (2x + 3t 2 (8x + 12 + 9x + lS ) = (3x + st3 (2x + 3) 1 1 2 (17x + 27) 3 . 2 where
x
�
4 . 8853 1 793 1
1 09.
T
V = 40(12/
��� - 0. 608
"'" IS, 660 . 4 gallons 9 V = 40 ( 1 ) 2 � 6 -0 . 608 "", 390.7 gallons 1
= 21r
m. = 21rJ2 "'" 8.89 seconds
--
23
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Chapter R: Review
111.
8 inches 8112 2/3 feet T = 2ffN) 2ff� 2ff( 41] "Jj6- 0. 9 1 seconds = 2Jj" = =
=
0
13.
0
1 5.
�
1 13 .
1 7.
Answers may vary. One possibility follows: If a = -5 , then .j;;i = �(_5)2 = J25 = 5 a Since we use the principal square root, which is always non-negative, .j;;i {a if a 0 = -a if a 0 which is the definition of l a l so .j;;i = l a l · *
1 9.
3 21.
<
'
23.
Ch apter R Review Exercises
3.
5. 7.
25.
AuB = {1,3,5,7}u{3,5,6,7,8} = {1, 3, 5,6,7,8} An C = {1,3,5, 7} n {2,3, 7,8,9} = {3, 7} A = {2,4,6,8,9} BnC = {3,7,8} B nC = {1,2,4,5,6,9}
9. (a)
(b) (c) ( d)
27.
29. 31.
none
{-lO} {-10, 0.65, 1.343434. . , i}
33.
{F7}
35.
{-10, 0. 65, 1.343434 . . , F7, i} -6+4 . (8 -3) = -6+4(5) = -6+ 20 = 14
(e) 11.
=
(
.
�
1.
4·3·3 3 4 9 = __0_ 3 16 3·4·4 =-4 �(_3)2 1 -31 3 4(x-3) = 4x-12 x>3
37.
=
I
•
d(P,Q) = d(-2,3) =\3- ( -2)\ = 1 3 + 2 1 = 1 51 = 5 4( -5) -20 =-10 4x =--=x = -5,y = 7 ::::> -x+ y -5+7 2 x = -5,y = 7 ::::> 5x- 1 / = 5 ( -5 t ( 7 f = 5·49 -5 = -49 x = -5 ::::> N = �( _5)2 = 1 -51 = 5 3 Domain = {xix 6} x-6 *
Since the triangles are similar, corresponding sides have the same ratio and corresponding angles have the same measure. Thus,
A = 90° , B = 45° , C = 45° 3 x 1 = J2 1·x = 3· J2 x = 3J2 (1.5)4 = 5. 0625 3 +5x-12 The coefficients are 3, 3x4, -2,5 +4X4 -2x 0, 5, and -12, and the degree is 5. (2x4 -8x3 +5x-1)+(6x3 +x2 +4) = 2x4 -2x3 +X2 + 5x+3
24
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Chapter R Review Exercises
39. (2x+ Y )(3x-5y) = 6x2 -1 0xy+3xy-5y 2 = 6x2 -7 xy -5y2 41. (4x+l)(4x-l) = (4X)2 + t2 = 16x2 -1 (x+l)(x+2)(x-3) = ( x2 +2x+x+2 ) (x-3) = ( X2 + 3x + 2 ) (x -3) = x3 -3x2 +3x2 -9x + 2x-6 =x3 -7x-6 3x2 +8x+25 45. x -3)3x3 - x2 + X + 4 3x3 -9x2 8x2 + X 8x2 -24x 25x+4 25x-75 79 Check: (x -3)(3x2 + 8x + 25) + (79) =3� +8x2 +25x-9x2 _24x-75+79 The = 3x3 _x2 +x+4 quotient is 3x 2 + 8x + 25 the remainder is 79 . -3x2 +4 x2 + 1)-3x4 +X2 +2 -3x4 -3x2 4x2 +2 4x2 +4 -2 Check: (x2 +1)(-3x2 +4)+(-2) = -3x4 +4x2 -3x2 +4-2 = -3x4 +x2 +2 The quotient is -3x 2 + 4 the remainder is -2 .
49.
X+l)XS xS +X4 -x4
43.
+1
-x2 _x2 -x x+l x+l
o
Check:
(x+l)(x4 _ x3 +X2 -x+l)+(O) =xs _x4 +x3 _x2 +X+X4 _x3 +x2 -x+l XS +1 The quotient is X4 -x3 + x2 -X + 1 ; the remainder is O. 51. x2 +5x-14 = (x+ 7 ) (x-2 ) 53. 6x2 -5x-6 =(3x+2)(2x-3) 55. 3x2 -15x-42 = (3x+6)(x-7) = 3(x+2)(x-7) 5 . 8x3 + 1 = (2X)3 + 13 = (2x + 1) ( (2x)2 -(2x)(1) + 12 ) = (2x+l) ( 4x2 -2x+l ) 59. 2x3 +3x2 -2x-3 = x2 (2x+3)-(2x+3) = ( 2x + 3) ( x2 -1) (2x + 3) ( x - l ) ( x +1) 61. 25x2 -4 = (5x)2 _22 = (5x+ 2)(5x-2) 63. prime 9x2 + 1over; a sum of perfect squares is always the set of real numbers =
;
47.
7
=
;
25
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Chapter R: Review
65.
67.
69.
71.
73.
x2 + Sx + 16 = (x + 4)( x + 4) = (x + 4)2 2X2 +l1x+14 = (2x+7)(x+2) = 2x+7 x2 -4 (x+2)(x-2) x-2 9x2 -1 3x-9 x2 -9 9x2 + 6x + 1 (3x+l)(3x-l) 3(x-3) (x+3)(x-3) (3x+l) 2 3(3x-l) (x + 3)(3x+ 1) x+l - x-I = (x+l)(x+l)-(x-l)(x-l) x-l x+l (x-l)(x+l) ( x2 + 2x + 1 ) - ( x2 -2x + 1 ) (x-l)(x+l) x2 +2x+l-x2 +2x-l 4x (x-l)(x+l) (x-l)(x+l) 3x+4 - --2x-3 x2 -4 x2 +4x+4 3x+4 2x-3 (x+2)(x-2) (X+2)2 (3x + 4)( x + 2)-(2x -3)( x-2) (X+2)2 (X-2) 3x2 + 6x+4x+S- ( 2X2 -4x-3x+ 6) (X+2)2 (X-2) 3x2 + 10x + S - ( 2x2 -7x + 6) (x+2)2 (x-2) 3x2 +10x+S-2x2 +7x-6 (X+2)2 (X-2) x2 +17x+2 (x+2)2 (X-2) x2 -1 2x �+ 6 = [ 2 �5�� 6 ) ( ::n x x-2 (X+l)(X-1) ) ( X-2 ) = x-I = [ (x-3)(x-2) x+l x-3
77. 79. 81.
85.
m J16:2
= 4Ji �-16 =�-S·2 = -2·lfi 5.J8 -2m = 5#2 -2J16:2 = lOJi -sJi = 212 =
( 25x-4/ 3 y-2 / 3 t 2 /2 / 4 3 1 ) -2 2 3 3 x = (5 y
--
_
75.
91.
93.
4 = 4 . J5 = -4J5J5 J5 J5 5 2 = 2 . 1+12 = 2 ( 1+12) 1-12 1-12 1 + 12 1- (12 t 2 ( 1+12 ) 2 ( 1+12 ) 1-2 -1 -2 ( 1+ = 12 ) 1+J5 = 1+J5 1+J5 1-J5 1-J5 · 1+J5 1+2J5 + ( J5 f 1- (J5t 6+2J5 -3-J5 = - -2 2 _
95.
-4
26
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Chapter R Review Exercises
97.
( 2+x2 )1 1 2 +x . "21 ( 2+x2 )- 1 / 2 ·2x = ( 2+x2 )1 / 2 + 2 { 2+x2X22 y 2 = ( 2+x2 )1 / 2 + x 22 1 / 2 ( 2+x ) ( 2+x2 )1 / 2 ( 2+x2 )11 2 +x2 {2 + x2 t 2 2+X2 +X11 2 2+2x21 ( 2+x2 ) 2 ( 2+x2 ) / 2 2(I+x2 ) (2+X2 t 2 ( x+4 )1 / 2 ·2x-x2 · -2I (x+4 )-11 2 x+4 x2 1 / (x+4)1 / 2 ·2x- 2(x+4) 2 x+4 1 2(x+4t 2 (X+4) / 2 . 2x-x2 2(x+4t 2 x+4 2(x+4)·2x-x2 4x2 +16x-x2 2 -x+4--'--2 - = 2(x+4t ----'2(x+4t x+4 ( 2(x+4t 3X2 +16X ) ( 1 ) = 3x2 +16x/ = 2 x+4 2(x+4)3 2 x(3x+16) 2(x+4)3 / 2
101.
I
99.
x2 -( x2 - 1 )
7 = [ X2 _x2 +1 ] (-12 )
1 03.
105. 1 07.
1 09.
�x2 -I x = [ � ] (:, ) = x' .JxCi 3(x2 +4 f / 3 +x . 4(x2 +4 t 3 ·2x = (x2 +4t 3 [3(x2 +4 ) +8x2 ] = { X2 + 4 )1 / 3 [3x2 + 12 + 8x2 ] ={X2 +4t 3 (1Ix2 +12 ) 281,421,906 = 2. 8 1421906xl08 2 = 144 + 256 = 400 = 202 , therefore we 12have2 +a16right triangle by the converse of the
Pythagorean Theorem. Total annual earnings per share (1st quarter earnings) (2nd quarter earnings) + ( 3 rd quarter earnings ) + (4th quarter earnings ) 1 . 2 - O. 75 -0. 3 0 + 0.20 $0. 3 5 per share. +
=
=
111.
=
Pond Area
=
area of outer circle - area of i n n er circle
= ;r(5)2 _;r(3)2 = 25;r - 9;r = 16;r 50.2 7 square feet Outer Perimeter =2;r ( outer radius ) = 27r(5) = 10;r 31.42 feet �
�
1 1 3.
Answers will vary.
27
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Chapter R: Review
1 . (a)
( b)
Recall that the Natural Numbers are also called the Counting Numbers . These are the numbers 1, 2, 3, 4, ... For the given set, the only natural number is 7. { 7 } Recall that the set of Integers is the set of positive and negative whole numbers and 0. { ...,-3,-2,-1,0,1,2,3, ... } For the given set, the only integers are ° and 7. Notice that this includes our result from part (a) because the set of natural numbers is a subset of the set of integers. {
(c)
b.
c. d.
0,7 }
3.
Recall that a rational number can be written as a ratio of two integers. In decimal form, this set contains terminating decimals and repeating decimals. For the given set, the only rational numbers are 0, 1. 2 , 7, and ..!.2 . Notice that this includes our result from part (b) because the set of integers is a subset of the set of rational numbers. 1 . 2, 7, �
{o,
(d)
Since the two triangles are similar, their corresponding angle measures are the same. Therefore, we have = 40° , = 95° , and C = 45° For similar triangles, the ratio of the lengths of corresponding sides is the same. -21 = -x3 or x = -23 unIts. A
4. a.
}
Recall that an irrational number is a real number that cannot be written as the ratio of two integers. In decimal form, this set contains decimals that do not terminate and do not repeat. For the given set, the only irrational numbers are J2 and { J2 ,1C }
b.
Recall that the set of real numbers is the union of the set of irrational numbers and the set of rational numbers. In parts (c) and (d) , we listed each entry from our set as either rational or irrational, therefore all the numbers in the given set are real numbers. 0, 1 . 2, J2, 7, ,1C
{
B
Remember to distribute the minus sign across the second polynomial. ( -2x3 + 4 x2 -6x + 10) - ( 6x3 -7 x2 + 8x -1 )
= -2x3 +4x2 -6x+ 1O-6x3 + 7x2 -8x+ 1 = -2x3 -6x3 +4x2 +7X2 -6x-8x+10+l = -8x3 + 11x2 -14x + 11
Since we are multiplying two binomials, we multiply by first distributing each term from the first binomial. (
1C .
(e)
3X- 1 y2 3 ( -3t ( 4)2 = 3 . _-31 ·16 = -1·16 = -16 1 2x-3y l = 1 2 ( -3 ) -3 ( 4 )1 = 1 -6-12 1 = 1 -1 81 = 18 �X2 + y2 = �( -3 ) 2 + ( 4 l = ·Jg +16 = .[i5 =5 5x3 -3x2 +2x-6 = 5 (-3 )3 -3 ( -3 l +2 ( -3) -6 = 5 ( -27 ) -3 {9) +2 ( -3) -6 = -135-27 -6-6 = -174
2 . a.
C hapter R Test
5. a.
�}
2x -3){ -5x + 2 ) = 2x (-5x + 2 ) - 3 ( -5x + 2) = 2x·-5x+2x . 2-3· ( -5x) -3·2 = -lOx2 +4x+ 15x-6 =-10x2 +l9x-6 x2 -6x+8 Since the leading coefficient is we are looking for two factors of 8 whose sum is -6. Since the product is positive and the sum is negative, the two factors must be negative. The factors we are looking for are -2 and -6. Thus, we can factor the expression as x 2 -6x 8 = ( x -2 ) ( x -4 ) . 1,
+
28
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Chapter R Test
b.
4x2 -25
d.
We notice that this is really the difference of two squares . That is, we can rewrite this as
(2x)2 -(5)2 . a2 _ b2 = (a-b)(a+b) . -25 = (2x -5)(2x + 5) 6x2 -19x-7 Ax2 + Bx + = 6 , B = -19 = -7 = (6)( -7) = -19 .
To factor this expression we
C.
This gives us
AC
8. To rationalize the denominator, we need to multiply the numerator and denominator by
5 - J3 3 · 5-J3 3·5-3·J3 5+J3 5-J3 52 -( J3 t 15-3J3 25-3 15-3J3 22 3. 3
to eliminate the radical in the denominator.
-42 . Therefore, we are
looking for two factors of -42 whose sum is Since the product is negative, the two factors have opposite signs. Since the sum is also negative, the factor with the larger absolute value must be negative.
1,-42 2,-21 3,-14 6,-7 2 -21 -19 . 6x2 + 2x -21x -7 2x(3x+ 1) -7(3x+ 1) (3x+l)(2x-7)
The factors
and
sum to
x-
6. Since we are dividing the polynomial by where c , we can use synthetic division.
=2 2)1 -3 8 -10 2 -2 12 -1 6 2 x2 -X + 6 2. 2 x3 -3x2 + 8x -10 = x2 -x+ 6 +-x-2 x-2 m =J9:3 = J9 . J3 =3J3
Therefore, the quotient is remainder is That is,
c
-3
1 0 . If
and the
-3
-2
a=0 b=0 or
or both.
1026 3 3 = 20 3 3 3 16·26-10·20 = 416-200 = 216
-6
x . y2( X-4 . yX .y X .y 2 = x-3-(-4) . y = x . y-4 lx 2(-2)
then either
16
b.
-3)
a·b = 0
0
1 1 . Looking at the diagram, we can see a rectangle, the pool, inside another rectangle. To [md the area of the deck, we can subtract the area of the pool from the area of the larger rectangle (pool and deck) . The width of the larger rectangle is + + + feet and the length is + feet. Note that we add twice because the deck is on each side of the pool. The area of the deck is then square feet. The fencing will go along the outside of the larger rectangle so we simply need to determine the perimeter of the larger rectangle to find the required amount of fencing. feet of fencing is needed.
7. a.
-3
x
9. We want to graph the set of values for such that they are greater than On a number line we plot with a parenthesis and then draw an arrow to the right. We use a parenthesis because the inequality is strict. I I I ( I I� I I �
-------
c.
Y -3"·2
Y
C with . We find that
Here we have the form A , and C
Y
. X3"·2
use the special form 4x 2
-2 1 3 ) 3 / 2 =(16)3 / 2 .( X4 / 3t 2 .( y-2 / 3 t 2 = ( J16t 4 3 . 2 3 = 43 x2 y_1 = 64x2 ( 16x4 / 3
=
16 + 16 + 26 + 26 = 32 + 52 = 84
-6 -(-2)
I
29
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Chapter 1 E quations and Inequalities Section 1 . 1
1. 3.
1 7.
Distributive { xl x "* 4 }
3x
5.
identity
7.
False. The solution is �3 .
9.
11.
13.
1 5.
-1 x = -5 3 12 3 ± X = 3 52 X = -45 The solution set is
=
x-4
3x - x = x - 4 - x 2x = -4 2x -4 2 2 x = -2 The solution is {-2} . 2t - 6 = 3 - t 2t - 6 + 6 = 3 -t + 6 2t = 9 -t 2t + t = 9 -t + t 3t 9 3t 9 3 3 t=3 The solution set is {3} . 6-x = 2x + 9 6 -x - 6 = 2x + 9 - 6 -x = 2x + 3 -x - 2x = 2x + 3 - 2x -3x = 3 -3x = -3 -3 -3 x = -1 The solution set is {-I} . 3 + 2n = 4n + 7 3 + 2n -3 = 4n + 7 -3 2n = 4n + 4 2n - 4n = 4n + 4 - 4n -2n = 4 -2n 4 set
7x = 21 21 7x = 7 7 x=3 The solution set is {3} . 3x + 15 = 0 3x +15 -l5 = 0 -15 3x = -15 -15 3x = 3 3 x = -5 The solution set is {-5}. 2x -3 = 0 2x - 3 + 3 = 0 + 3 2x = 3 2x 3 2 2 x = -23 The solution set is
3x + 4 = x 3x + 4 - 4 = x - 4
19.
::&:
=
-
21.
{%}.
23.
( ) C)
-2
-
-2
n = -2 The solution set is {-2} .
{%}. 30
Section 1 . 1 : Linear Equations
25.
27.
2(3 + 2x) = 3(x - 4) 6 + 4x = 3x - 1 2 6 + 4x - 6 = 3x - 1 2 - 6 4x = 3x - 1 8 4x - 3x = 3x - 1 8 - 3x x = -1 8 The solution set is {- 1 8} .
31.
2x - 20 = 3x 2x - 20 - 2x = 3x - 2x -20 = x x = -20 The solution set is {-20} .
8x - (3x + 2) = 3x - 1 O 8x - 3x - 2 = 3x - 10 5x - 2 = 3x - I O 5x - 2 + 2 = 3x - 1 0 + 2 5x = 3x - 8 5x - 3x = 3x - 8 - 3x 2x = -8 2x -8 2 2 x = -4 The solution set is {-4} .
6
2
35.
(% } (� � } X+2 = 2
- X
3x + 4 = I - x 3x + 4 - 4 = l - x - 4 3x = -3 - x 3x + x = -3 - x + x 4x = -3 4x -3 4 4 3 X = -4
The solution set is
(� } (� j} P =6
P+
4p = 3p + 2 4p - 3p = 3p + 2 - 3p p=2 The solution set is {2} .
3 I I -x + 2 = - --x 2 2 2
29.
2 1 I -p = -p +3 3 2
33.
37.
0.9t = 0.4 + 0. 1 t 0.9t - O. l t = 0.4 + O. l t - O. l t 0.8 t = 0.4 0.8 t 0.4 -- = 0.8 0.8 t = 0.5 The solution set is {0.5 } . x+l x+2 =2 + 7 3 X I X 2 2t + = 2 1 ( 2)
(;
;
}
7 ( x + l ) + ( 3 ) ( x + 2 ) = 42 7x + 7 + 3x + 6 = 42 l Ox + 1 3 = 42 1 0x + 1 3 - 1 3 = 42 - 1 3 l Ox = 29 l Ox 29 = 10 10 29 x=10
{-%}.
The solution set is
{��}.
31
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ. AJI rights reserved. Thi s material i s protected under aJl copyright laws a s they currently exist. No portion of this material may be reproduced, in any form or by any mean s , without permi ssion in wri ting from the publ isher.
Chapter 1 : Equations and Inequalities
2 4 -+- = 3 y y
39.
y
(� �) +
45.
= y (3)
2 + 4 = 3y 6 = 3y 6 3y -=3 3 2=y Since y 2 does not cause a denominator to equal zero, the solution set is {2} .
=
41.
47.
1 2 3 -+- = 2 x 4
Z(Z2 + 1 ) = 3 + Z3 Z3 + Z = 3 + Z3 Z3 + Z - Z3 = 3 + Z3 Z3 z=3 The solution set is {3 } .
_
2x + 8 = 3x 2x + 8 - 2x = 3x - 2x 8=x Since x 8 does not cause any denominator to equal zero, the solution set is { 8 } .
=
43.
x(2x - 3) = (2x + 1)(x - 4) 2X2 - 3x = 2X2 - 7 x - 4 2X2 - 3x - 2x2 = 2X2 - 7x - 4 - 2x2 -3x = -7x - 4 -3x + 7x = -7x - 4 + 7x 4x = - 4 4x - 4 4 4 x = -1 The solution set is {- 1} .
49.
(x + 7)(x - 1) = (x + 1)2 x2 + 6x - 7 = x2 + 2x + 1 x2 + 6x - 7 - x2 = x2 + 2x + 1 - x2 6x - 7 = 2x + 1 6x - 7 + 7 = 2x + 1 + 7 6x = 2x + 8 6x - 2x = 2x + 8 - 2x 4x = 8 4x 8 4 4 x=2 The solution set is {2} .
2 x -- + 3 = - x-2 x-2 x 2 _ + 3 (X - 2) = _ (X - 2) x-2 x-2 x + 3 (x - 2) = 2 x + 3x - 6 = 2 4x - 6 = 2 4x - 6 + 6 = 2 + 6 4x = 8 4x 8 4 4 x=2 Since x 2 causes a denominator to equal zero,
(
)
( )
=
we must discard it. Therefore the original equation has no solution.
32
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Section 1. 1 : Linear Equations
51.
2x 4 3 x2 -4 = X2 -4 - x + 2 4 2x 3 (x+2)(x-2) (x+2)(x-2) x+2 ( (X + 2) } X + 2} (x - 2) = ( (x+ ��x-2) _3x+2_ J (X+ 2)(x-2) 2)(X _ 2x = 4-3(x-2) 2x = 4 - 3x+6 2x = 10-3x 2x+3x = 10-3x+3x 5x = 1O 5x 10 5x = 25 Since x 2 causes a denominator to equal zero, we must discard it. Therefore the original equation has no solution. =
53.
x 3 x+2 2 2(X+2)C: 2 ) = 2(X+2)(i) 2x = 3(x+ 2) 2x = 3x+6 2x -3x = 3x + 6 -3x -x = 6 -x 6 -1 -1 x = -6 Since x -6 does not cause any denominator to equal zero, the solution set is {-6}.
55.
5 3 2x-3 x+5 5_) ( 2x 3)( x + 5) = (_3_) ( -3) ( + 5) (_2x-3 x+5 5(x+5) = 3(2x-3) 5x+ 25 = 6x-9 5x+ 25 -6x = 6x-9 -6x 25-x = -9 25 -x-25 = -9-25 -x -34 -x -34 -1 -1 x =34 Since x 34 does not cause any denominator to equal zero, the solution is {34}. 2x
_
x
=
=
=
33
© 2008 Pearson Educati on , Inc . , Upper Saddle River, NJ. All rights reserved . This material is protected under all cop
yri ght laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in wri ting from the publisher.
Chapter 1 : Equations and Inequalities
57.
6t + 7 -3t + 8 -= 4t - 1 2t - 4
(�:�}4t - l) (2t - 4} = (�:�!}4t - 1)(2t - 4} ( 6t + 7)( 2t - 4} = (3t + 8)( 4t - 1 } 1 2t2 - 24t + 14t - 28 = 1 2t2 - 3t + 32t - 8
12t2 - l Ot - 28 = 1 2t2 + 29t - 8 12t2 - 1 0t - 28 - 1 2t2 = 1 2t2 + 29t - 8 - 1 2t2 - l Ot - 28 = 29t - 8 - l Ot - 28 - 29t = 29t - 8 - 29t -28 - 39t = -8 -28 - 39t + 28 = -8 + 28 -39t = 20 -39t 20 -= -39 -39 20 t = -39 . 20 . ' to equa1 zero, the soIutlOn ' d d oes not cause any enommator set Smce t = -39
IS
59.
-7 4 ) (X + 5} (X - 2} = ( � + (_ ) (X + 5)(X - 2) x-2 x + 5 (x + 5} (x - 2}
{ -3920 } -
.
-3 4 7 ..,- -,+ -,---,...x - 2 x + 5 (x + 5}(x - 2)
-- =
4(x + 5} = -3 (x - 2} + 7 4x + 20 = -3x + 6 + 7 4x + 20 -3x + 1 3 4x + 20 + 3x = -3x + 1 3 + 3x 7x + 20 = 13 7 x + 20 - 20 = 1 3 - 20 7x = -7 7x -7 7 7 x = -1 x = -1 does not cause any denominator to equal zero, the solution set is
=
Since
{-I} .
34
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Section 1. 1 : L inear Equations
2 3 5 -- + -- = -y+3 y-4 y+6
61.
(_y 2+_3 + _y -3_4 ) (y + 3)(y - 4)(y + 6) = (_y +5_6 ) (y + 3)(y - 4) (y + 6)
2 (y - 4)(y + 6) + 3 (Y + 3)(y + 6) = 5 (Y + 3)(y - 4 ) 2 (l + 6y - 4y - 24 ) + 3 (y 2 + 6 y + 3 y + 1 8) = 5 ( y2 - 4 y + 3y - 1 2 )
Since 63.
2 (y2 + 2y - 24) + 3 (y 2 + 9y + 1 8) = 5 (l - y - 12 ) 2l + 4y - 48 + 3y2 + 27y + 54 = 5l - 5y - 60 5y2 + 3 1y + 6 = 5l - 5y - 60 5y2 + 3 1y + 6 - 5 y 2 = 5l - 5y - 60 - 5y2 3 1y + 6 = -5y - 60 3 1y + 6 + 5y = -5y - 60 + 5y 36y + 6 = -60 36y + 6 - 6 = -60 - 6 36y = -66 36y -66 36 36 11 y = -6 1 y=-
�
does not cause any denominator to equal zero, the solution set is
{- 161 } .
x x+3 -3 = x2 X x2 +X x2 - 1 x x+3 -3 (x + l )(x - l ) x(x - l ) x(x + l ) +3 X ( X + l )(X - l ) = -3 X(X + l ) (X - l) x x - l) x (x + l) (x)(x) - (x + 3) (x + 1) = -3 (x - l) x2 _ ( X2 + x + 3x + 3 ) = -3x + 3
(
)
[
)
x2 _ ( X2 + 4x + 3 ) = -3x + 3 x2 _ x2 - 4x - 3 = -3x + 3 -4x - 3 = -3x + 3 -4x - 3 + 4x = -3x + 3 + 4x -3 = 3 + x -3 - 3 = 3 + x - 3 -6 = x
Since x = -6 does not cause any denominator to equal zero, the solution set is
{-6} .
3S
© 2008 Pearson Education, Inc . , Upper Saddle River, NJ. A l l rights reserved. This material i s protected under a l l copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permi ssion in wri ting from the publi sher.
Chapter 1 : Equations and Inequalities
3 .2x + � = 1 9.23 65.87 1
65.
3.2x + � - � = 1 9.23 - � 65.871 65.87 1 65. 87 1
( )(
3.2x = 1 9.23 - � 65.87 1
( (
)( ) )( )
1 1 _ 3 .2X ) = 1 9.23 - � _ 3 .2 65.87 1 3.2 1 X = 1 9.23 - � _ ", 5.91 65 .87 1 3 .2 The solution set is approximately { 5 .9 1 } . 18 14.72 - 2 1 .58x = - x + 2.4 2.1 1 18 18 18 14.72 - 2 1 .58x - -- x = - x + 2.4 - - x 2. 1 1 2. 1 1 2.1 1 18 14.72 - 2 1 .58x - - x = 2.4 2. 1 1 18 14.72 - 2 1 .58x - - x - 1 4.72 = 2.4 - 14.72 2.1 1 18 -2 1 .5 8x - - x = - 1 2.32 2. 1 1
67.
[
1 -2 1 .58 -
( ( � 2. 1 1
]
) �)
-2 1 .5 8 - � X = -1 2.32 2. 1 1 -2 1 .5 8 -
2. 1 1
X = - 1 2.32
X = -1 2.32
The solution set is approximately { 0.4 1 } . 69.
[ [
1
-2 1 .58 - J!. 2. 1 1 1
18 -2 1 .58 - 2. 1 1
ax - b = c, a ;to O ax - b + b = c + b ax = b + c ax b + c a a b+c X = -a
] ]
"' 0.4 1
x x - + - = c a ;to 0, b ;to 0, a ;to -b a b '
71.
ab
(� i ) +
= ab . c
bx + ax = abc ( a + b ) x = abc ( a + b ) x abc -'-----"-- = -a+b a+b abc X = -a+b
36
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Section 1 . 1 : Linear Equations
7
3.
(
, 1 1 2 -- + -- = -x-a x+a x-1 l l 2 __ + __ (x _ a ) ( x + a ) ( x - 1) = _ (x - a)(x + a) (x - 1) x-a x+a x-1 (x + a ) ( x - l ) + (x - a ) ( x - 1) = 2 (x - a ) (x + a)
)
( )
x2 - x + ax - a + X2 - x - ax + a = 2 ( X2 + ax - ax - a2 ) 2X2 - 2x = 2 ( X2 _ a2 )
such that x 75.
77.
'#
±a, x '# 1
2X2 - 2x = 2X2 - 2a2 -2x = -2a2 -2x -2a2 -2 -2 x = a2
79
x + 2a = 16 + ax - 6a, if x = 4 4 + 2a = 1 6 + a( 4) - 6a 4 + 2a = 16 + 4a - 6a 4 + 2a = 1 6 - 2a 4a = 1 2 4a 1 2 4 4 a=3 1 1 1 -= -+R R, R 2 1 RR' R = RR' R � + _ 2 R 2 R, R 2 R, R = RR + RR, 2 2 R, R = R(R + R, ) 2 2 R, � R(R + R, ) 2 � + R, R + R, 2 R, R 2 =R R + R, 2
(�)
(
.
mv2 R m 2 RF = R F=
F
( ;]
F
mv2 R = F
]
81.
S=� 1-r S(l - r) = � (1 - r) 1-r S - Sr = a S - Sr - S = a - S -Sr = a - S -Sr a - S -S -S S-a r = -S
( )
37
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Chapter 1 : Equations and Inequalities
83.
Amount in bonds Amount in CDs Total x
x - 3000
91.
20, 000
x + ( x - 3000 ) = 20, 000 2x - 3000 = 20, 000 2x = 23, 000 x = 1 1, 500 $ 1 1 ,500 will be invested in bonds and $8500 85.
x + ( x + 0.53 ) = 3 .57 2x + 0.53 = 3 .57 2x = 3.04 x = 1 . 52
87.
x - 0. 1 5x = 425, 000 0.85x = 425, 000 x = 500, 000 The original price of the house was $500,000.
will be invested in CD's . Yahoo! searches Google searches Total x
x + 0 . 53
The amount of the reduction (i.e., the savings) is 0. 1 5($500,000) $75,000. =
3 . 57
93.
Let x represent the price the bookstore pays for the book. Then 0.35x represents the markup on the book. The selling price of the book is $92.00 . publisher price + markup = selling price x + 0.35x = 92.00 1 . 35x = 92.00 x � 68 . 1 5 The bookstore paid $68. 1 5 for the book.
Yahoo! was used for 1 .52 billion searches and Google was used for 2.05 billion searches. Dollars Hours Money per hour worked earned Regular x 40x 40 wage Overtime 1 .5x 8(1 .5x) 8 wage
95.
Adults Children
Tickets Price per sold ticket
x 5200 - x
7.50 4.50
7.50x + 4.50 ( 5200 - x ) = 29, 96 1 7.50x + 23, 400 - 4 . 50x = 29, 96 1 3.00x + 23, 400 = 29, 96 1 3.00x = 656 1 x = 2 1 87 There were 2 1 87 adult patrons.
40x + 8 ( 1 .5x ) = 442 40x + 1 2x = 442 52x = 442 442 x = - = 8.50 52
89.
Let x represent the original price of the house. Then 0 . 1 5x represents the reduction in the price of the house. The new price of the home is $425,000. original price -reduction = new price
Sandra's regular hourly wage is $8.50. Let x represent the score on the final exam.
97.
80 + 83 + 7 1 + 6 1 + 95 + x + x = 80 7 390 + 2x = 80 7 390 + 2x = 560 2x = 1 70 x = 85 Brooke needs a score of 85 on the final exam.
Money earned
7.50x 4 . 50(5200 - x)
Let w represent the width of the rectangle. Then w + 8 is the length. Perimeter is given by the formula P = 21 + 2w.
2(w + 8) + 2w = 60 2w + 1 6 + 2w = 60 4w + 1 6 = 60 4w = 44 w=l1 Now, 1 1 + 8 1 9 . =
The width of the rectangle is 1 1 feet and the length is 1 9 feet.
38
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Section 1 .2: Quadratic Equations
99.
Let x represent the number of worldwide Internet users in March 2006. Then 0 .2 1 9x represents the number U . S . Internet users, which equals 1 52 million
1 5.
0.2 1 9x = 1 52 X "" 694.06 In March 2006, there were about 694.06 million
101.
x=3
The solution set is {-1, 3 }
Internet users worldwide. Answers will vary.
1 7.
Section 1 .2
1.
x2 - 5x - 6 = (x - 6) (x + l) 1 9.
5. 7.
9.
add;
(%I
13.
3 t2 - 48 = 0 3(/2 - 1 6) = 0 3(t + 4)(t - 4) = 0 t + 4 = 0 or / - 4 = 0 t=4 / = - 4 or The solution set is {-4, 4 } . x (x - 8) + 1 2 = 0 x2 - 8x + 1 2 = 0 (x - 6) (x - 2) = 0 x - 6 = 0 or x - 2 = 0 x=2 x = 6 or The solution set is { 2, 6} .
25 4
False; a quadratic equation may have no real solutions. x2 - 9x = 0 x(x - 9) = 0 x = 0 or x - 9 = 0 x = 0 or x = 9
21.
The solution set is {O, 9 } . 11.
2X2 - 5x - 3 = 0 (2x + l)(x - 3) = 0 2x + 1 = 0 or x - 3 = 0
x2 - 25 = 0 (x + 5)(x - 5) = 0 x + 5 = 0 or x - 5 = 0 x=5 x = - 5 or The solution set is {-5, 5 } .
4x2 + 9 = 1 2x 4x2 - 1 2x + 9 = 0 (2x - 3)2 = 0 2x - 3 = 0 3 X =2
The solution set is 23.
Z2 + Z - 6 = 0 (z + 3)(z - 2) = 0 z + 3 = 0 or z - 2 = 0 z=2 z = - 3 or The solution set is {-3, 2 } .
{%} .
6(p2 - 1) = 5p 6p2 _ 6 = 5p 6p2 _ 5p - 6 = 0 (3p + 2)(2p - 3) = 0 or 2p - 3 = 0 3p + 2 = 0 3 2 p=p = -- or 2 3
The solution set is
{-I, %} .
39
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Chapter 1 : Equations and Inequalities
6 6x - 5 = x
25.
( 6x - 5 ) x =
33.
(�)
2x + 3 = ± ./9 2x + 3 = ±3 2x + 3 = 3 or 2x + 3 = -3 2x = -6 2x = 0 or x = -3 x = o or The solution set is {-3, O}
x
6x2 - 5x = 6 6x2 - 5x - 6 = 0 (3x + 2)(2x - 3) = 0 or 2x - 3 = 0 3x + 2 = 0 2 3 X=x = -- or 2 3
35.
Neither of these values causes a denominator to equal zero, so the solution set is
{-�, %}.
27.
(
4 (x - 2) 3 -3 --'----:--'- + - = --,--....,.. x-3 x x (x - 3)
J
(
37.
39.
J
4 ( x - 2) � -3 + x ( x - 3) = x ( x - 3) x x-3 x ( x - 3) 4x (x - 2) + 3 (x - 3) = -3
41.
=
-8)
J
= (-4) 2 = 1 6
(�. �J (�J (�{-�)J ( -�J 1 16
=
1 9
=
x2 + 4x = 2 1 x2 + 4x + 4 = 2 1 + 4
=
Neither of these values causes a denominator to equal zero, so the solution set is %, 2 .
{- }
43 . x2 _ .!. x - � = 0 2 16 I 3 x2 - - X = 16 2 1 1 3 1 x2 - - x + - = - + 16 16 16 2
x2 = 25 x = ±..fi5 x = ±5
(x-�J
The solution set is {-5, 5} . 31.
CT (
.
(x + 2 f = 25 x + 2 = ± ..[25 x + 2 ±5 x = -2 ± 5 x = 3 or x = -7 The solution set is {- 7, 3} .
4x2 - 8x + 3x - 9 = -3 4x2 - 5x - 6 0 (4x + 3) (x - 2) = O 4x + 3 = O or x - 2 = O 3 x=2 x = - - or 4
29.
(2x + 3) 2 = 9
=
�
fI = + -'1"4 - 2
x - '!' = + 4
(x _ l) 2 = 4
.!.
1 1 X = -±4 2 X = -3 or x = --I 4 4
x - I = ± J4 x - I = ±2 x - I = 2 or x - l = -2 x 3 or x = - 1 The solution set is { -I, 3} .
The solution set is
=
{-�,%}.
40
© 2008 Pearson Education , inc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publisher.
Section 1 .2: Quadratic Equations
45.
3x 2 + x - -1 = O 2 x2 + .! x - .! = 0 3 6 x 2 + -1 x = -1 3
51.
-(- 5) ± �(- 5)2 - 4(2)(3) �� x = ----� ( 27)-�2� 5 ± � 5 ± Ji 5 ± 1 == 4 4 4 +1 5-1 5 X= or X = 4 4 6 4 X = - or X = 4 4 3 x = - or x = 1 2
--
6
1 1 1 1 x2 + - x + - = - + 3 36 6 36 2 7 = (X + 36
i) X+ = ± i &
.J7
The solution set is { 1, %}.
1 x + - = +- 6 6 ± .J7 1 x = ----'-6
. . {-1-.J7
The solutIon set IS 47.
6
}
53.
- 1 + .J7 . ' 6
4/ - y + 2 = 0 a = 4, b = - 1, c = 2 y
x2 - 4x + 2 = 0 a = 1, b = - 4, c = 2 -(-4) ± �(- 4)2 - 4(1)(2) x = ---��----2(1) 4 ± F6=8 2 .J8 4 ± 2-1i ± 4 = 2 ± -li = = 2 2 The solution set is { 2 -Ii, 2 + -Ii } .
=
-(- 1) ± �(_1)2 - 4(4)(2) 2(4) 1 ± .J1=32 8
No real solution. 55.
4x2 = 1 - 2x 4x2 + 2x - 1 = 0 a = 4, b = 2, c = -1
�'--
- 1) -2 ± �22- 4(4)( --x=2(4) - 2 ± .J4+16 - 2 ± J20 8 8 - 2 ± 2 J5 - 1 ± J5 4 8 . set IS. - 1 -4 J5 ' -1 +4 J5 . The solutIOn
-
49.
2x2 - 5x + 3 = 0 a = 2, b = - 5, c = 3
x2 - 4x - 1 = 0 a = 1, b = - 4, c = - 1 - 4(1)(-1) ± �(- 4)2-(- 4)�'-X = -- --2(1) 4 ± 06+4 2 J20 4 ± 2 J5 = 2 ± J5 = 4± 2 2 The solution set is { 2 - J5 , 2 + J5 } .
{
57.
=
}
4x2 = 9x 4x2 - 9x 0 x(4x - 9) 0 x = O or 4x - 9 = 0 9 x = O or X=4 =
=
The solution set is { O, �}. 41
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Chapter 1: Equations and Inequalities
59.
9t 2 -6t+l =0 1 a = 9, b = -6, = 2 -4(9)(1) t = _(_6)±�(_6) 2(9) 6±.J36-36 6±0 ---- = -- =18 3 18
{�}.
The solution set is 61.
65.
C
� X2
4 2 =O 4 -!x-! 4 (% x2 - � X - �) = 4 (0) 3x2 -x-2 = 0 a = 3, b = -1, = -2 1 ) �( 1 ) 2 -4( 3)(-2) x = --'--( --=-'- ----:2-'- (:--:3) 1±.Ji+24 1±J25 1±5 6 6 6 1-5 1+5 x =-6- or x = -6X =-66 or X = --46 x = 1 or x = --23 The solution set is {- �, I } .
67.
C
_
_
±
C
_:...'---'...:.
_
2x(x+ 2) = 3 2x2 +4x-3 = 0 a = 2, b = 4, c = -3 2 -4( 2)(-3) = -4± .J16+24 x = -4±�42(2) 4 -4±./40 -4±2M -2±M 2 4 4 . set IS. { -2- M -2+ M } . The soluhon 2 ' 2 1 2 =0 4---x x2 = x 2 (0) x2 ( 4-!-� X x2 ) 4x2 -x-2 = 0 a = 4, b = -1, = -2 --'--�'-'--(_1)2(4) --':-2..,.-4(-'-.- 4)(-2) -'--'--'x = --'-_ (_1)± 1 ±.J1D2 l± 53 8 8 Neither of these values causes a denominator to equal zero, so the solution set is
1+53 } { I-53 8 ' 8 .
69.
63.
5x2 -3x = 1 5x2 -3x-l = 0 a = 5, b = -3, = -1 -( -3)± �( _3)2�-4(5)( -1) ��� X = ��-�� 2(5) 3 ± .J9+2O 3 ± .J29 10 10 . set IS. { 3-.J29 3+.J29 } . The solutIOn 10 ' 10
� x-2 + !x = 4 (� x-2 + !x ) X(X -2) = 4x(x-2) 3x(x)+(x-2) = 4x2 -8x 3x2 +x-2 = 4x2 -8x 2 a 1, b = -9, 0==2 x -9x+2 -(-9)±�(-9)2 -4(1)(2) X = --'---'-���--�� 2(1) 9 ± ..J8l=8 9 ± m 2 2 Neither of these values causes a denominator to =
C
C
equal zero, so the solution set is
{ 9- 2m ' 9+ 2m } .
42
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Section 1.2: Quadratic Equations
71.
73.
75.
77.
x2 - 4. 1x + 2.2 = 0 a = l, b = -4.1, c = 2.2 - ( -4.1 ) ± �( -4. 1 )2 - 4 ( 1 )( 2.2 ) X= 2 (1) 4. 1 ± "'1 6.8 1 - 8.8 4. 1 ± .J8.Qi 2 2 x� 3.47 or x�0 .63 The solution set is { 0.63, 3.47 } .
81.
The solution set is 83.
x2 + 13x - 3 = 0 a = l, b = 13, c = -3 -13 ± �( 13t - 4 ( 1 )( -3 ) x= 2 (1) -13 ± J3+12 -13 ± J15 2 2 x� 1 .07 or x� -2.80 The solution set is { -2.80, 1 .07 } .
85.
1r X2 - X -1r = 0 a =1r , b = -l, C = -1r - ( -1 ) ± �( _ 1 )2 - 4 (1r ) ( -1r ) X= 2 (1r ) 1±� 21r x� 1 . 1 7 or x� -0.85 The solution set is { -0.85, 1 . 1 7 } .
{-�,�}.
x2 +,, �2x = -21 x2 +,, �2x --21 = 0 2 x2 + .fix � = 2 ( 0 ) 2X2 + 2 x - 1 = 0 a = 2, b = 2 , c = -l -( 2.fi) ± �( 2.fi)2 - 4( 2)( -1 ) x= 2 ( 2) -2 .fi ± J8+8 -2 .fi ± J16 4 4 -2.fi ± 4 -.fi ± 2 2 4 - 2 -.J2 + 2 The solution set 2 ' 2
(
3x2 + 81r x + .J29 = 0 a = 3, b = 81r , C = .J29 -81r ± �( 81r )2 - 4 ( 3 ) ( .J29) -x = ----�--��---2 (3) 2 = -81r ± �641r6 - 1 2.J29 x -0.22 or x� -8. 16 The solution set is { -8. 1 6, -0.22 } .
{-%, %}.
2 + z = 6z2 0 = 6z2 - z - 2 0 = ( 3z - 2 )( 2z + 1 ) 3z - 2 = 0 or 2z + 1 = 0 z = -23 or z = --21 The solution set is
87.
{�}.
lOx2 - 1 9x - 15 = 0 ( 5x + 3 )( 2x - 5 ) = 0 5x + 3 = 0 or 2x - 5 = 0 x = -53 or X =-25 The solution set is
�
79.
16x2 - 8x + l = 0 ( 4x - l )( 4x - l ) = 0 4x - l = 0 x = "41
) .fi .fi -
. . {-.fi
x2 -5 =0 x2 = 5 x = ± Fs The solution set is {-Fs, Fs} .
1S
43
}
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
89.
x2 + x = 4 x2 + x - 4 = 0 a = 1, b = 1, c = -4 -(1) �(1)2 - 4(1)(-4) x = ± 2(1) - 1 ± v'i+!6 -1 ±07 2 2 I -/l7 ' The solutlOn set 2 .
91 .
IS .
{
-
-
-
95.
97.
2
}
.
x -2- ----:-7x -+-l -x - 2 + x + l = x2 - x - 2 7x + 2 x -.,... ..,...l -,-(x - 2)(x + l) x - 2 + -x+l == x -7X2)+(x1 + l) (X - 2)(X + l) x-2 x+l x(x + l) + 2(x - 2) = 7x + 1 x2 + x + 2x - 4 = 7 x + 1 x2 + 3x - 4 = 7 x + 1 x2 - 4x - 5 = 0 (x + l)(x - 5) = 0 x + 1 = 0 or x - 5 = 0 x = -l or x = 5 The value x = -1 causes a denominator to equal zero, so we disregard it. Thus, the solution set is { 5} .
( _X_+_2_ ) (X _ 2)(X+1) (
93.
1 + -/l7
)
2x2 - 6x + 7 = 0 a = 2, b = -6, = 7 b 2 -4ac = (_6)2 - 4(2)(7) = 36 - 56 = -20 Since the b 2 - 4ac < 0, the equation has no real solution. 9x2 - 30x + 25 = 0 a = 9, b = -30, c = 25 b 2 - 4ac = (-30)2 -4(9) (25) = 900 - 900 = 0 Since b 2 - 4ac = 0, the equation has one repeated real solution. 3x2 + 5x - 8 = 0 a = 3, b = 5, c = -8 b 2 -4ac (5)2 - 4(3) ( -8) = 25 + 96 = 121 Since b2 - 4ac 0, the equation has two unequal real solutions.
99.
c
20.2x2 + 3 14.5x + 3467.6 = 8000 20.2x2 + 3 14.5x - 4532.4 = 0 a = 20.2, b = 3 14.5, = -4 532.4 -(314.5) �(3 14.5)2 - 4(20.2)( -4532.4) x = ----± �-� 2(20.2) -3 14.5 ±.J 465, 128.17 40.4 � or x",, 9. 1 Disregard the negative solution since we are looking beyond the 2000- 2001 academic year. Thus, according to the equation, the average annual tuition-and-fee charges will be $8 000 approximately 9. 1 years after 2000- 2001, which is roughly the academic year 2009- 2010. c
__ ���--� �
=
>
44
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1.2: Quadratic Equations
101.
1 03.
1 05.
Let w represent the width of window. Then I = w + 2 represents the length of the window. Since the area is 143 square feet, we have: w(w + 2) = 143 w2 + 2w- 143 = 0 (w + l 3)(w - l l) = 0 � or w = 1 1 Discard the negative solution since width cannot be negative. The width of the rectangular window is 1 1 feet and the length is l3 feet. Let I represent the length of the rectangle. Let w represent the width of the rectangle. The perimeter is 26 meters and the area is 40 square meters. 2/ + 2w = 26 1 + w = l3 so w = 13 -I I w = 40 1 (l 3 - /) = 40 l3/ _ /2 = 40 12 -l31 + 40 = 0 (/ - 8)(/ - 5) = 0 1 = 8 or 1 = 5 w=5 w=8 The dimensions are 5 meters by 8 meters. Let x length of side of original sheet in feet. Length of box: x - 2 feet Width of box: x - 2 feet Height of box: 1 foot V =I · w · h 4 = (x - 2){ x - 2){ 1) 4 = x2 _ 4x + 4 0 = x2 - 4x 0 = x(x - 4) x = 0 or x = 4 Discard x = 0 since that is not a feasible length for the original sheet. Therefore, the original sheet should measure 4 feet on each side.
1 07.
1 09.
When the ball strikes the ground, the distance from the ground will be O. Therefore, we solve 96 + 80t - 1 6t2 = 0 -16t2 + 80t + 96 = 0 t2 - 5t - 6 = 0 (t - 6)(t + l) = 0 t = 6 or t = -1 Discard the negative solution since the time of flight must be positive. The ball will strike the ground after 6 seconds. b. When the ball passes the top of the building, it will be 96 feet from the ground. Therefore, we solve 96 + 80t - 1 6t2 = 96 -16t2 + 80t = 0 t2 - 5t = 0 t (t - 5) = 0 t = 0 or t = 5 The ball is at the top of the building at time t = 0 when it is thrown. It will pass the top of the building on the way down after 5 seconds. Let x represent the number of centimeters the length and width should be reduced. 12 - x = the new length, 7 - x the new width. The new volume is 90% of the old volume. (12 - x)(7 - x)(3) = 0.9(12)(7)(3) 3x2 - 57 x + 252 = 226.8 3x2 - 57x + 25.2 = 0 x2 - 19x + 8.4 = 0 - 4(1)(8.4) = 19 ±J327.4 x = -(-19) ± �(-19)2 2 2(1) x 0.45 or x 1 8 .55 Since 1 8.55 exceeds the dimensions, it is discarded. The dimensions of the new chocolate bar are: 1 1 .55 cm by 6.55 cm by 3 cm.
a.
=
=
�
�
45
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
111.
Let x represent the width of the border measured in feet. The radius of the pool is 5 feet. Then x + 5 represents the radius of the circle, including both the pool and the border. The total area of the pool and border is AT = 1t (x + 5)2 . T he area o f th e pool is Ap = 1t( 5)2 = 251t . The area of the border is AB = AT - Ap = 1t (x + 5)2 - 251t . Since the concrete is 3 inches or 0. 25 feet thick, the volume of the concrete in the border is 0.25AB = 0.25 (1t (x + 5)2 - 251t ) Solving the volume equ ation: 0.25 (1t (x + 5)2 - 251t ) = 27 1t( x2 + lOx + 25 - 25 ) = 108 1tX2 + 101tx - 108 = 0 -108) x = -101t ± �(101t2()21t )- 4(1t )("-3 1. 42 ± �1001t 2 + 4321t 6. 28 x:::e 2. 71 or x:::e -12. 71 Discard the negative solution. The width of the border is roughly 2.71 feet. Let x represent the width of the border measured in feet. The total area is AT = (6 + 2x)(1 0 + 2x). The area of the garden is AG = 6 · 10 60. The area of the border is AB =AT-AG = (6 + 2x)(10 + 2x) - 60 . Since the concrete is 3 inches or 0.25 feet thick, the volume of the concrete in the border is O.25AB = 0.25 ( 6 + 2x)(10 + 2x) - 60) Solving the volume equation: 0.25 ( (6 + 2x)(10 + 2x) - 60) = 27 60 + 32x + 4x2 - 60 = 108 4x2 + 32x - 108 = 0 x2 + 8x - 27 = 0 - 4(1)(- 27) = - 8 ±$ x = - 8 ±� 822(1) 2 x:::e 2.56 or x:::e -10.5 6 Discard the negative solution. The width of the border is approximately 2.5 6 feet. ---.:�---.:--:.......:..
--
1 13.
1 1 5.
Let x = the length of a traditional 4: 3 format TV . Then �4 x = the width of the traditional TV. The diagonal of the 37-inch tr aditional TV is 37 inches, so by the Pyt hagorean theorem we hav e: x2 + � X = 372
( J
x2 +� X2 = 1369 16 16 x2 + : x2 = 16(1369) 6 16x2 + 9x2 = 2 1, 904 25 x2 = 2 1, 904 x2 = 876. 1 6 x = ± .J876. 16 = ±29.6 Since the length cannot be negative, the length of the traditional 37-inch TV is 29. 6 inches and the width is �4 (29. 6) = 22. 2 inches. Thus, the area of th e traditional 37- inch TV is (29. 6)(22. 2) = 657. 12 square inches.
(
:-
-
)
Lety = the length of a 37-inch 16:9 LCD TV . Then � y = the width of the LCD TV . 16 The diagonal of a 37-inch LCD TV is 37 inches, so by the Pythagorean theorem we have: / + Y = 372 6 � Y 2 + 25 6Y 2 = 1369 256 y2 +�/ = 256(1369) 256 256/ + 81/ = 350,464 337/ = 350, 464 35 0, 464 Y 2 = 337 350, 464 Y ± 337 :::e ± 32 2. 48 Since the length cannot be negative, the length o f the LCD TV is 350, 464:::e 32.248 inches and the 337 350, 464 width is � 16 337 :::e 1 8. 140 inches. Thus, the area of the 37-inch 16:9 format LCD TV is
(: J
=
(
)
=
46
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
( 350,337464 )(�16
1 1 7.
1 1 9.
121.
350, 464 337
Section 1.2: Quadratic Equations
)
1 23.
197, 136 '" 584.97 square m. ches. ---'-337 The traditional 4:3 format TV has the larger screen since its area is larger. 1 "2 n (n + 1) = 666 n (n + I) = 1332 n2 + n - 1332 = 0 (n - 36)(n + 37) = 0 n = 36 or n = -37 Since the number of consecutive integers cannot be negative, we discard the negative value. We must add 36 consecutive integers, beginning at 1, in order to get a sum of 666. The roots of a quadratic equation are -b+ ..J'b 2::- -4 a-cb2 - 4ac and x = -'--: :Xl = -b - �2a 2a 2 b2 - 4ac + -b + �b2 - 4ac Xl + x2 = -b - �2a 2a -b - �b2 - 4ac - b + �b'2::- --4-a-c 2a -2b 2a b a In order to have one repeated solution, we need the discriminant to be O. b2 - 4ac = 0 12 -4(k )(k ) = 0 1 - 4k 2 = 0 4k 2 = 1 k 2 =..!.4
For ax2 + bx +c = 0 : -b + � b2-4ac b2 - 4ac and x = ---'--Xl = -b - �2a 2a -2 For ax2 - bx +c = 0 : -(-b)_ �( _b) 2 - 4ac 2a b - �b2 - 4ac 2a - -b + - 4ac "
and X2
= -X2
•
"
1 25.
(
./g;
1
_ (_ b) + �(_ b) 2 - 4ac 2a b + �b2 - 4ac 2a - 4ac - -b
(
-./g:
1
x2 = 9 and X = 3 are not equivalent because they do not have the same solu tion set. In the first equation we can also hav e X = -3. b. X =.J9 and X = 3 are equivalent becau se .J9 = 3 . (x - l)(x - 2) = (x - l) 2 and x - 2 = x- l are not equivalent because they do not have the same solution set. The first equation has the solution set { I} while the second equation has no solutions. Answers will vary. Knowing the discriminant allows us to know how many real solutions the equation will have. Answers will vary. a.
c.
1 27.
k = +-V4 fI k = -21 or k = --21
1 29.
47
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
Section 1.3 1.
3.
27.
Integers: { -3, O} Rationals: {-3, 0, %} 3 = 3 2-.J3 = 3(2-.J3) 2 +J3 2 +J3 ' 2 -J3 22_ (J3 ) 2
) ) = 3( 24 -J3 -3 = 3( 2 -J3
5. 7.
9. 11.
13.
Tru e; the set of real numbers is a subset of the set of comp lex numbers. (2 - 3i) + (6 + 8i) = (2 + 6) + (-3 + 8)i = 8 + 5i
(2 - 5i) - (8 + 6i) = (2 - 8) + (-5 - 6)i = -6 - 1 1 i
17.
2i(2 - 3i) = 4i - 6i2 = 4i - 6(-I) = 6 + 4i
23.
25.
35.
(-3 + 2i) - ( 4 -4i) = (-3 - 4) + (2 - (- 4))i = -7 + 6i
3 (2 - 6i) = 6 - 1 8 i
21.
31.
(1 + 1/ = 1 + 2i + i2 = 1 + 2i + (-I) = 2i
{ -2i, 2i}
1 5.
19.
6-i = 6 - i. 1 - i = 6 -----::6i - i + i2 1+i 1+i 1-i 1 - i + i - i2 + (-I) =-5 - 7i = -5 - -1 7. = 6 -1 7i- (-1) 2 2 2
39.
41.
(3 - 4i)(2 + i) = 6 + 3i - 8i - 4i2 = 6 - 5i -4(-1) = 1 0 - 5i (-6 + i)(- 6 - i) = 36 + 6i - 6i - i2 = 36 - (-1) = 37 10 ._3 + 4i 30 + 40i 10 =-_ -3 - 4i 3 - 4i 3 + 4i 9 + 12i - 12i - 16i2 30 + 40i 30 + 40i 9 - 16(-1) 25 30 40 . 6 8 . +1 =+-1 = 25 25 5 5
45.
-
6i 3 -4i5 = i3 (6 - 4i2 ) = i2· i(6 - 4(-I)) = -1· i(lO) = -lOi (1 + i)3 = (l + i)(1 + i){ l + i) = { l + 2i + i2 )(1 + i) = (1 + 2i - 1)(1 + i) = 2i{ l + i) = 2i + 2i2 = 2i + 2( -1) = - 2 + 2i i6 + i4 + i2 + 1 = V )3 + ( i2 ) 2 + i2 + 1 = _( 1) 3 + (_ 1)2 + (-1) + 1 = -1 + 1 - 1 + 1 =0
47.
R = 2i
49.
�- 25 = 5 i
51.
2 + i =2 + i. -i -i 1 - 2i = 1 - 21. = -2i-(--1)(-I) =-1
i-15 = =- =-- = il5 i14+1 il4 . i (i 2 )7 . i 1 = -1 = -I ' -i =-i =-i = 1. =--(-1) -i -i i f _ i2 (-I i
�(3 + 4i)(4i - 3) = �12i - 9 + 16i2 - 12i = �-9 + 16(-1) = �- 25 5i =
48
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1.3: Complex Numbers; Quadratic Equations in the Complex Number System
53 .
55.
57.
59.
61.
x2 + 4 = 0 x2 = -4 x = ±.f:4 x = ±2i The solution set is { -2i, 2i} .
67.
x2 - 6x + 13 = 0 a = 1, b = -6, c = 1 3, b2 - 4ac = (- 6l - 4(1)(13) = 36 - 52 = -16 ±� -16 = -6 ± 4i = 3 + 21. x = -(-6)2(1) 2 The solution set is { 3 - 2i, 3 + 2i} . x2 - 6x + 1 0 = 0 a = 1, b = -6, c = 10 b2 -4ac = (_ 6)2 - 4(1)(10) = 36 - 40 = -4 ± 6 ± 2i = 3 ± 1. X = -(-6)2(1).j=4 = -2 The solution set is { 3 - i, 3 + i} .
69.
8x2 - 4x + l = 0 a = 8, b = -4,c = 1 b2 - 4ac = (_ 4)2 - 4(8)(1) = 1 6 - 32 = -16 ±� -16 =-4 ± 4i = -1 ± -I 1. x = -(-4)2(8) 16 4 4 ' set . "4I -"41/. , "41 +"4I 1. . The sol utlOn
{
71.
}
5x2 + 1 = 2x 5x2 - 2x + l = 0 a = 5, b = -2, c = 1 b2 -4ac = ( _2) 2 - 4(5)(1) = 4 - 20 = -16 ±� -16 =-2 ± 4i =-1 ± -I 2. X = -(-2)2(5) 10 5 5 2 . 1 + 2 1. } . The so lutt' on set . 51 -51, 5 5 IS
x2 + x + 1 = 0 a = 1, b = 1, c = 1, b2 - 4ac = 12 - 4(1)(1) = 1 - 4 = -3 - 1 ±H = - l ±J3 i =--1 +-1 J3 . X= 2(1) 2 2- 2 The solution set is
x2 - 1 6 = 0 ( x + 4)( x - 4) = 0 x = -4 or x = 4 The solution set is { -4, 4} .
IS
63.
65.
73.
{
{-.!.2 - J32 i, -.!.2 + J32 i}.
x3 - 8 = 0 (x - 2) ( x2 + 2x + 4 ) = 0 x - 2 = 0� x = 2 or x2 + 2x + 4 = 0 a = 1, b = 2, c = 4 b2 - 4ac = 22 - 4(1)(4) = 4 - 16 = -12 - 2 ± 2.J3 i .J3 X = - 2 ±� 2(1) = 2 = -I ± i The solution set is { 2, -1 -J3i, -1 +J3i } .
x4 = 16 x4 - 16 = 0 ( X2 - 4 )( x2 + 4 ) = 0 (x - 2)(x + 2) ( x2 + 4 ) = 0 x - 2 = 0 or x + 2 = 0 or x2 + 4 = 0 x = 2 or x = -2 or x2 = -4 x = 2 or x = -2 or x = ±.f:4 = ±2i The solution set is { -2, 2, -2i, 2i} . x4 + 13x2 + 36 = 0 ( X2 + 9 )( x2 + 4 ) = 0 x2 + 9 = 0 or x2 + 4 = 0 x2 = -9 or x2 = -4 x = ±.f:4 x = ±H or x = ±2i x = ±3i or The solution set is {- 3i, 3i, -2i, 2i} . 3x2 - 3x + 4 = 0 a = 3, b = - 3, c = 4 b2 - 4ac = (_ 3)2 - 4(3)(4) = 9 - 48 = -39 The equation has two complex solutions that are conjugates of each other.
49
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
75.
77.
2 X2 + 3x = 4 2 X2 + 3x - 4 = 0 a =2 , b = 3,c = -4 b2 -4ac = 32 - 4(2 )(-4) = 9 + 32 = 41 Th e equationh as two unequal real solutions.
Th e oth er solution is 2 + 3i =2 - 3 i.
81.
z + z = 3 - 4i + 3 - 4i = 3 - 4i + 3 + 4i = 6
85.
87.
89.
1.
5.
9x2 - l2 x + 4 = 0 , a = 9, b = -12 , c = 4 b2 - 4ac = (-12 )2 - 4(9)(4) = 144 - 144 = 0 Th e equation has a repeated real solution.
79.
83.
Section 1.4
7.
9.
z · z = (3 - 4i)(3 - 4i) = (3 - 4i)(3 + 4i) = 9 + 12 i - 12 i - 16i2 = 9 - 16(-1) =2 5
11.
= VI = 3l 8-+4ii = 31 8-+4ii . 33 ++ 4i4i 54 + 72 i + 3i + 4i2 54 + 75i - 4 9 + 16 9 + 12 i - 12 i - 1 6i2 = 50 + 75i =2 + 3i 25 Th e impedance is 2 + 3i ohms. Z
13.
z + z = (a + b i) + (a + b i) = a +bi+a-bi =2 a z - z = a + b i - (a + b i) = a + bi - (a - b i) = a +bi-a +bi =2 b i
15.
z + w = (a + b i) + (c +d i) = (a + c ) + (b +d ) i = (a +c ) - (b +d )i = (a - b i) + ( -d i) = a + bi +c +d i =z +w Answers will vary. c
91.
True quadratic in form "'/2 t - l = 1 (..j2 t_ l) 2 = 12 2 t-l = 1 2 t =2 t=1 Check: �2 (1) - 1 = Ji = 1 The solution set is { I} . ...} 3t + 4 = -6 Since the principal square root is never negati ve, the equation has no real solution. � 1 -2 x - 3 = 0 � 1 -2 x = 3 (� 1_2 x) 3 = 33 l -2 x =2 7 -2 x =2 6 x = -13 Check: � 1 -2 ( -13) - 3 = m - 3 = 0 The solution set is { -1 3} . � 5x - 4 =2 W 5x - 4t =24 5x - 4 = 16 5x =2 0 x=4 Check: � 5(4) - 4 =� =2 The solution set is { 4} . � X2 +2 x = -1 (� x2 +2 x t = (_ 1)5 x2 +2 x = -1 x2 +2 x + l = 0 (x + l)2 = 0 x +l = 0 x = -l Check: �(_ 1)2 +2 (-1) =�.h -2 =v:1 = -1 The solution set is { -I} .
50
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
1 7.
1 9.
21.
23.
x = 8Fx ( x/ = ( 8Fxt X2 = 64x X2 - 64x = 0 x( x - 64) = 0 x = 0 or x = 64 Check 0: 0 = 8Fa Check 64: 64 = 8J64 64 = 64 0=0 The solution set is { O, 64} .
Check:
�(-�J - ( -�) - 4 = ( - %) + 2 )�� + % -4 = � �2 = �2
The solution set is
.J15 - 2x = X (.J 15 - 2x t = x2 1 5 - 2x = x2 x2 + 2x - 1 5 = 0 (x + 5)(x - 3) = 0 x = -5 or x = 3 Check -5: ..;r-:1 5=---- 2(:--- 5)- = 55 = 5 ;to -5 Check 3: �15 - 2(3) = J9 = 3 = 3 Disr egar d x = -5 as extr aneous. The solution set is { 3} .
25.
{-�}.
-5 = -5
3 +.J 3x + 1 = x .J3x + 1 = x - 3 (.J 3x + 1 t = ( x - 3)2 3x + 1 = x2 - 6x + 9 0 = x2 - 9x + 8 0 = (x - l)(x - 8) x = 1 or x = 8 Check 1 : 3 + �3(1) + 1 = 3 +14 5 ;to 1 Check 8: 3 + �3(8) + 1 = 3 +55 = 8 = 8 Discar d x = 1 as extr aneous. The solution set is { 8} . =
x = 2� x2 = ( 2.J x - l t x2 = 4(x - l) x2 = 4x - 4 x2 -4x + 4 = 0 (x _ 2)2 = 0 x=2 Check: 2 = 2� 2=2 The solution set is { 2} .
27
�X2 - x - 4 = x + 2 ( �x2 - X - 4 r = ( x + 2/ x2 - x - 4 = x2 + 4x + 4 -8 = 5x --85 = x
•
.J2x + 3 - .[;+1 = 1 .J2x + 3 = 1 + .Jx + l ( .J2x + 3 t = ( 1 + .Jx + 1 t 2x + 3 = 1 + 2.[;+1 + x + 1 x + l = 2.[;+1 (x + 1)2 = ( 2.[;+1t x2 + 2x + 1 = 4(x + 1) x2 + 2x + l = 4x + 4 x2 - 2x - 3 = 0 (x + l)(x - 3) = 0 x = -1 or x = 3 Check-I : �2(-1) + 3 - .J-l + 1 =J1 -JO = 1 - 0 = 1 = 1 Check 3: �2(3) + 3 -.J3+i =J9 -14 = 3 - 2 = 1 = 1 The solution set is {-I, 3} .
51
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
29.
31.
�3x+l-�x-l = 2 ..}3x + 1 = 2 + �x-I (�3x+ 1 t = (2 + �X -1 t 3x + 1 = 4 + 4�x - 1 + x -1 2x- 2 = 4�x-1 (2x- 2)2 = ( 4�x-1t 4x2 -8x +4 = 16(x-l) x2 -2x+1 = 4x-4 x2 -6x+5 = 0 (x-1)(x-5) = 0 x = 1 or x = 5 Check 1: �3(1 ) +1 -� =J4-JO = 2-0 = 2 = 2 Check 5: �3(5 ) + 1 -� = M -J4 =4- 2 = 2 = 2 The solution set is {l, 5} .
33.
(3x+1(2 = 4 ( (3X+ 1(2f = (4)2 3x+ 1 = 16 3x = 15 x=5 Check: ( 3(5)+ It2 = 161/2 = 4
The solution set is { 5} . 35.
(5x - 2)1I3 = 2 ( (5x -2Y'3 r = (2)3 5x -2 = 8 5x = 10 x=2 Check: (5{2)- 2t3 = 8113 = 2 The solution set is { 2} .
37.
�3 -2../x = ../x ( �3 -2../x r = (../x)2 3-2../x = x -2../x = x-3 ( _2../x)2 = (X _3)2 4x = x2 -6x+9 0= x2 -1Ox+9 0 = (x-l)(x-9) x = 1 or x = 9 Check 1: Check 9: �3 -2.Ji = .Ji �3-2J9 = J9 �3 - 2 = 1 �3 - 2 ·3 = 3 .Ji = 1 .../-3 3 1=1 Discar d x = 9 as extr aneous.
( x2 +9t2 = 5 ( ( x2+9 )1/2)2 = (5 )2 x2 +9 = 25 x2 = 16 x = ±M = ±4 Check -4: ( ( _4)2 9 Y'2 = 251/2 = 5 Check 4: ( (4)2 +9Y'2 = 25112 = 5 The solution set is {-4, 4} . X3 /2 -3 Xll2 = 0 XI/\X-3) = 0 Xl/2 = 0 or x -3 = 0 x = 0 or x = 3 Check 0: 03 /2 _3 . 0"2 = 0-0 = 0 Check 3: 33 /2 _3 . 3112 = 3F3 -3F3 = 0 The solution set is {O, 3} . x4 -5x2+ 4 = 0 (x2 -4 )(x2-1 ) = 0 x2 -4 = 0 or x2 - 1 = 0 x = or x = ±1 The solution set is {-2, -1 , 1, 2} . +
39.
#
41.
The solution set is { I} .
±2
52
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
2 u +1 = 0 1 u = -2 1 s + 1 = -2 s = --23
3 x4 _2 x2 - 1 = 0 ( 3x2 + l )( x2 - 1 ) = 0 3 x2 + 1 = 0 or x2 - 1 = 0 3x2 = -l or x2 = 1 Not real or x = ± l The solution set is { -I, I } . 45. x6 + 7x3 - 8 = 0 ( x3 + 8 )( x3 - 1 ) = 0 x3 + 8 = 0 or x 3 - 1 = 0 x3 = -8 or x 3 = 1 x = -2 or x 1 The solution set is { -2 , 1} .
43.
53 .
(X +2 ) 2 + 7 (x +2 ) + 12 = 0 Let u = x +2 , so that u 2 = (X +2 ) 2 . u 2 + 7u + 12 = 0 (u + 3)(u + 4) = 0 u + 3 = 0 or + 4 = 0 = -4 u = -3 or x +2 = -3 or x +2 = -4 x = -5 or x = -6 The solution set is { -6, -5} . u
(3 x + 4) 2 - 6 (3x + 4) + 9 = 0 Let u =3 x + 4 so that u 2 = (3 x + 4r u 2 - 6u + 9 = 0 (u _3 )2 = 0 u -3 = 0 u =3 3x + 4 = 3 x = -3"1 The solution set is
51.
or
{
s =2
x - 4x..lx = 0 x ( I - 4..1x ) = 0 x = 0 or 1 - 4..1x = 0 1 = 4..1x 1- =..Ix 4 (�)2 = (..Ixt -1-16 - x Check: x = 0: 0 - 4(0).JO = 0 0=0 X = I�: ( � ) - 4 C� ).Jk = 0 /6 - 4 C � )( � ) = 0 1� - /6 = 0 0=0 The solution set is {o, 1� }.
u
49.
or s + I =3
The solution set is - � ,2 } .
=
47.
or u -3 = 0 or u =3
55.
{-�}.
2 s( + 1)2 - 5(s + 1) = 3 Let u =s + 1 so that u 2 = (s + i I . 2 u 2 - 5u =3 2 u 2 - 5u -3 = 0 (2 u + 1)(u -3 ) = 0
x +..Ix =2 0 Let u =..Ix so that u 2 = X. u 2 + u =2 0 u 2 + u -2 0 = 0 (u + 5)(u - 4) = 0 u + 5 = 0 or u - 4 = 0 u = -5 or u = 4 =4 ..Ix = -5 or ..Ix or x = 16 not possible Check: 1 6 +.Ji6 = 20 1 6 + 4 =2 0 The solution set is { 16} .
53
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
57.
59.
(1/2 - 2t l/4 + 1 = 0 Let u = t1l4 so that u2 = t1/2. u2 -2u + 1 = 0
(U_ 1)2 = 0 u-1 = 0 u= 1 / (1 4 = 1 t=1 Check: 1 1/2 - 2(1t4 + 1 = 0 1-2+1=0 0=0 The solution set is {1} . 4Xl/2 _9x1l4 + 4 = 0 Let u = xl/4 so that u2 = x1/2 . 4u2 -9u+4 = 0 �r-(_-9)- 2 - _ - 4(-4 )- ( 4-) 9 ±.JU u = -(-9) ± 2(4) = 8 9±.JUx114 = -..".:8 x= 9±
)
-9 9 - +4� O f-;n), ( ;n 4( 81-18.JU + 17) -n (9 -.JU)+ 256 = 0 324 -n.JU + 68 - 648 + n.JU + 256 = 0 0=0 . set . 9_.JU 4 9 +.JU 4 . The solutlOn 8 ' 8 IS
61 .
( f1J Check x = ( 9 + f1 J 4 ( 9+ + 4� O -9 9 + [ ;nJ}" [( ;nJr
{( ) ( ) ]
�5x2 - 6 = x (�5X2 - 6 r = X4 5x2 - 6 = x4 0 = x4 -5x2 + 6 Let u = x2 so that u2 = X4. o =u2 -5u+ 6 0 = (u -3){u-2) u = 3 or u = 2 x2 = 3 or x2 = 2 x = ±.J3 or x = ±.fi Check:
x = -.J3: �5 (- .J3t - 6 = -.J3 �15 - 6 = -.J3 jf§ -:t=-.J3 x =.J3: �5(.J3f - 6 =.J3 �15 - 6 =.J3 jf§ = .J3 .J3 = .J3 = - .fi: � ( - .fif - 6 = - .fi �10- 6 = -.fi � -:t=-.fi
4(9 +.JUt -n (9+.JU)+ 256 = 0 4 ( 81 + 18.JU + 17) -n ( 9+.JU) + 256 = 0 324 + n.JU + 68 -648 -n.JU + 256 = 0 0=0
x
5
54
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. A l l rights reserved. This material i s protected under a l l copyright l aws a s they currently
exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
x =J2: �5 (J2t - 6 =J2 �10 - 6 =J2 * =J2 J2 = J2 The solution set is {J2, .J3 } . 63.
65.
u = -1 or 1_ _ = _1 or x+l 1 = -x- lor x = -2 or
u= 2 1 _ x+l_ = 2 1 = 2x +2 -2x = 1 X = --21
Check:
x2 +3X + �X2 + 3x = 6 Let u = �x2 + 3x so that u2 = x2 + 3x. u2 +u = 6 u2 +u-6 = 0 (u+3)(u-2) = 0 u = -3 or u=2 �X2 +3x = -3 or �X2 + 3x = 2 Not possible or x2 +3x = 4 x2 + 3x-4 = 0 (x + 4)(x- 1) = 0 x = -4 or x = 1 Check x = -4: ( _4)2 +3( -4)+ �(-4)2 + 3( -4) = 6 16-12 + .J16-12 = 6 16- 12+ .J4 = 6 6=6 Check x = 1: (1)2 +3(1)+ �(1)2 + 3(1) = 6 1+3+ � = 6 4 + .J4 = 6 6=6 The solution set is {-4 , I}. 1 = _1_ + 2 (x+ l)2 x+l 1 so that u2 = (-1 2 Let u = x+l x+l ) u2 = u + 2 u2 -u- 2 = 0 (u+1)(u- 2) = 0
1 =--+ 1 2 -2 (-2 + 1)2 + 1 1 = -1 + 2 1=1 1 1 x = --:2 = 1 +2 ( - -!- +I t ( - -!- + 1 ) 4 = 2+2 4=4 The solution set is {-2, --!-} . 3x-2 -7x-1 - 6 = 0 Let u = X-I so that u2 = x-2. 3u2 - 7u- 6 = 0 ( 3u + 2) (u - 3) = 0 u = --23 or u=3 x-I = 3 x = --23 or ( X-I t = ( -1rl or ( X-I t = (3t or X = -31 X = --32 x = -2:
67.
_I
Check:
x = - "23 : 3 ( -"23 )-2 - 7 (-"23 )-1 -6 = 0 3 (�) -7 (- i) - 6 = 0 i3 + 143 _ 6 = 0 0=0 x = � : 3 ( � r2 -7 Uf -6 = 0 3(9)-7(3)-6 = 0 27 -21-6 = 0 0=0
--
The solution set is
{-%,�}.
55
© 2008 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
69.
2X2/3 - Sxll3 -3 = 0 Let u = xl/3 so that u2 = x2/3. 2u2 - Su -3 = 0 ( 2u + 1)( u -3) = 0 u= --21 or u = 3 Xl l3 21 or Xl/3 = 3 ( x1l3)3 = (-1 J or (X l3 )3 = (3)3 X = --81 or x = 27
Check
( ) ( ) 2 ( � ) -s (-i ) -3=0
1 2 -"81 2/3 -5 -"81 1/3 -3 = 0 Check = -g:
The solution set is
1 S -+--3=0 2 2 3-3=0 0=0 Check = 27: 2( 27t3 -5( 27Y/3 -3=0 2(9)-5(3)-3=0 18-1S-3=0 3-3=0 0=0 The solution set is 27 .
73.
x
{-i,
71.
- � +1
( - �) +1 G� ) + (- �) = 8 ( 21 S ) 0)
16-8 = 8 8=8 _2 2 2(-2) Check v = -2'. (--2 + 1 ) + (-2)+1 = 8 4+4 =8 8=8
I
x
( iS ]2 2 ( - � ) _ =8 v = _i:S _ +
{-2,-1}.
x3 -9x = 0 X(X2 -9) = 0 x(x-3)(x+3 ) = 0 x = O x -3 = 0 x + 3 = 0 x = 3 x = -3 The solution set is {-3,0,3} . 4x3 = 3x2 4x3 -3x2 = 0 x2(4x -3) = 0 x2 = 0 or 4x -3 = 0 x=0 4x = 3 X = -43 m
}
75.
V)2 + v+l 2v = 8 (v+l _v_)2 + 2 (v+l _v_) = 8 (v+l 2 _v_ so that u2 = (_V Let u = v+l v+l_)
The solution set is
u2 + 2u = 8 u2 + 2u-8 = 0 (u +4)(u -2 ) = 0 u = -4 or u = 2 _v_ = _4 _v_=2 or v+l v+l v = -4v-4 or v = 2v+ 2 v = --4 or v = -2
77.
x3 + x2-20x = 0 x(X2 +x-20) = 0 x(x + S)(x-4) = 0 x = O x + S = O x-4 = 0 x=-S x=4 The solution set is {- S, 0, 4} . m
S
{O,%} . m
56
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
79.
81.
83.
85.
0 0 0 0 I} . x3 - 3x2 -4x + 1 2 = 0 x 2 (x -3 ) -4( x -3 ) = 0 (x -3 ) (x2 -4) = 0 (x-3}(x - 2}(x+ 2 ) = 0 x - 3 = 0 or x -2 = 0 or x + 2 = 0 x=3 X = 2 x = -2 The solu tion set is {-2 , 2 ,3} . 2x3 + 4 = x2 + 8x 2x3 - x2 -8x + 4 = 0 x2 ( 2x - 1) - 4 ( 2x - 1) = 0 ( 2x - 1 ) (X2 -4) = 0 ( 2 x - l} (x -2 )(x + 2 ) = 0 2x -1 = 0 or x - 2 = 0 or x + 2 = 0 2x = 1 x = 2 x = -2 x = 2"1 The solu tion set is {- 2, �, 2} . 5x3 + 45x = 2 X2 + 18 5x3 _ 2X2 +45x-18 = 0 x2 ( 5x - 2 ) + 9 ( 5x -2 ) = 0 (5x - 2 ) (X2 + 9) = 0 5x-2 = 0 or x2 + 9 = 0 5x = 2 x2 = -9 X = -25 no r eal solu tions +2 + 1 ) -1 + + 1 ) ( x2 + =0
x3 x - x l = x2 (x ( x 1) = - 1) = (x (x+ l)(x - l)(x + l) = O or x - I = x 1 x = -1 x=1 The solu tion set is {-l,
The solu tion set is
87.
89.
x ( x2 - 3x t 3 - x t3 [ +
+2(x2 -3xf/3 =0 (x2 3 X 2(X2 -3x)J = 0 ( x2 - 3x(3 ( x + 2x2 - 6x) = 0 (X2 -3xt3 (2x2 -5x) = 0 (x2 -3xt 3 = 0 or 2x2 - 5x = 0 x2 - 3x = 0 or 2X2 - 5x = 0 x(x-3) = 0 or x(2x-5 ) = 0 x = 0 or x= 3 or x = 0 or x =-25 The solu tion set is {O, %, 3} . X_4XI/2 +2=0 Let u = Xl/2 so that u 2 = x2 . u 2 - 4u+2=0 -0 )- (--2-) ( -4- "")2 - 4-4)±�,..-�----( ---� u = -2 4 = ±2J8 = 4±22.J2 =2 -+ .J2 u = 2 + .J2 or u 2 - .fi Xl/2 = 2+.J2 or XI/2 =2_.fi (x1/2r =( 2 +.J2t or (xIl2t =(2-.fir X=(2+.J2t or x = (2-.fir Check x = ( 2 +.J2t : (2+.J2t - 4(2+.J2)+2=0 4 +4.J2 + 2 - 8-4.J2 +2 = 0 0=0 C heck X=(2-.J2t: (2 -.J2t -4 ( 2 -.J2) + 2 = 0 4 -4.J2+2-8+4.J2+2=0 0=0 The solu tion set is {(2-.J2r, ( 2 +.J2r}�{0.34,11 .66} . =
{�} .
57
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
93.
Let U = x2 so that u2 = X4 . u2 +.J3u - 3 -3 ) 2 - 4- (-1)-(--3)_ -J32± M u _ -J3 ± �.-(J32(1) =0
-
-
=-../32+M 2 =-J3+2 M -;::I= 5 =±�r----;J3=-;-M
u
x
x
Check
or
or x or
x= �-J3 ;M:
0
---'---:-2-=-J3-M 2 =-J3-2 M =±�-J3 ; MI 5 u
x
r-Not--:r =-eal---=-=
=
(�-J3;M)4+J3(�-J3;M)2 -3=0 (-J3 ;MJ +J3(-J3 ;M)_3=0 3-2J3M+15 +J3(-J3)+J3M 3=0 4 2 18-2.J4s +-3+.J4s -3=0 2 4 9- .J4s +-3+.J4s - 3=0 2 2 9-..[45-3+..[45 -3=0 2 3-3=0 0=0 Check x= �-J3 ;M: (_�-J3 ;MJ+J3(_�-J3;MJ -3=0 ( -J3 ;M)\J3( -J3 ;M)_3=0 3-2J3M - + 4 + J3( J32) J3M-3=0 18-2..[45 +-3+.J4s -3=0 4 9-..[45+-3+2.J4s-3=0 2 2 9-..[45-3+.J4s -3=0 2 3-3=0 0=0 The solu tion set is {_�-J3;M, �-r--;::: J33:--;-;:M5= }", {-l.03, l.03} . ��.::....:.--'-----'--
-
�-
t=_ I + I - � 2IT : - .JI2IT-+4IT2IT(l - .J2ITI +4IT2 )2 = IT +l---'--+1+4IT2 ) _- IT+l- � IT(1 - 2� 2IT 4IT2 2- 2� 4IT +4IT2 2IT2 +1-2IT� l- � 2IT +2IT2 2IT2 +1-2IT� Check
-
+
IT( 1 +t)2 = IT + 1 +t Let U = 1 +t so that u2 = (1 +t) 2 . ITU2 =IT+U ITU2 -U - IT = - 4(IT)(-IT) = 1±� u= -(-1)±�( -1)2 2IT 2(IT) 1±.Jl+41T2 l +t 2IT 1±.Jl+41T2 t = -1+---::2IT-Check t= _ 1+ 1+ � 2IT : ff [ I + � r 'ff+I + � +1+4IT2 ) _- IT+l+ � IT[1 +2� 2IT 4IT2 2+2� 4IT +4IT2 27r2 +1+2IT� 1+ � 2IT +2IT2 2IT2 +1+2IT�
15
l+ _ {_l+l- � l+ 2IT ' � 2IT }
The solu tion set is
� { -1.85, O.1 7} .
58
© 2008 Pearson Education, Inc ., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1.5: Solving Inequalities
95.
97.
-k = 12 k2 -k -12 = 0 (k-4 ) (k+3 ) = 0 k = 4 or x+3 =4 or x-3 x+3 =4x-12 or 3x = 15 or x = 5 or e
101.
k = -3 x+3 =_3 x-3 x+3 = -3x+9 4x = 6 x = -46 = 1 . 5 Neither of these values causes a denominator to equal zer o, so the solu tion set is {I. 5, 5} . Solve the equ ation Jj + too = 4 . _s_ +.[;4 -4=0 (1100) C too + Jj -4J= (0)(1100) s +275/S -4400= 0 Let u = J;, so that u2 = s. u2 + 275u -4400 =- -0 - - - - u = -275 �r-i7-52-:- 2- 4(1)(- 4400) = -275 J93,225 u"" 15.1 6382 or u"" -290.1 638 Since u = J; , it must be positive, so s = u2 "" (15.1 638 )2 "" 229. 94 The distance to the water 's sur face is appr oximately 229. 9 4 feet. T = 27rV{T3'2 Let T = 1 6 . 5 and solve for 1 6. 5 = 27rfu 127r6 . 5 = V3i(T 2 J = (fuJ C�: (�)2 = � 27r 32 = 32 C�:J "" 220.7 The length was appr oximately 220.7 feet.
S ection 1.5 1.
x -2 �
I
3. 5. 7.
1
9.
1 1 00
11.
13.
±
1 5.
1 7.
a.
d.
t
-> - . a c
a.
c
[0,2] 0 x 2 [2,(0) x 2 [0,3 ) 0 x<3 3<5 3+3 < 5+3 6<8 3<5 3-5< 5-5 -2 < 0 3 <5 3(3 ) < 3(5 ) 9 < 15 3<5 -2(3 )> -2(5 ) -6> -10 4>-3 4+3>-3+3 7>0
Inter val: Inequ ality: Inter val: Inequality: Inter val: Inequality:
c.
1 9.
o
-2
negative multiplication pr oper ties (for inequalities) Tr ue. This follows fr om the addition pr op er ty for inequalities. False. Since both sides of the inequality ar e being divided by a negative number , the sense, or dir ection, of the inequality must be r ever sed. b That .
b.
t.
[
I
1S,
±
99.
x - - 2 = O.
Answer s will var y. One example: Fx
�
�
�
�
59
© 2008 Pearson Education , Inc., Upper Saddle River, NJ. All rights reserved. This material i s protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
b.
4 > -3 4 - 5 > -3 - 5 - 1 > -8 4 > -3 3(4»3(-3) 12 > -9 4 > -3 d. -2( 4) < -2(-3) -8 < 6 2x + 1 < 2 2x + l + 3 < 2 + 3 2x + 4 < 5 b. 2x+ 1 < 2 2x+ 1 - 5 < 2- 5 2x- 4 < -3 2x+ 1 < 2 3(2x + 1) < 3 (2) 6x + 3 < 6 2x+ l < 2 d. -2 (2x+l) > -2 (2) -4x- 2 > -4 [0, 4]
35.
I
37.
c.
21.
a.
25.
[ 4, 6)
I I
27.
[
43.
If x � -4, then 3x � -12.
45.
If x > 6, then - 2x < - 1 2.
47.
If x�5, then - 4x::o;- 20.
49.
If 2x > 6, thenx > 3.
51.
If -i1 x::o; 3, thenx � - 6.
55.
[
4
57.
(-00,-4) )
I III
-4
x +l < 5 x + l- 1 < 5- 1 x<4 {x l x < 4} or (-00, 4)
o
I'"
[ 2
I
-3 < x < - 2
( -
3
] 5
)
-2
I'"
I o
59.
I '"
l1li
I
I
I o
I-
)
I"
4
o
1- 2x::O; 3 - 2x::O;2 x�- l The solutionset is {x l x � - I} or [- 1, 00). I
I
I
[ I
-\
0
I
I
I
� I '"
3x-7 > 2 3x > 9 x>3 The solution set is {x l x > 3} or (3, 00) . I
2::O;x::O;5 0
33.
Ifx > - 4, thenx+ 4 > 0.
I'"
6
-3
41.
I I
31.
)
)
I
If x < 5, thenx - 5 < O.
[ 4, 00) 0
29.
I'"
4
4
0
x < -3
I
]
1-. I-
4
o
39.
53 .
[ 0
I
I
[
I
I ..
c.
23.
x�4
I o
I
I
(
I
3
I -'1
I'"
3x - 1 � 3 + x 2x�4 x�2 The solutionset is {xl x � 2} or [2, 00) . I
I
I
I o
I
[ 2
I
II-
60
I'"
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1.5: Solving Inequalities
61.
-2(x+3) < 8 -2x-6 < 8 -2x < 14 x > -7
69.
The solutionset is I (I I -7
63.
4-3( I -x):O::; 3 4-3+3x:O::; 3 3x+l:O::; 3 3x:o::; 2 x <- 23
0
1 > x+8 2(x-4) 1 -x-2 2 > x+8 _.!.x2 > 10 x < -20
2
The solutionset is
)
The solution set is o
(_ ;,1 �) .
{x x < -20} or ( -20) . I
{x l}:o::; x :O::; 3} or [},3]. 3
{x l - 121 < x < �} or I)
01
1
I-
2
75.1 < 1--x21 < 4 0 < --x21 < 3 > x > - 6 or -6 < < 0 The solutionset is {x 1-6 < x < O} or ( -6, 0) . x
I
{x l x ��} or [� , (0 ) . 4 3
I-
]
o
[
5
II 2
I-
1 0
] I I I-
3
I I (I
-<Xl,
-20
I
IE
3
2x-l 0 73. -3 <--< 4 -12 < 2x-l < 0 -I I < 2x < I - 112- < x <-21
"3
{xI3:o::; x :O::;5} or [3,5].
2
o
I·
]
o
�� 2 1-�4 2x �4-x 3x �4 x >- 43
I
71.-5 :0::; 4 -3x :0::; 2 -9:0::; -3x �-2 3 � x �-23
{ x l x :o:}:; } or ( -OO,}].
The solutionset is 1
67.
I-
I
o
I I I [I
I ..
I "I
I I
The solutionset is
The solutionset is
65.
I.
:0::;
The solutionset is
{xlx > -7} or (-7, 00) . I
0:0::; 2x -6 4 6:0::; 2x:o::; 10 3:O::; x :O::; 5
1
1
( 1 I 1
-6
I) I I .. 0
77.(x+2)( x-3»(x-l)(x+l) x2 -x -6 > x2 -1 -x-6> -1 -x > 5 x < -5 The solution set is {xl x < -5} or (-00, -5) .
"I"
I
1
"'1
) I I I I I-
-5
6
0
© 2008 Pearson Education, Inc . , Upper Saddle River, Nl. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
79.
x(4x + 3) � (2x + l)2 4x2 + 3x � 4x2 + 4x + l 3x � 4x + l -x � 1 x:::: - 1
87.
The solution set is
{xlx::::-l} , , , , [ , -I 0
81.
x + 1 < -3 -21 -< -3 4 6 � 4x + 4 < 9 2 � 4x < 5 -21 � x < -45
{l
or
Since
,
-
[
83.
I
1
,-
,
4
"2
89.
(4x + 2 t < 0 1 -4x + 2 < 0 4x + 2 < 0 x < --21
{l
91.
The solution set is x x < -�} or ( -00, -�) . I_I 85.
, )1
o
2
I"
_\
93.
0 < -x2 < 35 O < �x an d �x < l5 S ince �x > 0 , t his mean s tha t x > o . Therefor e, -X2 < 35 5X � < 5X 10 < 3x 10 -3 < x
95.
( ) (�)
{l }
The solution set is x x > �I or ('3° , 00 ) . 1
1
,
,
3
( 1
10
T
,
4
1
,
>
{} )
The solution set is x � � x < %} or [�, %). , 0
1 __
2x - 4 > 0 , this means that 2x - 4 0 . Therefore, 1 1 -2x - 4 < -2 1 1 2(x - 2) < -2 2(X - 2 2( - 2» < 2(x - 2> l < x-2 3<x The solution set is {x l x > 3} or (3, 00) .
[-1,00) .
, .
0 < ( 2x - 4 )-1 < '21 1 1 0 < -2x - 4 < -2 1 1 o < __ a nd __ < 1. 2x - 4 2 2x - 4
,
,
o
,
(�)
, ( 3
1.1-
If -1 < x < 1, then -1 + 4 < x + 4 < 1 + 4 3 < x+4 < 5 So, a = 3 an d b = 5. If 2 < x < 3, then -4(2) < -4(x) < -4(3) -12 < -4x < -8 S o,a = -12 a nd b = -8. If 0 < x < 4, then 2(0) < 2(x) < 2(4) 0 < 2x < 8 0 + 3 < 2x + 3 < 8 + 3 3 < 2x + 3 < 1 1 S o, a = 3 a nd b = 11. If -3 < x < 0, th en -3 + 4 < x + 4 < 0 + 4 l < x+4 <4 1 1 1 > -x + 4 > -4 1 -41 < -x+4 < 1 S o, a = "41 an d b = 1 .
'.1-
62
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1.5: Solving Inequalities
97.
If 6 < 3x < 12, then 12 3x < -36 < 3 3 2<x<4 22 < x2 < 42 4 < X 2 < 16 So, 4 and b 16. a
99.
101. 1 03.
=
45, 000 900, 000
1 07.
=
.J3x + 6 We need 3x + 6 � 0 3x � -6 x � -2 To the domain is {xl x � -2} or [-2, (0 ). 21 young adult's age 30 Let x = age at death. x - 30 � 49.66 x � 79.66 Therefore, the average life expectancy for a 30-year-old male in 2005 will be greater than or equal to 79.66 years. Let x = age at death. b. x - 30 � 53.58 x � 83.58 Therefore, the average life expectancy for a 30-year-old female in 2005 will be greater than or equal to 83.58 years. By the given information, a female can expect to live 83.58 - 79.66 3.92 years longer. Let P represent the selling price and C represent the commission. Calculating the commission: C 45, 000 + 0.25(P - 900, 000) 45, 000 + 0.25P - 225, 000 0.25P - 1 80, 000 Calculate the commission range, given the price range: 900, 000 � P � 1, 100, 000 0.25(900, 000) � 0.25P � 0.25(1, 100, 000) 225,000 ::; 0.25P� 275, 000 <
0.05 5% to
a.
1 09.
95, 000
=
I, I 00, 000
=
0.086 8.6%' =
inclusive. As a percent of selling price, the commission ranges from 5% to 8.6%, inclusive. Let W = weekly wages and T tax withheld. Calculating the withholding tax range, given the range of weekly wages: 700 ::; W � 900 700 - 620 � W - 620 ::; 900 - 620 80 � W - 620 � 280 0.25(80) � 0.25(W - 620) � 0.25(280) 20 � 0.25 (W - 620)� 70 20 7 S . 30 S 0.25 (W 620 ) + 78.30 � 70 78.30 98.30 � T � 148.30 The amount withheld varies from $98.30 to $148.30, inclusive. Let K represent the monthly usage in kilowatt hours and let C represent the monthly customer bill. Calculating the bill: C 0.08275K + 7.58 Calculating the range of kilowatt-hours, given the range of bills: 63.47 � C � 214.53 63.47 � 0.08275K + 7.58 � 214.53 55.89 � 0.08275K ::; 206.95 675.41 � K � 2500.91 The usage varies from 675.41 kilowatt-hours to 2500.91 kilowatt-hours, inclusive. Let C represent the dealer's cost and M represent the markup over dealer's cost. If the price is $ 1 8,000, then 1 8, 000 C + MC C(1 + M) 000 Solving for C yields: C 18, I+M Calculating the range of dealer costs, given the range of markups: 0.12 � M � 0. 1 8 1 . 12 ::; I + M ::; 1 . 1 8 _1_ >- _1_ >- _1_ 1 . 12 1 + M 1 . 1 8 18, 000 > 1 8, 000 > 18, 000 1 . 12 - I + M - 1 . 1 8 1 6, 07 1 .43 � C � 1 5, 254.24 The dealer's cost varies from $ 15,254.24 to $ 1 6,071 .43, inclusive. =
+
<
+
-
=
c.
=
1 05.
=
111.
=
=
=
=
=
=
225 , 000 - I SO, 000 S 0.25P - I SO, 000 S 275, 000 - I SO, 000
45, 000 � C::; 95, 000 The agent's commission ranges from $45,000 to $95,000, inclusive.
63
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
1 13.
1 1 5.
Let T represent the score on the last test and G represent the course grade. Calculating the course grade and solving for the last test: + 89 + T G = 68 + 82 + 87 5 G = 3265+ T 5G = 326 + T T = 5G - 326 Calculating the range of scores on the last test, given the grade range: 80 � G < 90 400 � 5G < 450 74 � 5G - 326 < 124 74 � T < 124 To get a grade ofB, you need at least a74 on the fifth test. b. Let T represent the score on the last test and G represent the course grade. Calculating the course grade and solving for the last test: G = 68 + 82 + 876 + 89 + 2T G = 326 6+ 2T G = 1633+ T T = 3G - 163 Calculating the range of scores on the last test, given the grade range: 80 � G < 90 240 � 3G < 270 77 � 3G - 1 63 < 107 77 � T < 107 To get a grade ofB, you need at least a77 on the fifth test. Since a < b , -a2 < -b2 and -a2 < -b2 a- + -a < -a + -b and -a + -b < -b + -b 2 2 2 2 2 2 2 2 a a + b a < -and +2 b < b 2 a +-b < b . Thus, a < 2
a.
1 1 7.
1 1 9.
If 0 < a < b, then ab > a 2 > 0 and b2 > ab > 0 (fabt > a 2 and b2 > (fabt fab > a and b > fab Thus, a < Fab < b .
( )
For O < a < b , -h1 = -21 -a1 + -b1 +a .h h . !h = 1.2 bab
( ) . 1 = 1.2 ( � ab ) h
h = a2a+bb
2a b -a (a + b) h - a = a2ab a = a +b +b 2ab _a 2 - ab a b -a 2 a +b a+b a(b a) = a +b > 0 Therefore, h > a . b( a + b) - 2ab b _ h = b a2ab +b = a+b a b + b2 - 2ab b2 - ab a +b a +b b(b a) = a +b > 0 Therefore, h < b , and we have a < h < b . Since O < a < b, then a - b < O and ab > O. Therefore, aab- b < O. So, a b <0 ab - ab 1- -1 < 0 b a -b1 < a1 Now, since b > 0 , then i > 0 , so we have 0 < [;1 < -;1 . _
121.
--
-
64
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1.6: Equations and Inequalities Involving Absolute Value
x2 � 0 x2 + 1 � 0 + 1 x2 + 1 � 1
, we have
123.
Since
1 25.
Therefore, the expression less than -5 . Answers will vary.
1 5. 1 -2 1 x = 4 2x = 4 x=2 x2 + 1
The solution set is {2} .
can never be 1 7.
;
27 x=2
27 x = -2 27 27 The so 1utIon · set 1S
Section 1.6
1. 1 - 2 1 = 2
7.
19.
True or 2x = - 6 or x = -3 The solution set is {-3, 3 } .
I}.
21.
}
-2'2 .
{-I, %} .
u
The solution set is
13. 1 - 2x l = 1 8 1 1 - 2x l = 8 - 2x = 8 or - 2x = - 8 x = - 4 or x=4 The solution set is {-4, 4 } .
25.
{
}
I - 2 1 = -21 No solution, since absolute value always yields a non-negative number.
23. 4 - 1 2x l = 3 - 1 2x l = - 1 1 2x l = 1 2x = 1 or 2x = -1 x = -1 or x = --1 2 2
or 1 - 4t = -5 or - 4t = -6 3 t = -l or t=2
1 - 4t = 5 -4t = 4
The solution set is
or or or or
.
=
1 1 - 4t I + 8 = 1 3 1 1 - 4t 1 = 5
{
-x + -2 = - 2 3 5 5x + 6 = - 30 5x = -36 36 x = -5 36 24 The so 1utIon · set 1S - 5 ' 5 .
1 2x l = 6
9. 1 2x + 3 1 = 5 2x + 3 5 or 2x + 3 = - 5 2x = 2 or 2x = - 8 x = 1 or x = -4 T�e solution set is {-4,
1 x1 +2� 1 = 2
-+- = 2 3 5 5x + 6 = 30 5x = 24 24 x =5
2x = 6 x=3
11.
or
.
3. {-5, 5} 5.
2 31x l = 9 2 Ixl=
1 x2 - 9 1 = 0 x2 - 9 = 0 x2 = 9 x = ±3
The solution set is
{- �, �} . {- 3, 3 } .
65
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
27.
I x2 - 2x 1 = 3
35.
or x2 - 2x = -3 x2 - 2x = 3 x2 - 2x - 3 = 0 or x2 - 2x + 3 = 0 ( x -3 ) (x + 1 ) = 0 or x = 2 +- .J42 - 1 2 x = 3 or x = -1 or x = 2 +- 2.,/-8 no real sol. The solution set is {-1, 3} . 29.
1 37.
I X2 + X - 1 1 = 1 x2 + x - 1 = 1 or x2 + x - 1 = -1 x2 + x - 2 = 0 or x2 + x = 0 ( x - 1)( x + 2) = 0 or x ( x + 1) = 0 x = 1, x = -2 or x = 0, x = -1 The solution set is {-2, - 1, 0, 1} .
31.
39.
I 3x�:=�- 2 I = 2
3x - 2 = -2 or -2x - 3 3x - 2 = 2 (2x - 3) or 3x - 2 = -2(2x - 3) 3x - 2 = 4x - 6 or 3x - 2 = -4x + 6 -x = -4 or 7x = 8 x=4 or x = -78 Neither of these values cause the denominator to equal zero, so the solution set is 4 .
I x2 + 3x I = I x2 - 2x I x2 + 3x = x2 - 2x or 3x = -2x or 5x = 0 or x=O or x=O or The solution set is
�
1
4
0
)
1
1
-4
1
0
1
(
1
4
lit ·
I X-21+2 <3 I x-2l d -1 < x-2 < 1 l < x< 3 {x l l < x < 3} or (1,3) �
1 ..
3
I"
E
1
0
41.
I·
�
1
-4
1 3x I > 12 3x < -12 or 3x > 12 x < - 4 or x > 4 {x l x < -4 or x > 4} or (-00, -4) u( 4, (0 ) 11 1
-2x - 3 = 2
33.
1 2x l < 8 -8 < 2x < 8 -4 < x < 4 {x l - 4 < x < 4} or (-4,4)
3
1
1 3t - 2 1 � 4 -4 � 3t - 2 � 4 - 2 � 3t � 6 --23 � t � 2
{t l -� � t � 2} or [-�, 2]
{�, }
1
EI
1
0
_1 3
x2 + 3x = - ( x2 - 2x ) x2 + 3x = _x2 + 2x 2X2 + x = 0 x(2x + 1) = 0 x = O or x = --21
43.
{ - �, o} .
2
1 2x - 3 1 � 2 2x - 3 � -2 or 2x - 3 � 2 2x � 1 or 2x � 5 x -< -21 or x >- -52 o
I
"2
5
"2
66
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1.6: Equations and Inequalities Involving Absolute Value
45.
(
,
€
.
�3
0
-1
47
( 2
'
"
57 .
{l % , ... ,
I)' 3 2
, � ..
5 . 9
(, 3 2
, ..
15xl � -1 Absolute value yields a non-negative number, so this inequality is true for all real numbers, (-00, 00). '
"
,-,-
o
1 \ -�I2x +< 31 2 +3
1 -1 < -3 --2 < 1 6( _ 1) < 6 2 + 3 - < 6(1) -6 < 2 (2x + 3) - 3 < 6 -6 < 4x + 6 - 3 < 6 -6 < 4x + 3 < 6 -9 < 4x < 3 --94 < x < -43
( \ �)
{x l - � < x < �} or (-�,�) , (, 9 4
%} or ( -oo, -%)u(%, oo) 0
]
2
0
12xl < -1 This is impossible since absolute value always yields a non-negative number. No solution.
, ... ,
1 - 2x I > 1 -3 1 1 2x l > 3 2x < -3 or 2x > 3 x < - 23 or x > 23 x x < - or x >
-1
o
,-
o
[ ,
,
, , , , , , , , , , ,-
1 -4xl + I -5 1 � 1 1 -4xl + 5 � 1 1 -4xl � -4 This is impossible since absolute value always yields a non-negative number. The inequality has no solution. , ,
5 . 1
55.
, ..
1 1 - 2x I > 3 1 - 2x < -3 or 1 - 2x > 3 -2x < -4 or - 2x > 2 x > 2 or x < -1 {x l x < -l or x > 2} or ( -00, -I) u (2, 00) ,
- 1 2x- l l � -3 1 2x- 11 � 3 -3 � 2x- l � 3 - 2 � 2x� 4 -1 � x � 2 { x 1 -1 � x ::; 2} or [-1, 2] ,
2
, .. , , � , - I 0
4 . 9
53 .
1 1 - 4x l - 7 < -2 1 1 - 4x l < 5 -5 < 1 - 4x < 5 -6 < -4x < 4 -6 > x > 4 -4 -4 -32 > x > -1 or - 1 < x < -32 { x 1 - 1 < x < 1} or -1, 1 )
61.
, �, -
o
,) , 3
,-
4
5 + l x - l l > "21
9 I x - 1 1 > - "2 Absolute value yields a non-negative number, so this inequality is true for all real numbers, (-00, 00). , ... ,
o
67
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in wri ting from the publisher.
Chapter 1: Equations and Inequalities
63.
65.
67.
69.
71.
73.
A temperature x that differs from 98.6° F by at least 1 .5°F . I x - 98.6° I � 1 .5° x - 98.6° S; -1 .5° or x - 98.6° � 1 .5° x S; 97. l° or x � 100. 1° The temperatures that are considered unhealthy are those that are less than 97. 1 of or greater than 100 . 1 "F, inclusive. The true average number of books read x should differ from 13.4 by less than 1 .35 books. I x - 13.4 1 < 1 .35 -1 .35 < x - 13.4 < 1 .35 12.05 < x < 14.75 Gallup is 99% confident that the actual average number of books read per year is between 12.05 and 14.75 books. x differs from 3 by less than I X -3 1 < 21 --21 < x - 3 < -21 -25 < x < -72
75.
I x - 21 S; 7 -7 S; x - 2 S; 7 -5 S; x S; 9 -15 S; x - l0 S; -1 1 1 -1 5 � -x - 10 � -1 1 -1 S; x -1 10 S; -15 --
a = -1 , b = -� 15 77.
�.
79.
x differs from -3 by more than 2. I x - (-3) 1 > 2 I x+3 1 > 2 x + 3 < - 2 or x + 3 > 2 x < -5 or x > -1 {x l x < -5 0r x > -I} I x - 11 < 3 -3 < x - 1 < 3 -3 + 5 « x - l) + 5 < 3 + 5 2 < x+4 < 8 a = 2, b = 8 I x + 41 S; 2 -2 S; x + 4 S; 2 -6 S; x S; -2 -12 S; 2x S; -4 -15 S; 2x - 3 S; -7 a = -15, b = -7
Given that a > 0, b > 0, and -.Ia <.Jb , show that a < b. Note that b - a = ( .Jb + -.Ia)( .Jb - -.Ia) . Since -.Ia < .Jb means .Jb - -.Ia > 0 , we have b - a = ( .Jb + -.Ia) ( .Jb - -.Ia) > ° . Therefore, b - a > ° which means a < b. Prove l a + b l S; l a l + I b l · Note that l a + bl 2 = l a + b l · l a + bl · Case 1 : If a + b � 0, then l a + b l = a + b, so l a + b l · l a + b l = (a + b)(a + b) = a2 + 2ab + b2 s; l a l 2 + 2 I a l · l bl + l bI 2 = ( I a l + I b l) 2 by problem 78 Thus, (l a + b l) 2 S; (I a l + I bl )2 l a + bl S; l a l + I b l · Case 2: If a + b < O, then l a + b l = -(a + b), so l a + b l · l a + b l = ( -( a + b ) )( -(a + b) ) = (a + b) (a + b) = a2 + 2ab + b2 S; l a l 2 + 2 l a l ·l bl + I bl 2 = (I a l + I bl )2 by problem 78 Thus, ( I a + b\) 2 S; ( I a l + I bl f l a + b l S; l a l + I bl
68
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1. 7: Problem Solving: Interest, Mixture, Uniform Motion, and Constant Rate Job Applications
81.
Given that a > 0, x2 < a x2 - a < 0
91.
If x < -Fa , then x + Fa < 0 and x - Fa < -2Fa < o . Therefore, ( x + Fa ) ( x - Fa ) > 0 , which is a contradiction. If -Fa < x < Fa , then 0 < x + Fa < 2Fa and -2Fa < x - Fa < 0 . Therefore, ( x + Fa )( x - Fa ) < 0 . If x > Fa , then x + Fa > 2 Fa > 0 and x - Fa > 0 . Therefore, ( x + Fa )( x - Fa » O , which is a contradiction. So the solution set for x2 < a is { xl - Fa < x < Fa } . 83.
85.
x2 < 1 -Ji < x < Ji -1 < x < 1 The solution set is {x l - l < x < I} .
93 - 95.
89.
Answers will vary.
Section 1.7
x2 � 9 x � -.J9 or x � .J9 x � 3 or x � 3 The solution set is {xi x � -3 or x � 3} .
1. 3.
-
87.
1 3x - 12x + 11 1 = 4 3x - 12x + ll = 4 or 3x -12x + l l = -4 3x -12x + l l = 4 3x - 4 = 12x + 11 2x + 1 = 3x - 4 or 2x + 1 = -(3x -4) -x = -5 or 2x + 1 = -3x + 4 or 5x = 3 x=5 x=5 or x = -35 or 3x -12x + l l = -4 3x + 4 = 12x + ll 2x + 1 = 3x + 4 or 2x + 1 = -(3x + 4) -x = 3 or 2x + 1 = -3x - 4 x = -3 or 5x = -5 x = -3 or x = -1 The only values that check in the original equation are x = 5 and x = -1 . The solution set is {-I, 5} .
5.
x2 � 1 6 -.J16 � x � .J16 -4 � x � 4 The solution set is {xl - 4 � x � 4} .
7.
mathematical modeling uniform motion True; this is the uniform motion formula. Let A represent the area of the circle and the radius. The area of a circle is the product of times the square of the radius: A = r
1[
1[ r 2
9.
x2 > 4 x < -14 or x > 14 x < -2 or x > 2 The solution set is {xi x < -2 or x > 2} .
11.
13.
Let A represent the area of the square and the length of a side. The area of the square is the square of the length of a side: A = S2 Let F represent the force, the mass, and a the acceleration. Force equals the product of the mass times the acceleration: F = rna Let W represent the work, F the force, and d the distance. Work equals force times distance: s
rn
W = Fd
69
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
1 5.
1 7.
C = total variable cost in dollars, x = number of dishwashers manufactured: C = 1 50x Let x represent the amount of money invested in bonds. Then 50, 000 - x represents the amount of money invested in CD's. Since the total interest is to be $6,000, we have: 0. 15x + 0.07(50, 000 - x) 6, 000 (100)(0. I5x + 0.07(50, 000 - x)) = (6, 000)(100) 15x + 7(50, 000 - x) = 600, 000 1 5x + 350, 000 - 7x = 600, 000 8x + 350, 000 = 600, 000 8x = 250,000 x = 3 1, 250 $3 1,250 should be invested in bonds at 15% and $ 1 8,750 should be invested in CD's at 7%. Let x represent the amount of money loaned at 8%. Then 12, 000 - x represents the amount of money loaned at 1 8%. Since the total interest is to be $1,000, we have: 0.08x + 0. 1 8(12, 000 - x) = 1, 000 (1 00) (0.08x + 0. 18(12,000 - x)) = (1, 000)(100) 8x + 1 8(12, 000 - x) 100, 000 8x + 216, 000 - I 8x = 100, 000 -IOx + 216, 000 = 100, 000 -lOx = -1 16, 000 x = 1 1, 600 $ 1 1,600 is loaned at 8% and $400 is at 1 8%. Let x represent the number of pounds of Earl Gray tea. Then 100 - x represents the number of pounds of Orange Pekoe tea. 5x + 3(100 - x) = 4.50(100) 5x + 300 - 3x = 450 2x + 300 = 450 2x = 1 50 x = 75 75 pounds of Earl Gray tea must be blended with 25 pounds of Orange Pekoe.
23.
=
19.
25.
_
-
-
=
21.
Let x represent the number of pounds of cashews. Then x + 60 represents the number of pounds in the mixture. 9x + 3.50(60) = 7 .50(x + 60) 9x + 210 = 7 .50x + 450 1.5x = 240 x = 160 160 pounds of cashews must be added to the 60 pounds of almonds. Let r represent the speed of the current. Rate Time Distance 1 6-r I Upstream 1 6 - r 20 3 60 - 3 Downstream 16 + r �� = t 1 6+4-r Since the distance is the same in each direction: 1 6 - r = -16 + r -4 3 4(16 - r) = 3(16 + r) 64 - 4r = 48 + 3r 16 = 7r 16 � 2 .286 r=7 The speed of the current is approximately 2.286 miles per hour. Let r represent the speed of the current. Rate Time Distance Upstream l 5 - r 1 510- r 10 Downstream l 5 + r 1510+ r 10 Since the total time is 1 .5 hours, we have: � 15-r +� 15 + r = 1 .5 10(15 + r) + 1 0(15 - r) = 1 .5(15 - r)(15 + r) 150 + l Or + 1 50 - l Or = 1 .5(225 - r2 ) 300 = 1 .5(225 - r2 ) 200 = 225 - r2 r2 - 25 = 0 (r - 5)(r + 5) = 0 r = 5 or r = -5 Speed must be positive, so disregard r = -5 . The speed of the current is 5 miles per hour.
27.
--
--
70
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1. 7: Problem Solving: Interest, Mixture, Uniform Motion, and Constant Rate Job Applications
29.
Let r represent Karen's normal walking speed. Rate Time Distance With walkway r + 2.5 r +502.5 50 Against walkway r - 2.5 r -502.5 50 Since the total time is 40 seconds: � r + 2.5 + � r - 2.5 = 40 50(r - 2.5) + 50(r + 2.5) = 40(r - 2.5)(r + 2.5) 50r - 125 + 50r + 125 = 40(r2 - 6.25) 100r = 40r2 - 250 0 = 40r2 - 1 00r - 250 0 = 4r2 - l Or - 25 - 4(4)(-25) r = -(-10) ± �(-10)2 2(4) 10 ± v'5OO 10 ± lO.J5 5 ± 5.J5 8 8 4 r "" 4.05 or r "" -1.55 Speed must be positive, so disregard r "" -1 .55 . Karen' normal walking speed is approximately 4.05 feet per second. Let w represent the width of a regulation doubles tennis court. Then 2w+ 6 represents the length. The area is 2808 square feet: w(2w + 6) = 2808 2w2 + 6w = 2808 2W2 + 6w- 2808 = 0 w2 + 3w-1404 = 0 (w+ 39)(w - 36) = 0 w+ 39 = 0 or w - 36 = 0 w = 36 w = -39 or The width must be positive, so disregard w = -39 . The width of a regulation doubles tennis court is 36 feet and the length is 2(36) + 6 78 feet.
33.
I I tI
--
--
-,-_-'...-!"":",--=,,--,--,-,_,-,:,,, .-
31.
Let t represent the time it takes to do the job together. of job done Time to do job Part in one minute Trent 30 30 20 Lois 20 Together t 1 1 1 30 + 20 = ( 2t + 3t = 60 5t = 60 t = 12 Working together, the job can be done in 12 minutes. 1 = length of the garden w = width of the garden The length of the garden is to be twice its width. Thus, 1 = 2w . The dimensions of the fence are 1 + 4 and w+4 . The perimeter is 46 feet, so: 2(/ + 4) + 2( w + 4) = 46 2(2w+ 4) + 2(w + 4) = 46 4w + 8 + 2w + 8 = 46 6w + 16 = 46 6w = 30 w=5 The dimensions of the garden are 5 feet by 10 feet. b. Area = 1 . w = 5 . 10 = 50 square feet If the dimensions of the garden are the same, then the length and width of the fence are also the same (I + 4) . The perimeter is 46 feet, so: 2(/ + 4) + 2(/ + 4) = 46 21 + 8 + 21 + 8 = 46 4/ + 16 = 46 41 = 30 1 = 7.5 The dimensions of the garden are 7.5 feet by 7.5 feet. d. Area = 1 · w = 7.5(7.5) = 56.25 square feet.
35.
a.
c.
=
71
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
37.
Let t represent the time it takes for the defensive back to catch the tight end. Time to Time Rate Distance 1 00 yards Tight 12 sec 1 00 25 25 t t 12 3 Ena -3 Def. t 11000 = 10 lOt 10 sec Back Since the defensive back has to 5 yards farther, we have: 25 3 t + 5 = 10t 25t + 1 5 = 30t 15 = 5 t t = 3 -)0 10t = 30 The defensive back will catch the tight end at the 45 yard line (15 + 30 45). Let x represent the number of gallons of pure water. Then x + 1 represents the number of gallons in the 60% solution. (%)(gallons) + (%)(gallons) = (%)(gallons) O(x) + 1(1) = 0.60(x + 1) 1 = 0.6x + 0.6 0.4 = 0.6x x = -46 = -23 2 "3 gallon of pure water should be added.
45.
run
_
run
47.
=
39.
41.
43.
_
_
Let t represent the time the auxiliary pump needs to run. Since the two pumps are emptying one tanker, we have: i+i = 1 4 9 27 + 4t = 36 4t = 9 t = = 2.25 The auxiliary pump must run for 2.25 hours. It must be started at 9:45 a.m. Let t represent the time for the tub to fill with the faucets on and the stopper removed. Since one tub is being filled, we have: /5 + - ;0 = 1 4t - 3t = 60 t = 60 60 minutes is required to fill the tub. Let t represent the time spent ing. Then 5 - t represents the time spent biking. Rate Time Distance Run 6 t 6t Bike 25 5 - t 25(5 - t) The total distance is 87 miles: 6t + 25(5 - t) = 87 6t + 125 - 25t = 87 -19t + 125 = 87 -19t = -38 t=2 The time spent ing is 2 hours, so the distance of the run is 6(2) = 12 miles. The distance of the bicycle race is 25(5 - 2) = 75 miles.
�
49.
( )
Let x represent the number of ounces of water to be evaporated; the amount of salt remains the same. Therefore, we get 0.04(32) = 0.06(32 - x) 1 .28 = 1 .92 - 0.06x 0.06x = 0.64 64 32 x - 0.64 0.06 - 6 - 3 - 101.3 10 1 "" 1 0.67 ounces of water need to be evaporated. Let x represent the number of grams of pure gold. Then 60 - x represents the number of grams of 12 karat gold to be used. x + � (60 - x) = � (60) x + 30 - 0.5x = 40 0.5x = 10 x = 20 20 grams of pure gold should be mixed with 40 grams of 12 karat gold. _
Let t represent the time it takes for Mike to catch up with Dan. Since the distances are the same, we have: 1 1 "6 t = "9 (t + l) 3t = 2t + 2 t=2 Mike will pass Dan after 2 minutes, which is a distance of � mile.
51.
-
runn
runn
72
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1 Review Exercises
53.
55.
\020 meters/sec. In 9.99 seconds, Burke will run \020 (9.99) = 83.25 meters.
3.
Burke's rate is
Lewis would win by 1 6.75 meters. Let x be the original selling price of the shirt. Profit = Revenue - Cost
5.
4 = x - 0.40x - 20 � 24 = 0.60x � x = 40
The original price should be $40 to ensure a profit of $4 after the sale. If the sale is 50% off, the profit is:
40 - 0.50(40) - 20 = 40 - 20 - 20 = 0 At 50% off there will be no profit. 57.
7.
=
9.
(not possible) The time traveled with the tail wind
= 9 1 9 � 1 .6709 1 hours . 550 Since they were 20 minutes
(t hOur ) early, the
11.
time in still air would have been: 1 .6709 1 hrs + 20 min = (1 .6709 1 + 0.33333) hrs 2.00424 hrs Thus, with no wind, the ground speed is ��9 458.53 . Therefore, the tail wind is 2 . 424 550 - 458.53 = 9 1 .47 knots . �
�(x - ±) = % - � (12) ( �) ( x - ± ) = ( % -� } 12)
The solution set is {V} . 13.
Chapter 1 Review Exercises
�
2- =8
The solution set is
x(l - x) = 6 x - x2 = 6 o = x2 - x + 6 b2 - 4ac = (_1) 2 - 4 (1) (6) = 1 - 24 = -23
6x - 2 = 9 - 2x 8x = 1 1 11 X=8
�
6 - x = 24 x = -18
x = -6 -x-I 5 5x = 6x - 6 6=x Since x 6 does not cause a denominator to equal zero, the solution set is {6} .
Therefore, there are no real solutions.
t
1.
H}.
=
0.25x + 9.6 = 10.6 + 0.58x -0.33x = 1 x � -3 .03 liters
59.
3x - -x = 1 4 3 12 9x - 4x = 1 5x = 1 X = -1 5
The solution set is
It is impossible to mix two solutions with a lower concentration and end up with a new solution with a higher concentration. Algebraic Solution: Let x the number of liters of 25% solution. (% ) ( liters ) + (% ) ( liters ) = (% ) ( liters ) 0.25x + 0.48 ( 20) = 0.58( 20 + x)
was:
- 2(5 - 3x) + 8 = 4 + 5x - 1 0 + 6x + 8 = 4 + 5x 6x - 2 = 4 + 5x x=6 The solution set is {6} .
(x - 1)(2x + 3) = 3 2X2 + x - 3 = 3 2x2 + x - 6 = 0 (2x - 3)(x + 2) = 0 X = -3 2
or x = - 2
The solution set is
{- 1 8} . 73
{-2,%} .
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
1 5.
2x+ 3 = 4x2 0 = 4x2 -2x-3 x -(-2) ± �(-2)2 = 2 ± ,J52 = 2 ± 2.Jlj = 1 ± .Jlj
25.
- 4(4)(-3)
2(4)
8
. The solutIOn set .
IS
17.
�X2 - 1 = 2 (�x2 _ d = (2)3 x2 -1 = 8 x2 = 9 x = ±3 Check x = -3 : �(_3)2 -1 = 2 :vB = 2 2=2
The solution set is 19.
{
8
4
1 - .Jl3 1 + .Jl3
}
-- ' -- . 4 4
27.
Check x = 3 :
�(3) 2 -1 = 2 :vB = 2
{-3,3} .
23 .
.Jx+l +.Jx-l = .J2x + l (.JX+l +.JX -I/ = ( .J2X + l t x + 1 + 2.Jx+l.Jx-I + x -1 2x + 1 2x + 2.Jx+1.Jx-l = 2x + 1 2.Jx+l.Jx- l = 1 (2.Jx+l.Jx- l) 2 = (1)2 4(x+ l)(x- l) = 1 4x2 -4 = 1 4x2 = 5 x2 = -45 x = -+ .J52 .J5 : Check x = 2 =
2=2
x(x+l)+2 = 0 x2 +x+2 = 0 x -1 ±�(1)2(1)2 - 4(1)(2) = -l ±2H
No real solutions. 21.
�2x+3 = 2 (�2x +3r = 24 2x +3 = 1 6 2x = 1 3 X = -132 13 : Check x = 2 4 2 ( 1� ) + 3 = �13 + 3 = � = 2 The solution set is { 1�} .
X4 -5x2 + 4 = 0 ( x2 -4 )( x 2 -1 ) = 0 x2 -4 = 0 or x2 - 1 = 0 x = ±2 or x = ±1 The solution set is {-2, -1, 1, 2} . .J2x-3 +x = 3 .J2x-3 = 3 - x 2x -3 = 9 - 6x + x2 x2 - 8x + 12 = 0 (x-2)(x-6) = 0 x = 2 or x = 6 Check x = 2: .j2(2) -3 + 2 = Jl + 2 = 3 Check x = 6: .j2(6) - 3 + 6 = .J9 +6 = 9 ,* 3 The solution set is {2} .
J1 +1 + J1 - 1 "N�) + I =
1 .79890743995 1 .79890743995
.J5 .' Check x = -2
�-� +I + �- � - I "N- �) + I
'
The second solution is not possible because it makes the radicand negative. The solution set is
{�} .
74
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1 Review Exercises
29.
The solution set is 31.
m or X = -m x = -l+n 1-n The solution set is {�, 1 + n n "# 1, n "# -1. 1 - n �},
2 X l/2 - 3 = 0 2Xl/2 = 3 ( 2X I /2 ) 2 = 32 4x = 9 x = '49 Check x = '49 : ( )1 () 2 '49 1 2 - 3 = 2 23 - 3 = 3 - 3 = 0
35.
{%} .
x-6 - 7x-3 - 8 = 0 Let u = x-3 so that u2 = x-6• u2 - 7u - 8 = 0 (u -8)(u + 1) = 0 or u = -1 u=8 or x-3 = -1 x-3 = 8 ( x-3 t 1 3 = (8rl1 3 or ( x -3 t l 3 = ( -1 r l/3 or x = -1 X = -21
3 7.
) () Check 21 : ( 2l � - 7 21 -3 - 8 = 64 - 56 - 8 = 0 Check - 1 : (-1 t - 7 ( -1 r3 - 8 = 1 + 7 - 8 = 0 The solution set is { -1, �} . 33 .
lOa2x2 - 2abx - 36b2 = 0 5a2x2 - abx - 1 8b2 = 0 (5ax + 9b}(ax - 2b) = 0 or ax - 2b = 0 5ax + 9b = 0 5ax = -9b ax = 2b 2b 9b X = -X= 5a a 2b } The solution set is { - 9b 5a , a , a "# O. �x2 + 3x + 7 - �x2 - 3x + 9 + 2 = 0 �x2 + 3x + 7 = �x2 - 3x + 9 - 2 ( �X 2 + 3x + 7 ) 2 = (�x2 - 3x + 9 _ 2) 2 x2 + 3x + 7 = x2 - 3x + 9 - 4 �x2 -3x + 9 + 4 6x - 6 = _4�X2 - 3x + 9 (6(x _ l) ) 2 = (-4�x2 _ 3x + 9) 2 36 ( x2 - 2x + 1 ) = 16 ( x2 - 3x + 9 ) 36x2 - 72x + 36 = 16x2 - 48x + 144 20x2 - 24x - 108 = 0 5x2 - 6x - 27 = 0 (5x + 9)( x -3) = 0 x = - 95 or x = 3 Check x = --95 ' ' -
x2 + m2 = 2mx + (nx) 2 x2 + m2 = 2mx + n2 x2 _ x2 n2x2 - 2mx + m2 = 0 ( 1 - n2 ) x2 - 2mx + m2 = 0 -m-i---4-(�1---n2--:-)-m-2 �r(--2_( -2m) -:-------±--'----, x=2 ( 1 - n2 ) 2m ± �4m2 --4m2 + 4m2n2 �-= --�� ) ( 2 1 - n2 2m ± � 2m ± 2mn 2 ( 1 - n2 ) 2 ( l - n2 ) _- 2m (l ± n) _- m (1 ± n) _- m (1 ± n) 2 ( 1 - n2 ) I - n2 (1 + n)(1 - n)
!!. _ 27 + 7 - !!. + 27 + 9 + 2 25 5 25 5 8 1 - 135 + 175 _ 81 + 135 + 225 + 2 25 25 �441 + 2 = .!.!. _ � + 2 = 0 _ = �121 25 25 5 5
75
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
x=3: �(3)2 +3(3) + 7 _ �(3)2 -3(3 ) +9 +2 = ../9 + 9 + 7 -../9 -9 + 9 + 2 = J25-J9+2 = 2+2 =4#0
Check
The solution set is 39.
41 .
47.
{-�}.
1 2x+31 = 7 2x + 3 = 7 or 2x + 3 = -7 2x = 4 or 2x = -10 x = 2 or x = -5 The solution set is {-5, 2}. 1 2-3x l +2 = 9 1 2-3x l = 7 2-3x = 7 or 2-3x = -7 -3x = 5 or -3x = -9 x = 53 or x = 3 The solution set is {-�, 3 } 2x3 = 3x2 2x3 -3x2 = 0 x2 (2x-3 ) = 0 2x-3 = 0 x = 0 or x=-23 The solution set is
45.
•
t
14 49.
--
43 .
2x-3 2 �-x --+ 5 2 2(2x -3) + 10(2) � 5x 4x- 6 +20 � 5x 14 � x x � 14 {x\ x �14} or [14,(0 )
-9 � 2x+3 -4 � 7 3 6 � 2x+3 � -28 33 � 2x � -31 33 -> x -> _ � 2 2 33 31 - -2 -< x -< -2 {x \ - 321 � x � 3;} or [_ 321 , 3;] [
33
3 .1
2
2
51 .
2 < 3 -3x <6 12 24 < 3-3x <72 21 < -3x < 69 7 x > -2 3 {xl -23 < x < -7} or (-23,-7) -
{o, %} .
>
)
-7
- 23
2x3 + 5x2 -8x -20 = 0 x2 (2x+ 5)-4(2x + 5 ) = 0 (2x+5)(x2 -4) = 0 2x + 5 = 0 or x2 -4 = 0 2x = -5 or x2 = 4 x = --25 or x = ±2 The solution set is { -%,-2,2 } .
53 .
\ 3x+4 1 < "2 -..!.. < 3x+4 <..!.. --< 3x <---< <-{xj-%<x<- H or (-%,-�) 1
2 9 2 3 2
3
-2
2
7 2 7 6
X
)
7
�
-6"
76
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1 Review Exercises 55.
1 2x - 5 1 � 9 2x - 5 � - 9 or 2x - 5 � 9 2x � 1 4 2x � - 4 or x�7 x � - 2 or { x l x � - 2 or x � 7} or (-00, - 2] u [7, 00 ) -
57.
7
2 + 1 2 - 3x l � 4 1 2 - 3x l � 2 -2 � 2 - 3x � 2 -4 � -3x � 0
75.
]
l
4
n
. 77.
or
63. 65. 67. 69.
1 fj -i 2 2
--+
2x2 + X - 2 = 0 a = 2, b 1, c - 2, b2 - 4ac = 12 - 4 (2 ) ( - 2 ) = 1 + 16 = 1 7
}
.
=
The solution set is
-I>x 79.
{ x l x < -I or x > f} or (-00, - 1) U G , oo ) 61.
fj
- 1 ± J0 - 1 ± J0 x = ---;-.,---4 2 ( 2)
3
7
.{I
=
7 x < - I or x > -
(
I fj . - I ± � - I ± fj i = -- + -/ = 2- 2 2 2 (1)
· Th e so IutlOn set IS - - - - i 2 2 '
1 - 1 2 - 3x 1 < -4 -1 2 - 3x l < -5 1 2 - 3x l > 5 2 - 3x < -5 or 2 - 3x > 5 7 < 3x or - 3 > 3x 3
=
x=
"3
-7 < x
x2 + x + I 0 a = I, b = I, c = I, b2 - 4ac = 1 2 - 4 ( 1 ) ( 1 ) 1 - 4 = -3 =
3
-I
= ( 4 + 1 2i + 9i2 ) (2 + 3i)
= (-5 + 12i) ( 2 + 3i) = - 1 0 - 1 5i + 24i + 36i2 = - 46 + 9i
i�x�o
59.
(2 + 3i)3 = (2 + 3i) 2 ( 2 + 3i)
I .
[
2
73.
I •
-'
,
x2 + 3 x x2 - x + 3 = 0 a = I, b = - I , c = 3, b2 - 4ac = (- 1 ) 2 - 4 ( 1) (3 ) = 1 - 1 2 = - 1 1 =
X=
(! . 6 r = 9
- ( - I ) ± F-11 1 ± JI1 i I JII / = = - +2 2 2 2(1)
. . The solutIOn set IS
(!'(-4)f = �
81.
(6 + 3 i) - (2 - 4i) = (6 - 2) + (3 - ( - 4)) i = 4 + 7i 4 (3 - i) + 3 ( -5 + 2i) = 12 - 4i - 15 + 6i = -3 + 2i
- 1 + J0 } { - I -4J0 , --. 4
{ 2I
--
JII 2
i,
I
JII }
-+-i . 2 2
x(J - x) = 6 -x2 + x - 6 = 0 a = - 1, b 1, c = - 6, b2 - 4ac = e - 4 ( - I ) (- 6) = 1 - 24 = - 23 =
9 - 3i 3 3-i 3 - = _.3 + i 3 + i 3 - i 9 - 3i + 3i - i2 9 - 3i 9 3 . = -- = - - - / 10 10 10
X=
- 1 ± �- 23 - 1 ± J2j i 1 + J2j . = - - -- / = 2 2 -2 2 (- 1 )
. . {I2
The solutIon set IS 77
--
J2j -2
i,
}
I J2j - + -i . 2 2
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as they currently
exi st. N o portion of th is mate r i al may be repro duced, i n any form or by any means, w i thout perm i s s i o n i n writing from the publ i sher.
Chapter 1: Equations and Inequalities
83. 85.
87.
21 + 2w P ·r·t = (9000)(0.07)(1) = $630 Using s = t , we have t = 3 and = 1 100 . Finding the distance in feet: s 1 100(3) = 3300 The storm is 3300 feet away. Let s represent the distance the plane can travel. With wind Against wind Rate 250+30 = 280 250-30 = 220 (s 1 2) (s 1 2) Time 220s 280s p J
93 .
=
=
J
v
v
s
r + 50 = 3r - l0 60 = 2r r = 30
=
89.
95.
Dist. "2 "2 Since the total time is at most 5 hours, we have:
(s I 2) + (s I 2) � 5 280 220 _s_ + _s_ � 5 560 440 l Is + 14s 5(6160) 25s � 30,800 s � 1232 �
91.
Let r represent the rate of the Metra train in miles per hour. Metra Train Amtrak Train r + 50 r Rate 1 Time 3 r + 50 Dist. 3r The Amtrak Train has traveled 10 fewer miles than the Metra Train.
The plane can travel at most 1232 miles or 616 miles one way and return 616 miles. Let t represent the time it takes the helicopter to reach the raft. Raft Helicopter Rate 5 90 Time t t 90t Dist. 5t Since the total distance is 150 miles, we have:
The Metra Train is traveling at 30 mph, and the Amtrak Train is traveling at 30 + 50 = 80 mph. Let t represent the time it takes Clarissa to complete the job by herself. Clarissa Shawna Time to do t t+5 job alone Part of job 1 1 t+5 done in 1 day t Time on job 6 6 (days) Part of job 6 done by each 6t t + 5 person Since the two people paint one house, we have: -
--
-
--
6 �+_ t t +_5 = 1 6(t + 5) + 6t = t(t + 5) 6t + 30 + 6t = t2 + 5t t2 - 7t -30 = 0 (t - 1 O)(t + 3) = 0 t = 10 or t = -3 It takes Clarissa 10 days to paint the house when
5t+90t = 150 95t = 150 t "" 1.58 hours 1 hour and 35 minutes The helicopter will reach the raft in about 1 hour and 35 minutes.
97.
""
working by herself. Let x represent the amount of water added. % salt Tot. amt. amt. of salt
10% 64 (0.10)(64) 0% x (O.OO)(x) 2% 64 +x (0.02)(64 + x) (0.10)( 64) + (0.00)( x) = (0.02)( 64 + x) 6.4 = 1.28 + 0. 0 2x 5.12 = 0.02x x = 256 256 ounces of water must be added.
78
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1 Review Exercises
99.
Let the length of leg 1 = x. Then the length of leg 2 17 - x. By the Pythagorean Theorem we have
1 05.
=
101.
1 03 .
x2 +(17 - X/ = (13) 2 x2 + x2 -34x + 289 = 169 2X2 -34x+ 120 = 0 x2 -17x + 60 = 0 (x -12)(x- 5) = 0 x = 12 or x = 5 The legs are 5 centimeters and 12 centimeters long. Let x represent the amount of the 15% solution added. % acid tot. amt. amt. of acid 40% 60 (0.4 0)( 60) 15% x (0.15)(x) 25% 60+ x (0.25)( 60+ x) (0.40)( 60) +(0.15)( x) = (0.25)( 60+x) 24 + 0.15x = 15 + 0.25x 9 = O . lx x = 90 90 cubic centimeters of the 15% solution must be added, producing 150 cubic centimeters of the 25% solution. Let t represent the time it takes the smaller
1 07.
pump to finish filling the tank. 3hp Pump 8hp Pump Time to do 8 12 job alone Part of job 1 1 12 8 done in 1 hr Time on job t + 4 4 (hrs) Part of job 4 +4 done by each t 12 8 pump Since the two pumps fill one tank, we have:
-
-
Let x represent the amount Scott receives. Then i x represents the amount Alice receives and 1 ' " receIves. The '2 x represents the amount TnCIa total amount is $900,000, so we have:
x+ '43 x + '21 x = 900,000 4 ( x+ i x+ � x ) = 4(900,000) 4x+3x + 2x = 3,600,000 9x = 3,600,000 x = 400,000 3 3 So, '4 x = '4 ( 400,000) = 300,000 and 1 1 '2 x = '2 ( 400,000) = 200,000 . Scott receives $400,000, Alice receives $300,000, and Tricia receives $200,000 . Let t represent the time it takes the older
machine to complete the job by itself. Old copier New copier Time to do t t-I job alone Part of job 1 1 t-1 t done in 1 hr Time on job 1.2 1.2 (hrs) Part of job 1.2 1 .2 done by each t t-l copier Since the two copiers complete one job, we have:
-
-
-
-
.!.2+� t t -l = 1 1.2(t -1) + 1.2t = t(t -1) 1 . 2t -1.2 + 1 . 2t = t2 -t t2 -3. 4 t +1 . 2 = 0 5t2 -17t + 6 = 0 (5t -2)(t - 3) = 0 t = 0 .4 or t 3 It takes the old copier 3 hours to do the job by itself. (0 . 4 hour is impossible since together it takes 1.2 hours.)
--
t+4 +± = 1 12 8 t+4 = -1 -12 2 t+4 = 6 t=2
=
It takes the small pump a total of 2 more hours to fill the tank. 79
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1: Equations and Inequalities
1 09.
The effective speed of the train (i.e., relative to the man) is 30 - 4 26 miles per hour. The time . 5 sec - 5 . _- _5 _ hr - 1 hr. 3600 720 60
4.
=
IS
mm
_ _ _
= 1 = 26 ( _ 720 ) 26 rmles . . = 720 26 . 5280 ", 190.67 feet = 720 The freight train is about 190.67 feet long. s
vt
C h apter 1 Test 1.
2x x 5 3 2 12 x 12 e3 -�) = 12 C52 ) 8x-6x = 5 2x = 5 X = -52 The solution set is
2.
3.
�2x-5 + 2 = 4 �2x- 5 = 2 (�2x -5 t = (2)2 2x-5 = 4 2x = 9 x = -92 Check: �2 ( � ) - 5 + 2 = 4 �9- 5 + 2 = 4 14 + 2 = 4 2+2=4 4=4 The solution set is
{�}.
5· 12x-31 + 7 = 10 1 2x-31 = 3 2x-3 = 3 or 2x-3 = -3 2x = 6 or 2x = 0 x = 3 or x = 0 The solutions set is {O, 3}. 3x3 + 2X2 -12x -8 = 0 6. x2 (3x + 2) - 4 (3x + 2) = 0 ( x2 - 4) ( 3x + 2) = 0 (x + 2)( x - 2)( 3x + 2) = 0 x + 2 = 0 or x - 2 = 0 or 3x 2 0 x = -2 or x = 2 or X = --32 The solution set is {-2,- � , 2 } .
{%}.
x(x-1) = 6 x2 - x = 6 x2 -x-6 = 0 (x-3)(x + 2) = 0 x -3 = 0 or x + 2 = 0 x = 3 or x = -2 The solution set is {-2, 3}. x4 -3x2 _4 = 0 (X2 -4)( x2 + 1) = 0 x2 -4 = 0 or x2 + 1 = 0 x2 = 4 or x2 = -1 x = ±2 or Not real The solution set is {-2, 2}.
+
=
7. 3x2 - x+l = 0 _-1)-2-- 4-(3-)(-1) X = ( -1) ± �r-(2(3) = 1 ± .J=li 6 (N t rea1) _
0
This equation has no real solutions.
80
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publi sher.
Chapter 1 Test
8.
-3 � 3x2- 4 � 6 2(-3) � 2 eX; 4 ) � 2(6) -6 � 3x - 4 � 12 -2 � 3x � 16 16 --23 -< x <- 3
13.
{Xl - �3 -< x -< 3 } or [-�3 ' 3 J �
[
�
]
Let x represent the amount of the $8-per-pound coffee. Arnt. of coffee Price Total $ (pounds) ($) 4 (20)( 4) 20 x 8 (8)(x) 20 + x 5 (5)(20 + x) 80 + 8x = (5)(20 + x) 80 + 8x = 100 + 5x 3x = 20 x = 203 = 6 1. Add 61- pounds of $8/lb coffee to get 261pounds of $5/lb coffee. 3
9 · 13x + 41 < 8 -8 < 3x + 4 < 8 -12 < 3x < 4 -4 < x < -43
{XI -4 < x <�} or (-4, �)
10.
3"
2 + 12x - 51 � 9 12x - 51 � 7 2x - 5 � -7 or 2x - 5 � 7 2x � -2 or 2x � 12 x � -l or x�6 {x l x � -1 or x � 6} or (-oo, -I] u [6, oo). I
-} 11.
12.
.
)
4
-4
[
H
6
-2 -2 . _-6 - 2i 3+i -6 - 2i 3-i =_ 3 - i 3 + i 9 + 3i - 3i - i2 9 - (-1) -3 - i = --3 --1 1. = -610- 2i = -5 5 5
--
4x2 - 4x + 5 = 0 ( �r-(------x = -4) ± 2(-44))2 - 4( 4) (5) 4 ± 8i .!. = 4 ± H4 8 = 8 = 2 ±i . 1 - 1,. 1 + 1 . Th· soIutlOn · set "2 "2 _
IS
IS
{
.}
81
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 2 Graphs Section 2 . 1 1.
5.
7. 9.
11.
1 5.
0 abscissa ( or x-coordinate); ordinate (or y coordinate) midpoint False; points that lie in Quadrant IV will have a positive x-coordinate and a negative y-coordinate. The point (-1 , 4) lies in Quadrant II. (a) (b) (c) (d) (e) (f)
19.
Quadrant II x-axis Quadrant III Quadrant I y-axis Quadrant IV y 6 A
= (-3, 2) •
D
=
(6 , 5)
21.
•
B = (6, 0) x
-6
c=
1 7.
(-2, -2) . E = (0, -3)
F
•
= (6, -3)
23.
-6 13.
The points will be on a vertical line that is two units to the right of the y-axis. y
5
25.
x
(2, -3) -5
82
d ( FL� ) = �(2 _0)2 + (1 - 0)2 = �h2 + 12 = .J4+i = .[5 d (� ,� ) = �(-2 -1)2 + (2-1)2 = �(_3)2 + 12 = J9+i = J1O d (� , �) = �(5 -3i +(4_(_4))2 = �22 +(8) 2 = .J4 + 64 = 168 = 2.J17 d (� ,P2 ) = �(6_(_3»)2 +(0 - 2)2 = �92 + (_2)2 = .J81+4 = J85 d(� ,� ) = �(6-4)2 +(4 _(_3»)2 = )22 + 72 = .J4+ 49 = .Js3 d(� , P2 ) = �(2.3 _(_ 0.2)2 + (1.1-0.3)2 = �(2 . 5)2 + (0.8)2 = .J6.25 + 0.64 = .J6.89 "" 2 . 62
Section 2. 1 : The Distance and Midpoint Formulas
2 7.
29.
d (�,P2 ) = �(0-a)2 + (0-b)2 = �(_a)2 + (_b)2 = .Ja2 +b2 A = (-2,5), B = (1,3), C = (-1, 0) d ( A, B) = �(1-(-2) )2 + (3 _ 5)2 = �32 + (_2)2 = J9+4 = JU d (B ,C) = �( _1 _1) 2 + (0-3)2 = �(_2)2 + ( _3)2 = J4+9 = JU d (A ,C) = �(-1 - (_2))2 + (0 _ 5)2 = �12 + (_5)2 = .Jl + 25 56
31.
A = (-5,3) , B = (6, 0) , C = (5 ,5) d (A,B) = �(6-(-5))2 +(0-3)2 = �1 12 +(_3)2 = .J121;9 = .JGo d ( B,C) = �(5 - 6) 2 +(5 - 0)2 = �(_1)2 + 52 = J1+25 = 56 d(A,C) = �(5 _ (_5))2 +(5-3)2 = .J102 + 22 = .J100 + 4 = .J104 = 256 y
6
C
=
(5, 5)
=
x
B
x
=
(6, 0)
--6
Verifying that D. ABC is a right triangle by the Pythagorean Theorem:
-5
[ d (A, C) f + [ d (B,C) f = [ d (A,B)]2 (.J104f + (56f = ( .JGof 104 + 26 = 130 130 = 130 The area of a triangle is A = � bh . In this
Verifying that D. ABC is a right triangle by the Pythagorean Theorem:
[ d (A,B) t + [ d ( B,C)]2 = [d ( A,C)]2 ( JUf + (JU)2 = (56t 13 + 13 = 26 26 = 26 The area of a triangle is A = � . bh . In this
problem,
A = � . [ d ( A,C)] . [ d (B,C)] = L.JI04 2 ·56 = l.2 . 256 . 56 = L2 2 . 26 = 26 square units
problem,
A = L2 [d(A,B)] . [ d(B,C)] = l..2 JU . JU = l.2 · 13 = 1f square units 83
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 2: Graphs
33.
A = (4,-3), B = (0,-3), C = (4, 2) d(A,B) = �(0-4) 2 +(_3 _(_3))2 = �(_4)2 + 02 = .J16+ 0 =M=4 d(B,C) = �(4-0l +(2 _(_3)) 2 = �42 + 52 = .J16+ 25 = .J4i d(A,C) = �(4-4)2 +(2 _(_3))2 = �02 + 52 = .J0+ 25 = 55 = 5
35.
5
=
=
,
37.
39.
The coordinates of the midpoint are:
41.
The coordinates of the midpoint are:
43.
The coordinates of the midpoint are:
(4, 2) x
-5
(X, Y) = ( Xl +2 X2 ' Yl +2 Y2 ) = ( -3 +6 2+0 2 ' 2 )
The coordinates of the midpoint are:
=(%,�) =(%,1)
5 =
(� % )
= (4,0)
y
c
(X, Y) = ( Xl +2 X2 ' Yl +2 Y2 ) (3 +2 ' -4+42 )
The coordinates of the midpoint are:
A = (4, -3)
Verifying that !l ABC is a right triangle by the Pythagorean Theorem:
[d(A,B) f + [ d(A,C) f = [d(B,C) f 42 + 5 2 = ( .J4i )2 16+ 25 = 41 41 = 41 The area of a triangle is A = tbh . In this
problem,
A = 2"1 [d(A,B)] . [d(A, C)] = � 4 . 5 = 10 square units .
(x, Y) = ( Xl +2 X2 ' Yl +2 Y2 ) = ( 4 +2 6 ' -3+2 1 ) = C� , -22 ) = (5, -1)
(X, Y) = ( Xl +2 X2 ' Yl +2 Y2 ) = ( -0 . 22+ 2.3 ' 0.3 2+1.1 ) = (�2 ' l±) 2 = (1.05, 0.7) (X, Y) = ( Xl +2 X2 ' Yl +2 Y2 ) b+O = (� 2 ' 2 )
84
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 2. 1 : The Distance and Midpoint Formulas
45.
(2, y)
that are a Consider points of the form distance of units from the point
(-2, -1) . 5 d = �(X -xl ) 2 +( Y - Yll 2 2 = �( _2_2)2 +( y)2 = �(_4)2 +( -1-y)2 =�1 6 +1+2y+ y2 =�/ +2y+17 5 =�/ +2y+17 52 =(�y2 +2Y +17r 25 = / +2y+17 0 =/ +2y-8 0=(y+4)(y-2) y+4 = 0 or y-2 = 0 y = -4 y = 2 Thus, the points (2, -4) and (2,2) are a distance of 5 units from the point (-2, -1) . Points on the x-axis have a y-coordinate of O. Thus, we consider points of the form (x, 0) that are a distance of 5 units from the point (4, -3) . d = �(X -xl ) 2 + ( Y - Yl ) 2 2 2 = �(4_x)2 +( _3_0)2 = �1 6 -8x+x2 +( _3)2 =�1 6 -8x+x2 +9 = �X2 -8x+25 5 = �X2 -8x + 25 52 = Ux2 -8x+25 r 25 = X2 -8x + 25 0 = x2 -8x 0 =x(x-8) x = 0 or x-8 = 0 x=8 Thus, the points (0,0) and (8,0) are on the x-axis and a distance of 5 units from the point (4, -3) .
49.
-1-
47.
X + Xl +-'- ( X, Y ) - (2 2 Yl 2-Y2 ) F: =(xpYl ) = (-3, 6 ) and (x,y) = (- 1,4) , so X = Xl +X2 2 and Y = Yl +2 Y2 -1 = -3+x2 2 4 = 6 +2Y2 8 = 6 + Y2 -2 = -3+X2 2 = Y2 1 = x2 Thus, P2 = (1, 2) The midpoint of AB is: 0600 D = ( ; , ; ) = (3, 0) The midpoint of AC is: 0404 E = ( ; , ; ) = ( 2, 2) M
.
.
51.
The midpoint ofBC is:
= ( 6 ; 4 , �) = (5, 2) d(C, D ) = �(0_4) 2 +(3-4)2 = �(_4)2 +(_1)2 = .J1 6 + 1 =m d( B, E) = �(2 _ 6 ) 2 +(2-0)2 = �(_4)2 +22 = .J1 6 +4 = .[20 = 2)5 (2-_-0-)2-+-(5---0-)2 d (A, F ) = �'= �22 +52 = .J4+ 25 =m F
85
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 2: Graphs
53.
Let � = ( 0,0) , �
= (s, s ) .
P2 = (O, s ) , � = ( s,O) , and
y
k----?I
(s, s)
59. x
The points F1 and P4 are endpoints of one diagonal and the points P2 and � are the endpoints of the other diagonal.
= ( O ; s , O ; s ) = (�,�) ( 0 ; s s � 0 ) (�, �) 2,3 = , =
Since d( Pz ,�) = d(�, �) , the triangle is isosceles. Since [ d( lL� ) t + [ d( Pz , � ) t = [ d(F1, P2 )] 2 the triangle is also a right triangle. Therefore, the triangle is an isosceles right triangle. Using the Pythagorean Theorem:
,
90 2 +902 = d 2 8100+8100 = d 2 16200 = d 2 d = .,)16200 = 90.,)2 "'" 127.28 feet
M 1, 4 M
The midpoints of the diagonals are the same. Therefore, the diagonals of a square intersect at their midpoints. 55.
57.
d(F1, P2 ) = �(-4- 2) 2 + (1 - 1)2 = �(-6? + 02 = 56 = 6 d(Pz,�) = �(_4_(_4) )2 + (-3 - 1) 2 = �02 + (_4)2 = .J16 = 4 d(F1,� ) = �(-4-2) 2 +(-3 -1) 2 = �(_6) 2 + (_4)2 = .,)3 6+ 16 =.J52 = 203 Since [ d( F1 ,P2 )]2 + [ d( P2 , �) ]2 = [ d( F1 , � ) ] 2 , the triangle is a right triangle.
61.
a.
First: (90, 0), Second: (90, 90) , Third: (0, 90) y
d(F1, P2 ) = �(0-(-2) / +(7 _(_1))2 = �2 2 + 82 = .,)4 + 64 = .J68 = 2JU d(Pz, � ) = �(3 _0 )2 +(2 -7) 2 = �3 2 +(_5 )2 = .J9+25 = ../34 d(F1, � ) = �(3_(_2))2 + (2 _ (_ 1))2 = �52 +32 = .,)25 +9 =../34
(0,90)
(0 , 0 )
(90,90)
(90,0)
x
b.
Using the distance formula:
c.
Using the distance formula:
d = �(310-90)2 + (15 -90)2 = �2202 + (-75)2 = .,)54025 = 5.,)2161 "'" 232 . 43 feet d = �(300- 0)2 + (300-90)2 = �3002 +2102 = .,)1 34100 = 30.,)149 "'" 366 .20 feet
86
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry
63.
The Neon heading east moves a distance 30t after t hours. The truck heading south moves a distance 40t after t hours. Their distance apart after t hours is: d
3. 5. 7.
= �(30t)2 + (40t)2 = �900t2 + 1600t2 = �2500t2 = 50t miles
9.
11.
13. 65.
a.
The shortest side is between � = (2 . 6, 1.5) and � = (2.7, 1 . 7) . The estimate for the desired intersection point is:
( Xl +X2 2 ' Yl +2 Y2 ) = ( 2.6+2g2.7)' 1.5 +2 1.7 ) (�2 2
1 5.
'
= (2.65, 1.6)
False; the y-coordinate of a point at which the graph crosses or touches the x-axis is always O. The x-coordinate of such a point is an x-intercept.
= x4 _J; 0= 04 -Fa 1 = 14 -0 0 = (_1)4 -.J=l 1 :;t: 0 0=0 0 :;t: 1-.J=l The point (0, 0) is on the graph of the equation. / = x 2 +9 32 = 02 +9 9=9 The point (0, 3) is on the graph of the equation. x2 + y2 = 4 02 + 2 2 = 4 (-2) 2 + 22 4 (fir + (fi r = 4 4=4 8 :;t: 4 4=4 (0, 2) and (J2, J2) are on the graph of the y
equation.
Using the distance formula: d
(-3,4)
=
=
b.
intercepts y-axis
= �(2 . 65 _1.4) 2 + (1 . 6 _1.3)2 = �(1.25)2 + (0.3) 2 = .J1.5625 + 0. 0 9 = .J1.6525 1 . 285 units
1 7.
y = x+ 2 X-intercept: 0 =x+2 -2 = X
y-intercept:
y = 0+ 2 Y=2 The intercepts are (-2,0) and (0,2) .
�
Y
Section 2 . 2 1.
2 ( x+3 ) - 1 = -7 2 ( x+3 ) = -6 x+3 = -3 x = -6
The solution set is
y = x + 2
x
-5
{-6} .
87
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exist. No portion of thi s material may be reproduced , in any form or by any means, without permission in writing from the publisher.
Chapter 2: Graphs
19. Y
= 2x + B y-intercept: x-intercept: y = 2(0) + B 0 = 2x + B 2x = -B y=B x = -4 The intercepts are (-4, 0) and (0, 8) .
Y 5
x
25. 2x + 3y = 6 y-intercept: x-intercepts: 2 (0) + 3y = 6 2x + 3 (0) = 6 3y = 6 2x = 6 y=2 x=3 The intercepts are (3, 0) and (0, 2) .
x
21.
Y 5
Y
= x2 - 1 y-intercept: x-intercepts: y = 02 - 1 0 = x2 - 1 y = -l x2 = 1 x = ±l The intercepts are (-1, 0) , (1, 0) , and (0, -1) . Yi
Y
=
r
-
x
1 -5
2 7 . 9X2 + 4y = 36 y-intercept: x-intercepts: 9(0) 2 + 4y = 36 9x2 + 4 (0) = 36 4y = 36 9x2 = 36 y=9 x2 = 4 x = ±2 The intercepts are (-2, 0) , (2, 0) , and (0,9) .
x
-5
23. y = -x2 + 4 y-intercepts: x-intercepts: y = _ (0) 2 + 4 ° = _x2 + 4 y=4 x2 = 4 x = ±2 The intercepts are (-2, 0) , (2, 0) , and (0,4) .
Y 10
x
88
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry
y
29.
39.
5
(b) = (-3, 4) •
(3, 4) •
b. x
•
-5
(c) = (-3, -4)
41.
•
(a) = (3,
-4)
43.
y
31.
(b) •
•
-5
(a)
(c)
(-2, -1 )
47.
(2, - 1 )
49. Y 5
33.
(c)
( - 5 . 2)
=
(a)
•
=
•
5
•
( - 5 , -2)
=
Symmetric with respect to the y-axis. Intercepts: (0, 0) Symmetric with respect to the x-axis.
a.
a.
a.
a.
b.
Intercepts: (-2,0) , (0,0) , and (2,0) Symmetric with respect to the origin. Intercepts: (-1,0) , (0,-1) , (1,0) Symmetric with respect to the y-axis. Intercepts: none Symmetric with respect to the origin. y
51.
(5 , 2)
-5
(b)
b.
b.
-5
(a) = (-3, •
5
4)
•
(5, -2)
-5
(c) = (3, •
4)
y
53 .
5
x
•
(-3, -4)
-5
= =
( ¥ . 2)
--5
5
(a) (c)
( 0, -9 )
•
(b) = (3, -4)
y
37.
x
x
y
35.
(- 1 ,0) , (0,1) , and (1 ,0)
Intercepts:
b.
x
5 =
45.
(2, 1 )
•
• =
=
Intercepts: (-1, 0) and (1, 0) Symmetric with respect to the x-axis, y-axis, and the origin.
a.
b.
5
(-2, 1 )
a.
55.
(0, 3) (0, 3) x
(0, -3)
(b) = (0, -3)
-5
/ = x+4
y-intercepts: / = 0+4 02 = x + 4 -4 = x / =4 y = ±2 The intercepts are ( -4, 0) , (0,-2) and (0,2) .
x-intercepts:
89
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 2: Graphs
Test x-axis symmetry: Let y = -y (_y) 2 = x + 4 y2 = x + 4 same Test v-axis symmetry: Let x = -x / = -x + 4 different Test origin symmetry: Let x = -x and y = -y . (_y) 2 = -x + 4 / = -x + 4 different Therefore, the graph will have x-axis symmetry. 57.
59.
61.
9x2 + 4y2 = 36 y-intercepts: x-intercepts: 9x2 + 4(0) 2 = 36 9(0) 2 + 4/ = 36 4/ = 36 9x2 = 36 y2 = 9 x2 = 4 y = ±3 x = ±2 The intercepts are (-2, 0) , (2, 0), ( 0, -3), and (0,3) Test x-axis symmetry: Let y = -y 9x2 + 4(_y) 2 = 36 9x2 + 4/ = 36 same Test v-axis symmetry: Let x = -x 9(_X) 2 + 4/ = 36 9x2 + 4y2 = 36 same Test origin symmetry: Let x = -x and y = -y 9(_x) 2 + 4 ( _y) 2 = 36 9x2 + 4/ = 36 same Therefore, the graph will have x-axis, y-axis, and origin symmetry. .
y = V; y-intercepts: x-intercepts: y=W=O O = V; O=x The only intercept is (0, 0) . Test x-axis symmetry: Let y = -y -y = V; different Test v-axis symmetry: Let x = -x y = h = -V; different Test origin symmetry: Let x = -x and y = -y -y = h = -V; y = V; same Therefore, the graph will have origin symmetry.
63.
x2 + y - 9 = 0 y-intercepts: x-intercepts: x2 -9 = 0 02 + y - 9 = 0 y=9 x2 = 9 x = ±3 The intercepts are (-3, 0) , (3, 0) , and (0,9) . Test x-axis symmetry: Let y = -y x2 - Y -9 = 0 different Test y-axis symmetry: Let x = -x (-X) 2 + y - 9 = 0 x2 + Y 9 = 0 same Test origin symmetry: Let x = -x and y = -y (-X) 2 _ y - 9 = 0 x2 Y - 9 = 0 different Therefore, the graph will have y-axis symmetry. -
y = x3 - 27 y-intercepts: x-intercepts: 0 = x3 - 27 y = 03 - 27 y = -27 x3 = 27 x=3 The intercepts are (3, 0) and (0, -27) . Test x-axis symmetry: Let y = -y -y = x3 - 27 different Test v-axis symmetry: Let x = -x y = (_X)3 - 27 Y = _x 3 - 27 different Test origin symmetry: Let x = -x and y = -y _y = (_x)3 _ 27 Y = x 3 + 27 different Therefore, the graph has none of the indicated symmetries.
-
90
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry
65. y = x2 - 3x - 4 x-intercepts: y-intercepts: 0 = x2 - 3x - 4 y = 02 - 3 ( 0) - 4 0 = (x - 4)(x + l) y = -4 x = 4 or x = -l The intercepts are (4, 0) , (-1, 0) , and (0, -4) . Test x-axis symmetry: Let y = -y -y = x2 - 3x - 4 different Test v-axis symmetry: Let x = -x y = (_X) 2 - 3( -x) - 4 y = x2 + 3x - 4 different Test origin symmetry: Let x = -x and y = -y -y = (_X) 2 - 3 ( -x) - 4 -y = x2 + 3x - 4 different Therefore, the graph has none of the indicated symmetries. 6 7· y = 3x x2 + 9 x-intercepts: y-intercepts: 3 (0) - -0 - O O=� y - -x2 + 9 02 + 9 9 3x = 0 x=O The only intercept is (0, 0) . Test x-axis symmetry: Let y = -y -y = � different x +9 Test v-axis symmetry: Let x = -x y = 3 ( -x) (_X) 2 + 9 3x different y = --x2 + 9 Test origin symmetry: Let x = -x and y = -y -y = 3(-x ) (_X) 2 + 9 3x -y = --x2 + 9 3x y = -- same x2 + 9 Therefore, the graph has origin symmetry.
69. y = _x3 x2 - 9 x-intercepts: y-intercepts: _x3 0 =O _03 = 0 = -y = -x2 - 9 02 - 9 -9 _x3 = 0 x=O The only intercept is (0, 0) . Test x-axis symmetry: Let y = -Y _x3 -y = x2 -9 x3 different Y ---zx -9 Test v-axis symmetry: Let x = -x _(_X)-:- 3 Y = -'--'(-x) 2 _ 9 x3 different Y ---zx -9 Test origin symmetry: Let x = -x and y = -y _ (_X)3 -Y = -'-=--'(_x) 2 _ 9 x3 -y = -x2 - 9 _x3 same y = -x2 - 9 Therefore, the graph has origin symmetry. -
-
x
91
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 2: Graphs
73.
y = .,Jx
83.
a.
y 5
-5
(0, 0)
( x2 + ( 0) 2 -x r = X 2 + (0) 2 (X2 -xt = x2 X4 _ 2x3 + x2 = x 2 X4 - 2x3 = 0 x3 (x - 2) = 0 x3 = 0 or x -2 = 0 x=o x=2 y-intercepts: ( (0)2 + i - o r = (0)2 +i (i ) 2 = i y4 = i
5
-5 75.
77.
79.
81.
( x2 + i -x t = x2 + i x-intercepts:
If the point (3, b) is on the graph of y = 4x + 1 , then we have b = 4 (3) + 1 = 1 2 + 1 = 13 Thus, b = 13 . If the point (a, 4) is on the graph of y = x2 +3x , then we have 4 = a2 + 3a 0 = a2 + 3a - 4 0 = (a + 4)(a - 1) a + 4 = 0 or a - 1 = 0 a=l a = -4 Thus, a = -4 or a = 1 . For a graph with origin symmetry, if the point ( a, b) is on the graph, then so is the point ( -a, -b) . Since the point (1, 2 ) is on the graph of an equation with origin symmetry, the point ( -1, -2) must also be on the graph.
b.
/ -i = 0 i (i - 1 ) = 0 or i - 1 = 0 i =l y = ±l The intercepts are (0, 0) , (2,0), (0, -1), and (0, 1) . Test x-axis symmetry: Let y = -y
(x 2 + (-y) 2 -x r = x2 +(_y) 2 (X 2 + y2 _ X r = x2 + y2 same Test y-axis symmetry: Let x = -x ((-x)2 +i -(-x) t = (-x) 2 + i (X 2 + i + x t = x2 + y2 different Test origin symmetry: Let x = -x and y = -y ( (-x)2 +(_y) 2 _(-X) t = (_x)2 +(_y)2 (X2 + y2 + x) 2 = x2 + / different
For a graph with origin symmetry, if the point (a, b) is on the graph, then so is the point (-a, -b ) . Since -4 is an x-intercept in this case, the point (-4,0) is on the graph of the equation. Due to the origin symmetry, the point (4,0) must also be on the graph. Therefore, 4 is another x-intercept.
Thus, the graph will have x-axis symmetry.
92
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 2.3: Lines
85. a.
87. Answers will vary. One example: y
x
5
-
89. Answers will vary
Section 2.3 x
1. undefined; 0 3.
y= b ; y-intercept
5. False; the slope is
l'
2y=3x+5 3 5 y=-x+2 2
7.
9.
-2 b.
Since y=
c.
m =m 2 1
R= I x I for all
R and y= I x I
For y=
(Fxt
'
x, the graphs of
11.
are the same.
; y-intercepts;
2
a. b.
Slope=
1-0
I =2 -0 2
--
If x increases by 2 units, y will increase by 1 unit.
the domain of the variable
x is x � 0 ; for y = x, the domain of the 13.
variable x is all real numbers. Thus,
(Fxt d.
= x only for x � O.
For y =
R , the range of the
m ' m =-1 2 i
1-2
1
a.
Slope=
b.
If x increases by 3 units, y will decrease
---
1-(- 2 )
3
by 1 unit.
variable y is
y � 0 ; for y= x, the range of the variable y is all real numbers. Also, if x � O . Otherwise,
R= x only
R = -x.
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Chapter 2: Graphs
15.
y -y 0 -3 3 = Slope =_2 I = X2-Xl 4-2 2 _ _
__
23.
__
P =(1, 2);m = 3
x x
17.
25.
y y 1-3 -2 Slope =_2- 1 =--X2-Xl 2-(-2) 4
P =( 2 , 4);m =-
�
_ _
2
y
y
-5
27. 19.
1-( ) Slope=Y2 YI =- -I =Q =O X2-Xl 2 -(-3) 5
P =( - 1, 3);m =0 y 5
y
P=(-1,3)
5
x
x
(-3, -1)
(2, -1)
-5
29.
-5
21.
_2-Y1 = -2-2 Slope =Y X2-Xl -1-(-1) _ _
---
-4 o
P =(0 , 3);
s lope undefined y 5
undefined.
P=(O,3)
y 5
x
(-1,2) x
-5
(note: the line is the y-axis) 5
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Section 2.3: Lines
31.
33.
39.
Slope = 4 =i; point: ( 1, 2 ) 1 Ifx increases by 1 unit, then y increases by 4 units. Answers will vary. Three possible points are: x=1 +1 =2 and y=2 + 4 = 6 ( 2,6) x=2 +1 =3 and y=6 + 4 =10 ( 3,10) x=3 +1 = 4 and y=10 + 4 =1 4 ( 4,14 )
--
4 1.
3 -3 Slope = -"2 =2" ; point: ( 2,-4 ) Ifx increases by 2 units, then y decreases by 3 units. Answers will vary. Three possible points are: x=2 +2 = 4 and y=-4 -3 =-7 ( 4,-7 ) x= 4 + 2 = 6 and y=-7 - 3 =-10 ( 6,-10) x= 6 + 2 = 8 and y=-10 -3 =-13 ( 8,-13 )
35.
37.
(-1, 3) and ( 1, 1) are points on the line. -2 1-3 =-=-I Slope = 1-(-1) 2 Usingy-y, =m(x-x,) y-I =-I(x-I) y-I=-x+l y=-x+2 x+ Y =2 or y=-x+ 2
43.
-2 Slope =-2 =-; point: (-2,-3 ) 1 Ifx increases by 1 unit, then ydecreases by 2 units. Answers will vary. Three possible points are: x=-2+1=-1 and y=-3 - 2=-5 (-1,-5) x=-I+l=O and y=-5- 2 =-7 ( 0,-7 ) x=0 +1 =1 and y=-7 - 2 = -9 ( 1,-9 ) (0, 0) and ( 2, 1) are points on the line. 1-0 1 Slope =--=2-0 2 y-intercept is 0; using Y =mx+ b : 1 y=-x+O 2 2y=x 0=x- 2y 1 x- 2Y =0 or y=-x 2
y-y, =m(x-x,), m =2 y-3 = 2(x-3) y-3 = 2x- 6 y= 2x-3 2x- Y =3 or y= 2x-3 1 y-y,=m(x-x,), m =-2 1 y- 2 =--(x- l) 2 1 1 y- 2 =--x+2 2 1 5 y=--x+2 2 5 1 x+ 2y= 5 or y=--x+2 2
45.
Slope 3; containing (-2, 3) y-y, =m(x-x,) y-3 =3(x-(- 2» y-3 =3x+6 y=3x+9 3x-y=-9 or y=3x+ 9
47.
Slope
=
=-
� ; containing (1, -1)
3 y-y, =m(x-x,) 2 y-(-I)=--(x-I) 3 2 2 y+l=--x+3 3 2 1 y=--x-3 3
2 1 2x+3y=-1 or y=--x-3 3
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Chapter 2: Graphs
49.
Containing ( 1, 3) and (-1, 2) 2-3 -1 1 m =-- =-=-1- 1 -2 2 Y -Y I = m(x-xl ) 1 y-3 =-(x-l) 2 1 1 y-3 =-x-2 2 5 1 y=-x+2 2 5 1 x-2y=-5 or y=-x+2 2
51.
Slope =-3; y-intercept =3 y= mx+b y=-3x+3 3x+y=3 or y=-3x+ 3
53.
x-intercept =2; y-intercept =-1 Points are (2,0) and (0,-1) -1 -0 -1 1 m =--=-=0 -2 -2 2 y=mx+b 1 y=-x-l 2 1 x-2y=2 or y=-x- I 2
55.
57.
59.
61.
Parallel to 2x- y =-2 ; Slope =2 Containing the point (0, 0) Y-YI =m(x-xl ) y-O=2(x-O) Y =2x 2x- Y =0 or Y =2x
63.
Parallel to x= 5 ; Containing (4,2) This is a vertical line. x= 4 No slope -intercept form.
65.
67.
Perpendicular to y=.!.. x+4; Containing (1, -2) 2 Slope of perpendicular =-2 y- Y I = m(x-x1) y-(-2) =-2(x-l) y +2 =-2x+ 2 � y=-2x 2x+Y =0 or y=-2x Perpendicular to 2x+y=2 ; Containing the point (-3 , 0) Slope of perpendicular =.!.. 2 y- Y l = m(x-x1) 3 1 1 y-O =-(x-(-3)) � Y = -x+2 2 2 1 3 x-2Y =-3 or y= -x+2 2
Slope undefined; containing the point (2, 4) This is a vertical line. x=2 No slope-intercept form. Horizontal lines have slope m =0 and take the form y=b . Therefore, the horizontal line passing through the point (-3,2 ) is y=2 . Parallel to y=2x; Slope =2 Containing (-1, 2) y- Y ! = m(x-xl ) y-2 =2(x-(-1)) y-2 =2x+2 � y=2x+ 4 2x-y=-4 or y=2x+ 4
69.
Perpendicular to x= 8; Containing (3, 4) Slope of perpendicular =0 (horizontal line) y=4
71.
Y =2x +3
; Slope = 2; y-intercept = 3
x
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Section 2.3: Lines
1
73. -y= x-1; 2
81. x+ y=1; y=-x+1
y=2x - 2
Slope
Slope =2; y-intercept =-2
=
- 1; y-intercept = 1 y
x x
-2
75.
Slope is undefined y-intercept - none
83. x=-4;
y= .!.. x+2 ; Slope= .!.. ; y-intercept =2 2
2
y S
x
-5
-s
1
77. x+2y=4; 2y=-x+4 � y=--x+2
85.
2
Slope= -.!.. ; y-intercept 2
=
y= 5 ; Slope =0; y-intercept = 5 y 8
2
(0,5)
x
-2 87.
-5
� ; y-intercept 3
=0
y
3
=
-
=
2
79. 2x-3y=6; -3y=- 2x+6� y=-x-2
Slope
y x=0 ; y= x Slope 1; y-intercept s
2
=-
y
x
5
-5
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Chapter 2: Graphs
89.
3 2y-3x=0; 2y=3x�y= 2x
b.
Slope =�; y-intercept 0 2
=
x
-10
x
95. a.
91. a.
x-intercept: 2x + 3 (0) = 6 2x=6 x=3 The point (3,0) is on the graph.
x-intercept: 7x + 2 (0) = 21 7x= 21 x=3 The point (3,0) is on the graph. y-intercept: 7 (0) + 2y= 21 2y=21 21 y= 2 21 is on the graph. The point 0, 2
( )
y-intercept: 2(0) + 3y= 6 3y= 6 y= 2 The point (0, 2) is on the graph.
b.
b.
x
97. a. 93. a.
x-intercept: -4x+ 5 (0) = 40 -4x=40 x=-l0 The point (-10,0) is on the graph.
x-intercept: .!.x+.!.(O) =1 2 3 1 -x=l 2 x= 2 The point ( 2,0) is on the graph. y-intercept:
y-intercept: -4 ( 0) + 5 Y = 40 5y=40 y=8 The point (0,8) is on the graph.
.!. ( 0) +.!.y=1 2
3 1 -y=l 3 Y =3 The point (0,3) is on the graph.
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Section 2.3: Lines
b.
111.
2 = --2 = 5 - 3 =-2 - 1 -3 3 3-,-0 - = -3 P2 = ( 1,3 ) , � = ( - 1,0 ) : m2 = 1 - ( -1 ) 2 Since ml . m2 = - 1 , the line segments 1;P2 and P2� are perpendicular. Thus, the points 1;, P2, and P:J are vertices of a right triangle. � = ( -2,5 ) ,
P2 = ( 1,3 )
:
ml
--
-
x
-5
-s
99. a.
b.
x-intercept: 0.2x - 0.5 ( 0 ) = 1 0.2x = 1 x=5 The point ( 5,0 ) is on the graph.
113.
1; = ( - 1,0 ) ,
P2 = ( 2,3 ) , P:J = ( 1,-2 ) , P4 = ( 4,1 )
1-3 3 = 3 - 0 = 3 = 1; m24 = = -1; 4-2 2 - (- 1) 1 - ( -2 ) 3 = 3 = 1 ; m13 = -2 - 0 = -1 m34 = _ 4 1 1 ( -1 ) Opposite sides are parallel (same slope) and adjacent sides are perpendicular (product of slopes is - 1 ). Therefore, the vertices are for a rectangle. ml2
_
y-intercept: 0.2 ( 0 ) - 0.5y = 1 -0.5y = 1 y = -2 The point ( 0,-2 ) is on the graph.
115.
Y
S
Let x number of miles driven, and let C = cost in dollars. Total cost (cost per mile)(number of miles) + fixed cost C = 0.20x + 29 When x 1 1 0, C = ( 0.20 )( 1 1 0 ) + 29 = $5 1 .00 . When x 230, C = ( 0.20 )( 230 ) + 29 = $75.00 . =
=
=
=
117.
-s
1 0 1.
=
The equation of the x-axis is y = O. (The slope is 0 and the y-intercept is 0.) The slopes are the same but the y-intercepts are different. Therefore, the two lines are parallel.
1 05.
The slopes are different and their product does not equal -1 . Therefore, the lines are neither parallel nor perpendicular.
109.
=
=
1 03.
107.
Let x number newspapers delivered, and let C cost in dollars. Total cost (delivery cost per paper)(number of papers delivered) + fixed cost C = 0.53x + 1,070,000
1 19. a.
b.
Intercepts: ( 0,2 ) and ( -2,0 ) . Thus, slope = 1 . y = x + 2 or x - y = - 2 Intercepts: ( 3,0 ) and ( 0,1 ) . Thus, slope 1 y = -- x + 1 or x + 3 y = 3 3
=
C 0.08275x + 7.58 ; 0 � x � 400 =
Y
o
1 - . 3
c.
-
d.
100
200
kW-hr
300 400
x
For 1 00 kWh, C = 0.08275(100) + 7.58 $ 1 5 .86 =
For 300 kWh, C = 0.08725(300) + 7.58 = $32.4 1
99
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Chapter 2: Graphs e.
12 1 .
(
diagram indicates that the slope is negative. Therefore, the only slope that can be used to obtain the 30-inch rise and still meet design requirements is m . In words, for 2 every 1 2 inches of the ramp must drop exactly 1 inch.
For each usage increase of 1 kWh, the monthly charge increases by $0.08275 (that is, 8.275 cents).
OF) = (0, 32); ( C OF) = (1 00, 212) 2 1 2 - 32 1 80 9 sIope = 1 00 - 0 1 00 5 2. °F - 32 = (OC - 0) 5 2. of - 32 = (0C) 5 °C = �(OF -32) 9 If of = 70 , then °C = � (70 - 32) = � (38) 9 9 O
C,
123. a.
O
=
,
run,
125. a.
The y-intercept is (0, 30), so b 30. Since the ramp drops 2 inches for every 25 inches -2 2 . Thus, of the slope is m = - = -25 25 the equation is y = -�x + 30 . 25 Let y O. 2 0 = --x + 30 25 2 -x = 30 25 25 �x = 25 (30) 2 25 2 x = 375 The x-intercept is (375, 0). This means that the ramp meets the floor 375 inches (or 3 1 .25 feet) from the base of the platform. No. From part (b), the is 3 1 .25 feet which exceeds the required maximum of 30 feet. First, design requirements state that the maximum slope is a drop of 1 inch for each 12 inches of This means 1mI:5: . 1 Second, the run is restricted to be no more than 30 feet 360 inches. For a rise of 30 inches, this means the minimum slope is 30 1 - = -. That . Iml � - 1 . Thus, the 12 360 1 2 =
run,
h.
h.
=
c.
( )
c.
d.
-/
127.
Let x number of boxes to be sold, and =
A money, in dollars, spent on advertising. =
We have the points (Xl' AI ) = (1 00, 000, 40, 000); (X2' A2 ) = (200, 000, 60, 000) 60, 000 - 40, 000 slope 200, 000 - 1 00, 000 20, 000 1 1 00, 000 5 1 A - 40, 000 = -(x - 1 00, 000) 5 1 A - 40, 000 = -x - 20, 000 5 1 A =-x + 20, 000 5 If 300,000, then A = .!.(300, 000) + 20, 000 = $80, 000 5 Each additional box sold requires an additional $0.20 in advertising. x =
2x - y = C Graph the lines: 2x - y = - 4 2x - y = a 2x - y = 2 All the lines have the same slope, 2. The lines are parallel. 2x-y=-4
run
�
run.
=
only possible slope i s ImI = IS,
� . The
1
100
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Section 2.4: Circles
129.
1 31 .
(b), (c), (e) and (g) The line has positive slope and positive y-intercept. (c)
The equation
- y -2 has slope 1 and yintercept (0, 2). The equation x - y = 1 has slope 1 and y-intercept (0, - 1). Thus, the lines are parallel with positive slopes. One line has a positive y-intercept and the other with a negative y-intercept.
133 - 135. 137.
Section 2.4
x
=
1.
add; 25
3.
False. For example, x 2 + y 2 + 2x + 2y + 8 0
5.
True; r2 = 9 � r
7.
No, a line does not need to have both an x intercept and a y-intercept. Vertical and horizontal lines have only one intercept (unless they are a coordinate axis). Every line must have at least one intercept. Two lines that have the same x-intercept and y intercept (assuming the x-intercept is not 0) are the same line since a line is uniquely defmed by two distinct points.
141.
Yes. Two distinct lines with the same y-intercept, but different slopes, can have the same x-intercept if the x-intercept is x = O . Assume Line 1 has equation y = m1 x +b and Line 2 has equation y = m2x +b,
is not a circle. It has no real solutions. =
3
Center (2, 1 ) Radius = distance from (0,1) to ( 2 , 1 ) = �( 2 0 )2 +(1 - 1)2 =.[4=2 Equation: (x - 2) 2 +(y_1) 2 = 4 =
-
Answers will vary.
139.
=
9.
Center midpoint of ( 1 , 2) and (4, 2) = �4,2;2 = (t,2 ) =
e
)
(t, 2 ) to (4,2)
(4 - ) + (2 - 2)2 = [9= V '4 '2 '2 ( )2 Equation: x - '25 +(y - 2) 2 = 9 Radius = distance from 5 2
3
'4
1 1.
Line 1 has x-intercept � and y-intercept b. m1 -
( X_h)2 + (y_k)2 = r2 ( X -0 )2 +(y_0)2 = 2 2 x 2 +y 2 = 4 General form: x 2 +/ - 4 0 =
y
Line 2 has x-intercept -� and y-intercept b. m2 Assume also that Line 1 and Line 2 have unequal slopes, that is m1 *' m2. If the lines have the same x-intercept, then b b
s
x
b b m1 m2 -m2b=-m1b -m2b+m1b 0 But - m2b+m1b=0=>b("'J - m2)=0 =>b=0 or m1 - m2 =0 => m1 = m2 Since we are assuming that m1 *' m2, the only way that the two lines can have the same x-intercept is if b= O.
-s
13.
=
(x_h)2 + (y_k)2 =r2 (x -0)2 + (y 2 )2 = 2 2 x2 +( y_ 2 )2 4 General form: x 2 +/ - 4y + 4 = 4 x 2 + / - 4y =0 _
=
101
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Chapter 2: Graphs y
19.
5
x
( ) () (X-�J + / = ±
General form: x 2 - x + -1 + Y 2 -1 4 4 x2 + / - x = 0
-5
1 5.
(X _ h )2 + (y _ k)2 = r2 2 2 X - "21 + (y - O) 2 = "21
y 2
(x - h/ + (y _ k)2 = r2 (x _ 4)2 + (y _ ( _ 3»2 = 52 (X _ 4)2 + (y + 3)2 = 25 General form: x2 - 8x + 16 + / + 6y + 9 = 25 x2 + / - 8x + 6y = °
x
y
-2
x
21 .
x2 + y2 = 4 x2 + y2 = 22 a. Center: (0, 0) ; Radius = 2
b. 17.
( _ h) 2 + (y _ k) 2 = r2 (x _ ( _2) 2 + (y _ 1)2 = 42 (X + 2)2 + (y _ 1)2 = 1 6 General form: x2 + 4x + 4 + y2 - 2y + 1 = 16 x2 + / + 4x - 2y - 1 1 = ° X
y
5
x
y 8
-4 c.
x
-4
x-intercepts: x 2 + (0/ = 4 x2 = 4 x = ±J4 = ±2 y-intercepts: (0) 2 + y 2 = 4 y2 = 4 y = ± J4 = ±2 The intercepts are (-2, 0) , (2, 0) , (0, -2) , and (0, 2).
1 02
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Section 2.4: Circles
23.
c.
2 (x - 3f + 2/ = 8 (X_3) 2 + / = 4 Center: (3, 0); Radius = 2 a.
b.
y
5
x
y-intercepts: (0 _ 1)2 + (y - 2)2 = 32 ( _ 1)2 + (y _ 2)2 = 32 1 + (y_ 2) 2 = 9 (Y_ 2) 2 = 8 y - 2 = ±../8 y - 2 = ±2.fi y = 2 ± 2.fi The intercepts are ( 1 -.J5, 0 ) , ( 1 0) ,
-4 c.
25.
x-intercepts: (x - 3) 2 + (0) 2 = 4 (X_ 3) 2 = 4 x - 3 = ±.J4 x - 3 = ±2 x =3±2 x = 5 or x = 1 y-intercepts: (0 - 3/ + / = 4 (-3) 2 + / = 4 9 + y2 = 4 y2 = -5 No real solution. The intercepts are (1, 0) and (5, 0) .
( 0, 2 - 2.J2 ) , and ( 0, 2 + 2.J2 ) .
27.
+.J5,
x2 + / + 4x - 4y - I = 0 x2 + 4x + / - 4y = 1 (x2 + 4x + 4) + (/ - 4y + 4) = 1 + 4 + 4 (X + 2)2 + (y_ 2)2 = 32 Center: (-2, 2); Radius = 3 a.
b.
x2 + / - 2x - 4 y - 4 = ° x2 - 2x + /- 4 y = 4 (x2 - 2x + 1) + (/ - 4 y + 4) = 4 + 1 + 4 (X_ 1)2 + (y _ 2)2 = 32 Center: ( 1 , 2); Radius = 3 a.
b.
x-intercepts: (x _ 1)2 + (0 - 2)2 = 32 (X_ 1)2 + ( _2)2 = 32 (x - I) 2 + 4 = 9 (x - ll = 5 x - I = ±.J5 x = I ±.J5
y
-5
y
c.
x-intercepts: (x + 2)2 + (0 - 2)2 = 32 (X + 2)2 + 4 = 9 (X + 2)2 = 5 x+2 = x = -2±.J5
±.J5
x
-5
103
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Chapter 2: Graphs
(0+2)2+ (y -2)2=32 4+(y-2)2=9 (y _ 2)2 =5 y-2=±../5 y=2±../5 The intercepts are (-2-../5,0), (-2+../5,0), (0,2-../5), and (0,2+../5). x2 + i -x+2y+1=0 x2 - X +i + 2 y -1 (X2 -x+�) + (i +2y+1)=-1+�+ 1 (x- lJ + (y + 1)2 =(lJ Center: (.!. -1 ) ' Radius .!. 2 2" y-intercepts:
29.
31 .
2x2 +2i - 12x + 8y-24= x2 + i - 6x+4y = 1 2 x2 6x+ y2 + 4 1 2 (x2 -6 x+ 9) + (i +4 y+ 4)=12+ 9+ 4 (X-3)2+ (y+2)2 52 °
Y
_
a.
b.
Center: (3,-2); Radius Y 5
=
=
=
5
=
a.
=
b.
+ (0+2)2=52 (x-3)2+4=25 (X-3)2=21 x-3=±Jii. x=3±Jii. y-intercepts: (0-3i + (y +2)2 52 9 + (y + 2) 2=25 (y + 2)2 = 16 y+2=±4 -2±4 2 or y = -6 y The intercepts are (3 - Jii., 0) , (3+ Jii., 0) , (0,-6), and (0,2). 2X2+8x+2i = x2+4x+y2=0 x2 + 4 x+4+ i = 0+4 (x+ 2)2 y2 =2 2 Center: (-2,0); Radius: = 2
c.
y
2
x-intercepts: (x - 3)2
=
x
c.
=
( -lJ (0+ 1)2 =(lJ ( x-lJ + 1=� (x - �J 43
x-intercepts: x
+
33.
No real solutions
y-intercepts:
(0 -lJ +(y+ 1)2 (lJ 1 1 -+(y+ ) 2 =l 4 2 4 (y+ 1 ) = y + 1=0 y =- 1
=
°
a.
b.
=
Y
+
y
r
5
°
x
The only intercept is (0, -1) .
-5
104
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Section 2.4: Circles
(x+2)2+(0)2=22 (X+2)2=4 (X+2)2 =±J4 x+2=±2 x= 0 x=-2±2 or x=-4 y-intercepts: (0 + 2)2+ /= 22 4+/ =4 /=0 y=O The intercepts are (-4,0) and (0,0). Center at (0,0); containing point (- 2,3). r=�(-2-0)2+(3_ 0)2 =.J4+9 =J13 Equation: (x-0)2+ (y- oi =( J13t x2+ y2=13 Center at (2, 3); tangent to the x-axis. r=3 Equation: (X_2)2+(y_3)2=32 (X_2)2+(y_ 3)2=9 Endpoints of a diameter are (1, 4) and (-3,2). The center is at the midpoint of that diameter: Center: C+;-3)' T)=(-1,3) Radius: r= �(1-(_1))2+ (4 -3)2 =..J4+i=.J5 Equation: (x-(-1))2+(y_ 3)2=(.J5t (x+1)2+(y_3)2=5 Center at (-1,3); tangent to the line y 2. This means that the circle contains the point (-1,2), so the radius is r 1. Equation : (x+1)2+ ( y-3)2= (1)2 (x+1)2+(y_3)2=1 (c); Center: (1,-2 ) ; Radius 2 (b) ; Center: (-1,2); Radius 2
c.
35.
37.
39.
41.
47.
x-intercepts:
45.
(x,y)
x= y
x2+ y2=9 x2+X2=9 2x2=9 x2=-29 x= f2. = 3.J2 Vi 2
2x. Thus,
=,' = [2 3� J = (3J2)' =18 'qu", uniG. The diameter of the Ferris wheel was 250 feet, so the radius was 125 feet. The maximum height was 264 feet, so the center was at a height of 264-125=139 feet above the ground. Since the center of the wheel is on the y-axis, it is the point (0, 139). Thus, an equation for the wheel is: (x_O)2+(y_139)2=1252 x2+(y_139)2=15,625 x2+ y2+ 2x+4y-4091= 0 x2+2x+ /+4y-4091= 0 x2+2x+1+ / + 4y+4= 4091+5 (x+1)2+(y+2)2=4096 The circle representing Earth has center (-1,-2) and radius .J4096 = 64 . So the radius of the satellite's orbit is 64+0.6= 64.6 units. The equation of the orbit is (x+1)2+(y+2)2=(64.6)2 x2+ y2+2x+4y-4168.16= 0 The length of one side of the square is the area is A
49.
51.
=
=
=
43.
Let the upper-right comer of the square be the point . The circle and the square are both centered about the origin. Because of symmetry, we have that at the upper-right comer of the square. Therefore, we get
=
=
105
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Chapter 2: Graphs
53.
x2+/ = 9 Center: (0, 0) Slope from center to (1, 2.J2 ) is 2..Ji-0 = 2..Ji =2.J2 . 1-0 1 =--.J24 . SIope f the tangent I·me · 2,,2 Equation of the tangent line is: .J2 (x-l) y-2,,2 =- 4 y-2..Ji=- ..Ji4 x+ .J24 4y- 8..Ji= -..Ji x+..Ji ..Ji x+ 4y= 9..Ji .J2 x+4y- 9..Ji= 0 Let (h, k) be the center of the circle. x-2y+4=0 2y=x+4 1 y=-x+2 2 The slope of the tangent line is .!. . The slope 2 from (h, k) to (0,2) is -2. 2-k =_2 O-h 2-k=2h The other tangent line is y= 2x- , and it has slope 2. The slope from (h, k) to (3, -I) is _.!.. 2 -1-k -=--21 3-h 2+2k = 3 - h 2k= I -h h =1-2k Solve the two equations in h and k : 2-k=2(1-2k) 2-k=2-4k 3k=0 k=O h =1-2(0)=1 The center of the circle is (1, 0). 0
IS
57.
r::: -1
y=2 .
Therefore, the path of the center of the circle has the equation
r:::
55.
Consider the following diagram:
59.
(b), (c), (e) and (g) We need h, k > 0 and
61.
Answers will vary.
(0,0) on the graph.
Section 2.5 1. 3.
5.
7
7.
y=kx y=kx 2=10k k=-102 =-51 y=-x51 A =kx2 47r=k(2)2 47r= 4k 7r=k A = 7rX2 F=�d2 10=�52 10=�25 k=250 F = 250d2
106
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Section 2.5: Variation
9.
z
=k(x2+y2)
21.
5=k(32+42) 5=k(25) k=2.=� 25 5 I Z=S(x2 +y2 ) 11.
23.
kd2
M= J-; 24= 24=
k (42)
J9
16k 3
=
16=k(I)2 k=16 Therefore,we have equation s = 16t2. 1f t = 3 seconds,then s = 16(3)2 = 144 feet.
If s = 64 feet,then 64=1612 t2 = 4 t= ±2 Time must be positive,so we disregard t = -2. It takes 2 seconds to fall 64 feet.
( )
k=24 2 =� 16 2
13.
kB 6A9=k(1000) 0.00649=k Therefore we have the linear equation p = 0.00649B . If B=145000 ,then p = 0.00649(145000)= $941.05 . P
r2 = ka3 d2 k(23) 22 = 42 k(8) 4= 16
25.
E
3
=kW = k(20)
k=� 20 Therefore,we have the linear equation IfW=
4=!: 2 k=8 r2 = 8a3 d2
27.
E
=
�w. 20
15,then E=�(15)=2.25 . 20
R
=kg 47AO =k(12) 3.95= k Therefore,we have the linear equation R 3.95g . If g =10.5 ,then R =(3.95)(10.5) '" $41.48. =
17.
I
A=-bh 2
107
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Chapter 2: Graphs
29.
D=� D=156, p=2.75; 156=_2.k7-5 k=429429 S D= . p 429 143 bags of candy D=-= 3 =� =600, =150; 600=� 150 k=90,000 . = 90,000 So, we have the equatIOn P 90,000 If = 200 , then = 200 450 cm3. W=�d2 If W=125, d= 3960 then k and k=1,960,200,000 125=-39602 . 1,960,200,000 So, we have the equation W= -'---'--'-d At the top of Mt. McKinley, we have d=3960+3.8=3963.8 , so W= 1,960,200,000 (3963.8)2 124.76 pounds.
= ksd3 36=k(75)(2)3 36 = 600k 0.06=k h
39.
P
a.
= 45 s =125, =0.06sd3. 45= (0.06)(125)d3 45=7.5d3 6=d3 d= :if6 1.82 inches =kmv2 1250=k(25)(10f 1250=2500k k=0.5 So, we have the equation = 0.5mv2. If m = 25 and v=15, then =0.5(25)(15)2=2812.5 Joules = kpd 100= k(25)(5) 75=125k0.75 0.6=k S we have the equatIOn = 0.6pd . If p = 40, d=8, and =0.50,then = 0.60.(40)(8) 50 =384 psi.
So, we have the equation h If h and then
0,
b. 31.
V
P
V
33.
V
K
K
--
-- =
S
43.
0,
t
.
S
t
--
t
S
45
�
37.
K
41.
P
V
P
�
-
47.
Answers will vary.
Chapter 2 Review Exercises
W=�d2 k_ 55=_ 39602 k=862,488,000
1.
fl a.
. W= 862,488,000 d2
b.
So, we have the equatIOn
d=3965,then W - 862,488,000 39652 54.86 pounds.
If
_ -
=(0,0) andP2=(4,2) d( fl ,Pz )=�(4_0)2+(2_0)2 = ../16 + 4 = Eo 2/5 The coordinates of the midpoint are: (X,Y)= (Xl +2 X2'Yl +2Y2 ) =(0+42 '�)=(� 2 2'�)=(2 2 '1) =
108
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Chapter 2 Review Exercises
1 = -= = 4-0 4 2 F or each run of 2, there is a rise of 1. F; = (1,-1) and P2 = (-2 , 3) d(F;,P2 ) = )(-2_1)2 +(3-(-1)/ = -h+16 =j2s = 5
c.
d.
3.
�y = 2-0 2 slope -
9.
LU
( )_( + 2 + ) = C+(- ) -1+3 ) = ( �1 , %) = (- �,1 ) �Y = 3-(-1) =4 = --4 slope =-2 -1 -3 3 For each run of 3, there is a rise of -4. F; = ( 4,-4) and 12 = ( 4,8) d(F;,P2 ) = )(4-4)2 +(8_(_4))2 = .JO+144 = .J144 = 12 LU
d.
5.
y = -y
The coordinates of the midpoint are: x Xl Y2 Yl - X,Y - 2 ' 2 2 2 ' 2
c.
11.
a.
b.
c.
d.
7.
Y
( ) = ( 4+42 '-4+8 ) = (�2'�2 ) = (4'2) 2
y = ±2 The intercepts are (-4, 0), (4, 0), (0, -2), and (0, 2). Test x-axis symmetry: Let Y
x2+4(-y)2 =16
= - = 8-(4-4-4) = -120 , undefined
slope �Y
y =-
x2 + 4/=16 same Test y-axis symmetry: Let x -x
LU
=
(_X)2 +4/=16 x2 +4/=16 same Test origin symmetry: Let x = -x and y = -y . (_x)2 +4(_y)2 =16 x2+4/=16 same
An undefined slope means the points lie on a vertical line. There is no change in x.
+4
2x = 3(-y)2 2x = 3y2 same Test y-axis symmetry: Let x = -x 2(-x) = 3/ -2x = 3/ different Test origin symm etry: Let x = -x and y = -y . 2(-x) = 3(_y)2 -2x = 3y2 different Therefore, the graph will have x-axis symmetry. x2+4y2=16 x-intercepts: y-intercepts: 2 x2 + 4 (0) =16 (Ol +4y2=16 x2 = 16 4/ =16 = ±4 y2 = 4 x
The coordinates of the midpoint are: + (x ,Y ) = xl + x2 ' Yl Y2 2 2
= x2
y-intercepts: 2(0) = 3/ y0 == Oy2
x =O The only intercept is (0, 0). Test x-axis symmetry: Let
a.
b.
= 3y2 x-intercepts: 2x = 3(0)2 2x = 0 2x
y
Therefore, the graph will have x-axis, y-axis, and origin symmetry.
109
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Chapter 2: Graphs
13.
= X4+2X2+1 y-intercepts: x-intercepts: y = (0)4+2(0)2+1 0 = x4+2x2+1 0 = (X2+1)(x2+1) = 1 x2+1 = 0 x2 = -1 no real solutions The only intercept is (0, 1). Test x-axis symmetry: Let y -y -y = X4+2X2+1 y = _x4-2X2-1 different Test y-axis symmetry: Let x = -x y = (-xt +2(_x)2+1 y = X4+2X2+1 same Test origin symmetry: Let x = -x and y = -y . -y = (_X)4+2(_X)2+1 -y = X4+2X2+1 y = -x4-2x2-1 different Y
17.
19.
=
21.
Y
=
x
Therefore, the graph will have y-axis symmetry.
1 5.
(x_h)2+(y_k)2 =r2 (x_(_2))2+(y-3/ = 42 (X+2)2+(y_3)2 = 16 (X_h)2+(y_k)2 = r2 (x-(_1))2+( y_(_2))2 = 12 (x+l)2+(y+2)2 = 1 x2+(y_1)2 = 4 x2+(y_1)2 = 22 Center: (0,1); Radius 2
x2+x+i +2y = 0 x-intercepts: x 2 +x+ (0)2 +2(0) = 0 x2+x == O0 x(x+l) x = 0, x = -1 y-intercepts: (0)2+ 0 +i + 2y = 0 iy(y+2) +2y == 00 y = O,y = -2 The intercepts are (-1,0), (0,0), and (0,-2). Test x-axis symmetry: Let y = -Y x2+X+(_y)2+2(-y) = 0 x2+x+i -2y = 0 different Test y-axis symmetry: Let x = -x (_X)2+(-x)+i +2y = 0 x2 X+ y2+2y = 0 different Test origin symmetry: Let x = -x and y = -y . (_x)2+(_x)+(_y)2+2(-y) = 0 x2-X+i -2y = 0 different
x2+(0-1/ = 4 x2+1 = 4 x2 = 3 x = ±.J3 y-intercepts: 02+ (Y -1/ = 4 (y_ l)2 = 4 y- l = ±2 = I±2 y y = 3 or y = -1 The intercepts are (-.J3, 0), (.J3, 0), (0, - 1 ) , and (0,3). x2+ y2_ 2x+ 4y-4 = 0 x2-2x+i +4y = 4 (X2-2x+l)+(i +4y+4) = 4+1+4 (x_ l)2+(y+2/ = 32 Center: (1, -2) Radius 3 x-intercepts:
23.
_
=
The graph has none of the indicated symmetries. 110
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Chapter 2 Review Exercises
Y 5
(
2 x-intercepts: (x _ 1) 2 + (0 + 2) 2 = (.J5 ) (x -l) 2 + 4 = 5 (x _ 1) 2 = 1 x -I = ± 1 x =1±1 x = 2 or x = ° 2 y-intercepts: (0 _ 1) 2 + (y + 2) 2 = J"S) 1 + (y + 2)2 = 5 (y + 2) 2 = 4 y + 2 = ±2 Y = -2 ± 2 y = O or y = -4 The intercepts are 0, 0 ) , 2, 0 ) , and 0,- 4 ) .
(
( ( (
x-intercepts: x _ l ) 2 + 0 + 2 ) 2 = 3 2 x _ l)2 + 4 = 9 x _l) 2 = 5 x -I = ± J"S x = 1 ± J"S 2 y-intercepts: 0 _ 1 ) + y + 2 ) 2 = 3 2 1 + y + 2)2 = 9 y+2 =8 y + 2 = ± .J8 y + 2 = ±2 J2 y = -2 ± 2J2 The intercepts are 1 - J"S , 0 ) , 1 + J"S , 0 ) ,
(
( (
( ( ( /
( 0, -2 - 2J2 ) , and ( 0, -2 + 2J2 ) . (
(
25.
(
(
(
)
Center: ( 1 , -2) Radius = J"S
2 7.
Slope =-2 ; containing (3,- 1 ) y - Y I = m x - XI ) y - (-I) = -2 x - 3 ) y +l = -2x + 6 y = -2x + 5 or 2x + y = 5
29.
vertical; containing (- 3,4) Vertical lines have equations of the form x = where is the x-intercept. Now, a vertical line containing the point (-3, 4) must have an x-intercept of -3, so the equation of the line is x = -3. The equation does not have a slope intercept form.
( (
a,
a
3x2 + 3y 2 - 6x + 1 2y = ° x 2 + i - 2x + 4 Y = ° x 2 - 2x + y 2 + 4y = ° x 2 - 2x + l + i +4y +4 = 1 + 4 x_l ) 2 + y + 2 ) 2 = J"S )2
)(
(
31.
y-intercept = -2; containing (5,- 3) Points are (5,-3) and (0,-2) - 2 - (-3) = 1 = --1 m= 5 0-5 -5 Y = mx +b 1 y = --x - 2 or x + 5y = - 1 0 5
33.
Parallel to 2x - 3y = -4 2x - 3y = - 4 -3y = -2x - 4 -3y -2x - 4 -3 -3 4 2 y = -x +3 3
(
y 5
-5
Slope = � ; containing (-5,3) 3 111
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Chapter 2: Graphs
35.
y-Yl=m (x-xl ) 2 Y -3=-(x-( 32 -5)) y-3=-(x+5) 32 1 0 y-3=-3 x+-3 2 1 9 or 2x-3y= - 1 9 y=-x+3 3 Perpendicular to x+Y = 2 x+y=2 y=-x+2 The slope of this line is -1 , so the slope of a line perpendicular to it is 1 . Slope 1 ; containing (4,-3) y-Yl=m(x-xl) y-(-3)=l(x-4) y+3=x-4 y = x- or x-y= 4x-5y=-20 -5y=-4x-20 4 y=-x+4 5 slope �; y-intercept= 4 x-intercept: Let y O. 4x-5(0)=-20 4x=-20 x=-5
39.
1 1 1 -x--y=-6 2 3 1 1 1 --y=--x-6 3 32 I y=-x+2 2 3 . ="21 sIope "2; y-mtercept x-intercept: Let y O. 1 1 O)=--1 -x--( 2 3 1 61 -x=-61 2 x=--3 =
=
=
7
37.
x
7
41.
2x-3y= 1 2 x-intercept: 2x - 3(0) =12 2x = 1 2 x=6
y-intercept: 2(0) - 3y =12 12
-3y= y=-4 The intercepts are (6,0) and (0, -4) .
=
=
Y 5
x
-5
11 2
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Chapter 2 Review Exercises
43.
1 1 -x+-y=2 3 x-i21 ntercept: y-intercept:1 1 1 2"x+3" 1(O)=2 2" (O)+3"1 y=2 -x=2 -y=2 2 x=4 3 y=6 The intercepts are (4,0) and (0,6)
49.
3.
.
45.
Y
Given the points A = (- 2,0),B = (-4,4),and e = (8,5). Find the distance between each pair of points. d(A,B)= �( -4_( _2))2 +( 4-0)2 =�4 +16 =Ea=215 d(B,e)=�( 8-( -4))2 +( 5-4)2 = �144+1 =Ms d(A,e)= �( 8-(-2))2 +( 5-0f =�100+25 =JW=515 [ d(A,B)J +[ d(A,e)J =[ d(B,e)J (Eat + (JW/ = (Ft45t 20+125=145 145=145 The isPythagorean Theorem is satisfied,so this a right triangle. Find the slopes: 4 =-4-(4-0- 2) =-=-2 -2 1 5-4 = 8-(-4)=12 5-0 = 5 = 1 =8-(-2) 10 "2 Since =-2 .�2 = -1, the sides and are perpendicular and the triangle is a rightAetriangle. 1-5 slope of -AB =--=-1 6-2 slope of A-e =-1-5 --= -1 8-2 slope of Be =-1-1 --=-1 8-6 lie on a line. Therefore,the points
=x3
b.
x
mAB
m BC
47.
m
slope �,3 containing the point (1,2) y
=
m ·m AC AB
5 1.
2
4
AC
x
113
AB
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 2: Graphs
53.
=kg 46.67=k(13) k=46.1367 =3.59 Therefore, we have the equation R=3.59g . If g=11.2 , then = 3.59(11.2) "" $40.2 1. R
Chapter 2 Test 1.
p
2.
3652=(k)(93)3 3652 k=933 Therefore, we have the equation T2= 3659332 a3 If 88 days, then 882=(396;2}a)3 a3 =(882)(�J a= 3 (882)(:::2 ) "" 36 million miles The graph of X= 0 is a vertical line passing through the origin. That is, X= 0 is the
d(�,P2)=�(5_(_1))2+(_1_3)2 =�62+(_4)2 =�36+16 =.J52=2Jl3 The coordinates of the midpoint are: x2 Yl +2 ) (x,Y)= (Xl +2' 3+(-1)) =(-1+5 2 ' 2 =(�'3) (2, 1) -1-3 -4-=--2 --m =--= 5-( -x X2 I -I) 6 3 If X increases by 3 units, will decrease by 2 units. =x2-9 Y2
=
T=
Y2- Yl
3. a.
b.
4.
Y
Y
x
57. a.
b.
c.
equation of the y-axis. The grapy of Y is a horizontal line passing through the origin. That is, Y the equation of the x-axis.
=0
x+ =0 y=-x Y
=
0 is 5.
x+ = 0
-1.
xy=0 y = 0 or x= 0 The graph of xy= 0 consists of the coordinate axes. x 2 + y 2= 0 =0 and x=0 The graph of x2 + y2=0 is consists of the d.
e.
y 5
The graph of y is line passing through the origin with slope =
=x
/
(0, -9)
-5
(1,-1)(4_2) ,
(9,-3)
Y
origin.
114
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publi sher.
Chapter 2 Test
6.
x2+Y=9 x-2intercepts: x +0=9 x2=9 x= ±3
�
y-intercept: (0)2+Y=9 y=9
The intercepts are (-3,0), (3,0), and (0,9). Test x-axis symmetry: Let
7.
y=-y
5
x2+(-y)=9 x2- Y=9 different Test y-axis symmetry: Let x=-x (_x) 2+Y = 9 x2+ y=9 same Test origin symmetry: Let x=-x and y=-y (_x) 2+ (_Y ) =9 x2-y=9 different Therefore, the graph will have y-axis symmetry. Slope -2; containing (3,-4) y-y,= m(x-x,) y-(-4)=-2(x-3) y+4=-2x+6 y =-2x+2
-5
10.
=
2x+3y=6 3y=-2x+6 2 y=--x+2 3 Parallel line Any line parallel to
2x+3y=6 has slope
m= -�.3 The line contains (1,- 1) : y-y, = m(x-x,) 2 y-(-l)=--(x-I) 32 2 y+1=--x+32 31 y =--x-3 3 Perpendicular line Any line perpendicular to 2x+3y =6 has slope m= �.2 The line contains (0,3) : y-y, = m(x-x,) 3 y-3=-(x-O) 23 y-3=-x 23 y=-x+3 2
y
x
8.
�+y2 2+4x-2y-4=0 x +4x2+ y2-2y=4 2 (x +4x+4)+(y2 -2y+l)2=4+4+1 ( x+2) + (y_1) = 32 Center: (-2, I); Radius = 3 Y
(x_h)2+(y_k)2=r2 (X_4 ) 2 + ( y_(_3) ) 2=5 2 ( X_4) 2 + ( y+3 ) 2= 25 (x_4) 2 + ( y+3 ) 2= 25 General form: x2-8x+I6+ y2+6y+9=25 x2+ /-8x+6y=0 115
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publi sher.
Chapter 2: Graphs
11.
Let R = the resistance, length,and r = radius. Then R= k·-. r2 Now, R 10 ohms,when 50 feet and r= 6 x10-3 inch, so 50 10= k. (6x1O-3r (6xIO-3)2 = 7.2xI 0-6 = 10· Therefore,we50have the equation R= (7.2x1 0-6 )�r . If =100 feet and r = 7 10-3 inch,then R= (7.2x1 0-6 ) ( 7 x11000-3)2 ",,14.69 ohms. I
I=
7.
=
1=
9.
k
I
11.
x
3.
5.
2x - 3 � 7 2x � 1 0 x�5 {x x � 5}
!
or (-00,5]
5
13.
Chapter 2 Cumulative Review 1.
! x - 2!=1 x - 2 =1 or x - 2 =-1 x=3 The solution set is x=1 { I, 3 } . x2=-9 x=±H x= ±3i The solution set is {-3i, 3i} .
3x - 5 =0 3x = x=3
55 The solution set is {%}. 2X2-5x-3 = 0 ( 2x + I )( x - 3 )= 0 x = - -I or x= 3 2 The solution set is {-�, 3} . x 2 + 2x +5 = 0 r-2 -2 ± �'2 --;-:_-4--2( 1 )--'-( 1'-'-)-(5--'-) -----' x= -2±� 2 -2±..J-16 No real2solutions
!
x-2! � 1
-1 �x-2 � 1 l�x�3 { x l l � x:::;3 }
17.
] 3
= �(_1_4)2 + ( 3 - (_2))2 = �(_5)2 +(5)2 =.J25+25 = .J50 = 5.fi 3+(-2) = (�!) MidPoint= (-1+4 2 ' 2 J 2' 2 = x3 o
15.
[ 1
or [1, 3]
2
4
d ( P, Q)
y
x
116
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exist. No portion oflhis material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 2 Cumulative Review
19.
Perpendicular to Y=2x+ 1 ; Contains (3,5) Slope of perpendicular -.!.2 Y-YI =m (x-xl ) y-5=--2 (x-3) 1 -3 y-5=--x+ 21 132 y=--x+ 2 -2 =
1
y
-10
-6
-£.
2
-4 --6
2
4
6
8 10 12
x
-8
-10
117
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Chapter
3
Functions and Their Graphs Section 3.1
1. 3.
5. 7.
27.
We must not allow the denominator to be 0. x + 4 *- ° => x *- -4 ; Domain: { xl x *- -4} . independent; dependent
29.
[0, 5] We need the intersection of the intervals [0, 7] and [-2, 5] . That is, domain of/ n domain of g . "I
[
-2
0
-2
o
[
1
1
1
5
]
-2
0
7
1
5
7
5
7
�
g
31.
11.
True
13.
False; if the domain is not specified, we assume it is the largest set of real numbers for which the value of/is a real number.
19.
Not a function
21.
Function Domain: {I , 2, 3, 4} Range: { 3}
23.
Not a function
25.
Function Domain: {-2, - 1 , 0, I } Range: { O, 1 , 4}
/ = 4 - x2 =
33.
=
x =/ Solve for y: y ±.,J; For x 1, Y = ±1. Thus, (1, 1) and (1, -1) are on the graph. This is not a function, since a distinct x -value corresponds to two differenty-values. =
=
Function Domain: {Elvis, Colleen, Kaleigh, Marissa} Range: {Jan. 8, Mar. 15, Sept. 1 7} Not a function
= -
Solve for y: y = ±�4 _ X2 For x 0, y ±2 . Thus, (0, 2) and (0, -2) are on the graph. This is not a function, since a distinct x value corresponds to two differenty-values.
(g- /) (x) or g (x) - /(x)
17.
1 . The graph passes the vertical line x test Thus, the equation represents a function. Graph y
]�f
�"41-4-4-4[�--���]+-+-+1·· f+g
15.
=
(-1,3)
..III
9.
Graph y x 2 . The graph passes the vertical line
35.
Graph y 2 X 2 - 3x + 4 . The graph passes the vertical line test. Thus, the equation represents a function. =
118
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Section 3.1: Functions
37.
39.
2X2+3y2=1 Solve fory: 2X2+3i =1 3i =1-2x2 1-2x2 Y2=-3 1- X2 Y=±� � For x= 0, y= ±Jf . Thus, (o,Jf) and (0,-Jf) are on the graph. This is not a function, since a distinct x-value corresponds to two differenty-values. f(x)=3x2+2x-4 f(0)=3(0)2+2(0)-4=-4 f(I)=3(1)2+2(1)-4=3 +2-4=1 f(-I)=3(-1)2+2(-I)-4=3-2-4=-3 f(-x)=3(_X)2+2(-x)-4=3x2-2x-4 -f(x)=-(3x2+2x-4)=-3x2-2x+4 f(x+l)=3(x+l)2+2(x+l)-4 =3 (X2+2x+1)+2x+ 2-4 =3x2+6x+3 +2x+2-4 =3x2+8x+l f(2x)=3(2x)2+2(2x)-4=12x2+4x-4 f(x+h)=3(x+h)2+2(x+h)-4 =3(x2+2xh+h2)+2x+2h-4 =3x2+6xh+3h2+2x+2h-4 f(x)=+ x +1 o2 0 f(O)=-=-=O 0 +1 1 f(l)_ 121+1 _ 21
c.
d. e. f.
f(2x)=(2x)2x2+1 4x2x2+1 x+h x+h = f( x+h)=(x+h) 2+1 x2+2xh+h2+1 f(x)=lxl+4 f(0)=1 01+ 4=0+4=4 f(1)=111+4=1+4=5 f(-1)=I- l l+4=1+4=5 f(-x)=I-x 1+4=1 xl+4 -f(x)=-(Ixl+4)= -I x1-4 f(x+l)=lx+lj+4 f(2x)=1 2x1+4=21 x 1+4 f(x+h)=lx+hl+4
g.
a.
h.
b.
c.
43.
d.
a.
b.
e.
c.
f.
d.
e. f.
g.
g.
h.
41.
-1 =--1 f(-1)_- (_1)-12+1 1+1 2 f(-x)= (-x):+1 x -x -f(x)=- (x2+)1 =x2+1 f(x+l)=(x+l)x+l2+1 x+l 2 x +2x+1+1 x+l
h.
a.
b.
119
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Chapter 3: Functions and Their Graphs
45.
f (x)== 2x + 1 3x - 5 a. b. c.
d. e. f.
g. h.
47.
49.
51.
53.
55.
2 (0) + 1 0 + 1 _.!. = = 3 (0) - 5 0 - 5 5 2 (1) + 1 f (1)== 3 (1) == 2 + 1 ==2==_ � -5 3-5 -2 2 2 (-1) + 1 - 2 + 1 2 .!. == = = f (-I)== 3 (-1) - 5 -3 - 5 - 8 8
f (0) =
f (-x) =
57.
2 (-x) + 1 - 2x + l 2x - l = = 3(-x) - 5 -3x - 5 3x + 5
( )
_f (x) = _ 2X + l == - 2x - l 3x - 5 3x - 5 2 ( x + 1) + 1 2x + 2 + 1 2x + 3 == == f (x + 1) = 3(x + l) - 5 3x + 3 - 5 3x - 2 2 (2x) + 1 4x + l = f (2x)== 3 ( 2x ) - 5 6x - 5 f ( x + h ) ==
59.
61.
----
2 ( x + h) + 1 2x + 2h + 1 = 3 ( x + h) - 5 3x + 3h - 5
f(x) = -5x + 4 Domain: {x I x is any real number }
x f(x) ==-x2 + 1 Domain: {x I x is any real number }
--
g(x) == x x2 - 1 6 x 2 - 1 6:;t 0 x 2 :;t 1 6 => x:;t ±4 Domain: {xl x:;t -4, x:;t 4}
h e x) == -J3x - 1 2 3x - 12 � 0 3x � 1 2 x�4 Domain: {xl x �4} 4 -Jx - 9 x-9 > 0 x>9 Domain: {xl x > 9} f(x) =
g .f2
p(x) == -== ,---; x - I vx - l x-I > 0 x>1 Domain: {xl x > I} f(x)==3x + 4 g (x) == 2x - 3 a. (f + g)(x) == 3x + 4 + 2x - 3 == 5x + 1 Domain: {xl x is any real number } . b. (f - g)(x) = (3x + 4) - (2x - 3) = 3x + 4 - 2x + 3 = x+7 Domain: {xl x is any real number } . c. (f · g)(x) = (3x + 4)(2x - 3) = 6x 2 - 9x + 8x - 1 2 == 6x 2 - x - 1 2 Domain: {xl x is any real number } . d.
( fg ) (x)== 3x2x +- 43
2x - 3:;t 0 => 2x :;t 3 => x :;t -3 2
-2 F(x) = -x3 X +x x3 + x:;t 0 x(x 2 + 1):;t 0 x:;t 0, x 2 :;t -l Domain: {x l x:;t O}
e.
f.
g.
h.
(f + g )(3) = 5(3) + 1 = 1 5 + 1 == 1 6 (f - g )( 4) ==4 + 7 == 1 1 (f . g )(2) = 6(2) 2 - 2 - 12 = 24 - 2 - 12 == 10 f (1) = 3(1) + 4 = 3 + 4 ==� = -7 2(1) - 3 2 - 3 -1 g
()
120
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Section 3.1: Functions
63.
f(x)=x-I g(x)= 2X2 (f+g)(x)=x-1+2x2= 2x2+x-1 Domain: { x I x is any real number } . (f-g)(x)=(x-1)2 - (2X2 ) = _2X +x-1 Domain: {xI x is any real number } . (f ·g)(x)= (x-1)(2x2)=2x3 _2x2 Domain: {xI x is any real number } . (gf}x)=� 2X2 Domain: {xlx;toO}. (f+g)(3)=2(3)2+3 -1=2(9)+3-1 =18+3-1= 20 (f-g)(4)=-2(4)2+4-1 =-2(16)+4-1 =-32+4-1=-29 (f.g)(2)=2(2)3 -2(2)2=2(8)-2(4) =16-8=8 (;)Cl)= ;(�)� = 2�1) = %=0 f(x)=../x g(x)=3x-5 (f+g)(x)=../x+3x-5 Domain: {xlx�O}. (f-g)(x)=../x-(3x-5)=../x-3x+5 Domain: {xlx�O}. (f.g)(x)= ../x(3x-5)=3x../x-5../x Domain: {xlx�O}. ../x (gf}x)= 3x-5 x� 0 and 3x-5;to 0 3x;to5 x;to-53 Domain: {xl x� 0 and x;to %} .
a.
f.
b.
g.
c.
h.
d.
67.
a.
e. f.
b.
g.
c.
h.
65.
(f+g)(3)=J3+3(3)-5 =J3+9-5=J3+4 (f -g)(4)= 14-3(4)+5 =2-12+5=-5 (f .g)(2)=3(2)-fi-5./2 = 6-fi-5-fi= ./2 _l_ =-2_l=_.!.2 (gf}l)= � 3(1)-5 = 3-5 f(x)=I+-x1 g(x)=-x1 1 1+-2 (f+g)(x)=1+-x1 +-= x x Domain: {xlx;toO}. 1 1 1 (f-g)(x)=1+---= xx Domain: {xlx;toO}. 1 1 (f .g)(x)= (1+-x1 r-=x X x2 Domain: {xlx;toO}. x+1 x+1 f( }x)= 1+.!.x = _x_ g .!.x .!.x = x . .:.1 =x+1 Domain: {xlx;toO}. 2 5 (f+g)(3)=1+-=3 3 (f-g)(4)=1 1 -1 =-+-=1 1 3 (f ·g)(2)=-+ 2 (2)2 2 4 4 ( ; )cl)=1 1= 2
e.
d.
a.
b.
+-
__
e.
f.
c.
g.
d.
h.
+
=>
121
© 2008 Pearson Education, Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.
Chapter 3: Functions and Their Graphs
69.
f(x) = 2x + 3 3x - 2 a.
4x -3x - 2
g.
--
4x (f + g)(x) = 2x + 3 + -3x - 2 3x - 2 2x + 3 + 4x 6x + 3 3x - 2 3x - 2 3x - 2 * 0 3x * 2 => x * Domain:
b.
g(x ) =
h.
i { x l x * i} .
71.
3 -� (f - g)(x) = 2x + 3x - 2 3x - 2 2x + 3 - 4x - 2x + 3 3x - 2 3x - 2 3x - 2 * ° 73.
c.
( ) (�)
2 = 8x + 12x (f . g)(x) = 2X + 3 3x - 2 3x - 2 (3x - 2) 2 3x - 2 * 0 75.
d.
(]
2x + 3 f (x) = 3x - 2 = 2x + 3 . 3x - 2 = 2x + 3 g � 3x - 2 4x 4x 3x - 2 3x - 2 * ° and x * ° 3x * 2 x * -2 3
{l %
(f . g)( 2) = 8(2i + 12(22 ) (3(2) - 2) 8(4) + 24 = 32 + 24 = 56 = 7 (6 _ 2) 2 (4)2 16 "2
( fg ) = 2(14(1)) + 3 = 2 4+ 3 = �4 ( ) I
1 (f + g)(x) = 6 - -x 2 1 6 - - x = 3x + l + g(x) 2 7 5 - - x = g(x) 2 g(x) = 5 - -7 x 2 f(x) = 4x + 3 f(x + h ) - f(x) 4(x + h) + 3 - (4x + 3) h h 4x + 4h + 3 - 4x - 3 h 4h =-=4 h
f(x) = 3x + 1
f(x) = x 2 - x + 4 f(x + h ) - f(x) h (X + h) 2 - (x + h ) + 4 - (x 2 - x + 4) h x 2 + 2xh + h 2 - x - h + 4 - x 2 + x - 4 h 2 2xh + h - h h = 2x + h - l
}
Domain: x x * and x * o . e.
f.
(f + g )(3) = 6(3) + 3 = 1 8 + 3 = E = 3 3(3 ) - 2 9 - 2 7 (f - g )( 4) = - 2(4 ) + 3 = - 8 + 3 = -=2 = _ .!. 3(4) - 2 12 - 2 1 0 2
122
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77.
f { x ) = 3x 2 - 2x + 6 f {x + h) - f {x) h 3 { X + h ) 2 - 2 { X + h ) + 6 - [3x 2 - 2x + 6 ] h 3 ( X 2 + 2xh + h 2 ) - 2x - 2h + 6 - 3x 2 + 2x - 6 h 3x 2 + 6xh + 3h 2 - 2h - 3x 2 6xh + 3h 2 - 2h h h = 6x + 3h - 2
[
79.
8 1.
83.
Section 3.1: Functions
85.
J
87.
f(x) = 2x - A and f(4) = 0 x-3 2(4) -A f(4) = 4-3 8 0= A 1 0 = 8-A A=8 f is undefined when x = 3 . --
Let x represent the length of the rectangle. Then, � represents the width of the rectangle 2 since the length is twice the width. The function
f(x) = x3 - 2 f(x + h ) - f(x) h ( x + h ) 3 - 2 - ( x3 - 2 ) h x3 + 3x2 h + 3xh 2 + h 3 - 2 - x3 + 2 h 3 2 2 3x h + 3xh + h --- = 3 2 + 3 h + h 2 ---X X h
for the area .
IS:
89.
x 2 = -x 1 2 A(x) = x · -x = 2 2 2
Let x represent the number of hours worked. The function for the gross salary is:
G(x) = l Ox
91. a.
b.
f(x) = 2x3 + Ax 2 + 4x - 5 and f(2) = 5 f(2) = 2(2) 3 + A(2) 2 + 4(2) - 5 5 = 16 + 4A + 8 - 5 5 = 4A + 19 -14 = 4A A = -14 = _ 7... 4 2
c.
93. a.
f(x) = 3x + 8 and f(O) = 2 2x - A f(O) = 3(0) + 8 2(0) - A 2=� -A -2A = 8 A = -4
P is the dependent variable; independent variable
a
is the
P(20) = 0.01 5(20) 2 - 4.962(20) + 290.580 = 6 - 99.24 + 290.580 = 197.34 In 2005 there are 1 97.34 million people who are 20 years of age or older. P(O) = 0.0 1 5(0) 2 - 4.962(0) + 290.580 = 290.580 In 2005 there are 290.580 million people. H (l) = 20 - 4.9 ( 1 /
= 20 - 4.9 = 1 5 . 1 meters H ( 1 . 1 ) = 20 - 4.9 ( 1 . 1 ) 2 = 20 - 4.9 ( 1 .2 1 ) = 20 - 5.929 = 14.071 meters H (1.2) = 20 - 4.9 ( 1 .2 ) 2 = 20 - 4.9 ( 1 .44 ) = 20 - 7.056 = 12 .944 meters H ( 1 .3 ) = 20 - 4.9 ( 1 .3 / = 20 - 4.9 ( 1 .69 ) = 20 - 8.28 1 = 1 1 .719 meters
123
© 2008 Pearson Education, Inc., Upper Saddle River, N J . All rights reserved. Thi s material is protected under all copyright laws as they currently exist. No portion of thi s material may be reproduced, in any form or by any mean s , without permission in writing from the publi sher.
Chapter 3: Functions and Their Graphs
b.
H (x) = 1 5 : 1 5 = 20 - 4.9x 2 -5 = - 4.9x 2 x 2 "" 1 . 0204 1 .0 1 seconds H (x) = lO : 1 0 = 20 - 4 . 9x 2 -10 = - 4 . 9x 2 x 2 "" 2 . 0408 x "" 1 . 43 seconds H (x) = 5 : 5 = 20 - 4.9x 2 -15 = -4 . 9x 2 x 2 "" 3 . 061 2 x "" 1 . 7 5 seconds H (x) = 0 0 = 20 - 4 . 9x 2 - 20 = -4 . 9x 2 x 2 "" 4 . 08 1 6 x "" 2 . 02 seconds
97.
99.
X""
c.
95.
( �}X) = ��:� H (x) = (p . I)(x) = P (x) . I (x) R (x) =
101. a.
P(x) =
=
b.
=
=
R (x) - C (x)
( - 1 .2x2 + 220x) - (0.OSx3 - 2X2
103. a.
b.
6Sx + SOO)
- 1 .2x2 + 220x - 0. OSx3 + 2x2 - 6Sx - SOO -0.OSX3 + 0. 8X2 + I S Sx - SOO
P(1 S)
=
-0.OS(1 S)3 + 0. 8(I S)2 + I S S(1 S) - SOO
= - 1 68.7S + 1 80 + 232S = $ 1 836.2S
c.
+
- s ao
When 1 5 hundred cell phones are sold, the profit is $ 1 836 . 25 .
h (x) = 2x h ( a + b) = 2 ( a + b) = 2a + 2b = h (a) + h (b) h (x) = 2x has the property. g (x) = x 2 g ( a + b) = ( a + b) 2 = a 2 + 2ab + b 2 Since
C (x) = 100 + � + 36, 000 x 10 C(500) = 100 + 500 + 36, 000 a. 10 500 = 100 + 50 + 72 = $222 450 + 36, 000 b. C (450) = 1 00 + 10 450 = 100 + 45 + 80 = $225 C ( 600) = 1 00 + 600 + 36, 000 c. 10 600 = 100 + 60 + 60 = $220 400 + 36, 000 d. C (400) = 100 + 10 400 = 1 00 + 40 + 90 = $230
c.
a 2 + 2ab + b 2 '" a 2 + b 2 = g ( a ) + g (b) , g (x) = x 2 does not have the property. F (x) = 5x - 2 F ( a + b) = 5 ( a + b) - 2 = 5a + 5b - 2
Since
5a + 5b - 2 '" 5a - 2 + 5b - 2 = F ( a ) + F (b) , F (x) = 5x - 2 does not have the property.
d.
G (x) = -1 x 1 "' -1 + -1 G (a) + G (b) G(a + b) = a+b a b G (x) = -1 does not have the property. x =
105.
Answers will vary.
124
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Section 3.2: The Graph of a Function
S ection 3 . 2 1.
n.
x 2 + 4/ = 1 6 x-intercepts: x2 + 4 (0) 2 16 x2 = 16 x = ±4 => (-4, 0) , ( 4, 0) y-intercepts: (0) 2 + 4y2 = 1 6 4y2 = 1 6 y2 = 4 y = ±2 => (0, -2) , (0, 2)
11.
13.
=
3. 5.
9.
a.
b. c.
d.
e. f.
g. h.
i.
j. k. I. m.
Not a function since vertical lines will intersect the graph in more than one point. Function Domain: {xl - 7t � X � 7t} ; Range: { Y I - 1 � y � l}
( % ) (% , 0) . (0, 1)
b.
Intercepts: - , 0 .
c.
Symmetry about y-axis.
15.
Not a function since vertical lines will intersect the graph in more than one point.
17.
Function Domain: {x l x > O} ; Range: { Y I y is any real number} b. Intercepts: ( 1 , 0) c. None
f (x) = ax 2 + 4 a (-1) 2 + 4 = 2 => a = -2 False; e.g. y = -1 . x
=
a.
vertical
7.
f(x) = - 2 when x = -5 and x 8.
a.
19.
f (O) = 3 since (0, 3) is on the graph. f(- 6) = -3 since (- 6, -3) is on the graph. f(6) = 0 since (6, 0) is on the graph. f(l l) = 1 since (1 1, 1) is on the graph. f(3) is positive sincef(3) ,., 3 .7. f( -4) is negative since f (-4) ,., - 1 . f (x) = 0 when x = -3, x = 6, and x = 10. f(x» O when - 3 < x < 6, and l 0 < x � 1 1 .
Function Domain: {x I x is any real number} ; Range: { Y I y � 2} b. Intercepts: (-3, 0), (3, 0), (0,2) c. Symmetry about y-axis. a.
21.
Function Domain: {x I x is any real number} ; Range: { Y I y � -3} b. Interc epts: ( 1 , 0), (3,0), (0,9) c. None a.
The domain offis {xl - 6 � x � l l} or [ - 6, 1 1] . 23.
The range offis { Y I - 3 � Y � 4} or [-3, 4] . The x-intercepts are -3 , 6, and 10. The y-intercept is 3. The line y = ! intersects the graph 3 times. 2 The line x = 5 intersects the graph 1 time. f (x) = 3 when x = 0 and x = 4.
f(x) = 2x 2 - x - l a.
b.
f(-I) = 2 ( _ 1) 2 - (-1) - 1 = 2 The point (-1, 2) is on the graph off f(-2) = 2( _2) 2 - ( -2) - 1 = 9 The point (-2, 9) is on the graph off
125
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publ isher.
Chapter 3: Functions and Their Graphs
c.
Solve for x : -1 = 2x 2 - x - 1 0 = 2X 2 _ X o = x ( 2x - 1 ) => = 0, x = 1
27.
(1 , -1 ) are on the graph of I · The domain oflis { x l x is any real number } .
a.
x
(0, -1) and
d. e.
I(x) = � X4 + 1
b.
x-intercepts: 1 ( x ) =0 => 2X 2 - x - 1 = 0
( 2x + 1 )( x - 1 ) = 0 => x = - "2'1 x = 1
25.
(-± 0) and ( 1, 0)
c.
y-intercept: 1 ( 0 ) =2 ( 0 )2 - 0 - 1 = -1 => ( 0, -1 ) I(x) = x + 2 x-6 a. 1(3) = 3 + 2 = -� * 14 3-6 3 The point ( 3, 14 ) is not on the graph off
b.
c.
d. e.
d.
1(4) = 4 + 2 = � = -3 4-6 -2 The point ( 4, -3 ) is on the graph off Solve for x : 2= x+2 x-6 2x - I 2 = x + 2 x = 14 ( 14, 2) is a point on the graph of 1 .
e.
Solve for x :
2 1 = 2x 4 X +1 X 4 + 1 = 2x 2 X 4 _ 2x 2 + 1 = 0 (x 2 _ 1) 2 = 0 Z x - 1 = 0 => x = ±I ( 1 , 1 ) and (-1 , 1 ) are on the graph of 1 .
The domain of1 is { x I x is any real number } . x-intercept: z l (x ) =O => 2x = O X4 + 1 z 2x = 0 => x = 0 => ( 0, 0 ) -
f.
y-intercept: 2 (0 )z 0 = __ = 0 => ( 0, 0 ) 1(0) = 04 + 1 0 + 1
44x-z + x + 6 h ( x ) = -z -
The domain oflis { x l x * 6 } . x-intercepts: l (x ) =O => x + 2 = O x-6 x + 2 = O => x = -2 => ( -2, 0 ) y-intercept: 1 ( 0 ) = 0 + 2 = - .!. => 0, - .!. 0-6 3 3
29.
44 ( 8 ) 2 - + (8) + 6 28 2 = _ 28 1 6 + 14 784 "" lOA feet
a.
h (8) = -
b.
h (1 2) =
v
-
f.
( I� ) is on the graph off
The point 2,
'
f.
2 1(-1) = 2(-1) = � = 1 (_1) 4 + 1 2 The point (-1 , 1 ) is on the graph off 2 _� I( 2 ) _- 2(2) 4 (2) + 1 1 7
( )
-
44 ( 1 2 ) 2 + (1 2) + 6 28 2 = _ 6336 + 1 8 784 "" 9.9 feet
126
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Section 3.2: The Graph of a Function
c.
From part (a) we know the point ( 8, 1 0.4 ) is on the graph and from part (b) we know the point ( 1 2, 9.9 ) is on the graph. We could evaluate the function at several more values ofx (e.g. x = O , x = 1 5 , and x = 20 ) to obtain additional points. 44 ( 0 ) 2 h (O) = - + (0) + 6 = 6 28 2 44 ( 1 5 ) 2 h (1 5) = + ( 1 5 ) + 6 � 8.4 28 2 44 ( 20 ) 2 + ( 20 ) + 6 � 3.6 h ( 20 ) = 28 2 Some additional points are ( 0, 6 ) , ( 1 5, 8 .4 )
31.
b.
c.
and ( 20, 3.6 ) . The complete graph is given below.
d.
h
10
(8, 10.4)
5 x
d.
=
a.
-
15
-32x-2 hex) = - +X 1 30 2 -32(1 00) 2 + 1 00 h(IOO) 1 3 02 = -320, 000 + 1 00 � 8 1 . 07 fieet 1 6, 900
44 ( 1 5 ) 2 + ( 1 5 ) + 6 � 8.4 feet 28 2 No; when the ball is 1 5 feet in front of the foul line, it will be below the hoop. Therefore it cannot go through the hoop. h (1 5 ) =
In order for the ball to pass through the hoop, we need to have h ( 1 5 ) = 1 0 . 44 ( 1 5 ) 2 + ( 1 5) + 6 10 = v2 44 ( 1 5 ) 2 -1 1 =
e.
f + 300 h(300) = -32(300 1 30 2 - 2, 880, 000 + 300 � 129.59 feet = 1 6, 900 -32(500) 2 h(500) = + 500 1 30 2 - 8, 000, 000 + 500 � 26.63 feet = 1 6, 900 -32x 2 Solving hex) = --2- + x = 0 130 2 -32x +x=O 1 30 2 x -322X + 1 = 0 1 30 x = O or -32x2 + 1 = 0 130 1 = 32x2 130 1 30 2 = 32x 130 2 = 528. 13 feet x = -32 Therefore, the golf ball travels 528.13 feet. -32x 2 YI = --- + X 1 30 2
-( )
1 50
1::::::=:::=::1 600
v 2 = 4 ( 225 ) v 2 = 900 v = 30 ft/sec The ball must be shot with an initial velocity of 30 feet per second in order to go through the hoop. 127
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Chapter 3: Functions and Their Graphs f.
=
Use INTERSECT on the graphs of -32x 2 + x and =90 . Y2 Yl = 130
33. C(x) 1 00 + � + 36000 x 10
-
1 50
o
2
I..-______....J
-5
a.
600
I SO
b.
-5
200 225
� 300 325 350
h.
c.
'1 5 0 500
� 650 700 750
K
35. a.
1 2 '1 . 2 6 1 2 9 . 1 '1 B 1. 6 6 131.B 129 . 59 125 118.05
b.
264
Vl B2 B2.01
� 132.0l 132.03 132.02
X
260 261 262 2n 2 6 '1 265 266
c. d. e.
Vl 132 132. 01 132.02
f.
� B2.03 B2.02
Y l - 1 32 . 029 1 1 2426 Y l = 1 32 . 03 1 242604 x
"0 261 262 263 2 6 '1 265 266
ERROR B25 '1 7 0 l55 lOO 269 250
Y1
rru
225 222 2 2 0 . '1 5 220 220.3B 2 2 1 . '1 3 22l
X:... 6 00
The ball travels approximately feet before it reaches its maximum height of approximately 1 32.03 feet. 260 261 262 263 2 6 '1 265 266
1m
The cost per passenger is minimized to about $220 when the ground speed is roughiy 600 ·1 es per hour. X
Vl
�:... 2 75
x
1'/ 1 S 1 00+X/ 1 0+360 ...
The ball reaches a height of 90 feet twice. The first time is when the ball has traveled approximately 1 1 5.07 feet, and the second time is when the ball has traveled about 4 1 3 .05 feet. The ball travels approximately 275 feet before it reaches its maximum height of approxunate I IJ 1 3 1 8 feet. X
=
TblStart =0; �Tbl 50 0 50 100 150 200 250 300
O�S=60
g.
11======:::=1 1 000
Yl
=
-
37.
The graph of a function can have any number of x-intercepts. The graph of a function can have at most one y-intercept (otherwise the graph would fail the vertical line test).
39.
(a) III; (b) IV; (c) I; (d) V; (e) II
132 132.01 B2.02 B2.03
� 132.02
Yl = 1 32 . 029585799
(f +g)(2)= f(2) + g(2)=2 + 1=3 (f +g)(4)= f(4) +g(4) 1 + (-3)=-2 (f-g)(6)=f(6)-g(6) = 0 - 1=-1 (g /)(6)=g(6)-f(6)= 1 - 0=1 (f .g)(2)=f(2) ·g(2)=2(1)=2 (gf }4)=g(f(44)) = _-31 = _ .!.3
128
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Section 3.3: Properties of Functions
y
41.
origin: x � and y � -y (-Y ) = S (-x) 2 - 1 -x
-y = Sx2 - 1 y = -Sx2 + 1 different The equation has symmetry with respect to the y-axis only. (6, 0)
43.
a.
b. c. d. e.
f.
g. h. i.
45.
10
5.
Time (in minutes)
2 hours elapsed; Kevin was between 0 and 3 miles from home. O.S hours elapsed; Kevin was 3 miles from home. 0.3 hours elapsed; Kevin was between 0 and 3 miles from home. 0.2 hours elapsed; Kevin was at home. 0.9 hours elapsed; Kevin was between 0 and 2.8 miles from home. 0.3 hours elapsed; Kevin was 2.8 miles from home. 1 . 1 hours elapsed; Kevin was between 0 and 2.8 miles from home. The farthest distance Kevin is from home is 3 miles. Kevin returned home 2 times.
The intercepts are ( -3, 0 ) , ( 3, 0 ) , and ( 0, -9 ) .
Answers (graphs) will vary. Points of the form (S, y) and of the form (x, 0) cannot be on the graph of the function.
7.
even; odd
9.
True
11.
Yes
13.
No, it only increases on (S, 1 0).
15.
f is increasing on the intervals ( -8, -2 ) , ( 0, 2 ) , (S, oo ) .
17.
Yes. The local maximum at x = 2 is 10.
19.
f has local maxima at x = - 2 and x = 2 . The
21.
Section 3 . 3 1.
2<x<S
3.
x-axis: y � -y
y = x2 - 9 X-intercepts: 0 = x2 - 9 x2 = 9 � x = ±3 y-intercept: y = ( 0 ) 2 _ 9 = -9
local maxima are 6 and 1 0, respectively. Intercepts: (-2, 0), (2, 0), and (0, 3). b. Domain: {xl - 4 � x � 4} or [-4, 4 ) ; Range: { Y I 0 � y � 3} or [ 0, 3 ] . a.
c. d.
(-Y ) = Sx2 - 1 -y = Sx2 - 1 y = -Sx2 + 1 different y-axis: x � -x y = S (-x) 2 - 1 Y = Sx2 - 1 same
23.
a.
b. c.
d.
Increasing: (-2, 0) and (2, 4); Decreasing: (-4, -2) and (0, 2). Since the graph is symmetric with respect to the y-axis, the function is even. Intercepts: (0, 1). Domain: {x I x is any real number } ; Range: { y l y > O } or ( 0, 00 ) . Increasing: ( -00 , 00 ) ; Decreasing: never. Since the graph is not symmetric with respect to the y-axis or the origin, the function is neither even nor odd.
129
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Chapter 3: Functions and Their Graphs
25.
a.
b.
c.
d.
27.
a.
b. c.
d.
29.
31.
a.
b.
a.
b. 33.
35.
37.
Intercepts: (-1t, 0), (1t, 0), and (0, 0) .
39.
Domain: {xl - 1t � X � 1t} or [-1t, 1t] ; Range: { Y I - I � y � l} or [-l, l] . Increasing:
1 41. g(x) = 2
( -%' %}
( %) and (%' 1t)
Decreasing: -1t, -
x g (-x) = _1 _ = � = g (x) (-xi x2 Therefore, g is even.
.
Since the graph is symmetric with respect to the origin, the function is odd. Intercepts:
I(x) = x + 1 x l I(-x) = -x + l - x l = -x + 1 x l I is neither even nor odd.
3
-x43. h(x) = -9
G, 0) . (%' 0) . and ( 0, �) .
3x 2 3 3 h e -x) = _( _X)- 9 = x - 9 = -h (x) 3x2 3( _X)2 Therefore, h is odd.
Domain: {x l - 3 � x � 3 } or [-3, 3] ; Range: { y 1 - 1 � Y � 2} or [-1, 2] .
45. I (x) = x 3 - 3x + 2 on the interval ( -2, 2) Use MAXIMUM and MINIMUM on the graph of Yl = x3 - 3x + 2 .
Increasing: (2, 3) ; Decreasing: ( -1, 1 ) ; Constant: ( -3, - 1 ) and ( 1, 2) Since the graph is not symmetric with respect to the y-axis or the origin, the function is neither even nor odd.
5
2
I has a local maximum of 3 at x = O. I has a local minimum of 0 at both x = - 2 and x = 2.
I has a local maximum of 1 at x = ;
-
5
5 Y
I has a local minimum of -1 at x = - ; .
.J----....
/ -2 11
.
.
-
. " ..
" i r,i r,",ur,', X=.�����77�
I
/
-'-...-. . �
'1 = 1 . !; !; [ - 1 1
2
x
-5
local maximum at: ( -1, 4) ; local minimum at: ( 1 , 0) I is increasing on: ( -2, -1 ) and ( 1 , 2) ; I is decreasing on: ( -1, 1 )
I(x) = 4x3 1(-X) = 4(-X)3 = - 4x3 = -/ (x) Therefore, I is odd. g (x) = -3x2 - 5 g (-x) = -3(-x)2 - 5 = - 3x2 - 5 = g (x) Therefore, g is even. F(x) =
130
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Section 3.3: Properties of Functions
4 7. f (x) = x5 _ x 3 on the interval ( -2, 2)
51.
Use MAXIMUM and MINIMUM on the graph of y\ = x5 _ x3 . -2
05
l
1\
" o x i l'l u l'l X= ·.77�S9S.
!
8
2
'V
�'
\
I
ttinil'tlu,....
X=.77�S978
V=.1BS9Q32
.. _.----_/ \.•.., �" --
-- -�
- 3 h";-;-: i r:-C ' i :: r,)� u :- r:::')
-+-
�= · 1 . B 6 � (I !I !I
= -.
\.
2
v 1 0n
-0 . 5
/
\
-0 . 5 0.5
-2
f (x) = 0.25x4 + 0.3x3 - 0.9X2 + 3 on the interval ( -3, 2) Use MAXIMUM and MINIMUM on the graph of y\ = 0.25x 4 + 0.3x 3 - 0.9x2 + 3 .
-2
8
!.'
8S9
-3
local maximum at: (-0.77, 0. 19) ; local minimum at: (0.77, -0. 19) ; fis increasing on: (-2, -0.77) and (0.77, 2) ; fis decreasing on: (-0.77, 0.77)
--
\
-3
\
----
\..-
" i rl i r,". u r,".
6
-
J../ �
_-If-_ ',' = 3
..../
�: = 1 . 1 3 B '1 E
�;,
49. f (x) = -0.2X3 - 0.6x 2 + 4x - 6 on the interval ( -6, 4) Use MAXIMUM and MINIMUM on the graph of y\ = -0.2x3 - 0.6x2 + 4x - 6 . 2
�
" '] :( i �) u r...
-:i
\' =. 9 '1 B 0 '1 2: B B
:i
8
-2
_--r-�"'"
'/ = <: . 6 '1 B 2: B 2: 1
-2
2
�./'
�-
X = . !I 6 � O !l B B ':
2
'X
2
local maximum at: ( 0, 3) ; local minimum at: (-1 .87, 0.95) , (0.97, 2.65) fis increasing on: (-1 .87, 0) and (0.97, 2) ; fis decreasing on: (-3, -1 .87) and (0, 0.97) 53.
f(x) = _2X 2 + 4 Average rate of change off from x = ° to x=2 f (2) - / (0) _ _2 (2) 2 + 4 - -2 (0) 2 +4 2 2-0 (-4) - ( 4) -8 = = = 4 2 2 b. Average rate of change off from x = 1 to x = 3: f (3) - f (l) _ _2 (3) 2 + 4 - _2(1) 2 + 4 2 3-1 (-14) - (2) -16 = - = -8 = 2 2 a.
X = 1 .76BB767 H o x i t'l u t'l
'1'= - 1 . 9 1) B 7 B '1
-30
.'.
local maximum at: (1 .77, - 1 .91) ; local minimum at: (-3.77, - 1 8.89) fis increasing on: (-3.77, 1 .77) ; f is decreasing on: ( -6, -3.77) and (1. 77, 4 )
(
)(
(
)(
131
)
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)
Chapter 3: Functions and Their Graphs
Average rate of change of/ from x = 1 to x=4: /(4)- /(1) _- (_2{4)2+4)-(-2{1)2+4) 3 4-1 (-28)-(2) = 3 =-303 = -10 55. g(x)= x3 -2x+1 Average rate of change ofg from x = -3 to x=-2 : g(-2)-g(-3) -2-(-3) [(_2)3 -2(-2)+IJ- [(-3)3 -2(-3)+IJ 1 (-3)-(-20) = 1 =!21 = 17 Average rate of change ofg from x = -1 to x =1 : g(I)-g(-I) 1-(-1) [(1)3 -2(1)+1J- [(_1)3 -2(-1)+1J 2 -2 = (0)-(2) 2 = 2 =-1 Average rate of change ofg from x=1 to x=3 : g(3)-g(l) 3-1 [(3)3 -2(3)+IJ- [(1)3 -2(1)+IJ 2 (22)-(0) 22 =11 2 2 5 7 . /(x)=5x-2 Average rate of change of/ from 1 to 3: �Y = /(3)- /(1) = 13-3 10 = 2 =5 3-1 3-1 Thus, the average rate of change of/from 1 to 3 is 5.
b.
c.
(1, / {1)) (3,/(3)) 5.
= (x-Xl ) y-3 5(x-l) y-3=5x-5 Y =5x-2 5 9. g(x)= x2-2 Average rate of change ofg from to 1: �Y g(I)-g(-2) = -1-2 = -3 = = 1-(-2) 1-(-2) 3 -1 Therefore, the average rate of change ofg from -2 to 1 is -1 . From (a), the slope of the secant line joining ( g(-2) ) and (1,g(1)) is -1. We use the =
a.
-2
a.
�
b.
b.
-2 ,
point-slope form to find the equation of the secant line:
Y- Yl = (x-Xl ) y-2=-1(x-(-2)) y-2=-x-2 y = -x h(x)=x2-2x Average rate of change of h from 2 to 4: �Y = h(4)-h(2) 8-0 = � =4-2 2 =4 4-2 Therefore, the average rate of change of h from 2 to 4 is 4. From (a), the slope of the secant line joining (2,h(2)) and (4,h(4)) is 4. We use the msec
61.
c.
=
From (a), the slope of the secant line joining is We use the point and slope form to fmd the equation of the secant line: Y - Yl msec
a.
�
b.
point-slope form to find the equation of the secant line:
=
Y- Yl = (X-Xl ) y-O=4(X-2) Y =4x-8 length = 24 -2x width = 24-2x height = X V(x)=x(24-2x)(24-2x)=x(24-2X)2 V(3)= 3(24-2(3» 2= 3(18)2 = 3(324) = 972 cu.in. msec
a.
63. a.
�
b.
;
132
;
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Section 3.3: Properties of Functions
c.
d.
h.
V(1 0) = 10(24 - 2(1 0)) 2 = 1 0(4) 2 = 10(16) = 1 60 cu.in. 2 Yl = x(24 - 2x)
2500
1 1 00
;; ;;;: 1 2 :=I o �;;:;;;:;===;;;;:; o
-300
1 1 00
Use MAXIMUM.
c.
69. a. o
"axiMUIYI
X=� �Y=102�
Use MINIMUM. Rounding to the nearest whole number, the average cost is minimized when approximately 10 lawnrnowers are produced per hour.
The minimum average cost is approximately $239 per mower.
P(2.5) - p(O ) 2.5 - 0 0. 1 8 - 0.09 2.5 - 0 0.09 2.5 = 0.036 gram per hour
avg. rate of change =
12
o
The volume is largest when x = 4 inches. 6 5 . a.
s e t) = -16t 2 + 80t + 6 110
o h.
c.
67.
On overage, the population is increasing at a rate of 0.036 gram per hour from 0 to 2.5 hours.
L::=:===�:::::.J o
h.
6
p( 6) - P(4.5) 6 - 4.5 0.50 - 0.35 6 - 4.5 0. 15 1 .5 = 0. 1 gram per hour
avg. rate of change = ----''--'-----'---. - .:...
Use MAXIMUM. The maximum height occurs when t = 2.5 seconds. 1 10
o
(\
"axiMUM
X=2.5 _Y=106
o
\ L
On overage, the population is increasing at a rate of 0. 1 gram per hour from 4.5 to 6 hours.
6 c.
From the graph, the maximum height is 106 feet.
C (x) = 0.3x 2 + 2 1x _ 25 1 + 2500 x 2500 2 a. Yl = 0.3x + 2 1x - 25 1 + - x
71.
0�
The average rate of change is increasing as time passes. This indicates that the population is increasing at an increasing rate.
I(x) = x 2 a.
2500
-300
-'---"--�
Average rate of change ofI from x = 0 to
x=1 : 1 (1) - 1 (0) 1 2 _ 02 ! = =I = 1 1 1-0
� 30
__ __ __ __ __
133
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Chapter 3: Functions and Their Graphs
b.
c.
d.
e.
f.
Average rate of change off from x = 0 to x = 0.5 : f(0.5) - f (O) = (0.5) 2 _ 02 = 0.25 = 0.5 0.5 - 0 0.5 0.5 Average rate of change off from x = 0 to x = 0.1 : f(O. l) - f (O) = (0. 1) 2 _ 02 0.01 = = 0. 1 0. 1 - 0 0. 1 0. 1 Average rate of change off from x = 0 to x = O.O l : f (O.Ol) - f (O) (O.Q1f _ 02 0.01 0.01 - 0 = 0.0001 = 0.0 1 0.0 1 Average rate of change off from x = 0 to x = O.OOl : f(O.OOl) - f (O) _ (o.OO lf _ 02 0.001 - 0 0.001 0.000001 = 0.001 = 0.001 Graphing the secant lines:
. _...... .. .c:;;.-l . ....---:"..".,--:::....tx
-I '\ 1
�
;/
"
y=
�"-..
/"
y = O.OO lx
I
X
g. h.
-I
The secant lines are beginning to look more and more like the tangent line to the graph offat the point where x = O . The slopes of the secant lines are getting smaller and smaller. They seem to be approaching the number zero.
73. f(x) = 2x + 5 a.
= f(x + h) - f(x) h 2( x + h) + 5 - 2x - 5 = 2h 2 = h h When x = l : =2 h = 0.5 => h = 0. 1 => =2 h = O.Ol => msec = 2 as h � 0, � 2 msec
=
b.
msec
msec
msec
��
c.
- I I ....---.,;:�..-.:::.. --
'\ -)
�
�
y=
:/
�
.
�
-I
m
x
-I
Using the point (1, 1 (1) ) = (1, 7) and slope, = 2 , we get the secant line: Y - 7 = 2 (x - 1) y - 7 = 2x - 2 y = 2x + 5 Graphing:
d.
I
\0
/ /'" "5 ------':'7'/." ��-+-���- 5 "-
x
l'"../l'"
"5
The graph and the secant line coincide. 134
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Section 3.3: Properties of Functions
b.
75. f(x) = x2 + 2x a.
b.
msec
= f(x + h) - f(x) h + h)2 + 2ex + h) - ex2 + 2x) (X = h
mse
c.
Using point (1, f (I)) = (1, 0) and slope 1 .02, we get the secant line: y - 0 = 1 . 02 ( x - I ) y = 1 .02x - 1 .02 Graphing:
d.
-I
=
c
-----
�.I
c
mse
c
.
Using point (1, f (1)) = (1, 3) and slope 4.01, we get the secant line: y - 3 = 4.0 1(x - l) y - 3 = 4.01x - 4.01 y = 4.0 1x - 1 .0 1 Graphing: =
d.
//
1
---..�
-1
1 7 9. f(x) = -x a.
/i;"
6
!.'
./f../
.. .
x
..-
b.
= f(x + h) - f(x) h 2 (x + h) 2 - 3 (x + h) + 1 - ( 2x2 - 3x + 1 ) h
� �-
,
...
/
/�---
....
�:.-
f-------
2
:(
-3
= f(x + h) - f(x) h
(x + h) X When x 1 , =h = 0.5 � =
msec
msec
1 ( 1 + 0.5 )( 1 ) 1 __ 1 .5 _ �3 "" -0.667 1 =( 1 + 0. 1 )( 1 ) "" -0.909 = 1 1. 1 = _ !2. 11 1 =( 1 + 0.01 ) ( I ) "" -0.990 = _ _1_ = _ 100 1 .01 101 1 = --1 = -1 � (1 + 0)(1) 1 =
2(x 2 + 2xh + h 2 ) - 3x - 3h + I - 2x 2 + 3x - l h 2X 2 + 4xh + 2h 2 - 3x - 3h + 1 - 2X 2 + 3x - I
4xh + 2h2 - 3h h = 4x + 2h - 3
msec
2
\
2
77 . fex) = 2x2 - 3x + 1 a.
c
=
=
mse
=
msec
msec
c.
=
msec
x2 + 2xh + h2 + 2x + 2h - X 2 - 2x h 2xh + h2 + 2h h = 2x + h + 2 When x 1 , h = 0.5 � = 2 · 1 + 0 . 5 + 2 = 4.5 = 2 · 1 + 0.1 + 2 = 4. 1 h = 0. 1 � = 2 · 1 + 0.01 + 2 4.01 h = 0.01 � as h � 0, � 2·1+0+2 =4 mse
When x 1 , h = 0.5 � msec = 4 · 1 + 2 (0.5) - 3 = 2 h = 0. 1 � = 4 . 1 + 2 (0. 1) - 3 = 1 .2 = 4 · 1 + 2 (0.01) - 3 1 .02 h = 0.Q1 � � 4 · 1 + 2 (0) - 3 = 1 as h � 0,
h = 0. 1 �
msec
=
_ _
h
h = 0.01 �
as h � 0,
mse
c
m se
c
135
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Chapter 3: Functions and Their Graphs
c.
Section 3.4
Using point (1, f (1)) = (1, 1) and 100 , we get the secant l'me: slope -101 y-1 = (x - I) 00 x + 100 y - 1 = - 11OI 101 201 1 00 Y = - 101 x + IOl
1.
=
-���
d.
Graphing:
x
3
-5
-1 -1
Answers will vary. One possibility follows: y
2
5.
(2, -6)
83.
85.
y
5
3.
81.
y = -[;
A function that is increasing on an interval can have at most one x-intercept on the interval. The graph of/could not "tum" and cross it again or it would start to decrease.
Y = x3 - 8 intercept: Let x = 0 , then y = (0)3 - 8 = -8 . x-intercept: Let y = 0 , then 0 = x3 - 8 x3 = 8 x=2 The intercepts are (0, -8) and (2, 0) . piecewise-defined y-
7.
False; the cube root function is odd and increasing on the interval (-00, 00) .
9.
C
11.
E
13. B
To be an even function we need f (-x) = / (x) and to be an odd function we need / (-x) = -/ (x) . In order for a function be both even and odd, we would need / ( x ) = -/ ( x) . This is only possible if f (x) = 0 .
15.
F
17.
/ (x) = x
y
x
136
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Section 3.4: Library of Functions; Piecewise-defined Functions
19.
/ (x) = x3
29.
l e x) = a. b.
x
e1
x
if x 7: 0 if x = 0
Domain: {x I x is any real number} x-intercept: none y- intercept: / (0) = 1 The only intercept is (0, 1) .
21.
c.
/ (x) = -1 x
y
5
( 2,
Graph:
y
�)
V
I
I
X
X �
5
d. 23.
l e x) = $
31.
y
. .5 ) 0 , a' ::::;:{L
b.
25. a.
b.
c.
27. a.
c.
/(- 2) = (_ 2) 2 = 4 /(0) = 2 /(2) = 2(2) + 1 = 5
c.
/ (2) = 2 (2) - 4 = 0
d.
/ (3) = (3) 3 - 2 = 25
Graph:
-5
5
x
-5
/ (0) = 2 (0) - 4 = -4 / (1) = 2 (1) - 4 = -2
x-intercept: none y- intercept: / ( 0 ) = -2 ( 0 ) + 3 3 The only intercept is (0, 3) .
I
b.
<
=
�x
(-I , -I )
-5
{3x- 2-X 2+ 3
if x 1 if x 2': 1 Domain: {x I x is any real number}
/(X) = a.
5
Range: {YI y 7: o } ; (-00, 0) or (0, 00 )
d.
Range: { Y I Y 2': I} ; [1, (0)
1 37
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Chapter 3: Functions and Their Graphs
33.
{:
c.
if - 2 � x < l if x = 1 f(x) = -x + 2 if x > 1 a.
b.
+3
Domain: {x l x � - 2} ; [-2, (0) -x + 2 = 0 x+3 = 0 -x = -2 x = -3 x=2 (not in domain) x-intercept: 2 y-intercept: f (0) = 0 + 3 = 3
x
-2 d.
The intercepts are (2, 0) and (0, 3) . c.
37.
Graph:
{
b.
-s
35.
Range: { Y I y < 4 and y = 5} ; (-00, 4) U {5}
Range: {y I y is any real number}
x f(X) = l l x3 a.
x
d.
Graph:
c.
if - 2 � x < 0 if x > 0
Domain: {x l - 2 � x < O and x > 0} or {x l x � -2, x * 0} ; [ -2, 0) U (0, 00) . x-intercept: none There are no x-intercepts since there are no values for x such that f ( x ) = 0 . y-intercept: There is no y-intercept since x = 0 is not in the domain. Graph: y
{� x
if x < 0 if x � 0 Domain: {x I x is any real number}
f(x) = ; a.
b.
x
l+x =0 x=O x = -1 x-intercepts: -1, 0 y-intercept: f ( 0 ) = 0 2 = 0
-1 d.
The intercepts are (-1, 0) and (0, 0) .
39.
Range: {yl y > O} ; (0, 00)
f(x) 2 int(x) Domain: {x I x is any real number} b. x-intercepts: All values for x such that 0 � x < 1 . y-intercept: f ( 0) = 2 int ( 0) = 0 The intercepts are all ordered pairs (x, 0) when 0 � x < 1 . =
a.
138
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Section 3.4: L ibrary of Functions; Piecewise-defined Functions c.
Graph:
49. a.
y
4
b.
-4
4
x
c. d.
Range: { Y I Y is an even integer}
41.
Answers may vary. One possibility follows: -x if - 1 ::; x ::; 0 f(x) = 1 -x if 0 < x � 2 2
43.
Answers may vary. One possibility follows: -x if x � 0 f(x) = -x + 2 if 0 < x � 2
{
{
45. a.
b.
c.
47.
d.
f(1 .6) = int (2(1 .6») = int(3.2) = 3 f( -1 .8) = int ( 2 ( 1 .8») = int( -3.6) = - 4 -
{
C
5b
{
i:s
35 if O < x � 300 C= OAOx - 85 if x > 300 a.
51.
f(1 .2) = int (2(1 .2») = int(2.4) = 2
Charge for 50 therms: C = 9.45 + 0.7958(50) + 0.36375(50) = $67.43 Charge for 500 therms: C = 9.45 + 0.36375(50) + 0. 1 1445(450) + 0.7958(500) = $477.04 For 0 � x � 50 : C = 9.45 + O.36375x + 0.7958x = 1 . 1 5955x + 9.45 For x > 50 : C = 9.45 + 0.36375 (50) + 0. 1 1445 (x - 50) + 0.7958x = 9.45 + 1 8 . 1 875 + 0. 1 1445x - 5.7225 + 0.7958x = 0.91 025x + 2 1 .9 1 5 The monthly charge function: 1 . l 5955X + 9.45 for 0 ::; x ::; 5 0 C0.9 1 025x + 2 1 .9 1 5 for x > 50 Graph:
6
1 00
C(200) = $35.00
b.
C (365) = 0.40 (365 ) - 85 = $61 .00
c.
C (301) = OAO(301) - 85 = $35.40
For schedule X: 0. 10x 755.00 + 0. 1 5(x - 7550) 4220.00 + 0.25(x - 30, 650) f(x) = 1 5, 1 07.50 + 0.28(x - 74, 200) 37, 675.50 + 0.33(x - 1 54, 800) 97, 653.00 + 0.35(x - 336, 550)
(0, 9.45)
'---+----'--""-'
-,-,-�,-,I ->. 1 _ 1-,-
Thenns
10 0
x
if 0 < x ::; 7550 if 7550 < x ::; 30, 650 if 30, 650 < x ::; 74, 200 if 74, 200 < x ::; 1 54, 800 if 1 54, 800 < x ::; 336, 550 if x > 336, 550
139
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Chapter 3: Functions and Their Graphs
53.
a.
Let x represent the number of miles and e be the cost of transportation. 0 .50X if 0 � x � 1 00 0.50(100) + OAO(x - 1 00) if 100 < x � 400 C(x) 0.50(100) + 0 0 40(300) + 0.25(x - 400) if 400 < x � 800 0.50(100) + 0040(300) + 0.25(400) + O(x - 800) if 800 < x � 960
{ {
0 .50X 1 0 + 0AOx C(x) = 70 + 0.25x 270
if 0 � x � 100 if 1 00 < x � 400 if 400 < x � 800 if 800 < x � 960
y
'" ;...
2� 0
U
55.
(800. 270)
300
250
200
1 50
1 00
Dist ance
(miles)
b.
For hauls between 1 00 and 400 miles the cost is: C(x) = 1 0 + OAOx .
c.
For hauls between 400 and 800 miles the cost is: C(x) = 70 + 0.25x . 57.
Let x the amount of the bill in dollars. The minimum payment due is given by x if x < 10 1 0 if 10 � x < 500 f (x) = 30 if 500 � x < 1 000 50 if 1000 � x < 1 500 70 if x � 1500 =
a.
b. c. d.
e.
y
•
f.
500
1000
1 500
Amount of B i l l (dollars)
W
= lOoe (10 0 45 + 1 0.J5 - 5)(33 - 1 0) 40 e W = 33 _ '" 22.04 (10A5 + 10.Ji5 - 1 5)(33 - 10) '" -3 0 e W = 33 22.04 W = 33 - 1 .5958(33 - 1 0) = - 4°e When 0 � < 1 . 79 , the wind speed is so small that there is no effect on the temperature. When the wind speed exceeds 20, the wind chill depends only on the temperature. v
x
140
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Section 3.4: Library of Functions; Piecewise-defined Functions
59.
Let x the number of ounces and C (x) the postage due. For O < x � l : C(x) = $0.39 =
For
1
<
63.
=
x � 2 : C (x) = 0 . 3 9 + 0 2 4 .
=
$0.63
The graph of y = x 2 is the reflection of the graph of y = x 2 about the x-axis.
\'"
�'
-5
For 2 < x � 3 : C(x) = 0.39 + 2 (0.24) = $0.87
y
For 3 < x � 4 : C(x) = 0.39 + 3 (0.24) = $ 1 . 1 1
/
_
6
,,
il'� = - x 2;
r
c+
x
"
6
;:�;�r-
sJ 2.3 1
0---.
2.07 1 .83 1
0---.
�Q) r 8� 1.59 US [
61.
5
\
-6
3.27 f3 03 .
l.l 1 L 0.87
,I
,/ ... .,,\
.....
The graph of y = -I x I is the reflection of the graph of y = 1 x I about the x-axis.
For 1 2 < x � 1 3 : C (x) = 0.39 + 1 2 (0.24) = $3.27
�
"
/y = x2
0---. 0---.
-0
�
0 .63
$:
0.3'
0---.
� �
o
2
1
1 4
1
1 6
1
1 8
1
1 1 1 IO 12
Weighl (ounces)
Multiplying a function by -1 causes the graph to be a reflection about the x-axis of the original function's graph.
1 ./
65.
14
Each graph is that of y = x 2 , but shifted horizontally. 6
The graph of y = (x _ 1)3 + 2 is a shifting of the graph of y = x3 one unit to the right and two units up. Yes, the result could be predicted. 8 y1 = (x _I)3+2
l....
y = x,
:J
-2
x
67.
-2
If y = ( - k) 2 , k > O , the shift is to the right k units; if y = (x + k) 2 , k > 0 , the shift is to the left k units. The graph of y = (x + 4) 2 is the same as the graph of y = x 2 , but shifted to the left 4 units. The graph of y = (x - 5) 2 is the graph of y = x 2 , but shifted to the right 5 units. X
I
l ./ /
,
""-
..._-,i---·"
, / . J
J
."..-'1'
x
8
-J
The graphs of y = xn , a positive odd integer, all have the same general shape. All go through the points (-1, -1) , (0, 0) , and (1, 1) . As increases, the graph of the function increases at a greater rate for I x I > 1 and is flatter around 0 for Ixl <1 . n
n
Y:;��ll-y ts
../
�.'
-I
r
:;-;-� l.'-;/ ('1/
---c
1
x
-1
141
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Chapter 3: Functions and Their Graphs
69.
For 0 x 1 , the graph of Y = xr , r rational and r > 0 , flattens down toward the x-axis as r gets bigger. For x > 1 , the graph of y xr increases at a greater rate as r gets bigger. <
33.
<
=
Section 3.5 1. 3. 5.
35.
horizontal; right -5 , -2 , and 2 (shift left three units)
(c); To go from Y = f (x) to y = 2f (x) , we multiply the y-coordinate of each point on the graph of y = f (x) by 2 . Thus, the point (0, 3) would become (0, 6) . f(x) = x 2 - 1 Using the graph of y = x 2 , vertically shift downward 1 unit. y
False; to obtain the graph of y = f (x + 2) - 3 you shift the graph of y = f (x) to the left 2 units and down 3 units.
x
(0, - 1)
7. B
-2
9. H 11.
I
13.
L
15. 17.
F G
19.
y = (X - 4) 3
21.
Y = x3 + 4
23.
y = { _X)3 = _x3
25.
Y = 4x3
27.
29.
31.
37.
g(x) = x 3 + 1 Using the graph of y = x 3 , vertically shift upward 1 unit.
x
(1) y = .fx + 2 (2) y = - ( .fx + 2 )
39.
(3) y = - ( h + 2 ) = - h - 2
h(x) = .Jx - 2 Using the graph of y = .fx , horizontally shift to the right 2 units. y
5
(1) y = -.fx (2) y = -.fx + 2 (3) y = -.Jx + 3 + 2 (c); To go from y = f (x) to y = -f (x) we reflect about the x-axis. This means we change the sign of the y-coordinate for each point on the graph of y = f(x) . Thus, the point (3, 0) would remain the same.
-5
142
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Section 3.5: Graphing Techniques: Transformations
41.
f(x) = (x_1)3 + 2 Using the graph of y = x 3 , horizontally shift to the right 1 unit [y= (x_1)3 ] then vertically shift up 2 units [y=(X_l )3 + 2 ] .
47.
f(x)=- $
Using the graph of about the x-axis.
y= $ , reflect the graph
y
'
10
x
x
-10
4 9. 43.
g(x)= 4$
Using the graph of a factor of 4.
g(x)= h Using the graph of y= $ , reflect the graph about the y-axis. y
y= $ , vertically stretch by
10
x
-10
51 .
45.
h(x)= L= (�)(�) Using the graph of y =.!.. , vertically compress x
h(x)=_x3 + 2
y= x3 , reflect the graph =_x3 J , then shift vertically upward 2 units [Y=_x3 + 2 ] . Using the graph of about the x-axis [Y y
by a factor of -1 . 2
-5
x
----L_--L_-+_--L_---:--* x
143
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Chapter 3: Functions and Their Graphs
53.
57.
f(x) = 2(x + 1)2 - 3 Using the graph of y = x 2 , horizontally shift to the left 1 unit y (x + 1) 2 vertically stretch
[
]
=
[
J,
by a factor of 2 y 2 ( x + 1) 2 and then vertically shift downward 3 units Y = 2 (X + l) 2 - 3 .
[
5
]
(-4, 0)
-5
]
,
[
]
about the x-axis Y = - (x + 1) 3 and vertically shift downward 1 unit
]
shift upward 1 unit Y = .Jx - 2 + 1 . y
f(x) = - (x + l) 3 - 1 Using the graph of y = x 3 horizontally shift to the left 1 unit Y = (x + 1)3 ] , reflect the graph
[
g(x) = ·h - 2 + 1 Using the graph of y = Fx , horizontally shift to the right 2 units Y = .Jx - 2 and vertically
[
x
-5
59.
[
]
[
y
x
55.
,
downward 2 units Y = � - 2 .
'
=
h e x) = H - 2 Using the graph of y = Fx reflect the graph about the y-axis [ Y = � J and vertically shift
y
[
Y
=
'
]
- (x + 1)3 - 1 .
5
S
x
-2
-s
144
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Section 3.5: Graphing Techniques: Transformations
61.
b.
g(x) = 2 1 1 - x l = 21 - ( - l + x ) I = 2 I x - I i Using the graph of y = I x I , horizontally shift to the right 1 unit [Y = Ix - II J and vertically stretch by a factor or 2 [Y = 2 1x - II J
G(x) = f(x + 2) Shift left 2 units.
y
'
(-2, 2)
.
(-6, -2) ( 1. 0)
P(x) = -f(x) Reflect about the x-axis. y
5
-
63.
(-4. 2)
hex) = 2 int(x - 1) Using the graph of y int(x) , horizontally shift to the right 1 unit [Y = int(x - 1)] , and vertically stretch by a factor of 2 [y = 2 int(x - 1) ] . =
5
x
(0. -2)
y
5
-5
x
c.
5
(2, -2)
-5
.....,
d.
....., x
H(x) = f(x + 1) - 2 Shift left 1 unit and shift down 2 units. y 3
-5
65. a.
x
F(x) = f(x) + 3 Shift up 3 units . y
7 e.
x
Q(x) = -1 f(x) 2 Compress vertically by a factor of .!. 2
.
y
5
-3
x
-5
145
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Chapter 3: Functions and Their Graphs
f.
c.
g(x) = f( -x)
Reflect about the y-axis. y
5
P(x) = -f(x)
Reflect about the x-axis. (-� , t )
y
2
x
x
(4, -2) -5
g.
h(x) = f(2x)
Compress horizontally by a factor of y
.!.2
d.
H(x) = f(x + l) - 2 Shift left 1 unit and shift down 2 units. y
.
5
x
x
(-2, -2) -5
67. a.
F(x) = f(x) + 3 Shift up 3 units.
e.
�
Q(x) = f(x)
Compress vertically by a factor of
y 8
( � , 4)
(-n, 3) -;t
b.
y
- It 2
-z
� 2
.!.2
.
( i , !)
(n, 3)
x
x It
-I
G (x) = f(x + 2) Shift left 2 units.
f.
y
g(x) = f(-x)
Reflect about the y-axis . y
2
( � - z, t)
(-� , l) x
2
x
(- � - 2, -1)
-2
146
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Section 3.5: Graphing Techniques: Transformations
g.
c.
h e x) 1(2x) =
Compress horizontally by a factor of ..!. . 2 y
2
The graph of y -I (x) is the same as the graph of y I ( x) , but reflected about the x-axis. Therefore, we can say that the graph of y -I ( x) must be decreasing on the interval (-1, 5 ) . =
=
'
=
( 4It , 1 )
d.
x 1t
The graph of y I ( -x) is the same as the graph of y I ( x ) , but reflected about the y-axis. Therefore, we can say that the graph of y I ( -x) must be decreasing on the interval (-5, 1) . =
=
=
69. a.
The graph of y I (x + 2) is the same as the graph of y I (x) , but shifted 2 units to the left. Therefore, the x-intercepts are -7 and 1 . The graph of y l (x - 2) is the same as the graph of y I (x) , but shifted 2 units to the right. Therefore, the x-intercepts are -3 and 5 . The graph of y 41 ( x) is the same as the graph of y I ( x ) , but stretched vertically by a factor of 4. Therefore, the x-intercepts are still -5 and 3 since the y-coordinate of each is O. The graph of y I (-x) is the same as the graph of y I ( x) , but reflected about the y-axis. Therefore, the x-intercepts are 5 and -3 . =
73.
a.
=
b.
y 1 I(x) 1 =
(-2, 1) (-1 , I )
y 2
( I, I)
=
=
c.
-2
=
=
d.
b.
=
(- I, I )
=
=
71. a.
y 1 ( 1 x l)
y 2
-2
( I , 1) 2
-2
The graph of y I (x + 2) is the same as the graph of y I (x) , but shifted 2 units to the left. Therefore, the graph of I ( x + 2) is increasing on the interval (-3, 3) . =
=
b.
The graph of y l (x - 5) is the same as the graph of y I (x) , but shifted 5 units to the right. Therefore, the graph of l (x - 5) is increasing on the interval (4, 1 0) . =
=
147
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exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in wri ting from the publisher.
Chapter 3: Functions and Their Graphs
75.
f(x) = x 2 + 2x f(x) = (x 2 + 2x + 1) - 1 f(x) = (x + l) 2 - 1 Using f(x) = x 2 shift left 1 unit and shift down 1 unit.
79.
f(x) = x 2 + x + l
( ±) + I - ± ( 1 }2 f(x) = x + "2 + "43
f(x) = x2 + X +
,
y
Using f(x) = x2 shift left ,
.!.
2
unit and shift up
-3 umt.. 4
x
5
-
77.
f(x) = x 2 - 8x + 1 f(x) = ( x2 - 8x + 1 6 ) + 1 - 1 6 f(x) = (x - 4)2 - 1 5
-2
81.
Using f(x) = x 2 , shift right 4 units and shift down 1 5 units. y
2
-1
x
f (x) = 2x 2 - 1 2x + 19 = 2 ( X 2 - 6x ) + 19
= 2 ( X 2 - 6x + 9 ) + 1 9 - 1 8
= 2 (x - 3) 2 + 1 Using f (x) = x 2 , shift right 3 units, vertically stretch by a factor of 2, and then shift up 1 unit.
9
y
.5
(4. -1 5)
Vl · (1, 3) .
(3.. 1 (4, ) 3) x
-5
148
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83.
f( x) =-3x2 - 1 2x - 1 7 =-3 ( X2 +4x) -17
Section 3.5: Graphing Techniques: Transformations 87. a.
=-3 ( X2 +4x+4 ) - 1 7+ 1 2
= - 3 ( x + 2) 2 -5 Using f( x) = x2 , shift left 2 units, stretch
b.
vertically by a factor of 3, reflect about the x axis, and shift down 5 units. y
t �
(-2. -5)
(-3,-s)!A.
85.
= T(t) - 2, the graph of T(t) should be shifted down 2 units. This change will lower the temperature in the house by 2 degrees. To
y
graph
:�
� ��F � 5 f>4LJ l
x
I (-1, -8)
From the graph, the thermostat is set at nOF during the daytime hours. The thermostat appears to be set at 65°F overnight.
�
9
f2 j
I-
60� 56r !
I
Y =(X-C) 2 If C = 0, y = x2. If c = 3, y =(x- 3) 2; shift right 3 units. If c =-2, y =(x+ 2) 2; shift left 2 units.
c.
'
I
I
I
I
8
I
I
12
I
I
I
I
I
B [2 16 20 24 Time (hours after midnight)
I
28
I
t
I
I
To graph y =T ( t+1 ) ,the graph ofT ( t ) should be shifted left one unit. This change will cause the program to switch between the daytime temperature and overnight temperature one hour sooner. The home will begin warming up at 5am instead of 6am and will begin cooling down at 8pm inst ad of9pm. o
T
so
G: t,.,
x
'!)
3
-2
<:
t0-
�
::: tJ� 60 56 o
I
I
4
Time
I
I
I
16
I
I
20
I
24
(hoU!'s after midnight)
149
I
28
,t
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Chapter 3: Functions and Their Graphs
89.
F
= -9 C + 32 5 288 256 224 192 160 128 96
d.
'F
Select the 1 0% tax since the graph of Y1 = 0.9p (x) � Y2 = - 0.05X2 + 100x - 6800 for all x in the domain.
(100,212)
Section 3.6 1.
= -9 (K - 273) + 32 5 Shift the graph 273 units to the right. F
a.
"F
288 256 224 192 160 128 % 64 32 --�.�--��--J-��� K o 270 290 310 330 350 370
91.
a.
b. c.
The distance d from P to the origin is d = � X2 + i . Since P is a point on the graph of y = x2 -8 , we have: ,..-----
d (X) = � X2 + (X2 - 8) 2 = � x4- 1 5x2 + 64 b. c.
d (O) = � 04 - 1 5(0)2 + 64 = $4 = 8 d (l) = � (1)4 - 1 5(li + 64 = "' 1 - 1 5 + 64 = Fo = 5J2 �7.07 40
d.
p (x) = -0,05x2 + 1 OOx - 2000 Y2
Yl
-5
e.
d is smallest when x �-2.74 or when x�2.74 .
a.
The distance d from P to the point ( 1, 0) is d = � (x - 1)2 +i. Since P is a point on the graph of y = Fx , we have:
Select the 1 0% tax since the profits are higher. The graph ofY1 is obtained by shifting the graph of p(x) vertically down 1 0,000 units. The graph ofY2 is obtained by multiplying the y-coordinate of the graph of p(x) by 0.9. Thus,Y2 is the graph of p(x) vertically compressed by a factor of 0.9.
3.
�
f
d (X) = (X - 1f+ (Fx = � x2 - x + 1 where x�O .
1 50
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Section 3.6: Mathematical Models: Building Functions 2
b.
c.
Graphing the area equation: 10
0�======l2 c.
d
o
is smallest when x = -} .
:�L�
5. By definition, a triangle has area
d.
A = .!. b h, b = base, h = height. From the figure, 2 we know that b = x and h y. Expressing the area of the triangle as a function of x , we have: A(x) = '21 xy = '21 x ( x3) = '21 x .
The area is largest when x � 1 .4 1 . Graphing the perimeter equation: 12
=
4
7.
a.
A(x) = xy = x ( 1 6 - x2) = -x3 + 1 6x
b.
Domain: {xI O<x<4}
c.
The area is largest when x � 2.31 .
"�xifllufll X=1.�1�21�� �V=11.U�708
•
The perimeter is largest when x � 1.4 1 .
30
11.
a.
C = circumference, A = total area, r = radius, x = side of square =>
r = S-2x Total Area = areasquare + areacircle = x 2 + r 2 S-2x) 2 = X 2 + 2S- 20x+4x2 A( X) = x 2 + 1t (-C = 21tr = 1 0 - 4x
7r
1t
7r
b.
9.
a.
In Quadrant I, X2 + y2 = 4� y = .J4-x2
A(x) = (2x)(2y) = 4x.J4 - x2 b.
------
1t
Since the lengths must be positive, we have: 1 O - 4x>O and x>O - 4x>-1 0 and x>O and x>0 x< 2.S Domain: {xl 0< x< 2.S}
p(x) = 2(2x) + 2(2y) = 4x+4.J4 - x2
151
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Chapter 3: Functions and Their Graphs c.
The total area is smallest when x � 1 .40 meters.
1 9.
8
a.
d2 =d\2 +d/ d2 = (2 - 30t) 2 + (3 - 40t) 2 d (t) = � (2 - 30t) 2 + (3 - 40t) 2 = � 4 - 1 20t + 900t2 + 9 - 240t + 1 600t 2 = �2500t2 - 360t + 13
0 ..._______ 2.5 o
d2 =3-401 d,=2-30t
13.
a.
b.
Since the wire of length x is bent into a circle, the circumference is x . Therefore, C (x) = x . Since C = x = 2n r, r =
b.
Xl'lin=0 Xl'lax=.15 Xscl=.05 Yl'lin=-l Yl'lax=4 Yscl=l Xres=l
-2n . x
21. 1 5.
a.
A = area, r = radius; diameter = 2r A(r) = (2r)(r ) = 2r2
b.
P
= perimeter per) = 2(2r ) + 2r 6 r
Area of the equilateral triangle 1 -J3 x=-J3 x 2 A=-x '-
2
2
4
x2 From problem 1 6, we have r 2 = 3
x=.onOOB5
V=.�0000001
r = radius of cylinder, h = height of cylinder, V = volume of cylinder H -h . 1es: H = -B Y SInU " 1ar tnang R r Hr = R ( H - h) Hr = RH - Rh Rh = R H -Hr RH -Hr H (R - r) h= = R R H ( - r) nH (R - r) r 2 V = nr 2 h = nr 2 = R
( :
.
Area inside the circle, but outside the triangle: A(x) = nr 2 -J3 x 2
__4
"ini ..... u .....
-
=
1 7.
The distance is smallest at t � 0.07 hours.
J
[�_ )
-J3 X2 = n�- -J3 x2 = 3 4 3 4
1 52
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Chapter 3 Review Exercises
23.
a.
The time on the boat is given by �. The
.
. on 1and'IS gIven by hme Island
12 - x -5
e.
3
.
_
f. Town
5.
12-x
d1
= �x2 + 22 = �x2 + 4
The total time for the trip is:
b. c.
d.
T(x) = 12 5- x +� = 12 - x + � 3 5 3 Domain: { xl 0 S; x $12 }
d.
T(4) = 12 5- 4 + f,i2;4 3 .J2O = -58 +--,:" 3 3.09 hours
e.
T(8) = 125- 8 + J'i,2;4 3 = -4 + J68 ,:" 3. 5 5 hours 3 5
f.
7.
Chapter 3 Review Exercises
,
This relation represents a function. Domain = {-I, 2, 4}; Range = { O 3}.
a.
b. c.
d.
f(x) = �x2_4 a. f(2) = � = -./ 4 - 4 = .JO = 0 b. f(-2) = �(_2) 2 - 4 = -./ 4 -4 = .JO = 0 c.
--
1.
f(x - 2) = 3(x - 2) (x _ 2)2 _1 3(x-2) 3x - 6 x2 - 4x + 4 -1 x2 -4x +3 f(2x) = 3(2x) = � (2X)2 -1 4x2 -1
= 4 6-1 _ = �3 = 2 1(2) = � (2)2 - 1 _ f(-2) = 3(-2) = 4--61 = -63 = -2 (_2)2 - 1 fe-x) = 3(-x) = -3x (-x)2_ 1 x2 - 1 3X = -3x -I(x) = - x2 - 1 x2- 1
f(-x) = �(-x)2_ 4 = �X2_ 4 -f(x) = _�X2 - 4 f(x - 2) = �(x _ 2)2 - 4 = �x2- 4x + 4 -4 = �x2 - 4x f(2x) = �(2x)2 _ 4 = �4x2 _4 = � 4 (x2 - 1) = 2�x2 -I
x2 - 4 f(x) = -x2 a. f(2) = 22 - 4 = 4 -4 4 = Q4 = o 22 (_2)2 - 4 = 4 - 4 = Q = O b. f(-2) = 4 4 (_2)2 x2- 4 c. fe -x) = (_X)2 - 4 = ( -x )2 X2d.
e.
( )
---------,
f(x - 2) - (x _ 2)2 - 4 - x2 -4x + 4 -4 (X _ 2)2 (x _ 2)2 x2 -4x x(x -4) (x - 2)2 (x - 2)2
1 53
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exist. No portion of this material may b e reproduced, in any form or by any means, without permission in wri ting from the publi sher.
Chapter 3: Functions and Their Graphs
f.
9.
f(2x) = (2X)2 - 4 = 4x2 - 4 4x2 (2X)2 4 (x2 -1) x2 _ 1
(f . g)(x) = f(x) ' g(x) = (2 -x) (3x+ 1) = 6x + 2 - 3x2 - X = -3x2 +5x+2 Domain: {xl x is any real number } f (x) = f (x) = 2 - x g g (x) 3x + l 3x+l * 0 3x*-I�x*--31
()
x f(x ) = ----Zx -9
The denominator cannot be zero:
x2 - 9 * 0 (x+3)(x - 3)*0 x * -3 or 3 Domain: {xl x* -3, x * 3} 11.
{ l -�}
Domain: x x *
f(x) = "'/2 - x
1 9.
The radicand must be non-negative:
2 -x�O x:s:2
Domain: {xlx:S:2} or (-00,2] 1 3.
f(x) = Fx r;r
(f -g)(x) = f (x) -g(x) = 3x2 + X + 1- 3x = 3x2 - 2x+l Domain: {x I x is any real number }
The radicand must be non-negative and the denominator cannot be zero: x> 0 Domain: {xlx>O} or (0,00 ) 15.
f(x) =
x x2 + 2x - 3
(f·g)(x) = f(x)·g(x } = (3x2 + X+I) (3x)
The denominator cannot be zero:
x2 + 2x - 3 * 0 (x + 3) (x - l) * 0 x* -3 or 1 Domain: {xlx * -3, x * I} 1 7.
f(x) = 3x2 +x+l g(x) = 3x (f+g)(x) = f (x)+g(x) = 3x2 + X + 1+ 3x = 3x2 + 4x + l Domain: {x I x is any real number }
= 9x3 + 3x2 +3x Domain: {x I x is any real number } f (x} 3x2 +x+l ( gf ) (x) = g(x) = 3x
f(x) = 2 -x g(x) = 3x+l (f+g)(x) = f (x}+ g (x) = 2 -x+ 3x + 1 = 2x + 3 Domain: {xl x is any real number }
3x*O� x * 0 Domain: {xl x * O}
(f -g)(x) = f (x) - g (x) = 2 -x - (3x + 1) = 2 -x - 3x - l = -4x + l Domain: {x I x is any real number } 1 54
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Chapter 3 Review Exercises
21.
I(x)= xx+l -I
g (x) = x..!.. (f + g)(x) = I (x)+ g (x) (x - l) x+l +-1 = x(x+l)+I --;-----7= -x-I x x (x - I ) x2 +x+x - l x2 + 2x - l x(x -l) x(x - l) Domain: {xl x"# O,X"# I }
�
---'-
Domain: {xl x"# O,X"# I }
()
d.
I (x ) = -3 when x
f.
-4
I(x) ° when ° x � 3 To graph y=/ (x - 3 ) ,shiftthe graphofl horizontally 3 units to the right. >
<
6
y
=
j(x-3)
(6,3) x
(� )
g. To graph y = I x , stretch the graph of I horizontally by a factor of 2.
(���)(�) ( \
y
6
(�) (�) = x(x+1) 1
=
y
= + x x- )
x+l I (x) = I ( x) = x - I = x-I g g (x) I x Domain: {xl x"# O,X"# I }
23.
I ( -2)= - 1
e.
(f - g)(x) = I (x) - g (x) x (x+l) - I (x - l) = x+l 1 = x - I -� x(x - l) x2 + X - X + 1 x2 +1 x (x - l) x (x - l) Domain: {xl X"# O,X"# I} (f. g )(x) = I(x), g (x) =
c.
.\'
=
J'( �x'\ �2 }
x-I
(6,3) x
(-8, -3)
I(x) = _2X2 + X+ 1 I (x+h) - f (X) h -2 (x+h) 2 + (x+ h)+ 1 - ( _2X2 +X+ 1) = h -2 (x2 + 2xh+ h 2 )+ X+ h + 1 + 2X 2 -x - 1 h _2X 2 - 4xh - 2h 2 +x+h+l+2x 2 - x - l
h.
-6
To graph y = -I (.�) , reflect the graph ofl vertically about the y-axis. y
(-4,3)
6
Y=
-.t(x) x
h
_- -4xh - 2h 2 +h -_ h ( -4x - 2h+l) h h = -4x - 2h + 1 25.
a.
Domain: { xl - 4�x�3 }; [ -4,3 ] Range: { YI - 3�y�3 }; [ -3,3 ]
b.
x-intercept: (0,0); y-intercept: (0,0)
-6
2 7.
a.
Domain: { xl - 4�x�4 }; [ -4, 4] Range: { YI - 3 � y�1 }; [ -3,1 ]
b.
Increasing: (-4,-1) and (3, 4) ; Decreasing: (-1, 3)
1 55
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Chapter 3: Functions and Their Graphs c.
29.
31.
33.
35.
3 7.
Local minimum is -3 when x = 3 ; Local maximum is 1 when x = -1 . Note that x = 4 and x = -4 do not yield local extrema because there is no open interval that contains either value. d. The graph is not symmetric with respect to the x-axis, the y-axis or the origin. e. The function is neither even nor odd. f. x-intercepts: -2, 0, 4 , y-intercept: 0 f(x) = x3 - 4x fe-x) = ( _x)3 - 4(-x) = _ x3 +4x = - (x3 - 4x ) = -f(x) f is odd.
fis increasing on: (-3, -0.91) and (0.9 1, 3); fis decreasing on: (-0.9 1, 0.9 1) . 39.
20
-2
-2
41.
-20
H=1.7976371
./= -3.S6�B66
_2
..
P.=.�B603BS
Y=.S�3HS9�
W
-10
3
Y=1.S3196�1
3
f(x) = 8x2 -x f(2) - f(l) 8(2)2 - 2 -[8(1)2 - 1 ] a. 2-1 1 = 32 - 2 - (7) = 23 f(l) - f(O) 8(1)2 - 1 - [ 8(0)2 - 0] b. 1-0 1 = 8 - 1 - (0) = 7 c.
f (x) = 2x3 - 5x+l on the interval (-3, 3) Use MAXIMUM and MINIMUM on the graph of YI = 2x3 - 5x+1 .
-20
20
local maximum at: (0.4 1, 1 .53) ; local minima at: (-0.34, 0.54) and (1 .80, -3.56) ; fis increasing on: (-0.34, 0.41) and (1 .80, 3); fis decreasing on: (-2, -0.34) and (0.4 1, 1 .80) .
X f(x) = - 1+x2 f(-x) = -x = � = -f(x) 1+(-x)2 1+x2 is odd. f
local maximum at: (-0.91,4.04); local minimum at: (0.91, -2.04);
3
H= -.3362�U
-10
G(x) = I -x+x3 G( -x) = 1 - (-x)+ ( _x)3 = 1+x -x3 * -G(x) orG(x) G is neither even nor odd.
20
r--:--
��� ----:-1J � �-7--10 20
1 1 hex) = -+-+1 X4 x2 1 +--+1 1 1 1 = hex) he-x) = -= -+-+1 _ _ X4 x2 x)2 X)4 ( ( h is even.
20
f(x) = 2x4- 5x3 + 2x+10n the interval (-2, 3) Use MAXIMUM and MINIMUM on the graph of YI = 2x4- 5x3+2x+1 .
43.
f(4) - f(2) 8(4)2 - 4 -[8(2)2 - 2 ] 4-2 2 1 28 4 (30) = = 94 = 47 2 2
f(x) = 2 - 5x f(3) - f(2) [2 - 5 (3)J -[ 2 - 5 (2)J 3-2 3-2 (2 - 1 5) - (2 - 1 0) 1 = -1 3 - ( -8) = -5
156
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exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publisher.
Chapter 3 Review Exercises
45.
47.
49.
51.
Intercepts: (-4,0), (4,0), (0, -4) Domain: {xI x is any real number} Range: {Yl y �-4} or [-4,00)
f(x) = 3x - 4x2 f(3) - f(2) _-[ 3 (3) - 4(3)2J -[ 3 (2) - 4(2)2J 3- 2 3-2 16) = (9 - 36 )-(61 = -27+ 1 0 = -17
57.
g(x) = -21 x I. Reflect the graphof y = 1 x 1
about the x-axis and vertically stretch the graph by a factor of 2. y 2 (0,0)
The graph does not pass the Vertical Line Test and is therefore not a function. The graph passes the Vertical Line Test and is therefore a function.
x
f(x) = Ixl -8
Intercepts: (0, 0) Domain: {x I x is any real number } Range: {YI y :s; O} or (-00, 0]
x
59.
-s
53.
f(x) = -Fx
hex) = .,Jx - 1 . Using the graph of Y = -Fx horizontally shift the graph to the right 1 unit. ,
y 5
y. 5
x
x
-5
Intercept: (1, 0) Domain: { xI x � 1} or [1, 00) Range: {YIY �O} or [0,00)
-5
55.
F(x) = Ix 1- 4. Using the graph of y =1 x l , vertically shift the graph downward 4 units.
61.
y 5
f(x) = .,Jl - x = �-(x -l) . Reflect the graphof Y = -Fx about the y-axis and horizontally shift the graph to the right 1 unit.
x
1 57
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Chapter 3: Functions and Their Graphs y
s
67.
x
-s
f(x)
{3Xx+1
if -2 < x $1 if x > 1
a.
Domain: { xl x > -2 } or (-2,00)
b.
x-intercept: ( 0,0 ) y-intercept: (0,0) Graph:
c.
Intercepts: (1, 0), (0, 1) Domain: { xl x $1} or ( -<Xl, 1]
=
y
Range: { Y I y � o } or [0,(0 ) 63.
h(x) (x _1)2 + 2. Using the graph of y x2 , horizontally shift the graph to the right 1 unit and vertically shift the graph up 2 units. =
=
( - 2. -6)
d.
x
69.
-5
Intercepts: (0, 3) Domain: { x I x is any real number }
b.
Range: { Y l y � 2 } or [ 2,(0 )
65.
c.
g(x) 3 (x _ 1)3 + 1. Using the graph of y x3, horizontally shift the graph to the right 1 unit vertically stretch the graph by a factor of 3, and vertically shift the graph up 1 unit. y =
=
Range: { y I y > -6} or (-6, 00)
fcx)" a.
-6
f
3x
if -4 $ x < ° if x ° if x > ° =
Domain: { xl x � -4 } or [ -4,(0 ) x-intercept: none y-intercept: (0, 1) Graph:
x
(2,4) -5
Intercepts: (0,-2),
(-4, -4)
5 x
(I-�,OJ ( 1- � J ( ,o
""
d.
-5
Range: { Y I Y � -4, y;to o }
0.3, 0 )
Domain: { xl x is any real number }
Range: { Y I y is any real number }
1 58
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Chapter 3 Test
71.
73 .
Ax + S and f( l) = 4 f(x) = 6x -2 A(1) + S = 4 6(1) - 2 A+S = 4 4 A + S = 16 A=11 4 S = 411: r2; V = -7rr3 3 Let R = 2r , S = new surface area , and 2 . V2 = new volume S2 = 47rR2 = 47r(2r)2 = 47r( 4r2) = 4( 47rr2) = 4S
d. e.
1.
r"" 2.S2
a.
{(2,S),(4,6),(6,7), (S,S)} This relation is a function because there are no ordered pairs that have the same first element and different second elements. Domain: {2, 4,6,S} Range: {S, 6,7,S}
b.
We are given that the volume is 100 cubic feet, so we have
V = 7rr2h = 100 => h = 100 7rr2
c.
The amount of material needed to construct the drum is the surface area of the barrel. The cylindrical body of the barrel can be viewed as a rectangle whose dimensions are given by
d.
{(1, 3) , ( 4,-2),( -3,S),(1,7)} This relation is not a function because there are two ordered pairs that have the same first element but different second elements. This relation is not a function because the graph fails the vertical line test. This relation is a function because it passes the vertical line test. Domain: {xl x is any real number } Range:
21(1"
0'
2.
{Y I Y � 2}
or
[2,00)
f (x) = --./4-Sx The function tells us to take the square root of 4 -Sx . Only nonnegative numbers have real square roots so we need 4 -Sx � 0 .
A = area top + areabottom + areabody = 7rr2 + 7rr2 + 27rrh = 27rr2 + 27rrh 1 0� = 27rr2 + 200 A (r) = 27rr2 + 27rr 7rr r A(3) = 27r(3)2 + 200 3 0 2 = lS7r + � "" 123.22 ft2 A ( 4) = 27r (4 l + 200 4 = 327r + SO "" IS0.S3 ft 2
4 - Sx �O 4 -Sx - 4 � 0-4 -Sx �-4 -Sx -< -4 -S -S x � 4S
( )
c.
Y=119.26S��.
Chapter 3 Test
Thus, if the radius of the sphere doubles, the surface area is 4 times as large and the volume is S times as large as for the original sphere.
b.
200
The surface area is smallest when feet.
( )
a.
Graphing:
Hfnl�ur.-. H:2.S1S3�7
3 = S �7rr3 = SV V2 = �7rR3 3 3 3 = �7r(2r)
75.
A( S) = 27r( S) 2 + 2�0 = S07r + 40"" 1 97.0S ft2
-- -
f(-1) = �4 -S(-1) = --./4 + S = ../9 = 3 1 59
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Chapter 3: Functions and Their Graphs
3.
e.
g ( x ) = x+2 Ix+21 The function tells us to divide x+ 2 by Ix+ 21 . Division by 0 is undefined, so the denominator can never equal O. This means that x '* -2 . Domain: {x 1 x '* -2}
x
(-1) + 2 � g( -1) \( -1) + 2 \ = 1 11 = 1
4.
To solve 1 (x) < 0 , we want to find x values such that the graph is below the axis. The graph is below the x-axis for values in the domain that are less than -2 and greater than 2. Therefore, the solution set is {x 1 -5 :5: x < -2 or 2 < x :5: 5} . In interval notation we would write the solution set as [-5, -2) u (2, 5].
6.
x-4 x2 + 5x - 36 The function tells us to divide x - 4 by x 2 + 5x - 36 . Since division by 0 is not defmed, we need to exclude any values which make the denominator O. x2 + 5x - 36 = 0 (x+9) {x - 4) = 0 x -9 or x = 4 Domain: {xlx,* -9, x,* 4} (note: there is a common factor of x - 4 but we must determine the domain prior to simplifying) -5 = -1 h( -1) = -:-(,--1�) _- 4__ --40 8 2 (-1) + 5 (-1) - 36
l (x) = -x4+2x3 + 4x 2 - 2 We set Xmin -5 and Xmax 5 . The standard Ymin and Ymax will not be good enough to see the whole picture so some adjustment must be made. =
h (x) =
Pl�tl
-
Pl�t2
, =
MQxi
b.
c.
d.
..... / ....
u ..... �= ·.8507827
_
a.
Pl�t3
'Yla-XA4+2XA3+4X 2 'Y 2 = 'Y 3 = 'Y�= 'Y5= Y&
2
=
5.
=
� Y= -.8&02529
����_5 w�XMax=5
Xsc1=l YMin=-10 YMax=20 Yscl=2 Xres=l
Mini ..... u ..... �=O
f\ ' Y=·2
\
/\
..... lL J We see that the graph has a local maximum of -0.8 6 (rounded to two places) when x = -0.85 and another local maximum of 1 5.55 when x = 2.35 . There is a local minimum of -2 when x O. Thus, we have Local maxima: 1 (-0.8 5) "" -0.8 6 1 (2.35) "" 1 5.55 Local minima: 1 (0) = -2 The function is increasing on the intervals (-5, -0.85) and (0, 2.35) and decreasing on the intervals (-0.8 5, 0) and (2.35, 5) . MQxi
To find the domain, note that all the points on the graph will have an x-coordinate between -5 and 5, inclusive. To find the range, note that all the points on the graph will have a y-coordinate between -3 and 3, inclusive. Domain: {x 1-5 :5: x :5: 5} or [-5, 5] Range: {yl-3 :5: y :5: 3} or [-3, 3]
u ..... �=2.350783&
Y=15.5
753
=
The intercepts are (0, 2) , (-2, 0) , and (2, 0) . X-intercepts: -2, 2 y-intercept: 2 1 (1) is the value of the function when x = 1 . According to the graph, 1 ( 1) = 3 . Since (-5, -3) and (3, -3) are the only points on the graph for which y I (x) = -3 , we have I (x) = -3 when x = -5 and x = 3 . =
1 60
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Chapter 3 Test
7.
a.
{
c.
j (X) = 2X + l x<- 1 x-4 x�- 1 To graph the function, we graph each "piece". rirst we graph the line y = 2x+ 1 but only keep the part for which x < -1 . Then we plot the line y = x - 4 but only keep the part for which x� -1 . y
3
y�x-4,
j (x+h) - j (x) = 2 (x+h) 2 +1 - 2x 2 +1 =
1 0.
x
a.
( )( ) ( 2 ( x2 + 2xh+ h 2 )+ 1 ) - ( 2X 2 + 1 )
= 2X2 + 4xh + 2h2 + 1 - 2X2 - 1 = 4xh + 2h2 The basic function is y = x3 so we start with the graph of this function. y
y=xJ
x
y�2x+l, x<-l
b.
To fmd the intercepts, notice that the only piece that hits either axis is y = x - 4 . y = x-4 y = x-4 0 = x-4 y = 0-4 4=x y = -4 The intercepts are (0, -4) and (4, 0) .
c.
To find g( -5) we first note that x = -5 so we must use the first "piece" because -5< - 1 . g ( -5) = 2 ( -5 )+ 1 = - 1 0 + 1 = 9
Next we shift this graph 1 unit to the left to obtain the graph of y = (x+ 1)3.
-
d.
8.
y
The average rate of change from 3 to 4 is given by L\ y j (4) - j (3) L\x 4-3 3 (4) 2 - 2 (4)+4 - 3 (3) 2 - 2 (3)+4 4-3 = 44 - 25 = .!.2. = 1 9 1 4-3
(
9.
Next we reflect this graph about the x-axis to obtain the graph of y = -(x+ 1)3.
To find g (2) we first note that x = 2 so we must use the second "piece" because 2 � -1 . g (2) = 2 - 4 = -2
)(
a.
f- g =
b.
(
2
)
x
)' �-(x+l)'
Next we stretch this graph vertically by a factor of 2 to obtain the graph of y = -2 (x+ 1)3 .
( 2X 2 + 1 ) - (3x - 2)
Y
10
= 2x 2 + 1 - 3x + 2 = 2x 2 - 3x+3 f. g = 2X 2 + 1 ( 3x - 2) = 6x3 - 4x 2 + 3x - 2
)
\
2
x
)' � -2 (X+I Y
The last step is to shift this graph up 3 units 161
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Chapter 3: Functions and Their Graphs
= -2( x + 1)3 + 3 .
to obtain the graph of y y
,
2
PlotZ
"axil"lul"I �=5.1�3�77� �Y=B.665367�
= Ixl
The basic function is y so we start with the graph of this function. y
y=
b.
x
8
8. 67%, 1997 (x'" 5 ). 2010, x = 2010-1992 =18 . r(18) = -0.115(18)2 + 1.183(18)+5. 623 = -10. 3 43 fy' -113.343
x
4 = Ix + 41.
-10. 343% .
The model predicts that the interest rate will be This is not a reasonable value since it implies that the bank would be paying interest to the borrower.
Next we shift this graph units to the left to obtain the graph of y y=Ix+4i
1 2 . a.
x
(-2,2)
2
Next we shift this graph up units to obtain the graph ofy
= Ix + 41 + 2 . y=ix+4i+2
Let x width of the rink in feet. Then the length of the rectangular portion is given by The radius of the semicircular portions is half the width, or 1. To find the volume, we first find the area of the surface and multiply by the thickness of the ice. The two semicircles can be combined to form a complete circle, so the area is given by
2x -20 . =
A 8
1 1 . a.
•
The highest rate during this period appears to be occurring in For we have 1(lS)
y
XMin=el XMax=12 Xscl=l YMin=el YMax=l el Yscl=l Xres=l
x
y = -2 (x+ Ij +3
b.
Plot)
'Y1a-.115 XZ+l.18 X+5.623 ,Yz= 'Y3= 'Y�= 'Y5= 'Y6=
x
=
=
=
w+7rr2 (2x-20)(x) + 7r(1J 7rx2 = 2x2 -20x+-4 =
t·
=
We have expressed our measures in feet so we need to convert the thickness to feet as well. in . ...!J!-. = ft = � ft 1 2 in Now we multiply this b y the area to obtain the volume. That is,
r(x) = -0.115x2 +1.183x+5.623 For the years 1992 to 2004, we have values of x between 0 and 12. Therefore, we can let Xmin 0 and Xmax 12. Since r is the interest rate as a percent, we can try letting Ymin 0 and Yrnax 10. =
r=
0.75
0.75 12
16
v(x) = 1 � ( 2X2 -20x+ 7r;2 J v(x) = £8 _ 5x4 + 7rX264
162
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Chapter 3 Cumulative Review b.
Ifx the = 90rink. is 90 feet wide, then we have 902 --5(90)-+ 7Z"(90)2 "" 1297.61 V(90) = 8 4 64 The volume of ice is roughly 1297. 6 1 ft3 .
9.
14x+ll �7 4x + 1 :s; -7 or 4x + 1 �7 4x :S;-8 4x�6 x:S;-2 x�-32 Solution set: { x I x -2 or x %} Interval notation: (-00, -2] [%,00 ) �
:s;
Chapter 3 Cumulative Review 1.
3.
5.
7.
u
3x-8 = 10 3x -8 + 8 = 10+ 8 3x = 18 -3x3 = -183 x=6 The solution set is {6} . x2 -8x-9 = 0 (x-9)(x+l) = 0 x -9 = 0 or x + 1 = 0 x=9 x = -1 The solution set is {-I, 9} . 12x+31 = 4 2x + 3 = -4 or 2x + 3 = 4 2x = -7 2x = 1 X = --72 X =-21 The solution set is {-�,�}. 2-3x > 6 -3x > 4 x < --43 Solution set: {x I x < --j} Interval notation: (-00, --j)
..
1 1.
]
3
-2
'2
x
3x-2y = 12 x-intercept: 3x -2( 0) = 12 3x = 12 x=4 The point (4,0) is on the graph. y-intercept: 3(0)-2y = 12 -2y = 12 y = -6 The point (0, -6) is on the graph. y
5
1 63
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in wri ting from the publi sher.
Chapter 3: Functions and Their Graphs
13.
x2 +(Y _ 3)2 = 16 This is the equation of a circle with radius = ..[I6 = 4 and center at (0,3). Starting at the center we can obtain some points on the graph by moving 4 units up, down, left, and right. The corresponding points are (0,7), (0, -1) , ( -4,3) , and (4,3), respectively. Yl
17.
f(x) = (x+2)2 -3
= x2 , shift the graph 2 units to the left [y = (x + 2 f ] and down 3 units [ Y =(X+2)2 -3 ] Starting with the graph of y
r
Y 5
x x
-3
15.
3x2 -4y = 12 3x2 -4(0) = 12 3x2 = 12 x2 = 4 x = ±2 y-intercept: 3(0)2 -4y = 12 -4y = 12 y = -3
(-2,' -3)
(0, -1)
-
19.
x-intercepts:
The intercepts are
if x�2 f(x) = fll2-x xl if x > 2 Graph the line y = 2 -x for x�2 . Two points on the graph are (0,2) and (2, 0) . Graph the line y = x for x > 2 . There is a hole in the graph at x = 2 . �t / / (4,4)
�5
(-2, 0), (2, 0) , and (0, -3) .
Check x-axis symmetry:
3x2 -4(-Y) = 12 3x2 +4y = 12 different Check y-axis symmetry: 3(-x)2 -4Y = 12 3x2 -4y = 12 same Check origin symmetry: 3( _X)2 -4(-y)= 12 3x2 + 4y = 12 different
5
'
, , , � (;, �)' -5
'
;,
X
•
The graph of the equation has y-axis symmetry. 164
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Chapter 4 Linear and Quadratic Functions c.
S ection 4.1 1.
d.
From the equation y = 2x - 3 , we see that the y intercept is -3 . Thus, the point (0, -3) is on the graph. To obtain a second point, we choose a value for x and find the corresponding value for y. Let x = 1 , then y = 2(1) - 3 = -1. Thus, the point (1, -1) is also on the graph. Plotting the two points and connecting with a line yields the graph below.
1 5.
average rate of change increasing
=
2
h(x) = -3x + 4 a. Slope -3 ; y-intercept 4 b. Plot the point (0, 4). Use the slope to find an additional point by moving 1 unit to the right and 3 units down. =
=
y
x
c.
3.
5. 7. 9. 11. 13.
d.
f(2) = 3(2)2 - 2 = 10 f(4) = 3(4)2 _ 2 = 46 i1y = f(4) - f(2) = 46 - 10 = 36 = 18 L1x 4-2 4-2 2
1 7.
f(-2) = (_2)2 - 4 = 4 - 4 = °
-3
=
f (x) = '41 x - 3
±; y-intercept
a.
Slope
b.
Plot the point (0, -3) . Use the slope to find an additional point by moving 4 units to the right and 1 unit up.
slope; y-intercept positive False. If x increases by 3, then y increases by 2.
=
=
-3
y
5
f(x) = 2x + 3 a. Slope 2; y-intercept 3 b. Plot the point (0,3). Use the slope to fmd an additional point by moving 1 unit to the right and 2 units up. =
average rate of change decreasing
=
-5 c.
d.
average rate of change increasing
=
1
'4
x
165
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AIl rights reserved. This material i s protected under all copyright laws as they currently
exist. No portion of this material may b e reproduced , in any form or by any means, without permi ssion in wri ting from the publi sher.
Chapter 4: Linear and Quadratic Functions
19.
F (x) = 4
a.
Slope 0; y-intercept = 4 Plot the point (0, 4) and draw a horizontal line through it.
25.
=
b.
.\'
5
(0.4)
0 1 2
x
-5
c.
d.
21
.
x -2 -1
average rate of change = 0 constant
27.
y = j (x) A vg. rate of change
�y &
=-
4 1
0 1 2
0 1 0
-2 -10
&
-4 - (-26) = -= 22 -1 - (-2) 1 22 2 - ( -4) 6 0 - ( -1) = -1 = 6
A vg. rate of change
=
x y = j (x) 8 -2 8 -1
�y
A vg. rate of change = &
8 - 8 = -0 = 0 -1 -(-2) 1 8 - 8 = -0 = 0 0 8 0 - ( -1) 1 8 - 8 -0 = 0 -= 1 8 1-0 1 8 - 8 -0 = 0 2 -= 8 2-1 1 This is a Imear functIOn wIth slope = 0, smce the
average rate of change is constant at O.
=
x y = j (x) -8 -2 -3 -1
2
�y
Avg. rate of change =-
This is not a linear functIOn, SInce the average rate of change is not constant.
-3 -3 1 - 4 = -= -1 - (-2) 1 -2 - 1 -3 -3 0 -2 1 0 - ( -1) = -= -5 - (-2) = -= -3 -3 -5 1 1-0 1 -8 - ( -5) = -= -3 -3 -8 2 1 2-1 ThIS IS a lInear functIOn wIth slope -3, SInce the average rate of change is constant at -3. 23.
x y = j (x) -26 -2 -4 -1
29.
�y &
-3 - ( -8 ) - -5 - 5 -1 -(-2) 1 0 - ( -3) = -3 = 3 0 - ( -1) 1
j (x) = 4x - l; g (x) = -2x +5 j (x) = 0 a. 4x - l = 0 X = -41
b.
j (x) > 0 4x - l > 0 x > -41
The solution set is
This IS not a linear functIon, SInce the average rate of change is not constant.
{x /x > ±} or (±, 00) .
166
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exist. No portion of thi s material may be reprod uced , in any form or by any means, without permi ssion in wri ting from the publi sher.
Section 4.1: L inear Functions and Their Properties
c.
d.
e.
= g(x) 4x-l = -2x+5 6x = 6 x=1 J ( x ):-::; g ( x ) 4x-l:-::; -2x+5 6x:-::; 6 x:-::; 1 J (x)
33.
a.
(x) =
g ( x ) when their graphs intersect. J Thus, x --4 . J ( x ):-::; g x when the graph ofJis above the graph of g. Thus, the solution is { x l x --4} or ( -00,-4) . =
b.
35.
The solution set is { x Ix:-::;
I} or (-00, 1].
a. b.
x
37.
()
< J ( x ) = g ( x ) when their graphs intersect. Thus, x = -6 . g(x) :-::; J ( x ) < h ( x ) when the graph ofJis
above or intersects the graph of g and below the graph of h. Thus, the solution is or x l :-::; x
{ -6 < 5} [-6, 5) . C(x) = 0. 2 5x+35 C(40) = 0. 2 5(40)+35 = $45 . Solve C(x) = 0. 2 5x+ 35 = 80 0. 2 5x + 35 = 80 0.25x = 45 x =� = 180 miles 0. 2 5 Solve C(x) = 0. 2 5x+35:-::; 100 0. 2 5x + 35:-::; 100 0. 2 5x:-::; 65 x:-::; � = 260 miles 0. 2 5 B(t) = 19. 2 5t+585.72 B(10) = 19. 2 5(10)+585.72 = $778. 22 Solve B(t) = 19. 2 5t+585. 7 2 =893. 7 2 19. 25t + 585.72 = 893. 72 19.25t = 308 308 = 16 years t = -19. 2 5 Therefore the average monthly benefit will be $893. 72 in the year 2006. Solve B(t) = 19. 2 5t+585. 7 2 > 1000 19. 25t + 585. 7 2 > 1000 19. 2 5t > 414. 2 8 414. 2 8 "" 21. 5 2 years t > --19.25 Therefore the average monthly benefit will exceed $1000 in the year 2012.
a.
b.
3 1 . a.
b.
c.
d.
e.
f.
(40, 50) 50 is = x = 40. The point (88, 80) is on the graph of y = J(x) , so the solution to J(x) = 80 is x = 88 . The point (-40, 0) is on the graph of y = J(x) , so the solution to J(x) = 0 is x = -40 . The y-coordinates of the graph of y = J(x) are above 50 when the x-coordinates are larger than 40. Thus, the solution to J(x) > 50 is {xix > 40} or (40, ) The y-coordinates of the graph of y = J(x) are below 80 when the x-coordinates are smaller than 88. Thus, the solution to J(x) :-::; 80 is { x l x:-::; 88} or (-00, 88]. The y-coordinates of the graph of y = J(x) are between 0 and 80 when the x-coordinates are between -40 and 88. Thus, the solution to 0 < J(x) < 80 is {xl-40 < x < 88} or (-40, 88) . is on the graph of The point y J(x) , so the solution to J(x) =
(0
c.
39.
a.
b.
.
c.
167
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exist. No portion of this material may b e reproduced , in any form or by any means, without permi ssion in wri ting from the publi sher.
Chapter 4: Linear and Quadratic Functions
41.
S ( p ) = - 200+50p; D( p ) = 1000 - 25p a. SOlveS ( p ) =D ( p ) , - 200+50p = 1 000 - 25p 75p = 1200 p = 1 200 = 16 75 S ( 1 6) = - 200+50 ( 1 6) = 600 Thus, the equilibrium price is $ 1 6, and the equilibrium quantity is 600 T-shirts. b. Solve D ( p » S ( p ) . 1 000 - 25p> - 200+ 50p 1 200>75p 1200>p 75 1 6>p The demand will exceed supply when the price is less than $ 1 6 (but still greater than $0). c. The price will eventually be increased.
T
€
� �
b.
c.
d.
4.nO()
3,O()()
2.(1)0 .1.000
(7300,730)
<�% ./
e.
--
43. a.
5,000
45.
We are told that the tax function T is for adjusted gross incomes x between $7,300 and $29,700, inclusive. Thus, the domain is { x i 7, 300�x�29, 700 } or [ 7300, 29700] .
T ( 1 8000 ) = 0. 1 5 ( 1 8000 - 7300 )+ 730 = 2335 If a single filer's adjusted gross income is $ 1 8,000, then his or her tax bill will be $2335. The independent variable is adjusted gross income, x. The dependent variable is the tax bill, T. Evaluate T at x = 7300, 1 8000, and 29700 . T ( 7300) = 0. 1 5 ( 7300 - 7300 )+ 730 = 730 T ( 1 8000) = 0. 1 5 ( 1 8000 - 7300)+ 730 = 2335 T ( 29700) = 0. 1 5 ( 29700 - 7300)+ 730 = 4090 Thus, the points ( 7300, 730 ) , ( 1 8000, 2335 ) , and ( 29700, 4090 ) are on the graph.
�.��
x
AdjusleJ Gn,ss Income ($)
We must solve T ( x ) = 2860 . 0. 1 5 ( x - 7300 )+ 730 = 2860 O. l 5x - 1095+ 730 = 2860 0. 1 5x - 365 = 2860 0. 1 5x = 3225 x = 2 1, 500 A single filer with an adjusted gross income of $2 1 ,500 will have a tax bill of $2860.
R ( x ) = 8x; C( x ) = 4.5x+17500 a. Solve R ( x ) = C ( x) . 8x = 4.5x + 1 7500 3.5x = 1 7500 x = 1 7500 � 5000 3.5 The break-even point occurs when the company sells 5000 units. Solve R ( x )> C ( x) b. 8x > 4.5x+ 1 7500 3.5x> 1 7500 x> 1 7500 = 5000 3.5 The company makes a profit if it sells more than 5000 units.
4 7 . a.
Consider the data points ( x, y) , where x the age in years of the computer and y the value in dollars of the computer. So we have the points ( 0,3000) and (3, 0) . The slope formula yields: m = �y = 0 - 3000 = -3000 = - 1 000 3-0 � 3 The y-intercept is ( 0,3000 ) , so b 3000 . Therefore, the linear function is V ( x ) = mx+b = -1000x+3000 . =
=
=
168
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exist. No portion of this material may be reproduced, in any form or by any mean s, without permi ssion in writing from the publi sher.
Section 4.1: L inear Functions and Their Properties
b.
The graph of V ( x ) = -1 000x+3000 j§ til
'0 .:3
b.
V(x)
CJ ::>
ca >
B
�
c::l
d.
b.
� QJ
..:l �
""
is
:0
c.
d.
49.
a.
2
Age
c.
d.
5 1 . a.
The graph of V ( x ) = - 1 2, 000x + 120, 000 �
1 000
X
V ( 2 ) = - 1 000 ( 2 )+3000 = 1 000 The computer's value after 2 years will be $ 1 000. Solve V ( x ) = 2000 -1 OOOx+ 3000 = 2000 -1 000x = - 1 000 x =1 The computer will be worth $2000 after 1 year.
19
( (x)
u
.::.:. 0 0
c.
The graph of C ( x ) = 90x+ 1 800
V(x)
b.
1 20,000 1 00,000 80,000
60,000
53.
40,000
20,000
2
4
Age
x
55.
V ( 4) = -12000 ( 4)+ 1 20000 = 72000 The machine's value after 4 years is given by $72,000. Solve V ( x ) = 72000 . -1 2000x+ 120000 = 72000 - 1 2000x = -48000 x = 4 The machine will be worth $72,000 after 4 years. Let x the number of bicycles manufactured. We can use the cost function C ( x ) = mx+b , with m 90 and b 1 800. Therefore C ( x ) = 90x+1 800 =
=
=
4
6
8 10 1 2 1 4
Number o f Bicycles
The cost of manufacturing 14 bicycles is given by C(14) = 90(14)+ 1 800 = $3060 . Solve C(x) = 90x+ 1 800 = 3780 90x+ 1 800 = 3780 90x = 1 980 x = 22 So 22 bicycles can be manufactured for $3780. Let x number of miles driven, and let C cost in dollars. Total cost (cost per mile)(number of miles) + fixed cost C(x) = 0.07x + 29 C(1 1 0) = (0.07)(1 1 0)+ 29 = $36.70 C(230) = (0.07)(230)+ 29 = $45.10 =
=
=
The graph shown has a positive slope and a positive y-intercept. Therefore, the function from (d) and (e) might have the graph shown. A linear function f(x) = mx+b will be odd provided fe-x) = -f(x) . That is, provided m ( -x)+b = -(mx+b) . -mx+b = -mx -b b = -b 2b = 0 b=O So a linear function f (x) = mx+b will be odd provided b = 0 . A linear function f (x) = mx+b will be even provided fe-x) = f(x) . That is, provided m (-x)+b = mx+b . -mx+b = mx+b -mxb = mx 0 = 2mx m =O So a linear function f(x) = mx+b will be even provided m = 0 .
1 69
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exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publi sher.
Chapter 4: Linear and Quadratic Functions
b.
Section 4.2 1.
scatter diagram
3.
Linear relation, m > 0
5.
Linear relation, m < 0
7.
Nonlinear relation
9.
a.
2 - (- 2)
The equation of the line is: Y - Y1
a
a
a
o
3
Answers will vary. We select (3, 4) and (9, 1 6). The slope of the line containing 12 = 2 . these pomts IS: m = 196_-34 = ""6 The equation of the line is: .
.
d.
Y - Yl
c.
= m (x - x1 ) y - 4 = 2(x - 3) y - 4 = 2x - 6 y = 2x - 2 20
e.
-6
Using the LINear REGression program, the line of best fit is: y = 2.2x + 1 .2
e.
3
13.
d.
1 y = -9 x + 4 2 6
c.
0 1.1:::::====:=..1 10 b.
a.
1 50
o ��=======� 1 0 o
a
Using the LINear REGression program, the line of best fit is: y = 2.0357x - 2.357 1
a
b.
a.
· ----'--' ...:.-'..---'_.=..l 0 -25 L 90
Answers will vary. We select (-20, 1 00) and (-1 0, 140). The slope of the line containing these points is: 140 -1 00 40 = 4 = - 1 0 - (-20) 1 0
The equation o f the line is:
6 a
Y - Yl
= m (x - x1 ) y - l 00 = 4 (x - (-20») y - l 00 = 4x + 80 y = 4x + 1 80
a
3
-3
a
a a
m= 11.
�
y + 4 = 4"9 x + "29
--,
_ _ _ __
a
= m (x - xl )
y - (- 4) = ( x - (- 2»)
2r.:0'--
a
Answers will vary. We select (-2,--4) and (2, 5). The slope of the line containing . IS. : = 5 - (- 4) = 9 . these pomts m 4"
a
-6 1 70
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exist. No portion of this material may b e reproduced, in any form or by any means , without permission in writing from the publisher.
Section 4.2: Building Linear Functions from Data c.
-25
d. �
a.
/ /a
- -�
oj Q ::::: o 0 '.0 "'Ci
0.. ......
§ ;,
�
.
o
U
a.
;g
0
t3 ' '-
'
.
10
10
20
y
280 ,3 '" 2(,0 v '0 240
.�
... <;
E i
•
•
220
•
200
38
.
30
90
b.
o
c.
•
30
•
40
•
•
50
I
42
•
•
•
•
46
50
54
58 62
Weight (gra ms)
66
x
Linear. Answers will vary. We will use the points (39.52, 2 1 0) and (66.45, 280) . m = 280 - 2 1 0 = � "" 2.59933 l6 66.45 - 39.52 26.93 y - 2 l 0 = 2.59933 l 6(x - 39.52) y - 2 1 0 = 2.59933 1 6x - 102.7255848 y = 2.599x + 1 07.274 v
d.
'') ' c;::
Disposable Income (thousands of dollars)
'J)
280
� 260 U ..... 0 240 � .0 220 E ;::l Z 200
Answers will vary. We select ( 1 8, 1 3) and (50, 39). The slope of the line containing . . 39 - 1 3 = 26 = 13 these pomts = --50 - 1 8 32 1 6 The equation o f the line is: C - C\ = m(1 - /\ ) C - 13 =
d.
o
40
1S:
c.
.
50
c:: ::; 20 ""
b.
1 7.
C
Using the LINear REGression program, the line of best fit is: C = 0.75491 + 0.6266 . The correlation coefficient is: "" 0.987 . r
90 Using the LINear REGression program, the line of best fit is: y = 3.86 1 3x + 1 80.2920 1 50
-25 ' 1 5.
e.
1 50
m
38
�� (1 - 1 8)
e.
C _ 1 3 = .!2 I _ l l7 16 8 C = .!2 I - .!l 16 8 As disposable income increases by $ 1 000, consumption increases by 13 16 · 1000 = $81 2.50.
f.
19.
Let 1 = 42 in the equation found in part (b). C = .!2 (42) _.!l = 32.5 16 8 The consumption of a family whose disposable income is $42,000 should be $32,500.
a.
42
46
50
54
S8
62
Weight ( g.ra m s )
66
x
x = 62.3 : y = 2.599 ( 62.3) + 1 07.274 "" 269 We predict that a candy bar weighing 62.3 grams will contain 269 calories. If the weight of a candy bar is increased by 1 gram, then the number of calories will increase by 2.599. 1�8� � __ __ __ __ __
a
a a a a
a
D
a
24 � ' __�__�__�� ' 28
16
1 71
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exist. No portion of this material may b e reproduced, in any form or by any means , without permi ssion in writing from the publ isher.
Chapter 4: Linear and Quadratic Functions b. c.
d. e.
U s i ng the L INear REGression program , the
For each
1
Y 50
23.
C(H) = 0.3734H + 7.3268
line of best fit is:
2
�
inch increase in h e ight, the
0.3734 inch . C(26) = 0.3734(26) + 7.3268 "" 1 7.0 inches
c ircumference i ncreases by
40
� () ::: Q
30
-:; 'u
To fi n d the he ight, w e solve th e fol l ow ing
20
c
equation :
•
•
1 7.4 = 0.3734H + 7.3268 1 0.0732 = 0.3734H 26.98 "" H
35
b.
The re l ation i s n ot a function because paired w ith both
o
"""
.
>,
�
S
60 58 -
•
25.
23 is
� v 50 ;::: c 48
c.
1.
•
= x2 - 9
To fmd the y-intercept, let x 2 y
•
46
y
=
-.
x
Price (dollnrs/pair)
U s i n g the LIN ear REGression program, t h e
0 - 9 -9 . =
2
-9 = 0 2 x =9 x = ±.J9 = ±3
The intercepts are
D = - 1 .3355 P + 86. 1 974 . The corre l ation coeffi c i ent i s : "" -0.949 1 . As the price of the j ean s increases by $ 1 , the l in e of b e s t fit i s :
f.
5.
1 .34 pairs per day. D ( p ) = - 1 .3355p + 86. 1 974 Domain :
7.
{ p i 0 < p :-:; 64}
N ote th at the p-intercept is rough ly
=0:
(0, -9), (-3, 0),
and
(3, 0) .
25 4
deman d for the j eans d ecreases by about e.
=0:
To find the x- intercept(s), let y
r
d.
0 implies that the
Section 4.3
• v ;;: 4 4 o iJ 42 40 o '-.}---1--I-----L---1.---''--.L---'- P o 1 8 20 22 24 26 28 �O '
A corre l ation coe ffi c i ent of
data do n ot have a l i near re lation s h i p .
•
•
45
would not make sense to find the line of best fit.
•
�6 �/l � ::: :... )4 2.. t. 52 § �
56 and 53.
40
x
The data d o not fo l l ow a l i n e ar pattern s o it
inches. a.
•
Age 0[' M o t h e r
1 7.4 inches w o u l d h ave a he ight of about 26.98
A c h i l d w ith a h ead c ircumference of
21.
•
•
64.54
9.
and that the number of pairs of j e ans i n demand cannot be negative .
parabola
b
2a Tru e ;
-
b -
2a
=- 4 =2 2(-1) --
g. D(28) = - 1 .3355(28) + 86. 1 974 "" 48.8034 Deman d i s about 49 pairs.
1 72
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Section 4.3: Quadra tic Functions and Their Properties
11.
C
13.
F
1 5.
G
25.
by a factor of
1 7.
H
1 9.
j(x) = 4"1 x 2
Using the graph of y = x 2 , compress vertically by a factor of 4"1 .
i, then shift up 1 unit.
-s
y
-
27. x
--5
21.
f(x) = 4"1 x 2 + 1 Using the graph of y = x 2 , compress vertically
�
(0, 1 )
2
x
:)
f(x) = x 2 + 4x+ 2 = (x 2 +4x+4)+2 - 4 = (x+2) 2 _ 2 Using the graph of y = x 2 , shift left 2 units, then shift down 2 units.
j(X) = x 2 - 2 Using the graph of y = x 2 , compress vertically by a factor of 2, then shift down 2 units.
x
y
( - 2 . -2) -5
x
29. -s
23.
�
j(X) = X2 +2 Using the graph of y = x 2 , compress vertically by a factor of
� then shift up 2 units.
j(x) = 2x 2 - 4x+l = 2 ( X 2 - 2x )+ 1 = 2(x 2 - 2x+l)+1 - 2 = 2(x - l) 2 - 1 Using the graph of y = x 2 , shift right 1 unit, stretch vertically by a factor of 2, then shift down 1 unit. y '
,
Y
x
-5
-2
5
(1, - 1 )
x
-5
1 73
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exist. No portion of this materi al may be reproduced, in any form or by any mean s , without permi ssion in wri ting from the publi sher.
Chapter 4: Linear and Quadratic Functions
31.
Thus, the vertex is (-1, -1) . The axis of symmetry is the line x = -1 . The discriminant is b2 -4ac = (2i -4(1)(0) = 4 0 , so the graph has x-intercepts. The two x-intercepts are found by solving: x2 + 0 x(x+2) = 0 x = 0 or x = -2 The x-intercepts are -2 and O . The y-intercept is f(O) = 0 .
f(x) = _x2 -2x = _ (x2 +2x) = -(x2 +2x+l)+1 = -(x+l)2 +1 Using the graph of = x2 , shift left 1 unit, reflect across the x-axis, then shift up 1 unit.
>
y
2x
y 5
x
33.
=
x
f(x) = 2"1 x2 +x-I = � ( x2 + 2x) -1 =�(X2 + 2x+I)-I-� =�(X+I) 2 _ % Using the graph of = x2 , shift left 1 unit, compress vertically by a factor of �, then shift down 2"3 unItS..
b. c.
y
37. a.
x
= -1
The domain is ( ) . The range is [-1, ) Decreasing on ( -1) . Increasing on (-1, ) . For f(x) = -x2 -6x, a = -l , b =-6 , c = O . Since a = -1 0, the graph opens down. The x-coordinate of the vertex is � x = -b2a = -(-6) 2(-1) = -2 = -3. The y-coordinate of the vertex is f ( ;: ) = f(-3) = -( _3)2 -6(-3) -9+ 18 = 9. Thus, the vertex is (-3, 9) . The axis of symmetry is the line x -3 . The discriminant is: b2 -4ac = ( _6)2 -4(-1)(0) = 36 0, soThethe graph has two x-intercepts. x-intercepts are found by _x2 -6x = 0 -x (x + 6) = 0 x = 0 or x = -6. The x-intercepts are -6 and The y-intercepts are f(O) = 0 . -00,
(0
(0
-00 ,
.
00
<
=
-5
35.
a.
=
For f(x) = x2 + 2x , a = 1 , b = 2 , c = O. Since a = 1 0 , the graph opens up. The x-coordinate of the vertex is
>
>
x =
-b
2a
=
-(2) 2(\)
=
-2 2
=
-
I
solving:
.
The y-coordinate of the vertex is f ( ;! ) = fe-I) = ( _ 1)2 + 2(-1) = 1-2 = -1.
o.
1 74
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exist. No portion of thi s material may be reproduced, in any form or by any means , without permi ssion in writing from the publi sher.
Section 4.3: Quadratic Functions and Their Properties
41.
a.
For f(x) = x 2 + 2x - 8 , a 1 , b = 2 , c = -8 . Since a 1 > 0 , the graph opens up. The x-coordinate of the vertex is -b = -2 = -2 = -l . x=2a 2(1) 2 The y-coordinate of the vertex is = f( -1) = ( _ 1)2 + 2(-1) - 8 f =
=
(;!)
b. c.
The domain is ( -00, 00 ) . The range is ( -00, 9 ] . Increasing on ( -00 , - 3) . Decreasing on ( 3 00 ) . -
39. a.
= 1 - 2 - 8 = -9. Thus, the vertex is ( 1 - 9) . The axis of symmetry is the line x = -1 . The discriminant is: b 2 - 4ac = 2 2 - 4(1)(-8) = 4 + 32 = 36 0 , so the graph has two x-intercepts. The x-intercepts are found by solving: x 2 + 2x - 8 = 0 (x + 4)(x - 2) = 0 x = -4 or X = 2. The x-intercepts are -4 and 2 . The y-intercept is f(O) = -8 . = -1 -
>
,
For f(x) = 2X 2 - 8x, a = 2 , b = -8 , c = 0 . Since a 2 > 0 , the graph opens up. The x-coordinate of the vertex is x = -b = -(-8) = 8 = 2 . 2a 2(2) '4 The y-coordinate of the vertex is = f(2) 2(2) 2 - 8(2) = 8 - 1 6 -8. f =
( ;!)
,
=
x
=
Thus, the vertex is (2, - 8) . The axis of symmetry is the line x 2 . The discriminant is: b 2 - 4ac = ( _ 8)2 - 4(2)(0) = 64 > 0, so the graph has two x-intercepts. The x-intercepts are found by solving: 2x 2 - 8x = 0 2x(x - 4) = 0 x = 0 or x = 4. The x-intercepts are 0 and 4. The y-intercepts is f(O) = 0 .
�
x
=
b. c.
43. x
a.
The domain is ( -00, 00 ) . The range is [-9, 00 ) . Decreasing on ( -00, - 1) . Increasing on (-1, 00 ) . For f(x) = x2 + 2x + 1 , = 1 , b = 2 , c = 1 . Since a = 1 > 0 , the graph opens up. The x-coordinate of the vertex is -2 -b = -2 = - l . x=2a 2(1) = 2 The y-coordinate of the vertex is = f(-l) f a
( ;! )
b. c.
The domain is Decreasing on
= ( _ 1)2 + 2( -1) + 1 = 1 - 2 + 1 = O. Thus, the vertex is (-1, 0) . The axis of symmetry is the line x = -1 .
( -00, 00 ) . The range is [-8, 00 ) . ( -00 , 2) . Increasing on (2, 00 ) . 1 75
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exist. No portion of this material may be reproduced, in any form or by any mean s, without permi ssion in wri ting from the publi sher.
Chapter 4: Linear and Qua dratic Functions
The discriminant is: b 2 - 4ac = 2 2 - 4(1)(1) = 4 - 4 = 0 , so the graph has one x-intercept. The x-intercept is found by solving: x 2 + 2x + l 0 =
(x + l) 2 = 0 x = -1.
-2
The x-intercept i s -1 . The y-intercept is 1(0) = 1 . b.
- 5 ( - 1 , 0) 1
b. c.
4 5 . a.
x =
1 1 1 1
-11
c.
x
47. a.
-5
The domain is ( -00, (0) . The range is [0, (0) . Decreasing on ( -00, - 1) . Increasing on (-1, (0) .
Thus, the vertex is
. =
�
=
= .!. - .!. + 2 = .!2 . 8
(�, - %)
The axis of symmetry is the line x . The discriminant is: b 2 - 4ac = 2 2 - 4(-2)(-3) 4 - 24 = -20 , so the graph has no x-intercepts. The y-intercept is 1(0) = -3 .
( ;� ) = I (�) = 2 (�y - � + 2
4 Thus, the vertex is
( ;�) = I (�) = -2 (�y + 2 (�) - 3 = - .!. + 1 - 3 = - � 2· 2
2(2)
8
00
For I (x) = -2x 2 + 2x - 3 , a = -2 , b = 2 , c = -3 . Since a = -2 < 0 , the graph opens down. The x-coordinate of the vertex is -b = --(2) -2 = -1 . x=2a 2(-2) -4 2 The y-coordinate of the vertex is I
The y-coordinate of the vertex is I
[l� , ) . Decreasing on (-oo,±) . Increasing on (±,oo) . ( -00 , (0 ) . The range is
The domain is
=
For l(x) = 2x 2 - x + 2 , a = 2 , b = - 1 , c = 2 . Since a = 2 > 0 , the graph opens up. The x-coordinate of the vertex is x = -b = -(- 1) = "41 . 2a
3
I 1 x= 4
x
(�, : ) .
I
The axis of symmetry is the line x = t . The discriminant is: b 2 - 4ac = (_ 1) 2 - 4(2)(2 ) = 1 - 1 6 = - 1 5 , so the graph has no x-intercepts. The y-intercept is 1(0) = 2 . 1
x =
I
2: x
b.
The domain is The range is
( -00 , (0 ) .
( - %] . -00,
1 76
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exist. No portion of thi s material may b e reproduced, in any form or by any means, without permi ssion in writing from the publi sher.
Section 4.3: Quadratic Functions and Their Properties
c.
( �) . Decreasing on (�, ) . For l( x) = 3 x 2 +6 x + 2 , a = 3 , b = 6 , c = 2 . Since a = 3 0 , the graph opens up. The x-coordinate of the vertex is x = -b = -6 = -6 = -1 . 2a 2(3) 6
Increasing on
51.
-�,
a.
<
down. The x-coordinate of the vertex is
�
49. a.
l(x) = -4x2 -6x+2, a = -4, b = -6 , c = 2 . Since a = -4 0 , the graph opens
For
-(-6) = -6 = --3 . x = --2ab = -2(-4) -8 4
The y-coordinate of the vertex is
>
1( ;� ) = / (- %) = -4 ( - %J -6 ( -%) + 2 = _ 2.+2.+ 4 2 2 = .!2.4 Thus, the vertex is ( _ %, 1; ) . The axis of symmetry is the line x = -% The discriminant is: b 2 -4ac = (_6)2 -4(-4)(2) = 36+32 68 ,
The y-coordinate of the vertex is
I( ;�) = 1(-1) = 3(-1)2 + 6(-1) + 2 = 3-6+2 = -l. (-1, -1)
.
Thus, the vertex is . The axis of symmetry is the line x -1 . The discriminant is:
= b2 -4ac = 62 -4(3)(2) = 36-24 = 12 ,
=
so the graph has two x-intercepts. The x-intercepts are found by solving: 0
so the graph has two x-intercepts. The x-intercepts are found by solving: 0
-4x2 -6x + 2 = x = - b ±�b 2 -4ac -(-6)±J(;8 2(-4) 2a 6±J68 6±2.JU 3±.JU -4 . The x-mtercepts are -3+.JU and -3-.JU . 4 4 The y-intercept is 1( 0 ) = 2 .
3x2 +6x+2 = ±�b 2--4ac x = --b-":":"' 2a �� -6±J!i. -6±2J3 -3±J3 3 6 6 J3 and -1 + 3"" . J3 . The x-mtercepts are - 1 - 3"" The y-intercept is 1( 0 ) = 2 .
-8
-8
IY
(0, 2)
(-1 + 1, 0) 4
b. c.
I
= -1 domain is ( -�, � ) .
The The range is
x
x
b.
[-1, ) . Decreasing on ( - 1) . Increasing on (-1, ) . �
The domain is The range is
-�,
c.
�
( -�, � ) .
( 1;] . _� ,
( -%, ) Increasing on ( -% ) .
Decreasing on
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.
-�,
1 77
© 2008
Pearson Educati on , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently
exist. No portion of thi s material may be reproduced, in any form or by any means , without permi ssion in wri ting from the publi sher.
Chapter 4: Linear and Quadra tic Functions
53.
55.
Consider the form y = a ( x -h)2 + k . From the graph we know that the vertex is (-1, -2) so we have h = -1 and k = -2 . The graph also passes through the point (x, y) = (0, -1) . Substituting these values for x, y, h, and k, we can solve for a:
59.
-1 = a(0-(-I)) 2 + (-2) -1 = a(1)2 -2 -1 = a-2 l=a The quadratic function is f ( x) = (x + 1)2 -2 = x2 + 2x -1 . Consider the form y = a (x -h) 2 + . From the graph we know that the vertex is (-3,5) so we have h = -3 and = 5 . The graph also passes through the point (x, y) = (0, -4) . Substituting these values for x, h, and we can solve for a: -4 = a(0-(-3»)2 + 5 -4 = a(3 l + 5 -4 =9a+S -9 = 9a -1 = a The quadratic function is f(x) = _(X+3)2 +5 = _x2 -6x-4 . Consider the form y = a (x -h l + . From the graph we know that the vertex is (1, -3) so we have h = 1 and = -3 . The graph also passes through the point (x,y) = (3,5) . Substituting these values for x, y, h, and we can solve for a: 5 = a(3-1)2 +(-3) S = a(2)2 -3 S = 4a-3 8 =4a 2=a The quadratic function is f (x) = 2 (x -1 / -3 = 2x2 -4x -1 .
=
61.
>
.
63 .
<
k
57.
= = -2 For f(x) = 2x2 +12x-3, a = 2, b = 12, c = -3. Since a = 2 0, the graph opens up, so the vertex is a minimum point. The minimum occurs at -12 = 4 -12 = -3. The nummum . . value IS x = -2ab = 2(2) f( -3) = 2(-3)2 + 12(-3) -3 = 18-36-3 = -21 . For f(x) = -x2 +10x-4,a = -I, b = lO , c = -4 . Since a = -1 0, the graph opens down, so the vertex is a maximum point. The maximum occurs -10 = -10 = 5 . The maximum at x = -2ab = 2(-1) -2 value is f(S) = _(5)2 + 10(5)-4 = -25 +50-4 = 21 . For f(x) = -3x2 + 12x+l,a = -3, b = 12, c = 1. Since a = -3 0, the graph opens down, so the vertex is a maximum point. The maximum occurs -b = --12 -12 = 2 . The maXImum . value at x = 2a 2(-3) = --6 is f(2) = -3(2) 2 +12(2)+1 = -12+24+1 = 13. se e form f(x) = a(x-h)2 + . The vertex is (0,2) , so h and 2. f(x) = a(x-W +2 = ax2 + 2 . Since the graph passes through (1, 8) , f(1) = 8 . f(x) = ax2 +2 8 = a(I)2 + 2 8 = a+2 6=a f ( x ) = 6x2 + 2 . a = 6, b = 0, c = 2 f( -3) 2 ( 3) + 12 ( -3) 1 8 - 3 6 - 1 8 .
k
y,
For f(x) = 2X2 + 12x, a = 2, b = 12, c = 0 . Since a = 2 > 0, the graph opens up, so the vertex is a minimum point. The minimum -b = --12 -12 occurs at x =2a 2(2) = -4 = -3. The minimum value is
k,
65.
<
k
67.
U
th
= °
k
k,
k
k=
1 78
© 2008
Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all cop yri ght l aws as they currently
exist. No portion of thi s material may b e reproduced, in any form or by any mean s , without permi ssion in wri ting from the publi sher.
Section 4.3: Qua dra tic Functions and Their Properties
69.
a
I (x) = g(x) _x2 +4 = -2x+1 0= x2 -2x-3 0= (x+ 1)(x-3) x + 1 = 0 or x -3 = 0 x = -l x=3 The solution set is {-I, 3}. c. 1(1) = -(-1)2 +4 = -1+4 = 3 g(l) = -2(-1)+1 = 2 + 1 =3 1(3) = _ (3)2 +4 = -9+4 = -5 g(3) = -2(3)+1 = -6+1 = -5 Thus, the graphs of1 and g intersect at the points (-1, 3) and (3,-5) .
and d.
h.
I(x) = g(x) 2x-1 = x2 -4 = x2 -2x-3 0 = (x+1)(x-3) x + 1 = 0 or -3 = 0 x = -l x=3 The solution set is {-I, 3}. c. 1(-1) = 2(-1)-1 = -2-1 = -3 g(-l) = (_ 1)2 -4 = 1-4 = -3 1(3) = 2(3) -1 = 6 -1 = 5 g(3) = (3)2 -4 = 9 -4 = 5 Thus, the graphs of1 and g intersect at the points (-1, -3) and (3, 5) .
h.
o
73.
a
and d.
x
71 .
a
and d.
x
I (x) = g(x) _x2 +5x = x2 +3x-4 0= 2X2 -2x-4 0= x2 -x-2 0= (x+1)(x-2) x + 1 = 0 or x -2 = 0 x = -l x=2 The solution set is {-I, 2}. c. / (-1) = -(-1) 2 +5(-1) = -1-5 =-6 g ( -1) = ( -1 )2 + 3 ( -1 ) -4 = 1 -3 -4 -6 1(2) _(2)2 +5(2) = -4+10 6 g(2) = 22 +3(2)-4 = 4+6-4 = 6 Thus, the graphs of1 and g intersect at the points (-1, -6) and (2, 6) .
h.
y
x
=
=
1 79
=
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reprod uced, in any form or by any mean s , without permi ssion in wri ting from the publi sher.
Chapter 4: Linear and Quadra tic Functions
75.
a.
For a = 1 : J(x) = a(x - rl )(x - r2 ) = 1(x-(-3» (x-1) = (x+3)(x-1) = x2 +2x-3 For a = 2 : J(x) = 2(x -(-3» (x -1) = 2(x+3)(x-1) = 2(x2 +2x-3) = 2x2 +4x-6 For a = -2 : J(x) = -2(x-(-3» (x -1) = -2(x+3)(x-1) = -2(x2 +2x-3) = _2X2 -4x+6 For a = 5 : J(x) 5(x -(-3» (x -1) = 5(x+3)(x-1) = 5(x 2 +2x-3) = 5x2 +10x-15 The are not affected by the by valuex-intercepts of a. The y-intercept is multiplied the value of a . The is unaffected valueaxisof of symmetry For this problem, the axisby theof symmetry is x = -1 for all values of a. The x-coordinate of theofvertex is not affected by the value a. The y-coordinate of the vertex is multiplied by the value of a . The x-coordinate of the vertex is the mean of the x-intercepts. 4 = -2 x = --2ab = --2(1) y = J(-2) = (-2)2 +4(-2)-21 = -25 The vertex is (-2, -25) . J(x) = 0 x2 +4x-21 = 0 (x+ 7)(x-3) = 0 x + 7 = 0 or x -3 = 0 x = -7 x=3 Thex-intercepts ofJare (-7, 0) and (3, 0).
c.
J(x) = -21 x 2 +4x-21 = -21 x2 +4x = 0 x(x+4) = 0 x = 0 or x+4 = 0 x = -4 The solutions J (x) = -21 are -4 and O. Thus, the points (-4,-21) and (0,-21) are on the graph off y
d.
( - 7, 0)
=
b.
c.
d.
e.
77. a.
b.
(0, -2 1) -26
a .
79.
(x, y) represent the line(3,y1=) x.is Then theLet distance from (x,a y)pointto theon point d = �(X _ 3)2 +(y _ 1)2 . Sincey =x, we can replace y variableas with x so thatof x:we have the distancetheexpressed a function d(x) = �(X 3)2 +(x _ l)2 � x 2 -6x + + x2 -2x + I = � 2X2 -8x + 10 Squaring both sides of this function, we obtain [d(x) t = 2X2 -8x+1O . Now, expression on the right is quadratic. Since the a = 2 0, it has a minimum. Finding the xcoordinate of the minimum point of [d(x) t will also give us the x-coordinate of the minimum of d(x) : x = -2ab = -(2(2)- 8) = �4 2 . So , 2 is the xcoordinate ofpoint the point on theSinceliney y==x,xthethatyis closest to the (3, 1). coordinate is also 2. Thus, the point is (2, 2) is point on the line y = x that is closest to (3, 1). the _
9
=
>
=
1 80
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ. A l l rights reserved. This material is protected under a l l copyright laws a s they currently exist. No portion of this material may b e reproduced, in any form or by any means, without permi ssion in writing from the publi sher.
Section 4.3: Qua dratic Functions and Their Properties
81.
R(p) = _4p2 + 4000p , a = -4, = 4000, = O. Since a = -4 0 the graph is a parabola that opens down, so the vertex is a maximum point. The -4000 = 500 . maXlmum occurs at p = -2a = --2(-4) Thus, the unit price should be $500 revenue. The maximum revenue is for maximum R(500) = -4(500)2 + 4000(500) = -1000000 + 2000000 = $1,000,000 C(x) = x2 -140x+ 7400 , = 1, = -140, = 7400. Since = 1 > 0, the graph opens the vertexmarginal is a cost minimum point. up,Thesominimum 140 = 70 mp3 occurs at x = -2a = -(-140) = 2(1) 2 players produced. The minimum marginal cost is I(;!) = 1(70) = (70/ -140(70)+ 7400 = 4900 -9800+ 7400 = $2500 = -3. 24, = 242.1, = -738. 4 The maximum the income levelnumber is of hunters occurs when -242.1 ;:,: 37. 4 years old x = -2a = 2-242.1 = -( -3. 24 ) -6. 4 8 The number of hunters this old is: H(37.4) = -3. 24(37.4)2 + 242.1(37. 4 ) -738.4 ;:,: 3784 hunters The maximum occurs when x = 37.4 , so the function increases on the interval (0, 37.4) and decreases on the interval (37.4, ) the number ofbetween huntersages is decreasing for45Therefore, individuals who are 40 and years of age. b
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.
c
x
-b
83. a.
a
b
c
c.
a
b.
a.
b
a
89.
a.
c
b.
b.
c.
00
b.
=
©
b
b
b
-b
M(23) = 1.00(23)2 -136. 74(23)+4764. 8 9 ;:,: 2149 male murder victims 1. 00x2 -136.74x + 4764. 8 9 1456 1. 00x2 -136.74x + 3308. 8 9 = 0 = 1. 00, = -136. 74, = 3308. 8 9 a
-b
c
-b
-b
87. a.
x ;:,: 31.4 or x ;:,: 105.3 Disregard 105.3 since it (20:$ falls outside the domain for the function x 90). Thus, 1456 forthe31.number 4 yearofolds.male murder victims is A minimum occurs when 7 4) = -136. 74 x = -2a = -(-136. 2(1.00) 2 = 68. 3 7 so the function decreases on the interval (20, 68. 3 7) and increases on the interval (68. 3 7, 90) . As age increases between 20 and 65, the number of murder victims decreases. R(x) = 75x -0. 2 X2 a = -0.2, = 75, = 0 The maximum revenue occurs when -75 = --75 = 187. 5 X = -2a = 2(-0. 2 ) -0.4 The maximum revenue occurs when x = 187 or x = 188 watches. The maximum revenue is: R(187) = 75(187) -0.2(187)2 = $703l. 20 R(188) = 75(188) -0. 2(188)2 = $7031. 20 P(x) = R(x) -C(x) = 75x-0.2x2 -(32x+1750) = -0.2X2 + 43x -1750 P(x) = -0. 2 x2 +43x-1750 a = -0. 2 , = 43, = -1750 -43 = --43 = 107. 5 X = -2a = 2(-0. 2 ) -0.4 The maximum profit occurs when x = 107 or x = 108 watches. The maximum profit is: P(107) = -0.2(107/ + 43(107) -1750 = $561. 2 0 P(108) -0.2(108)2 + 4(108) -1750 = $561.20 Answers will vary. <
-b
85.
= -(-1 36.74) ± �(-136.74)' - 4(\ .00)(3308.89) 2(1 .00) .----1 36.74 ± -15462.2676 2 1 36.74 ± 73.91 ;:,: 2
C
=
c
d.
181
2008 Pearson Education , Inc . , Upper Saddle River, NJ.
All rights reserved. Th i s material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced , in any form or by any means , without permi ssion in writing from the publi sher.
Chapter 4: Linear and Qua dratic Functions
91.
93.
Iftwox even is even,numbers then axare2 added and bxto arean odd even.number When the result is odd. Thus, f(x) is odd. If x is odd, then ax2 and bx are odd. The sum of three odd numbers is an odd number. Thus, f(x) is odd. y
= x 2 + 2x - 3 ; y = x 2 + 2x + 1 ;
y
Section 4.4 1.
R = 3x
b.
= x 2 + 2x
d.
Eachthemember of thischaracteristics: family will be a parabola with following (i) opens upwards since a 0; ( . . ) vertex occurs at x = - b - 2 = - 1 ,. 2a 2(1) (iii) There is at least one x-intercept since
5. a.
>
II
- =
b.
The graph of the quadratic function f( x ) = ax2 + bx + c will not have any x-intercepts whenever b2 - 4ac 0 . No. We know that the graph of a quadratic function f ( x) = ax2 + bx + c is a parabola with vertex ( - fa , f ( -fa) ) . If a 0, then the vertex is a minimum point, so the range is [f ( -fa)' ) If a 0, then the vertex is a maximum point, so the range is ( -00, f ( - fa)] . Therefore, it is impossible for the range to be
= - "61 (300) + 1 00 = -50 + 100 = $50
100 - x If x = -5p + 1 00, then p = -. 5 -x R(x) = x C 00 ) = _ � x 2 2 0x 5 R(1 5) = - � (1 5) 2 + 20(1 5) = - 45 + 300 = $255
<
c.
>
.
P
+
-
b 2 - 4ac � 0 .
(0
The maximum revenue is R(300) = -� (300i + 100(300) = -15000 + 30000 = $ 1 5, 000
(- 1 ,-4)
97.
- 20000 + 20000 3 40000 $ 1 3, 333.33 = 3 "" x = -b = - 1 00 = -100 = 300 = 300 1 2a 2 ( - t ) (-t) =
c.
95.
i
R(200) = - (200) 2 + 1 00(200)
- 20 - 20 1 00 = = x = -b = = 50 2a 2 ( -t) ( -j) 2
The maximum revenue is R(50) = _ � (50) 2 + 20(50)
= -500 + 1 000 = $500
<
d.
p = 1 00 - 50 = 50 = $ 1 0 5 5
(-00, (0 ) .
1 82
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publi sher.
Section 4.4: Quadratic Models; Building Quadratic Functions from Data
7.
Let w = width and 1 = length of the rectangular area. Solving = 2w+ 21 = 400 for 1 : 1 = 4002-2w = 200 _ w . Then A(w) = (200-w)w= 200w-w2 = _w2 +200w -200 -200 = 100 yards -b -w=-= -2 2a 2(-1) = -A(100) = -1002 + 200(100) = -10000 + 20000 = 10,000 yd2 Let x = width and = length of the rectangle. Solving = 2x + = 4000 for = 4000-2x . Then A(x) = (4000-2x)x = 4000x-2x2 = -2X2 + 4000x -4000 =-4 = 1000 meters -4000 x = 2ab = 2(-2) = ---= maximizes area. A(1000) = -2(1000)2 +4000(1000) . = -2000000 + 4000000 = 2,000,000 can be enclosed is area thatmeters. largest The 2, 000,000 square 8 x2 +x+200 -32x2 = -+x+200 h(x) = 625 2 (50) a = - 6258 ' b = 1, = 200. The maximum height occurs when -1 = 625 "" 39 feet from x = -2ab = 2 ( -8/625 ) 16 base of the cliff. The maximum height is h ( 62516 ) = 2 ( 62516 )2 + 62516 +200 7025 "" 219.5 feet. = 32"
a.
c.
P
b. c.
9.
y
11.
P
y
d.
y:
y
200 the MAXIMUM function Using 25 0 o ����====�� o
e.
t\
o X=�9.t) 6 2�9 � 'i =H9.5�125 . o
2 00
the ZERO function Using 25 0
1\
__
a.
Solving when h(x) = 0 : __6258_x2 +x+200 = 0 -4(-8/625)(200) x = -1±�122(-8/625) X "" -l±� -0. 0 256 X "" -91. 90 or X "" 1 70 negative, the beapproximately cannot the distance Since water the strikes projectile 1 70 feet from the base of the cliff. 250
\ 2�YO o X=170.0 23B 6 � 'i= O .....k. o
c
f.
200
__6258_x2 +x+200 = 100 - 6258 x 2 + x 100 = 0 -1 J6.12 -x = �12 -4(-8/625)(100) -0.0256 2(-8/625) = -----'. 135. 70be negative, the 7 or x ",,cannot x ",, -5the7 . 5distance Since whenofit thefromwater is 1 00 feet135.above projectile istheapproximately base the feet 7 cliff. +
±
b.
625
1 83
© 2008 Pearson Education , Inc . , Upper Saddle River, Nl. All rights reserved. This material is protected under all copyright l aws as they currently exist. No portion of thi s material may b e reproduced, i n any form or by any means , without permission in writing from the publi sher.
Chapter 4: Linear and Quadratic Functions
13.
Locate the point where theof thecable touchesthetheorigin road. atThen the equation parabola is of the form: y = ax2, where a O. Since (200, 75) is on the parabola, we can findthethepointconstant Since 75 = a(200)2 , then a = 20075 2 = 0. 001875 . When x = 100 , we have: y = 0.001875(100)2 = 18. 7 5 meters . >
a :
(-200,75)
y
1 5.
-
(200 75)
1 9.
a.
x
200
-200
b.
x = the depth of the gutter and y = the width of theLet gutter. Then A= is the cross-sectional area of the gutter. Since the aluminum sheets for the gutter are 12 inches wide, we have 2x + y = 12 . Solving for y : y = 12 -2x . The area is to be maximized, so: A= xy = x(12 -2x) = -2X2 + 12x . This equation is a parabola opening down; thus, it has a -12 -12 = 3 . maXimum when x = -2a = -2(-2) = -4 Thus, a depth of 3 inches produces a maximum cross-sectional area. Let x = the width of the rectangle or the diameter of the semicircle and let y = the length of the rectangle. The perimeter of each semicircle is 1t2X The perimeter of the track is given 1tX + y + y = 1500 . 1tX + 2'"' by: 2'"' Solving for x : 1tx+2y = 1500 = 1500-2y --"x = -1500-2y 1t The area of the rectangle is: -2 i + 1500 y . A = xy = ( 1500-2Y ) y = 1t 1t 1t This equation is a parabola it has a maximum when opening down; thus, xy
.
1 7.
-1500 = 375. y = 2a = ( -1t-2 ) = -1500 -4 2 1t 750 ::::: 238.73 = Thus, x = 1500-2(375) 1t 1t The dimensions for the rectangle with maximum area are 7501t 238. 7 3 meters by 375 meters. d = 1.1v+0. 06v2 If v = 45, then = 1.1(45) + 0.06(45)2 = 49. 6 + 121. 5 = 171 A car traveling 45 miles per hour on dry, level concrete will require 171 feet to stop. If 200, then 200 = 1.1v+0. 06v2 0. 06v2 + l.1v -200 = 0 -1.1± �(1.1)2 -4(0.06)(-200) v=--��2�(0�.�06�)-----1.1±v'49.21 0.12 ::::: vSpeed ::::: -67.cannot 62 orbevnegative, 49. 2 9 so the maximum speed thatis you be traveling accident aboutcould 49 miles per hour.and avoid the The term 1.1 v might represent the reaction time distance. We are given: Vex) = Ioc(a -x) = _1oc2 + ala . The reaction rate is a maximum when: -ak = ak = a X = 2a = 2(-k) 2k "2 ' I(x) = -5x2 +8, h = 1 Area = � ( 2ah2 + 6c) = 1( 2(-5)(1)2 + 6(8) ) = 1 (-10+48) = 338 sq. units I(x) x2 +3x+5, h 4 Area = � ( 2ah2 +6c ) = � ( 2(1)(4i +6(5) ) = '34 (32+30) =-2483- sq. umts. -b
-b
:::::
d
d
=
c.
21.
-b
•
23.
1tX
25.
=
=
1 84
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All ri ghts reserved. Thi s material i s protected under all copyright l aws as they currently exist. No portion of thi s material may b e reprod uced , in any form or by any mean s , without permi ssion in wri ting from the publi sher.
Section 4.5: Inequalities In volving Quadratic Functions
27.
a.
b.
I ncome 40,000
30.()O()
1 0,000 o
b.
c.
d.
•
•
20.00()
•
c.
•
d.
•
•
���__��__���__�� A ge
e.
-34.3(46) 2 + 3 1 57(46) - 39, 1 1 4 "" $33, 529
Using the QUADratic REGression program
31.
l(x) = -34. 3x 2 +3157x-39,1 14 4�0�,0�0�0 -.
1.
80
3.
h
a.
1 00 ..c /;J) -
;) :r:
'
80
60
40 20
•
•
40
•
•
80
•
•
1 20
•
D ist ance
•
1 60
•
Answers will vary. follows: Ifthusthe price is $140, no oneOnewillpossibility buy the calculators, making the revenue $0.
Section 4.5
__ __ ____
o i.!::=:=�====� o 29.
h(x) = -0.0037x2 +1. 03x+5. 7
=
QuadRe9 ':::j= ax 2: +bx+c. a=-34 . 26392857 b=3156.595357 c.=-39114 . 53634
e.
-b
Qu adRe9 ':::j= ax 2: +bx+c. a = - . 003712 1 212 b=1 . 031818 1 82 c. = 5 . 666666667
From the graph, the data appear to be quadratic with a < 0 . -3157 ",, 46 x = -2ab = 2(-34. 3) An will earn the most income at an ageindividual of 46 years. The maximum income will be: /(46)
3 "" 139. 2 x = 2a = 2(-0.-1.00037) The ballitswillmaximum travel 139.height. 2 feet before it reaches The maximum height will be:2 h(139. 2 ) = -0. 0 037(139. 2 ) + 1. 03(139.2) + 5. 7 "" 77. 4 feet Using the QUADratic REGression program
-3x-2 < 7 -3x < 9 x > -3 The solution set is { x l x > -3 } or ( -3,00 ) . f(x) > 0 when the graph offis above the x axis. Thus, { xi x < -2 or x > 2 } or, using interval notation, ( -00, -2 ) ( 2,00 ) . f(x)�0 when the graph offis below or intersects the x-axis. Thus, { xl -2 x 2 } or, using interval notation, [ -2, 2] . a.
b. •
200
u
�
�
x
From the graph, the data appear to be quadratic with a < 0 . 1 85
© 2008 Pearson Education , Inc . , Upper Saddle River, Nl. All rights reserved. Thi s material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publi sher.
Chapter 4: Linear and Qua dra tic Functions
5.
b.
<
-00 ,
7.
10
g( x) � f ( x )
when the graph of g is above or intersects the graph off Thus { xl -2 :::; x :::; 1 } or, using interval notation, [ -2, 1 ] . f(x) > g ( x ) when the graph offis above the graph of g. Thus, { xl x -2 or x > 1} or, using interval notation, ( -2 ) ( 1, ) x2 -3x-10 0 We graph the function f(x) = x2 -3x-10 . The intercepts are y-intercept: f(O) = -10 x-intercepts: x2 -3x -10 = 0 (x-5)(x+2) = 0 x = 5, x = - 2 The vertex is at x = -2ab = -(-3) 2(1) = i.2 Since f (i2 ) = - 494 ' the vertex is (i2 ' 494 ) . 10 10 a.
u
(0
10 <
.
<
-00,
<
11.
_
<
9.
u
(0
.
<
=
-20 the x-axis for -2 x 5 . The graph is below Since the inequality is strict, the solution set is { x 1 - 2 x 5 } or, using interval notation, ( -2, 5 ) . x2 -4x > 0 We graph the function f(x) = x2 -4x . The intercepts are y-intercept: f(O) = 0 x-intercepts: x2 -4x 0 x(x-4) = 0 x = O, x = 4 The vertex is at x = -2ab = -(-4) 2(1) = i2 = 2 . Since f ( 2 ) = -4 , the vertex is ( 2,-4 ) . <
-10 The graph is above the x-axis when x 0 or x > 4 . Since the inequality is strict, the solution set is { x 1 x 0 or x > 4 } or, using interval notation, ( 0) ( 4, ) x2 -9 0 We graph the function f(x) = x2 -9 . The intercepts are y-intercept: f(O) = -9 x-intercepts: (x + 3)(x -3) = 0 x = -3, x = 3 The vertex is at x = -2ab = -(0) 2(1) = o . Since f(O) -9 , the vertex is ( 0,-9 ) . 10
<
-10 the x-axis when -3 x 3 . The graph is below Since the inequality is strict, the solution set is { x 1 - 3 x 3 } or, using interval notation, ( -3, 3 ) . x2 +x > 12 x2 +x-12 > 0 We graph the function f(x) = x2 x -12 . y-intercept: f(O) = -12 x-intercepts: x2 + x -12 = 0 (x+4)(x-3) 0 x = -4, x = 3 The vertex is at x = -2ab -(1) 2(1) = _ .!.2 . Since f ( - .!.2 ) = - 494 ' the vertex is ( _.!.2 ' 494 ) . <
<
<
13.
<
<
+
=
=
=
_
© 2008
1 86 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright l aws as they currently
exist. No portion of this material may b e reproduced, in any form or by any means, without permi ssion in writing from the publisher.
Section 4.5: Inequalities In volving Quadratic Functions
10
x-intercepts: x2 - 7x-8 = 0 (x+l)(x-8) = 0 x = -1, x = 8 The vertex is at x = -2ab = -(-2(1)7) 22 . Since 1 (22 ) = - �4 ' the vertex is (22 ' - �4 ) . 10 10
10
-20 the x-axis when x < -4 or The graph is above x > 3 . Since the inequality is strict, the solution set is { xl x < -4 or x > 3 } or, using interval notation, (-00,-4) (3, 00) . 2X2 < 5X+3 2X2 -5x-3 < 0 We graph the function 1 (x) = 2x2 -5x -3 . The intercepts are y-intercept: 1(0) = -3 x-intercepts: 2X2 -5x -3 = 0 (2x+l)(x-3) = 0 x = --21 ' x = 3 The vertex is at x = -2ab = -(-5) 2(2) = 2.4 . Since 1 (2.4 ) = - 498 ' the vertex is ( 2.4 ' 498 ) . 10
=
U
1 5.
U
1 9.
.:.... ..:. --'----'-..!...-'.-'---.:.....:...
_
10 -10 The graph is below the x-axis when .!.2 < x < 3 . Since the inequality is strict, the solution set is { xl-� < x < 3 } or, using interval notation, (-�, 3) . x(x-7) 8 x2 -7x 8 x2 -7x-8 > 0 We graph the function J(x) = x2 -7x-8 . The intercepts are y-intercept: 1(0) = -8 -
1 7.
-25 the x-axis when x < -l or The graph is above x > 8 . Since the inequality is strict, the solution set is { xl x < -l or x > 8 } or, using interval notation, (-00,-1) (8, 00 ) . 4x2 +9 < 6x 4x2 -6x+9 < 0 We graph the function I(x) = 4x2 -6x+9 . y-intercept: 1(0) = 9 ) 2 -4(4)(9) x-intercepts: x = -(-6) ±�(-6 2(4) = 6 ± ..)8- 108 (not real) Therefore,Jhas no x-intercepts. � � The vertex is at x = -2ab = -(-6) 2(4) = 8 = 4 . Since 1 (�4 ) = 247 ' the vertex is (�4 ' 274 ) . 25
-5 never is graph The is no real solution. below the x-axis. Thus, there
>
>
1 87
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Education , Inc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any mean s , without permi ssion in writing from the publi sher.
6 ( X2 -1 ) > 5x 6x2 -6> 5x 6x2 -5x-6 > 0 We graph the function f(x) = 6x2 -5x -6 . y-intercept: f(O) = -6 x-intercepts: 6x2 -5x-6 = 0 (3x+2)(2x-3) = 0 x = --23 ' X =-32 The vertex is at x = -2ab = -(-5) 2(6) = 2-12 . Since f (2-12 ) = _ 16924 ' the vertex is (2-12 , _ 16924 ) . 10
Chapter 4: Linear and Quadra tic Functions
21.
�
-
l ()
j
10 <
�
-00 ,
25.
(0 .
b.
<
.
.
c.
1
-00
23.
U
�
=
The graph is above the x-axis when x -�3 or x > �2 . Since the inequality is strict, solution set . { X I X -"32 or x > "23 } or, usmg mterva notation, ( , -�)u( 3 �2 ' (0). The domain of the expression f(x) = ..Jx2 -16 includes all values for which x2 -16 � 0 . We graph the function p( x) = x2 -16 . The intercepts ofp are y-intercept: p(O) = -6 x-intercepts: x2 -16 = 0 (x +4)(x -4) = 0 x = -4, x = 4 The vertex ofp is at x = -2ab = -(0) 2(1) = O . Since p(O) = -16 , the vertex is ( 0,-16 ) . <
�
�
a.
10
-10
IS
The graph ofp-20is above the x-axis when x -4 or x > 4 . Since the inequality is not strict, the solution set of x2 -16 � 0 is {x I x -4 or x 4} . Thus, the domain offis also {x I x -4 or x 4} or, using interval notation, ( -4] [4, ) f(x) = x2 -1; g(x) = 3x + 3 f(x) = 0 x2 -1 = 0 (x -1)(x + 1) = 0 x 1; x = -1 Solution set: { -1, I } . g(x) = 0 3x+3 = 0 3x = -3 x = -1 Solution set: {-I } . f(x) = g(x) x2 -1 = 3x+3 x2 -3x-4 = 0 (x-4)(x+I) = 0 x = 4;x = -I Solution set: {-I, 4} . f(x) > 0 We graph the function f(x) = x2 -1 . y-intercept: f(O) = -1 x-intercepts: x2 -1 = 0 (x+I)(x-I) = O x = -I, x = I The vertex is at x = -2ab = -2(1)(0) = O . Since f(O) - 1 , the vertex is (0,-1). d.
=
1 88
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Section 4.5: Inequalities In volving Quadratic Functions
g.
10
10
f(x) :2: 1 x2 - 1 :2: 1 x2 - 2 :2: 0 We graph the function p(x) x2 - 2 . The intercepts of p are y-intercept: p(O) = -2 x-intercepts: x2 - 2 = 0 x2 = 2 x = ±.fi = O . Since The vertex is at x = -2ab = -(0) 2(1) p(O) = -2 , the vertex is (0, -2). 10 =
-10 The graph is above the x-axis when x < -1 or x > 1 . Since the inequality is strict, the solution set is { xl x < -l or x > I} or, using interval notation, ( -1) u (1 , ) . g(x) � 0 3x + 3 � 0 3x � -3 x � -1 The solution set is { x I x � -I} or, using interval notation, ( - 1] . f(x) > g(x) x2 - 1 > 3x + 3 x2 - 3x - 4 > 0 We graph the function p( x) = x2 - 3x - 4 . The intercepts of p are y-intercept: p(O) = -4 x-intercepts: x2 - 3x - 4 = 0 (x - 4)(x + l) = 0 x = 4, x = -1 = � . Since The vertex is at x = -2ab = -(-3) 2(1) 2 CXl
-CXl,
e.
10
-CXl ,
f.
p
27.
(�2 ) = _ 254 ' the vertex is (�2 ' 25 ) . _
10
4
10 -10 The graph of p is above the x-axis when x < -l or x > 4 . Since the inequality is strict, the solution set is { xl x < -l or x > 4 } or, using interval notation, (-CXl, -1) u (4, CXl) .
-10 The graph ofp is above the x-axis when x < -.fi or x > .fi . Since the inequality is not strict, the solution set is { xl x � -.fi or X :2: .fi } or, using interval notation, (-CXl, -.fi J u [.fi, CXl) . f(x) = _x2 + 1; g(x) = 4x + 1 a. f(x) = 0 _x2 + 1 = 0 l - x2 = 0 (l - x) (l + x) = O x = 1; x = -1 Solution set: {-I, I} . g (x) = 0 b. 4x + l = 0 4x = -1 X = --41 Solution set:
{-�} .
1 89
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Chapter 4: Linear and Quadratic Functions
c.
d.
x-intercepts: _x2 -4x = 0 -x(x + 4) = 0 x = O; x = -4 The vertex is at x = -2ab = -(-4) 2(-1) = � -2 = -2 . Since p( -2) = 4 , the vertex is (-2, 4).
f(x) = g (x) _X2 + 1 = 4x + 1 0 = X2 + 4x 0 = x(x + 4) x = 0; x - 4 Solution set: {-4, O} . f(x) > 0 We graph the function f(x) = _x2 + I . y-intercept: f(O) = 1 _x2 + 1 = 0 x-intercepts: x2 - I = 0 (x + I)(x - I) = O x = -I;x = 1 -(0) = O . Since The vertex is at x = -2ab = 2(-1) f(O) = I , the vertex is (0, 1).
10
10
-10 The graph ofp is above the x-axis when -4 < x < O . Since the inequality is strict, the solution set is { x 1 - 4 < x < O} or, using interval notation, (-4, 0) . f(x) ;::: 1 g. _x2 + 1 ;::: 1 _x2 ;::: 0 We graph the function p(x) = _x2 . The -(0) = O . Since vertex is at x = -2ab = 2(-1) p(O) = 0 , the vertex is (0, 0). Since a = -1 < 0 , the parabola opens downward. 10
10
10 -10
c.
The graph is above the x-axis when -I < x < I . Since the inequality is strict, the solution set is { xl - I < x < I } or, using interval notation, (-I, I) . g(x) :o; O 4x + I :O; 0 4x :O; -I x :o; --41
10
-1 0
{ l -±} or, using interval notation, ( - ±] .
The solution set is x x :O;
29.
-00 ,
f.
f(x) > g(x) _x2 + 1 > 4x + 1 -x2 - 4x > 0 We graph the function p(x) = _x2 - 4x . The intercepts of p are y-intercept: p(O) = 0
The graph of p is never above the x-axis, but it does touch the x-axis at x = O. Since the inequality is not strict, the solution set is {O}. f(X) = X2 _ 4; g(x) = -x2 + 4 a. f(x) = 0 x2 - 4 = 0 (x - 2)(x + 2) = 0 x = 2;x = -2 Solution set: {-2, 2} .
1 90
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in wri tin g from the publi sher.
Section 4. 5: Inequalities Involving Quadra tic Functions
b.
c.
d.
g (x) = 0 _x2 + 4 = 0 x2 - 4 = 0 (x + 2)(x - 2) = 0 x = -2;x = 2 Solution set: {-2, 2} . f(x) = g(x) x2 - 4 = -x2 + 4 2X2 - 8 = 0 2(x - 2)(x + 2) = 0 x = 2;x = -2 Solution set: {-2, 2} . f(x) > 0 x2 - 4 > 0 We graph the function f(x) = x2 - 4 . y-intercept: f(O) = -4 x2 - 4 0 x-intercepts: (x + 2)(x - 2) = 0 x = -2; x = 2 -b -(0) The vertex is at x = 2a = -2(-1) = O . Since /(0) = -4 , the vertex is (0, -4) .
-b -(0) The vertex is at x = 2a = -2(-1) = O . Since g(O) = 4 , the vertex is (0, 4). 10
-101
-10
e.
The grapl;1 is below the x-axis when x < -2 or x > 2 . Since the inequality is not strict, the solution set is { xl x :"0: -2 or x � 2 } or, using interval notation, ( 00 -2] u [2, ) f(x) > g(x) x2 - 4 > _x2 + 4 2X2 - 8 > 0 We graph the function p( x) = 2X2 - 8 . y-intercept: p(O) = -8 2X2 - 8 = 0 x-intercepts: 2(x + 2)(x - 2) = 0 x -2; x 2 -b -(0) The vertex is at x = = - = o . Since 2a 2(2) p(O) = - 8 , the vertex is (0, -8). -
f.
=
10
�
10
-10
=
-101
I t '.
,
(0
.
=
10
10
-101
The graph is above the x-axis when x -2 or x > 2 . Since the inequality is strict, the solution set is { xl x < -2 or x > 2 } or, using interval notation, (-00,-2) u (2, (0 ) . g(x) :"O: O _x2 + 4 :"0: 0 We graph the function g(x) = _x2 + 4 . y-intercept: g(O) = 4 _x2 + 4 = 0 x-intercepts: x2 - 4 = 0 (x + 2)(x - 2) = 0 x = -2; x = 2
\ l ,
1 10
-10
<
The graph is above the x-axis when x < -2 or x > 2 . Since the inequality is strict, the solution set is { xl x < -2 or x > 2 } or, using interval notation, ( 00 -2) u (2, (0) . g. f(x) � 1 x2 - 4 � 1 2 x -5 � 0 We graph the function p(x) = x2 - 5 . y-intercept: p(O) = -5 x-intercepts: x2 - 5 = 0 x2 = 5 x = ±J5 -
,
191
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Chapter 4: Linear and Qua dratic Functions 10
The vertex is at x -b -(0) O . Since 2a 2(1) p(O) = -5 , the vertex is (0, -5). =
=
=
10
10
-10
10
The graph is above the x-axis when x < -1 or x > 2 . Since the inequality is strict, the solution set is {xl x < -1 or x > 2} or, using interval notation, ( -00, -1) u (2, )
-10
The graph ofp is above the x-axis when x < -.J5 or x > .J5 . Since the inequality is not strict, the solution set is { x i x � -.J5 or x � .J5 } or, using interval notation, (-00, -.J5] u [ .J5, ) (0
31.
(0
e.
.
f(X) = X2 _ X-2; g(x) = x2 +x-2 f(x) = 0 x2 -x-2 = 0 (x-2)(x+ 1) = 0 x = 2,x = -1
a.
Solution set:
b.
c.
d.
10 -10
.
'
g(x) � O x2 +x-2 � 0 We graph the function g(x) = x2 + X -2. y-intercept: g(O) = -2 x-intercepts: x2 +x-2 = 0 (x+2)(x-l) = 0 x = -2;x = 1 The vertex is at x = -2ab = -(1) 2(1) = _ .!.2 . Since f ( -�) -�, the vertex is (-�,-�} 10 =
{-I, 2} .
g(x) = O x2 +x-2 = 0 (x+2)(x-l) = 0 x = -2;x = 1 Solution set: {- 2, I} f(x) = g(x) x2 -X -2 = x2 + X -2 -2x = 0 x=o Solution set: { O } . f(x) > 0 x2 -x-2 > 0 We graph the function f(x) = x2 -X -2 . y-intercept: f(O) = -2 x-intercepts: x2 -x-2 = 0 (x-2)(x+l) = 0 x = 2;x = -1 The vertex is at x = -2ab = -(-1) 2(1) = .!.2 . Since f (.!.2 ) = _2. the vertex is (.!.2 ' _2. ) . 4
.
The graph is below the x-axis when -2 < x < 1 . Since the inequality is not strict, the solution set is { x 1 - 2 � x � I } or, using interval notation, [-2, 1] .
f.
g.
f(x) > g(x) x2 -X -2 > x2 + X -2 -2x > 0 x
�
4
1 92
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exist. No portion of thi s material may be reproduced, in any form or by any mean s, without permi ssion in writing from the publisher.
Section 4.5: Inequalities Involving Quadratic Functions
-
t
-(80) = 2.5 = . 2a 2(-16 ) Since 1 (2.5)=4 , the vertex is (2.5, 4). 5
x-intercepts: x2 - x - 3 =0 - - - - ( 3).:.... .. 1 ) 2 _ 4 (1) (----.:. _ ( -1) ± �rx=---'-----'--'--'-----"-.,-----'-.. 2 (1) 0+12 1±� 1± 2 2 X"" -1.30 or X"" 2.30 -b -(-1) The vertex is at x= = = Since 2a 2(1) 2 -� . p � = -� the vertex is 2 4' 2' 4 10
�. (� )
()
-5 The graph ofIis above the t-axis when 2 < t < 3 . Since the inequality is strict, the solution set is {t 12 < t < 3} or, using interval notation, (2, 3) . The ball is more than 96 feet above the ground for times between 2 and 3 seconds.
10
� ---
+� . ---
-10 The graph ofp is above the x-axis when 11 X< or X> . Smce the 2 2 inequality is not strict, the solution set is
35.
{ I 1-� 1+�} . [-00, f3] [ f3 oo}
33.
a.
x
R(p) == _4p2 +4000p= 0 - 4 P (p-1000) = 0 p==0, p=1000 Thus, the revenue equals zero when the price is $0 or $1000. Find the values ofp for which _4p2 + 4000p> 800,000 _4p2 + 4000p - 800,000>0 We graph I(p)= _4p2 +4000p- 800, 000 . The intercepts are y-intercept: 1(0) =-800,000 p-intercepts: -4p2 4000 P - 800000 = 0 p2 -1000p 200000= 0 - - - "-- - - - - - _ (-1000) ± �r-( -1000) 2 - 4 (1) ( 200000-) p= 2 (1) 1000 ± .j200000 2 1000 ± 200J"i 2 =500 ±100J"i p"" 276.39; p"" 723.61 . -(4000) -b =500 . The vertex is at p= = 2a 2(-4) Since I (500)=200, 000 , the vertex is (500, 200000).
+ +
The ball strikes the ground when
s(t)=80t-16t2 =0. 80t-16t2 =0 16t(5 -t )=0 1=0, t=5 b.
a.
b.
�
-- - or x ;::: -- - or, usmg 2 2 interval notation, 1u 1+ , x
-b
The vertex is at =
The ball strikes the ground after 5 seconds. Find the values of t for which 80t-1612 > 96 -16t2 80t - 96> 0 We graph the function I(t)=-16t2+ 80t- 96 . The intercepts are y-intercept: 1(0) =-96 t-intercepts: -16t2 80t- 96=0 -16(t2 -5t+6)=0 16(t-2)(t- 3)=0 t=2, t =3
+
+
193
© 2008
Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 4: Linear and Quadratic Functions
250,000
Ol�--+-----���
1000 h.
-50,000
The graph offis above the p-axis when 276.39 < p < 723.61 . Since the inequality is strict, the solution set is {p!276.39 < p < 723.61} or, using interval notation, (276.39,723.61) . The revenue is more than $800,000 for prices between $276.39 and $723.61.
The graph offis above the c-axis when 0.112 < c < 81.907 . Since the inequality is strict, the solution set is {c! 0.112 < c < 81.907} or, using interval notation, (0.112,81.907) . Since the round is to be on the ground Note, 75 km = 75,000 m. So, x=75,000, v=897, and g=9.81.
y=0
.
( 2 )( 89 7 )
2 c(75 ,000)_(I+C2) 9.81 7S,000 =0
75,000c -34,290.724(1 + c2)=0 75,000c -34,290.724 -34,290.724c2=0 -34,290.724c2+ 75,000c -34,290.724=0 We graph
a.
f C c) = -34,290.724c2 + 75,000c - 34,290.724 .
Since the round must clear a hill 200 meters high, this mean y> 200 . Now x=2000, v=897, and g=9.81 .
c(2000) -(1+ c2)
The intercepts are y-intercept: f(O) =-34,290.724 c-intercepts:
( 9.�1)( 28�007 J > 200
-34,290.724c2 + 75,000c - 34,290.724 0 2 c = -(75,000)± �(75,000) -4(-34,290.724)(-34,290.724) =
2000c -24.3845(1 + c2)> 200 2000c -24.3845 -24.3845c2> 200 -24.3845c2+ 2000c -224.3845> 0
2(-34,290.724)
- 75,000± �921,584,990.2 -68,581.448 c"" 0.651 or c"" 1.536 =
We graph
f(c)=-24.3845c2+ 2000c -224.3845 .
It is possible to hit the target 75 kilometers away so long as c "" 0.651 or C"" 1.536 .
The intercepts are y-intercept: f(O)=-224.3845 c-intercepts:
39.
-24.3845c2 + 2000c -224.3845=0 -2000± �(200W - 4(-24.3825)(-224.3845)
We graph the function f(x)= (x -4)2. y-intercept: f(O)=16 x-intercepts: (x - 4)2=0
c= ------�--�--�----��----�
2(-24.3825) -2000± �3,9 78,113.985 = -48. 769 C"" 0.112 or c "" 81.90 7
x-4=0 x=4
The vertex is the vertex is (4, 0).
The vertex is at -
c= b= 2a
-(2000) =41.010 . 2(-24.3845)
(X_4)2 � 0
10
Since
f(41.010) "" 40,785.273, the vertex is
_IOI
(41.010, 40785.273).
�!\...a.....J
10
50,000
-10
The graph is never below the x-axis. Since the inequality is not strict, the only solution comes from the x-intercept. Therefore, the given inequality has exactly one real solution, namely
100
x=4 .
-5000 194
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Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 4 Review Exercises
41.
Solving x2 + x + 1> ° We graph the function f(x)=x2 + X + 1. y-intercept: f(O)=1 x-intercepts: b 2 - 4ac = 12 - 4(1) (1)=-3, so f has no x-intercepts. The vertex is at x= -b =
-(1)= _.!.. Since 2a 2(1) 2
( )
f -.!. =� the vertex is 2 4' 10
(-.!.2' �4 ) .
increasing
c.
5.
10
G ( x)=4 Slope = 0; y-intercept = 4 Plot the point (0, 4) and draw a horizontal line through it.
a.
b.
y
-10
The graph is always above the x-axis. Thus, the solution is the set of all real numbers or, using interval notation, (-00, (0 ) .
5
(-3.4)
f( x)=2x-5
-5
Slope = 2; y-intercept = -5
a.
b.
7.
y
3.
-54
b.
Slope =
�; y-intercept
Y
= f( x)
Avg. rate of change
-2
L'ly =
&
3-(-2) --5 - -5 0-(-1) 1 8-3 5 -=-=5 1 8 1-0 1 13-8 2 13 =�=5 2-1 1 18-13=�=5 3 18 3-2 1 This is a linear function with slope 5, since the average rate of change is constant at 5.
°
x
3
=
h(x)= x- 6 a.
x
-1
increasing
c.
constant
c.
Plot the point (0,-5). Use the slope to find an additional point by moving 1 unit to the right and 2 units up. 2
(4,4) x
Chapter 4 Review Exercises
1.
(0.4)
=
-6
Plot the point (0, -6 ) . Use the slope to find an additional point by moving 5 units to the right and 4 units up. 195
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Chapter 4: Linear and Quadratic Functions
9.
f(x) = (x - 2)2 + 2 Using the graph of y = x2 , shift right 2 units, then shift up 2 units.
b -4 -=2 4 = x = - -= . 2a 2(1) 2 The y-coordinate of the vertex is = f(2) = (2)2 - 4 ( 2 )+ 6 = 2 . f -
( :J
y
Thus, the vertex is (2, 2). The axis of symmetry is the line x = 2 The discriminant is: b2 - 4ac=(-4)2 - 4 (1) (6)=-8<0 , so the graph has no x-intercepts. The y-intercept is f(O) = 6 . .
-2 -
11.
8
2
x
f(x) = _(x_ 4)2 Using the graph of y =x2 , shift the graph 4 units right, then reflect about the x-axis. y 5
-2
(4, 0)
b. c.
13.
f(x)=2(x+1)2+4 Using the graph of y = x2 , stretch vertically by a factor of 2, then shift 1 unit left, then shift 4 units up.
17.
a.
8
-2
x
The domain is ( -eX), eX) . The range is [2, eX) . Decreasing on ( -eX), 2) . Increasing on (2, eX) ) . 1 f(x)=-x2 -16 4
a = .!.' b = a, c =-16. Since a = .!. > a' the 4 4 graph opens up. The x-coordinate of the b . -0 a vertex IS x = -- = --- = '-- = O. 2a
<±)
-4
15.
-2
-1
2
4
�
( )
The y-coordinate of the vertex is f -� = f(O) = .!. (0)2 -16 = -16. 2a 4 Thus, the vertex is (0, -16). The axis of symmetry is the line x 0 . The discriminant is: c-16) = 16 > 0 , so b2 - 4ac = (0)2 - 4
x
=
(±)
f(x) = (X_2)2 +2 =x2 - 4x+4+2 =x2 - 4x+6 a = 1, b = -4, c = 6. Since a = 1 > 0, the graph opens up. The x-coordinate of the vertex is a.
the graph has two x-intercepts. The x-intercepts are found by solving:
196
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Chapter 4 Review Exercises
1 - x2 -16=0 4 x2 -64 = 0 x2 =64 X= 8 or x=- 8 The x-intercepts are -8 and 8. The y-intercept is 1(0) = -16 .
b. c.
19.
a.
x
-2
b.
The domain is ( -00,00) . The range is (-00, 1] .
c.
Increasing on -00,
( �) Decreasing on (�, ) 00
The domain is (-00, 00) . The range is [-16, 00) .
21.
l(x)=2.x2 +3x+l 2 a = 2." b= 3 c =1. Since a = 2. > 0' the 2 2 graph opens up. The x-coordinate of the b 3 3 I vertex IS. X=--=- -=--=--. 2a 2(9/2) 9 3 The y-coordinate of the vertex is
Decreasing on (-00, 0) . Increasing on (0, 00 ) . l( x)=- 4x2 +4x a =- 4, b = 4, c = O. Since a =- 4 < 0, the graph opens down. The x-coordinate of the . b 4 4 1 vertex IS x=--=---=--=- . 2a 2(- 4) -8 2 The y-coordinate of the vertex is
( :a )=/ (-�)=� (-�r + ( - �)+1 3
1 -
I = -1+1=-I 2 2
( :J= (�)=-4 (�J +4 (�) =-1+2=1 Thus, the vertex is (�, 1) . I -
a.
.
Thus, the vertex is
I
(-�, �) .
The axis of symmetry is the line x= -! . 3
The discriminant is: b2 - 4ac = 3 2 - 4 c1)=9 -18=-9 < 0 , so
( �)
The axis of symmetry is the line x= ! . 2 The discriminant is: b2 - 4ac = 4 2 -4(-4)(0) =16 > 0 , so the graph has two x-intercepts. The x-intercepts are found by solving: - 4x2+4x= 0 - 4x(x-1)=0 x= 0 or x =1 The x-intercepts are 0 and 1. The y-intercept is 1(0) =- 4(0)2 + 4(0)=0 .
the graph has no x-intercepts. The y intercept is 1(0)=2.(0 ) 2 + 3 (0) + 1= 1. 2
-2
I I I I 1 -12 I
2
x
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Chapter 4: Linear and Quadratic Functions
b.
c.
(-00, 00) . The range is [� ,00 J. Decreasing on ( -oo,-t J . Increasing on ( t 00) . -
23.
a.
b.
The domain is
c.
,
f(x) = 3x2+4x-1 a = 3, b = 4, c = -1. Since a = 3 > 0, the
25.
graph opens up. The x-coordinate of the
(-00,00) . The range is [-�, 00J. Decreasing on ( -00, -1J Increasing on (-1,00).
The domain is
f(x)=3x2-6x+4 a = 3, b = -6, c = 4. Since a = 3 > 0, the graph
. b 4 4 2 vertex x = - - = - -- = - - = - - . 2a 2(3) 6 3
opens up, so the vertex is a minimum point. . . The rrurumum occurs at x = - b = - -6 = -6 = 1 .
The y-coordinate of the vertex is
The minimum value is
IS
-
2a
( :J
f=
3
3
Thus, the vertex is
_
3
(-%,-�) .
27.
f(l) = 3(1) 2 -6(1)+ 4
f(x)=-x2+8x-4 a = -1, b = 8, c = -4. Since a = -1 < 0, the
3
The discriminant is:
x =- b =- 8 =- 8 = 4 2a 2(-1) -2
graph has two x-intercepts. The x-intercepts are found by solving:
The maximum value is
-
b2-4ac = (4)2-4(3)(-1) = 28 > 0 , so the
2a
( :J
f-
- 4±..fi8 2(3)
29.
-4±2.J7 6
-2±.J7 3 The x-intercepts are -2-.J7 3 -2+.J7 ",,0.22 . 3
""
x
I
--
= =
-
.
2
f( 4 ) = -( 4 ) +8(4)-4
-16+32-4 = 12
f(x) = -3x2+ 12x + 4 a = -3, b = 12, c = 4. Since a = -3 < 0, the
graph opens down, so the vertex is a maximum point. The maximum occurs at
-1.55 and
x=-
� = -� = -E = 2 . 2a 2(-3) -6
( :a )
The maximum value is
The y-intercept is f(O) = 3(0)2+ 4(0)-1 = -1 . =-
6
graph opens down, so the vertex is a maximum point. The maximum occurs at
The axis of symmetry is the line x = -� .
3x2+4x-1=0. x -b±�b2- 4ac
2(3)
=3-6+4=1
� �_1 = 2 _
=
-
32
f-
Y
=
2
f(2) = 3(2) +12(2)+4
= -1
_
2 + 24 + 4 = 16
-5 I
198
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Chapter 4 Review Exercises
31.
{l
x2+ 6x -16<0 We graph the function f(x) =x2+6x-16 . The intercepts are y-intercept: f(O) =-16 x-intercepts: x2 + 6x - 16 = 0 (x+ 8)(x- 2)=0 x=- 8, x = 2 The vertex .IS at x = -b = -(6) =-3 . Since 2a 2(1) f(-3)=-25 , the vertex is (-3,-25).
solution set is x x � interval notation, 35.
a.
b.
10
c.
-30
The graph is below the x-axis when -8<x<2 . Since the inequality is strict, the solution set is {x 1 -8<x<2} or, using interval notation, (-8, 2) . 33.
37.
3
10
b.
c.
(7 )
d.
-25 The graph is above the x-axis when x<-.!. or 3 x > 5 . Since the inequality is not strict, the
u
-00,
00
.
CompanyA: C(x)=0.06x+7.00 Company B: C(x)= 0.08x 0.06x+ 7.00 = 0.08x 7.00 = 0.02x 350=x The bill from Company A will equal the bill from Company B if 350 minutes are used. 0.08x<0.06x + 7.00 0.02x<7.00 x<350 The bill from Company B will be less than the bill from Company A if fewer than 350 minutes are used . That is, 0 � x<350 . The revenue will equal the quantity x sold times the price p. That is, R=xp. Thus, R(x) =x - J.-X+150 =_J.-x2+150x 10 10
(
3x2�14x+5 3x2 -14x-5�0 We graph the function f(x) =3x2 -14x -5 . The intercepts are y-intercept: f(O) =-5 x-intercepts: 3x2 -14x-5 = 0 (3x+1)(x-5)=0 1 x=- -' x =5 3 . The vertex IS at x = -b = -(-14) = 14 = 7 2a 2(3) 6 3 . 64 . 64 the vertex IS. 3'-3 S·mce f 7 = -3'
()
a.
(
-� or x�5} or, using - � ] [5, )
)
R(100) =- J.- (100)2+150(100) = 14,000 10 The revenue is $14,000 if 100 units are sold. a =_J.- , b=150, c = O. Since a =_J.-<0, 10 10 the graph opens down, so the vertex is a maximum point. The maximum occurs at -b - (150) -150 = = 750 . Thus, the x= = 2a 2(-1/10) -1/5 quantity that maximizes revenue is 750 units. The maximum revenue is R(750)=_J.- (750)2 +150(750) 10 =-56, 250+112,500 = $56,250 From part (c), we know revenue is maximizes when x= 750 units are sold. The price that should be charged for this is 1 P =- - (750) +150 = $75 . 10
199
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Chapter 4: Linear and Quadratic Functions
39.
Consider the diagram
43.
A(x) = x(1 0 - x) = _x 2 + 1 0x
i
The maximum value occurs at the vertex: b 10 = 5 10 = - = - -x = -2a 2(-1) - 2
x
I � Y� I
The area function is:
1
The maximum area is:
A(5) = _(5)2 + 10(5) = - 25 + 50 = 25 square units 10
Total amount of fence = 3x + 2Y = 1 0, 000
Y = 1O, 000 - 3x = 5000 - � x 2 2 Total area enclosed = (x ) (y) = (x) 5000 - X
( %)
(0,10 - x) .._--...
3 = - -3 x 2 + 5000x is a A(x)=5000x - _X2 2 2 quadratic function with a = - � < O . 2
(x, 0)
45.
a.
So the vertex corresponds to the maximum value for this function. The vertex occurs when
S S
,.-..
b 5000 =-5000 x= - -= 2a 2 (-3/2) 3 '
'-"'
:0 � (";l
( ) -%( 50300 r + 5000( 50300 ) = _ � ( 25, 000, 000 ) + 25, 000, 000 2 3 9
The maximum area is:
5000 = A 3
b.
----
41.
Y 39
c.
• •• • •
38 37
36
35
34 0
= - 12, 500, 000 + 25, 000, 000 3 3 12, 500, 000 3 "" 4, 166, 666.67 square meters
10
•
•
24
•
• •
25
Humerus
26
(mm)
27
x
Yes, the two variables appear to have a linear relationship. Using the LINear REGression program, the line of best fit is: y = l.3902x + 1 . 1 140 LinRe9 '::/=ax+b a=1.390171918 b=1. 113952697 r�=.9050023758 r=.9513161282
C(x) = 4.9x 2 - 617Ax + 1 9, 600 ; a = 4.9, b = -6 17A, c = 1 9, 600. Since a = 4.9 > 0, the graph opens up, so the vertex is
d.
a minimum point. a. The minimum marginal cost occurs at
Y = 1 .39017 (26.5) + 1. 1 1 395 "" 37.95
rum
x = _..!!.... = _ - 617040 = 617 040 = 63 . 2(4.9) 2a 9.8 Thus, 63 golf clubs should be manufactured
b.
in order to minimize the marginal cost. The minimum marginal cost is
C ( 63 ) 4.9 ( 63 )2 - ( 617.40 )( 63 )+ 19600 =
=
$151.90
200
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Chapter 4 Test
Chapter 4 Test
1.
4.
f (x) =-4x+3 Slope -4; y-intercept 3. a. b. The slope is negative, so the graph is decreasing. Plot the point (0, 3) . Use the slope to find c. an additional point by moving 1 unit to the right and 4 units down. =
=
f(x) = g(x) 2 x +3x=5x+3 x2 - 2x-3 =0 (x+1)(x-3)=0 x+1= ° or x -3 = ° x= -l or x= 3 The solution set is { - I, 3} . y
y
.x
2.
5.
f(x) = 3x2 -2x - 8 y-intercept: f(O) =-8 x-intercepts: 3x2 - 2x - 8 = ° (3x+ 4)(x - 2) = 0 4 x=--·x=2 3' The intercepts are (0,-8),
3.
x
y
(-}' 0), and (2, 0) .
x
G(x) =-2x2 + 4x+1 y-intercept: G(O) =1 x-intercepts: _2 X2+ 4x+1 =0 a =-2, b = 4, c =1 x=
-4 6.
42 - 4( ± �-'--2)( 1) ..:... -b ± -Jb2 - 4ac = -4 ---'----"-..:.
(
J.
(
a.
b.
2( -2) 2a -4 ± J24 -4 ± 2.J6 2 ± .J6 -4 -4 2 .J6 2 -The intercepts are (0, 1) , -2 , 0 , and 2+ .J6, 0 2
f(x) = (x-3 )2_2 Using the graph of y = x2 , shift right 3 units, then shift down 2 units.
J
f(x) = 3x2 -12x+ 4 a=3, b=-12, c=4. Since a= 3>0, the graph opens up. The x-coordinate of the vertex is -12 b -12 x=--=---=--=2 . 2a 2 (3) 6 The y-coordinate of the vertex is = f (2)=3 (2)2 -12 (2)+ 4 f -
( :J
=12 - 24 + 4 =-8 Thus, the vertex is (2, -8) .
201
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Chapter 4: Linear and Quadratic Functions
c.
d.
10
The axis of symmetry is the line x= 2 . The discriminant is: b2 - 4ac =( _12)2 - 4( 3)( 4)=96 > 0 , so the graph has two x-intercepts. The x-intercepts are found by solving: 3x2 -12x+ 4 =0 . ±.J 2 - 4ac -(-12)± .J% x= -b b 2(3) 2a 12 ± 4 ..[6 6 ± 2 ..[6 3 6
-10 The graph is above the x-axis when x < 4 or x > 6 . Since the inequality is not strict, the solution set is {xl x ::; 4 or x � 6 } or, using interval notation, ( -ex), 4 ] u [6, ex)) .
-----
. The x-mtercepts are 6-Z.J6 � 0.37 and ---
6±Z.J6
---
3
3
9.
.
b.
' � 3.63 . The y-mtercept IS
f(O) = 3 (0) 2 -12(0)+ 4 = 4 . e.
)' 8
(4,4)
10.
7.
f(x) =_2X2+12x+ 3 a =-2, b =12, c = 3. Since a =-2 < 0, the graph opens down, so the vertex is a maximum point. The maximum occurs at b 12 12 x=--==--=3 . 2a -4 2(-2) The maximum value is f( 3)= _2( 3)2+12 ( 3)+3 =-18+ 36+ 3 = 21. --
8.
a.
x2-lOx+ 24�0 We graph the function f(x) = x2 -lOx+ 24 . The intercepts are y-intercept: f(O) = 24 x-intercepts: x2 -lOx+ 24 = 0 (x- 4)(x -6) = 0 x= 4, x= 6 . 10 -b -(-10) The vertex IS at x=-= --- = -=5 . 2 2a 2(1) Since f(5) =-1 , the vertex is (5, -1).
C(m) = 0.15m+129.50
C(860) = 0.15(860) +129.50 =129+129.50 = 258.50 If 860 miles are driven, the rental cost is $258.50. c. C(m) = 213.80 0.15m+129.50=213.80 0.15m = 84.30 m =562 The rental cost is $213.80 if 562 miles were driven. rex) = -0.115x2+1.183x+ 5.623; a =-0.115, b =1.183, c = 5.623 Since a =-0.115 < 0, the graph opens down, so the vertex is a maximum point. a. The maximum interest rate occurs at b -1.183 -1.183 5 = X=--= � .14 . 2a 2(-0.115) -0.23 The maximum interest rate was about r(5.14) =-0.115(5.14)2+1.183(5.14)+ 5.623 �8.67 Thus, the interest rate was highest in 1997, and the highest rate at this time was about 8.67%. b. The year 2010 corresponds to x=18 . r (18 ) =-0.115(18 )2+1.183( 18 )+5.623 �-10.34 The model estimates the rate in 2010 to be -10.34%. This rate does not make sense since an interest rate cannot be negative.
202
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Chapter 4 Cumulative Review
Chapter 4 Cumulative Review
1.
9.
P=(-1,3);Q=( 4,-2) Distance between P and Q:
d(P,Q) = �(4-( _1))2 + (_2 _ 3)2 = �(5)2 + ( _5)2 = ../25 + 25 = J50 = 5.fi
(
{ I �}
( ) ) 2 2' 2
Domain: z z*
Midpoint between P and Q: -1 + 4 3 - 2 = � � =( 1 . 5 0.5) ' ' 3.
2
5x + 3�0 5x�-3 3 x�- 5
11.
.
[
3 5
{xl x�-�} or [-�, (0).
b.
)
Slope of perpendicular = = m (x - xJ
�
( 4)
x+4
I f(I)=_ _= � * �, so 1) is not on
1+4 5 4
the graph off
c.
Perpendicular to Y = 2x + 1 ; Containing (3,5) Y-YI
x f(x) = a.
The solution set is
5.
3z - 1 6z-7 The denominator cannot be zero: 6z - 7 *0 6z*7 7 z*6 h(z) =
-�2 13.
y-5 = - ( X - 3)
y - 5 = --1 x + -3 2 2 13 y = --1 x + 2 2
-2 =-I, so (-2, - 1) is a f ( -2) = -2 =2 -2 + 4
point on the graph off Solve for x:
2 = _x_ x+4 2x +8= x x= -8 So, (-8, 2) is a point on the graph off
f ( x ) = x3 -5x + 4 on the interval (-4, 4) Use MAXIMUM and MINIMUM on the graph of Y I = x3 - 5x + 4 .
10
4
y
Haxi ..... ul'1'l H= '1.<:90997
V=S.30331'tB
-10 10
-41 --�--�--�-- 4 7.
HinilYlum H=1.290!l!lIiB
Yes, this is a function since each x-value is paired with exactly one y-value.
v= -3.303315
-10
Local maximum is 5.30 and occurs at x "" -1 .29 ; Local minimum is -3.30 and occurs at X "" 1 .29 ; fis increasing on (-4, -1 .29) or (1 .29, 4) ; fis decreasing on (-1 .29, 1 .29) . 203
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Chapter 4: Linear and Quadratic Functions
15.
a.
b.
c.
d.
y
10
Intercepts: (-1, 0) , (0,-1) , (1,0) x-intercepts: -1, 1 y-intercept: -1 The graph is symmetric with respect to the y-aXIS. When x = 2 , the function takes on a value of 1. Therefore, f (2) = 1.
e.
The function takes on the value 3 at x = -4 and x = 4.
f.
f ( x) < ° means that the graph lies below the x-axis. This happens for x values between -1 and 1. Thus, the solution set is { x! -1 < x < I} or (-1, 1) .
g.
point is multiplied by 2 .
Domain: {x!-4:<:;x:<:;4} or [-4, 4] Range: {y!-I:<:;y:<:;3} or [-1, 3]
x
-5
-10
j. k.
Since the graph is symmetric about the y axis, the function is even. The function is increasing on the open interval (0, 4) .
The graph of y = f ( x) + 2 is the graph of y = f ( x) but shifted up 2 units.
-5
5
x
-5 h.
The graph of y f (-x) is the graph of y = f ( x) but reflected about the y-axis. =
y
5
x
-5
-5 i.
The graph of y = 2 f ( x) is the graph of y f ( x) but stretched vertically by a factor of 2. That is, the coordinate of each =
204
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Chapter 5 Polynomial and Rational Functions Section 5.1
21.
1. (-2, 0 ) , (2, 0) , and (0, 9)
x-intercepts: let y = 0 and solve for x 9x2+ 4 ( 0) = 36 9x2= 36 x2 = 4 x=±2
23.
y-intercepts: let x = 0 and solve for y 9 ( 0) 2+4y= 36 4y =36 y= 9 3.
down; 4
5.
smooth; continuous
f(x) =(x + 1) 4 Using the graph of y =X4 , shift the graph horizontally, 1 unit to the left. y
(-1, 0)
7. touches
9. False; the x-intercepts of the graph of a
.x
-5
polynomial function are also called zeros of the function.
11. f(x) = 4X+X3 is a polynomial function of
25.
degree 3 .
1-x2 1 1 2 2 2 function of degree 2.
13. g(x) = --=- --x2 is a polynomial
15.
G(x) = 2(x _1)2 (x2 + 1) = 2(x2 - 2x+ 1)(x2 + 1) = 2(X4 +X2 - 2x 3 _2x+X2 +1) = 2(x 4 -2x 3 + 2X2 -2x+ l) = 2X4 - 4x 3 + 4x 2 - 4x+2 is a polynomial function of degree 4.
f(x)=l-�=l-x-l is not a polynomial x function because it contains a negative exponent.
f(x) = x5 -3 Using the graph of y = x5 , shift the graph vertically, 3 units down.
x
17. g(x) =X3/2 -x2+ 2 is not a polynomial
function because it contains a fractional exponent. 1 2
19. F(x) =5x 4 -7tX3 +- is a polynomial function
of degree 4.
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Chapter 5: Polynomial and Rational Functions
27.
1 4 f(x) = -x 2
33.
f(x) = 2(x + 1) 4 + 1 Using the graph of y = X 4 , shift the graph horizontally, 1 unit to the left, stretch vertically by a factor of 2, and shift vertically 1 unit up.
Using the graph of y = X 4 , compress the graph vertically by a factor of .! .
2
.l
(0,0)
-5
-5
29.
35.
f(x) = _x 5 Using the graph of y = x5 , reflect the graph about the x-axis.
f(x) = 4 - (x - 2) 5 = _(x - 2) 5 + 4 Using the graph of y = x 5 , shift the graph horizontally, 2 units to the right, reflect about the x-axis, and shift vertically 4 units up.
-5
31.
37.
f(x) = a(x - (-1») (x - l)(x - 3) For a = 1 : f(x) = (x + l)(x - l) (x - 3) = ( X 2 - 1 ) (x -3) = x3 - 3x 2 - X + 3
39.
f(x) = a(x - (-3») (x - O)(x - 4) For a = 1 : f (x) = (x + 3) (x) (x - 4) = ( x 2 + 3x ) (x - 4) = x 3 - 4x 2 + 3x 2 - 12x = x 3 _ x 2 - 12x
f(x) = (x - l) 5 + 2 Using the graph of y = x5 , shift the graph horizontally, 1 unit to the right, and shift vertically 2 units up.
.x
206
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Section 5.1: Polynomial Functions and Models
41.
f(x) = a(x - (- 4))(x - (-1)) (x - 2)(x - 3) For a = 1 : f(x) = (x + 4)(x + l)(x - 2)(x - 3) = ( X 2 + 5X + 4 )( X 2 - 5x + 6 ) =X4
49.
-5x3 +6x2 +5x3 -25x2 +30x+4x2 -20x+24
b.
b.
c.
c.
47.
The real zeros of f(x) = 3(x - 7)(x + 3) 2 are: 7, with multiplicity one; and -3, with multiplicity two. The graph crosses the x-axis at 7 (odd multiplicity) and touches it at -3 (even multiplicity).
n
e.
The function resembles values of l xl .
a.
b. c.
51.
= 3x3 for large
The real zeros of f(x) = 4 ( x 2 + 1 ) (x - 2)3 is: 2, with multiplicity three. x 2 + 1 = 0 has no real solution. The graph crosses the x-axis at 2 (odd multiplicity).
53.
n
e.
The function resembles values of I xI .
The function resembles values of Ix I .
a.
55.
Near 5 : f(x) "" 8 1 (x - 5)3 The function resembles values of Ix I .
e.
a.
b.
= -2x6 for large
Near -4 : f(x)",, -729 (x + 4) 2 ;
e.
a.
y
The real zeros of f(x) = (x - 5)\x + 4) 2 are: 5, with multiplicity three; and -4, with multiplicity two. The graph crosses the x-axis at 5 (odd multiplicity) and touches it at -4 (even multiplicity).
n
d.
= 4x5 for large
-1 = 6-1 = 5
d.
c.
-1 = 5-1 = 4 y
e.
b.
Near 2: f(x) "" 20(x - 2)3
d.
( 1 )2
1
n
c.
-1 = 3-1 = 2 y
2
Near -"2: f(x)",, -36. 125 x + "2
d.
b.
Near -3 : f(x) "" -30(x + 3) 2 ; Near 7: f(x)",, 300(x - 7)
d.
The graph touches the x-axis at-'!' (even multiplicity).
f(x) = a(x - (- 1)) (x - 3) 2 For a = 1 : f(x) = (x + l)(x - 3) 2 ( X + 1) ( X 2 - 6x + 9 ) = x3 - 6x 2 + 9x + x 2 - 6x + 9 = x3 - 5x 2 + 3x + 9 a.
( ±J (x2 + 4r is:- ± , With
multiplicity two. x 2 + 4 = 0 has no real solution.
=
45.
The real zero of
f(X) = - 2 X +
= x 4 - 15x 2 + 10x + 24
43.
a.
-1 5-1 = 4 =
y
=
x5 for large
t
f(x) = 3 ( x 2 + 8 )( x 2 + 9 has no real zeros. x 2 + 8 = 0 and x 2 + 9 = 0 have no real
solutions. The graph neither touches nor crosses the x axis. No real zeros n
-1 = 6 - 1 = 5
The function resembles values of I xI .
y
= 3x6 for large
The real zeros of f(x) = _2x 2 (x 2 - 2) are: -.J2 and .J2 with multiplicity one; and 0, with multiplicity two. The graph touches the x-axis at 0 (even multiplicity) and crosses the x-axis at -.J2 and.J2 (odd multiplicities).
207
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Chapter 5: Polynomial and Rational Functions
c.
d.
e.
c. d.
e.
n
-1 = 4-1 = 3 The function resembles y = _2X 4 for large values of Ix I .
f.
Degree is 3. The function resembles y = x3 for large values of Ixl . 2 Near 0: f(x) "" -3x2 ; Near 3: f(x) "" 9(x - 3) Graphing by hand: y
=
59. 61. 63.
f(x) = (x _ 1) 2 a.
b. c.
y-intercept: f(O) = (0 _ 1) 2 = 1 x-intercept: solve f(x) = 0 (x _ 1) 2 = 0 � x = 1 The graph touches the x-axis at x = 1 , since this zero has multiplicity 2. Degree is 2; The function resembles y = x 2 for large values of I xl .
69.
e.
f.
f (x) = 6x 3 (x + 4) x-intercepts: --4, 0; y-intercept: 0 a. b. crosses x-axis at x --4 and x 0 Degree 4 ; The function resembles c. y = 6x4 for large values of Ixl . d. 3 Near -4: f(x) "" -384(x + 4) ; Near 0: e. f(x) "" 24x 3 f. graphing by hand =
d.
Near 1 : f(x) "" (x - 1)2 Graphing:
(-5,750)
(1,0)
-5
5
b.
=
y 800
(-3, -162)
f(x) = x 2 (x - 3) a.
=
.x
-5
67.
+(4. 16)
16l-
Could be; zeros: -1, 1, 2 ; minimum degree 3 Can't be; not continuous at x = -1 c, e, f c, e
57.
65.
Near -J2: f(x) "" 1 1 .3 1 ( x + .J2 ) ; Near 0: f(x) "" 4x 2 ; Near J2: f(x)",, -1 1 .3 1 ( x - .J2 )
3
x
_200-(-2, -96)
y-intercept: f(O) = 02 (0 - 3) = 0 x-intercepts: solve f(x) = 0 0 = x 2 (x - 3) x = 0, x = 3 touches x-axis at x = 0 ; crosses x-axis at x=3 208
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.1: Polynomial Functions and Models
71.
f ( x ) = -4 x 2 ( x + 2 ) a. x-intercepts: 0, -2 ; y-intercept: 0 b. crosses x-axis at x = -2 ; touches x-axis at x 0 Degree 3 ; The function resembles y = -4x 3 for large values of I xl . d. 2 e. Near -2: f(x)",, - 1 6(x + 2) ; Near 0: f(x) "" -8x 2 f. graphing by hand
75.
=
c.
=
f(x) = 4 x - x3 = x ( 4 - x 2 ) = x ( 2 + x )( 2 - x ) a. x-intercepts: 0, -2, 2 ; y-intercept: 0 b. crosses x-axis at x = 0, x = -2, x = 2 Degree 3; The function resembles y = _x 3 for large values of I xl . d. 2 Near -2: f(x) "" -8(x + 2) ; Near 0: e. f(x) "" 4 x ; Near 2: f(x) "" -8(x - 2) f. graphing by hand
c.
y
=
y
50
(-3,36) (-1, -4) x
73.
f(x) = (x -l) ( x - 2 ) (x + 4 ) a. x-intercepts: 1, 2, -4 ; y-intercept: 8 b. crosses x-axis at x = 1, 2, -4 Degree 3; The function resembles y = x3 for large values of I x l . d. 2 e. Near -4 : f(x) "" 30(x + 4 ) ; Near 1: f(x) "" -5(x - 1) ; Near 2: f(x) "" 6(x - 2) f. graphing by hand c.
77.
f(x) = x 2 (x - 2)(x + 2) a.
=
b. c.
d.
e. x
y-intercept: f(x) = 02 (0 - 2)(0 + 2) = 0 x-intercepts: solve f(x) = 0 0 = x 2 (x - 2)(x + 2) x = 0, 2, - 2 crosses x-axis at x = 2, x = -2 ; touches x axis at x 0 Degree is 4. The function resembles y = X 4 for large values of I xl . 3 Near -2: f(x) "" -1 6(x + 2) ; Near 0: f(x) "" -4 x 2 ; Near 2: f(x) "" 16(x - 2) =
209
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
f.
graphing by hand
(-1, -3)
-20
79.
81.
y
(1, -3)
=
=
f(x) = (X + 2) 2 (x - 2) 2 a. y-intercept: 1 6 x-intercepts: -2, 2 b. The graph touches the x-axis at x = -2 and x = 2 since these zeroes have multiplicity is 2. Degree is 4. The graph of the function c. resembles y = X 4 for large values of I xl . d. 3 e.
f.
f ( x ) = (x _ l) 2 (x - 3)(x + l ) a. y-intercept: f(O) = (0_1) 2 (0 - 3)(0 + 1) = -3 x-intercept: solve f(x) = 0 (x _ 1) 2 (x - 3) (x + 1) = 0 x = 1 or x = 3 or x = -1 b. The graph touches the x-axis at x 1 , since this zero has multiplicity 2. The graph crosses the x-axis at x 3 and x -1, since these zeros have multiplicity 1 . Degree is 4. The graph of the function c. resembles y = X 4 for large values of I xl . d. 3 e. Near - I : f(x) -1 6(x + 1); Near l: f(x) "'" -4(x - l ) 2 ; Near 3 : f(x) "'" 16(x - 3) f. Graphing: "",
y
Near -2: f(x) "'" 16(x + 2) 2 ; Near 2: f(x) "'" 1 6(x - 2) 2 Graphing: )'
30
-5
83. x
(-2,0)
=
(0, -3)
1
-10
f(x) = (x + 2 f ( x - 4) 2 a.
-)0
X
b. c.
d. e.
y-intercept: f(O) = (0 + 2) 2 (0 - 4) 2 = 64 x-intercept: solve f(x) = 0 (x + 2) 2 (X _ 4) 2 = 0�x = -2 or x = 4 The graph touches the x-axis at x = -2 and x = 4 since each zero has multiplicity 2. Degree is 4. The graph of the function resembles y = X 4 for large values of I xl . 3
Near -2: f(x) "'" 36(x + 2) 2 ; Near 4: f(x) "'" 36(x - 4) 2
210
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.1: Polynomial Functions and Models
f.
Graphing by hand;
87.
y
l(x) = -x 2 ( x 2 - I )( x + I ) = _x 2 (x - I)(x + I)(x + 1) = _x 2 ( x - I )( x + I )2 a.
oX
-10
85.
b.
I(x) = x 2 (x - 2)(x 2 + 3) a. y-intercept: 1(0) = 0 2 (0 - 2)(0 2 + 3) = 0 x-intercept: solve I(x) = 0 x 2 (x - 2)(x 2 + 3) = 0 => x = 0 or x = 2 x 2 + 3 = 0 has no real solution
b.
c.
d. e. f.
y-intercept: 1(0) = _0 2 (0 2 - 1)(0 + 1) = 0 x-intercept: solve I(x) = 0 _x 2 ( x - l )( x + I )2 = 0 X = 0 or x = 1 or x = -1 The graph touches the x-axis at x = 0 and x = -1 , since each zero has multiplicity 2. The graph crosses the x-axis at x 1 , since this zero has multiplicity 1 . Degree is 5 . The graph of the function resembles y = _x 5 for large values of I xl . =
c.
d.
The graph touches the x-axis at x = 0 , since this zero has mUltiplicity 2. The graph crosses the x-axis at x = 2 , since this zero has multiplicity 1 . Degree is 5 . The graph of the function resembles y = x 5 for large values of I xl .
e.
f.
4 Near - I : l(x)",2(x +l) 2 ; Near 0: I(x)",x 2 ; Near 1 : I(x)",-4(x - 1) Graphing by hand: y :I
4 Near 0: I(x)",-6x 2 ; Near 2: I(x) ",28(x - 2) Graphing by hand:
(
y
160
(3, 108)
89.
1
3/)
2' 32
-1
I(x) = x3 + 0.2X 2 - 1 .5876x - 0.3 1752 a.
�
-1
x
b.
Degree 3; The graph of the function resembles y = x3 for large values of IxI . Graphing utility =
2
-3 J
I :\1
J
13
-2 c.
x-intercepts: -1 .26, --D.2, 1 .26; y-intercept: -0.3 1752
211
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
d.
Above on (-1 .26, -0.20) and ( 1 .26, 00) ; Below on (-00, -1 .26) and (-0.20, 1 .26).
e.
2 turning points; local maximum: (-0.80, 0.57); local minimum: (0.66, -0.99) Graphing by hand
f.
(-0.5,0.40)
d.
e.
f.
y 2
Above on ( -3.56, 0.5) and (0.5, 00); below on (-00, -3.56) .
2 turning points; local maximum: (-2.2 1 , 9.9 1); local minimum: (0.50, 0) Graphing by hand (-2.21,9.9\) Ib (-\,5.76) x
x
g. h.
91.
g.
Domain: { x I x is any real number } ; Range: {y I y is any real number } .
h.
I is increasing on (-00, -0.80) and (0.66, 00) ; I is decreasing on ( -0.80, 0.66)
I(x) = x3 + 2 . 56x 2 - 3.3 1x+0.89 a. Degree 3; The graph of the function resembles y x 3 for large values of IxI . h. Graphing utility
93.
=
(-3.9, -6.58) -10 Domain: { x I x is any real number } ; Range: {y I y is any real number } . I is increasing on (-00, -2.21) and (0.50, 00) ; I is decreasing on ( -2.2 1, 0.50) X 4 - 2.5x2 + 0.5625 Degree 4; The graph of the function r resembles y X 4 for large values of l xl . Graphing utility
f(x) a.
=
=
=
=
h.
12
-3 c.
5
\
x-intercepts: -3.56, 0.50; y-intercept: 0.89 c.
\
/ .......
V
-2
J
/
V
3
x-intercepts: - 1 .50, -0.50, 0.50, 1 .50; y-intercept: 0.5625
212 © 2008 Pearson Education, Inc., Upper Saddle River, NJ.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.1: Polynomial Functions and Models
d.
e.
f.
Above on ( -00, -1 .5 ) , ( -0.5, 0.5 ) , and ( 1 .5, 00 ) ; below on ( -1 .5, -0.5 ) and ( 0.5, 1 .5 ) .
d.
e.
3
turning points: local maximum: (0, 0.5625); local minima: (-1 . 12, -1), ( 1 . 12, -1) Graphing by hand (-1.75,2.29) (0,0.56)
f.
Above on ( -00, -1 .07 ) and ( 1 .62, 00) ; below on ( -1 .07, 1 .62 ) .
1 turning point; local minimum: (-0.42, -4.64) Graphing by hand y 5
x
(1.75,2.29)
(-1.5,0)
x
g. g. h.
a.
b.
Domain: { x I x is any real number } ; Range: {YI Y� -I } .
h.
f is increasing on ( - 1 . 12, 0 ) and ( 1 . 12, 00 ) ; f is decreasing on ( -00, -1 . 12 ) and ( 0, 1 . 12 )
97.
f is increasing on ( -0.42, 00 ) ; f decreasing on ( -00, -0.42 )
IS
f ( x ) = -2x 5 - .J2x 2 - x - .J2 a. Degree 5 ; The graph of the function resembles Y = -2x 5 for large values of l xl . b. Graphing utility 3 =
Degree 4 ; The graph of the function resembles Y = 2x 4 for large values ofI xI . Graphing utility: =
-2�--+-��----�12
2
c.
c.
Domain: {x l x is any real number } ; Range: {YI y� -4 .64 } .
d.
-6 x-intercepts: -1 .07, 1 .62; y-intercept: -4
-4
x-intercept: -0.98; y-intercept: -.J2 -1 .41 Above on ( -00, -0.98 ) ; below on ( -0.98, 00 ) ""
213
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
c.
f.
d.
No turning points; No local extrema Graphing by hand
12
!�J
y 10
( - 1 . 25, 3.73)
g. h.
99.
a.
( - 0.98. 0) 2
x
o
- 10
c.
Domain: { xI x is any real number } ; Range: {y I y is any real number } . f is decreasing on (-00, 00)
Graphing, we see that the graph may be a cubic relation.
f.
II
10
•
5
b.
c.
•
•
•
2
4
•
•
6
•
•
101.
a.
:j
9
o
For the decade 2001 to 2010 we have x = 9 . H ( 9 ) = 0. 1 6 ( 9 ) 3 - 2.32 ( 9 ) 2 + 9.33 ( 9 ) -2.21 = 1 0.48 The model predicts that the increase in the number of major hurricanes will continue during this decade. This agrees with the scientists. This tends to support the prediction. The model predicted about 1 0 major hurricanes. 2005 is halfway through the decade and half the predicted number of major hurricanes have already occurred. Graphing, we see that the graph may be a cubic relation . T
8
70 65
x
60
For the decade 1 96 1 - 1 970, we have x = 5 . H ( 5 ) = 0. 1 6 ( 5 )3 - 2.32 ( 5 ) 2 + 9.33 ( 5 ) - 2.21 = 6.44 The model predicts that 6 or 7 major hurricanes struck the United States between 1961 and 1 970. The graphing utility agrees to two decimal )laces.
•
55
50 45
40
b.
..... ub l C�eg ,=,=ax 3 +b x � +cx+d a= . 1 59090909 1 b= -2 . 32034632 c=9 . 33008658 d= -2 . 2 1 42857 1 4
c.
•
•
•
•
•
•
•
........J ........... L. ..... .1 . ..........J ........... L ... . ... L. ...... 1 . ...........1... ... _ 12 1 5 1 8 2 1 24 x 9 0 3 6
/). T T ( 12 ) - T ( 9 ) 57.9 - 5 1 . 1 /).x 12 - 9 12 - 9 = � "" 2.27 3 The average rate of change in temperature from 9am to noon was about 2.27°F per hour. /). T T ( 1 8 ) - T ( 1 5 ) 63.0 - 63.0 = = =0 /).x 18-15 1 8 - 15 The average rate of change in temperature from 9am to noon was OaF per hour.
214
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.2: Properties of Rational Functions
d.
e.
e.
At Spm we have x = 17 . r (l7) = -0.01 03 (l7)3 + 0.32(l7)2 - 1 .37 (l7) + 4S.39 = 63.9761 The predicted temperature at Spm is :::; 64°F . The graphing utility agrees with the given model.
f.
1 09.
r-- ub l C�eg ':I=ax � +bx 2: +cx+d a= - . 0 1 02974 1 86 b= . 3 1 7376 1 42 4 c= - 1 . 374242424 d=45 . 392857 1 4
f.
1 03.
True only if d 0, otherwise the statement is false. Answers will vary. One possibility: j (x) = -S ( x _l)3 (x - 2) x - x + =
( i J( �)
Section 5.2
70
\
g.
False, since j (-x) = (_X) 3 + b (_x) 2 + c (-x) + d = -x3 + bx 2 - CX + d :;t: -j (x). (unless b = d = O)
1.
True
3.
y = 1 x
o �__________� · 2S 40
The y-intercept is 45.4°F . The model predicts that the midnight temperature was 4S.4°F . Answers will vary.
1 05.
Answers will vary , j (x) = (x + 2)(x -l) 2 and g ( x) = (x + 2) 3 (x _ 1) 2 are two such polynomials.
107.
j(x) = x3 + bx 2 + cx + d a. True since every polynomial function has exactly one y-intercept, in this case (0, d). b. True, a third degree polynomial will have at most 3 x-intercepts since the equation x3 + bx 2 + cx + d = 0 will have at most 3 real solutions. c. True, a third degree polynomial will have at least I x-intercept since the equation x3 + bx2 + CX + d = 0 will have at least I real solution. d. True, since j has degree 3 and the leading coefficient I .
y 5
( 2, � ) 5
5.
y=1
7.
proper True
9. 11.
13.
x
In R(x) = � , the denominator, q(x) = x - 3 , x-3 has a zero at 3 . Thus, the domain of R(x) is all real numbers except 3. {x I x :;t: 3} - 4x 2 , the denominator, (x - 2)(x + 4) q(x) = (x - 2)(x + 4) , has zeros at 2 and -4. Thus, the domain of H(x) is all real numbers except 2 and -4. {x I x :;t: 2, -4} In H(x) =
215
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
15.
. In F(x) = 3x(x - l) , the denomlllator, 2X2 - 5x - 3 q(x) 2X2 - 5x - 3 = (2x + l)(x - 3) , has zeros at
27.
=
- .! and 3 . Thus, the domain of F(x) is all 2 real numbers except and 3 . x I x "* - , 3
-i
17.
19.
21.
{
i
b.
}
In R(x) = +- the denominator, x -8 q(x) = x3 -8 = (x - 2)(x2 + 2x + 4) , has a zero at 2 ( x2 + 2x + 4 has no real zeros). Thus, the domain of R(x) is all real numbers except 2. {x l x "* 2} '
29.
c.
Intercept: (0, 0) Horizontal Asymptote: Y = I
d.
Vertical Asymptotes: x = - 2, x = 2
e.
Oblique Asymptote: none
F (x) = 2 + .! ; Using the function, y=.! , shift x x the graph vertically 2 units up. y 7
---
x
In R(x) = 3(x2 - x - 6) the denominator, 4(x2 - 9) q(x) 4(x2 - 9) = 4(x - 3)(x + 3) , has zeros at 3 and -3 . Thus, the domain of R(x) is all real numbers except 3 and -3. {x I x "* 3, -3} ,
a.
b. c.
d. e.
25.
Domain: { x l x "* - 2, x "* 2} ; Range: { Y I Y � 0 or Y > I}
x = O
. In H(x) = 3x2 + x , the denomlllator, x2 + 4 q(x) = x2 + 4 , has no real zeros. Thus, the domain of H(x) is all real numbers.
=
23.
a.
a.
b. c.
d. e.
31.
1 ; USlllg th e fiunctton, 1 · Y x (x _ l) 2 shift the graph horizontally I unit to the right. R ( x) -
.
=
2 '
y 4
Domain: { x l x "* 2} ; Range: {y I Y "* l} Intercept: (0, 0) Horizontal Asymptote: y = 1
3 2
Vertical Asymptote: x = 2 Oblique Asymptote: none
x
Domain: { x l x "* O} ; Range: all real numbers Intercepts: (- 1 , 0) and ( 1 , 0) Horizontal Asymptote: none Vertical Asymptote: x = O Oblique Asymptote: y = 2x
x= l
216
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.2: Properties of Rational Functions
33.
(_1_);
H(x) � -2 Using the function x+l x+l � , shift the graph horizontally I unit to the y x left, reflect about the x-axis, and stretch vertically by a factor of 2. =
=
39.
R(x) = x2 - 4 1 � x2 x2
_1_ =
_
( )+1; Using the
4 1 x2
= _
_
function y = , reflect about the x-axis, stretch x2 vertically by a factor of 4, and shift vertically I unit up.
=
y
y= l
x
x
35.
R(x)
-1 X2 + 4x + 4
x = o
1 �2 ; Using the
41.
--
= --
(X + 2)
8 =
=
2
2 + (x - 3) 2
( (X -1 3)2 J + l
1;
m
=
1
=
=
-
4 ,
(x - 2) ( x 2 + 2x + 4 ) H(x) = x3 - 8 (x - 2) (x - 3) x 2 - 5x + 6 x 2 + 2x + 4 , where x :t- 2, 3 x-3 The degree of the numerator in lowest terms is n 2 . The degree of the denominator in lowest terms is m . Since n = m + , there is an oblique asymptote. Dividing: x+5 x - 3) x2 + 2x + 4
---1 =
1
_(X2 - 3x) 5x + 4 - (5x - 1 5)
=
x
=
=
Using the function y 1 , shift the graph right x2 3 units, stretch vertically by a factor of 2, and shift vertically 1 unit up. y
1.
=
43.
37.
=
, the line y � 3 is a horizontal asymptote. The denominator is zero at x so x - 4 is a vertical asymptote. n
x
G(x) 1 + 2 (x - 3) 2
1.
=
-2 y
2
3x ; The degree of the numerator, -x+4
=
=
I I
=
p(x) 3x, is n = The degree of the denominator, q(x) x + 4, is m Since
function y � , shift the graph horizontally 2 x units to the left, and reflect about the x-axis. x =
R(x)
3
1919
H(x) x + 5 + -- , x :t- 2, 3 x-3 Thus, the oblique asymptote is y x + 5 . The denominator in lowest terms is zero at x so x 3 is a vertical asymptote. =
=
y= l
=
=
x
217
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3
Chapter 5: Polynomial and Rational Functions
45.
47.
49.
T(x) = x3 ; The degree of the numerator, X4 - 1 p(x) = x3 , is n = 3 . The degree of the denominator, q(x) = X4 - 1 is m = 4 . Since n < m , the line y = 0 is a horizontal asymptote. The denominator is zero at x = -1 and x = 1 , so x = -1 and x = 1 are vertical asymptotes.
51.
--
Q(x) =
G(x) = x3 - 1 ; The degree of the numerator, x - x2 p(x) = x3 - 1, is n = 3 . The degree of the denominator, q(x) = x - x2 is m = 2 . Since n = m + 1 , there is an oblique asymptote. Dividing: --
-x - 1
_x2 + X)X3 x3 _ x2
5 _ x2
; The degree of the numerator, 3x4 p(x) = 5 - x2 , is n = 2 . The degree of the denominator, q(x) = 3x4 is m = 4 . Since n < m , the line y = 0 is a horizontal asymptote. The denominator is zero at x = 0 , so x = 0 is a vertical asymptote. --
-1
x-I x - I = -x - 1 - -1 , x otol G(x) = -x - 1 + -x x - x2 Thus, the oblique asymptote is y -x - I . G(x) must be in lowest terms to find the vertical asymptote: G(x) = x3 - 1 = (X _ 1)(X2 + x + 1) = x2 + x + 1 -x(x - 1) -x x - x2 The denominator is zero at x = 0 , so x = 0 is a vertical asymptote. =
R(x) = 3x4 + 4 ; The degree of the numerator, x3 + 3x p(x) = 3x4 + 4, is n = 4 . The degree of the denominator, q(x) = x3 + 3x is m = 3 . Since n = m + 1 , there is an oblique asymptote. Dividing: 3x x3 + 3x)3x4 +4 3x4 + 9x2 +4 - 9x2 - 9x2 +4 -R(x) = 3x + x 3 + 3x Thus, the oblique asymptote is y = 3x . The denominator is zero at x = 0 , so x = 0 is a vertical asymptote. ---
53.
3.99 x l O'4 (6.374 x l 06 + h) 2 ' g ( O ) = 3.99 x l O 4 ::::: 9.8208 ml s2 (6.374 x l 06 + 0)2
g (h) = a.
b.
c.
d.
3.99 x 1014 (6.374 x 1 06 + 443)2 ::::: 9.8 195 mls2 3.99 x 101 4 g (8848) = (6.374 x 1 06 + 8848)2 ::::: 9.7936 mls2 ' g (h) = 3.99 x l O 4 (6.374 x l 06 + h) 2 g (443) =
3.99 x 1014 4 0 as h oo -» h2 Thus, the h-axis is the horizontal asymptote. �
218
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Section 5.3: The Graph of a Rational Function
e.
3 .99 x l 014 = 0 , to solve this g ( h ) - --------::(6.374 x 1 06 + h )
57.
2
equation would require that 3.99 x l 014 = 0 , which is impossible. Therefore, there is no height above sea level at which g = O. In other words, there is no point in the entire universe that is unaffected by the Earth's gravity! 55.
a.
59.
Rror = 1 0R2 1 0 + R2 RTOT
- -
�O
- - - - - - - - - - - - - -
Section 5.3
5
1. b.
Answers will vary. If x = 4 is a vertical asymptote, then x = 4 is a zero of the denominator. If x = 4 is a zero of a polynomial, then ( x - 4 ) is a factor of the polynomial. Therefore, if x = 4 is a vertical asymptote of a rational function, then ( x - 4) must be a factor of the denominator. A rational function cannot have both a horizontal and oblique asymptote. To have an oblique asymptote, the degree of the numerator must be exactly one larger than the degree of the denominator. However, if the numerator has a higher degree, there is no horizontal asymptote.
Horizontal asymptote: y = Rror = 10 As the value of R2 increases without bound, the total resistance approaches 10 ohms, the resistance of R) .
x-intercepts: Set the numerator equal to 0 and solve for x. x-I = 0 x=1 y-intercept: Set x equal to 0 and solve for y. 0 - 1 -1 1 y - 2 -- - -- -4 -0 - 4 -4 The intercepts are (1, 0) and 0, .
( �)
c.
3. Plot2
P l o t3
'Y 1 S 1 7 'Y 2 S2XA ( 3/2 ) / ( X+ .f ( X ) ) 'Y 3 = 'Y � = ,Y s = ,Y & =
5.
Xl"l i n= - H l Xl"lax= 1 50 Xsc l = 1 0 Yl"l i n= -5 Yl"lax=25 Ysc l =5 Xres= l
False; polynomials are always continuous. False; the graph of a rational function never crosses a vertical asymptote.
We would need R) "" 1 03 .5 ohms.
219
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Chapter 5: Polynomial and Rational Functions
R(x) = p(x) , where th e degree of p(x) = n q(x)
In pro blems 7-44. we will use th e term inology:
and the degree of
q(x) = m . 7.
R(x) = x + 1 p(x) = x + 1; q(x) = x(X + 4) = X2 + 4x; n = l; m = 2 x(x + 4) Step I :
Domain: {xl x *' - 4, x *' o} Since 0 is not in the domain, there is no y-intercept.
Step 2:
The function is in lowest terms. The x-intercept is the zero of p(x) : x = - 1 The x-intercept is -1 . Near -1 , R (x) "" -.!.( x + 1) . Plot the point (-1, 0) and show a line with 3 negative slope there.
Step 3 :
R(x) = x + 1 is in lowest terms. x(x + 4) The vertical asymptotes are the zeros of q(x) : x = - 4 and x = 0 . Plot these lines using dashes.
Step 4:
Since n < m , the line y = 0 is the horizontal asymptote. Solve R (x)= 0 to find intersection points: x+1 =0 x(x + 4) x+1 =0 x = -l R(x) intersects y = 0 at ( I 0). Plot the point (-1, 0) and the line y = 0 using dashes. -
,
Step 5: •
Interv al
Number Chosen
Val ue of R
( - 00, -5
-4
•
-4)
R( - 5 )
=
R( - 2)
Location of Graph B e l ow x - ax i s Poi nt on Graph
Steps 6 & 7 : Graphing
(
- 5, - � 5 y 2
)
•
( - 4, - 1 ) -2
-4 5
-1
=
�
Above x - axi s
(-2, n
( - 1 , 0) 1
-1
I{- I) = -��
o
•
(0, I
,
(0 )
R - '2
R( I )
Below x - axi s
Above x - ax i s
(-4, -D
=
�
(I, �)
x=o
1 (- 1, 0)
�I . I
(-5 _1 ) '
5
I
:
-2
220
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5. 3: The Graph of a Rational Function
9.
3x + 3 p(x) = 3x + 3; q(x) = 2x + 4; n = l; m = 1 R(x) = -2x + 4 Step I : Domain: {xl x * - 2} . . 3 (0) + 3 3 . 0, -3 . The y-mtercept R(O) = = - . Plot the pomt 2 (0) + 4 4 4
( )
IS
Step 2:
3x + 3 = 3 (x + l ) is in lowest terms. The x-intercept is the zero of p(x) , x = -1 . R(x) = -2x + 4 2 ( x + 2 ) Near - 1 , R (x) � (x + 1) . Plot the point ( - 1, 0) and show a line with positive slope there. 2 3x + 3 = 3 (x + l)) is in lowest terms. R(x) = -2x + 4 2 ( x + 2 The vertical asymptote is the zero of q(x) : x = - 2 . Graph this asymptote using a dashed line. ::::<
Step 3:
Step 4 :
Since n = m , the line y = � is the horizontal asymptote. 2 Solve to find intersection points: 3x + 3 3 2x + 4 2 2 (3x + 3) = 3 (2x + 4) 6x + 6 = 6x + 4 0*2 R(x) does not intersect y = � . Plot the line y = � with dashes. 2 2
Step 5: Interval
•
Number Chosen
( - x, -3
-
Val ue of R
R( - 3)
Poi nt on Graph
(-3, 3)
-2
•
2)
=
x =
•
( - 2, - 1 ) -� 2
( D
3
R -
Locati on of G raph Abov e x-axis
Steps 6 & 7 : Graphing:
-I
=
-�
Below x - axis
(-� -�) 2'
2
( - 1 , x)
0
R(O)
•
=�
Above x - axis
(0, n
-2 1 Y
<-yI 5
(O. �)
- - - - 1- --JZ:.;;;";;--�
x
-5
221
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Chapter 5: Polynomial and Rational Functions
11.
--
3 p(x) = 3; q(x) = x 2 -4; n = O; m = 2 R(x) = 3 = 2 ( x 2)( x + 2) x -4 Step I :
Domain: {xl x * - 2, x * 2}
( )
The y-intercept is R(O) = + = � = - � . Plot the point 0, - � . 4 -4 4 o -4 Step 2:
R is in lowest terms. The x-intercepts are the zeros of p(x) . Since p ( x ) is a constant, there are no x intercepts.
Step 3:
R(x) = -3- is in lowest terms. The vertical asymptotes are the zeros of q(x) : x = - 2 and x = 2 . x2 - 4 Graph each of these asymptotes using dashed lines.
Step 4:
Since n < m , the line y = 0 is the horizontal asymptote. Solve to find intersection points: _3_ = 0 x2 - 4 3 = 0 ( x2 - 4 ) 3*0 R(x) does not intersect y = 0 . Plot the line y = 0 with dashes.
Step 5 : Interval
· 41
N umber Chosen Value of R
Location of Graph Point on Graph
-2
•
( - co, - 2) -3
R (- 3 )
=
2
( - 2, 2)
°
5
3
•
R (O)
=
-4 3
Below x-axi s
Above x - axis
(- 3 , D
( 0, - D
�
(2, co)
3
R (3 )
=
�
Above x - axi s
(3, D
Steps 6 & 7 : Graphing: x =
x
�����r-�+-� y = 0
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.3: The Graph of a Rational Function
13.
4 2
P(x) = x �x + 1 p(x) = x -1 Step 1 :
X4 + X2 + 1; q(x) = x2 - 1; n = 4; m = 2
Domain: {xl x * - 1, x * I } The y-intercept is P(O) =
04 °�0-2 1+ 1 = - 1 = -1 . Plot the point ( 0, -1) . J...
4 2 2 P(x) = x 4 �x 2 + 1 is in lowest terms. The vertical asymptotes are the zeros of q(x) : x = -1 and x = 1 .
P(x) = x + x + 1 is in lowest terms. The x-intercept is the zero of p(x) . Since p ( x ) is never 0, x -1 there are no x-intercepts.
Step 2: Step 3 :
x -1 Graph each of these asymptotes using dashed lines.
Step 4 :
Since
n >
m + 1 , there is no horizontal or oblique asymptote.
Step 5:
-I
..
•
( - :0, - 1 )
Interv al N umber Chosen
•
( - 1. I )
2
-2
°
Val ue of P
P ( - 2) = 7
Location of G raph
Above x-axis
P (O) = - I
( - 2, 7)
Point on G raph
(l , e<;)
•
P(2)
=7
B elow x-axis
Abov e x - axi s
( 0, - I )
( 2, 7)
Steps 6 & 7 : Graphing:
\ iy
(-2, 7j-J!
I I
15.
x3 2
X =
I -1 1
U,7l I I
I Ix
X
=
1
x
3
2
( x - I) ( x 2 + + 1 ) H(x ) = --1 = p(x) = x - 1; q(x) = x -9; n = 3; m = 2 x + 3) x - 3) x -9 Step 1 :
_
( (
Domain: {xl x * - 3, x * 3} The y-intercept is H(O) =
0:
- 1 = 2 = .!. . Plot the point ° -9 -9 9
(0)).9
223
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Chapter 5: Polynomial and Rational Functions
Step 2 :
H(x) i s i n lowest terms. The x-intercept i s the zero o f p(x) : Near
1,
I.
H ( x ) "" -�( x - I ) . Plot the point (1, 0) and indicate a line with negative slope there. 8
Step 3:
H(x) is in lowest terms. The vertical asymptotes are the zeros of q(x) : x = -3 and x = 3 . Graph each of these asymptotes using dashed lines.
Step 4 :
Since n = m + 1 , there is an oblique asymptote. Dividing: x 9x -1 H(x) = x + x 2 - 9 x3 + Ox 2 + Ox - I x2 - 9 x3 - 9x 9x - 1 The oblique asymptote i s y = x . Graph this asymptote with a dashed line. Solve to find intersection points: x3 - I = X -x2 - 9 x3 - I = x3 - 9x - I = -9x
)
X = -1 9 The oblique asymptote intersects H(x) at Step 5 : N u mber Chosen
-4
H ( - 4)
Val ue of H
Location of Graph on
x = -3 1
Y I 1 10 1 1 r
=
H(O)
- 9.3
( -4,
-
'/ I '/ '/ '/ '/ r I '/ 1 /
/ 1 ( 1. 0)
I
I 1 ,I I I Ix
=
/
= \i
(0, �)
Y
I
Above x- axis
9.3)
II (S, 7.7S) · W
•
(- 3, 1)
0
Below x- axi s
G raph
Steps 6 & 7: Graphing:
•
( - <:N, -3)
Interval
Poi n t
-3
. �
(i, i)·
=
3
•
( 1 , 3) 2
H(2)
=
-
1 4 .
Below x-axis
(2, - 1 .4)
(3 . (0)
�
4
H(4)
=9
Above x-axis
(4, 9)
X
x
3
224
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Section 5.3: The Graph of a Rational Function
17.
x2 R(x) = --::-2(x + 3)(x - 2) X +x-6 Step I : Domain: {xl x "" - 3, x "" 2} The y-intercept is R(O) = Step 2:
Step 3: Step 4:
0 2 = = 0 . Plot the point (0, 0) . � -6 o2 + 0 - 6
x2 is in lowest terms. The x-intercept is the zero of p(x) : ° x2 + x - 6 Near 0, R (x) ;::: _.!. x 2 . Plot the point (0, 0) and indicate a parabola opening down there. 6 R(x) =
X2 is in lowest terms. The vertical asymptotes are the zeros of q(x) : 2 x +x-6 x = -3 and x = 2 . Graph each of these asymptotes using dashed lines. Since n = m , the line y = I is the horizontal asymptote. Graph this asymptote with a dashed line. Solve to find intersection points: x2 = 1 x2 + x - 6 x2 = x2 + x - 6 0 = x-6 x=6 R(x) intersects y = 1 at (6, 1 ).
R(x) =
----
Step 5: N umber Chosen
-6
R
Locati on of Graph
Poi nt on G raph
Steps 6 & 7 : Graphing:
•
( - ce. -3)
Interval
Val ue of
-3
..
I
( - 6jJ) :
R( - 6)
=
1 .5
( - 6, 1 .5)
2
y = l ----, I
( - 3, 0) -I
Above x-axis
Y
o
•
R(- I )
= -
.!. 6
Below x-axi s
(- l , -k)
(0, 2) I
R( I)
= - 0.25
Below x-axi s
(I,
-0.25)
2
•
(2. C/O)
3
R(3)
=
•
1.5
Above x - axis (3 . 1 .5)
I
�D_.I
(6, 1 )
X
X =
225
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Chapter 5: Polynomial and Rational Functions
19.
x x = --- p(x) = x; q(x) = x2 - 4; n = l; G(x) = -x 2 - 4 (x + 2)(x - 2) Step 1 :
Step 2 :
m
=2
Domain: {xl x ,c - 2, x ,c 2} The y-intercept is G(O) = -:}!-- = � = 0 . Plot the point (0, 0) . o -4 -4 G(x) = + is in lowest terms. The x-intercept is the zero of p(x) : ° x -4 Near 0 , G ( x ) "" - ..!. x . Plot the point (0, 0) and indicate a line with negative slope there. 4
Step 3 :
G( x) = + is in lowest terms. The vertical asymptotes are the zeros of q( x) : x = - 2 and x = 2 . x -4 Graph each of these asymptotes using a dashed line.
Step 4 :
Since n < m , the line y = ° is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: _x_ = O x2 - 4 x=O G(x) intersects y = ° at (0, 0).
Step 5 :
-2
�
Interval
( - co, - 2)
Val ue of G
G(-3)
N umber Chosen
-3
=
•
-�
Steps 6 & 7 : Graphing:
x =
(
-3,
( - 2, 0) -I
0( - 1 )
=
�
Above x-axis
Location of Graph Below x-ax'is Poi nt o n G raph
o
•
-D
( n -
1,
2
•
(0, 2) I
G( l )
3
0(3)
= -�
Below x-axis
(1, - D
( 2 , �)
•
= ;,�
Abov e x - ax i s
I
(3, �)
y 2
I I I I x = 2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher .
Section 5.3: The Graph of a Rational Function
21.
R(x) =
3 3 p(x) = 3; q(x) = (x - 1)(x 2 - 4); n = O; m = 3 (x - 1)(x 2 - 4) (x - 1)(x + 2)(x - 2)
Step 1 :
Domain: {xl x "# - 2, x "# 1, x "# 2} 3 = � Plot the point O, � . The y-intercept is R(O) = 4 (0 - 1)(0 2 - 4) 4
Step 2:
R(x) =
Step 3:
3 is in lowest terms. (x - l)(x 2 - 4) The vertical asymptotes are the zeros of q( x) : x = - 2, x = 1, and x = 2 . Graph each of these asymptotes using a dashed line.
Step 4:
Since n < m , the line y = 0 is the horizontal asymptote. Graph this asymptote with a dashed line. Solve to find intersection points: 3 =0 (x - l)(x 2 - 4) 3 "# 0 R(x) does not intersect y = 0 .
.
( )
3 is in lowest terms. There is no x-intercept. (x - l)(x 2 - 4)
R(x) =
Step 5 : In terv al
(-00, -2)
Value of R
R( -3)
Number Chosen Loc a ti on of
Point
on
-2 •
•
Graph
Graph
Steps 6 & 7 : Graphing: x =
-3
=
( -2. 1)
0
R(O)
-2'0
= (0
Above x-a xis
Below x-axis
(o· n
(-3, -�)
•
2 •
(1. 2) 3 '2
R(�)
=
-
24 T
Below x - axis
(f - �) 2'
7
(2, 00) 3
R(3)
•
3 = 10
A bove x - axis
(3 , tt )
-2I 1I 1II 1I (3- , --:24) II AI I
-4
2
7
227
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
23.
2 (x - I)(x + I) H(x) = x - 1 = p(x) = x 2 - I; q(x) = x4 - 1 6; n = 2; m = 4 X4 - 16 (x 2 + 4)(x + 2)(x - 2) Step 1 :
Domain: { x l x * - 2, x * 2 } The y-intercept is H(O) =
Step 2:
Step 3 : Step 4:
H(x) =
�2 _ 1 = 2 =
o
� . Plot the point (O,�)16 .
- 16 -16 16
�2 _ 1
is in lowest terms. The x-intercepts are the zeros of p(x) : -1 and 1 x -16 2 ( x + I ) ; Near 1 , H ( x ) "" -2 (x - I) Near -1 , H ( x ) "" 15 15 Plot ( -1, 0 ) and indicate a line with positive slope there. Plot ( 1, 0 ) and indicate a line with negative slope there.
is in lowest terms. The vertical asymptotes are the zeros of q(x) : x = - 2 and x = 2 x -16 Graph each of these asymptotes using a dashed line. H(x) =
�2 - 1
Since n < m , the line y = 0 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: x2 - 1 = 0 X4 - 1 6 x2 - 1 0 x = ±1 H(x) intersects y = 0 at (-1 , 0) and (1, 0). --
=
Step 5: •
-2
•
-1
•
(-2, - 1)
Interval
(-00, - 2)
Value of H
H(-3) "'" 0.12
H( - 1.5)
Point
(-3. 0.1 2)
(- 1 .5. -0.11 )
Number Chosen
-3
Location of Graph Above x-axi s on
Graph
Steps 6 & 7: Graphing:
- 1. 5
""
- 0 .1 1
Below x-axis
(- 1. 1)
0
H(O)
= ft,
Above x-axis
(0, t\)
•
2
•
( 1 , 2) 1.5
H(1.5) "'"
-
0. 1 1
B elow x-axis
(1.5,
-
0. 1 1
)
•
(2. 00)
3
H(3) "'" 0. 1 2
Above x-a xis
(3, 0.12)
y
x=2 ,
228
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Section 5.3: The Graph of a Rational Function
25.
2 F(x) = x - 3x - 4 (x + 1)(x - 4) p(x) = x 2 - 3x - 4; q(x) = x + 2; n = 2; m = l x+2 x+2 Step
1:
Domain: {xl x � - 2}
2 - 4 = - 2 . Plot the point ( 0, -2) . The y-intercept i s F( O) = 0 - 3(0) - 4 = 2 0+2 Step 2:
2 . are the zeros f p ( x ) : -I and 4 . F ( x ) = X - 3x - 4 . . Iowest terms. The x-mtercepts x+2 Near -1 , F (x) '" -5 (x + I) ; Near 4, F (x) � (x - 4) . IS m
0
'"
6
(- 1 , 0)
Plot and indicate a line with negative slope there. Plot ( 4, 0) and indicate a line with positive slope there. Step 3: Step 4:
2 F(x) = x - 3x - 4 is in lowest terms. The vertical asymptote is the zero of q(x) : x = - 2 x+2 Graph this asymptote using a dashed line.
,
Since n = m + 1 there is an oblique asymptote. Dividing: x-5 6 2 F(x) = x - 5 + - X + 2 X - 3x - 4 x+2 x 2 + 2x - 5x - 4 - 5x - 1 0
)
6
The oblique asymptote is y = x - 5 . Graph this asymptote using a dashed line. Solve to fmd intersection points: x 2 - 3x - 4 = x - 5 x+2 x 2 - 3x - 4 x 2 - 3x - 1 0 -4 � - 1 0 The oblique asymptote does not intersect F(x) . ----
=
Step 5: Interval
Number
-2
•
Chosen
Val ue of F
•
( - co, -2)
on
Graph
F( - 3) = - 1 4 ( - 3, - 1 4)
•
( - 2, - I ) - 1. 5
-3
Locati on of Graph Below x- axi s Poi nt
- .I
F( - 1 .5)
=
5.5
Above x - axis
( - 1 .5, 5.5)
4
•
( - 1 , 4)
0
F(O)
=
Bel ow x - axi s
(0, -2)
(4, co)
5
-2
•
F(5)
=
0.86
Above x - axis
(5, 0.86)
229
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Chapter 5: Polynomial and Rational Functions
Steps 6 & 7: Graphing:
y =x-s / /
x
1 1 1/
//
A / 1
1
rAl
1 1 -16 (-3, -14) 1 x = -2
,,: /.
2 7.
/
2 R(X) = X + X - 12 = (X + 4)(X - 3) p(x) = x 2 + x - 12; q(x) = x - 4; n = 2; m = l x-4 x-4 Step 1 :
Domain: {xl x * 4} 2 The y-intercept is R(O) = 0 + 0 - 12 = -12 = 3 . Plot the point (0, 3) . 0-4 -4
Step 2:
2 R(x) = x + x - 1 2 is in lowest terms. The x-intercepts are the zeros of p(x) : x-4 7 + 4) ; Near 3, R (x) "" -7 (x - 3) . Near -4 , R (x) "" -(x
-4
and 3 .
8
Plot (-4, 0) and indicate a line with positive slope there. Plot (3, 0) and indicate a line with negative slope there. Step 3: Step 4:
2 R(x) = x + x - 1 2 is in lowest terms. The vertical asymptote is the zero of q(x) : x = 4 x-4 Graph this asymptote using a dashed line. Since n = m + 1 , there is an oblique asymptote. Dividing: x+5 8 R(x) = x + 5 + - X _ 4 X2 + x - 1 2 x-4 x 2 - 4x 5x - 1 2 5x - 20
)
8
The oblique asymptote is y = x + 5 . Graph this asymptote using a dashed line. Solve to find intersection points: x 2 + x - 12 = x + 5 x-4 2 x + x - 1 2 = x 2 + X - 20 -12 * - 20 The oblique asymptote does not intersect R(x) .
----
23 0
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.3: The Graph of a Rational Function
Step 5 : Interval
Number Chosen Value of
-4
•
R
•
( - 00, -4) -5
R( - 5)
0
R(O)
= -�
Location of Graph Bel ow x - axis
Poi nt on G raph
Steps 6 & 7 : Graphing: y 18
3
(0, 3)
'} �
4
•
(3. 4)
3.5
R(3.5)
=
(4, 'Xl) 5
R(5)
- 7.5
=
•
\8
Above x - axis
B e l ow x-axis
( 3. 5 , - 7.5)
(5. 1 8)
x=4 1 t; 5, 18)/; / / .v = x + 5 I
I I I 1 /
/(
1
29.
-
=
•
Above x - axis
(-<; ) - ,
(-
3
) 4, 3
/
/
(0, 3) x
2 F(x) = x + x - 12 = (x + 4)(X - 3) p(x) = x 2 + x - 12; q(x) = x + 2; n = 2; m = l x+2 x+2 Step 1 :
Domain: {xl x * - 2} 2 -12 = - 6 . Plot the point (0, -6) . The y-intercept is F(O) = 0 + 0 - 12 = 2 0+2
Step 2:
Step 3:
2 . F( x) = X + x - 1 2 . m. Iowest terms. The x-mtercepts are the zeros f p ( x ) : x+2 7 7 Near -4 , F (x) � -(x + 4) ; Near 3, F (x) � -(x - 3) . 2 5 Plot (-4, 0) and indicate a line with positive slope there. Plot (3, 0) and indicate a line with positive slope there. 1S
0
-4
and 3 .
2 F(x) = x + x - 1 2 is in lowest terms. The vertical asymptote is the zero of q(x) : x x+2 Graph this asymptote using a dashed line.
=
-2
23 1
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
Step 4:
Since n = m + 1 , there is an oblique asymptote. Dividing: x-I -10 2 F(x) = x - l + - + 2 X X + x - 12 x +2 x 2 + 2x - x - 12 -x- 2 -10 The oblique asymptote is y = x - I . Graph this asymptote using a dashed line. Solve to find intersection points: x 2 + x - 12 = x - I x+2 2 x + x - 1 2 = x2 + X - 2 -12 :;t - 2 The oblique asymptote does not intersect F(x) .
)
Step 5:
-4
•
Interval
( - 7.:, -4)
Value of F
F ( - 5)
Poi nt on G raph
(-5, -D
N umber Chosen
-5
=
•
( - 4, - 2 )
-2
•
-3
F ( -3) = 6
8
-3
Location of Graph Below x-axis
Above x-axis
( - 3, 6)
( - 2, 3) 0
F (O)
= -6
Below x- axi s
(0, - 6)
Steps 6 & 7: Graphing:
3
•
•
(3, 00)
4
F (4)
4
= 3
Above x-axis
T4, IT
x
31.
p(x) = x(x - l) 2 ; q(x) = (X + 3)3 ; n = 3; m = 3
R(x) Step 1 :
Domain: {xl x :;t -3}
:
The y-intercept is R(O) = 0(0 - 1) = � = O . Plot the point ( 0, 0 ) . (0 + 3) 27
232
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Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.3: The Graph of a Rational Function
Step
2:
2
R(x)
= X(X - 1 ) (x + 3) 3
Near
0, R ( x )
",, _1
is in lowest terms . The x-intercepts are the zeros of
p(X) : ° and 1
1 , R ( x ) "" � ( x - l ) 2 . 64
x ; Near
27 (0, 0) and indicate a line with positive slope there . Plot (1, 0) and indicate a parabola that opens up there . Plot
Step
3:
2
R(x)
= x(x _ 1 ) (x + 3)3
is in lowest terms. The vertical asymptote is the zero of q(x) : x = - 3
1
Graph this asymptote with a dashed line. Step 4:
Since
, the line
n = m
y =
is the horizontal asymptote. Graph this asymptote with a dashed line.
Solve to find intersection points : x(x _ 1) 2
=1 (X + 3)3 x 3 - 2 x 2 + X = x 3 + 9x 2 + 27 x + 27 0 = l 1x 2 + 26x + 27 b 2 - 4 ac = 26 2 - 4 (1 1)(27) = -5 1 2
Step 5 :
no real solution R(x) does not intersect y = 1
. -3
•
•
( - ce, -3)
I nterval
N umber Chosen
-4
( - 3. 0 ) -I
R(- I) =
Point on G raph
( - 4, 1 00)
( - I , - 0.5)
Graphing:
lO
x = -3 I Y
i I
)
-0.5 R(D
Below x - axis
y
(0. 1 ) 2
R ( -4) = 100
6 & 7:
•
I
Val ue of R
Location o f Graph Abov e x - axis
Steps
o
•
=
0.003
Above x - axis
0, 0.003)
( I , cc)
2
R (2)
=
•
0.016
A bove x-axis
(2, 0.0 1 6)
0.01
See enlarged
view at right. -x
x
10
Enlarged view
233
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
33.
2 R(x) = X + x - 12 = (x + 4)(x - 3) = x + 4 p(x) = X2 + x - 12; q(x) = X 2 - x - 6; X 2 - x - 6 (x - 3)(x + 2) x + 2 Step 1 : Domain: {xl x -cf. - 2, x -cf. 3} 2 -12 = 2 . Plot the point (0, 2) . The y-intercept is R(O) = 0 + 0 - 12 = -6 o2 - 0 - 6 Step 2:
n
= 2;
m
=2
In lowest terms, R(x) = x + 4 , x -cf. 3 . Note: R (x) is still undefined at both 3 and -2 . x+2 The x-intercept is the zero of y = x + 4 : -4 Near -4 , R (x) ,., -.!. ( x + 4) . Plot (-4, 0) and indicate a line with negative slope there. 2
Step 3:
In lowest terms, R(x) = x + 4 , x -cf. 3 . The vertical asymptote is the zero of f (x) = x + 2 : x = - 2 ; x+2 Graph this asymptote using a dashed line. Note: x = 3 is not a vertical asymptote because the reduced form must be used to find the asymptotes. The graph has a hole at 3,
( �) .
Step 4:
Since n = m , the line y = 1 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: x 2 + x - 12 = 1 x2 - x - 6 x2 + x - 1 2 = x2 - x - 6 2x = 6 x=3 R(x) does not intersect y = 1 because R(x) is not defined at x = 3 .
Step 5: Interval
Number Chosen
Val ue
of R
•
( - :c, -4) -5
R ( - 5)
=
5
Location of G raph Above x-axi s Poi nt on G raph
(-S, H
-4
•
( -4, - 2 ) -3
R( -3)
= -1
Below x-axis
( - 3 , - 1)
-2
•
3
•
( - 2 , 3)
0
R(O)
=
2
Above x-axi s
(0, 2)
•
(3, cc)
4
R(4)
4
= :3
Above x-axis
(4, �)
Steps 6 & 7 : Graphing:
234
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Section 5.3: Th e Graph of a Rational Function
35.
2 R(x) = 6x - 7x - 3 = (3X + l)(2X - 3) = 3x + l p(x) = 6x 2 - 7x - 3; 2x 2 - 7x + 6 (2x - 3)(x - 2) x - 2 Step
1:
Domain:
{x I X # % , X # 2}
The y-intercept is
Step
2:
Near
Step
3:
2 R(O) = 6(0) - 7(0) - 3 = -3 = _ .!. . 2 2(0) - 7(0) + 6 6 2
In lowest terms, R (x) Note :
q(X) = 2X 2 _ 7x + 6; n = 2; m = 2
3x + 1 , x # -3 . = -2 x-2
Plo t the point
( 0, ) 1 -2
The x-intercept is the zero of y
.
= 3x + 1 : - -1 ; 3
x = � is not a zero because reduced form must be used to find the zeros. 2 - R (x) 3x + 1 ) . P lot the point 0 and indicate a line with negative slope there.
±,
�
In lowest terms,
( -±, )
-�(
R(x) = 3x + 1 , x # � . 2 x-2
The vertical asymptote is the zero of f (x) = x - 2 :
x
=2;
Graph this asymptote using a dashed line . Note :
x=� 2
is not a vertical asymptote because reduced form must be used to fmd the asymptotes.
The graph has a hole at Step 4:
Since
(%' ) -1 1
n = m , the line y = 3
.
is the horizontal asymptote. Graph this asymptote using a dashed line.
Solve to find intersection points :
6x 2 - 7x - 3 = 3 -:--2X 2 - 7x + 6 6x 2 - 7 x - 3 = 6x 2 - 2 1x + 1 8 2 1 4x = 1 X = -3 2 R(x) does not intersect y = 3 Step 5 : Interval
Number Chosen Val ue
of R
.
. (-x. - 5) -I
R(- I )
=� .3
Location or G raph Abov e x-axis
Point on G raph
(
-
I,
n
because
R(x)
l
- 3'
•
is not defined at
x=�. 2
3
(- ! , �) 0
3
R(O)
2
=
2
2"
•
l
-2
Below x-axis
( 0, - � )
•
G, 2) 1 .7
R( 1 .7)
=
- 20.3
Below x-axis
( 1.7,
- 20.3)
•
(2, w)
6
R(6)
=
4.75
Above x - axis
(6, 4.75)
235
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Chapter 5: Polynomial and Rational Functions
Steps
6 & 7:
Graphing:
2 Yx =:1I �
8
3 ,- - )1 -= --
- 10
(- j , oy
(O, -n
-12
3 7.
(6, 4', 75 )
1----X
I I I I
I G · -11) I
2 R(x) = x + 5x + 6 = (x + 2)(x + 3) = x + 2 x+3 x+3 Step
1:
Domain:
{xl x *' -3}
The y-intercept is R(O) = Step
2:
p(x) = x 2 + 5x + 6; q(x) = x + 3; n = 2; m = 1
In lowest terms, R(x)
0 2 + 5(0) + 6 = � = 2 . 3 0+3
= x + 2, x *' -3 .
P lot the point
(0, 2) .
The x-intercept is the zero of y =
x + 2 : -2;
Note: -3 i s not a zero because reduced form must b e used t o find the zeros. Near -2 , R (x) = x + 2 . Plot the point (0, -2) and indicate the line y = x + 2 there. Step
3:
In lowest terms,
R(x) = x + 2, x *' -3 .
There are no vertical asymptotes . Note :
x = -3
is not a vertical
asymptote because reduced form must be used to find the asymptotes. The graph has a hole at
( -3, -1) . Step
4:
Since
n = m+1
there is an oblique asymptote. The line y
= x+2
is the oblique asymptote. Solve to
find intersection points :
x 2 + 5x + 6 = x + 2 x+3 x 2 + 5x + 6 = (x + 2){x + 3) x 2 + 5x + 6 = x 2 + 5x + 6 0=0 The oblique asymptote intersects
© 2008 Pearson
R(x)
at every point of the form
(x, x + 2)
except (-3, - 1).
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.3: The Graph of a Rational Function
Step
5:
-4
Nu mber Chosen
of R
R( -4)
Graph
Steps 6 & 7 : Graphing :
=
( - 4,
( 3, -
-2
Below x - axis
Location of Graph Poi nt o n
•
( - 00, - 3 )
Interval
Val ue
-3
•
-2)
- 2.5
-
-2
•
2)
R( - 2.5)
(- 2,
et.:
°
=
-�
B e l ow x - axis
(- 2 .5, - D
R(O)
=
)
�
2
Above x- ax i s
(0, 2 )
y 5
x
(-3, - 1 ) 1 x2 + 1 39. f(x) = x + - = -x x Step 1 :
Domain:
-5
2 p(x) = x + 1; q(x) = x; n = 2; m = 1
{ x l x ;t O }
There is no y-intercept because Step 2 :
f(x) =
x2 + 1 x
0 is not in the domain.
is in lowest terms. There are no x-intercepts since x 2 + 1 =
0
has no real solutions.
Step
3:
X2 + 1 . . f( x ) = -- IS 111 I owest terms. The vertical asymptote is the zero of q(x) : x = x asymptote using a dashed line.
Step
4:
Since n = m + 1 , there is an oblique asymptote. Dividing:
0
Graph this
1
f(x) = x + x The oblique asymptote is y=x.
Graph this asymptote using a dashed line. Solve to find intersection points : x2 + 1 -- = X X 2 x + 1 = x2 h ': O The oblique asymptote does not intersect f(x) .
237
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Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
Step 5 :
o
Interval
(-00, 0)
Value ofl
/(- 1 ) = -2
Location of Graph
B elow x-axis
Ntunber Chosen
41.
(0, 1
-1
Point 011 Graph
Steps 6 & 7: Graphing :
•
)
(0
1(1 ) = 2
Above x-axis
(-1, -2)
(1 , 2)
x = o y '5
x3 + 1 = ( x + 1) ( x 2 - X + 1 ) p(x) = x3 + 1; q(x) = x; n = 3; m = I(x) = x 2 + -1 = -l. x x x Step 1 :
Domain: {xl x * o} There is no y-intercept because 0 is not in the domain.
Step 2 :
I(x) = x3 + 1 is in lowest terms. The x-intercept is the zero of p(x) : -1 x Near -1 , I (x) "" -3 (x + 1) . Plot the point (-1, 0) and indicate a line with negative slope there.
Step 3 :
I(x) = x3 + 1 is in lowest terms. The vertical asymptote is the zero of q(x) : x = 0 x Graph this asymptote using a dashed line.
Step 4:
Since n > m + 1 , there is no horizontal or oblique asymptote.
238
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All rights reserved . This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.3: The Graph of a Rational Function
Step 5 : Interval
Number C hosen Val u e
-1
. .
oft
Location of Graph Poi nt on Gra p h
•
( -00, - ] ) -2
t( -2)
=
o
( - 1 , 0) -
3.5
•
1 2
f( -U
=
1
- 1.75
B e l ow x - a xis
A b ove x- axis
(-i, - L75)
( - 2, 3.5 )
•
(0, 00)
f(l)
=
2
Above x- axis
(1 , 2 )
Steps 6 & 7 : Graphing:
x=O
x
-5
1 X4 + 1 43. I ( x ) = x + - = -x3 x3 Step 1 : Step 2 : Step
3:
Step 4:
{ x l x "" o}
Domain:
There is no y-intercept because 0 is not in the domain. x4 + 1 . . 4 I ( x ) = -- IS m lowest terms. There are no x-intercepts since X + 1 = x3 X4 + 1 . I ( x ) = --- IS m · I asymptote . I owest terms. The verhca x3 Graph this asymptote using a dashed line. Since
n
=
m
IS .
0
has no real solutions.
the zero 0 f q ( x ) : x = 0
+ 1 , there is an oblique asymptote. Dividing: 1
I(x) = x + ) x
The oblique asymptote is y = x. Graph this asymptote using a dashed line. Solve to find intersection points: X4 + 1 --
=X x3 4 X4 + 1 = X
1 "" 0 The oblique asymptote does not intersect I(x) . 239
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
Step 5 :
Interval Nwnber Chosen Value off
(-00, 0) -1 j(- 1 ) = -2
Point on Graph
(- 1 , -2)
7:
Graphing:
x = o y 3
H /l -3
/
/
/
(0, 00) 1 f(1) = 2
Above x-axis
( 1 , 2)
v = x
�
/
�
•
B elow x-axis
Location of Graph
Steps 6 &
o
•
/
/
x
-3
45. One possibility:
47.
49.
51.
x2 R(x) = --2 x -4
a.
( x - I ) ( x - 3) ( x2 + a ) One possibility: R(x) - ----,------: -'- -:- -'(x + l )2 (x - 2)2 (Using the point (0, 1) leads to a = 4 / 3 . ) a.
b.
A = xy 1 000 = xy 1 000 y= x If the length of a perpendicular side is x feet,
The degree of the numerator is I and the degree of the denominator is 2. Thus, the horizontal asymptote is y = 0 . The concentration of the drug decreases to as time increases.
the length of the parallel side is
y = 1 000 x
feet. Thus,
0
C (x) = 2 . 8 . x + 5 .
Graphing:
= 1 6x +
0.4 b. ,=======1 1 2
c.
The cost of the proj ect is the sum of the cost for the parallel side, the two other sides, and the posts.
l OOO + 4 (25) x
5000 + 1 00 x
The domain is x > O . Note that x is a length so it cannot be negative. In addition, if x = 0 , there is no rectangle (that is, the area is 0 square feet) .
Using MAXIMUM, the concentration is highest when t ::::: 0 . 7 1 hours.
240
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[nc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
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Section 5. 3: Th e Graph of a Rational Function
c.
d. The surface area is a minimum when
5000 c(x) = 1 6x + -- + 1 00 x
X "" 2 1 .54 inches . 1 0' 000 . "" 2 1 .54 mches y= (2 1 .544 ) 2
W I NDOW Xl"l i n=0 Xl"la.x=300 Xscl =S0 YM i n=0 Yl"lax= 1 0000 Ysc l = 1 000 Xres= l
The dimensions of the box are: 2 1 .54 in. by 2 1 .54 in. e.
d. Using MINIMUM, the dimensions of cheapest cost are about 1 7.7 feet by feet (longer side parallel to river) .
56.6 55.
IliPiI'\UI'\
�
»<>1 7 . 6 7 7 6 6 8 . V = 6 6 S . 6 8 S � 2
a.
Answers will vary. One possibility is to save costs or reduce weight by minimizing the material needed to construct the box.
500 h=n r2 qr) = 6(2n r2 ) + 4(2n rh)
500 = n r 2 h
•
2 53 Note : x = 1 7 - = - feet 3 3 1 000 3000 feet . = Y = 53 / 3 53 53.
a.
=>
= 1 2n r2 + 8n r and
= 1 2n r2 + b.
The surface area is the sum of the areas of the six sides.
2 1 .54 in. by
( ��� J
4000 r
Graphing:
S = xy + xy + xy + xy + x2 + x2 = 4xy + 2x2 The volume is
x . X · Y = x 2 Y = 1 0, 000 Thus,
S(X) = 4X
=>
-
O �-===���� I O
1 0, 000 y = -x2
( 1O��00 ) + 2X2
o
Using MINIMUM, the cost is least for r "" 3 .76 cm.
40, 000 x 2x3 + 40, 000 x
= 2x2 +
b.
57.
x2 - 1 - x-I
y
Graphing : ......,...--
0t.-======.I 60
r-------
."..,...-
x
x3 - 1 y = - x-I
o
c.
y=
Using MINIMUM, the minimum surface area (amount of cardboard) is about 2785 square inches . x
241
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
Section 5.4
X4 - 1 y = - x-I
1.
3 - 4x > 5 -4x > 2 x < --1 2 1 {x l x < - -}
x
2
x5 - 1 y = - x-I
) 1
3. x
x 1 is not a vertical asymptote because of the =
following behavior: When x itd :
1,
"hole" with coordinates
-64
50
8
Negative
Positive
Positive
{ x l x < -2 } , or, using (-00, -2) .
x3 - 4x 2 > 0 x 2 (x - 4) > 0 f(x) = x3 - 4x 2 = x 2 (x - 4) x = 0, x = 4 are the zeros of f . Interval (-00, 0) (0, 4) Number
5
Value off
-5
-3
25
Conclusion
Negative
Negative
Positive
notation,
5 9 . Answers will vary.
)
00
1
The solution set is
(1, n ) .
(4,
-1
Chosen
will have a
6
Value off
interval notation, 5.
0
(5, 00)
Conclusion
The solution set is
In general, the graph of
. an rnteger,
-3
Chosen
x 3 - 1 = (x - 1) ( x 2 + x + I ) = X 2 + x + I y = -x-I x-I ( x 2 + 1 )( x 2 - 1 ) x 4 - 1 = -'---'-'-----'y = -x-I x-I ( X2 + I ) (x - I)(x + I) x-I 3 2 =x +X +x+1 ( x 4 + X3 + X 2 + x + I ) (x - I) x 5 - 1 = -'--' ----------'-y = -x-I x-I = x 4 + x3 + x 2 + X + 1 n ;;::
(x - 5f (x + 2) < 0 f(x) = (x - 5) 2 (X + 2) x = 5, x = - 2 are the zeros of f . Interval (-00, - 2) (-2, 5) Number
x 2 - 1 = (x + I) (x - I) = x + I y = -x-I x-I
xn - 1 y = -- , x-I
2
{ xl x > 4 } , or, using interval
(4, 00 ) .
61. Answers will vary, one example is
R (x) =
2 (x - 3) (x + 2) 2 (x _ 1) 3
2 42
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All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any fonn or by any means, without pennission in writing from the publisher.
Section 5.4: Polynomial and Rational Inequalities
7.
13.
x3 -9x S 0 x(x-3)(x+3)sO f(x) x3 -9x x -3, x O,x 3 are the zeros of f .
=
=
=
=
=
Interval Number
(-3,0)
(0, 3)
-1
I
4
Value off
-28
8
-8
28
Conclusion
Negative
Positive
Negative
Positive
15.
(3, 00)
°
1.5
2.5
4
Value off
-6
0.375
-0.375
6
Conclusion
Negative
Positive
Negative
Positive
x3 _2X2 -3x>0 x ( x2 -2x-3 ) >0 x(x+1)(x-3)>0 f(x) x3 -2x2 -3x x -1, x 0, x 3 are the zeros of f . =
=
=
(-«>, - 1)
(-1,0)
(0,3)
(3, 00)
-2
-0.5
I
4
Value off
-1O
0.875
-4
20
Conclusion
Negative
Positive
Negative
Positive
Interval
=
Number Chosen
The solution set is { x l -1 < x < 0 or x>3 }, or, using interval notation, ( -1,0) or (3,00) .
17.
(x-1) ( x2+x+ 4) �0 f ( x) ( x -1) ( X2 + X + 4) x 1 is the zero of f. x2 + X + 4 0 has no real solution. Interval (-00, 1) (1, 00) Number 0 2 Chosen Value off -4 10 Conclusion Negative Positive The solution set is { x l x �1 } , or, using interval notation, [1, (0 ) .
X2(X2 -1) > 0 x2(x-1)(x+ 1)>0 f(x) x2 (x -1)(x + 1) x -1, x 0, x 1 are the zeros of f
=
=
(2, 3)
=
=
11.
(1,2)
The solution set is { x l xs1 or 2s xs3 }, or, using interval notation, ( -00,1] or [2,3] .
2x3> -8x2 2x3 +8x2>0 2X2 ( x+ 4) >0 f( x) 2x3 +8x2 X O,x -4 are the zeros off Interval (-00, -4) (-4,0) (0, 00) Number -5 1 -1 Chosen Value off -50 6 10 Conclusion Negative Positive Positive The solution set is { x 1-4 < x < 0 or x>O } , or, using interval notation, ( -4,0) or (0,00) . =
(-00, 1)
Chosen
The solution set is { x 1 xs -3 or 0s xs3 } , or, using interval notation, ( -00, -3] or [0,3 ] . 9.
=
Number
(3, 00)
-4
Chosen
=
Interval
=
(-«>,-3)
(x-1)(x-2)(x-3)s 0 f(x) (x -1)(x -2)(x -3) x 1, x 2, x 3 are the zeros of f .
=
=
=
Interval Number
=
=
(-00, -1)
(-1,0)
-2
-0.5
(0,
I)
0.5
(I, 00) 2
Chosen Value off
12
-0.1875
-0.1875
12
Conclusion
Positive
Negative
Negative
Positive
The solution set is { x I x < -1 or x>1 } , or, using interval notation, ( -00, -1) or (1,00) .
243
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Chapter 5: Polynomial and Rational Functions
1 9.
X 4>1 X4 -1>0 (x2 -1)(x2 + 1)>0 (x -l)(x + 1)(x2 + 1)>0 f(x) (x -l)(x + 1)(x2 + 1) x 1, x -1 are the zeros of f ; x2 + 1 has no real solution
23.
=
=
=
=
(
-
Number
<Xl,
-1)
-2
(-1,1) 0
(I,
<Xl
2
Chosen 15
-I
15
Conclusion
Positive
Negative
Positive
25.
x+1 >0 x-I f(x) x+l x-I The zeros and values where f is undefmed are x -1 and x 1 . Interval (-00, -1) (-1,1) (1, ) Number 0 2 -2 Chosen 1 Value off -1 3 3 Conclusion Positive Negative Positive The solution set is { x I x < -l or x>1 } , or, using interval notation, (-00, -1) or (1, 00) .
00 )
(0, I)
-2
-0.5
0.5
Chosen Value off
-1.5
1.5
-1.5
1.5
Conclusion
Negative
Positive
Negative
Positive
2
(x-2)2 � 0 x2 -1 (X_2)2 � 0 (x + l)(x -1) f(x) (x-2i x2 - 1 The zeros and values where f is undefmed are x -1, x 1 and x 2 .
=
=
(I,
(-1,0)
The solution set is { xlx:-::; -1 or O < x:-::;l } , or, using interval notation, (-00, -1] or (0,1] .
The solution set is { x I x < -l or x>1 } , or, using interval notation, (-00, -1) or (1, 00) . 21.
=
( -<Xl, - I)
Number
)
Value ofl
=
Interval
=
Interval
(x-1)(x+1) :-::; 0 x + 1) f(x) (x -1)(x x The zeros and values where f is undefmed are x -1, x 0 and x 1 .
=
=
=
(0
=
=
(-00, -I)
(- \, I)
(\, 2)
(2, 00)
-2
0
1.5
3
Valueofl
\6 -
-4
0.2
0.\25
Conclusion
Positive
Negative
Positive
Positive
Interval Number Chosen
3
The solution set is { x I x < -l or x>1 } , or, using interval notation, (-00, -1) or (1, 00) .
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.4: Polynomial and Rational Inequalities
6x-5<6x 6 0 6x-5--< x 6x2 -5x-6 ----< 0 x (2x-3)(3x+2)< 0 x 2) I(x) (2x-3)(3x+ x The zeros and values where I is undefined are x --2 ' x 0 and x 23 . 3
27.
31.
=
=
=
=
=
= -
( -�) (-�, 0) (0,%) (%,00)
Interval
-00,
Number
-1
-O . S
1
2
Valueoff
-s
4
-s
4
Conclusion
Negative
Positive
Negative
Positive
Chosen
=
1 -4.5 Value ofl 12 12 Conclusion Positive Negative Positive The solution set is { x 1-2< x�9 } , or, using interval notation, (-2,9].
{I
}, or, using interval notation, ( -00,-%) or (0,%) .
The solution set is x x< - % or 0< x< %
29.
3x-5�2 -x+2 3x-5 -2� 0 x+2 3x-5-2(x+2)�O x+2 x-9 --�O x+2 x-9 I(x) x+2 The zeros and values where I is undefined are x -2 and x 9 . Interval (-00, -2) (-2,9) (9,00) Number 10 0 -3 Chosen
33.
x+4 1 --< x-2 x+4 _1�0 x-2 x+ 4- (x-2)�O x-2 6 �O -x -2 6_ I(x) _ x-2 The value where I is undefined is x 2 . Interval (-00,2) (2,00) Number 0 3 Chosen Value ofl -3 6 Conclusion Negative Positive The solution set is { x 1 x< 2 } , or, using interval notation, (-00,2) .
2 1 --< x-2 3x-9 1 2 -- ---<0 x-2 3x-9 _3x_- 9_ -_ 2_ -'.(x__- 2-"-)< 0 (x-2)(3x-9) x-5 ----< (x-2)(3x-9) - 0 x-5 I(x) (x -2)(3x -9) The zeros and values where I is undefined are x 2, x 3, and x 5 . =
=
=
=
Interval
=
=
(-eX),2)
(2,3)
( 3,5)
(5, eX» )
0
2 .5
4
6
Number Chosen Valueoff
5 -18
10 3
1 -6
1 -
Conclusion
Negative
Positive
Negative
Positive
36
The solution set is { x 1x< 2 or 3< x< 5 } , or, using interval notation, (-00,2) or (3,5) . 245
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
35.
2x+5 x+l -->-x+l x-I 2x+5 x+l >0 x+ 1 x-I (2x+5)(x-l) - (x+l)(x+l) >0 (x + 1)(x -1) 2x2 +3X-5- ( X2 +2x+l) ----------�------� >O (x + 1)(x -1) x2 + x-6 >0 (x + 1)(x -1) (x+3)(x-2) >0 (x+l)(x-l) f(x) (x+3)(x-2) (x + 1)(x -1) The zeros and values where f is undefined are x -3, x -1, x 1 and x 2 .
Interval
_
=
=
J
Valueof Chosen
Conclusion
(-00, - 3)
-4
0.4
Positive
(-3,-1)
-2
--
Negative
( -1,I)
0
6
Positive
4
3
1.5
(1,2 ) ( 2, 00)
3
-1.8
0.75
39.
=
=
-4.5
(-4, - 3)
-3.5
(-3,0)
-I
(0, I)
0.5
(1,00)
2
Positive
-
Negative
49 108
Positive
-
-0.75
Negative
- 63 44 120
Negative
-
Positive
-
7
,
(3 _ X)3(2x+1)< 0 x3 -1 3(2x+1) (3_x) -7-- -....!.< ..,- 0 ( -1) ( X2 + X + 1) 3 f(x) (3 _x) (2x+1) (x -1) ( X2 + X + 1) The zeros and values where f is undefined are x 3,x - 21 and x 1 . =
Negative Positive
=
=
Interval
-
=
,
Number
Valueoff
Conclusion
Chosen
=
=
(-5,-4)
216 --7 243 ---44
X
x2(3 + x)(x+4) >0 (x + 5)(x -1) _ (3 + x)(x + 4) f(x) x2(x+5)(x-1) The zeros and values where f is undefined are x -5, x -4, x -3, x 0 and x 1 . =
-6
Conclusion
--'- -'----
The solution set is { xlx< -3 or -1<x
2 } , or, using interval notation, (-00,-3) or (-1,1) or (2,00). 37.
(-00, -5)
Value of f
=
=
Number Interval
Chosen
The solution set is { x I x< -5 or -4 � x � -3 or x 0 or x> 1 } or, using interval notation, (-00,-5) or [-4,-3] or O or (1,00).
=
=
Number
( -00, -k) (-k,l)
-I
32
Positive
0
-27
Negative
4
5/7
Positive
(3,00)
2
(I, 3)
=
-1/7
Negative
{l� } using interval notation, ( -�, 1) or (3,00).
The solution set is x - < x3 ,or,
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246
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©
Section 5.4: Polynomial and Rational Inequalities
41.
43.
Let x be the positive number. Then x3 >4x2 x3 -4x2>0 x2(x- 4)>0 f(x) == x2(x -4) x == 0 and x== 4 are the zeros of f . Interval (-00,0) (0,4) (4, 00) Number 5 1 -1 Chosen -5 25 -3 Value off Conclusion Negative Negative Positive Since x must be positive, all real numbers greater than 4 satisfy the condition. The solution set is { x I x>4 } , or, using interval notation, (4,00) . The domain of the expression consists of all real numbers x for which x4-16�0 (x2 + 4)(X2 -4) �0 ( x2 + 4)( x -2)( x + 2) �0 p(x)== ( X2 + 4 )( x-2)( x+2) x== -2 and x== 2 are the zeros of p. Interval (-00, -2) (-2,2) (2, 00) Number 0 3 -3 Chosen Value ofp 65 -16 65 Conclusion Positive Negative Positive The domain of f is { x l x:S;-2 orx�2 } , or, using interval notation, (-00, -2] or [2,(0 ) .
45.
The domain of the expression includes all values for which x-2 x+4 �o f(x)== xx-2 +4 The zeros and values where the expression is undefined are x== -4 and x== 2 . Interval (-00,- 4) (-4,2) (2,00) Number 3 -5 0 Chosen 1 1 7 Value off "7 -"2 Conclusion Positive Negative Positive The solution or domain is { x I x < -4 or x �2 } , or, using interval notation, ( -00,-4 ) or [2,(0 ) . f ( x) :s; g ( x) X4 -I:S;_2x2 +2 X4 +2x2 -3:S;0 ( X2 +3 )( X2 -I):S;O ( X2 +3 )( x-l)( x+I):S;0 h ( x )== ( x2 +3 )( x-I )( x+l) x== -1 and x== 1 are the zeros of h. Interval (-00,-1) (-I,I) (I, ) Number 2 -2 0 Chosen 21 Value of h 21 -3 Conclusion Positive Negative Positive f( x) :s; g ( x) if -1:S;x:s; 1 . That is, on the interval [ -I,I] . --
47.
(0
f(x)
=
X4
-
1 .r·
x
g(x)
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
=
-
2x2
+
2
247
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Chapter 5: Polynomial and Rational Functions
49.
f(x)� g(x) x4-4�3x2 X4 -3x2 - 4� 0 ( X2 -4)( x2 + 1)� 0 (x -2)(x + 2) (X2 + 1)� 0 h ( x) ( x -2)( x + 2)( x2 + 1) x -2 and x 2 are the zeros of h. Interval (-00, -2) (-2,2) (2, 00) Number 3 0 -3 Chosen 50 Value of h 50 -4 Conclusion Positive Negative Positive =
=
=
53.
f(x) � g (x) if -2� x� 2. That is, on the interval [ -2,2]. g(x)
=
3x2
Interval (-00, 0) (0,250) (250, (0) Number 260 1 -1 Chosen Value off -5020 4980 -10/13 Conclusion Negative Positive Negative The number of bicycles produced cannot be negative, so the solution is { xl x �250 }, or, using interval notation, [250, (0) . The company must produce at least 250 bicycles each day to keep average costs to no more than $100. K �16 2(150)(S + 42) �16 S2 300S + 12,600 >16 S2 300S+12,600 16>0 S2 300S+12,600-16S2 >0 S2 Solve -16S2 + 300S + 12,600 0 and S2 0 . The zeros and values where the left hand side is undefined are S 0 , S 39 , S -20 Since the stretch cannot be negative, we only consider cases where S� O. Interval (0, 39) (39, 00) Number 1 40 Chosen Value ofleft side 12884 -0.625 Conclusion Positive Negative The cord will stretch less than 39 feet. b. The safe height is determined by the minimum clearance (3 feet), the free length of the cord (42 feet), and the stretch in the cord (39 feet). Therefore, the platform must be at least 3 + 42 + 39 84 feet above the ground. a.
=
=
x
51.
=
>:;
We need to solve C(x)�100 . 80x+5000�100 x 80x+5000 100x � 0 x x 5000-20x�O x - x-"",) � 0 _20---'.(2_50_ _ x f(x) 20(250-x) x The zeros and values where the expression is undefined are x 0 and x 250 . _
=
=
-'-----""'
=
55.
=
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
>:;
.
Answers will vary, for example, x2 < 0 has no real solution and x2� 0 has exactly one real solution.
248
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Section 5.5: The Real Zeros of a Polynomial Function
57.
No,the student is not correct. For example, x=-5 is in the solution set,but does not satisfy the original inequality. -5+ 4 = 2= '!'JO -5-3 -8 8 When mUltiplying both sides of an inequality by a negative,we must switch the direction of the inequality. Since we do not know the sign of x + 3 , we cannot multiply both sides of the inequality by this quantity.
17.
1 9.
1 1 =---+1-1=0 8 8 Thus, .!.2 is a zero of f and x -.!.2 is a factor off
Section 5.5 1. 3.
5. 7. 9.
11.
1 3.
1 5.
f(x)=4x6 -64x4+ x2-15; c =-4 f(-4)= 4( _4)6 -64{ _4) 4 + (_4)2 -15 =16,384 -16,384 + 16-15=1;>,: 0 Thus,-4 is not a zero of f and x + 4 is not a factor of f . f(x)=2X4 _x3 +2x-l; c=-21
f(-1)=2(-1)2 -(-I)=2+1=3 Using synthetic division: 3)3 -5 0 7 -4 9 12 36 129 3 4 12 43 125 Quotient: 3x3 + 4x2 +12x+43 Remainder: 125 Remainder,Dividend -4
21.
False; the potential rational zeros are ±!2 and ±1 . f(x)=4x3-3x2-8x + 4; c=2 f(2)= 4(2)3 -3(2)2 -8(2) + 4 =32-12-16 + 4 =8;>,: 0 Thus,2 is not a zero of f and x-2 is not a factor of f . f(x)=3x4-6x3-5x+l0; c=2 f(2)=3(2)4-6(2)3 -5(2) + 10 =48 -48 -10 + 10= 0 Thus,2 is a zero of f and x -2 is a factor of f .
f(x)=-4x7+ X3-x2+2 The maximum number of zeros is the degree of the polynomial,which is 7. Examining f (x)=-4x7+ x3 -x2 + 2 ,there are three variations in sign; thus,there are three positive real zeros or there is one positive real zero. Examining f(-x)=-4(-xf + (_X)3 _(_X)2 +2, = 4x -x3 -x2 + 2 there are two variations in sign; thus,there are two negative real zeros or no negative real zeros. f(x)=2x6 -3x2 -X + 1 The maximum number of zeros is the degree of the polynomial,which is 6. Examining f (x)=2x6 -3x2 -X + 1 , there are two variations in sign; thus,there are two positive real zeros or no positive real zeros. Examining f(-x)=2(-X)6 _3( _x)2 -(-x)+l, =2x6 -3x2 + X + 1 there are two variations in sign; thus,there are two negative real zeros or no negative real zeros. 7
23.
f(x)=3x6 + 82x3 + 27; c =-3 f(-3)=3(-3)6 +82(-3)3 +27 =2187-2214 + 27=0 Thus,-3 is a zero of f and x + 3 is a factor of f ·
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·
249
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Chapter 5: Polynomial and Rational Functions
f(x)=3x3 -2X2 + X + 2 The maximum number of zeros is the degree of the polynomial, which is 3. Examining f( x)=3x3 -2X2 + X + 2 , there are two variations in sign; thus, there are two positive real zeros or no positive real zeros. Examining f(-x)=3(-X)3 _2(_X)2 +(-x)+2, = -3x3 -2X2 -X + 2 there is one variation in sign; thus, there is one negative real zero. 27. f(x) = _x4 + x2-1 The maximum number of zeros is the degree of the polynomial, which is 4. Examining f(x) =_x4 + x2 -1 , there are two variations in sign; thus, there are two positive real zeros or no positive real zeros. Examining f( -x)= _(_x) 4 +(_X)2 -1 =_x4 + x2 -1 , there are two variations in sign; thus, there are two negative real zeros or no negative real zeros. 29. f(x)=x5 + x4+ x2+ x+l The maximum number of zeros is the degree of the polynomial, which is 5 . Examining f(x)=x5 + X4 + x2 + X + 1 , there are no variations in sign; thus, there are no positive real zeros. Examining f(-x)=(_x)5 +(-xt +(_X)2 +(-x)+I , =_x5 + X4 + X2 -x+l there are three variations in sign; thus, there are three negative real zeros or there is one negative real zero. 31 . f(x)= x6 -1 The maximum number of zeros is the degree of the polynomial, which is 6. Examining f(x)= x6 -1 , there is one variation in sign; thus, there is one positive real zero. Examining f(-x)=( _x)6 -1 =x6 -1 , there is one variation in sign; thus, there is one negative real zero.
25.
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
33.
f(x)=3x4-3x3 + x2- x+l p must be a factor of 1: p =±l q must be a factor of 3: q = ±1, ±3 P +1'-+ 1 The possible rational zeros are: -= q 3
35.
f(x)=x5 -6x2 +9x-3 p must be a factor of -3: p = ±1,±3 q must be a factor of 1: q = ±1 The possible rational zeros are: pq = ±1,±3
37.
f(x)=-4x3-x2+ x+ 2 p must be a factor of 2: p = ±1, ±2 q must be a factor of -4: q = ±1,±2,±4 The possible rational zeros are: p = +1, +2, +.!. ' + .!. - - -2 -4 q
39.
f(x)=6x4_x2 +9 p must be a factor of9: p=±I, ± 3,±9 q must be a factor of 6: q = ± 1, ± 2, ±3, ±6 The possible rational zeros are: 1 1 -,±3, 1 ± -, 3 ±9,±9 -p ±1, ±-,±-,± 2 3 6 2 2 q
41 .
=
f(x)=2x5 -x3 + 2X2 + 12 must be a factor of 12: p = ±1, ± 2, ± 3,±4,±6,±12 q must be a factor of 2: q = ±1, ±2 The possible rational zeros are: 1 3 p -= ±1, ± 2, ± 4, ±-,±3,±-,±6,±12 q 2 2 P
43.
f(x)=6x4 + 2x3 -x2 + 20 P must be a factor of 20: p = ±1, ± 2,±4,±5,±1O,±20 q must be a factor of 6: q = ± 1, ±2, ±3, ±6 The possible rational zeros are: p q ±1, ± 2, ± '21 ' ± '31 ' ± '32 ' ± '61 ,±4'± '34 ,±5,± 5'2 ' 20 ± �3 ,± �6 ,±10,± .!Q. 3 ,±20,± 3 =
250
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Section 5.5: The Real Zeros of a Polynomial Function
45.
Step 4: Using synthetic division: Wetryx-l : 1)2 -1 2 -1 2 3 2 3 2 x -1 is not a factor Wetryx--2I : )2 -1 2 -1 o I 2 0 2 0 x - ..!.2 is a factor and the quotient is 2x2 + 2 . Thus, f(x) 2x3 _x2 + 2x-I X- � (2X2 +2)
f(x) x3 + 2X2 -5x-6 Step 1: I (x) has at most 3 real zeros. Step 2: By Descartes' Rule of Signs, there is one positive real zero. f(-x) (_X)3 +2(-X)2 -5(-x)-6 _x3 + 2X2 +5x-6 thus, there are two negative real zeros or no negative real zeros. Step 3: Possible rational zeros: p ±I, ±2, ±3, ±6; q ±I; P =±1, ± 2, ± 3, ± 6 =
=
=
=
�
=
q
Step 4: Using synthetic division: Wetryx+3 :
)
3 1
-
1
2
-5
-6
-3
3
6
-1
-2
0
=
Since the remainder is 0, x - (-3) x + 3 a factor. The other factor is the quotient: x2 -x-2. Thus, f(x) (x + 3) ( x2 - X -2) (x+3)(x+I)(x-2) The real zeros are -3,-1, and 2, each of multiplicity 1. =
IS
=
49.
f(x)=2x3- x2 +2x-l Step 1 : I(x) has at most 3 real zeros. Step 2: By Descartes' Rule of Signs, there are three positive real zeros or there is one positive real zero. fe -x) 2(-X)3 - (-xi + 2(-x)-1 -2x3 _ x2 -2x-I thus, there are no negative real zeros. Step 3: Possible rational zeros: p ±1 q ±1, ± 2 p +-1, -+ ..!. 2 q
f(x)=2x3 -4x2 -10x+20 =2( X3 _2X2 -5x+1O) Step 1: I(x) has at most 3 real zeros. Step 2: By Descartes' Rule of Signs, there are two positive real zeros or no positive real zeros. f( -x)=2( _X)3 - 4( _X)2 -10(-x) + 20 , -2x3 - 4x2 + lOx + 20 thus, there is one negative real zeros. Step 3: Possible rational zeros: p ±1,±2,±5,±10; q ±l; P ±I,±2,±5,±10 =
=
=
=
q
=
=
=
Step 4: Using synthetic division: Wetryx-2: 2)1 -2 -5 10 2 0 -10 o -5 0 Since the remainder is 0, x -2 is a factor. The other factor is the quotient: x2 - 5 .
=
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
2 x - ( x2 + I) =
=
=
( �)
( )
Since x2 + 1 0 has no real solutions, the only real zero is x ..!.2 , of multiplicity 1.
=
47.
=
=
251
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Chapter 5: Polynomial and Rational Functions
We can find the remaining real zeros by solving x2 - 5 0 x2 = 5 x=±J5 Thus, f(x)=2(x -2)( x-J5 )( x+J5 ) . The real zeros are 2, J5 , and -J5 , each of multiplicity 1.
53.
=
51 .
f(x)=2x4 + X3 -7x2 -3x+3 Step 1: f (X) has at most 4 real zeros. Step 2: By Descartes' Rule of Signs, there are two positive real zeros or no positive real zeros. f(-x)=2(-xt +( _X)3 -7( _X)2 -3(-x)+3 =2x4 -x3 -7x2 + 3x + 3 thus, there are two negative real zeros or no negative real zeros. Step 3: Possible rational zeros: =±1, ±3; q ±1,±2; p E.= +- .!..2'-+1' +- �2' +-3 q Step 4: Using synthetic division: We try x+l: -1)2 1 -7 -3 3 -2 1 6 -3 2 -1 - 6 3 0 x + 1 is a factor and the quotient is 2x3 -x2 -6x + 3 . Factoring by grouping gives 2x3 -x2 -6x + 3 =x2 (2x-1) -3 (2x-1 ) = (2x -1) ( x2 -3 ) Set each of these factors equal to 0 and solve: 2x -1=0 x2 -3 0 2x=1 x= 21 Thus, f(x) (2x-1)(x + 1)( x -../3)( x +../3 )
f(x)= X4 + x3 -3x2 -x+2 Step 1: f (X) has at most 4 real zeros. Step 2: By Descartes' Rule of Signs, there are two positive real zeros or no positive real zeros. f(-x)=(_X)4 +( _x)3 -3( -xf -( -x)+2 thus, =X4 _ x3-3x2 +x+2 there are two negative real zeros or no negative real zeros. Step 3: Possible rational zeros: ±1, ± 2; q ± l; p p =±1, ± 2 ==
==
q
Step 4: Using synthetic division: Wetry x+2: -
)
2 1
1
3
-1
2
-2
1
2
2
-2
-1
-1
1
o
-
x + 2 is a factor and the quotient is x3 _x2 - x+ l .
=
Wetry x+l on x3 _x2 - x+l -1)1 -1 -1 1 -1 2 -1 0 -2 x + 1 is a factor and the quotient is x2-2x + 1 . Thus, f(x)=(x+ 2)(x+ l)( x2 -2x+l) =(x+ 2)(x+l)(x-l/ The real zeros are -2, -1, each of multiplicity 1, and 1, of multiplicity 2.
=
-
=
=
( �}x + 1)(x-../3)(x +../3)
2 x-
The real zeros are .!..2 , of multiplicity 1.
-1 ,
.J3 , and -.J3 , each 252
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.5: The Real Zeros of a Polynomial Function
55.
f(x)=4x4+5x3+9x2+10x+2 Step 1: f(x) has at most 4 real zeros. Step 2: By Descartes' Rule of Signs, there are no positive real zeros. f( -x)= 4( -x t +5( _X)3 +9(_X)2 +IO( -x)+2 =4x4-5x3 +9x2 -10x+2 thus, there are four negative real zeros or two negative real zeros or no negative real zeros. Step 3: Possible rational zeros: p = ±I, ± 2; q = ±1, ±2 ,±4; +- .!.. '+- .!..' +1 + .E.= q 4 2 - ' -2
57.
Step I: f (x) has at most 4 real zeros. Step 2: By Descartes' Rule of Signs, there are three positive real zeros or there is one positive real zero. f(-x)=(-xt -( _x)3 + 2(_X)2 -4(-x) -S = X4+ x3 +2X2 + 4x -S thus, there is one negative real zero. Step 3: Possible rational zeros: p = ± I, ± 2, ± 4, ± S; q = ± I; P =±I, ±2,± 4,±S q
Step 4: Using synthetic division: We try x+l: -1)4 5 9 10 2 -4 -I -S -2 4 S 2 o x+1 is a factor and the quotient is 4x3+X2+Sx+2. Factoring by grouping gives 4x3+X2+Sx+2 =x2(4x+I) + 2(4x+I) =(4x+l)(x2 +2) Set each of these factors equal to 0 and solve: 4x +1=0 x2+ 2=0 4x=-1 x2=-2 x=±H. x=--41 no real sol. Thus, f e x)=(4x+l)(x+I) ( x2 +2 )
Step 4: Using synthetic division: We try x+l:
)
-1 1
1
2
-4
2
-4
-8 8
-2
4
-8
0
We try x-2 on x3 _2X2 +4x-8
)
2 1
1
-2
4
-8
2
0
8
o
4
o
x -2 is a factor and the quotient is x2+ 4 . Thus, f (x)=(x +1)(x-2) ( x2 + 4) . Since x2 +4= 0 has no real solutions, the solution set is {-I, 2}.
( }
multiplicity I.
-1 -1
x+1 is a factor and the quotient is x3 _ 2x2 +4x-S.
= 4 x+ � x+I) ( x2 + 2) The real zeros are --1 and -I each of 4
X4-x3+2x2- 4x-S=O The solutions of the equation are the zeros of f (x)= x4 -x3 + 2X2 - 4x-S .
59.
,
3x3+4X2 -7x+2 =0 The solutions of the equation are the zeros of f(x)=3x3 +4X2 -7x+ 2. Step I: f (x) has at most 3 real zeros. Step 2: By Descartes' Rule of Signs, there are two positive real zeros or no positive real zeros. f(-x)=3(-X)3 +4(_X)2 -7(-x)+2 =-3x3 +4x2 +7x+2 thus, there is one negative real zero.
253 © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
Step 3: Possible rational zeros: p= ±1, ± 2; q= ±1, ± 3 p +1, +2 +-1 +-2 -= - - ' - 3' - 3 q
63.
Step I: f (x) has at most 4 real zeros. Step 2: By Descartes' Rule of Signs, there are two positive real zeros or no positive real zeros. f(-x)= (_X)4+4(_X)3 +2(_X)2 -(-x)+6 = X4 -4x3+2x2 +x+6 thus, there are two negative real zeros or no negative real zeros. Step 3: Possible rational zeros: p=±I, ±2, ±3, ±6; q=±I; E.-= ± 1, ± 2, ± 3, ± 6 q
Step 4: Using synthetic division: 2 We try x--: 3
%) 3
-7 2 2 4 -2 3 6 -3 o x -�3 is a factor. The other factor is the quotient 3x2+6x-3 . Thus, f(x)= x (3X2 + 6x-3 ) 4
( -%) =3 (x - %J ( X2 +2x -1)
Step 4: Using synthetic division: We try x+3: -
Using the quadratic formula to solve x2+2x-1=0 : -4- (1-)( ---1) x= -2 ± ..;r---42(1)
{
61.
1
)
%}.
{
65.
}
5, \Ir;5, 3"1 .
-\Ir;
6 3 -6
4
2
-1
1
-1
2
-3 -3
0
2008 Pearson Education, Inc. , Upper Saddle River, NJ.
-2 1
1 -2
-1 2
2 -2
1
-1
1
0
x+2 is a factor and the quotient is x2 -X + I . Thus, f(x)= (x+3)(x+2 ) (x2 -x+I) . Since x2 -x+1= 0 has no real solutions , the solution set is {-3, -2 } .
3x3 _x2 -15x+5= 0 Solving by factoring: x2(3x -1)-5(3x-l)= 0 (3x -1)( X2 - 5)= 0 (3x-1)(x-J"S )(x+J"S )= 0 The solution set is
3)1
x+ 3 is a factor and the quotient is x3+x2 -x+2. We try x+2 on x3 + x2 -X +2
--.!..--....:..:..-'-� -
-2 ± .J8 2 12 12 = -2 ±2 2 =-I± The solution set is -I- 12 , -1+12,
X4+4X3+2X2 -x+6= 0 The solutions of the equation are the zeros of f(x)=x4+4x3 +2X2 -x+6 .
x3 - � x2+ � X+1= 0 :::::> 3x3 -2x2+ 8x + 3 = 0 3 3 The solutions of the equation are the zeros of f(x)=3x3 -2x2 + 8x+ 3. Step 1: f (x) has at most 3 real zeros. Step 2: By Descartes' Rule of Signs, there are two positive real zeros or no positive real zeros. f(-x)=3(_X)3 -2(-X)2+8(-x)+3 , = -3x3 -2x2 -8x+ 3 thus, there is one negative real zero.
254
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©
Section 5.5: The Real Zeros of a Polynomial Function
Step 3: To find the possible rational zeros: p=±l, ± 3; q=± 1, ± 3 P -=±1, ± 3,±-31
1)2 -19 57 -64 20 2 -17 40 -24 2 -17 40 -24 -4 x -1 is not a factor 1 Wetryx--: 2 )2 -19 57 -64 20 1 -9 24 -20 2 -18 48 -40 o . . x --21 . a .,lactor and the quotient 2x3 -18x2 +48x-40. Thus, f(x)= x - � (2x3 -18x2+ 48x -40)
q
Step 4: Using synthetic division: We try x+-1 : 3 -� )3 -2 8 3 -1 -3 3 -3 9 0 x+-1 is a factor. The other factor is the 3 quotient: 3x2 -3x+9 Thus, f(x)= x+ (3x2 -3x +9 )
�
IS
( ) =2 ( x- �) ( x3 -9x2+24x-20)
.
( �) =( x+ � ) (3 ) ( x2 - X + 3 )
=(3x+1) ( x2 - X + 3 ) Since x2 - x+ 3=0 has no real solutions, the solution set is
{-�}.
67.
2X4 -19x3 +57x2 -64x+20=0 The solutions of the equation are the zeros of f(x)=2X4 -19x3 +57x2 -64x+20 . Step 1: f(x) has at most 4 real zeros. Step 2: By Descartes' Rule of Signs, there are four positive real zeros or two positive real zeros or no positive real zeros. fe -x)=2 ( _x) 4 -19 ( _X)3 +57 ( _X)2 -64 ( -x)+20 =2X4+19x3 +57x2+ 64x+20 Thus, there are no negative real zeros. Step 3: To find the possible rational zeros: p=±1, ±2,±4,±5,±10,±20; q=±1, ±2; 5 1 P -= ±l, ±-,±2,±4,±5,±-,±10,±20 2 2 q Step 4: Using synthetic division: We try x-I :
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
IS
Now try x -2 as a factor of x3 -9x2+24x-20 . 2)1 -9 24 -20 2 -14 20 -7 10 0 x -2 is a factor, and the other factor is the quotient x2 -7x+10. Thus, x3 -9x2 +24x-20=( x-2 ) ( x2 -7x+l0) =( x-2 )( x-2 )( x-5 ) f(x)=2 X- X-2 )2 ( x-5 )
( �}
The solution set is
{�,2,5}
.
255
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Chapter 5: Polynomial and Rational Functions
69.
I(x)= x3+2X2 -5x -6= (x + 3)(x+1)(x -2) x-intercepts: -3, -1,2; Near -3 : l( x)",,( x+3)( -3+1)( -3 -2)=1O( x+3) Near -1 : l( x)""( -1+3)(x+l)(-1-2)=-6( x+l) Near 2: I( x)""(2+3)(2+1)(x-2)=15 (x-2) Plot the point ( -3,0) and show a line with positive slope there. Plot the point (-1,0) and show a line with negative slope there. Plot the point (2,0) and show a line with positive slope there. intercept: I( 0)=03 +2 ( 0)2 -5 ( 0) -6= -6 ; The graph of I crosses the x-axis at x -3, -1 and 2 since each zero has multiplicity 1.
y-
=
Interval Number Chosen Value of I Location of Graph Point on Graph
(-00, -3) -4
-18 Below x-axis (-4, -18)
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
( -3, -1) -2 4 Above x-axis ( -2, 4)
(-1,2 ) 0 -6 Below x-axis ( 0, -6)
(2, 00) 3 24 Above x-axis (3,24)
256
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Section 5.5: The Real Zeros of a Polynomial Function
71 .
1(x) = 2x3 - X 2 + 2 x - 1 =
�
x-int""p! ; No", �. Plot the point
(x - �) ( J(X) .
2X 2 + 2 )
(x- �) H�r ++%(x- �)
(�, 0) and show a line with positive slope there.
y-intercept: 1(0) = 2 (0)3 _ 02 + 2(0) - 1 = -1 The graph of 1 crosses the x-axis at x .!2 since the zero has multiplicity 1 . =
Interval Number Chosen Value of 1 Location of Graph Point on Graph
( �)
(� )
0 -1
1 2
Below x-axis
Above x-axis
-oo,
(0, -1)
,oo
(1, 2)
x
257 © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 5: Polynomial and Rational Functions
73.
!Cx)=x4+ x2-2=(x+l)(x-l)(x2+2) x-intercepts: -1, 1 Near -1: ! (x)""(x + 1)(-1 -1) (( -1) 2 + 2 ) =-6 (x + 1) Near 1: ! (x)""(1 + 1)( x-I)(12 + 2) =6(x -1) Plot the point (-1,0) and show a line with negative slope there. Plot the point (1,0) and show a line with positive slope there. y-intercept: ! (0) 04 + 02 -2 =-2 The graph of ! crosses the x-axis at x -1 and 1 since each zero has multiplicity 1. =
=
Interval Number Chosen Value of ! Location of Graph Point on Graph
(-00,-1) -2 18 Above x-axis (-2,18)
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
(-1,1) 0 -2 Below x-axis (0,-2)
(1,00) 2 18 Above x-axis (2,18)
258
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Section 5.5: The Real Zeros of a Polynomial Function
75.
= 4X4
-2 (2x +1)(2x-1)( x2 2) . x-mtercepts: --21 '-21 -� I(X)-(2X+l)(+ �H[H)'.+-�(2X+l) N''' � . I( x) -(2m+1)<2X-l {G)' ++�(2X-l) I ( x)
+
7x2
+
=
No.,
(-�, 0) and show a line with negative slope there. Plot the point (�, 0) and show a line with positive slope there.
Plot the point
= 4( ) 4
-2 =-2
y-intercept: 1 (0) 0 +7 (0)2 The graph of 1 crosses the x-axis at x Interval Number Chosen Value of 1 Location of Graph Point on Graph
2 and .!2 since each zero has multiplicity 1. ( �) (-.!2'.!2) (�,oo) -91 91 -02 Above x-axis Below x-axis Above x-axis (1,9) (-1,9) (0,-2) =
-.!
-oo,-
x
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
259
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Chapter 5: Polynomial and Rational Functions
77.
I(x) =X4 +X3 -3x2 -x+2 =(x+2)(x+1)(x-1)2 x-intercepts: -1, 1 Near -2: l(x)::::(x+2)(-2+1)(-2-1)22 =-9(x+2) =4(x+1) Near -1: l(x)::::(-1+2)(x+1)(-1-1) 2 Near 1: l(x)::::(1+2){1+1)(x-1) =6(x-1)2 Plot the point (-2,0) and show a line with negative slope there. Plot the point (-1,0) and show a line with positive slope there. Plot the point (1,0) and show a parabola opening up there. y-intercept: 1(0) =04 +03 -3(0)2 -0+2 = 2 The graph of 1 crosses the x-axis at x -2 and-1 since each zero has multiplicity 1. The graph of 1 touches the x-axis at x 1 since the zero has multiplicity 2. Interval (-1,1) (-2,-1) (-00,-2) 0 -1.5 Number Chosen -3 -1.5625 2 32 Value of 1 Below xa xis Above x-axis Above xa xis Location of Graph Point on Graph (0,2) (-3,32) (-1.5,-1.5625) -2,
=
=
(1,00) 212 Above x-axis (2,12)
y 16 (2, 12)
(0,2)
(-2.0) (
-
L5,
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
-5
-1. 5625) (-1.
0)
5
x
260
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Section 5.5: The Real Zeros of a Polynomial Function
79.
f(x) = 4x5 - 8x4 -X+2 =(x- 2) (.J2x -1)(.J2x+1)(2X2 +1) . x-mtercepts: -2 '-2 '2 -.J2.J2 N",
N'''
-�. J(X)-(-�-2)(�(-�H(�X+l)H �J +}2(�+4)(�X+l) � J(X)-(�-2)(�X-l)(�(�H[2(�J +} 2(�-4)(�X-l) .
Near 2: f (x) "" (x-2)(.J2(2) -1)(.J2 (2)+1) ( 2(2)2 +1) = 63 (x-2) Plot the point (-� ,0) and show a line with positive slope there. Plot the point (� ,0) and show a line with negative slope there. Plot the point (2,0) and show a line with positive slope there. y-intercept: f (0)=4 (0)5 _8 (0)4 -0+2=2 The graph of f crosses the x-axis at x = - .fi2 , x .fi2 and x =2 since each zero has multiplicity 1. (-00,- �) ( .J22 .J22 ) (�'2) (2,00) Interval Number -1 3233 2 -9 -13 Value ofChosen f Graph Below x-axis Above x-axis Below x-axis Above x-axis Location Point onofGraph (1, -3) ( -1, -9) (3,323) (0,2) =
_
'
°
x
(1, -3) (-1,-9) -16
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
261
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Chapter 5: Polynomial and Rational Functions
81.
f(x)=X4 -3x2 -4 a3 =0,a2 = -3,a1 = O,ao =-4 I+Max{I-41,1 1,1-31 ,1 I} Max{I,I-41 +1 1 +1-31 +1 I} =1 +Max{4,0,3,0} =Max{I,4 + +3 + O} =1+4 =5 =Max{I,7} =7 The smaller of the two numbers is 5. Thus, every zero of f lies between - 5 and 5. f(x)=x4+x3 - x-I a3 =1,a2 =0,a1 =-1, ao = - 1 Max{I,I-II +I-II +1 1 +1 1 1 } I+Max{ I-II,I-II,1 1,1 1 I} =Max{ 1,1 +1+ +I} =1 +Max{I,1,0,I} =1+1=2 =Max{I,3} =3 The smaller of the two numbers is 2. Thus, every zero of f lies between - 2 and 2. f(x)=3x4 +3x3 _x2 -I2x-12 =3 (x4 +X3 _�x2 -4X-4) Note: The leading coefficient must be 1 . a3 =I,a2 =- "31 ,a1 =-4,ao =-4 Max{1'1-41+1-41 +1-�I+ll l} =Max{I,4 +4 +� +I} =1+Max {4,4'�'1} =1+4 =5 =Max{1, 283 }= 283 The smaller of the two numbers is 5. Thus, every zero of f lies between - 5 and 5. f (x)=4x5 -x4 + 2x3 -2x +x-l=4 (x5 - -41 x4 +-21 x3 - -21 x2 +-41 X "--41 ) Note: The leading coefficient must be 1. a4 =-4"1 ,a3 ="21 ,a2 =- "21 ,a1 =4"1 ,ao =-4"1 Max{I,I-±I +I±I +I-±I +I±I +I-±I} I+Max{I-±I,I±I,I-±I,I±I+±I} °
°
°
°
°
83.
°
°
°
85.
87.
2
=1+-1 =-3 The smaller of the two numbers is l2 . Thus, every zero of f lies between _l2 and l2 . 2
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
2
262
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Section 5.5: The Real Zeros of a Polynomial Function
89.
91.
f ( x ) = 8x4 _ 2x 2 + 5x - l; [ 0, 1 ] f(O) = -1 < ° f( = 1 0 > °
97.
and l) The valueandof thenegative functionat theis positive at onethe endpoint other. Since function is c ontinuous, the Intermediate Value Theorem guarantees given interval . at least one zero in the
95.
- 5 � r � -4 f ( x ) = 2x3 + 6x 2 - 8x + 2 [-5, -4] 10
Consider the function Subdivide the interval into equal subintervals: [-5, -4.9]; [-4.9, -4.8] ; [-4.8, -4.7]; [-4.7, -4.6]; [-4.6, -4.5]; [-4.5, -4.4]; [-4.4, -4.3]; [-4.3, -4.2]; [-4.2, -4. 1 ] ; [-4. 1 , -4] f ( -5 ) = -58; f ( -4.9 ) = -50.038 f ( -4.9 ) = -50.038; f ( -4.8 ) = -42.544 f ( -4.8 ) = -42.544; f ( -4.7 ) = -35.506 f ( -4.7 ) = -35.506; f ( -4.6 ) = -28.912 f ( -4.6 ) = -28.9 1 2; f ( -4.5 ) = -22.75 f ( -4.5 ) = -22.75; f ( -4.4 ) = - 1 7.008 f ( -4.4 ) = - 17.008; f ( -4.3 ) = - 1 1 .674 f ( -4.3 ) = - 1 1 .674; f ( -4.2 ) = -6.736 f ( -4.2 ) = -6.736; f ( -4. 1 ) = -2. 1 82 f ( -4. 1 ) = -2. 1 82; f ( -4 ) = 2 f [-4. 1 , -4].
f(x) = 2x3 + 6x 2 - 8x + 2; [ -5, - 4 ] f(-5) = -58 < ° andf(- 4) = 2 > °
The valueandof thenegative functionat theis positive at onethe endpoint other. Since function isguarantees continuous,at theleastIntermediate Value Theorem one zero i n the given interval. 93.
2x3 + 6x 2 - 8x + 2 = 0;
f(x) = x5 _ x4 + 7x3 - 7x 2 - 1 8x + 1 8; [ 1 .4, 1 . 5 ] f(I .4) = - 0. 1 754 < ° andf(1 .5) = 1 .4063 > °
The valueandof thenegative functionat theis positive at onethe endpoint ot h er. Since function isguarantees continuous,at ltheeastIntermediate Value Theorem one zero i n the given interval. 8x4 - 2X 2 + 5x - 1 = 0; 1 Consider the function f ( x ) = 8x4 - 2X2 5x - 1 Subdivide thes: interval [0, 1 ] into 1 0 equal subinterval [0,0. 1 ] ; [0. 1 ,0.2]; [0.2,0.3]; [0.3,0.4] ; [0.4,0.5];
So has a real zero on the interval Subdivide the interval [-4. 1 , -4] into 10 equal subintervals: [-4 . 1 , -4.09] ; [-4.09, -4.08] ; [-4.08, -4.07];
°�r�
+
[-4.07, -4.06] ; [-4.06, -4.05]; [-4.05, -4.04]; [-4.04, -4.03 ] ; [-4.03, -4.02] ; [-4.02, -4.01]; [-4 .0 1 , -4] f ( -4. 1 ) = -2. 1 82; f ( -4.09 ) � - 1 .7473 f ( -4.09 ) � -1 .7473; f ( -4.08 ) � -1.3 1 62 f ( -4.08 ) � - 1 .3 1 62; f ( -4.07 ) � -0.8889 f ( -4.07 ) � -0.8889; f ( -4.06 ) � -0.4652 f ( -4.06 ) � -0.4652; f ( -4.05 ) � -0.0453 f ( -4.05 ) � -0.4653; f ( -4.04 ) � 0.37 1 1 f r = -4.04 , [-4.05, -4.04],
[0.5,0.6] ; [0.6,0.7] ; [0.7,0.8]; [0.8,0.9]; [0.9, 1 ] f ( O ) = -1; f ( 0. 1 ) = -0.5 1 92 f ( O. 1 ) = -0.5 1 92; f ( 0.2 ) = -0.0672 f ( 0.2 ) = -0.0672; f ( 0.3 ) = 0.3848 f [0.2,0.3] .
So has a real zero on the interval Subdivide the interval [0.2,0.3] into 1 0 equal subintervals: [0.2,0.2 1 ]; [0.2 1 ,0.22] ; [0.22,0.23] ; [0.23,0.24] ;
So has a real zero on the interval therefore correct to two decimal places.
[0.24,0.25]; [0.25,0.26] ; [0.26,0.27]; [0.27,0.28] ; [0.28,0.29] ; [0.29,0.3] f ( 0.2 ) = -0.0672; f ( 0.2 1 ) � -0.02264 f ( 0.2 1 ) � -0.02264; f ( 0.22 ) � 0.02 1 9 f [0.2 1 ,0.22], r = 0.2 1
So has a real zero on the interval , correct to two decimal pltherefore aces. © 2008 Pearson Education, Inc., Upper Saddle River, NJ.
263
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Chapter 5: Polynomial and Rational Functions
99. /(x)=x3 +x2+x-4 1 ( 1 ) = - 1; / ( 2 ) = 1 0 So 1 has a real zero on the interval [ 1 ,2] . Subdivide the interval [ 1 ,2] into 1 0 equal subint e rvals: [ 1 , 1 . 1 ] ; [ 1 . 1 , 1 .2] ; [ 1 .2, 1 .3]; [ 1 .3 , 1 .4]; [ 1 .4, 1 .5]; [ 1 .5 , 1 .6] ; [ 1 .6, 1 .7] ; [ 1 .7, 1 .8] ; [ 1 . 8, 1 .9] ; [ 1 .9,2] 1 ( 1 ) = -1; /( 1 . 1 ) = -0.359 1( 1 . 1 ) = -0.359; /( 1 .2 ) = 0.368 So 1 has a real zero on the interval [ 1 . 1 , 1 .2]. Subdivide the interval [ 1 . 1 , 1 .2] into 1 0 equal subintervals: [ 1 . 1 , 1 . 1 1] ; [ 1 . 1 1 , 1 . 12] ; [ 1 . 1 2, 1 . 1 3 ] ; [ 1 . 1 3, 1 . 14] ;
[2.58,2.59] ; [2.59,2.6] 1 ( 2.5 ) - 1 .75; / ( 2.5 1 ) � - 1 .2576 1 ( 2.5 1 ) � -1 .2576; / ( 2.52 ) � -0.7555 1 2.52 ) � -0.7555; / ( 2.53 ) � -0.2434 1 2.53 ) � -0.2434; / 2.54 ) � 0.2787 1 [2.53,2.54], r 2.53
=
1 03.
[ 1 . 14, 1 . 15]; [ 1 . 1 5, 1 . 1 6] ; [ 1 . 1 6, 1 . 1 7] ; [ 1 . 17, 1 . 1 8] ; [ 1 . 1 8, 1 . 19] ; [ 1 . 1 9, 1 .2] 1 1 . 1 ) = -0.359; / 1 . 1 1 ) � -0.2903 1 ( 1 . 1 1 ) � -0.2903; / 1 . 1 2 ) � -0.2207 1 1 . 1 2 ) � -0.2207; / 1 . 1 3 ) � -0. 1 502 1 1 . 1 3 ) � -0. 1 502; / ( 1 . 1 4 ) � -0.0789 1 1 . 14 ) � -0.0789; / 1 . 1 5 ) � -0.0066 1 ( 1 . 1 5 ) � -0.0066; / 1 . 1 6 ) � 0.0665 1 [ 1 . 1 5, 1 . 1 6], r = 1.15 ,
101.
-2k + 1 O = 0 -2k = -1 0 k=5
( ( ( ( ( ( ( ( ( So has a real zero on the interval therefore correct to two decimal places. I (x) =2x4 -3x3 -4x2-8 1( 2 ) = - 1 6; /( 3 ) = 37 So 1 has a real zero on the interval [2,3]. Subdivide the interval [2,3] into 1 0 equal subintervals: [2,2. 1 ] ; [2. 1 ,2.2] ; [2.2,2.3]; [2.3,2.4]; [2.4,2.5]; [2.5,2.6] ; [2.6,2.7]; [2.7,2.8]; [2.8,2.9] ; [2.9,3] 1( 2 ) = - 1 6; /( 2. 1 ) = -14.5268 1 ( 2. 1 ) = -14.5268; / ( 2.2 ) = -12.4528 1 ( 2.2 ) = -12.4528; /( 2.3 ) = -9.6928 1( 2.3 ) = -9.6928; /( 2.4 ) = -6. 1 568 1( 2.4 ) = -6. 1 568; /( 2.5 ) = -1 .75 1( 2.5 ) = -1 .75; /( 2.6 ) = 3.6272 So 1 has a real zero on the interval [2.5,2.6) . Subdivide the interval [2.5,2.6] into 1 0 equal subintervals: [2.5,2.5 1 ] ; [2.5 1 ,2.52] ; [2.52,2.53]; [2.53,2.54];
1 05.
1 07.
From the Remainder remainder is 0 °Theorem, we know that the 1 ( 1 ) = 2( 1 ) 2 - 8( 1 / +1-2 =2 -8+1- 2 =-7 The remainder is -7. We want to prove that x-e is a factor of xn -en , for any positive integer By the Factor Theorem, x -e will be a factor of 1(x) provided I (e) = O . Here, I(x) =xn _en , so that 1( e) = en -en = 0 . Therefore, x -e is factor of xn -en . x3 -8x2 +1 6x - 3 = 0 has solution x = 3 , so 3 2 x - 3 is a factor of 1(x) = x - 8x + 1 6x - 3 . Using synthetic division n.
a
1 09.
3) 1 - 8 1 6 - 3 3 -15 3 -5 1 0
Thus, 3 1(x) =x -8x2 + 1 6x - 3 =(x - 3 ) (X2 -5x+ 1 ) . Solving x2 -5x+ 1 = 0 5 ±m X = 5 ± -.122 5 - 4 = --The sum of these two 2roots is 5 + m 5 - m .!Q + = =5. 2 2 2
[2.54,2.55]; [2.55,2.56];[2.56,2.57]; [2.57,2.5 8] ;
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
( ( ( So has a real zero on the interval therefore = , correct to two decimal places. From the Remainder and Factor Theorems, x - 2 is a factoroflif 1( 2 ) = 0 . ( 2 ) 3 _ k( 2 )2 + k ( 2 ) + 2 = 0 8 - 4k + 2k + 2 = 0
264
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Section 5.5: The Real Zeros of a Polynomial Function
111.
Let X be the length of a side of the original cube. After removing the I -inch slice, one dimension will be x - 1 . The volume of the new solid will be: (x - l) · x · x . Solve the volume equation: (x - I) · x · x = 294 x3 _ x 2 = 294 x 3 - x 2 - 294 = ° The solutions to this equation are the same as the real zeros of 1 ( x ) = x3 - x 2 - 294 . By Descartes' Rule of Signs, we know that there is one positive real zero. p = ±1, ±2, ±3, ±6, ±7, ± 1 4, ±2 1, ±42, ±49, ±98, ±l47, ±294 q = ±1 The possible rational zeros are the same as the values for p. p = ±1, ±2, ±3, ±6, ±7, ±14, ±2 1, ±42, ±49, ±98,
I ( X ) = X II + all _ I x II -I + all _ 2 x 11 - 2 + . . . + al x + ao ;
1 1 3.
where a n - I , all _ 2 , . . . al , ao are integers. If r is a real zero of 1 , then r is either rational or irrational. We know that the rational roots of 1 must be of the fonn ao q
E. q
where p is a divisor of
and q is a divisor of 1 . This means that
= ± 1 . So if r is rational, then r = E. = ±p . q
Therefore, r is an integer or r is irrational.
q
±147, ±294 Using synthetic division: 7 )1 -1 ° - 294 7 42 294 1 6 42 ° 7 is a zero, so the length of the edge of the original cube was 7 inches.
We begin with the interval [0, 1 ] . 1 (0) = - 1 ; 1 (1 ) = 1 0 Let mi = the midpoint of the interval being considered. So ml = 0.5 n
m il
I(mn )
New interval
0.5 1 [0,0.5] 1 ( 0.5 ) = 1 .5 > ° 2 0.25 [0,0.25] 1 ( 0.25 ) = 0. 1 5625 > ° [0. 1 25,0.25] 0. 1 25 3 1 ( 0. 125 ) 0:; -0.4043 < ° [0. 1 875,0.25] 4 0. 1 875 1 (0. 1 875 ) 0:; -0. 1 229 < ° [0. 1 875,0.2 1 875] 0.2 1 875 5 1 ( 0.2 1 875 ) 0:; 0.0 1 64 > ° [0.203 1 25,0.2 1 875] 0.203 1 2 5 6 1 ( 0.203 125 ) 0:; -0.0533 < ° [0.2 1 09375,0.2 1 875] 0.2 1 09375 7 1 ( 0.2 109375 ) 0:; -0.0 1 85 < ° Smce the endpomts of the new mterval at Step 7 agree to two decnnal places, r = 0.2 1 , correct to two decimal places. 2008 Pearson Education, Inc., Upper Saddle River, NJ.
265
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©
Chapter 5: Polynomial and Rational Functions
(b)
1 (x) = X4 + 8x3 - x 2 + 2; - 1 � r � 0 We begin with the interval [-1 ,0). I( -1) = -6; 1 (0) = 2 Let m j = the midpoint of the interval being considered. So m1 = -0.5 n New interval mn I ( mn ) 1 -0.5 [-1 , -0.5] 1 (-0.5) 0.8 125 > 0 2 -0.75 [-0.75, -0.5] I( -0.75) ", - 1 .62 1 1 < 0 -0.625 3 [-0.625, -0.5] I( -0.625) ", -0. 1 9 1 2 < 0 4 -0.5625 [-0.625, -0.5625] 1 (-0.5625) ", 0.3599 > 0 -0.59375 5 [-0.625, -0.59375] 1 (-0.59375) ", 0.0972 > 0 6 -0.609375 [-0.609375, -0.59375] 1 (-0.609375) ", -0.0437 < 0 7 -0.60 1 5 625 [-0.609375, -0.601 5 625] I( -0.60 1 5625) ", 0.0275 > 0 Smce the endpomts of the new mterva1 at Step 7 agree to two decnna1 places, r = -0.60, correct to two decimal places. =
(c)
1 (x) = 2x3 + 6x2 - 8x + 2; - 5 � r � -4 We begin with the interval [-5,-4). 1 (-5) = -58; 1 (-4) = 2 Let mj = the midpoint of the interval being considered. So m1 = -4.5 n New interval mn I ( mn ) 1 -4.5 [-4.5,-4] I( -4.5) = -22.75 < 0 2 -4.25 [-4.25,-4] 1 (-4.25) ", -9. 1 56 < 0 3 -4. 1 25 [-4. 125,-4] I ( -4. 1 25) ", -3.2852 < 0 4 -4.0625 [-4.0625,-4] 1 ( -4.0625) ", -0.5708 < 0 5 -4.03 125 [-4.0625, -4.03 1 25] I( -4.03 125) ", 0.7324 > 0 6 -4.046875 [-4.0625, -4.046875] 1 (-4.046875) ", 0.0852 > 0 7 -4.0546875 [-4.0546875, -4.046875] 1 (-4.0546875) ", -0.24 1 7 < 0 8 -4.0507 8 1 25 [-4.05078 1 25, -4.046875] 1 (-4.05078 125) ", -0.0779 < 0 9 -4.048828 1 25 [-4.05078 1 25, -4.048828 1 25] 1 (-4.048828125) ", 0.0037 > 0 10 -4.0498046875 [-4.0498046875, -4.048828 125] 1 (-4.0498045875) ", -0.0371 < 0 Smce the endpomts of the new mterval at Step 1 0 agree to two decImal places, r = -4.05, correct to two decimal places.
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266
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Section 5.5: The Real Zeros of a Polynomial Function
(d)
l (x) = 3x3 - 1 0x + 9; - 3 :<::: r :<::: -2 We begin with the interval [-3,-2]. I ( -3) = -42; 1 (-2) = 5 Let rni = the midpoint of the interval being considered. So rnl = -2.5 n New interval rn n I ( rnn ) 1 -2.5 [-2.5, -2] 1 (-2.5) = -1 2.875 < 0 2 -2.25 [-2.25, -2] 1 (-2.25) ,., -2.67 1 9 < 0 3 -2. 125 [-2.25, -2. 1 25] I( -2. 1 25) ,., 1 .4629 > 0 4 -2. 1 875 [-2. 1 875, -2. 1 25] I( -2. 1 875) ,., -0.5276 < 0 5 -2. 1 5625 [-2 . 1 875, -2. 1 5625] I( -2. 1 5 625) ,., 0.4866 > 0 6 -2. 1 7 1 875 [-2. 1 7 1 875, -2. 1 5 625] I( -2. 1 7 1 875) ,., -0.0157 < 0 7 -2. 1 640625 [-2. 1 7 1 875, -2. 1 640625] I( -2. 1 640625) ,., 0.2366 > 0 8 -2. 1 6796875 [-2. 1 7 1 875, -2. 1 6796875] I ( -2. 1 6796875) ,., 0. 1 108 > 0 9 -2. 1 6992 1 875 [-2. 1 7 1 875, -2. 1 6992 1 875] 1 (-2. 1 6992 1 875) ,., 0.0476 > 0 1 0 -2. 1 708984375 [-2. 1 7 1 875, -2. 1 708984375] I( -2. 1 708984375) ,., 0.0 1 60 > 0 Smce the endpomts of the new mterval at Step 10 agree to two deCImal places, r = -2. 1 7, correct to two decimal places.
(e)
1 :<::: r :<::: 2 I (x) = x3 + X 2 + x - 4; We begin with the interval [ 1 ,2]. 1 (1) = - 1; 1 (2) = 1 0 Let rni = the midpoint of the interval being considered. So rnl = 1 .5 n New interval rn n I ( rnn ) 1 [ 1 , 1 .5] 1 .5 1 (1 .5) = 3 . 1 25 > 0 2 1 .25 [ 1 , 1 .25] 1 (1 .25) ,., 0.7656 > 0 [ 1 . 1 25 , 1 .25] 1 . 1 25 3 1 (1 . 125) ,., -0. 1855 < 0 4 1 . 1 875 [ 1 . 1 25, 1 . 1 875] 1 (1 . 1 875) ,., 0.2722 > 0 [ 1 . 1 25, 1 . 1 5625] 5 1 . 1 5 625 1 (1 . 1 5625) ,., 0.0390 > 0 6 1 . 140625 [ 1 . 140625, 1 . 1 5625] 1 ( 1 . 1 40625) ,., -0.0744 < 0 7 1 . 1484375 [ 1 . 1 484375, 1 . 1 5625] 1 (1 . 1484375) ,., -0.01 80 < 0 [ 1 . 1484375, 1 . 1 5234375] 1 . 1 5234375 8 1 (1 . 1 5234375) ,., 0.0 140 > 0 [ 1 . 1 50390625, 1 . 1 5234375] 9 1 . 1 50390625 1 (1 . 1 50390625) ,., -0.0038 < 0 Smce the endpomts of the new mterval at Step 9 agree to two deCImal places, r = 1 . 1 5 , correct to two decimal places.
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267
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Chapter 5: Polynomial and Rational Functions
(f)
1 ( x)
2X4
=
+ x2
°�
- 1;
r
�1
We begin with the interval [0, 1 ] . 1 ( 0 ) - 1; 1 ( 1 ) = 2 =
mi
Let
m1
So
=
=
the midpoint of the interval being considered.
0.5
n
mn
1 ( 0.5 )
1
0.5
2
0.75
3
0 . 625
4
0. 6875
5
0.7 1 875
6
0.703 1 2 5
7
0.7 1 09375
8
0 .70703 1 2 5
9
0.708984375
New interval
I (mn )
=
-0.625
<
[0.5 , 1 ]
°
1 ( 0.75 ) "" 0. 1 953 > °
[0.5,0.75]
1 ( 0.625 ) "" -0.3042 < ° 1 ( 0.6875 ) "" -0.0805
<
[0. 625 ,0.75] [0. 6875,0.75]
°
1 ( 0.7 1 875 ) "" 0.0504 > °
[0.6875,0.7 1 875]
1 ( 0.7 1 09375 ) "" 0.0 1 64 > °
[0.703 1 25 , 0.7 1 09375]
1 ( 0.703 1 25 ) "" -0.0 1 68 < ° 1 ( 0.70703 1 25 ) "" -0.0003
<
[0.703 1 25,0.7 1 875]
°
1 ( 0 .708984375 ) "" 0.0080 > °
[0.70703 1 25 , 0 .7 1 09375] [0.70703 1 2 5 , 0.7089843 75]
Smce the endpomts of the new mterval at Step 9 agree to two decImal places, 0.70, correct to two decimal places.
r =
(g)
I (x)
2X4 - 3x 3 - 4x 2 - 8;
=
We begin with the interval [2,3] 1 ( 2 ) = - 1 6; 1 ( 3 ) 37 =
Let So
mi
m1
=
=
the midpoint of the interval being considered.
2.5
n
mn
1
2.5
2
2.75
3
2 . 625
4
2 . 5 625
5
2.53 125
6
2. 546875
7
2 . 5 3 90625
1 (2.5)
=
- 1 .75
I (mn )
<
°
New interval [2 .5,3]
1 ( 2.75 ) "" 1 3 .7422 > °
[2 . 5 ,2.75]
1 ( 2 . 5 625 ) "" 1 .4905 > °
[2 . 5 , 2 . 5 625]
1 ( 2 . 625 ) "" 5 . 1 3 5 3 > °
[2 .5,2.625]
1 ( 2 . 5 3 1 25 ) "" -0. 1 787 < °
[2. 5 3 1 25,2.5625]
1 ( 2.5390625 ) "" 0.2293 > °
[2. 5 3 1 25 , 2. 5 3 90625]
1 ( 2. 546875 ) "" 0 . 643 5 > °
[2. 5 3 1 25, 2 . 546875]
Smce the endpomts of the new mterval at Step 7 agree to two decnnal places, 2 . 5 3 , correct to two decimal places.
r =
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268
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Section 5.5: The Real Zeros of a Polynomial Function
1 ( x) = 3x 3 - 2X 2 - 20;
(h)
2�r�3
We begin with the interval [2,3] . 1 ( 2 ) = -4; 1 ( 3 ) = 43 Let
mi
=
the midpoint of the interval being considered.
So � = 2. 5 n
mn
New interval
I (mn )
1 ( 2 . 5 ) = 1 4.375 > 0
1
2.5
2
2.25
3
2 . 1 25
4
2 . 1 875
5
2 . 1 5 62 5
6
2 . 1 40625
7
2 . 1 3 28 1 25
8
2 . 1 2890625
9
2 . 1 30859375
1 ( 2 .25 ) "" 4.0469 > 0 1 ( 2 . 1 2 5 ) "" -0.244 1
<
[2,2 . 5 ] [2,2 .25] [2 . 1 25,2.25]
0
1 ( 2 . 1 875 ) "" 1 . 8323 > 0
[2 . 1 25,2. 1 875]
1 ( 2 . 1 40625 ) "" 0.2622 > 0
[2 . 1 25 , 2 . 1 40625]
1 ( 2 . 1 5 625 ) "" 0.777 1 > 0
[2 . 1 25,2. 1 5 625]
1 ( 2 . 1 328 1 25 ) "" 0.0080 > 0 1 ( 2 . 1 2 890625 ) "" -0. 1 1 83
<
1 ( 2 . 1 30859375 ) "" -0.0552
[2. 1 25 , 2. 1 328 1 25 ] [2. 1 2890625, 2. 1 328 1 25 ]
0 <
0
[2 . 1 30859375, 2. 1 328 1 25]
Smce the endpomts of the new mterval at Step 7 agree to two deCImal places, 2 . 1 3 , correct to two decimal places.
r =
1 1 7. l ( x) = 4x 3 - 5x 2 - 3x + l By the Rational Zero Theorem, the only possible I 1 . I zeros are : P = ± 1, ± - , ± rationa 2 4 q -
Since
.!.
is not in the list of possible rational 3 zeros, it is not a zero of 1 . 1 1 9. / ( x) = x7 + 6x5 - x4 + x + 2 By the Rational Zero Theorem, the only possible .
rationa I zeros are .. Since
P q
- + 1 , +2 _
_
.
�
is not in the list of possible rational 3 zeros, it is not a zero of 1 .
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269
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Chapter 5: Polynomial and Rational Functions
-i is a zero, its conjugate i is also a zero, and since 1 + i is a zero, its conjugate 1 - i is also a zero of I .
1 9. Since
Section 5. 6 1 . (3 - 2i) + ( -3 + Si) = 3 - 3 - 2i + Si
= 3i
(
(
I(x) = (x - 2)(x + i ) (x - i) x - (1 + i)) x - (I - i ) )
( 3 - 2i ) ( -3 + 5i ) = -9 + 1 5i + 6i - 1 0i 2
( (
= (x - 2) X 2 - i 2
= -9 + 2 1i - 1 0 ( - 1 )
= (x - 2) x 2 + 1
= 1 + 2 1i
(
) ( (x - 1) - i) ( (x - 1) + i)
)(
) )
x 2 - 2x + l - i 2
) ( x 2 - 2x + 2
3. one
= X 3 - 2X 2 + X
5. True
= x5 _ 2X4 + 2X 3 _ 2X4 + 4X 3 - 4x 2 + X 3 _ 2x 2 + 2x - 2x 2 + 4x - 4
7. Since complex zeros appear in conjugate pairs,
4 + i , the conjugate of 4 - i , is the remaining zero of I .
2 1 . Since -i is a zero, its conjugate i is also a zero.
(
I(x ) = (x - 3)(x - 3)(x + i) x - i )
-i , the conjugate of i , and 1 - i , the conjugate of 1 + i , are the remaining zeros of I .
( (
= x 2 - 6x + 9
1 1 . Since complex zeros appear in conjugate pairs,
= X 2 - 6x + 9
-i , the conjugate of i , and - 2i , the conjugate
)( )(
x2 - i2
X2 + 1
)
)
= x4 + X 2 - 6x3 - 6x + 9x 2 + 9 = x4 - 6x 3 + 1 0x 2 - 6x + 9
of 2i , are the remaining zeros of I . 1 3. Since complex zeros appear in conjugate pairs,
23. Since 2i is a zero, its conjugate - 2i is also a
-i , the conjugate of i , is the remaining zero.
zero of I . x - 2i and x + 2i are factors of I . Thus, (x - 2i)(x + 2i) = x 2 + 4 is a factor of f .
1 5. Since complex zeros appear in conjugate pairs,
2 - i , the conjugate of 2 + i , and -3 + i , the conjugate of -3 - i are the remaining zeros.
Using division to find the other factor:
,
an
2
= x5 _ 4X4 + 7x 3 - 8x 2 + 6x - 4
9. Since complex zeros appear in conjugate pairs,
For 1 7-22, we will use
-
x-4 2 x + 4 x 3 - 4x 2 + 4x - 1 6
)
= 1 as the lead coefficient of
the polynomial. Also note that
x3
+ 4x -16
( x - ( a + b i) ) ( x - ( a - bi)) = ( (x - a ) - b i ) ( (x - a ) + bi) = (x _ a) 2 _ ( bi) 2
-16 x - 4 is a factor, so the remaining zero is 4 . The zeros o f I are 4, 2 i , - 2 i .
1 7. Since 3 + 2i is a zero, its conjugate 3 - 2i is also
a zero of I .
J ( x) (x - 4)(x - 4) (x - (3 + 2 i)) (x - ( 3 - 2 i)) =
= (X 2 - 8x + 1 6 ) ( x - 3) - 2 i) ( x - 3) + 2i) = (X2 - 8x + 1 6) ( x2 - 6x + 9 - 4 i 2 ) = (X2 - 8x + 1 6) (X2 - 6x + 1 3 ) = X 4 _ 6x 3 + 1 3x2 _ 8x3 + 48x 2
- 1 04x + 1 6x2 - 96x + 208 = X 4 - 1 4x 3 + 77x2 - 200x + 208
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©
Section 5. 6: Complex Zeros; Fundamental Theorem of Algebra
25. Since
- 2i
2i is also a x - 2i and x + 2i are factors of f . (x - 2i)(x + 2i) = x2 + 4 is a factor of f .
29. Since
is a zero, its conjugate
- 4i
4i is also a x - 4i and x + 4i are factors of h . (x - 4i)(x + 4i) = x 2 + 1 6 is a factor of h . is a zero, its conjugate
zero of f .
zero of h .
Thus, Using division to find the other factor:
Thus, Using division to find the other factor:
2X 2 + 5x - 3 x2 + 4 2x4 + 5x3 + 5x2 + 20x - 1 2 2X4 + 8x2
3x + 2X2 - 33x - 22 ) X + 1 6 3x + 2x + 1 5x3 + l Ox - 528x - 352
3
2
)
5
4
3x 5
2
+ 48x3
5x3 - 3x 2 + 20x + 20x - 33x 3 - 22x2 - 528x - 528x
-12 - 12
- 352 - 352
2X2 + 5x - 3 = (2x - 1)(x + 3) The remaining zeros are The zeros of f are
.!. 2
and
-3 .
3x3 + 2X2 - 33x - 22 = x 2 (3x + 2) - 1 1(3x + 2) = (3x + 2)(x2 - 1 1)
2i, - 2i, - 3, .!. . 2
= (3x + 2) ( x - 01 ) ( x + 01)
is a zero, its conjugate 3 + 2i is also a zero of h . x - (3 - 2i) and x - (3 + 2i) are
27. Since
3 - 2i
- � , 01, and - .J1l . 3 4i, - 4i, - 01, .J1l, - � . 3
The remaining zeros are
factors of h . Thus,
The zeros of h are
(x - (3 - 2i))( x - (3 + 2 i)) ((x - 3) + 2 i)(( x - 3 ) - 2 i) =
= x2 - 6x + 9 - 4i2 = x2 - 6x + 1 3
31.
f(x) = x3 - 1 = ( x - 1) ( X2 + X + 1 ) x2 + x + 1 = 0
is a factor o f h . Using division to find the other factor:
of
x2 - 3x - 1O x2 - 6x + 1 3 x4 - 9x3 + 2 1x2 + 2 1x - 130 X4 6x3 + 1 3x2
X=
)
_
- 3x3 + 8x2 + 2 1x - 3x3 + 1 8x2 - 39x
The zeros are :
- l Ox2 + 60x - 1 30 - 1 0X2 + 60x - 1 30
are :
- 1 ± �12 - 4 (1) (1) -1 ± N = 2 (1) 2
= .!. + -J3 i 2 2
_
The solutions
f (x) + -
and
_
.!. -J3 i 2 2 _
1 -J3 . 1, - - + - 1 2 2 '
1 -J3 . I 2 2
- - - -
.
+ +�J +� l X
i
x2 - 3x - 10 = (x + 2)(x - 5) The remaining zeros are -2 and 5 . The zeros o f h are 3 - 2i, 3 + 2i, - 2, 5 .
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Chapter 5: Polynomial and Rational Functions
33.
f(x) = x3 - 8x 2 + 25x - 26 Step
f(x)
1:
has
37.
3 complex zeros.
Step
Step 2: By Descartes Rule o f Signs, there are three positive real zeros or there is one positive real zero.
Step
p = ± I, ± 2, ± 1 3, ± 26; q = ±I;
2
)1
x-2
q
Step
x-2 :
25 - 26 2 - 12 26 - 6 13 0 is a factor. The other factor is the
x 2 - 6x + 1 3 . x 2 - 6x + 1 3 = 0
Using synthetic division:
-3 ) 1
2 22 50 - 75 - 3 3 - 75 75 0 - 1 25 - 25
x+3
is a factor. The other factor is the
quotient: are:
x3 - x 2 + 25x - 25 .
x3 _ x2 + 25x - 25 = x 2 (x - l) + 25(x - l) = (x - l) ( x 2 + 25 )
-(- 6) ± �(- 6) 2 - 4(1)(1 3) . 2(1) 6 ± H6 6 ± 4i = = = 3 ± 2i 2 2 The zeros are 2, 3 - 2i, 3 + 2i . f ( x ) = ( x - 2 ) ( x - 3 + 2i ) ( x - 3 - 2i ) X=
35.
4:
We try x + 3 :
-8
The solutions of
Possible rational zeros:
1: = ±1, ± 3, ± 5, ± 1 5, ± 25, ± 75
Using synthetic division:
quotient:
3:
p = ±1, ± 3, ± 5, ± 1 5, ± 25, ± 75; q = ±1;
1: = ±1, ± 2, ± 1 3, ± 26 q We try
= (x - 1)(x + 5i)(x - 5i) -3, 1, - 5 i, 5 i . f ( x ) = ( x + 3 )( x - l )( x + 5i )( x - 5i )
The zeros are
f(x) = x4 + 5x 2 + 4 = ( x 2 + 4 ) ( X 2 + 1 ) = (x + 2i)(x - 2i)(x + i)(x - i) The zeros are :
- 2i, - i, i, 2i .
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
1
Thus, there are three negative real zeros or there is one negative real zero.
Possible rational zeros:
Step 4 :
has 4 complex zeros.
f( -x) = (_x)4 + 2( _x) 3 + 22( _x) 2 + 50( -x) - 7 5 = x4 - 2 x3 + 22x 2 - 50x - 75
there are no negative real zeros.
3:
f(x)
1:
Step 2 : By Descartes Rule of Signs, there is positive real zero.
f( -x) = (_X)3 - 8( _X) 2 + 25( -x) - 26 , thus, = - x3 - 8x 2 - 25x - 26 Step
f(x) = X4 + 2x3 + 22x 2 + SOx - 75
272
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Chapter 5 Review Exercises
39.
f(x) = 3x4 _ x 3 - 9x 2 + 1 59x - 52 1:
Step
f(x)
has
x 2 - 4x + 1 3 = 0 -(- 4) ± �(- 4) 2 - 4(1)(1 3) X= . 2(1) 4 ± .J=36 4 ± 6i = 2 ± 3 i = = 2 2
The solutions of
4 complex zeros.
Step 2: By Descartes Rule of Signs, there are three positive real zeros or there is one positive real zero.
The zeros are
f(- x) 3(_X) 4 _ (_X) 3 _ 9(_X) 2 + 1 59(- x) - 52 = 3x4 + X3 _ 9x 2 - 1 59x - 52 =
Thus, there is Step
3:
1
( j} X - 2 + 3i ) ( x - 2 - 3i)
negative real zero.
4 1 . If the coefficients are real numbers and
Possible rational zeros:
complex zeros must appear in conjugate pairs. We have a conjugate pair and one real zero. Thus, there is only one remaining zero, and it must be real because a complex zero would require a pair of complex conjugates.
q
Using synthetic division: Ch apter 5 Review Exercises
x+4 :
1.
-1 -9 1 59 - 52 - 1 2 52 - 1 72 52 0 3 - 13 43 - 1 3
- 4) 3
3.
j )3
1 x-3
on
which is not a nonnegative integer. 5.
3x 3 - 1 3x 2 + 43x - 1 3 :
-13
y
= x3 ,
shift left
.!. power, 2
2 units.
y 15
x
1 3 3x 2 - 1 2x + 39 . -
f(x) = (x + 2) 3 Using the graph of
43 - 1 3 -4 13 3 - 1 2 39 0
x
f(x) = 4x 5 - 3x 2 + 5x - 2 is a polynomial of degree 5 . f(x) = 3x 2 + 5x l / 2 - 1 is not a polynomial because the variable x is raised to the
x + 4 is a factor and the quotient is 3x3 - 1 3x2 + 43x - 1 3 . We try
is a
43. If the coefficients are real numbers, then
E. = ±1, ± 2, ± 4, ± 1 3, ± 26, ± 52,
We try
2+i
zero, then 2 - i would also be a zero. This would then re quire a polynomial of degree 4.
= ±1, ± 2, ± 4, ± 1 3, ± 26, ± 52; = ± 1, ± 3; q
4:
- 4, .!. , 2 - 3 i, 2 + 3 i . 3
f (x) = 3 (x + 4) x -
p
Step
are:
- is a factor and the quotient is
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
- 15
273
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Chapter 5: Polynomial and Rational Functions
7.
(e) Near -4 : f ( x ) "'" -4 ( -4 + 2 ) ( x + 4 ) = 8 ( x + 4 )
f(x) = - (x - I)4 Using the graph of y = X4 , shift right 1 unit, then reflect about the x-axis.
(a line with slope
8)
Near -2 : f ( x ) "'" -2 ( x + 2 ) ( -2 + 4 ) = -4 ( x + 2 )
y
5
(a line with slope -4 )
Near 0 : f ( x ) "'" x ( 0 + 2 ) ( 0 + 4 ) = 8x
( 1 , 0)
(a line with slope
x
8)
(f) Graphing: y
20 ( - 2, 0) ( - 3 , 3)
9. f(x) = (x _ 1)4 + 2
x
Using the graph of y = X4 , shift right 1 unit, then shift up 2 units .
- 20 1 3.
f(x) = (x - 2) 2 (X + 4) (a) y-intercept:
x-intercepts : solve f(x) = 0 : (x _ 2) 2 (x + 4) = 0 => x = 2 or x = -4
x
11.
f (O) = ( 0 _ 2 ) 2 ( 0 + 4 ) = 1 6
(b) The graph crosses the x-axis at x = -4 since this zero has multiplicity 1 . The graph touches the x-axis at x = 2 since this zero has multiplicity 2 .
f(x) = x(x + 2)(x + 4) (a) y- intercept: f ( 0 ) = ( 0 ) ( 0 + 2 ) ( 0 + 4 ) = 0 x-intercepts : solve f(x) = 0 :
(c) The function resembles y
x(x + 2)(x + 4) = 0
=
x 3 for large
values of I x l .
x = 0 or x = -2 or x = -4 (b) The graph crosses the x-axis at x = -4 , x = -2 and x = 0 since each zero has multiplicity 1 .
(d) The polynomial is of degree 3 so the graph has at most 3 - 1 = 2 turning points.
(c) The function resembles y = x 3 for large
(e) Near -4 : f ( x ) "'" ( _4 _ 2 )
values of I x l .
2
( x + 4) = 3 6 ( x + 4)
(a line with slope 3 6)
(d) The polynomial is of degree 3 so the graph has at most 3 - 1 = 2 turning points.
Near 2 : f ( x ) "'" ( x _ 2 )
2
( 2 + 4) = 6 ( x - 2)2
(a parabola opening upward)
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Chapter 5 Review Exercises
(f) Graphing:
1 7.
f ( x) ( x - l) 2 (x + 3) (x + l) =
f (0) = (0 - 1)2 (0 + 3) (0 + 1 ) =3 x-intercepts : solve f(x) = 0 : (x _ l) 2 (x + 3) (x + l) = 0 x = 1 or x = -3 or x - 1
y 50
(a) y-intercept:
( - 2, 32) (0, 1 6) ( - 4, 0)
x
=
(b) The graph crosses the x-axis at x -3 and x = - 1 since each zero has multiplicity 1 . The graph touches the x-axis at x = 1 since this zero has multiplicity 2. =
15.
f (x) = -2 x 3 + 4 x 2 = _2X 2 (X - 2) (a) x-intercepts :
( c ) The function resembles y = X4 for large values of I x l .
0, 2; y-intercept: 0
(b) crosses x axis at x x axis at x 0
=
2
and touches the
(d) The polynomial is of degree 4 so the graph has at most 4 - 1 3 turning points.
=
(c) The function resembles y
=
=
-2x 3
-3 : f (x) "" (_3 _ 1) 2 (x + 3) ( -3 + 1) = -32 (x + 3) (a line with slope -32 )
for large
(e) Near
values of I x l (d) The polynomial is of degree 3 so the graph has at most 3 - 1 = 2 turning points. (e) Near 0:
-1 : f ( x ) "" (_1 _ 1) 2 (- 1 + 3)(x + l) = 8 (x + 1) (a line with slope 8) Near
f ( x ) "" -2 x 2 (0 - 2) 4x2 =
(a parabola opening upward) Near
1: f (x) "" (x _ l) 2 (1 + 3)(1 + 1) = 8(x - l) 2
Near
2: f (x) "" _2 (2) 2 (x - 2) -8 (x - 2) =
(a line with slope - 8 )
(a parabola opening upward)
(f) Graphing by hand
(f) Graphing: y
( - 4, 75)
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80
275
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Chapter 5: Polynomial and Rational Functions
1 9.
R(x) = 3.
x+2 x+2 = (x + 3)(x - 3) x2 - 9
The degree of the numerator, The
{ x I x "# -3, x "# 3} . p(x) = x + 2,
is
- 9, is
m=2.
the numerator, p(x) = x2 + 3x + 2, is degree of the denominator,
1 .
n = 2 . The
q (x) = (x + 2/ = x2 + 4x + 4, is m = 2 . n = m , the line
Since
Since
y = ! = 1 is a horizontal 1
x+1 . · asymptote. S mce the d enormnator 0 f y = --
Since the denominator is zero at -3 and 3, x = 3 and x = 3 are vertical asymptotes.
is zero at
1:
2:
Domain:
is a vertical asymptote.
{ x l x "# o}
2x - 6 2 (x - 3) R(x) = -- = x x Near
3:
-2 , x = -2
x+2
2x - 6 p(x) = 2x - 6 ; q(x) = x; n = l; m = l x There is no y-intercept because
Step
=
. x2 + 3x + 2 ( x + 2) ( x + l) = -x + l 1S. m = x+2 (x + 2) 2 ( x + 2) 2 lowest terms. The denominator has a zero at -2. Thus, the domain is {x I x "# -2 } . The degree of R(x) =
n < m , the line y = 0 is a horizontal asymptote.
R(x) =
Step
n
degree of the denominator,
q (x) = x2
Step
21.
The denominator has zeros at -3 and Thus, the domain is
23.
is in lowest terms.
0 is not in
the domain.
is in lowest terms. The x-intercept is the zero of
3 : R (x) ;:; � (x - 3) . 3
2x - 6 2 (x - 3) R(x) = -- = x x
Plot the point
(3, 0)
p(x) : 3
and show a line with positive slope there.
is in lowest terms. The vertical asymptote is the zero of
q (x) : x = o .
Graph this asymptote using a dashed line. Step 4 :
Since
n = m , the line
y = � = 2 is the horizontal asymptote. Solve to find intersection points: 1
2x - 6 -- = 2 x 2x - 6 = 2x -6 "# 0 R(x) does not intersect
y
=2.
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
Plot the line y = 2 with dashes.
276
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Chapter 5 Review Exercises
Step
5:
Interval
Number Chosen
•
(-y;, 0) -2
5
R(-2)
Location of Graph
Above x-axis
6 & 7:
=
(-2.5)
Graphing:
3
•
(0,3) R(l)
4 =
-4
R(4)
(I. -4)
�IO v'
=
�
Above x-axis
Below x-axis
(4, n
1'=-72
-1x=O0
25.
•
(3, cr.)
I
Value ofR Point on Graph
Steps
o
•
H(x) x(xx +-22) p(x) = x + 2; q(x) = x(x_2) = x2_2x; n = l; m = 2
Step 1: Domain: { x l x 0, x 2 } . There is no y-intercept because 0 is not in the domain. Step 2: H(x) x(xx +-22) is in lowest terms. The x-intercept is the zero of p(x) : -2 Near -2 : H ( x ) "".!.( x + 2 ) . Plot the point ( -2, 0) and show a line with positive slope there. Step 3: H(x) = x(xx +-22) is in lowest terms. The vertical asymptotes are the zeros of q(x) : x = 0 and x Graph these asymptotes using dashed lines. Step Since m the line = 0 is the horizontal asymptote. Solve to find intersection points: *-
*-
8
4:
n <
,
x+2 = 0 x(x - 2) x+2 =0 x = -2 H(x)
=
2.
y
intersects y = 0 at (-2, 0). Plot the line
y
=0
using dashes.
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Chapter 5: Polynomial and Rational Functions
Step
5:
-2
•
Interval
(-x, -2)
Value of H
H(-3)
Point on Graph
(-3, -1s)
Number Chosen
-3
•
I
= -is
Location of Graph Below x-axis
Steps 6 Graphing:
o
•
(-2,0) -I
H(-I)
=
(0,2) I
H(I)
k
Above x-axis
2
•
= -3
Below x-axis
(-I. n
•
(2, x)
3
H(3)
=
�
3
Above x-axis
( n 3,
(1, -3)
& 7:
. X --'---b",....�!--I. -I--.L....L.-I_ - y
5
=
0
(1, -3)
27.
+ 3)(x - 2) p(x) x2 + x - 6; q(x) x2 - x - 6; R(x) x2 + x - 6 (x x2 - x - 6 (x - 3)(x + 2) 1: {x I x :;z!: - 2, x :;z!: 3} . 0: + 0 - 6 - 6 (0, 1) . o -0-6 -6 x2 + x - 6 . . I p(x) : -3 2: x2 - x - 6 -3 : R(x)::::< - 62.(x + 3) . (-3, 0) =
=
=
=
Step
Domain: = 1 . Plot the point They-intercept is R(O) IS owest terms. The x-intercepts are the zeros of and 2. Step R(x) Near Plot the point and show a line with negative slope there. Near 2: R(x)::::< 2.( x - 2) . Plot the point (2, 0) and show a line with negative slope there. Step 3: R(x) = x2x2 +- x -- 66 IS owest terms. The vemca' asymptotes are the zeros of q(x) : x -2 and 3 . Graph these asymptotes with dashed lines. =
=
m
_
X
=
©
2008 Pearson Education,
x
4 .
.
m
I
I
=
278
Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material i s protected under a l l copyri ght l aws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.
Chapter 5 Review Exercises
Step
4:
Since n = m , the line y = �1 = 1 is the horizontal asymptote. Solve to find intersection points: x2 + x - 6 = 1 x2 - x - 6 x2 + x - 6 = x2 - 6 2x = 0 x =O R(x) = 1 (0, -X
Step
intersects y at 1). Plot the line = 1 using dashes.
5: Interval
Number Chosen
.
(-co, -3)
.
-4
Value of R
R(-4)
Point on Graph
(-4,0.43)
=
0.43
Location of Graph Above x-axis
y
-3
3
2
-2
•
•
•
•
(-3, -2)
(-2.2)
(2,3)
-2.5
0
2.5
R( -2.5) "" -0.82 R(O) Below x-axis
(
-
2 . 5 , -0.82)
Steps 6 Graphing: & 7:
=
I
R (2.5)
=
- 1.22
Above x-axis
Below x-axis
(0, I)
(2.5, - 1.22)
•
(3, x)
4
R(4)
=
13
Above x-axis
(4, D
x=3 I
(o,ol
_ _ L_-.:. _
I 29.
x
-5
3 3 F(x) = x = ( x + 2x) ( x - 2 ) p(X) = X3 ; q(x) = x2 - 4; n = 3; m = 2 x2 - 4 --
Step 1: Domain: { x l x "# - 2, x "# 2} . The y-intercept is F(O) = � = � 0 . Plot the point (0, 0) . 02 3 Step 2: F(x) = -i-is in lowest terms. The X-intercept is the zero of p(x) : O. x Near 0: F ( x ) "" -� x3 . Plot the point (0, 0) and indicate a cubic function there (left tail up and right tail down). Step 3: F(x) = x2x3 is in lowest terms. The vertical asymptotes are the zeros of q(x) : x = 2 and x = 2 . Graph these asymptotes using dashed lines. -4
-4
=
-4
4
-
--
-4
279
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Chapter 5: Polynomial and Rational Functions
Step 4: Since x= m + 1 there is an oblique asymptote. Dividing: x3 =x+ x 4x 2 x -4x -4 2 -4 4x The oblique asymptote is = x. Solve to find intersection points: x3 --=x x2 -4 x3 =x3 -4x 4x =0 O = x at (0, 0). Plot the line = x using dashed lines. F(x) xin=tersects Step n
,
--
--
y
y
y
5:
Interval
( -ce, -2)
Value of F
F( - 3)
Number Chosen
Location of Graph
Point on
Steps
6 & 7:
-2
•
Graph
Graphing:
-3
=
-
'7
S-
( -1) 3.
-
)
F(-I)
(0,2)
=
!
3
Above x-axis
(-I, D
x=-2 y
)
F(l) =
-! 3
Below x-axis
( -D 1,
(2, 00)
3
F(3)
=
•
¥
Above x-axis
(3,¥)
x=2
:� (3,.¥)
I V= x I �. :/ .� · I . I ,/
! ! ! J
r
2
•
•
(-2,0)
Below x-axis -
()
•
/
1,./
X
(0,0)
280
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Chapter 5 Review Exercises
31 .
2X4
4
R(x) (x _1)2 p(x) = 2x ; q(x) (x -1)2 n = 4; m = 2 Step 1: Domain: {x l x;e1} . (0)4 = .Q = O . Plot the point (0,0) . The y-intercept is R(O) (02-1) 2 1 Step R(x) (x2x_ 1)4 2 is in lowest terms. The x-intercept is the zero of p(x) : O. Near 0: R(x) Plot the point (0,0) and show the graph of a quartic opening up there. Step 3: R(x) = (x _ 1) 2 is in lowest terms. The vertical asymptote is the zero of q(x): x 1 . Graph this asymptote using a dashed line. Step Since n > m + 1 there is no horizontal asymptote and no oblique asymptote. Step =
;
=
=
2:
=
""
2X4.
2X4
4:
=
,
5:
Interval
Value ofR
Location of Graph
Steps
6 & 7:
on
•
(-00, 0)
Numbcr Choscn
Point
o
•
-2
R(-2)
=
2
Rm
�
=�
Above x-axis
( 3�)
-
-
Abovc x-axis
--
G'D
x =
�t4
(1,
2
I
Graph__ -2 '9
Graphing:
•
(0. 1)
•
(0)
R(2)
=
32
Abovc x-axis
(2.32)
1
V
(2,32)
x
-10
28 1 © 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exi st. No portion of thi s material may be reproduced, in any form or by any means , without permi ssion in writing from the publi sher.
Chapter 5: Polynomial and Rational Functions
33 .
2 -4 (x + 2)(x -2) x+2 G(x) = x2x-x-2 (x-2)(x+l) x+l Step 1: Domain: { x I x -1, x 2 } . The y-intercept is G(O) = 20-2 -4-2 -2-4 = 2. Plot the point ( 0,2) . Step 2: In lowest terms, G(x) = xx+l+ 2 , x 2. The x-intercept is the zero of = x + 2 : -2; Note : 2 is not a zero because reduced form must be used to find the zeros. Near -2 : G( x) "" -x -2. Plot the point (-2,0) and show a line with negative slope there. Step 3: In lowest terms, G(x) = xx+l+ 2 , 2 . The vertical asymptote is the zero of f(x) = x + 1 : x = -1 ; Graph this asymptote using a dashed line. Note: x = 2 is not a vertical asymptote because reduced form must be used to find the asymptotes. The graph has a hole at ( 2, �) . Step Since n = m , the line y = !1 = 1 is the horizontal asymptote . Solve to find intersection points: x2 -4 ---2x -x-2 = 1 x2 -4 = x2 - -2 x = 2 G(x) does not intersect y = 1 because G(x) is not defined at x = . Plot the line y = 1 using dashes. Step 7:-
7:-
=
0
o
7:-
y
X 7:-
4:
X
2
5:
Interval
(-00, -2)
Value ofG
G(-3)=�
Point on Graph
(-3·n
Number Chosen
-2
•
•
-3
6 & 7:
Graphing:
•
(-2, - 1 )
(-],2)
°
-1.5
G(-1.5)=-1 Below x-axis
Location of Graph Above x-axis
Steps
-1
(-l.5, -1)
G(O) = 2
Above x-axis
(0,2)
2
•
(2, 'YO)
•
3
G(3)
=
1.25
Above x-axis
(3, 1.25)
x= y = 1·
x ����+-��-L���
-5
(-2,0) -5 282 © 2008 Pearson Educati on , Inc . , Upper Saddle Ri ver, NJ. A l l rights reserved. This material i s protected under a l l copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publi sher.
Chapter 5 Review Exercises
35.
x3 +X2 <4x+4 X3 +X2 -4x-4 < 0 X2 ( X + 1) -4 ( x + 1) < 0 (x2-4) (x+l) <0 (x-2)(x+ 2)(x+l) < 0 f(x) (x-2)(x+ 2)(x+ 1) x -2, x -1, and x 2 are the zeros of f . =
=
=
Interval
=
(-2,-1)
(-00 , -2)
Number
(-1, 2)
(2, 00)
-3
-3/2
0
3
Value off
-10
0.875
-4
20
Conclusion
Negative
Positive
Negative
Posi tive
Chosen
The solution set is {X I x < -2 or -1 < x < 2} , or, using interval notation, ( -2) (-1,2) . u
-co,
41
)
-4
37.
(
-2
o
)
4
2
I�
6 --�1 x+3 -x+3�O => _x+36 1�0 => 6-1(x+3) � O x+3 x+3 -(x-3) f(x) = x+3 The zeros and undefined are values x 3 andwherex -3the .expression is
39.
_ _
Interval
Number
Chosen
=
(
-co,
=
(3, )
-3) (-3,3)
-4
co
=
4
0
1
-off The solution set is { x 1- 3 < x 3 } , or, using interval notation, (-3,3] . Value
-7
Conclusion
Negative
1
Positive
=
7 Negative
-co,
�
-4
( 1 1 -2
0
2
]
4
1
2x-6 < 2 I-x 2x-6 -2 <0 I-x 2x -6 -2(1-x) I-x4x-8 < 0 -- < 0 I-x 4(x-2) f(x) I-x The zeros and undefined are values x = 1, andwherex the2 . expression is Interval ( 1) (1, 2) (2, ) Number 0 1. 5 3 Chosen Value off -8 4 -2 Conclusion Negative Positive Negative The solution set is { x 1 x < 1 or x > 2 } , or, using interval nota tion, ( 1) (2, ) --
�
1
I
-4
l1li
co
-co,
u
I 1 I ) ( I
-2
o
2
co
I� 1
4
.
�
283 © 2008 Pearson Education, Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any fOlm or by any means, without permi ssion in wri ting from the publi sher.
Chapter 5: Polynomial and Rational Functions
41.
(x-2)(x-l)�0 x-3 -2)(x f(x) = (x x-3 -1) The zeros and undefined are values x = 1, xwhere = 2, andthexexpressi = 3 . on is Interval
( -<Xl, I)
(1, 2)
(2,3)
(3,<Xl)
0
1.5
2.5
4
- -6 I
3 -2
6
Positive
Negative
Positive
Number Chosen Value off
-2 3
Conclusion
Negative
45.
47.
The solution set is { xiI:$; x :$; 2 or x > 3 }, or, using interval notation, [1,2] (3,00). -2
4 3.
1
E]
o
2
(
u
4
I�I I. 6
x2 -8x+12 > 0 x2 -16 f(x) = x2 x-8x+12 2 -16 (x-2)(x-6) >0 (x + 4)(x -4) The zeros and values where the expression is undefined are x = -4, x = 2, x = 4, and x = 6 . Interval
Number
Chosen
Value off
49.
-
-5
Negative
Positive
<Xl
<
1
-I
-8
Negative
7
Positive
<
)
-4
<
o
u
u
() ( 1 �I 4
8
48
48
-8
192
184
o
736
736
o
2944
2944
o
11,776
47,104
11,776 47,105
51.
Conclusion
5
0
f(4) =47,105 f (x) = 12x8 _x7 +8x4 -2x3 +x+3 There are 4 sign changes in f ( x) , so there are 4 posit positiivvee real real zeros, zeros. 2 positive real zeros, or no f ( -x) = 12( _x)8 - (-xf +8( _X)4 -2( _X)3 +( -x)+3 =12x8 +X7 +8x4 +2x3 -x+3 There are 2 sign changes in f (-x) , so there are 2 negative real zeros or no negative real zeros. ao =-3 , a8 =12 = ±1,±3 = ±1,±2,±3,±4,±6,±12 -= ±1, ±3, ±-,21 ±-,23 ±-31 , ±-,41 ±-,43 ±-,61 ±-121
Positi ve
0
)
4 12
12
77 93 -(-4,2) 43 -7 3 (2,4) --1 (4,6) 3 5 (6, ) 33 The solution set is { x l x -4 or 2 x 4 or x > 6 } , or, using interval notation, (-00, -4) (2,4) (6,00) . ( -<Xl, - 4)
f(x) =8x3 -3x2 +x+4 Since g(x) = x -1 then c = 1. From the Remai n der Theorem, t h e remai n der R when c): i s i s divided by x) x) g( f( f( 2 3 /(1) = 8(1) -3(1) + 1 + 4 =8-3+1+4 So R==1010 and g is not a factor of f . f(x) =x4 -2x3 +15x-2 Since g(x) = x + 2 then c = -2 . From the Remai Theorem, by g(thx)e remai is f(nc)der: R when f(x) nisderdivided f( -2) = (_2)4 -2(_2)3 + 15( -2) -2 =16-2(-8)-30-2 =0 So R = 0 and g is a factor of f .
5 3.
p
q
P q
•
2 84 © 2008 Pearson Education, Inc . , Upper Saddle River, N J . All rights reserved. Thi s material i s protected under all copyright laws a s they currently exist. No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.
Chapter 5 Review Exercises
55.
= x3 -3x' 2 -6x+8 f(x) By Descart RuleorofnoSigns, there arezeros. two posit ive realeszeros posit i ve real f(-x) = (_X)3 -3(_X)2 -6(-x) + 8 , there is one 2 =_x3 -3x negative real zero. +6x+8 Possi ble rational p =±1,±2, ±4, ±8;zeros:q =±l; q =±1, ±2, ±4, ±8 UsiWentgrysynthetic x+2 : division: -2)1 -3 -6 8 -2 10 -8 -5 4 x + 2 is a factor. The other factor is the quotient: x2 -5x + 4 . Thus, f(x) = (x + 2) ( x2 -5x + 4 ) . = (x+2)(x-1)(x-4) The zeros are -2, 1, and 4, each of multip licity 1. 4x3 +4x' 2 -7x+2 fBy(x)= RuleorofnoSigns, there arezeros. two positDescartes ive real zeros posit i ve real f(-x) = 4(_X)3 +4(-x) 2 -7(-x)+2; = -4x3 +4x2 + 7x+2 thus, there is one negative real zero. Possi b le rationalq =±1, zeros:±2, ±4; p =±1,±2; 1, -+ 2' -+�2 ' -+�4 qUsi=ng-+synthetic We try x+2 : division: -2)4 4 -7 2 -8 8 -2 4 -4 x + 2 is a factor. The other factor is the quotient: 4x2 -4x+l. Thus, f(x) =(x+2) ( 4x2 -4x+1 ) . = (x+ 2)(2x -1)(2x-l) The zeros are -2, of multip licity 1 and 1- ' of multiplicity 2.
59.
=
±
P
±
±
o
57.
= X4 -4x3 + 9x2 -20x + 20 f(x) By Descart es'zeros RuleoroftwoSigposit ns, there are zeros four or posit i v e real i v e real no positive real zeros. fe-x) (_X)4 _4(_x)3 +9(-x)2 -20(-x) +20 =x 4 +4x3 +9x2 +20x+20; Thus, there are no negative real zeros. Possi p = ±b1,le±rat2,ional ± 4, zeros: 5, ± 10, 20; q= ± 1; J!... = 1, ± 2, ± 4, ± 5, ± 10, ± 20 q Usingtrysyntheti We x-2 : c division: 2)1 -4 9 -20 20 2 -4 10 -20 -2 5 -10 x -2 is a factor and the quotient is x3 -2X2 + 5x -10 = x2 (x -2) + 5 ( x -2) = ( x -2) ( x2 + 5 ) Thus, f(x) = (x-2)(x-2) (x2 +5 ) = (X _ 2) 2 ( x2 +5 ) 2 + 5 = 0 has no real solutions, the only Since x zero is 2, of multiplicity 2. 3 -l1xof2 th+eXequat 2x 4 sol2xutions -6 =ion0 are the zeros of The = 2X4 +2X' 3 -l1x2 +x-6. f(x) By Descart eszeros RuleorofthSieregns,is onethereposit are ithree posit i v e real ve real zero. f(-x) = 2(_X) 4 + 2(_x)3 -11( _X)2 + (-x) -6 = 2x i4s -2x -11xive2 -Xreal-6zero. Thus, there one 3negat Possi rational±6;zeros:q =±1,±2; p =±1,ble±2,±3, 1 ±-3 -= ± 1 , ± 2, ± 3, ± 6, ± -, 2 2 q Usintgrysyntheti We x+3: c division: -3)2 2 -11 1 -6 -6 12 -3 6 2 -4 -2 0 x + 3 is a factor and the quotient is o
61.
P
+
P
o
285
© 2008 Pearson Education, Inc . , Upper Saddle River, NJ. All rights reserved. This material i s protected under all copyright laws as they currently exist. No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in wri ting from the publisher.
Chapter 5: Polynomial and Rational Functions
63.
2x3-4x2+ X -2 = 2X2( X -2 )+ I (x -2 ) = ( x -2 ) ( 2X2+ I ) Thus, f(x) = (x + 3)(x -2) ( 2X2 + 1 ) . Since 2x2 + I = 0 has no real solutions, the solution set is {-3,2} . + 7x3io+X2 The2X4 solut ns of-7x-3 the equat= i0on are the zeros of f(x) =2x4 +7X3 +x2 -7x-3. By Descart posit ive reales'zero.Rule of Signs, there is one f(-x) =2(-X)4 +7(-x)3+ (-xi -7(-x)-3 2 + 7 x -3 4 = 2x -7 x3 + x threezero. negative real zeros or there isThus,one there negatareive real Possi= ±1,ble±rat3;ional=zeros: ±1, ± 2; -= ±I, ±3, ± -21 , ± -32 Usintgrysyntheti We x+3: c division: -3)2 7 1 -7 -3 -6 -3 6 3 2 1 -2 -1 0 x + 3 is a factor and the quotient is 2x3+ x2 -2x -1 = x2 ( 2x + 1 ) -1( 2x + 1 ) . = ( 2x + 1 )( X2 -1 ) Thus, f(x) = (x + 3) ( 2x + 1 ) ( x2 -1) = (x + 3 )( 2x + 1 )( x -1 )( x + 1 ) The solution set is {-3, -1, -�, I}. f(x) =X3-x2-4x+2 -1, = -4, = 2 Max { 1, 1 2 1 + 1 -4 1 + I -I I} = Max { I, 7} =7 1 + Max { 1 2 1 , 1 -4 1 ' I -I I} = 1 + 4 = 5 The smallofer off tlhieestwobetwnumbers real zero een -5 andis 5,5.so every p
67.
a
69.
q
65.
a
2
=
a 1
a
2
= --,
a
ao
1
<
q
P
f(x) = 2x3 -7x2 -10x+35 = 2 (x3_ � X2 -5x+ 3; ) 7 =-5, =-35 2 2 3 Max { 1 ' 1 ; 1 + 1 -5 1 +1 - � I} =Max { I, 26} = 26 I+Max {1 3; 1 ' 1 -5 1 ' 1 - � 1} =1+ 3; = 3; =18.5 The zero smallerof fof tlhiees twobetwnumbers real een-18.i5s and18. 518., so5every . f(x) = 3x3 -x-I; [0, 1 ] f(O) = -1 0 andf(l) = 1 > 0 The valueandof thenegatfunctive iatontheis posit iveSiatnceonethe endpoint ot h er. function isguarantees continuous,at least the Intermedi aiten Val ue Theorem one zero t h e given interval. f(x) =8x4 -4x3-2x-l; [0, 1] f(O) = -1 0 andf(1) = 1 > 0 The valueandof thenegatfunctiveiatonthiseposit iveSince at onethe endpoint ot h er. function isguarantees continuous,at ltehaste Intermediat e tValue Theorem one zero i n he given interval. f ( x ) = x3 -X 2 f ( 1 ) =-2; f( 2 ) =4 So by the Intermediate Value Theorem, f hasSubdivide a zero onthetheinterval interval[I,2[1,J i2n].to 10 equal subi ntervals: [1.1,1. [[1.I,1.1J; 5( 1,1.) 6];-2;1 [1. 6(,1.1.12];7)];[1.=[1.-1.2 ,1.7,1.7369];8];[1.[31.,1.8,1.4];9];[1.[41.,1.9,25];] f f ( 1.1 ) = -1.769;1 ( 1.2 ) = -1.472 f( 1. 2 ) = -1. 4 72;1 ( 1. 3) = -1.103 f( 1. 3) -1.103;1 ( 1. 4 ) = -0. 6 56 f (1.4) = -0.656;1 ( 1. 5 ) = -0.125 f( 1. 5 ) = -0.125;1 ( 1. 6) = 0.496 So f has a real zero on the interval [1. 5 ,1.6]. Subdivide the interval [1. 5 ,1. 6] into 10 equal subi n tervals: [ 1. 5 ,1. 5 1]; [1. 5 1,1. 5 2]; [1. 5 2,1. 5 3J; [1. 53,1.54J;
71.
<
73 .
-
=
o
=
286
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. This material i s protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any mean s, without permi ssion in writing from the publi sher.
Chapter 5 Review Exercises
75.
[1.[1.554,1. 55 5];8]; [1.[1. 555,1. 55 6];[1. 556,1.9,1.567];] 7,1. 8,1. 9]; [1. / ( 1. 5 ) = -0.125;1 ( 1. 5 1 ) "" -0.0670 / ( 1. 5 1 ) "" -0.0670;1 ( 1. 5 2 ) "" -0. 0082 / ( 1.52 ) "" -0. 0082;1 ( 1. 5 3 ) "" 0.0516 So / has a real zero on the interval [1. 5 2,1.5 3], pltherefore aces. the zero is 1. 5 2, correct to two decimal / ( x) = 8x4 -4x3-2x-l / ( 0) = -1; / ( 1 ) = 1 , So by the Intermediate Value Theorem, / has a zero onvidthee thineterval [0,1].[0,1] into 10 equal Subdi i n terval subi s :0. 2]; [0. 2 , 0 . 3]; [0.3,0.4]; [0.4,0.5 ]; [0,[0. 50,.1];n0t.e6rval [0.1, ]; [0.6,0.7]; [0.7,0. 8]; [0. 8,0.9]; [0.9,1] / ( 0) = -1;1 ( 0.1 ) = -1. 2 032 /( 0.1 ) = -1.2032;1 ( 0. 2 ) = -1.4 192 / ( 0.2 ) = -1.4 192; 1 ( 0. 3 ) = -1. 6432 / ( 0.3 ) = -1. 6432;1 ( 0. 4 ) = -1. 8 512 / ( 0.4 ) = -1. 8 512;1 ( 0. 5 ) =-2 / ( 0. 5 ) = -2;1 ( 0. 6 ) = -2. 0272 / ( 0. 6) = -2.0272; 1 ( 0. 7 ) = -1. 8 512 / ( 0. 7 ) = -1. 8 512;1 ( 0. 8 ) = -1. 3 712 /( 0. 8 ) = -1. 3 712;1 ( 0. 9 ) = -0.4672 / ( 0.9 ) = -0.4672;1 ( 1 ) = 1 So / has a real zero on the interval [0.9,1]. Subdinterval vide tsh:e interval [0. 9 ,1] into 10 equal subi [0.[0.994,,00.9.91];5];[0.[0.9 91 ,5,00.9. 92];6];[0.[0.992,6,00.9.93];7]; [0.[0.993,7,00.9.94];8]; [0.98,0.99]; [0.99,1] / ( 0.9 ) = -0.4672;1 ( 0. 9 1 ) "" -0.3483 / ( 0.9 1 ) "" -0.3483;1 ( 0. 9 2 ) "" -0.2236 / ( 0.92 ) "" -0. 2 236;1( 0. 9 3 ) "" -0.0930 / ( 0.93 ) "" -0. 0930;1 ( 0. 94 ) "" 0. 0437 So / has a real zero on the interval [0.93,0.94], therefore the zero is 0. 9 3, correct to two decimal places.
77.
79.
81.
Since zeros appear in conjugate pairs, 4zero-iof, complex the/ conj u gat e of 4 + i , is the remaining . / ( x) = ( x-6)( x-4-i)( x-4+i) = x3 -14x2 +65x-l02 Since egatx zeros appear i1-in conj, thuegateconjpaiugaters, -i,of 1 the+ icompl conj u e of i , and , are the remaining zeros of / . / ( x) = ( x -i)( x+ i)( x -1-i)( x-I + i) = X4 _ 2x3+ 3x2 -2x + 2 /(x) =x3-3x2 -6x+S. By Descart RuleorofnoSigns, there arezeros. two posit ive reales'zeros posit i v e real fe-x) = ( _X )3-3 ( _X )2 -6( -x) +8 2 = _x3 -3x + 6xive+8real zero . thus, there is one negat Possibl ±4, ±S;zeros:q=±I; p=±I,e±2,rational q =±I, ±2, ±4, ±S Using syntx-Iheti: c division: We 1)1 -3 -6 8 1 -2 -8 -2 -S 0 x -1 is a factor and the quotient is x2 -2x -8 Thus, lex) = ( x-1) ( x2 -2x-S) (x-l)(x-4)(x+2). The multicomplex p licity 1.zeros are 1, 4, and -2, each of /(x) = 4X3+4x2 _ 7x+2 . By Descart RuleorofnoSiposit gns, ithere arezeros. two posit ive reales'zeros v e real fe-x) = 4 ( _x)3+4(_x)2 -7 ( -x) +2 2 + 7 xiv+e 2real zero. = -4x3is +one4xnegat Thus, there Possi ±I,±2,±4; p=±1,b le±2;rationalq =zeros: -q = +- 1' +- 21 ' -+-41 ' -+2 P
try
=
83.
P
287 © 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced , in any form or by any means, without permission in wri ting from the publisher.
Chapter 5: Polynomial and Rational Functions
Usintgrysyntheti We x+2 : c division: -2)4 4 -7 2 -8 8 -2 4 -4 x + 2 a factor and the quotient is 4x2 -4x 1 . Thus, f(x) = (x+2) ( 4x2 -4x+l ) = ( x + 2) ( 2x - 1) ( 2x -1 ) =(x+2)(2x-l)2 =4(x+2) ( x+ -tt The complex zeros are -2, of multiplicity 1, and ..!. , of multip licity 2. 2 ) =x4 -4x' 3 +9x2 -20x+20. f(ByxDescartes RuleoroftwoSigns,positthiveererealare zeros four or positive real zeros no positive real zeros. f(-x) = (_x)4 -4( _X)3 + 9( _x)2 -20( -x) + 20 2 thus, there=x4are+4xno3 +9x negative+20x+20 real zeros. Possi p = ±l,b le±rati2, o±nal4, ±zeros: 5,±10,±20; q = ±1; q = ±1, ± 2, ± 4, ± 5,±10,±20 Usi Wentgrysyntheti x-2 : c division: 2)1 -4 9 -20 20 2 -4 10 -20 -2 5 -10 0 x -2 is a factor and the quotient is x3 _2X2 +5x-l0. Thus, f(x) = (x-2) ( x3 -2x2 +5x-l0) . We can factor x3 -2X2 + 5x -10 by grouping. x3 -2X2 + 5x -10 = x2 (x -2) + 5 ( x -2) =(x-2) ( x2 +5 ) = (x -2) ( x + Fsi)( x -Fsi) f(x) = (x -2) 2 ( x + Fsi) ( x -Fsi) The complex zeros are 2, of multiplicity 2, and Fsi and -Fsi , each of multiplicity 1. is
85.
87.
o
+
!!..
= 2X4 +2X' 3 -llx2 +x-6 . fBy(x)Descartes 3 positive real zeros or thereRuleisofoneSigposins, tthere ive realare2 zero. fe -x) = 2( _X)4 + 2( _x)3 -ll( _x) +( -x)-6 -llx2 -Xreal-6zero. thus, th=ere2xis4 -2x one 3negative Possi p =±I,b le±2,ratio±3,nal±6;zeros:q= ±I,±2; 3 1 p ±1, ±-,±2, -= 2 q 2 ±3,±-±6 Usintgrysyntheti We x-2 : c division: 2)2 2 -11 -6 4 12 2 6 2 6 3 0 x -2 is a factor and the quotient is 2x3 + 6x2 + X + 3 . Thus, f(x) = (x-2) ( 2x3 +6x2 +x+3) . We can factor 2x3 + 6x2 + X + 3 by grouping. 2x3 + 6x2 + X + 3 = 2x2 ( X + 3) + (x + 3) = (x + 3) ( 2X2 + 1 ) = (x+ 3) (.fix + i) (.fiX-i) f(x) = (x -2)( x + 3) ( .fix + i) ( .fix -i) = 2(X-2)(x+3) ( x+ f i)( X- f i) The complex zeros are 2, -3, .fi2 , and .J22 . each of multip licity 1. 250 = nr 2 h h = nr2502 '. A(r) = 2n r 2 + 2nrh = 2nr2 + 2nr ( �:�) = 2nr2 + 500r A(3) = 2n . 32 + 5003 500 223. 22 square cm = 18n+-'" 3 A(5) = 2n . 5 2 + 5005 = 50n+l00 '" 257. 0 8 square cm --
89. a.
.
l
-
l
=>
b.
c.
288
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,
Chapter 5 Test d.
Use MINIMUM on the graph of 500 Yl = 21tX2 +--X Xl"lin=0 Xl"lax=10 Xscl=l Yl"lin=0 Yl"lax=500 Ysc 1 =50 Xres= 1
c.
MiniMUM lI=�.�1J:91BB �V=219.6BB7B
p
!(x) = ( x-3) 4_2 Using the graph of = X4 , shift right 3 units, then shift down 2 units. -
d.
y
-
f(x)
=
(x-3)4
2
x
10 -10
2.
a.
b.
q
q:
•
Chapter 5 Test
.\' ]0
p
p:
The areamiatels smallest approxi y 3.4 1 cm.when the radius is 1.
g ( x) = 2x3 + 5x2 -28x-15 Wea =list-15all andintegers that are factors of t h at are al l the i n tegers facto ors of a3 = 2 . ±1,± 3, ±5, ±15 ±1,±2 Now we form all possible ratios f: 1 ± 1' ±2"3 ' ±2",±3,±5 5 '±215 , ±15 q: ±2", If16gpossi has abilratiitiesonallisted. zero, it must be one of the We can find the rational zeros by using the fact that if ris a zero of g, then g (r) = . That is, wevalues evalufrom ate theourfuncti on foronal different list ofrati zeros. If we get g (r ) = 0 , we have a zero. we useandlongstartdivision toonreduce the polpolThenyynomi nomial again t h e reduced al. We will start with the2 positive integers: g(l) = 2(1)3 +5(1) -28(1)-15 , = 2+5-28-15 = -36 g (3) = 2(3)3 + 5(3)2 -28(3) -15 = 54+45-84-15 0 that 3 is a zero. This means So, we=know that ( -3) must be a factor of Using long division we get 2X2 2 +llx+5 X-3)2X3 +5x -28x-15 - ( 2x3 -6x2 ) l1x2 -28x -(llx2 -33x) 5x-15 -(5x-15) Thus, we can now write g ( x) = ( -3) ( 2X2 +llx + 5) The get: quadratic factor can be factored so we
The degree,maxinm=um3 . number of real zeros is the Fileadirst nweg coeffi write ctheientpolynomial is 1. 1 so that the g(x) = 2 ( x3 +%x2 -14x- ; ) For the expression in parentheses, we have a2 = 2"'5 al = -14 and ao = - 215 max {1, lao l + lall+ l a2 1} = max{I, I-lil + I-141 + 1�1} = max {I, 24} = 24 1 + max { lao l ,lall , l a2 1} = 1 +max{l- lil , I - 141, 1�1} = 1+14smal=le15r of the two numbers, 15, is the The bound. Therefore, between -15 and every 15. zero of g lies
0
g.
x
o
x
2 89
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Chapter 5: Polynomial and Rational Functions
e.
f.
g.
g( x) = ( x -3 )( 2x +l )( x + 5 ) Tolastfindtwo thefactors remaiequal ningtozeros0 andofsolve. g, we set the 2x + 1 = 0 x + 5 = 0 2x=-1 x= -5 x= --21 Therefore, the zeros are -5, --21 , and 3. Notice thesearatil rationalonalzeros the list how of potenti zeros.were all in x-intercept of a graphInarethetheprevisameous as theThezeros of thesfunction. part, we found the zeros to be -5 , --21 , and 3. Therefore, the x-intercepts are -5 , -.!.2 , and 3. To fmd the y-intercept, we simply find g( O) . g( O) = 2 ( 0)3 + 5 ( 0)2 -28 ( 0) -15 = -15 So, the y-intercept is -15 . Whether theisgraph crosses orthetouches atlicity. an x-iEachntercept determi n ed by mul t i p ynomizeroal occurs once,odd somultiplici thefactor multy,tipofthelicithetgraph y polof each is 1. For wil cross thethex-axix s ataxithes atzero. Thus, the graph crosses each of the three x-intercepts. The power functi o n that the graph of g resembles for large values of I xl is given by the term with the highest power ofx. In this case, the power function is y = 2x3 . So, the graph of will resemble the graph of y = 2x3 for large values of I xl .
h.
The three intercepts are -5 , -.!.,2 and 3. Near -5 : g ( x) = ( x -3 ) ( 2x + 1 ) ( x + 5 ) -8 ( -9)( x+5 ) = 72 ( x +5 ) (a line with slope 72) Near --21 : ( x) = ( x -3 )( 2x + 1 )( x + 5 ) ( - �} 2X+l) (%) 63 63 = --634 ( 2x+l) = --x-2 4 63 ( a line with slope - 2 or -31. 5 ) Near( 3: g x) = ( x -3 )( 2x + 1 )( x + 5 ) = ( x-3)( 7 )( 8) = 56 ( x-3) (a line with slope 56) We couldvalues first forevaluate thep determi functionneatthe several x to hel scale.ng all this information together, we Putti obtain the following graph: �
g
�
i.
y
60
:1
-50
g
3.
x
(1. -30)
x3 -4x2 +25 x-100= 0 x2 ( x -4) + 25 ( x - 4) = 0 ( x - 4) (X2 + 25 ) = 0 x - 4 = 0 or x2 + 25 0 x= 4 x2 = -25 x = ±..J-25 x =±5i The solution set is {4, -5 i,5 i} . =
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Chapter 5 Test
4.
3x3 + 2x -1 = 8x2 -4 3x3 -8x2 + 2x + 3 = 0 If we let the left side of the equation be 1 ( x) , thenliwest alarel insitegers mply finding zeros ofoffa = 3 We that arethefactors and all the integers that are factors of ao3 = 3 . : ± 1, ±3 ; ± 1, ±3 Now we form all possible ratios f: 1 -: ±-,±1,±3 3
5.
p
p
q:
q
=
p q
Plotl
Plot2
Plotl
3:1B3XA3-SX2+2X+ 'Y2= 'Y3= 'Y�= ,Ys= 'Y6=
We start bytor.factoring the numerator and denomina -14x+24 = 2(x-3)(x- 4) g(x) = 2Xx22 +6x-40 (x+1 O)(x-4) The domain oflis {x!x;t:-1 O,x;t:4}. with x;t: 4 . In lowest terms, g ( x) = 2(x-3) x+10 The graph hasisonethe verti cfactor al asymptote, x -10, since x + l onl y of t h e O denominator ofatg xin=lowest terms. Thea holgraphe in itshe still undefined 4, but t h ere is graph there instead of an asymptote. Sincedegree the degree ofdenominator, the numeratorthe isgraph the same as thehori of the has a znontal asymptoteTheequalleadito nthgecoeffi quotiecntientofinthe ltheeadinumerator g coefficients. the leadingthcoeffici the denominatoris 2is and1. Therefore, e graphehasnt in the horizontal asymptote = f 2 . x2 +2x-3 r ( x) = --- x+lng the numerator. Start by factori r(x) = (x+3)(x-1) x+l The domain of the function is {x! x -I} . Asymptotes: Sinceonethevertfunction is in lowestx =terms, the graph hasThe i cal asymptote, -1 . degree ofdenomi the numerator itshonee graph morewithl anhavethe degree of t h e n ator so anlongoblique asymptote. Tocouldfindalsoit, weuseneedsyntheti to usec di v ision note: we ( dilinvear). ision in this case because the dividend is x+1 2 x + 1)x + 2x -3 _ ( X2 +x) x-3 -(x+l) The oblique-4asymptote is = x+ 1 .
�
V
It appears that there is a zero near x = 1 . 1(1) = 3(1)3 _8(1)2 +2(1)+3 = 0 Therefore, x= 1 is a zero and (x -1) is a factor of 1 (x). We can reduce the polynomial expression by using synthetic division. 1)3 -8 2 3 3 -5 -3 3 -5 -3 0 Thus, 1 (x) = (x -1) ( 3x2 -5x -3 ) . We can find the remaining zeros by using the quadratic formula. 3x2 -5x-3 = 0 a =3,b=-5, c = -3 _( -5) ± --'-�r-('--_5- )"--2 _---4-'--(-3-'-'-)-( ---'.-3 ) ... x = --'-'-2(3) 5 ± J25+36 5 ± .J61 Thus, the sol6 ution set is6 {I, 5-fI ",-0.468, 5+fI ",2.135 } .
y
6.
=
;t:
y
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Chapter 5: Polynomial and Rational Functions
7.
From problem 6 we know that the domain is { x I x -I} and that the graph has one vertical asymptote, y=x+1. x = -1 , and one oblique asymptote, x-iTonfind tercepts: theequal x-intercepts, wesolveneedthetoresul set theting numerator to 0 and equation. (x+3)(x-1) = 0 x + 3 = 0 or x -1 = 0 x=-3 The x-intercepts arex=l-3 and 1. The points (-3,0) and (1,0) are on the graph. v-intercept: r( O) = 02 +2(0)-3 0+1 =-3 They-intercept is -3 . The point (0,-3) is on the graph. Test for symmetry: +2(-x)-3 = x2 -2x-3 r(-x)= (_X)2(-x)+l -x+1 Since ( -x ) =/; r ( x ) , the graph is not symmetric with respect to the y-axis. Since r (-x ) -r ( x) , the graph is not symmetric with respect to the origin. Behavi or nearne ifthetheasymptotes: Toasymptote, determi crosseson the oblique we solvegraph the equati r(x) = x+1 x2 +2x-3 ---- =x+ 1 ' x=/;-1 x+1 x2 + 2x -3 = x2 + 2x + 1 -3is=a1contradi false ction so the graph does The result not cross the oblique asymptote. The zeros of the1, divide numerator and denominator, -3, -1 , and the x-axis into four subi-00,ntervals. ( -3),( -3, -1),( -1,1),(1, 00) We canncheck a poigraphnt iisn each subior nbelterval to x determi e if the above o w the axis.
Interval Number Value ofr Location Point
=/;
(-00,-3) -5 -3 below ( -5,-3)
(-3,-1) -2 3 above (-2,3)
(-1,1) 0 -3 below (0,-3)
(1,00)
3 3 above (3,3)
x
-1 Sinceappltheypolynomi agate l hasPairealrs coeffi cients,to find we can the Conj u Theorem theconjremai nin3g-izero., mustIf 3also+ i beis aa zero, then itsthe u gate, zero. Thus, four zeros are -2, 0, 3 -i , and 3 + i . The Factor Theorem says that if f ( ) = 0 , then ( x ) is a of the polfunction: ynomial. This allows us to write thefactorfollowing x=
8.
r
c
=/;
f
-c
(x) = (x-(-2))(x -O)(x -(3 -i))(x -(3+i) ) a
where is any real number. If we l e t = 1 , we get = f ( x ) (x+2)(x){x- 3+i){x -3 - i) = (x2 +2x)(x - 3 + ) ( x - 3-i) a
a
i
=(X2 +2x) (x2 -6x+10) =x4 -6x3 +lOx2 +2X3 -12x2 +20x =x4 -4x3 - 2X2 +20x
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Chapter 5 Cumulative Review
9.
Sidenomi nce thenatordomaimustn exclcontaiudesn the4 andfactors 9, the(x -4) and (x -9) . However, because there is only one vertical asymptote, x = 4 , the numerator must also contain the factor (x -9) . The horizontal asymptote, = 2 , indicates that tthhee degree degree ofof thethe numerator mustandbethatthethesameratiaso denominator ofaccomplish the leadinthig coeffi to befactor 2. Weincan s by inciclentsudingneedsanother the numerator, (x - a ) , where a *- 4 , along with a factor of 2. Therefore, we have (x) = 2(x-9)(x-a) ( x-4 )( x-9 ) If we let a = 1 , we get ( x -9)( x - 1) = 2X2 -20x + 18 . ( x) = 2(x-4)(x-9) x2 - 1 3x+36 we haves area conti polynomial polSinycenomial nuous, function we simplandy need to show that 1 ( a) and 1 (b) have opposite signs (where a and b2 are the endpoints of the interval) . 1(0) = _2(0) -3(0)+8 = 8 1(4) =-2(4)2 -3(4}+8 = -36 Since 1(0) = 8 > 0 and 1(4} =-36 <0 , the Intermedi te Value guarantees0 andthat4. there is ataleast one Theorem real zero between x+2 < 2 Weofx-3alnote thatnumbers the domai n of3.the variable consists l real except Rearrange the terms so that the ri g ht side is x+2 _2 <0 x-3 For 1 (x) = x+2 -2 , we find the zeros ofland x-3 tthheis,values ofxtoatwriwhitec1h/is undefmed. To do we need as a si n gle rati o nal expression.x+2 x+2 x-3 I (x) = x-3 -2 = x-3 -2 . x-3 x+2-2x+6 -x+8 x-3 x-3 The oflitwos x =val8uand/is undefined at x = 3 . Wenumber usezerothese es to divide the real line into three subintervals.
3
Interval (-00,3) (3,8) I (8,00) osen 8 I io n I negative positive ne!?fitive Since we want to know where 1 (x) is negative, x for iwhity ischstrixcconcl 8 areudesolthat utions.valuesTheofinequal the solution set is {x I x < 3 or x> 8} . In interval notation we write (-00,3) or (8,00). Num. ch Value
.
1 1.
4
- -
3
1 I
I
I
9
-
I
6
Chapter 5 Cumulative Review 1.
r
1 0.
off
4
0
C Ol1clus
y
r
8 ..
•
3.
o.
= (1,3), Q = ( -4,2) dp,Q = �(_4_1} 2 +(2 _ 3) 2 = �(_5}2 +(_1)2 =)25+1 =56 x2 -3x < 4 x2 -3x-4 < 0 (x-4)(x+l)< 0 f(x)=x2-3x-4 x=-1, x=4 are the zeros of f. Interval (-00, - 1) (-1, 4) (4,00) Number -2 0 5 Chosen Value ofl 6 -4 6 Conclusion Positive Negative Positive The solution set is {xl-1 x 4 } or (-1,4) in interval notation. ( ) p
<
oC
--
-1
4
<
�
293
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Chapter 5: Polynomial and Rational Functions
5.
Paralle l to y= 2x + 1 ; Slope 2, Containing the Usipoinntg(3,the5poi) nt-slope formula yields: y- = m (x -XI ) y-5 = 2(x-3) = 2x-6 y= 2x-1 YI
=
y-5
y
13.
7.
9.
Thiordered s relatipaiorns is(3,not6) aandfunction because the (3, 8 ) have the same first element, but different second elements. 3x + 2:0::; 5x-1 3 :O:2:; x -:0:23 :; x x�-23 The solution set is {xI x � %} or [%,OC)) interval notation.
15.
I
o
I
1
[
2
°.
a.
m
•
x-axis: Replace yby -y: -y= x3 -9x, which is not equivalent to y= x3 -9x . y-axis: Replace x by -x: y (_x)3 -9 (-x) = _x3 +9x which is not equivalent to y= x3 -9x . Origin: Replace x by -x and by -y: -y= (_X)3 -9 ( -x) y= -x3 +9x which is equivalent to y= x3 -9x. Therefore, the graph is symmetric with respect to origin. Not function, since thewhen graphxfails= the Vertical LineaTest, for example, f(x) = x+5 x-I Domain { xl x I } . 7 7 6; f(2) = 2+5 "1 (2,6) is2-1 not on the graph off. The point ( 2,7 ) is on the graph. f(3) = 33-1+ 5 �2 4' (3,4) is on the graph off. Solve for x x+5 =9 x-I x+5 =9(x-l) x+5 =9x-9 14= 8x 14 7 x=-=8 4 Therefore, (� ,9) is on the graph of f . ,to
b.
=
=
c.
=
= '
,to
d.
3
x-intercepts: 0= x3 -9x 0= x ( x2 -9) 0= x(x+3)(x-3) x = 0, -3, and 3 ( 0, 0), (-3, 0), (3, 0) y-intercepts: y= 03 -9(0) = 0=>(0,0) Test for symmetry: 294
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Chapter 5 Cumulative Review
1 7.
f(x) = 2x 2 - 4x + a = 2, b = -4, = 1 . Since a = 2 > 0, the graph opens up. The x-coordi nate-4 of the vertex is b x = -2a = - -2(2) = 1 . The y-coordinate of the vertex is f ( - :a ) = f ( l ) = 2 (1)2 - 4 (1) + 1 = -1 . Thus,axithes ofvertex is (1, -1). The symmetry is the line x = 1 . The discriminant is: b2 - 4ac = (-4 / - 4 (2) (1) = 8 > 0 , so the graph has two x-intercepts. The x-intercepts are found by solving: 2X2 - 4x + l = 0 x = ---'-- (---4)2(2)'--: ± -.J8-:-
f is increasing on - 3) and (2, ; f is decreasing on (-3, 2) ; fhas a local maximum at x = -3, and the local maximum is f ( -3) = 5 . fhas a local minimum at x = 2, and the local minimum is f (2) = -6 . f(x) = {2X-3x++1 4 if -3if x x 2 2 Domain: {xl x > -3} or ( -3,00 ) x-intercept: ( -± 0) y-intercept: (0, 1 )
c
(0
)
e.
f.
21.
<
<
�
a.
b.
c.
'
y 5
\,-2)
4 ± 2J2 2 ± J2 4
5
2 2-J2 and -2 + J2 . . are -The x-mtercepts 2 2 The y-intercept is f(O) = 1 . y
( - 00 ,
d.
l
Range: { YI Y 5 } or (-00, 5 ) f(x) = x2 - 5 x + 1 g(x) = -4x - 7 (f + g)(x) = x2 - 5x + l + ( -4x -7) = x 2 -9x - 6 The domain is: {x l x is a real number} . ( fg ] (x) = fg(x)(x) = x2-4x- 5-x 7+ l The domain is: {x l x;t-�} .
d.
1
23.
x
<
a.
-1
b.
(1 - �, 0)
-2
1 9.
a.
b. c.
x-intercepts: (-5, 0) ; (-1, 0) ; (5, 0) ; y-intercept: (0,-3) is not symmetri theTheorigraph gin, x-axis or y-axis.c with respect to The function is neither even nor odd. 295
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C hapter 6 Exponential and Logarithmic Functions b.
Section 6 . 1 1.
3.
5. 7.
f (3) = _4(3)2 + 5 (3) = -4(9) + 1 5 = -36 + 15 = -21
a.
c.
d. c.
f. a.
b. c.
d. 11.
d.
Domain: False b.
9.
c.
-1 f(x) = -Xx22- - 25 x2 - 2 5 7:- 0 (x + 5 ){x - 5) 7:- 0 x 7:- -5, X 7:- 5 {xl x 7:- - 5, x 7:- 5} (J g)( 1) = f (g (1)) = f (0) = -1 (J g)( -1) = f (g ( -1)) = f (0) = -1 (g o f )(-1) = g (J(-1)) = g (-3) = 8 (g o f)(O) = g (J (O)) = g ( -1) = 0 (g g )( -2) = g ( g ( -2) ) = g ( 3) = 8 (J 0 f )(-1) = f (J(-1)) = f (-3) = -7 °
°
13.
(g /)(2) = g(f (2» = g(2 · 2) = g(4) = 3(4)2 + 1 = 48 + 1 = 49 ( f f)( l ) = f(f ( l» = f (2(1» = f (2) = 2(2) =4 (g g)(O) = g(g(O» = g ( 3(0)2 + 1 ) = g( l ) = 3(1)2 + 1 =4 °
°
°
1 f (x) = 4x 2 - 3 g(x) = 3 --x 2 2 a. ( f g)(4) = f (g(4» = f 3 - � (4) 2 = f (-5 ) = 4(-5)2 - 3 = 97 b. (g /)(2) = g( f (2 » = g( 4(2)2 - 3) = g(13) = 3 _ .!. (13)2 2 169 = 3_ 2 - 163 2 °
°
g(J(- l» ) = g (l) = 4 g(J(0» ) = g (0) = 5 f (g(- l» ) = f (3) = -1 f ( g ( 4» ) = f (2) = -2
(
)
°
f(x) = 2x g(x) = 3x2 + 1 a. ( f g)( 4) = f(g(4» = f ( 3( 4 )2 + 1 ) = f(49) = 2(49) = 98 °
c.
-
( f / )(1) = f(f(l» = f(4(1)2 - 3) = f(l) = 4(1)2 _ 3 =1 °
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Section 6. 1 : Compos ite Functions
(g g)(O) = g(g(O» = g ( 3- � (0)2 ) = g(3) = 3-�(3)2 2 =3- 2.2 3 2 f(x) = Fx g(x) = 2x (fog)(4) = f(g(4» = f (2(4» ) = f(8) = .J8 =2J2 (g 1)(2) = g(f(2» = g (J2 ) = 2J2 (f 1)(1) = f(f(l» = f ( .fi ) = f(1) = = .fi1 (g g)(O) = g(g(O» = g (2(0» ) = g(O) = 2(0) =0 f(x) = l x l g(X) =-x2-1+1 (f og)(4) = f(g(4» = f ( 42 1+ J = f C� ) = 1 1� I 17
d.
1 5.
b.
0
c.
d.
a.
b.
c.
d.
1 7.
0
1 9.
(g f)(2) = g(f( 2» = g(1 2 1 ) = g(2) 1 0
1)(1) = f(f(I» = f (l l l ) = f(l) =11J=1 (g g)(0) = g(g(O» = g ( 02 1+ J = g(l)
(f 0
0
f( x) = _x+13_ g(x) = � (f og)(4) =f(g(4» = f ( V4) = V43+1 (g f)(2) = g(f(2» = g C� l ) = g (%J = g(1) = Vi = 1 (f 01)(1) = f(f(l» = f C!J = f (%) 3 3 6 l2 +1 �2 5 (g g)(0) = g(g(O» = g ( JjO) = g(O) = JjO =0
a.
0
b.
0
0
c.
a.
d.
0
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Chapter 6: Exponential and Logarithmic Functions
21.
23 .
2 5.
29.
a.
0
=
c.
d.
31.
a.
°
2 7.
(g f)(x) = g(f(x» = g(2x+3) = 3(2x+3) 6x+9 Domain: {xl x is any real number} . (f f)(x) == ff((f2x+3 (x» ) = 2(2x+3) +3 = 4x+6+3 =4x+9 Domain: {xI x is any real number} . (g g)(x) = g(g(x» = g (3x) = 3(3x) =9x Domain: {xl x is any real number} . f(x) =3x+1 g(x) = x2 The domain offis {xI x is any real number } . The domain of g is {xI x is any real number} . (f g )(x) = f(g(x» = f{X2 ) = 3x2 + 1 Domain: {xI x is any real number } . (g f)(x) = g(f(x» = g (3x+ l) =(3x+I)2 = 9x2 +6x+l Domain: {xI x is any real number } . ( f f)( x) = f(f( x» = f(3x+l) = 3(3x+ I) + 1 =9x+3+1 =9x+4 Domain: {xI x is any real number } . (gog)(x) = g(g(x» = g ( X2 ) = ( X2)2 =x4 Domain: {xI x is any real number } .
b.
The domain of g is {xl x"* O} . The domain off is {xl x"* I} . Thus, g(x) "* 1 , so we solve: g(x) = 1 �=1 xx= 2 Thus, x"* 2 ; so the domain of f og is {xl x"* O, x"* 2} . The domain of g is {xl x "* O }. The domain off is {xl x"* I} . Thus, g(x) "* 1 , so we solve: g(x) = 1 -�=1 xx=-4 Thus, x"* -4; so the domain of fog is {xl x"* -4, x"* O} . The domain of g is {xI x is any real number } . The domain offis {xl x� O} . Thus, g(x) � O , so we2x+3 solve:� x�--23 Thus, the domain of f og is { xl x � -%} . The domain of g is {xl x � I} . The domain off is {xI x is any real number } . Thus, the domain of fog is {xl x�l} . f(x) = 2x+3 g(x) = 3x The domain off is {xI x is any real number} . The domain of g is {xI x is any real number } . (f g)(x) = f(g(x» = f(3x) = 2(3x) +3 = 6x+3 Domain: {xI x is any real number } .
b.
c.
0
0
0
0
0
d.
0
298 © 2008 Pearson Education , inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.
Section 6. 1 : Composite Functions
33.
g(x) = x2 + 4 { xI x } g {x I x } (f g)(x) = f(g(x» = f { x2 + 4 ) = { X2 + 4 t = X 4 + 8x2 + 1 6 { xI x } (g o f)(x) = g( f (x» = g { X2 ) = { x2 f + 4 = X4 + 4 { xI x } (f f)(x) = f(f(x» = f ( x2 ) = ( x2 ) 2 = X 4 {x I x } (g o g)(x) = g(g(x» = g { X2 + 4 ) = {x2 + 4f + 4 = x 4 + 8x2 + 16 + 4 = x4 + 8x2 + 20 {x I x }
The domain offis is any real number . The domain of is is any real number . a.
b.
c.
d.
0
Domain:
is any real number .
Domain:
is any real number .
0
3 f(x) = x-I
c.
is any real number .
Domain:
Domain: 3 5.
b.
f (x) = x2
d.
.
a.
Domain
e)
g(x) = --x4 f {x I x ;to I} .
The domain of
g
(f o g)(x) = f(g(x» =f 4 4 x_ __ x_ = -4-x =� = -4-x x x 4 4+x {xl x :;o!: -4, x :;o!: O} . (g f)(x) = g(f(x» x =g _ x-I 4 x x-I - 4(x - I) x {x I x :;o!: 0, x :;o!: I} .
( �)
-i-l
The domain of g is b.
(;)
Domain
C )
-
{xl x ;to O} . a. (f g)(x) = f(g(x» =f 3 2- - 1 x 3 2-x x 3x 2-x {xl x :;o!: O, x :;o!: 2} . 0
0
The domain of is is {xl x ;to O} .
= -
The domain of is
Domain
x f(x) = x-I
is any real number .
g(x) x2 f {x I x ;to I}
0
Domain
3 7.
( )
(g f)(x) = g(f(x» _ _2_ 2(x - I) = g _3 x-I = 3 = 3 x-I {xl x ;to I} ( f f )(x) = f(f(x» = f �I 3(x - 1) 3 3 3 3 (x I) 4-x x-I -1 x-I {xl x ;to 1 , x ;to 4} . 2x = X (g o g)(x) = g(g(x» = g -x = -22 = 2 x {xl x ;to O} .
0
__
Domain
( )
Domain 299
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Chapter 6: Exponential and Logarithmic Functions
(f f)(x) = f(f(x)) x) - f (_ x-I x x x x-I x-I x_ 1 x-(x-I) x-II =Xx-I x-I x-I Domain {x l x I} . (g g)(x) = g(g(x)) = g ( -x4 ) 44 -4x -4 x =x Domain {xl x ;l': o} . f(x) = Fx g(x) =2x+3 The domain offis {x l x � o} . The domain of g is {x I x is any real number} . (f g)(x) = f(g(x)) = f( 2x + 3) = .J2x + 3 Domain { x l x � -%} . (g f)(x) = g(f(x)) g (Fx ) = 2Fx + 3 Domain {x l x � o} . (f f)(x) = f(f(x)) = f (Fx ) = lfx)11 2 = ( x1 / 2 = Xl / 4 =� Domain {x l x ? o} . (g g)(x) = g(g(x)) = g (2x+3) = 2(2x+3)+3 = 4x+6+3 =4x+9 Domain {xI x is any real number} .
c.
0
_
41.
a.
_
;I':
d.
39.
0
a.
0
b.
0
c.
0
d.
0
f(x) = x2 + I g(x) = .Jx-I The domain offis {xI x is any real number} . The domain of g is {xl x � I} . (f g)(x) = f(g(x)) = f(.Jx-I ) =(.Jx-It +1 =x-I+I =x Domain {xl x � I } . (g f)(x) = g(f(x)) = g(X2 + I ) = �X2 +1-1 =N =I xl Domain {x I x is any real number} . (f f)(x) = f(f(x)) = f( x2 +1 ) = ( x2 + It + I =x4 +2x2 +1+1 =x4 +2x2 +2 Domain {x I x is any real number} . (gog)(x) = g(g(x)) = g (.Jx -I ) =�.Jx -I-I Now, .Jx -I -I ? 0 .Jx-I � I x-I � I x�2 Domain {x l x � 2} . b.
c.
=
0
0
0
d.
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Section 6. 1 : Compos ite Functions
43.
x - 5 g ( x ) =- x+2 J(x) =x+1 x- 3 The domain ofJis {xl x *- -I} . The domain of g is {xlx*-3} . ) (fog)(X) = J(g(X)) = f ( X+2 x-3 x+2 - 5 ( X+2 - 5 ) (X- 3) x- 3 x-3 x+2 +1 X+2 +1 ) (X- 3) ( x- 3 x- 3 x+2 - 5(x- 3) x+2-5x+15 x+2+l(x - 3) x+2+x-3 4x-I7 -4x+I7 --2x-I 1 2x-I Now, 2x -1 *- 0, so X *--.2 Also, from the domain of g, we know x *- 3 . Domain of Jo g : { x I x *- �, x*-3 } . (gof)(x) = g(f(x)) = ( X-x+15 ) X- 5 +2 ) (x+l) x- 5 +2 --( x+1 x+1 x5 X- 3 ( --5 -3 ) (x+l) x+1 x+1
-7----+---
a.
--7---7"---
or
b.
4 5.
0
0
g
47.
0
0
49.
x- 5+2(x+l) =x-5+2x+2 x - 5 - 3(x+l) x- 5 - 3x- 3 3x-3 3x-3 =-2x+8 -2x- 8
c.
(gog)(X) = g(g(X)) = g ( X+2 x-3 ) x+2 +2 ( X+2 +2 ) (X - 3) x-3 x-3 x+2 -3 ( X+2 -3 ) (X-3) x-3 x-3 x+2+2(x-3) x+2+2x-6 x+2-3(x 3) x+2-3x+9 3x-4 or - -3x-4 -2x+ 11 112x-11 Now, -2x + 11 *-0, so X*--.2 Also, from the domain of g, we know x*-3 . Domain of gog : {x i x *- I2I ,x *-3 } . (f g)(x) = J(g(x)) = J (�x ) = 2 (� X) = x (g f)(x) = g(f(x)) = g(2x) = '21 (2x) = x (f g)(x) = J(g(x)) = J (.rx ) = (.rx)3 = x (g f)(x) = g(f(x)) = g ( X3 ) =.fx3 = x (f g)(x) = J(g(x)) = J (� (X+6) ) = 2 (� (X+6) ) -6 = x+6-6 (g f)(x) == xg(f(x)) = g (2x-6) = '21 «2x-6)+6) = -21 ( 2x) =x
d.
or ---2x -8off,*- we0, soknow x *- -4.x *- -1Also,. from theNow,domain Domain of goJ : {x l x *- -4, x *- -I }. (f O f)(X) = J(f(X)) = J ( Xx+1- 5 ) x-5 - 5 ( X- 5 - 5 }x+ " " +I � (� x+1 x+1 +t}X+I)
0
0
I)
x-5- 5(x + l) =x- 5- 5x- 5 x-5+I(x+l) x - 5+x+1 -4x- I O =-2(2x+5) 2x+5 x- 2 2x- 4 2(x- 2)
Now, -2 *-0,we know so x *-x2.*- -1Also,. from the domainx off, Domain of J J : {x I x *- -1, x *- 2} . 0
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Chapter 6: Exponential and Logarithmic Functions
51.
(f g)(x) = f (g(x» = f � (X - b» 0
( ) = a (� (X -b» ) + b
63 .
a.
b.
= x-b+b =X (g f)(x) = g( f (x» = g (ax + b) 1 ax + b) - b) = -( a = -1 (ax) a =X H(x) = (2x + 3)4 0
53.
c.
d.
0
Sinumbers nce the anddomaithendomai offisntheof gsetisoftheallsetrealof all real numbers, the domains of both f o g and g f are all real numbers. ( f g )(x) = (g f )(x) aex + ad + b = aex + be + d ad + b = be + d ad + b = be + d. fog = go 0
H(x) = .Jx2 + 1
Answers may vary. One possibility is f(x) = ../x, g(x) = x2 + 1
S(r(t » = S % t3
Answers may vary. One possibility is Thus,
0
61 .
67.
f(x) = 2X2 + 5 g(x) = 3x + a ( f g)(x) = f(g(x» = f(3 x + a) = 2(3x + a)2 + 5 x = O , ( f o g)(0) = 23 . 2(3 · 0 + a)2 + 5 = 23 2a2 + 5 = 23 2a2 - 1 8 = 0 2(a + 3)(a - 3) = 0 a = -3 a = 3
ret) = -23 P , t � 0
( ) = 47t ( 32 t 3 )2 = 47t (� t6 )
H(x) = 1 2x + 1 1
f(x) = 2x3 - 3x2 + 4x - 1 g(x) = 2 ( f g)(x) = f (g(x» = f (2) = 2(2)3 - 3(2)2 + 4(2) - 1 = 16 - 12 + 8 - 1 = 11 (g o f)(x) = g( f (x» = g ( 2X3 - 3x2 + 4x - 1 ) = 2
f when
S(r) = 47tr2
f(x) = l x l, g(x) = 2x + 1
59.
0
Thus,
Answers may vary. One possibility is 65.
57.
0
0
f(x) = X4 , g(x) = 2x + 3
55.
(f g )(x) = f (g(x» = f (ex + d) = a(ex + d) + b = aex + ad + b (g f)(x) = g( f (x» = g(ax + b) = e(ax + b) + d = aex + be + d
16 = -7tt 16 6 . Set) = -7tt
9
6
9
N(t) = 100t - 5t2, 0 � t � 10 C(N) = 15,000 + 8000N C(N(t » = C ( 100t - 5t 2 ) = 15,000 + 8000 ( l OOt - 5t 2 ) = 1 5,000 + 800, OOOt - 40, 000t 2 C(t) = 15, 000 + 800, OOOt - 40, 000t2 .
Thus,
0
When Solving:
or
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Section 6.2: One-to-One Functions; Inverse Functions
69.
P = - -41 x+100' 0 �x �400 -41 x = 100x = 4(100-p)
Section 6.2 1.
P
3.
= �4(100-p) 25 +600 = 2.JI00-p 25 + 600, 0 � � 100 Thus, C(p) = 2.JI0025 + 600, 0 � p � 100. h = 2r p
5.
P
73 .
7.
9.
/(x) the number of Euros bought for x dollars; g ( x) the number of yen bought for x Euros / (x) = 0. 8 382x g (x) = 140.9687x (go /)(x) = g ( J(x)) = g (0. 8382x) = 140. 9 687(0. 8382x) = 118.15996x (go /)(1000) = 118.15996(1000) = 118,159.96 yen Githatven that/and g are odd functions, we know /(-x) =-/(x) and g(-x)=-g(x) for all x in the domain of/and g, respectively. The composite function (fog)(x) =/ ( g(x)) has the following property: (f 0 g)(-x) = / ( g ( -x) ) = / ( ( x )) since g is odd = -/ ( g ( x ) ) since / is odd = -(f 0 g)(x) Thus, / 0 g is an odd function. =
11.
=
a.
b.
13.
c.
1 5.
d.
75.
1 7.
1 9.
21.
-g
The setareofnoordered pairspairsiswia functi osame n because there ordered t h the element and different second elements. fIrst The function is not defined when x2 + 3x -18 = 0 . Solve: x2 +3x-18 = 0 (x + 6)(x -3) = 0 x = -6domaior nxis= 3{x I x "* -6, x"* 3} . The y=x False. If/andthegdomai are inverse functions, thennthof/e range of/is n of and t h e domai g is the range of g. The functioninisputsone-to-one because totherethe aresameno two distinct t h at correspond output. The function iinputs, s not one-to-one because there arethat two different 20 Hours and 50 Hours, correspond to the same output, $200. The functioninputs, is not one-to-one because there areto two distinct 2 and -3, t h at correspond the same output. The functioninputs is one-to-one becausetotherethe aresameno two distinct that correspond output. The zfunction/is one-to-onethe because every hori ontal l i n e i n tersects graph at exactly one point. The are horifunction/is zontal linesnot(forone-to-one example,because y = 1) there that intersect the graph at more than one point. The function/is one-to-onethe because every hori z ontal li n e i n tersects graph at exactly one point.
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Chapter 6: Exponential and Logarithmic Functions
23.
Tothe find nterchangein thethe elements domaithen iwinverse, th the ielements range: in Annual Rainfall
460.00 202.01 1 96.46 1 91 .02 1 82.87
25.
33.
Location
Mt Waialeale , Hawaii Monrovia, Liberia Pago Pago, American Samoa Moulme i n , Burm a Lae , Papua N ew G u inea
Domain: {460. 00, 202. 0 1, 196.46, 191.02, 182. 8 7} Range: Waialeale, Monrovia, Pago Pago, Moulmei{Mtn, Lea} Tothe find nterchangein thethe elrange: ements in domaithen iwinverse, th the ielements Monthly Cost
of Life Insurance
$7.09 $8.40 $1 1 .29
27.
2 9.
31.
Age
30 40 45
Domain: {$7. 09, $8. 40, $11.2 9} Range: {30, 40, 45} Interchange the entri(2,-1), es in (11,0), each ordered { (5,-3), (9,-2), (-5,l)}pair: Domain: {5, 9, 2, 11, -5} Range: {-3,-2,-1, 0, 1} Interchange the entries { (1,-2), ( 2,-3), (4,2)} pair: (0,-10in), each (9,1),ordered Domain: {1, 2, 0, 9, 4} Range: {-2,-3,-1 O, 1, 2} f(x) =3x+4; g(x) = -(31 x-4) f ( g(X) ) = f G (X-4) ) = 3 G (X-4) ) +4 = (x-4) +4 = x g ( !(x) ) = g(3x+4) = '31 ( (3x + 4) -4) = '31 (3x) = x Thus, f and g are inverses of each other.
37.
x f(x) = 4x-8; g(x) =-+2 4 f ( g(x» ) = f(�+ 2 ) =4 (� +2) -8 = x+8-8 =x g ( !(x» ) = g(4x-8) 4x-8 +2 = -4 =x-2+2 =x Thus,!and g are inverses of each other. g(x) = 4x+8 f ( g(x» ) = f ( 4x+8 ) = (4x+8 t -8 = x+8-8 =x g ( !(x» ) = g(X3 -8) = 4(x3 -8)+8 = (;3 =x Thus,!and g are inverses of each other. f(x) =-;x1 g(x) =-x1 11 x f ( g(x) ) = f (-x1 ) =-=l·-=x 1 g ( !(x» ) = g (-x1 ) =-=1·-=x 11 x1 x Thus,!and g are inverses of each other. x
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Section 6.2: One-to-One Functions; Inverse Functions
39.
4x-3 -x) = I(x) = 2x+3 g( ; x+4 2-x 4X-3 2 ( ) +3 2-x = l(g(X))=/ ( 4x-3 ) 4x-3 +4 2-x -2-x (= 2 (�) +3) c2-X) ( 4X-3 2-x +4) ( 2-X) 2(4x-3) +3(2 -x) 4x-3+4(2- x) 8x-6+6-3x 4x-3+8-4x 5x =x5 2X+3 ) - 3 4 (� 2X+3 g(J(x)) = g ( x+4 ) = 2- 2x+3 x+4
43.
41 .
2
/
/
' / y =
y 2
x
/
45.
/
/
/
-2
Graphing the inverse: )!
3
/
47.
( 2- 2X+3 x+4 ) (X +4)
4(2x + 3) -3(x + 4) 2(x+4)-(2x+3) 8x+12-3x-12 5x2x+8-2x-3 = x5 Thus, Iand g are inverses of each other. Graphing the inverse:y
Graphing the inverse:( 1 , 2)
/
/
/
/
/
/
/
/
/
/
/
= X
I- I
x
-3
I(x) = 3x y =3x x = 3y Inverse y = -x3 r l (x) = �3 x Verifying: 1(1-1 (x) ) = I (± X) = 3 (± X) = x r l (I(x)) = rl (3x) = �3 (3x) = x Domain ofl Range of I-I All real numbers Range ofl Domain of I-I All real numbers =
=
=
X
/
/
/ /v / ..
=
/ (2 , 1) f- 1 / /
/
( - 2, -2)
/
X
-2
-5
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Chapter 6: Exponential and Logarithmic Functions
49.
4x + 2 f(x) == 4x+2 x = 4y + 2 Inverse 4y = x-2 x-2 y= -41 x y=--4 -2 rl(X) =::._4 2-2 Verifying: f { !-I (X» ) = f (� -�) = 4 (� -�) + 2 =x-2+2=x 4x+2 rl(J(x» ) = r l (4x+2) = 4 _ 2-2 =x+ -21 --21 =x Domain off Range of f-I All real numbers Range off Domain of f- I All real numbers
Domain off Range of f-I All real numbers Range off Domain of f-I All real numbers =
Y
=
Y+S f- t /"
/
/
/
/
/
/
/ /y
=
�
Y 5
/
/
/
/
/
/ y = x
x
-5 53 .
= x
f - l (x )
=
/
=
f(x ) = 4x + 2
51.
=
=
=
=
-�
f(x) = x3 -1 =X3 -1 x = i -1 Inverse i =x+l y=rx+i rl(x) =rx+i Verifying: f { !-I (x) ) = f( rx+!) = (rx+!t -1 =x+l-l =x r l (J(x» ) = rl ( x3 -1) = 4(x3 -1 ) + 1 =.(1 =X
f(x) = x2 + 4, x � 0 y = x2 +4, x � O x = i + 4, y � 0 Inverse y 2 = x-4, x � 4 y =.)x-4, x � 4 rl(x) =.)x-4, x�4 Verifying: f(rl(x» ) = f (.)x-4 ) = (.)x-4 r +4 =x-4+4 =x rl (J(x» ) = rl ( X2 +4 ) = �( X2 +4 ) -4 =H =l xl =x, x � 0 Domain off Range of r l {x I x � O} or [0, (0) Range off Domain of rl {x I x � 4} or [4, ) =
Y
=
=
(0
y 8
=
f(x) =
x2 +
X �
//
0
/
/
/
4,
/
/ f- l (x) - 2 /7 /�2 /
,Y
= X
--Ix
-4
///
=
x
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Section 6.2: One-to-One Functions; Inverse Functions
55.
Verifying: x +1 f V-I (X) ) = f e x ) 2x+1 _ 2 x l·x
f(x) = -x4 y=x4 x = y4 Inverse xy=4 y = -4x rl(x) =�x Verifying: f(rl(x) ) =f (�) = � = 4 {�) =X x -
x 2x+1-2x x = -1 = X l) rl (J(x)) = f -I (_ x-2 1 2 (_ x-2 )+1 x-2 ( 2(�)+1 }X- 2) 1 )(X-2) (_ x-2 2+ (x-2) = 1 = -x1 =x Domain off Range of f- I All real numbers except 2. Range off Domain of f- I All real numbers except O.
x Domain off Range of f -I All real numbers except O . Range off Domain of f-I All real numbers except =
=
=
o.
=
=
=
X
/
57.
5
=
=
x=o Y
-5
1 f(x) = x-21 y= - x-21 x = y-2 Inverse xy-2x = 1 xy = 2x+1 2x+1 y=- x 2x+1 rl(x) = x
x=2 'Y=x I / I / / I / I / /
----Y= 2 �====tr�����X y = O /
--
/
/
/
/
/
/
/
/
307 © 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently exist. No portion of thi s material may be reproduced, in any form or by any means , without permi ssion in wri ting from the publisher.
Chapter 6: Exponential and Logarithmic Functions
59.
f(x) = 3+x2 2 y = -3+x2 x = -3+ y Inverse x(3+ y) = 2 3x+xy = 2 xy = 2-3x 2-3x y=- x 2-3x r l(x) = x Verifying: 2-3X 2 f(rl(x) ) = f (-x ) = 3+ 2-3x x 2x 2·x (3+ 2 �3X )x 3x+2-3x = -2x2 = X 2) 2 3( 2 rl (I(x)) = rl (_3+x_) = 2� 3+x ( 2-3(�)}3+X) C�J(3+X) 2(3+x)-3(2) _ 6+2x-6 2 2 2x =-2 =X Domain off Range of f-I All real numbers except -3 . Range off Domain of f-I All real numbers except
61.
_
=
=
=
=
=
=
=
f(x) = � x+23x y = - x+2 x=� y+2 Inverse x(y+2) = 3y xy+2x =3y xy-3y = -2x y(x-3) = -2x -2x y=x-3 -2x rl (x)= x-3 Verifying: f(rl(x) ) = f( ::� ) 3 (�) _ (3 ( �)}X-3) -2x +2 ( -2x +2 )(X-3) x-3 x-3 -6x -6x - = --=x ----2x+2x-6 -6 rl (I(x)) = rl (� x+2 ) -2 (� ) ( -2 (� ) }X+2) 3x - (--3 3X --3 x+2 x+2 ) (x+2) -6x -6x --3x-3x-6 =-=x -6 Domain off Range of f-I All real numbers except -2 . Range off = Domain of f-I All real numbers except 3.
O.
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Section 6.2: One-to-One Functions; Inverse Functions
63.
f(x) = � 3x-l 2x y = 3x-l 2y Inverse x = -3y-l 3xy-x = 2y 3xy-2y =x y(3x-2) = x x y = -3x-2 x rl(x) = _ 3x-2_ Verifying: x ) = 2 (6) f ( rl(x) ) = f (x ) -1 3x-2 3 (_ 3x-2 ( 2 (6) ) (3X-2) ( 3 ( 3x2x� 2 ) -1)2x(3x -2) ---- =-=x 3x-(3x-2) 2 2x M rl (J(x)) = f (� 3x-l ) = 3 (� 3x-l ) -2 (� 3x- l ) (3X-l) (3 ( 3��2x1 ) -2) (3X-l) 3(2x) -2(3x -1) 2x = -2x =x ---6x-6x+2 2 Domain off Range of f- I All real numbers except -31 . Range off Domain of f-I All real numbers except -23 .
65.
=
=
f(x) = 3x+4 2x-3 3x+4 y = -2x-3 3y+4 x = -2y-3 Inverse x(2y-3) = 3y+4 2xy -3x = 3y + 4 2xy -3y = 3x+4 y(2x-3) = 3x+4 y = 3x+4 2x-3 3x+4 rl(x) = 2x-3 Verifying: 3X+4 ) + 4 3 x 2x-3 ( ) = f ( r\x) ) = f (x+2 2 ( 3X+4 ) _ 3 2x-3 ( 3 (�) +4}2X-3) (3(3x2 ( �:+ 4):�+)4(2x-3}-3)2X-3) 2(3x + 4) -3(2x -3) 9x+12+8x-12 -- = 17x -6x+8-6x+9 17 =x 3X+4 ) +4 3 ( 3X+4 rl (J(x)) =f-I ( 2x-3 ) = 2 2x-3 ( 3X+4 2x-3 ) _ 3 (3 (�) +4}2X-3) (3(23x( �:+ 4):�+)4(2x-3}-3)2X-3) 2(3x + 4) -3(2x -3) 9x+12+8x-12 17x =x --= 6x+8-6x+9 17 Domain off Range of f-I All real numbers except -23 . Range off Domain of f-I All real numbers except -23 . =
=
=
=
=
=
309
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Chapter 6: Exponential and Logarithmic Functions
67.
l(x) = 2x+3 x+2 2x+3 y= - x+2+ 3 2y x = y+2 Inverse xy + 2x = 2y + 3 xy -2y = -2x+3+3 y(x-2) = -2x y= -2x+3 x-2 -2x+3 rl(x) = x-2 Verifying: -2X+3 ) +3 ( 2 +3 x-2 1(1 -1 (x) ) = 1 ( -2X x-2 ) = -2x+3 x-2 +2 ( 2 ( -:�� 3 ) +3}X-2) ( -2X+3 +2 ) (X-2) x-2 +3(x-2) 2(-2x+3) -2) -2x + 3 + 2(x -4x+6+3x-6 -x =x ----- = -2x+3+2x-4 -1 2X+3 ) + 3 ( 2 � rl (l(x))= rl ( 2X+3 x+2 ) = 2x+3 x+2 _ 2 (�) +3}X+2) ( -22X+3 ( x+2 -2) (X+2) -2(2x + 3) + 3(x + 2) 2x + 3 -2(x + 2) -4x-6+3x+6--- = --x =x -2x+3-2x-4 -1 Domain ofl Range of I- I All real numbers except -2. Range ofl Domain of I-I All real numbers except 2.
69.
x2 -4- ' x > O I(x) = 2x2 2 x -4- ' x > O y =-2x2 X = -/2y--42- ' y > O Inverse 2x/ = / -4, x < -21 2xy 2 - / =-4, x< -21 / ( 2x-l) = -4, x< -21 / ( 1-2x) = 4, x < -21 4 ' x < -1 y2 = 1-2x 2 1 4 y = �1_ 2X ' x <-2 2 ' x < -1 y = .Jl-2x 2 1 2 x < I (x) = � ' 1-2x 2 Verifying: -4 2 J (k )� I ( I I(X) ) = / ( .J1-2x 2 ), 2 ( .J1-2x 4 _ -4 ) (1-2X) 4 4 (_ _ 2_x----;�-_ 1 1-2x +-:_ _ 4 4 )) (1-2X) 2 (_ 1-2x ) ( 2 ( _ 1-2x 4-4(1-2x) -4-4+8x- =-=X 8x 2(4) 8 8 I- I (I(x)) = rl [ X:�4 ) = (2X2 -4 1-2 -2x2 ) 2 2 -I
_
=
_
=
=
=
310 © 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.
Section 6.2: One-to-One Functions; Inverse Functions
Domain ofl Range of I - I {xl x> O} or (0, Range ofl Domain of I- I =
=
00
83.
)
=
Because the ordered pair (-1, 0) is on the graph, /(-I) = O . Because the ordered pair (1,2) is on the graph, 1(1) = 2 . Because the ordered pair (0,1) is on the graph, r l(1) = O . Because the ordered pair (1,2) is on the graph, r l(2) = 1. Since 1(7) 13 , we have r l (13) = 7; the iwhen nput ofthetheoutput function is function the outputis ofthethine putinverse of the of the inverse. nce theanddomai of a function is the range the n iSinverse, the nrange of the function is the ofdomai of the inverse, we get the following for I- I : Domain: [-2,00) Range: [5,00) nce theanddomain of a function is the range the n iSinverse, the range of the function is the ofdomai of the inverse, we get the following for g-I : Domain: [0,00) Range: all real numbers Since I (x) is increasing on the interval (0,5), it is one-to-one on the interval and has an inverse, rl (x) . In addition, we can say that r l (x) is increasing on the interval ( I ( 0), I (5)) . I(x) = mx+b, m "" 0 y= mx+b x = my + b Inverse x-b= my y = -m1 (x-b) rl(x) = �m (x-b), m "" O
7 1 . a.
85.
b.
77.
79.
81 .
�
�
(x)
87. a.
d.
75.
�
�
c.
73.
If { a ,b) is on the graph off, then (b, a ) is on the graph of I-I . Since the graph of I-I lies in quadrant I, both coordinates of (a,b) are positive, which means that both coordinates of (b, a) are positive. Thus, the graph of I-I must lie in quadrant I. Answers may vary. One possibility follows: I(x) = / x / ' x 0 is one-to-one. Thus, I(x) = x, x 0 y = x, x 0 rl = x, x 0 d = 6. 9 7r -90. 3 9 d + 90. 3 9 = 6. 9 7r d---+90. 3 9 = r 6. 9 7 we would write Therefore, 39 r( d ) = d +90. 6. 9 7 r (d(r)) = (6. 9 7r-90. 3 9)+90. 3 9 6.---97r + 90.6.3997 -90.39 -- = 6. 97r r 6.97 6.97 d d ( r (d )) = 6. 9 7 ( +90. 3 9 ) _90. 3 9 6.97 39 = dd +90. 3 9-90. = 39 "" 56. 0 1 r(300) = 300+90. 6.requi 97 red to stop was 300 feet, Ifthe distance the speed of the car was roughly 56 mph. 6 feet 72 inches W (72) = 50+ 2. 3 (72-60) = 50 + 2. 3 (12) = 50 + 27.6 = 77.6 The ideal weight ofa 6-foot male is 77.6 kg. = 50+ 2. 3(h -60) -50 = 2. 3h -138 +88 = 2. 3 h +88 2. 3 we would write Therefore, h(W) = 2.+883
=
b.
--
=
c.
89.
a.
b.
=
W
W
W
W
=h W
311 © 2008 Pearson Education , Inc . , Upper Saddle River, NJ. A l l rights reserved. Thi s material i s protected under a l l copyright laws a s they currently exist. No portion of this material may be reproduced, in any fonn or by any mean s , without permission in wri ting from the publisher.
Chapter 6: Exponential and Logarithmic Functions
c.
» )+88 h( W(h» )= (50+2. 3(2.h-60 3 +88 2.3h 50+ 2. 3 h -138 = 2. 3 = 2. 3 = h W (h(W») =50 + 2. 3 ( W+88 -60) 2.3
= 5 0 + W + 88 - 138 = W
d.
91.
a.
b.
c.
0
h(80) = 8 +88 = � 73.04 The heiofght80ofkga malis roughly e who is73atinhisches.ideal weight From the resttheriction given in the problem statement, domai n i s { g I 30,650 ::; g ::;74,200} or [30650, 74200]. T(15, 1 00) = 4220 + 0. 2 5(30, 650 -30, 650) =4220 T(74,200) = 4220 + 0. 2 5(74, 200 -30, 650) = 15,107. 5 Since T is linear and increasing, we have that the range is {T 1 4220::; T ::; 15,107. 5 } or [4220, 15107. 5 ] . T = 4220 + 0. 2 5 ( g -30, 650) T -4220 = 0. 2 5(g-30,650) T -4220 = g -30, 650 5 T -4220 +300.2' 650 =g 0. 2 5 Therefore, we would write g(T) = + 30 650 Domain: {T 1 4220::; T ::; 15,107. 5 } Range: { g 1 30,650::; g ::; 74, 200} The graph of is symmetri c about the y-axis. Sithencerock t represents the number of seconds after begins to fall, we know that t ;;: : 0 . The graph i s stri c tl y decreasing over i t s domain, so it is one-to-one. H = 100 -4. 9t2 H + 4.9t2 = 100 4.9t2 = 100 - H t2 = t = )'OO-H 2.3
2.3
4220 T0.25
93.
a.
b.
�
c.
95.
Therefore, we would write t ( H) = lOO4.-9 H . need tthe;;:: 0pri) ncipal square (rootNote:sincewe weonlyknow H (t ( H)) = 100 -4 . 9 ( l O��H J ) = 100_4. 9 ( 100-H 4.9 = lOO - l OO + H =H 9t 2 ) t(H(t)) = 100- ( 100-4. 4. 9 = )4. 9t 2 = J? = t ( since t ;;:: 0) 4. 9 2.02 t(80) = lOO4.-80 9 Itfalwil 80l take meters.the rock about 2. 02 seconds to �
ax + b I (x) = ex +d ax + b y = - ex + d ay + b x ey +d x(ey + d) ay + b exy + dx ay + b exy - ay = b - dx y( ex - a) b - dx b - dx y = - ex - a +b F I (X) -dx ex - a I I-I
= -- Inverse = = =
'
H
97. 99.
100 - H 4.9
4.9
= ax + b = ---dx + b . Now, = prOVl' ded that -ex + d ex - a This is only true if a = - d . Answers will vary. No, not every odd functi on is one-to-one. For 3 example, I(x) = x - x is an odd function, but it is not one-to-one.
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Section 6.3: Exponential Functions
Section 6.3 1.
3.
43 = 64 ; 82/3 = (.�8r = 22 = 4 ; T2 = �3 =..!.9 . False. To obtain the graph of = (x -2)3, we would shift the graph of = x3 to the 2 units. True False 32.2 "" 11. 2 12 32.23 "" 1 1 . 5 87 32.236 "" 11. 664 3.Js "" 11.665 23. 14 "" 8. 8 15 23. 1 4 1 "" 8. 821 23. 1 41 5 "" 8. 824 211 "" 8. 825 3.f·7 "" 21. 2 17 3.142.7 1 ",, 22.2 17 3.1412.7 1 8 "" 22.440 22.459 "" 3.320 e-O·85 "" 0.427 y
y
5. 7. 9. 11.
23.
right
a.
b.
d.
25.
a.
-
b. C.
d.
1 5.
a.
b. C.
d.
1 7. 1 9.
21.
ne ""
y
= f (x)
27.
f (x + l) f (x)
�3 = 2 -1 3 12 = 2 0 6 6 18 = 3 1 12 2 12 23 3018 since the ratio of an exponenti Not not constant. ve termsalisfunction consecuti -
H (x + H (x )
--
a
el . 2
x
Y = H (x)
a
C.
13.
1) 1 =4 -1 -41 -( 1 1 4) i1 =4 0 1 �4 = 4 1 4 64 =4 2 16 16 3 64 siwintceh the=rati4 . oSoofthe ails function an exponenti Yes, constant terms consecutive base is 4. f (x + l) x Y = f (x) f (x) -1 23 ( 3 3/ 2) =3 · -23 =2 �3 =2 0 3 .!3. = 2 1 6 6 24 =2 2 12 12 3 24 on siwintceh the=rati2 . oSoofthe aisl functi an exponenti Yes, constant terms consecutive base is 2. = H (x) HH(x(x)+ l) x i2 =2 -1 2 -46 = -32 0 4 231 1068 on since the ratio of an exponenti Not not constant. termsalisfuncti consecutive x
-
29. 31.
y
B
D
313
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Chapter 6: Exponential and Logarithmic Functions
Horizontal Asymptote: y
33. A 35. E 37.
1 Using the graph of y 2x , shift the graph up 1 uniDomai t. n: All real numbers Range: {y \ y > O} or (1, 00) Horizontal Asymptote: y 1 f(x)
==
==
°
2X +
==
==
43.
f(x)
==
TX
-2
Using the graph of y 3x , reflect the graph about theDomai y-axis,n: Alandl realshiftnumbers down 2 units. Range: {y \ y > -2} or (-2, 00) Horizontal Asymptote: y -2 ==
==
39.
f (x) 3x-1 ==
Using the graph of y 3x , shift the graph right 1 uniDomain: t. All real numbers Range: { y\y> O} or (0, 00) Horizontal Asymptote: y 0 ==
x
y = -2
==
45.
-5
2 + 4x-1 Using the graph of y 4x , shift the graph to the riDomain: ght one uniAllt real and numbers up 2 units. Range: {y \ y > 2} or (2, 00) Horizontal Asymptote: y 2 f(x)
==
==
-3
41.
-1
f (x) 3 ==
(0, i)
x
3
( 1, ��)
{�J
-
Using the graph of y (�J vertically stretch the graph bymula tfactor ofy-coordi 3. That is, for each point on the graph, i p l y the Domain: All real numbers nate by 3. Range: { y \ y > O} or (0, 00) ==
y ' 8
==
\
'
-4
-2
6
x
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Section 6.3: Exponential Functions
47.
2 Using the graph of y = 3x , stretch the graph hori zontaln: Ally lbyreala factor of 2, and shift up 2 units. Domai numbers Range: > 2} or (2, Horizontal Asymptote: y = 2 f(x) =
+ }' 1 2
{y I y
00
)
y
x
8
-
53.
x
-2
2
f(x) = S - e-x
Using the graph of y = eX , reflect the graph about the y-axis, reflect about the x-axis, and shift up 5 units. Domain: {yAlI lyrealS}numbers Range: or ( 5) Horizontal Asymptote: y = 5 <
49.
f(x) = e-x
Using the graph of y = eX , reflect the graph about tDomai he y-axis.n: All real numbers Range: {y I y > O} or (0, ) Horizontal Asymptote: y 00
=
y
y
8
°
x
55. x
-2
51.
-00 ,
2 Using the graph of y = eX , reflect the graph about thereflecty-axis,aboutstretch horizontal ly byupa factor of2, the x-axis, and shift 2 uni t s. Domai n : All real numbers Range: {y I y 2} or ( 2) Horizontal Asymptote: y = 2 f(x) = - e-x /2
<
f(x) = ex+ 2
Using the graph of y = eX , shift the graph 2 units toDomai the left.n: All real numbers Range: {y I y > O} or (0, ) Horizontal Asymptote: y 00
=
-00 ,
y
5
°
- - - - - - -
- - - - - - -
x
y
=2
-5
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Chapter 6: Exponential and Logarithmic Functions
57.
7x = 7
3
67.
Wesideshaveof thea single termTherefore, with the same basesetontheboth equation. we can exponents equal to each other: The solution set is { } x=3.
3 .
59.
-3x + 42 = 4x
The solution set is { } 42 = 7x
rx = 1 6 4 =2
6=x
rX
6 .
The solution set is { } -x = 4
x = -4
69.
-4 .
61. (.!.)X GJ (�J GJ
3x
2 . 71.
= 22
x = -1
( 2 )X
2
( )
4X . 2 X = 1 6 2 2 2 · 2x = 24 2 2
22 x . 2X = 2 2 2 x+x =
3 x=2
28
28
( )( )
x 2 + 2x = 8
The solution set is {%} .
x 2 + 2x - 8 = 0 x+4
x-2 =0
or 2 The solution set is { } x+4 = 0
3 J" = 9 x 3 3x = 3
x = -4
( 2 )X
x-
=0
x=2
-4, 2 .
3 3 X = 3 2x 3 x = 2x
73.
3 eX = e x + 8
x = 3x + 8
The solution set is { } -2x = 8
3 x - 2x = 0
x = -4
x x2 - 2 = 0
or
x+1 =0
-1, 7 .
2x = 3
x=0
= 3 6.<
x=7
4
)
-7
( t
or The solution set is { }
2x - 1 = 2
(
7
x-7 = 0
The solution set is { }
65.
2
-
= 27 2 x 3 x = 3
( )( )
x=2
=
2
-7
x 2 - 6x - 7 = 0 x-7 x+l =0
=
2 2x-1 2 2 .< -1
2
x 2 - 7 = 6x
52
6 3.
3x 3x
25
5
( f ( f 8-x + 1 4 = 1 6 x 3 +1 4 4 2 = 2 3 2 - x+ 42 = 2 4 x
-4 .
x2 - 2 = 0 x2 = 2
Ji The solution set is {-.J2, 0, Ji } . x=±
31 6 © 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of thi s material may be reproduced, in any form or by any means , without permi ssion in writing from the publisher.
Section 6. 3: Exponential Functions
75.
ex
2
e3x
= = 2 = X2 = 3x-2 X2 -3x+ 2 = 0 (x-2){x-l) = 0 x -2 = 0 or x-I = 0 x=2 x=1 The solution set is {I, 2} . 1(4)= 24 =16 The point (4,16) is on the graph off I(x) =-161 2x =�16 2x = �24 2x = 2-4 x =-4 The point ( -4, 116 ) is on the graph off g ( -1) 4-1 + 2 = .!.4 + 2 = 2.4 The point ( -1, %) is on the graph of g. g(x) = 66 4x +2 = 66 4x = 64 4x = 43 x=3 The point (3,66) is on the graph of g. H(-2) =3 (�r2 -2 =3(2/ -2 =10 The point (-2,10) is on the graph of H. eX
eX
2
e3 x e
2
b.
e2
.
e3 x - 2
-
b.
a.
8
3
77. a.
79.
H(x)= --13 3 (�J _ 2 = _ 1: 3 ( �r = � (�J = 1 (�J 2 (�Jx == (3�J The point ( 3, - 1:) is on the graph of H.
83.
8 5.
=
87.
b.
If 4x = 7 , then ( 4x f = T2 4-2x = �7 2 4-2x = �49 If TX = 2 , then (rTf = r2 32x = �22 3 2x = .!.4 We need a function of the fonn I ( x) = k· with > 0, I . The graph contains the points ( -l, �) . (0,1) , (1,3) , and (2,9) . In other words, 1(-1)=-31 , /(0)= 1, 1(1)=3, and 1(2)=9. Therefore, 1 (0) = k· 1 = k· 1 = k·l 1=k and 1 (1) = 3 Let' s use = 3, = 1. Then I (x) = 3x• Now we need to verify that this function yields the other a
a
aP· x
*"
a dO) aD
8 1 . a.
a P-( I )
= aP a
P
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,
Chapter 6: Exponential and Logarithmic Functions
known points on the graph. 1 (-1) =
TI
= .!.3 ;
9 3.
1(2) = 32 = 9
89.
So we have the function 1 ( x) = 3x . need a function of the form 1 ( x) = k . aP'x , with a > 0, a 1 . The graph contains the points (-1' - �} (0,-1), (1, - 6), and (2, -36) . Inother words, 1(-1) = - "61 ' 1 (0) = -1 , 1 ( 1)= - 6 , and
We
{
I
y
0.5
f
"*
1(2) = -36 . 1 (0) = k . a P '( O) -1 = k · ao -1 = k · l -1 = k 1(1) = -a P - ( l) -6 = -aP 6 = aP a = 6, p = 1 . I (x) = _6x
Domain: ( ) Range: {Y I - l :O:; y < O} or [-1, 0) Intercept: (0, -1) -00 , 00
Therefore,
95.
and
97.
•
1(2) = -62 = -36
99.
So we have the function 1 (x) = _6x . 1 ( x ) = { e x if x < 0 eX if x 0 y 4
101.
x
p(x) = 1 6, 630(0.90 )X a. p(3) = 16, 630(0.90) 3 � $12, 123 b. p(9) = 16, 630(0.90)9 � $6, 443
D (h) = 5e-O.4 h D (I) = 5e-0.4 ( I ) 5e-O .4 � 3.35 1 3.35 D ( 6) = 5e-O.4 ( 6) = 5e-2 .4 � 0.45 6 0.45 F (t) l _ e-o . 1t a. F(10) = I - e-O·1 (1 0) = 1 - e-1 � 0.632
After hours, mil igrams wil be present. milligrams After hours, milligrams wil be present. =
b.
-2
c.
- 00 , 00
�
00
oflight oflight
=
�
Domain: ( ) Range: {Y I Y I} or [1, Intercept: (0, 1)
p(n) = 100(0.97) n p(10) ::: 100(0.97io � 74% a. b. p(25) 100(0.97)25 � 47% =
Let's use Then Now we need to verify this function yields the other known points on thethatgraph. I ( -1) = _6-1 = -.!.6 91.
-e: �fx < 0 I (x) = -e x x � 0
)
The probabi lity thatPMa iscar0.632. will arrive within 10 minutes of 12:00 F (40) = I - e-o. I ( 40) = 1 - e-4 � 0.982 The probabi lity thatPMa iscar0.982. will arrive within 40 minutes of 12:00 As t � oo, F (t) = I - e-O·1t � 1 - 0 = 1
31 8 © 2008 Pearson Education , Inc. , Upper Saddle River, NJ. All rights reserved. Thi s material i s protected under all copyright l aws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publi sher.
Section 6.3: Exponential Functions
d.
Graphing the function: 1
�
V'
o
F
r =![,- e{H] r, = \2: [I_ e-(',")o ' ] =12 [I- e"'] = S .4 1 4 amperes after 0. 3 second
3.
40 (7) "" 0. 5 0 , so about minutes are needed for the probability to reach 50%. o
e.
1 07 .
7
1
V = . 5 0 3 � 1�7
1 03 .
�
1
40
20X e-20 P(x) =-- x! 1 0 P(15) = 20 15!5 e-2 "" 0. 0516 or 5.16% The probabi5: 00litPMy thatand156:cars00 PMwillisarri5.16%. ve between P( 20) = 20220!0 e-20 "" 0. 0 888 or The probabi5: 00litPMy thatand206:cars00 PMwillisarrive between
b.
,
t
- I
a.
b.
amperes after 0. 5 second = \�O[I- e{',")'] = 12[I- e-'] = 103 76 amperes after 1 second As � co, e-(�} � O . Therefore, as, 120 [ 1- e-( 150 ) 1 � 12[1-0]=12, t � co, JI =10 which amperes.means the maximum current is 12 See the graph at the bottom of the page. = I �O [I- e-(�)o' 1 = 24 [I- e" "] "" 3. 343 amperes after 0. 3 second = I�O [ I_ e-(�)" ' 1 = 24 [I- e" " ] "" 5. 309 amperes after 0. 5 second = I�O [I_ e{M] = 24[I- e" ' ] "" 9.443 amperes after 1 second As � co, e-(�} � O . Therefore, as, t �co, 11 = -1 205- [l- e-( 150) 1 � 24[1-0]=24 , which amperes.means the maximum current is 24 See the graph on the next page.
c.
8 . 88%
d.
8 . 88%.
1
,
1
a.
b.
1
c.
e.
,
,
t
- I
f.
319 © 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved . Thi s material is protected under all copyri ght laws as they currently exist. No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in writing from the publi sher.
Chapter 6: Exponential and Logarithmic Functions
y 24
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
22 20 18 16
12(/ )
14 12 10
- - - - - - - - - - - - -
(0.5. 7.585)
(0.3, 5.4 1 4)
�
4 2
o
109.
(1
.
0 . 10 . 376 ) - - - -
( 1.0, 9.443)
1 1 (/) .
=
12(1
-
e
-
3.0
2.0
1 1 5.
S
inh x= -l ( x - ) f(-x) = sinh(-x) =-21 ( e
a.
e-
=_
n
b.
e
x
�( x _ e
n
-x
-e
e
x
)
e-
x)
= -sinh x =-f(x) Therefore, f(x) = sinh x is an odd function. Let >-; = � ( ex _e x ) -
.
3 .5
x
1 13.
O .5
- - - - - - - - - - - - - - - - - - - - - - - - -
0. 1 0.2 0 . 3 0.4 0.5 0 . 6 0.7 0.8 0.9 l.O
1 1 1 2+ -2!1 +-+-+ 3! 4! . . +-n! 1 = 4; 2+ -2!1 + -3!1 + -"" 4! 2.7083 = 6; 2+ -2!1 + -3!1 + -4!1 + -5!1 + -6!1 "" 2. 7 181 = 8; 2+ -2!1 + -3!1 + -4!1 + -5!1 + -6!1 + -7!1 + -8!1 "" 2.7 182788 1 1 1 1 + -1 + -1 + -1 + -1 + -1 n =10; 2+-+-+-+2! 3! 4! 5! 6! 7! 8! 9! 10! "" 2. 7 182818 "" 2. 7 18281828 f(x) = aX f(x+ h)- f(x) ax+h _ a h hh aX ( a -1 ) hh = ax [ a h-I ) f(x) = aX f(-x) = a-x =_a1X =_f(x)1_
24( \
- e - 2f )
n
111.
=
- 3.5
1 1 7.
f(x) = 2(21X ) +1 f(1) = 2(2 ) + 1 = 2 2 + 1 = 4 + 1 = 5 f(2) = 2(22 ) + 1 = 24 + 1 = 16 + 1 = 17 f(3) = 2(23 ) +1 = 28 +1 = 256+1 =257 f(4) = 2(254 ) +1 = i6 +1 = 65,536+ = 65,537 2(2 ) + 1 = 232 + 1 = 4,294,967,296+ 1 f(5) == 4,294,967,297 = 641x6,700,417 Answers will vary. 1
1 1 9.
320 © 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material i s protected under all copyright l aws as they currently exist. No portion of this materi al may be reproduced , in any form or by any means , without permission in writing from the publisher.
I)
Section 6. 4: Logarithmic Functions
121.
()
Given the function J x =
a
X
with
,
>1,
a
If x > 0 , the graph becomes steeper as a increases. If x < 0 , the graph becomes less steep as increases.
Section 6.4 1.
3x - 7 � 8
-
{ l
3} .
The solution set is x x �
3.
-x-I
x+4
>0
()
J x =
x-I
x+4 Jis zero or undefined when x = l or x = -4 . Interval
(-00, -4)
(-4, 1)
(1, 00)
Test Value
-5
0
Value ofJ
6
Conclusion
positive
-
{ l
The solution set is x x
(
<
interval notation, -00, -4 5.
(� } ( ' -1
y
) ( 1, 00 ) .
29.
log l 1 6 = - 4 since I 2
31.
r.;:: 1 . 11 r.;:: log l O ...,, 1 0 = - smce 1 0 2 = ...,, 1 0 . 2
33.
10g .j2 4 = 4 since
35.
in J; =
37.
J(x) = in(x - 3) requires x - 3 > 0 .
39.
6 positive
}
s ince
13.
2x = 7 . 2 is equivalent to x = log 7 . 2 . 2
The domain of
41.
F
()
J X = 3 - 2 10g 4
.:: > 5 2 x > 10
43.
1 9.
loga 3 = 6 is equivalent to
21.
log3 2 = x is equivalent to 3x = 2 .
a
6
( )
I J(x) = in __ x+l P
3
log 8 = 3 is equivalent to 2 = 8 . 2
(.fi) 4 = 4 .
{ x l x > 3}
( )
or 3, 00 .
{ l
o} .
-5
requires
is x x :t=
[% ] { l
%
} (
-5 > 0 .
)
The domain ofJis x x > l O or 1 0, 00 .
= 8 is equivalent to x = in 8 .
1 7.
= 24 = 1 6 .
2
or, using
2 = 1 . 6 is equivalent to 2 = loga 1 .6 .
a
4
F(x) = log x 2 requires x 2 > 0 . 2 x 2 > 0 for all x :t= 0 .
•
11.
(�r
.!. since e l / 2 = J; .
':: - 5 > 0 2
= loga x , then x = aY
5 2 = 25 .
The domain of J is
a
9 = 3 2 is equivalent to 2 = log3 9 .
© 2008 Pearson
log5 25 = 2
) ( , l)
1, 0 ,
9.
e
27.
x>3
u
False. If
1 5.
log 1 = 0 since 2° = 1 . 2
e
x-3 > 0
1
-4 or x > I
7.
X
25.
2
1 -4 negative
=4 .
in 4 = x is equivalent to
a
2x
5x � 1 5 x�3
X
23.
=3.
1 requires __ > O . x+l
1 _ is undefined when (x) = _ x+l
x
Interval
(-00, - 1)
(-1, 00)
Test Value
-2
0
Value ofp
-1
1
Conclusion
negative
positive
{ l
= -1 .
} (
)
The domain ofJis x x > - I or - 1, 00 .
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exist. No portion of this material may be reprod uced, i n any form or by any means, without permission in writing from the publisher.
Chapter 6: Exponential and Logarithmic Functions
45.
g(X) =IOgs ( X ; I )
X
requires
;
I
59.
>O.
= x x+ 1 is zero or undefmed when - l or x = O .
p (x)
x
=
Interval
(-00, - 1)
( - 1, 0)
(0, 00)
Test Value
-2
1 -2
1
Value ofp
-1
-1
2
negative
positive
2 positive
Conclusion
61.
}
The domain of g is x x < - l or x > O ;
47.
{ l
(-oo,-I)u( O,oo) . f(x) = .J
.
{(x) .
=
(l)X 2
in x requires In x � 0 and x > 0
eO
In x � O x�
x�1
The domain of h is x x �
{ l
I}
or [ 1, (0
).
-3 6 3.
_ ---In
51.
53 .
55.
65.
10
3 "" 30.099 0.04 1n 4 + ln 2
log 4 + log 2
"" 2 .303
2 ln 5 + log 50 log 4 - In 2
57. If the graph of f(x) (2, 2) , then f(2)
= a2 =
log" 2
2
=
loga 2
=
) og l X 2
D
67.
A
69.
E
71.
f(x) b.
= loga x
=
B
a.
",, -5 3 . 99 1
(2. - 1 ) f- I (x)
=
In(x + 4)
Domain: (- 4,
(0
Using the graph of 4 units to the left. X = -4 I I I I I I I
contains the point
2 . Thus,
2
) y
=
In x , shift the graph
y 4 X
a = ±J2 Since the base a must be positive by definition, we have that a = J2 .
-4 c.
Range: (-00,
(0
)
Vertical Asymptote : x
=-4
322 © 2008 Pearson Education , Inc . , Upper Saddle River, NJ . All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of thi s material may be reproduced, in any form or by any mean s , without permission in writing from the publisher.
Section 6. 4: Logarithmic Functions
f(x)= ln(x+4) y = ln(x+4) x = In(y + 4) y+4= ex -4 rl(x) = -4
d.
y
=
e
e
f.
Inverse
Range off (-00, 00)
f.
Shift the graph of y
=
y 5
e
X
down
4
y = O
units .
75 .
-
a.
...,
- �)
Using the graph of y y 5
= In x,
shift up
7
-2
Domain: (0, 00)
Using the graph of y
shift down y 2
Domain: (0, 00)
2
3
f(x) = 2+ln x y = 2+ln x x = 2+ln y x-2 = ln y = r l(x) = e
x=
© 2008
!. , and
2
3
8
x = ()
c.
Range : ( - 00 , 00)
d.
f(x) = In(2x)-3 = In(2x)-3 x = In(2y)-3 x+3 = In(2y) 2y = y=-21 e r 1 ( x) =!.2
°
Vertical Asymptote :
x=
°
y
e
Inverse
x +3
x+ 3
Inverse
e
x-2
x 2 e -
e.
compress the
'0 . )
x=o
Vertical Asymptote :
X
( 1 , - 2.3)
-5
y
2
X
units.
-
Range: (-00, 00)
= ln x,
units.
-
d.
, shift the graph
graph horizontally by a factor of
y = -4
5
f(x) = 2+ln x
c.
X
f(x) = In(2x) -3
b.
b.
e
X
e.
a.
units to the right. y 8
=
X
x
73.
Using the graph of y
Range off (-00, 00)
X +3
e.
Range off (-00, 00 )
f.
Using the graph of
y=
e
X
, reflect the graph
about the y-axis, and reflect about the x-axis.
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exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publi sher.
Chapter 6: Exponential and Logarithmic Functions
79.
1
f(x) = - log ( 2x ) 2 Domain: a.
b.
(0, 00)
Using the graph of
y
= log x , compress the
graph horizontally by a factor of -
77.
3
b.
Domain: (4,
00
)
Using the graph of
y
(5, n
= log x , shift the graph
�
4 units to the right and 2 units up. x = 4 y 2
i� I
10
-2 c.
x
d.
-2 c.
d.
(
-00, 00
)
f(x) = log(x - 4) + 2
y
= log( x - 4) + 2 x = log(y - 4) + 2 x - 2 = log(y - 4)
y_4=10x-2 y l Ox-2 r l 10x-2 =
+4
(x) =
+4
e.
Range off
f.
Using the graph of
(
-00, 00
)
Vertical Asymptote: x =
0
1 y '21 ( ) 1 ( y) ( ) 102x ,!,, 10 2x r l ,!,, 102x ( ) y lOX , f(x) = '2 log ( 2x ) log 2x
Inverse
Y=
2
(x) =
)
y lOX , =
e.
2 Range off
f.
Using the graph of
- 00 , 00
=
compress the
graph horizontally by a factor of .!. , and 2
shift the graph
compress vertically by a factor of y 7
4
4
x
y = O
-4 2008 Pearson
-00 , 00
x
2y =
Inverse
2 units to the right and 4 units up. y 6
©
(
10
x = - log 2 2 2x = log 2Y
Vertical Asymptote: x = 4
-
x = o
Range :
0. 0)
=
Range :
, and
2
compress vertically by a factor of .!. . 2 Y 2
f(x) = log ( x - 4 ) + 2
a.
.!.
-5
-3
0 , 5) (O, n
5
.!. . 2
x
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exist. No portion of this material may be reproduced, in any form or by any means, without pennission i n wri ting from the publisher.
Section 6. 4: Logarithmic Functions
81.
f(x) = 3 + log 3 ( x + 2 )
a.
b.
Domain: (-2,
00
)
Using the graph of y = log 3 X , shift 2 units to the left, and shift up 3 units. x = -2 ( 1 Y 1 , 4) 1 .) _
..,
1
( - 1 , .J) 1
-
-5
c.
d.
5
c.
x
d.
1-5
Range:
(
)
Horizontal Asymptote: y = -3 f(x) = eX +2 - 3 y = eX +2 - 3
y = 3 + log 3 ( x + 2 )
r l (x) = In(x + 3) - 2 e.
Inverse
f.
x - 3 = log 3 ( y + 2 ) 3 y + 2 = 3 x3 y = 3 x- _ 2 3 rl (x) = 3 x - - 2 Range off
f.
Using the graph of y = 3 x , shift 3 units to
)
I I
X
x
------
f(x) = eX +2 - 3 a.
Domain:
b.
Using the graph of y = eX , shift the graph
-00 , 00
the left, and shift down 2 units . I y 5
I
I
( - 2, -2)
8 5.
-4
Using the graph of y = ln x , shift 3 units to
-5
the right, and shift down 2 units. Y 6
y = - 2 �- - - - -
Range off (-3, 00)
II
e.
-00, 00
Inverse
y = In(x + 3) - 2
f(x) = 3 + log 3 ( x + 2 )
(
)
x + 3 = ey +2 y + 2 = ln(x + 3)
Vertical Asymptote : x = -2
x = 3 + log 3 ( y + 2 )
83 .
00
x = ey +2 - 3
-00, 00
(
Range: (-3,
5
I I I
= -3
3 f(x) = 2 x / + 4 a.
Domain:
b.
Using the graph of y = 2 x , stretch the graph
(
- 00 , 00
)
horizontally by a factor of 3 , and shift units up. y 10
4
)
two units to the left, and shift 3 units down. -3
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ.
7
325 All rights reserved. Thi s material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.
Chapter 6: Exponential and Logarithmic Functions
(4, 00) =4 f(x) = 2xI3 + 4 = 2xl3 + 4 x = yl3 + 4 x-4 = 2yI3 (x-4) 2::. = = (x-4) rl(x) = (x-4) (4, 00) = x, 4
c.
Range :
Horizontal Asymptote :
d.
93.
y
eX
Inverse
y
log
95.
2
3 log
3 10g
2
Range off
f.
Using the graph of
log 4
The solution set is
2
e.
e5
The solution set is
Y
3
=5 = x=5 {5} . 64 = x 4x = 64 4x = 43 x=3 {3} . 243 = 2x + I = 243 = 2x+ 1 = 5 2x =4 x=2 {2} . = 10 3x = Inl0 X=--10 { In; O } .
In e x
97.
log 3
3 2 X+ l 32X+ l
shift units log y 2 to the right, and stretch vertically by a factor of 3 . y 8
35
The solution set is 99. x
In
3
-2
87.
log 3
The solution set is
x=2 X =32 x=9 {9} . ( 2x + 1) = 2x + 1 = 23 2x+l =8 2x =7 X =-27
101.
The solution set is 89.
log
2
3
The solution set is 91 .
1 03 .
log
.
IS
2
.
d x 2 + 1 ) =2
x2 + 1 = x2 + 1 = 9 x2 =8 x = ±J8 = ±2.J2 { -2.J2, 2.J2 } . 32
{-i} .
=
*'
=8 2x+5 = 1n 8 2x = -5+ln 8 -5+ln 8 x= --2 {-S+ In 8} e2 x + 5
· set The s oIutJon
4 2 x2 =4 x = 2 (x -2, {2} .
log x
3 e x
The solution set is
base is positive)
The solution set is
© 2008
326 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means , without permi ssion in writing from the publisher.
Section 6.4: Logarithmic Functions
105.log 28X=-3 8x = 2-3
c.
(23f = r3 23x 2 -3 3x=-3 x=-1 The solution set is {-I} . =
In(-x ) if x <0 113.f (x)= { In x if x> 0
107. SeO.2x = 7
7
S
G(x )= 2 log (2x+1)= 2 3 2x+l = 32 2x+l = 9 2 x=8 x= 4 The point (4,2) is on the graph of G.
y
)
,.,
7
.
0.2x= In S
s(
S(0.2x) = In�)
x
7
x=SIn S
The solution set is {SIn
�} .
Domain: { x I x -:;; O} Range: ( -00, 00) Intercepts: (-1,0), (1,0)
109.2·102 - X=S
102-x =� 2
S 2-x= log2
S -x=-2+1og2 5 x= 2-log2
The solution set is {2 111.a.
-IOg%} .
{-InX if 0 < x <1 if x � 1 Inx
x
-1
G(x)=log (2x+l) 3 We require that 2x + 1 be positive. 2x+l > 0 2x>-1 1 x>-2 Domain:
b.
115.f(x) =
-3
Domain: {xlx>O}; ( 0, 00 ) Range: {Y I Y � O}; [0,(0) Intercept: (1,0)
{XIX>-�} or (-�,oo)
117.pH = -loglO [H+ ]
G( 40)= log (2·40+1) 3 = log 81 3 =4 The point (40,4) is on the graph of G.
a.
b. c.
pH =-loglO [0.1] =-(-1) 1 =
pH =-loglo [0.01] = -(-2) = 2
pH = -loglO [0.001] = -(-3) 3 =
327
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Chapter 6: Exponential and Logarithmic Functions
d.
b.
As the H+ decreases, the pH increases.
[
e.
3 . 5 = - logl o H+
[ ]
-3 . 5 = 10g l O H+
]
In(0.2) = - 0. 1 t
[ H+ ] = 1 0 -35.
t=
'" 3.16xlO-4
= 0.0003 1 6
f.
[
J
P a.
=
loglo [ H+ ]
123.
'" 3 . 98 1 x l 0-8 0.0000000398 1
760e-01. 45h 320 = 760e-{)1· 45h 320 -01. 45h -=e 760 In
( ) 320
760
h=
( )
760e-{)1· 45h -{).145h -=e 760 667 In ( ) = -O.l45h 760 In ( 667 ) 760 h = -0. 1 45 '" 0.90
a.
1-
�� [
1-
e-(1015)r ]
L
= 5 , and
12
e-2r = -7
12 -2t = In(7 / 1 2) In(7 / 1 2)
'" 0.2695 -2 It takes approximately 0.2695 second to obtain a current of 0.5 ampere. t=
Substituting E = 1 2 , R = 1 0 , [ = 1 .0 , we obtain:
e-o.lr
I e-0lr
0.5 = -0 5 = .
![l-e-(RIL)rJ
� = 1_e-2r
Approximately 0.90 kilometers. =
5e-{).4h 2 = 5e-0.4h 0.4 = e-O.4h In(O.4) = - O.4h h = In(O.4) '" 2 .29 - 0.4 =
0.5 =
667 = 667
121. F(t)
e-{)J. r will never equal zero; thus, F(t) = 1 - e-Olr will never equal 1 00% because
Substituting E = 1 2 , R = 1 0 , [ = 0.5 , we obtain:
'" 5 .97
-0. 1 4 5 Approximately 5 . 97 kilometers. b.
It is impossible for the probability to reach
D
125. [ =
320 760
'" 1 6 .09 - 0. 1 Approximately 1 6.09 minutes.
Approximately 2.29 hours, or 2 hours and 1 7 minutes.
= -0. 1 45h In
In(0.2)
I.
H+ = 1 0 - 7.4
=
119.
c.
[ ]
7 .4 = - logl O H+ -7.4 =
l-e-{)I· r - 0.2 = _e-O.lr 0.2 = e-{)·lr 0.8 =
1 .0 =
.
_e-Ol. r 0.5 = e-Ol. r In (0. 5 ) = -O.lt In(0.5) ", 6.93 t=
�� [
.!..Q 1 12
=
e-2r -2t
--
-0.1
e-(1015)r ]
=
5, and
e -2 r
1 =
t=
Approximately 6.93 minutes.
1-
L
6 In(1 / 6) In(1 I6) -2
'" 0.8959
328
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Section 6.5: Properties of Logarithms
It takes approximately 0.8959 second to obtain a current of 0.5 ampere. Graphing:
2.0 t
a..
1.6 1.2
� 0.8
(0.8959,1) 135.
(0.2695,0.5 ) Seconds
127.
129.
( ) 3
1.
10=10 log (109 ) =10 ·9 =90 decibels
3. 5. 7. 9.
133. R a.
= e""
1.4= i (O .03) 1.4= eO. 03k
In(I.4)=0.03k k = In(I.4) 0.03 "" 11.216 c.
sum rloga M False log3 371 =71 In e
-4 =
4
-
ios,7 = 7
13.
log8 2 + log8 4= 10g8 ( 4 . 2)= log8 8=1
17.
=
= eI 1.2 1 6 (O. 17 ) e1 .9 0672 "" 6.73 1 00 = el 1 .216 x 100= el1.2 16 x In(100)=11.216x ) x= In(100 11.216 "" 0.41
R
x.
n.
--
b.
=
Section 6.5
L (IO-3 )=10 10g ( 1 0 12 ) -
a
x
7 L (IO-7 ) 10 10g 10-1 2 10=10 log (105 ) =10·5 =50 decibels =
=
=
f
r/)
5= el 1 .216 x In 5 11.216x x � 11.216 "" 0. 1 43 At a percent concentration of 0.143 or higher, the driver should be charged with a DUI. e. Answers will vary. If the base of logarithmic function equals 1, we would have the following: f(x)= log I (x) I- I ( )=IX 1 for every real number In other words, I -I would be a constant function and, therefore, I-I would not be one to-one.
d.
log26 ·log6 4= log6 4 108,6 6 = log6 (2 2yag, = log6 2 2108,6 = log6 ios, 6' = log6 6 2 =2
329
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Chapter 6: Exponential and Logarithmic Functions
19. 21.
310g,S-log,4 =310g'4� =-4S
43.
e log,I6 Let a=loge' 16, then (e2r =16. e 2a =16 e 20 =4 2 (e 20),,2=(4 2t2 eO =4 a=ln4 log,I6 = eIn4 = 4 . Th e' •
45.
47.
us,
23.
ln6=In(2·3)= ln2+ln3= a+b
25.
ln1.S=ln�=ln3ln2=b- a 2 1n8=ln23 =3·ln2=3a 1nif6=In6\15 1 =-In6 S
27. 29.
In( )= In e+Inx=1+Inx
37.
In(xex)=lnx+lnex =lnx+x
39.
41.
=3log2x-log2(x-3)
[
]
(X+21 =log[x(x+2)] - log(x+3)2 log X(x+3) = logx + log(x +2) -2log(x+3)
[
]
2 3 In X(X+4)22 =.!.3 In (X-2)(x+2 1) _X_
[
1I
(X+ 4)
]
=l[In(x -2)(x+1)-In(x+4)2] 1 - 2) +In(x+1)- 2ln(x+4)] =-[In(x 3 1 2 +4) I =-In(x 3 +1)--In(x 3 3 - 2)+-In(x 49.
In Sx.Jl+3x
(X- 4)3
=In( Sx.Jl+3x) -In(x- 4)3 = lnS+ lnx+In.Jl+3x -3ln(x-4) =lnS+lnx+ln(1+3xY'2-3ln(x-4) = lnS+lnx+'21 ln(1+3x)-3ln(x-4)
=� ln(2 . 3) =S1 (ln2+ln3) =.!.(a+b) S
35.
(�)
log2 x -3 = log2 x3 -log2 (x-3)
ex
loga ( U2V3 ) =loga u 2+loga v3 =2loga u +3logo v
51.
3logs u +4logs v= logs u 3 +logs v4 =logs ( U3V4)
53.
log3 Fx -log3 x3 = log3
(J) = log3 ( X��2 ) =log3 X-S/2 S =--log 2 3X
In (X 2�)=lnx 2+ln� =lnx 2+ 1n(I - xt2 =2lnx+-21 In(l-x) 330
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Section 6.5: Properties of Logarithms
55.
log 4 (x 2 -1)- 51og 4 (x+l) =log 4 (X 2 -1)- log 4 (x+1r =log 4 -15 (x+l) =lOg 4 (X+l)(X;I) (x+l)
63.
[ X2 ] [ ] =IOg' [ ::I�' ] (
57.
21og2 (x+l)-log2(x+3)-log2 (x-I) =log2(x+1) 2-log2 +3)-log2 (x-1) +1) 2 log2 (x-1) =log2 (x(x+3 ) 2 (x+l) =lOg2 (x+3)(x-l)
(X
lnC�J+ln( X;I)_ln(X2 -1) X X+lx ] _ln(X 2-1) =In [ x-I = In[�-;x-I (X2 -1) ] = h{ (x- ;)(:' -I) 1 In [ (x-1)(:: �)(X+1)] =lnCX�I)2) 2 __.
65.
log3 21= log21 log3 '" 2.771
67.
log71 = log71 '" -3.880 log l/3 71= log(I/3) -log3
69.
log7 5.615 logvrz 7= � log,,2
71.
lne '" 0.874 log e=-
73.
logx lnx or y= -y=log 4 =log4 ln4
'"
81og2 .J3x- 2 -log2
2
75.
+2) y=log2(x+2) = In(xln2+2) or y= log(x log2 ----=-'--= ---'-
3
=log2 (.J3x-2 r -(log24-log2x) +log2 4 = log2 (3x-2)4 -log24+log2 +log2 4 =log2 (3x -2)4 +log2 =log2[x(3x- 2t ] 61.
X
-2
(�)+log2 4 X
lnn
1[
=In(x-lr =-2ln(x-l)
59.
]
[
X
5 -2
-�
2loga(5x 3 ) logo (2x +3) =logo (5x 3 r -logo (2x+3)1/2 =logo (25x6 )-log a.J2x+3
77.
log(x+l) y=logx- I(x+l)= In(x+l) In(x -1) or y= log(x-1) 4
OII--<;;-�--��15 "...
[:;; ] 2x+3
=IOga
-4 331
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Chapter 6: Exponential and Logarithmic Functions
79. I(x)= Inx ; g(x) = eX ; hex) = x2 (f g )(x) = I( g(x)) = In (ex) =x Domain: {x I x is any real number} or a.
91.
0
( -00,00)
(g f)(x) = g (I(x)) = elnx = x Domain: { x I x > o} or (0,00) (Note: the restriction on the domain is due to the domain of Inx ) c. (f g)(5)= 5 [from part (a)] d. (f h)(x) = I( h(x)) =In (x2 ) Domain: { x I x "* O} e. (f h)( e) = In( e2 ) =21n e = 2 . 1 = 2 lny=lnx+lnC Iny=In(xC) y=Cx Iny = Inx+ln(x+1)+lnC Iny = In(x(x+l)C) y = Cx(x+l) Iny = 3x+ inC In y = In e3 x+In C Iny = In(ce3 x) y= Ce3 x In(y-3) = -4x+ inC In(y-3) = lne-4 x+lnC In(y-3) = In( Ce-4X ) y-3 = Ce-4 x y= Ce-4 x+3 1 4 )+InC 3 lny = -1 In(2x+1) --In(x+ 2 3 3 Inl = 1n(2x+1)1 1 2 -In(x+ 4)1/ + inC +I)1I2 Iny3 = ln C(2X (x+4 113
b.
0
=3 93.
0
0
83.
85.
87.
89.
) C(2x + 1)" 2 Y (x+4)1/3 C(2X+ 1)112 1/3 Y = (X+4)1/3 3
_
[
95.
97.
99.
]
n
n _---'---'n
-----=-'=--n
loga(x+.Jx 2 -1)+ loga(x -.Jx 2 -1 ) : = loga[(x+N-1)(X-.JX2 -dJ =loga [x 2 -(x2 -1)] = loga [x 2 _x 2 +1] = log a1 =0 2x+In (1+e -2x) = In e 2x +In (1+e -2x ) = In ( e 2x (1+ e -2x ) )
I(x) = logax means that x = a[(x). Now, raising both sides to the -1 power, we [(X) obtain X-I = (a[(X»)-I = (a-I)[(x) = --;;1
()
()
(X) [ 1 X-I = -a means that log lla X-I I (x) . Thus, log llaX-I = I(x) - log llaX= I(x) -I (x) = log lla X
]
[
log23 ·log3 4 ···log n ( +1) ·log n+1 2 log3 log 4 . . . log( +1) log 2 = -- . log 2 log3 log log ( +1) log 2 log2 =1 __
0
81.
log23 ·log3 4 ·log 4 5 ·log s 6 ·log 6 7·log 7 8 log3 . log4 . log 5 . log6 . log7 . log8 = log2 log3 log4 log5 log6 log7 log8 log2 3 = -- = -log2 log 2 3 log2 log2
=
<Jc(2x+ 1)1/6 (x+4)1/9
332
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Section 6.6: Logarithmic and Exponential Equations
101.
f(x) =
logo x
3. x 3 = x2 - 5
Using INTERSECT to solve: 3 Y I = x ; Y = x2-5 2 10
log" 1-loga x = -loga
=
x
= - f(x)
103.
=
If A loga M and B = logo N , then and = N. D a
=
logo
a
a
A
=M
I�tHs�ctio)� X= -1.�33�2B V= -2.9�S:2BS:
Thus,
A-D
5.
105.
� = logx2
2
The solution set is {1 6} .
Y2 = 2 10g x
2
7.
2
-3
log2 (5x) = 4
5x = 2 4
16 16 X=-
5x =
5
5 The solution set is fs6}.
y=2logx
-3
The domain of � = loga x2 is { xl x :t O } . The domain of >-; = 2 10g " x is { xl x > O } . These two domains are different because the logarithm property loga xn = ·loga x holds only when log" x exists. Answers may vary. One possibility follows: Let x = 4 and y = 4 . Then log2(x + y) = log2(4+4) = log28 = 3 . But log2X + log2y = log24 + log24 = 2+ 2 = 4 . Thus, log2(4 + 4) :t log24 + log24 and, in general, log2(x + y) :t log2X +log2y .
9.
log4 (x + 2) = log4 8 x +2 = 8
x=6
n
The solution set is {6} . 11.
-1 log3 X = 2 log 3 2 2 log3 X II2 = log3 22 x 1 l2 = 4 x = 16
13. Section 6.6
1.
=
X = 16
loga M -loga N
� f(
107.
log4 X
-10
-1 . 43 , so the solution set is { - 1 .43 } .
x = 42
= A-B =
x :::;
The solution set is {1 6} . 3log2 X = -log2 27 log2 x3 = log2 27 -1 x 3 = 2 TI
1
x2 - 7 x - 30 = 0
x3 = 27
(x + 3)(x - 1 O) = 0
1
or x - l 0 = 0 x = -3 or x = 10 The solution set is { -3, 1 O } . x+3 = 0
X=3
The solution set is
{�}.
333
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Chapter 6: Exponential and Logarithmic Functions
23.
15. 3 log (x - 1) + log 4 = 5
2 2 log (x - l) 3 + log 4 = 5 2 2 log 4(x - l) 3 = 5 2 4(x - l) 3 = 2 5
(
log8(x + 6) = I - logg(x + 4) log8(x + 6) + logg(x + 4) = 1
[
(X - I) 3 =
]
logg (x + 6)(x + 4) = I
)
(x + 6)(x + 4) = 8 1 x 2 + 4x + 6x + 24 = 8
32 4
x2 + l Ox + 1 6 = 0
(x - l) 3 = 8
(x + 8)(x + 2) = 0 x = -8 or x = -2
x-I = 2
(
) { -2 } .
{}
solution set is
17. log x + log(x + 1 5) = 2
25. ln x + In(x + 2) = 4
log x(x + 1 5) = 2
In x(x + 2) = 4
(
)
(
x2 + 2x - e4 = 0 r- - - - - - - 2 ± 2 2 - 4 (I )( - e 4 ) ----''-----.:... ---'---.:. .:.. .. x=
x 2 + 1 5x - 1 00 = 0 (x + 20)(x - 5) = 0
2(1)
)
Since log -20 is undefined, the solution set is
19.
=
log(2x + 1) - log(x - 2) =
( 21 ) 2x + x-
1
(
) ( ) log 3 [( x+l )( x + 4 ) ] = 2 ( x + l )( x + 4 ) = 32
-2 1 2 1 X= -=8 -8
x 2 + 4x + x + 4 = 9
{�)}.
x2 + 5x - 5 = 0
�
- 5 ± 5 2 - 4(1)(- 5) ...:. x = ----''---....:.---'...:.2(1)
21. log (x + 7) + log (x + 8) = 1 2
[ (x + 7)(x + 8) ] = 1 2
-5±
.J45
2
(x + 7)(x + 8) = i X=
x 2 + 8x + 7x + 5 6 = 2
-5 -3
J5
2 "" -5.854
x 2 + 1 5x + 54 = 0
-5±3 2 or X =
J5
- 5 + 3 J5 --2
"" 0 . 854 Since log 3 -8 .854 + 1 = log 3 -7 .854 is undefined, the solution set is
(x + 9)(x + 6) = 0 x = -9 or x = -6
( - 9 + 7 ) = log 2( - 2 ) 2 the solution set is { -6 } .
�
27. log 3 x + 1 + log 3 x + 4 = 2
-8x = -2 1
Since log
2
( ) is undefined, the solution set is { - I + �I + e4 } "" { 6 . 456 } .
2x + l = 1 0x - 20
log
2
Since In -8 .456
=1
= 101 x-2 2x + l = 1 O(x - 2)
2
� = - 2 ±2 .,Jl;l = - I ± �I+e4
�
2x + l
The solution set is
-2±
x = - 1 - I + e 4 or x = - 1 + I + e 4 "" -8 . 456 "" 6 . 456
log(2x + 1) = 1 + log(x - 2)
log
�
_
x = - 20 or x = 5
{ 5} .
)
x(x + 2) = e 4
x( x + 1 5) = 1 02
(
( )
Since log8 - 8 + 6 = log8 - 2 is undefined, the
x=3 The solution set is 3 .
{
is undefined,
-
S+3 2
.J5
(
)
}
"" 0.854 .
{
(
)
}
334
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Section 6.6: Logarithmic and Exponential Equations
37.
( xx:-x+ J=-1 ::2 ::=(�r X
logl13
The solution set is
{-logg1.2}=
x +x --=3 x 2 -x x2 +X=3 ( X2 -x ) x2 + X=3x2 -3x _2X2 +4x=0 -2x(x-2)=0 -2x=0 or x-2=0 x=O or x=2
x=0 , {2}.
39.
x=2 ,
The solution set is IOg2
In ( 3 1 - 2 X
{
43.
y-5 =8 2 x-5 =23
2" =10 "" 3.322 x= log2 1 0= lnl0 ln2
{
}
(%J =i- X In(%J = In ( i- x )
x ln(3/S)= (l-x)ln7 xln(3/5)=ln7-xln7 xln(3/5) +x ln 7= In 7 x(ln(3/5)+ In 7 )= In 7 x= In(3/5ln7+ln7 "" 1.356 ) In7 The solution set is In(3/5) +ln7 "" {1.356}.
{8}.
The solution set is Iog21
)= In (4x )
(1-2x)ln3=xln4 In3-2x In 3=x In 4 In3= 2x In3+x In 4 ln3=x(2 ln3+ln4) x= ln3 "" 0.307 2ln3+ln4 In3 "" {0.307}. The solution set is 2ln3+ln4
x= �, the solution set is {�}.
35.
5 (23.:c )=8 23x =�S
log2
Since each of the original logarithms is defined for
The solution set is
}
(1.2) ",, {-0.088}. - log 8
log
(%) ) "" 0.226 x= .!. (�)= In(8/5 5 3ln2 3 {� (%)}={ ln3�;) } "" {0.226} .
x-I x-2 x+6 x+3 (x-l) (x+3)=(x-2) (x+6) x2 +2x-3=x2 +4x-12 2x-3= 4x-12 9= 2x X=-29
x-5=3 x=8
{
3x= log2
Since each of the original logarithrns are not defined for but are defined for the solution set is
33.
8- x =1.2 -x=logg1.2 ) "" -0.088 x=-logg1.2=- loglog(1.2 8
O}=f�l�} "" {3 .322}.
{
}
335
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exist. No portion of this material may be reproduced, in any form or by any mean s , without permi ssion in writing from the publi sher.
Chapter 6: Exponential and Logarithmic Functions
45.
47.
1.2x= (0.5)-X In 1.2x = In (0.5fx xln(1.2)=-xln(0.5) xln(1.2)+xln(0.5)=0 x(ln(1.2)+10(0.5))=0 x=o The solution set is {O} .
53.
a
---
nl-x= eX 10 nl-x=10 eX (l-x)lnn=x Inn -xln n=x Inn=x+xlnn 10 n=x(1+ 10 n) Ion ",,0.534 x= l+lnn
Therefore, we get
�
4x=-2+17 = 10g 4 (-2+17 ) (we ignore the first solution since 4x is never
--
The solution set is
49.
51.
55.
25x -8·5"" =-16 ( 52 r -8·5'< =-16 ( 5x )2 -8 . 5'< =-16 Let u = 5'< u2 -8u=-16 u2 -8u+16=0 (U-4)2 =0 u=4 5'<= 4 x= 10g 5 4
}
The solution set is { 10g 5
57.
3x+4=0
4} "" {0.861} .
3·4x +4·2x+8=0 3.(22 r +4·2x +8=0 2 3·(2.< ) + 4·2'< +8=0 Let u= 2'<. 3u2 +4u+8=0 =3, b=4, c =8 2 -4(3)(8) =-4±../-80 = not real u= -4 ± )42(3) 6 a
No solution
The solution set is
4
Therefore, we get
y
x=0
{ (-2+17)} "" {-0.315} .
.
( t +3.3x -4=0 (3x -1)(3x+4)=0 or
x
The solution set is 10g
22x+ ZX -12=0 (2x t +ZX -12=0 (2x -3 )(2"" +4)=0 2"' -3=0 2x+4=0 or 2x=3 or 2"" =-4 10 ( 2"" )=103 No solution xln2=103 103 1.585 X=-"" 102 "" 1.585 . The solution set is
3X -1=0
or
negative)
"" {0.534}. {�} l+lnn
{:�
16'<+4X+1 -3=0 (42r +4·4x -3=0 (4x)2 +4.4x-3 =0 Let u = 4'<. u2 +4u -3=0 =l,b= 4,c= -3 2 -4(1)(-3) -4±J28 = U= -4±)42(1) 2 = -4±217 2 =-2±17
{O} .
The equation has no real solution.
336
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Section 6.6: Logarithmic and Exponential Equations
59.
4x - 1O·4 -x =3 Multiply both sides of the equation by 4x . 2 (4x ) -10 . 4- x . 4x =3 . 4x (4.<)2 -10=3 . 4" (4x )2 -3 . 4x -10=0 (4x -S) (4x +2)= 0 or 4x +2=0
67.
2
Thus,
x=log4 S The solution set is {log4 S} "" {1.161} . 61.
Inx=-x Using INTERSECT to solve: Yl = In x; Y2 = -x
69.
logs(x+1) -log4 (x-2)=I Using INTERSECT to solve: Yl =In(x+1) IIn(S) - In(x-2) 11n(4) Y2 =1
X ""
0.S7 , so the solution set is {0.S7}.
Inx=x3 -1 Using INTERSECT to solve: 3 Y = Inx; Y2 =x -1 l
4
\ ol===:='::��
-2
5
-4
63.
X ""
-2
Thus, 0.39 or x=1 , so the solution set is {0.39, I}. X ""
Int�n�ction X=�.7B71BQ� V=1
Thus,
Int�n�ction V=Q X=1 I LI
2.79 , so the solution set is {2.79} .
71.
eX =-x Using INTERSECT to solve: Yl = eX ; Y2 =-x
eX +Inx= 4 Using INTERSECT to solve: Yl = eX + Inx; Y2 = 4 5
2
-2
L
Int�rs�c on X=1.31531't9
U
-2
Thus, 65.
X ""
,.{
V=�
4
Thus, x 1.32 , so the solution set is {1.32}. ""
-0.S7 , so the solution set is {-0.S7}. 73.
eX =x2 Using INTERSECT to solve: 2 Yl = eX ; Y2 =x
e-x = Inx Using INTERSECT to solve: Yl =e-x ; Y2 =lnx 2
-2t-�+----r:-=---1 4 -3
Int�rs�ction X= ·.7Q3�67� V=.�9'tB66�1
Thus,
X ""
Int�rs�ction X=1.3Q97996 V=.�69B7't1't
3
Thus,
-1
-0.70 , so the solution set is {-0.70}.
-2
X ""
1.31 , so the solution set is {1.31} .
337
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Chapter 6: Exponential and Logarithmic Functions
75.
log2 (x +1)-log4 X=1 log2 X 1 log2 (x +l)-log2- -= 4 log2 (x +1) log22 _x 1 21og2 (x +l) - log2 X= 2 log2 (x+1)2 -log2 X= 2 2 log2 X :l) = 2 (x +l)2 = 22 X x2 +2x +l= 4x x2 -2x +l=0 (x-l)2 =0 x -I=0 x=1 Since each of the original logarithms is defined for x= 1 , the solution set is {I} . =
_ __
(
77.
81.
_
)
83.
2 eX (eX e-X)=4ex e2x -1=4ex (ex)2 -4ex -1=0 - -4(1- )(---1) -4-) 2 eX -(-4) ± �r--( 2(1) J20 = 4±2 = 4±2215 = 2±15 x= In ( 2-15) or x = In( 2 + 15) x"'" In(-0.236) or x "'" 1.444 Since 1u(-0.236) is undefined, the solution set is { In( 2 + 15)} {1.444} . _
=
logl6 X + log4 X + log2 X 7 log2 X log2 X + Iog X=7 --log2 16 +-log2 4 2 log2 X +-log2 X + Iog X= 7 -2 4 2 log2 X + 2log2 X +4 log2 X= 28 7 log2 x= 28 log2 x 4 x= 24 =16 Since each of the original logarithms is defmed for x=16 , the solution set is {16}. U!i. )2-x = 2x 2 (21/3 )2-X 2x2
= ---�---------
=
79.
2 eX +e- x =2 eX (eX +e-X)= 2ex e2x +1= 2ex (ex)2 2ex + 1 =0 (eX_l)2 =0 eX -1=0 eX =1 x=O The solution set is {O}.
"'"
85.
logs x +log3 X=1 lnx + lnx =1 lnS ln3 1 (lnx) _ +_ 1 =1 lnS ln3
(
=
1(2-x) x2 =2 23 1-(2-x) x2 3 2 -x =3x2 3x2 +x-2=0 (3x-2)(x +l)=0 x=-32 or x=-1
)
lnx= 1 1 lnS + ln3 (lnS)(ln3) lnx= ln3 + lnS lnx= (lnS)(ln3) lnlS
=
In
( 5)(1n3») { (1n5)(1n3»)} The solution set is e 1n15 ""'{1.921}. x=e hl15 "",1.921
The solution set is {-I, �} . 338
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Section 6.6: Logarithmic and Exponential Equations
87.
a.
b.
c.
d.
f(x)=3 log2 (x+3)=3 x+3= 23 x+3=8 x=5 The solution set is {5}. The point (5,3) is on the graph off g(x)= 4 log2 (3x+I)=4 3x+1= 24 3x +1=16 3x=15 x=5 The solution set is {5}. The point (5,4) is on the graph of g. f(x)= g(x) 10g2 (x+ 3) = log2 (3x +I) x +3=3x +I 2= 2x I =x The solution set is {I}, so the graphs intersect when x=I . That is, at the point (1,2) .
e.
f(x )-g(x )=2 log2 (x + 3) -log2 (3x+1)=2 x+3 log2--=2 3x +l x+3 =22 3x+1 x +3=4(3x+l) x+3=12x +4 -I=llx I --=X 11
{- }
The solution set is I\ . 89.
a.
Y
f(x)= 3>+1 g(x)=2x+2
8
-2
x
f(x)= g(x)
b.
f (x) + g (x )= 7 log2 (x+ 3)+ log2 (3x+I)= 7 log2 [(x +3)(3x+I) ]= 7 (x +3)(3x +l)=27 3x2+lOx+3=128 3x2+IOx-125= ° (3x +25)(x-5)=0 +25= ° or x-5=0 -25 x=5 x= 25 The solution set is {5} .
2
1n(3X+1 )= 1n ( 2x+2 ) (x +I)1n3=(x +2)1n2 x 1n3+1n3=x1n2+21n2 x 1n3-x1n2= 21n2- 1n3 x(1n3-ln2)=2ln2-1n3 X= 21n2-1n3 1n3-1n2 ",0.710 1n2-1n3 '" 6.541 f 21n3-1n2 The intersection point is roughly (0.710,6.541) .
(
c.
)
Based on the graph, f(x)> g(x) for x> 0.710 . The solution set is {xlx>0.710} or (0.710,00) .
339
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Chapter 6: Exponential and Logarithmic Functions
91.
a.,
b.
y
-] c.
95. f(x)
=
3X
a.,
f (x) = 2x -4
Using the graph of y = 2x , shift the graph down 4 units.
x
-4 -5
f (x) = g(x) 3X= 10 x= log 310
The intersection point is (log3 10,10) . 93.
a.
97.
b.
b.
Based on the graph, f (x) The solution set is {x I x
0 when x < 2. < 2} or (-00, 2).
a.
298(1.009y-2 006 = 310 (1. 009) 1- 2006 = �
<
298
( 298
In(I. 009) 1-2006 = In � J
(t-2006)ln(I. 009) = In(�J
x
2 c.
1 49 In(155 / l 49) t-2006= In(1 .009) In(15511 49) t= +2006 In (1 .009)
� 2010. 41
According to the model, the population of the
f (x) = g (x) l 2x + = 2-x +2 x+l =-x+2 2x= 1 1 x=2 f(�) = i/2 +1 = 23/2 = 2J2
The intersection point is
U.S. will reach 310 million people around the middle of the year 2010. b.
298(1.009) 1-2 006 = 360 (1.009) 1-2006 = 360 298
( 298 ) (t- 2006)ln(1. 009) = In ( 1 80 ) 1 49 In(1. 009) 1- 2006 = In 360
(�, 2J2 ) .
t-2006=
1n(1 80 / 1 49) 1n(1.009) In(1 801l 49) + t= 2006 In(1 .009)
�2027.10
According to the model, t he population of the
U.S. will reach 360 million people in the beginning of the year 2027. 340
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Section 6.7: Compound Interest
99.
a.
Section 6.7
14, 5 1 2(0.82/ = 9, 000 (0.82Y = 9, 000
( (
) )
1 4, 5 1 2 10g(0.82) 1 =10g 19,4,000 512 9, 000 I log(0.82) = IO g 1 4, 5 1 2 1 = log(9, 00011 4, 5 1 2) log(0.82)
:::::2 .4
According to the model, t he car will be worth b.
$9,000 after about 2.4 years. 14, 5 1 2(0.82/ = 4, 000 (0.82) 1 = 4, 000
P = $500, r = 0.06, 1 = 6 months = 0.5 year J = Prl = (500)(0.06)(0.5) = $15.00
3.
P = $ 1 00, r = 0.04, n = 4, 1=2 )(2) :::::$ 108 .29 = 100 1 + 0. 4 A =p I+
5.
P = $500, r = 0.08, n 4, 1 = 2.5 ( 4 )(2 . 5 ) nl A = P 1+ = 500 1 + 0. 8 :::::$609.50
7.
P = $600, r = 0.05, = 365, 1 = 3 nl 0.05 (365 )(3) :::::$697.09 A = P 1 + - = 600 1 + 365 n
( ) ( )
9.
According to the model, th e car will be worth
$4,000 after about 6.5 years. 14, 5 12(0.82/ = 2, 000
( ) ( )
::::: 1 0.0
According to the model, the car will be worth
$2,000 after about 1 0 years.
10 1.
Solution A: change to exponential expression; square root method; meaning of ± ; solve. Solution B: loga Mr = r loga M ; divide by 2; change to exponential expression; solve. The power rule loga Mr = r loga M only applies when M > O . In this equation, M = x - 1 . Now, x = -2 causes M = -2 - 1 -3 . Thus, if we use the power rule, we lose the valid solution
( )
n
(
)
P = $ 1 0, r = O. l 1, 1 = 2 A = Perl = 1 0e( o .II)(2) :::::$1 2.46 r
13.
A = $ 1 00, r = 0.06, n = 12, 1 = 2 - nl 0.06 (- 1 2)(2) :::::$88.72 P = A 1 + -r = 1 00 1 + n 12
15.
A = $1000, r = 0.06, = 365, 1 = 2.5 nl P= A I+ ( -365 )(2. 5 ) ::::: $860.72 = 1 000 1 + 0.06 365
17.
A = $600, r = 0.04, = 4, 1 = 2 nl (-4 )(2) ::::: P= A I+ = 600 1 + 0. 4 $554.09
19.
A = $80, r = 0.09, 1 = 3.25 P A e-rl 80e( -O .09 )(3.2 5 ) :::::$59.71
( )
21.
(
( .;;r
)
n
( .;;r ( )
=
=
( �)
( ';;)
P = $ 1 00, = 0. 1 0, 1 = 2.25 : A = Perl = 1 00 e( o . IO )(2.25 ) ::::$125.23
(0.82/ = 2, 000
1 4, 5 1 2 10g(0.82)1 = 10g 12,4,000 512 2, 000 Ilog(0.82) = log 1 4, 5 1 2 1 = log(2, 00011 4, 5 1 2) log(0. 82)
=
11.
::::: 6 .5
c.
( .;;r ( � r
r
1 4, 5 1 2 10g(0.82)1 =10g 14,4,000 512 4, 000 I log(0. 82) 10g 1 4, 5 1 2 1 = log(4, 000 / 1 4, 5 1 2) log(0. 82) =
1.
n
( �)
=
A = $400, r = O. 1 O, 1 = 1 P = A e-rl = 400eC-O IO)(I) :::::$361 .93
x = -2 .
341
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Chapter 6: Exponential and Logarithmic Functions
23.
6% compounded quarterly: )(I) = $ 1 0, 613 .64 A = 10, 000 1 + 0. 6
(
33.
�r
6 + % compounded annually: A = 10, 000 ( 1 + 0.0625 ) 1 = $ 1 0, 625 6 + % compounded annually yields the larger
25.
amount. 9% compounded monthly:
(
A = 10, 000 1 +
0��9 )
(12 )(1)
35.
3P = p ( l + f t
3P = P ( 1 + r )5 3 = (l + r) 5 ifi, = I + r r = ifi, - 1 "" 0.24573 The required rate is 24.573%. a.
= $ 1 0, 938.07
27.
)
amount. Suppose P dollars are invested for 1 year at 5%. Compounded quarterly yields:
( � F)(I) "" 1 .05095P .
b.
The interest earned is 1= 1 .05095P - P = 0.05095P 1= Prt Thus, 0.05095P = P· r·l 0.05095 = r
The effective interest rate is 5 .095%. Suppose P dollars are invested for 1 year at 5%. Compounded continuously yields:
37.
A = Pe(O.05)(I) 1 .05 1 27 P ""
The interest earned is 1= 1 .05 l 27P - P = 0.05 1 27P 1= Prt Thus,
2P = PeO.08t 2 = eO.08t In 2 = 0.08t I t = n 2 "" 8.66 0.08 It will take about 8.66 years to double.
Since the effective interest rate is 7%, we have:
1= Prt 1= P · 0.07 · 1 1= 0.07P
Thus, the amount in the account is A = P + 0.07P = 1 .07P
Let be the required interest rate. Then, x
0.05 127P = P · r·l .05 1 27 = r
( �)(4 1 .07 = ( I + � J
1 .07P = P 1 +
The effective interest rate is 5 . 1 27%. 31.
12
It will take about 8.69 years to double.
A = p 1 + 0. 5
29.
( 0��8rt 2 = ( 1 + 0.08 ) t 12 In 2 = ln ( l + 0��8rt In 2 = 12t ln ( l + 0��8) In 2 "" 8.69 t= l 2 ln ( l + 0��8) 2P = P I +
--
8.8% compounded daily: 3 65 0.088 A = 10, 000 1 + = $ 1 0' 9 19.77 365 9% compounded monthly yields the larger
(
l)
t
l 2P = P ( I + f ) 2P = P { I + r )3 2 = (1 + r ) 3
)(1)
4� ,,1 .07 = 1 + -r 4 �1 .07 - 1 = � 4 r = 4 ( �1 .07 -1) "" 0.06823 Thus, an interest rate of 6.823% compounded quarterly has an effective interest rate of 7%.
42 = I+ r r = 42 - 1 "" 0.25992
The required rate is 25.992%. 342
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Section 6.7: Compound Interest
39.
( 0��8r
9% compounded semiannually: (2)(20 A= 2000 1+ 0. 9 ) =$11,632.73
l 150=100 1+ 1.5 '" (1.006667 YZ I ln1.5 '" 12t In (1.006667) lnLS t- 12 ln(1.006667) - 5 .09 _
51. Will:
8.5% compounded continuously: A 2000e(0.085)(20) =$10,947.89 Will has more money after 20 years. Let x the year, then the average cost C of a
Henry: =
_
5.09 61.02 150=100eoosl 1.5= eO.os I ln1.5=0.08t '" 5.07 t= fiLS 0.08 Compounded continuously, it will take about 5.07 years (or 60.82 months). 25,000=1O,000eo.061 2.5= eO.061 ln2.5=0.06t t= ln2.5 0.06 '" 15.27 It will take about 15.27 years (or 15 years, 3 Compounded monthly, it will take about years (or months).
41.
( �)
53.
a.
b.
=
4-year private college is by the function
C(x)= 29,026(1.055)X-2005 . C( 2015)=29,026(1.055)20 15-2005 =29,026(1.055)10 '" 49,581 In 2015, the average cost of college at a 4year private college will be about $49,581. A = Perl 49, 581= PeO.04{lO) P = 49,581 '" 33,235 e'04{lO) An investment of $33,235 in 2005 would pay for the cost of college at a 4-year private college in 2015. °
months).
55.
43. A=90,000(1+0.03)5 =$104,335 The house will cost $104,335 in three years. 3 45. P =15,000e(-0.05)( ) '" $12,910.62 Jerome should ask for $12,910.62.
( r 20 A=319 (1+ 0.�32 r ) =319(1.016)40 ",602
A = p I+�
The government would have to pay back approximately billion in
47. A=15(1+0.15)5 =15(1.15)5 "'$30.17 per share for a total of about $3017. 49. 5.6% compounded continuously: A=1000e(0.056)(I) =$1057.60
57.
59.
Jim will not have enough money to buy the computer. compounded monthly:
5.9% A=1000 1+ 0 9 =$1060.62
( .�; r
The second bank offers the better deal.
$602 2025. P =1000, r=0.03, =2 A=1000(1-0.03)2 = $940.90 n
P =1000, A =950, =2 950=1000(1-r)2 0.95=(I-r)2 ±../0.95 =1-r r=I±../0.95 r '" 0.0253 or r '" 1.9747 Disregard r '" 1.9747. The inflation rate was 2.53%. n
343
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Chapter 6: Exponential and Logarithmic Functions
61.
r=0.02
69.
a.
.!. P=P(1-0.02Y
CPIo =152.4, CPI =195.3, n=2005 -1995=10 10 195.3=152.4 1+� 100 10 195.3 r = 1+ 152.4 100
( ) ( )
2 0.5P=P(0.9SY 0.5=(0.9Sy t=logo.98 (0.5) = lnO.5 '" 34.31 lnO.9S
The purchasing power will be half in 34.31 years.
63.
a.
_r_=10 195.3 -1
A =$10,000, r= 0.10, n=12, t=20
152.4 195.3 r=100 10 -1 '" 2.51% 152.4
100
( O��O )(-12)(20) ",$l364.62
P=lO,OOO 1+
b.
b. A =$10,000, r=0.10, t=20
P=1O,000e(-0.10)(20) '" $l353.35
65.
67.
( 0 1OS )(-1)(10) ",$463l.93 P=IO,OOO 1+-':
)
l n2
In (1.12) '" 6.12 years b.
ln3
4ln (l.015
(
)
)
( ) ( ) ( ) ( ) In (� ) 152.4 n= '" 27.3 In (l+ 2.51 ) 100
ln2 t=-�--� 1. 1n 1+ 0. 2
(
)
CPlo =152.4, CPJ=300, r=2.51 2.51 n 300=152.4 1+ 100 300 2.51 n = 1+ 100 152.4 300 2.51 n In 152.4 =In l+ 100 2.51 300 In =nln l+ 100 152.4
(
A =$10,000, r=O.OS, n=1, t=10
a.
(
years
The CPI will reach 300 about 27 years after 1995, or in the year 2022.
) '" lS.45 years
71.
r=3.1%
( )
3l n 2·CPJo =CPIo 1+-'100 2=(l.031Y
c.
n = logI.031 2 =
ln 2
", 22.7
lnl.031 It will take about 22.7 years for the CPI index to
( �) t=----:;-----:n'ln(l+ � )
lnm=nt'ln l+
double.
lnm
73 - 75. Answers will vary.
344
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Section 6.8: Exponential Growth and Decay; Newton's Law; Logistic Growth and Decay
c.
Section 6.8
1.
P(t)= SOO eO .02 1 P(O) = SOOe(0.02 )(0) = SOO insects b. growth rate: k 0.02 2 % P(10)= SOO e(0 . 02 ). ( 10) "" 61 1 insects c. d. Find t when P= 800 : 800 = SOO eO . 02 1 1.6= eO. 02 1 a.
=
t when N(t) = 10, 000 : 1 0 , 000= 1 000e(�I l 8 ) 1 1 0= e(�1 1 .8 )1 ln 1 0= (ln1.8)t ln10 t=--"" 3.9 days ln1.8
Find
=
7.
N(t) = No ekl
a.
Note that 18 months 2No = Noek(1.5) 2 = e1.Sk
b.
In 1.6= 0.02t
e.
3.
ln1.6 t=--"" 23.S d ays 0.02 Find t when P= 1000 : 1000= SOO e 0 02 1 2= eO. 02 1 ln2= 0.02t ln2 t= -- "" 34.7 days 0.02 decay rate:
k
=
( )
� (2) P(2)= 1O, 000e 1.5 ",25, 1 98
The population 2 years from now will be
2S,198.
9. Use A = Aoekl and solve for k:
0.5Ao = Ao/(1690) 0.5 e1690k In 0.5 = 1 690k InO.5 k= 1 690 When Ao = 1 0 and t = 50 :
-0.0244= -2.44%
=
grams
(In0.5)(50) A = 1 0e 1690
In 0.8= -0.0244t
lnO.8 "" 9.1 years -0.0244 Find t when A = 2S0 : 2S0 = SOOe-O . 0244 1 O.S= e-0.0244 1 lnO.S= -0.0244t ln O. S "" 28.4 years t= -0.0244
N(t)= No ekl b. If N(t) =1800, No = 1 000, 1800= 1 000ek ( l )
'" 9.797 grams
11. Use A = Aoekl and solve for k:
t=
5.
=
k= � 1 .5 If No =1 0, 000 and t=2 , then
b. A(1 0)= SOO e(-0 0244 ) ( 10) "" 391.7 Find t when A = 400 : c. 400= SOO e-0 .02 44 1 0.8= e-O· 0244 1
d.
1 .S years, so t 1.S.
In2 = 1 .5k
A(t) = Ao e-O. 0244 1 = SOOe-O . 0244 1
a.
=
half-life
=
S600 years
0.5Ao = Aoek(5600 ) 0.5 = e5600k In 0.5 = 5600k InO.5 k= 5600 Solve for t when A 0.3 Ao : =
.
(InO.5) 0 3 Ao = Aoe 5600 1
a.
(InO.5)
0.3 = e 5600 1 lno .5 Ino.3= t 5 600 5600 t= ( In 0.3 ) '" 9727 InO.5
and t= 1 , then
( ) --
1.8=/ k=ln1.8 8 3 If t= 3 , then N(3) = 1 000 e(hl 1 . ) ( ) = S832
The tree died approximately 9727 years ago.
mosquitoes.
345
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Chapter 6: Exponential and Logarithmic Functions
13.
a.
u= T+(uo -T)ekt with t=5 , T=70 , uo =450 , and u=300 : 300=70 +(450-70)ek(5) 230=380e5k 230 e5 -= k 380
u= T +(uo -T)ekt with t=3 , T=35 , uo =8 , and u=15 : 15=35+(8 -35)ek(3) -20=_27e3k 20 -=e3k 27
15. Using
Using
(��)=5k 23 k ='!'m( ) "" -0.1004 5 38
G�)=3k
m
T=70, uo =450, u=135 : 135=70 + (450-70) e 65=380e
(1n(20/27))(5)
u=35+(8-35)e -3-
5
At
( )
A= Aoekl and solve for k : 2.2=2.5ek(24) 0.88=e24k mO.88=24k mO.88 k= 24 When A o =2.5 andt=72 :
(1n(23/38))t
(In(2�138)) t (111(23/38)) t
90=380e
(blO.88)(72) "" 1.70 A=2.5e 24
-5-
After 3 days (72 hours), the amount of free chlorine will be 1.70 parts per million.
m
c.
The pan will be minutes.
(bl(20/27))(10)
17. Use
160=70+(450-70)e -5-
(38090 )=m(23/38) t 5 5 .m( 90 ) "" 14.3 t= m(23/38) 380
t=10 :
u=35+(8-35)e 3 ",,25.10C After 10 minutes, the thermometer will read approximately 25.1°C
m(23/38) t m � = 380 5 5 .m 65 "" 18 minutes t= 380 m(23/38) The temperature of the pan will be 135°F at about 5:18 PM. T=70, Uo =450, u=160 :
90 -=e 380
"" 18.630C
After 5 minutes, the thermometer will read approximately 18.63°C .
5
5
( )
b.
m(20/27) k= 3 At t=5 :
In(23/38) t
10(23/38) t
In(23/38) t
65 -=e 380
m
Find
t when A =1 :
(100.88)t (blO.88)t OA=e 24 1=2.5 e 24
minutes
160°F after about 14.3
mOA=
As time passes, the temperature of the pan approaches 70°F.
(m��88 }
t=�.mOA"" I72 mO.88
Ben will have to shock his pool again after 172 hours (or 7 .17 days) when the level of free chlorine reaches 1.0 parts per million.
346
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Section 6.8: Exponential Growth and Decay; Newton 's Law; Logistic Growth and Decay
19. Use
A = Ao ekt
and solve for k:
0.9 0.9 =�= = 0. 1 286 b. P(O) 1+6·1 7 1 + 6e-O. 32 ( O ) In 2000, about 12.86% of U. S . households =
0.36 = OAoi( 3O ) 0.9 = e30 k ln O.9 = 30k k = ln O.9 30 Note that 2 hours 1 20 minutes. When A = 0040 and t = 1 20 :
o
owned a DVD player.
c.
=
d.
( bI30O.9 )(l 2O) "" 0.26
A = OAOe
n o. s
O. lAo = Ao eC 8 0. 1 =
(e bI O8 .)
O 1 + 6e- . 32 1
8
e-0 ' 321 = -1 48
-0.32t = In(1 I48) t In (l / 4 8) "" 12.1 -0 . 32 Since 2000 + 1 2. 1 = 20 12. 1 , 80% of =
households will own a DVD player in 2012.
25.
a.
b. C.
As
t � 00, e -0 .439t
�
O.
Thus, P(t)
�
1000 .
The carrying capacity is 1000 grams of bacteria. Growth rate = 00439 = 43.9%.
1 000 1 + 32.33 e-0 . 439 ( 0 )
1000 = 30 33.33 The initial population was 30 grams of P(O)
=
=
bacteria. d.
}
e.
P(9)
=
1000 "" 6 1 6.6 1 + 32.33 e-0. 439 ( 9 )
After 9 hours, the population of bacteria will be about 6 1 6.6 grams. We need to find t such that P = 700:
700 =
( �}
1 000
1 + 32.33e
1 + 32.33 e-0 43 91 = .!.Q 7 4 91 32.33 e -0 3 = �
--
t = 8 · ln O. l "" 26.6 ln O.5
-0.4 3 91
7
The farmers need t o wait about 26.6 days before using the hay. a.
0.8.
8
1
ln O. l = ln .5
23.
=
=
0.9
6 e -0.321 -_ -1
( �}
21.
We need to find t such that P
=
0. 10 = OAO eC30) ( 0.9 ) t 0.25 = e 30 ln O.25 = ln 9 3 30 t = -- · ln O.25 "" 395 ln O.9 It will take approximately 395 minutes (or 6.58 hours) until 0. 1 0 M of sucrose remains. k Use A = Ao e t and solve for k : 0.5Ao = Ao ek ( 8 ) 0.5 = i k ln O.5 = 8k k = ln O.5 8 Find t when A = O.lAo : ill
owned a DVD player.
1 + 6 e-0.32 t �
t when A = 0. 1 0 :
n o. 9
=
0.8
After 2 hours, approximately 0.26 M of sucrose will remain. Find
t 2005 - 2000 = 5 0.9 0.9 "" 004070 P( 5 ) = 1 + 6 e -0 3 2( S ) 1 + 6 e- 1 .6 In 2005, about 40.7% of U.S. households
e -04 3 91
3 226.3 1
= --
-OA39t In(3/ 226.3 1) t = In(3 / 226.3 1) "" 9.85 -0.439 The population of bacteria will be 700 grams after about 9.85 hours. =
The maximum proportion is the carrying capacity, c = 0.9 = 90%.
347
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Chapter 6: Exponential and Logarithmic Functions
f.
We need to find t such that
a-rings is
57.91°F .
p = � (1000) = 500 : 2
1000 1+32.33e-0.439 1 1+32.33e-0. 439 1 = 2 32.33e -0 .4 39 t = 1 1 e-0. 439 1 = 32.33 -0.439t In (1I 32.33) In (1I32.33) t = "" 7.9 -0.439
�
\.. .
o
--
100
. '...., .
''' ''""\
o oX =3 0.06� '116 � V = 5 � 1 00
6 y = 1+e- (5 .085 -0.115 6 ( I 00» "" 0.00923 At 100° F , the predicted number of eroded
The predicted number of eroded or leaky O-rings is 5 when the temperature is about
30.07°F .
6
= y 1+e-(5.o85 -0. 115 6 (60» ",, 0.81
Section 6.9
At 60° F , the predicted number of eroded or leaky primary a-rings will be about 1.
1.
6
a.
y = 1+e-(5 .085 -0.115 6 (30» "" 5.01
At 30°F , the predicted number of eroded or leaky primary a-rings will be about 5. d.
�
6
or leaky primary a-rings will be about O.
c.
�
I�t�n�ction X='1:<.9B7BB9 � V=3
o
43.99°F .
The population of bacteria will reach one half of is carrying capacity after about 7.9 hours.
b.
\
The predicted number of eroded or leaky a-rings is 3 when the temperature is about
=
a.
when the temperature is about
6
500 =
27.
1
y.1 _
6
Use INTERSECT with
o
Y = 1, 3, 2
a
- 1 b=�========d 7
6 1+e-(5 .085 - 0. 115 6 x)
::::J 1 00 O �=====:=E:= o
6
a
o
b.
Using EXPonential REGression on the data
c.
Y = 0.0903(1.3384Y
yields:
y = 0.0903(1 .3384)"
= 0.0903 (eb1 (1 . 338 4) r
= 0.0903eh1 ( 1. 338 4)x N (t) = 0.0903e0 29 15 1 and 5 :
d.
1'; = 0.0903e0 29 15 x 1
e.
N (7) = 0.0903e(O .29 15 )- 7 "" 0.69
I n t Q n �( ti�rl � ' i::--:l= 5 7.910363 �V=1 �
1 00 o The predicted number of eroded or leaky
348
bacteria
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Section 6.9: Building Exponential, Logarithmic, and Logistic Models from Data
f.
We need to find t when
0.0903 e(O . 291 5 ). t = 0.7 5 e(O.291 5)t = �
N = 0.75 :
0.0903
0.29 1 5 t t 3.
D
a
-I ·
1
"" n
(
0 . 75
D.0903 0.29 1 5
)
J
""
7.26 hours
5.
D
40
.7 b.
(
c.
r
f. g.
l� .
.
d.
. . .�7
We need to find t when
A (t) = 0.5 . Ao
e.
1 00.3263 e( -0. 1 314)t = (0.5) ( 1 00.3263) e(-0. 1 3 1 4)t = 0.5 -0. l 3 14t = ln O . 5 In 0.5 "" 5.3 weeks t= -0. l 3 14 0. ( A (50) = 100.3263 e - 1 3 1 4 ) . 50 ",, 0. 14 grams We need to find t when
x
1995, = 3
Q
- I �'���'�'�'�'�'�'�' 1 0 450
Using EXPonential REGression on the data
= 75 1 .4698 (0.95 1 88Y
y = 75 1 . 4 698 (0.95 1 88Y
r
= 75 1 .4698 ( e1n(0 .95 188) = 75 1 .4698 e h1(0.95 188 )x A ( t ) 7 5 1 .470e( -0. 0493 2 )t =
1 10
40
)
Let x = 0 correspond to correspond to 1998, etc.
yields : y
= 1 00.3263 (0.8769Y
y = 100.3263 (0.8769Y
-I ·
J
a
�
= 100.3263 eh1(0 . 876 9) = 100.3263 e ln(0 876 9)x A ( t ) = 100.3263 e( -0 13 14) t d. � = 1 00.3263 e( -0.13 14)x
e.
a.
( (
800
Using EXPonential REGression on the data yields : y
c.
In (� 0.0903
20 1 00.3263 -0. l 3 14t = In 20 1 00.3263 20 I t = n 1 00.3263 "" 1 2 . 3 weeks -0. 1 3 1 4
� 1� 1 0�________�
a.
b.
=
e(-O . 1 3 1 4 )t =
�
= 75 1 .470 e( -O ·0493 2 )x 800
� 1 .� I O 450
Note that 201 0 is represented by t =
15 . ) A (1 5) = 75 1 .470e( -0 .0493 2 .1 5 "" 359 billion
cigarettes . f.
A ( t ) = 20 .
A ( t ) = 230 . 75 1 .470 e( -0 .0493 2 ) t = 230 e( -0 . 0493 2 )t = 230 75 1 .470 -0.04932t = In �
W e need t o find t when
( ) In ( ) = 7 5 1 .470 230
100.3263 e( -0.13 14)t = 20
75 1 .470
"" 24 years -0.04932 Now 1 995 + 24 2019. The number of t
=
cigarettes produced in the U.S. will decrease to 230 billion in the year 20 1 9.
349
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Chapter 6: Exponential and Logarithmic Functions
7.
a.
2 ;.: 4""' 0""' 0 _____---,
X � CXJ , 9. 1 968 e-0 .01 603 x � O , which means 1 + 9. 1 968 e--{)· 0 1 6 03 x � 1 , so 6.5 � 799, 475, 916.5 Y = 799, 475, 9 1.01603 1 + 9 . 1968 e --{) x
d. As
' 0
a o
a
b. c.
a
. � 1 90 1 50 : ...: ..:...� .:... .......:.., .".0;-:--'90 -,:,.
Therefore, the carrying capacity of the United States is approximately 799,475,917 people.
Using LnREGression on the data yields:
y = 32, 74 1 .02 - 6070.96 In x t; = 32, 74 1 .02 - 6070.96 ln x
e.
f.
"
: �..;....:... � . '--=....; ........:. ..:... ... 1 90 1 50 � 900
=
y = 1 650 : 1 650 = 32, 74 1 .02 - 6070.96 ln x -3 1, 09 1 .02 = -6070.96 ln x -3 1, 09 1 .02 = ln x -6070.96 5 . 1 2 1 3 "" ln x e 5. 1 2 1 3 X "" 1 68 If the price were $ 1 650, then approximately 1 68 computers would be demanded.
a.
X
- 10 ·
c.
a.
o
a
o
a
( ) ( )
0
-
Y = 799, 475, 9 106.5 1 + 9. 1968e--{)· 1 603 x y. _ 799, 475, 9 1 6.5 1 - 1 + 9. 1 96 8e-0 .0 1 603 x
. yIe Ids ·.
Therefore, the United States population will be 300,000,000 around the year 2007.
3 00 000 000
·
·
·
·
·
·
·
·
a
1 .6649 e--{).01 603 x "" 9. 1 968 In 1 .6649 "" -0.01 603x 9. 1 968 1 .6 649 In 9. 1- 968� "" X "'-0.01 603 X "" 1 07
70,000,000
·
1 975, x = 10 20 correspond to
---
a
� · · · · · · · · · · 1 1O
70,000,000
Let x = 5 correspond to correspond t o 1 980, x = 1 990, etc .
a
Using LOGISTIC REGression on the data
-10 ·
799, 475, 9 1 6.5 = 1 + 9. 1968e --{) · 01 603 x 300, 000, 000 799, 475, 9 1 6.5 1 = 9. 1 968 e --{)·01 603 x 300, 000, 000 1 .6649 "" 9. 1 968 e-0.01 603 x
80,000
a
a
b.
11.
Let x = 0 correspond to 1 900, x = 10 correspond to 1 9 10, x = 20 correspond to 1 920, etc. 3 00 000 000
(
799 , 4 7 5, 9 1 6.5 3 00, 000, ooo 1 + 9. 1 968e -0 0 1 603.
d. We need to find x when
9.
�----'-----:--
--
2400
""
=
2004 corresponds to x 104, so 799, 475, 9 1 6.5 y= 1 + 9. 1 968e -0 . 01 603( 104 ) "" 292, 1 77, 932 people Find x when y = 300, 000, 000 799, 475, 9 1 6 .5 = 300 000 000 1 + 9. 1 968 e-0.01 60 3 x
The year
·
·
b.
Using LOGISTIC REGression on the data . Yle Id s :
68, 684.7826 ---::-:-::-:=_ Y = ---'-1 + 1 8.94 1 6 e-0 . 19743 x
1 10 350
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)
Chapter 6 Review Exercises
y. _
c.
I
b.
(g o f)(-2)= g(f(- 2)) =g (3(-2) - 5) = g(-l 1) = 1 - 2(-1 1)2 =- 241 (f f)(4) = f(f(4)) c. = f (3(4) -5) =f(7) =3(7) - 5 =1 6 d. (g o g)(-I)=g(g(-I)) =g ( I- 2(-1)2 ) = g( - 1) =1 - 2(-1)2 =- 1
68,684.7826 1+1 8.94 1 6e-o.19 74 3x
-
80,000
0
� 00 , 1 8.94 1 6e-o· 197 43x � 0 , which o means 1 + 1 8.94 1 6e- .19 7 4 3x � 1 , so
d. As
x
y=
68,684.7826 � 68,684.7826 1 +18.94 1 6e- o· 1974 3x
Therefore, the carrying capacity of the cable TV market in the U .S . is about 68,685,000 subscribers. e.
The year 20 15 corresponds to
x
=45 , so
68,684.7826 "" 68,505 1 + 18.94 1 6e- o.19 74 3(4 5) In 20 15, cable TV will have approximately 68,505,000 subscribers in the U.S.
y=
3.
.
d. The year 2009 corresponds to x=25 , so
268.9893 "" 245 .04 . 1 +219.0605e- 0. 3 086 ( 25) For 2009, the function predicts that the
y=
The answer in part (e) of Example 1 predicts 1 ,5 12,0 10,000 cell phone subscribers in the U. S . in 2009. This prediction is more than 6 times our prediction in part (d). The function in Example 1 assumes exponential growth which is unlimited growth. Our logistic function in part (d) assumes that the growth is limited.
x
x
x
= f (.J6) =�.J6 +2 d. (g g)( -1) = g(g( - 1)) =g ( 2(-1) 2 +1 ) =g(3) =2(3) 2 +1 =19 0
x
f( )=3 - 5 g( )=1 - 2 2 (f g)(2) = f(g(2)) =f ( 1 - 2(2)2 ) = f(-7) = 3(-7) - 5 = -26
a.
0
0
0
Ch apter 6 Review E xercises
1.
a.
0
number of cell phone subscribers will be approximately 245,040,000 subscribers. e.
g(x)=2x2+1 f(x)=.Jx +2 (f g)(2) = f(g(2)) = f ( 2(2) 2 +1 ) = f(9) =.J9+2 =� b. (g f)( -2) =g(f(- 2)) =g ( �- 2+2 ) = g(O) =2(0)2 +1 =1 c. (f f)(4) = f(f(4)) =f ( .J4 +2 )
351
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exist. No portion of this material may b e reproduced, in any form or by any means , without permi ssion in wri ting from the publi sher.
Chapter 6: Exponential and Logarithmic Functions
5.
g( x) = 3x - 2 f ( x) = ex ( f g )(2) = f (g(2)) = f ( 3(2) - 2 ) = f (4) = e4 b. ( g o f)(- 2) = g ( f (- 2 )) = g e-2 = 3 e-2 - 2 3 -2 =e2
a.
0
(g g )(X) = g(g(X)) = g (3x + 1) = 3(3x + l) + 1 = 9x + 3 + 1 = 9x + 4 Domain: { x I x is any real number } .
0
( )
7.
9.
0
( )
c.
( f f)(4) = f ( f (4)) = f e 4 = e e4
d.
(g o g)(-I) = g(g(-I)) = g ( 3( -1) - 2 = g(-5) = 3(-5) - 2 = -17
0
1
g(x) = 3 x l f (x) = 3x2 + x + 1 The domain of f is { x I x is any real number } . The domain f g is { x I x is any real number } . 0
( f g)(x) = f (g(x)) = f 3x I ) = 3 3x 2 + l 3x l + 1 = 27x 2 + 3 I x l + l Domain: { x I x is any real number } .
)
0
(1 (1 1 ) ( )
(g f )(x) = g( f (x)) = g 3x 2 + X + 1 = 3 3x 2 + x + l
f (x) = 2 - x
g(x) = 3x + l The domain of f is { x I x is any real number } . The domain of g is { x I x is any real number } . 0
( f g)(x) = f (g(x)) = f (3x + 1) = 2 - (3x + l) = 2 - 3x - l = 1 - 3x Domain: { x I x is any real number } .
Domain:
( ) 1( )1 = 3 1 3x 2 + x + 1 1
{ x I x is any real number } .
0
( f f )(x) = f ( f (x)) = f 3x 2 + x + l
( ) = 3 ( 3x 2 + X + 1 r + ( 3x 2 + X + 1 ) + 1 = 3 ( 9x4 + 6x3 + 7X 2 + 2x + l ) + 3x 2 + x + l + l
0
(g f )( x) = g( f (x)) = g(2 - x) = 3(2 - x) + 1 = 6 - 3x + l = 7 - 3x Domain: { x I x is any real number } .
= 27x4 + 1 8x3 + 24x 2 + 7x + 5 Domain: { x I x is any real number } . 0
(g g)(x) = g(g(x)) = g 3x I ) = 1 3 3x l l =9 xl Domain: { x I x is any real number } .
1
0
( f f )(x) = f ( f (x)) = f (2 - x) = 2 - (2 - x) =2-2+x =x Domain: { x I x is any real number } .
(1 1
352
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Chapter 6 Review Exercises
11.
15. The function/is one-to-one because every
g(x) = -1 x The domain of / is {xl x otd} . The domain of g is {xl x * o} . /(x) = x + l x-I
0
horizontal line intersects the graph at exactly one point.
-4
-
"f(�)" t:
( 3 /
x
17.
l+x I-x
Domain
(J
0
{xl x * 0, x * I} .
,
/
-
/
l
/
t
/
/
-4
2x+3 5x-2 2x+3 y = -5x-2 x = 2y + 3 5y -2
le x) =
+1 =f ( X+I = :2-i ) x+x-I1 x-I +I)(X-I) ( X+I x-I X+I _I)(X_I) ( x-I x+ I +x-I 2x2 x+I-(x-l)
Domain of /
-I
(g o g) (x) " g (g (x)h Domain
13.
a.
b.
Range of /
19.
=-=X
{xl x * I} .
(�) " d)
=
Range of r l
=
All real numbers except
=
1 Domain of /-
=
All real numbers except
x-II y= x-II x --I y-
/(x) = _1-
=
-2 -.52 5
.
Inverse
x(y - I) = 1 xy - x = 1 .xy = x + l y=-
" X
x+1x rl (x) = x+ x
{xl x * O} .
I
The function is one-to-one because there are no two distinct inputs that correspond to the same output. The inverse is
Inverse
+ 2x+3 5x-2 rl(x) = 2x+3 5x-2
/ ) (x) = / ( J (x) )
Domain
x
/
x(5 y - 2) = 2 y + 3 5.xy - 2x = 2y + 3 5xy - 2 y = 2x + 3 y (5x - 2) = 2x 3 Y=
{xl x * - 1 , x * I}
=
/
4
(J g) (x) = / ( g (x))
Domain
y = x/
y
Domain of /
=
=
Range of
rl
All real numbers except 1 Range of / = Domain of /-
{(2, 1) , ( 5, 3) , ( S , 5 ) , (10, 6 )} .
=
1
All real numbers except °
353
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Chapter 6: Exponential and Logarithmic Functions
21.
3 I (x)=l7'3 x 3 y=l7'3 x 3 Inverse x =---u3 y /3 1 � =3 3 yl /3 x
(2,00)
(1,2) -3 3 2 2 --41 2 Value ofp Conclusion positive negative positive Thus, the domain of H (x )=log2 (X2 -3x + 2) is {x l x < 1 orx>2 } or (-00,1) u(2,00) . Interval (-00,1) Test Value 0
FI (X)= 273 x Domain of I Range of I- I All real numbers except 0 Range of I Domain of I- I All real numbers except 0
35.
In e.f2 = J2
39.
log3
=
=
=
=
23.
a.
b.
25. 27. 29.
1(4)=34 = 8 1 g(9)= log3 (9)= log3 (32 )=2
C.
1(-2)=T2 =.!9
d.
g
( 2\ )=log3 ( 2\ )
log3 ( T3 )=-3
52 = is equivalent to 2= logs logs =13 is equivalent to 5 13= z
z
u
43.
W
log3 + log3 v2 -log3 =log3 + 2 log3 -log3 log (x 2 .Jx 3 +1 )= logx 2 +log (x 3+1(2 =2 l0gx + log (x 3+1) U
V
U
W
W
�
In
[ x� )=In(Ah l )-In(X-3) j
I (x )= log(3x-2) requires: 3x-2> 0 x>-2 3 Domain: x x >% or
45.
(x ) log2 (X2 - 3x + 2 ) requires p(x)=x 2 -3x +2> 0 (x-2)(x-l)> 0 x =2 and x = 1 are the zeros of p . H
2
3 =lnX +ln (X2 +1t -In(x-3) = lnx + ln (x2 +1)- In(x - 3)
U
{ l } (%,00)
31.
u
=
41. =
[ : )=log3 uv2 -log3
3 3log4 X2 + "21 log4 "rX = log4 (X2 ) +log4 (X 1 /2 )1/2 = 1og4 X6 + 1og4 X 1 /4 = log4 (x6 . XI /4 ) log4 X2S /4 25 log x =4 4 =
=
354
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Chapter 6 Review Exercises
47.
In ( X� I +lnC: J-ln(X 2 -1) JI x ( ( x I 2 =In --:;- . x + 1 J- n x -1) X-I =In X2x +_1I 1 =In X - I ' x + I (x - l)(x + l) I =In-(x + 1)2 =In(x+lr 2 =-2ln(x+1) I Iog(x+3)+ log(x 2) ] 210g2 +310gx --[ 2 = log2 2 + logx3 -.!.. log [ (x + 3 )(x-2) ] 2 = log (4x3)- log( x+3)(x -2) ( 2 4x3 = log [(x + 3)(x 2)]112
[ ] (
49.
d.
)
e. f.
f(x)=2 x-3 y = 2x -3 =2y- 3 Inverse y-3= log2 x Y =3+log2 x l F (X) = 3+ log2 x
Range off (0, 00 ) Using the graph of y= log2 X , shift the graph vertically 3 units up.
- J
-5 -5
57.
ln x
= log3 X = ln3 1 :/
b.
18
I'
x = o
±
f(X)= ( 3-X ) 3.
3
x
5
�
-1 1
y=0
y 5
ln19 2.124 log4 19=-ln4
53. r;
-3
55.
Range: (0, 00 ) Horizontal Asymptote: X
-
(
51.
c.
Domain: (-00, 00 ) Using the graph of y = 3x reflect the graph about the y-axis, and compress vertically by a factor of -2I .
,
f(x) =2x-3 3.
b.
Domain: (-00, 00 ) Using the graph of y=2x shift the graph horizontally 3 units to the right.
,
y 7 t-
-1
c.
7
-3
Range: (0, 00 ) Horizontal Asymptote:
y =0
x
355
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Chapter 6: Exponential and Logarithmic Functions
d.
� (TX ) = � ( TX ) = � ( TY )
d.
f(x) = y x
Inverse
e.
f.
=
log 3 (2x)
-
= - log 3 (2x) rl (x) = - log 3 (2x) 2x > 0 y
x>O Range off
= l - e-x x = l - e-Y Inverse x - l = -e- Y l - x = e- Y -Y = ln ( l - x ) y = ln ( l x ) r l ( x) = - In ( l - x ) y
2x = TY -y
f (x) = l - e-x
e.
(0, 00)
0 -1 x<1
I-x >
-x >
Range off
Using the graph of y
= log 3 x , compress the
graph horizontally by a factor of reflect about the x-axis. y 5
f.
.! , and 2
-
(-00, 1)
Using the graph of y
= ln x , reflect the
graph about the y-axis, shift to the right unit, and reflect about the x-axis. y� I
1
5 x
-5
I
x = O
7
( � , - 2)
(- 1 .72,- 1 )
-5
1
f(x) = l - e- x
59. a.
Domain:
b.
Using the graph of y
x :: l
61. f(x) = -In ( x + 3 ) 2
(-00 , 00) = eX , reflect about the
y-axis, reflect about the x-axis, and shift up unit.
1
a.
Domain:
b.
Using the graph of y
Y
(-3, 00)
to the left
= ln x , shift the graph
3 units and compress vertically by
a factor of
t. y
5
5
(0, 0.55) c.
Range:
(-00, 1)
Horizontal Asymptote : y
-5 =
c.
1
Range:
(-00 , 00)
Vertical Asymptote : x =
-3
356
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Chapter 6 Review Exercises
6 7.
.!.2 In (x+3) y=.!.ln(x+3) 2 1 x = -In(y +3) 2 2x= In (y+3) y+3=e2 x y=e2x -3 F I (x) = e2 x _3
d.
I(x) =
(0
e.
Range off ( - 00,
f.
Using the graph of
3 units.
( r
Inverse
6 1 1 = -1 x= --
�
4
The solution set is
y = eX , compress
1- ' and shift
x ln 5 = x ln 3 + 2 ln 3
x In 5 - x ln 3 = 2 ln 3
x(ln5 -ln3)=2 ln3
y
2 ln 3 ln5-ln3 4.301 21 3 The solution set is { 4 . 3 0 1} . 5 3 x=
41-2x = 2
12 X=-
5
The solution set is
4
The solution set is
}
""
32
../x - 2 = 9
x - 2 = 92
=81 x=83 log 3 ../83 2 = log 3 J8i
x-2
1
x +X=-
2
+ 2x - l=0
Check:
2 _-4-(2-)(-_-1) -2 ± �r-2x= 2(2) -2 ± ..f0. -2 ± 2.J3 -1 ± .J3 2 4 4
{ �.J3 , �.J3 }
{I: } .
73 . log 3 "/x - 2 = 2 ../x - 2 =
{�} .
y2+X = .J3 3x2+x = 3 1 / 2 2x 2
- In
-5x = - 1 2
22 -4x = i 2 - 4x = 1 - 4x=-1 1 X=-
2
n
In
4x = 9x - 1 2
( 22 y- 2 X = 2
65 .
{
""
4 (32 t = (33 t34x = 39x-12
-5
63 .
{±} .
In (5x ) = In (3 H2 ) x ln 5 = ( x+ 2) ln 3
)
horizontally by a factor of down
=-3
log 64 x x-3 = 64 l/3 = 4- 1 3 x-3
-
= log 3 9
=2
The solution set is
{83} .
The solution is -I
-I
""
{-1.366, 0.366}. 3 57
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Chapter 6: Exponential and Logarithmic Functions
83.
a.
2
2 3 = 2 2 x +5.1 3 = 2X 2 + 5x 0 = 2X 2 + 5x - 3 o = (2x - 1)(x + 3) x = -1 or x = -3 2 The solution set is 77.
79 .
81.
Y x= 2
10
{- 3, �} .
I (x)= \og2 (x - 2)+ 1
( 6, 3) -10
log 6 (x + 3) + log 6 (x + 4) = 1 log 6 ( x + 3)(x + 4») = I (x + 3)(x + 4) = 61 x 2 + 7x + 12 = 6 x 2 + 7x + 6 = 0 (x + 6)(x + I) = 0 x = - 6 or x = -1 Since log 6 (-6 + 3) = log 6 (-3) is undefined, the solution set is {-I} . e1 - x = 5 l - x = 1n 5 -x = -I + 1n 5 x = 1 - 1n 5 -0.609 The solution set is { 1 - ln 5 } {-0.609} . ""
f ( x) = log 2 ( x - 2) + 1 Using the graph of y = log 2 X , shift the graph right 2 units and up 1 unit.
""
9x + 4 · 3 x - 3 = 0 ( 32 y + 4 · 3x - 3 = 0 ( 3x ) 2 + 4 · 3 x - 3 = 0 Let u = 3.1 . u 2 + 4u - 3 = 0 a = l, b = 4, c = -3 -----4) ± �r( ( 4 ) 2 - 4(1)( -3) ����--��� u= 2(1) 17 .fi8 -4 ± = = -4 ± 2 = -2 ± 17 2 2 � or Y = -2 + 17 J' can't be negative x = log 3 ( -2 + 17 ) _
b.
f (6 ) = log 2 ( 6 - 2 ) + 1 = log 2 (4) + 1 = 2 + 1 = 3 The solution set is {3 } . The point (6, 3) is on the graph off
c.
f ( x) = 4 log 2 (x - 2) + 1 = 4 log 2 (x - 2) = 3 x - 2 = 23 x-2 = 8 x = 10 The solution set is { I O} . The point (10, 4) is on the graph off
d.
f (x ) = 0 log 2 (x - 2) + 1 = 0 log 2 ( x - 2) = -1 x - 2 = T1 x - 2 = -1 2 x = -5 2 Based on the graph drawn in part (a), f ( x ) > 0 when x > 2. . The solution set is 2 x I x > or
{ %} (%,00).
The solution set is { log 3 ( -2 + 17) } "" {-0.398} . 358
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Chapter 6 Review Exercises
e.
) y = log 2 ( X - 2 ) + 1 x = log2 ( Y - 2 ) + 1 x - I = log2 ( )
f ( x ) = log2 (x - 2 + 1
y _ 2 = 2x-
1
y-2
10
;
;
;
;
;
;
I-I (x) =
2x-1 2 +
1n 0.05 =
;
t = 20 1 5 - 2005 = 1 0
)
o 5 P = Pa il = 6, 45 1, 05 8, 790e . O I 1 ( I O)
( ) 760
"'" 7, 237, 27 1, 5 0 1 people
300
97. A =
87. P = 25 e o. 1 d
P = 25 eo. 1 ( 4) = 25eo.4 "'" 3 7 . 3 watts
a.
2 = e O. 1 d
b.
n=
99.
0. 1
1 70
a.
a
log 1 0, 000 - log 90, 000 10g(l - 0.20)
( )
o
()
n = _--'-�'--_...o....!. 10g(1 - 0. 1 5)
( �)
10g 0 . 5
log 0 . 8 5
log 0 . 8 5
nt
"'" $4 1, 668.97
b.
. .
1 50
a a
a
a
a
a
a
a
a
. .. .. . . .
.
.
a
.
a
a
a
. . .
16
Using EXPonential REGression on the data
(
){ 0.995 1 Y Yj = ( 1 65 . 73 )( 0.995 1 Y
yields : y = 1 65 .73
0 i ' I
( ;r
a
""' 9 . 85 years
log 0.5i - log i
91. P = A I +
)1 0 "", 4 8 3 . 67
The government would have to pay back approximately $483 .67 billion in 20 1 5 .
In 2 "'" 6.9 decibels
10g
11
= 3 1 9 ( 1 .042 5
1n 2 = O . ld
89.
( r ( 0.0�25 ) ( 0)
p I+�
A = 3 19 1 +
50 = 25eO I d
d=
"'" 24 203
95. Pa = 6, 45 1, 058, 790 , k = 0. 0 1 1 5 , and
"'" 3229.5 meters
b.
=
(In O. 5 ) t 5600 InO. 05 5600 , (�)
The man died approximately 24,203 years ago.
-1 0
(
I
0.05 = e
t
85 . h(300) = 30(0) + 8000 IOg
a.
In5600 1 ( In5600o. 5 )
0.05 Ao = Aoe
y x=2
;
In O. 5 5600( 0. 5 )
k=
=
r 1 (x)
y = x/ ;
0.5 = e 5600 k In 0 . 5 = 5 600 k
Inverse
2x-1 + 2 = 2 x- 1 + 2
Y
o 0.5 Ao = Ao ek( 5600 ) A = A e kl
93.
( �r
= 85, 000 1 +
c.
""' 4 . 27 years
0. 4
2
1 70
(J 8)
359
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Chapter 6: Exponential and Logarithmic Functions
e.
d. Find x when y= 1 10 .
(165.73)(0.9951)' = 110 (0.9951Y = � 165.73 xlnO.9951 = In � 165.73 110 In 165.73 "" 83 x= lnO.9951 Therefore, it will take approximately 83
Find
46.9292 = 10 1 21.2733e-0. 73061 46.9292 = 10 (1 + 21.2733e- 0 7306 1 ) 46.9292 + 06 = 1 21.2733e-O · 73 1 10 46.9292 -1 = 2 1.2733e 10 3.69292 = 21.2733e- 0 73061 +
( ) ( )
-0.73061
3.69292 e-0.7 3 061 = 21.2733 3.69292 -0.7306t = In 21.2733 3.69292 In 2 1.2733 -'---� = I -0.7306 1 ",,2.4
( (
seconds for the probe to reach a temperature of 110°F.
101.
--, _ ...::;.; 5 0::... ,... _ __
a.
t when C = 10 .
---
-
- 1 �=======.l 9 o
b.
c.
y.
f.
46.93 1 + 2 1.273e- . 061 46 .93 _ + 1 21.273e-0.73061
I -
-I
d.
Therefore, after approximately 2.4 days (during the 10th hour on the 3rd day), 10 people had caught the cold.
The data appear to have a logistic relation Using LOGISTIC REGression on the data yields : C=
0 73
50
/,
o
1� 21.2733e-0.73 061 � 0 , which . means 1 + 21.2733e- 0 73 061 � 1 , so 46.9292 C= � 46.9292 + 1 2 1.2733e- 0.7 3 061
As
) )
oc) ,
1 when C= 46 . 46.9292 = 46 1 + 2 1.2733e-0 7306 1 46.9292 = 46 (1 + 21.2733e-0 73 06 1 ) 46.9292 = 1 + 21.2733e- 0. 73 061 46 46.9292 1 = 21.2733e-O·73061 46 0.0202 = 21.2733e-0 73 061 0.0202 = e-O · 73061 21.2733 _0_.0_2_0_2_= e-0.73061 21.2733 0.0202 = In 21.2733 -0.73061 0.0202 In 21.2733 -- = 1 -0.7306 I "" 9.5 Therefore, after approximately 9.5 days (during the 12th hour on the 10th day), 46
Find
( (
Therefore, according to the function, a maximum of about 47 people can catch the cold.
--'-
In reality, all 50 people living in the town might catch the cold.
) )
"';:"
people had caught the cold.
360
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Chapter 6 Test
Chapter 6 Test
1.
I( ) = x+2 g(x) = 2x + 5 x-2 The domain of I is {x I x * 2} . The domain of g is all real numbers. (jog)(x) = I(g(x») = 1 (2x + 5) X
3.
a.
(2x+ 5)+2 (2x+ 5) - 2 2x+7 2x+3
b.
c.
2.
a.
(go/ )(x) = g(j(-2») = g C�: �) = g( O) = 2(0) + 5 =5 (jog)(x) = I{g( -2)) = 1(2(-2) + 5) = / (1) = 11 +- 22 = 2 -1 = -3 Graph y = 4x2 + 3 y
-5
=
=
=
=
4.
•
:
3x = 243 3 x = 35 x=5 6. 10gb 16 = 2 b2 = 16 b = ±M = ±4 Since the base of a logarithm must be positive, the only viable solution is b = 4 . 7. log5 =4 x = 54 x = 625 8. + 2 22.086 9. log 20 1 .301 5.
5
x
The function is not one-to-one because it fails the horizontal line test. A horizontal line (for example, y = 4 ) intersects the graph twice. b. Graph y = .Jx + 3 - 5
(-3,- 5)
X
:
y
-5
The function is one-to-one because it passes the horizontal line test. Every horizontal line intersects the graph at most once. 2 j(x) = -3x - 5 2 y = -3x - 5 2 Inverse x = -3y - 5 x(3y - 5) = 2 3.xy- 5x = 2 3.xy = 5x + 2 5x + 2 y = -3x 5x l r (x) = 3x+ 2 Domain of I Range of I-I All real numbers except -53 . Range of I Domain of I-I All real numbers except O. If the point (3, -5) is on the graph off, then the point (-5, 3) must be on the graph of rl
5
e3
x
�
�
10.
-8
In 21 2.771 log 3 21 = -In3 �
361
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exist. No portion of this material may be reproduced, in any fonn or by any means, without pennission in writing /Tom the publisher.
Chapter 6: Exponential and Logarithmic Functions
11. 12.
In
133 "" 4.89 0 f(x) = 4x+1 -2 Domain: (-00, 00 ) " b. Using the graph of y = 4 , shift the graph 1 unit to the left, and shift 2 units down.
13.
!(x) = I - log5 ( x - 2 ) Domain: (2, 00 ) b. Using the graph of y = log5 , shift the graph to the right 2 units, reflect vertically about the y-axis, and shift up I unit. a.
X
a.
y
y
6
(3,1) -
5
5
x
y
-----
=-
I I I I
2
-5 c.
d.
e.
f.
c.
Range: (-2, 00 ) Horizontal Asymptote: y = -2 f( x) = 4"+1 - 2 y = 4 x+1 _ 2 x = 4y+I - 2 Inverse x + 2 = 4 y+ 1 y + 1 = log4 ( x + 2) y = log4 ( x + 2) - 1 l r (x) = log4 (x + 2) - 1 Range off (-2, 00) Using the graph of y = log4 x , shift the graph 2 units to the left, and shift down I unit.
d.
e. f.
x=2
Range: ( - 00, 00) Vertical Asymptote: x = 2 !(x) = I - log5 ( x - 2 ) y = I - 10gs ( x - 2 ) x = 1 -10g5 ( y - 2 ) Inverse x - I = -log5 ( y - 2 ) l - x = log5 ( y-2 ) y - 2 = 51-x y = 51-x + 2 rl(x) = 51-x + 2 Range off (-00, 00 ) Using the graph of y = 5x, reflect the graph horizontally about the y-axis, shift to the right I unit, and shift up 2 units. y
7
y 5
(2,0)
x
x
-3
I
1-5 I x =-2
-3
(0, 7)
7
362
x
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 6 Test
5 H2=::125 5 H2=::53 x + 2=::3 x=:: 1 The solution set is {I}. 15 . log(x + 9)=::2 x + 9 =:: 102 x + 9 =:: 100 x=:: 91 The solution set is { 91 }.
14.
19 . log2 (x - 4) + log2 (x + 4)=::3 log2 [( - 4)( x + 4) ]=::3 log2 ( x2 - 16 )=::3 x2 - 1 6 =:: 23 x2 -16 =:: 8 x2 =:: 24 x=::±J24=::±216 Because x -216 results in a negative arguments for the original logarithms, the only viable solution is x=::216. That is, the solution set is {216} "" {4.899}. 3 20. log2 2 4x x - 3x - 18 23 log2 (x +23)(xx - 6) == log2 ( 22 x3 ) - log2 [(x - 6)(x + 3)] =::log2 22 + log2 x3 - [log2 (x - 6) + log2 (x + 3)] =::2 + 31og2 - log2 (x - 6) - log2 (x +3) 21 . A == Ao ela 34 =:: 50/ (30) 0 . 68=::e30k In O .68 30k k== In O .68 30 Cno.68)1 Thus, the decay model is A == 50e 30 . We need to find t when A== 2 : (Ino.68)1 2 == 50e 30 (Ino.68)1 0.04 e 30 In��6 8 In 0. 04 == In 0.04 t == In��68 "" 250.39 X
=::
16 . 8-2e-x=::4 -2e-x=::-4 e-x=::2 -x=::In2 x=:: - ln2 "" -0.693 The solution set is { -In 2 } "" { -0.693} . 17 . 1og ( x2 + 3 )=:: log (x + 6} x2 + 3=::x + 6 x2 - x -3 == 0 - - - - - -=- 2 - 4(1 )(-3) l± .Ji3 ± �r( 1 ) -(-1) =:: -x=:: 2(1) 2 The solutIOn set 1S -"" { -1 . 303, 2.303} .
[
=::
7x+3== eX In7x+3=::In eX (x + 3)In7=:: x xIn7 + 3In7=::x xIn7 - x=::-3In7 x(In7 - 1)== -3In7 x== In-31n-I7 7
==
31n7
I-In 7
[
)
X
. . {1--JI32- ' I+JI3} 2
18 .
)
=::
==
",, -6.172
31n7 } ",, { -6.172}. The solution set is { 1-l n7
( } ( )
There will be 2 mg of the substance remaining after about 250.39 days.
363
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 6: Exponential and Logarithmic Functions
22.
a.
b.
Note that 8 months -23 year. Thus, 2 P = 1000 , r = 0.05 , n = 12 , and t = 3 =
Let n represent the number of people who must shout. Then the intensity will be nx 10-4. If D = 125 , then 125 = 1010g nx l 125 = 10 log ( nx 108 ) 12.5 = log ( nx108 ) nx 108 = 1012 .5 n = 104.5 "" 3 1, 623 About 3 1,623 people would have to shout at the same time in order for the resulting sound level to meet the pain threshold .
( O��:)
( (3 So, A = 1000 ( 1 +-1'-2 ) 12) 2/ ) = 1000 ( 1 + O��5r 005
b.
",,$ 1033.82 Note that 9 months -43 year. Thus, A = 1000 , r = 0.05 , n = 4 , and t = -43 So, 0.05 1000 = Ao 1 +4 1000 = Ao ( 1.0125 ) 3 Ao = 1000 3 ""$ 963.42 ( 1.0125 ) r = 0. 0 6 and n = 1 . So, 2Ao = Ao l + 0. 6 ( 2Ao = Ao(1.06Y 2 = (1 .06Y ln 2 t = log . 06 2 = l ln1 .06 "" 1 1 .9 It will take about 1 1.9 years to double your money under these conditions. =
(
c.
)(4)(3/4)
( � ) 1)
Chapter 6 Cumulative Review 1.
3.
1
a.
C ) C ) 10-12
80 = 10l0g o�12 8 = IOg :-12 8 = logI - log 8 = logI - (-12) 8 = logI + 12 -4 = log I I = 10-4 = 0.0001 If one person shouts, the intensity is 10-4 watts per square meter. Thus, if two people shout at the same time, the intensity will be 2 x 10-4 watts per square meter. Thus, the loudness will be = l O log ( 2 x 108 ) "" 83 D = 10l0g decibels
2
a.
--
23.
The graph represents a function since it passes the Vertical Line Test. The function is not a one-to-one function since the graph fails the Horizontal Line Test. x +l =1 2 ! ! "'" ' 2 + 2 = 4 4 = 2 l ' 2 ' 2 is not on the graph. ! + J3 = ! +� = l ' J3 is on b. 4 4 ' 2' 2 2 2 the graph. 2x -4y = 16 x-intercept: y-intercept: 2 ( 0 ) -4y = 16 2x -4 ( 0 ) = 16 -4y = 16 2x = 16 x=8 y = -4
5.
(!J2 ( ) +! ! (!!J ( ) 2 ( )2 (! )
y 10
x �;4 J ( 21O�
-10
364
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 6 Cumulative Review
Given that the graph of I(x) = ax2 + bx + c has vertex (4, -8) and passes through the point b =4, (0,24) , we can conclude -2a I( 4) = -8 , and 1(0) = 24 . Noti ce that 1 (0) = 24 a (0) 2 + b ( O ) + c = 24 c = 24 Therefore, I (x) = ax2 + bx + c = + bx + 24 . b = 4 , so that b = -8a , Furthermore, -2a and 1 (4) = -8 a ( 4) 2 + b ( 4) + 24 = -8 l6a + 4b + 24 = -8 l 6a + 4b = -32 4a + b = -8 Replacing b with -8a in this equation yields 4a -8a = -8 -4a = -8 a=2 So b = -8a = -8(2) = -16 . Therefore, we have the function I (x) = 2X2 - 16x + 24 . 2 g(x) = 9. I (x) = x2 + 2 x- 3 I(g(x)) = I � 3 = _2 x-3 + 2 4 ---=-2 + 2 (x-3) The domain ofIis {xI x is any real number} . The domain of g is {xl x:;to 3} . So, the domain of I(g(x)) is {xl x:;to 3} . 4 + 2 = -4 + 2 = 3 l (g(S)) = (5 _43) 2 + 2 = 4 22
7.
11.
a.
g(x) = 3 x + 2 Using the graph of y = 3 x , shift up 2 units . y
10
y
=
2
-
- - �-=-:::� x
ax2
-5
Domain of g: (- 00, 00) Range of g: (2, 00 ) Horizontal Asymptote for g: y 2 g(x) = 3X +2 Y = 3x + 2 x = Y + 2 Inverse x-2 = Y y = log 3 (x-2) g-I (x) log 3 (x-2) Domain of g-I : (2, 00 ) Range of g -I : (-00, 00 ) Vertical Asymptote for g-I : x = 2 =
b.
=
c.
e ) ( )
v=x / /
.'
2
/
/
/
/
/
/
/
/
x g -l(x)
-5
365
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in wri ting from the publisher.
Chapter 6: Exponential and Logarithmic Functions
13.
log3 (x + 1) + log3 (2x - 3) = log9 9 log3 ( x + l)(2x - 3)) = 1 (x + l)(2x -3) = 31 2X2 - x - 3 = 3 2X2 - x - 6 = 0 (2x+ 3)(x-2) = 0 x = 23 or x = 2 --
( ) (-�) is undefined
Since log3 -% + 1 = log3 the solution set is {2} . 15.
a.
2;,;:o:...-_____-.
a
0�=====l80 b.
c.
o
Answers will vary. Answers will vary.
366
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exist. No portion of this material may be reproduced , in any form or by any means, without permission in writing from the publisher.
Chapter 7 Trigonometric Functions Section 7.1 1. 3.
5.
25.
C
= 27rr standard position �,!!.-
27.
7. True 9. True
29.
11.
13.
15. 31.
17.
19. 33.
21.
23.
(
)
1 .� .� 0 40010'25"= 40 + 10 ._ 60 + 25 60 60 "" (40 + 0 . 1667 + 0.00694)° "" 40.17°
(
)
(
)o
.�. � 0 1°2'3" = 1 + 2 . � + 3 60 60 60 "" (1 + 0.0 333 + 0.00083t "" 1 .03° .� . � 9°9'9" = 9 + 9· � 60 + 9 60 60 = (9 + 0.15 +0.0025t "" 9. 15° 40.32°= 40°+ 0.32° = 40°+ 0.32(60') = 40°+19.2' = 40°+19'+0.2' = 40°+19'+0.2(60") = 40°+19'+ 12" = 40°19'12" 18.255°= 18°+ 0.255° = 18°+ 0.255(60') = 1 8°+15.3' = 1 8°+15'+0.3' = 18°+15'+ 0.3(60") = 1 8°+15'+ 18" = 18°15'18" 19.99°= 19°+ 0.99° = 19°+ 0.99(60') = 19°+59.4' = 19°+59'+0.4' = 19°+59'+0.4(60") = 19°+59'+ 24" = 19°59'24"
35.
300 = 30 . � 180 radian = 2:.6 radian
37.
41t rad'lans 1t d' 2400 = 240·ra Ian 3 1 80 = -
367
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in wri ting from the publisher.
Chapter 7: Trigonometric Functions
180 degrees",362.1 1° . 6. 32 rad·lans = 6.32·7t
39.
- 600 = -60 · � 180 radian = _2:3 radian
69
41.
1800 = 180 . � 180 radian = 7t radians
71.
43.
-135 0 = - 135 · � 180 radian
45.
-900 = -90 . � 1 80 radian = - 2:2 radians
47.
49.
51.
53.
55.
57.
59.
61.
63.
= _
s
37t radians 4 73.
--
75.
-_.-
67.
r = 5 miles; = 3 miles; = r() () = !..r = �5 = 0.6 radian s
s
77.
_.-
· r = 2 inches; () = 300= 30·� 1 80 = 2:6 radian ' s
r()
= = 2 · 2:6 = 2:3'" 1.047 inches
-
81.
-400 = - 40·� 1 80 radian 27t rad Ian . = -9 '" -0.70 radian
() = 3.!. radian·' A = 2 fe A
= '!'r2() 2
=�r2G) 2 = .!.6 r2 12 = r2 r = = 213'" 3 .464 feet 2
1250 = 125 ·� 1 80 radian 257t radlans . 36 '" 2.1 8 radians 3.14 radians = 3.14·180 7t degrees'" 179.910 2 radians = 2 · 180 7t degrees'" 1 14.59°
.Jl2
= --
65.
s
s
-
-
() = 3.!. radian; = 2 feet; = r() 2 = 6 feet r = -() = (1/3)
s
180 degrees = 600 -7t3 = 7t_3 . 7t 57t = 57t 180 degrees = - 225 0 4 4 7t 7t = 7t_. 180 degrees = 900 2 2 7t 7t 7t 180 12 = 12 7t degrees = 150 180 degrees = -90 0 --7t2 = --7t2 . 7t - 7t6 = - -7t6 . 180 7t degrees = -30 0 1 77t radian'" 0.30 radian 170 = 17·� 180 radian = 180 --
r = 10 meters; () = .!.2 radian; = r() = 10·-21 = 5 meters
368
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.1: Angles and Their Measure
83.
85.
87.
r =5 miles; A =3 mi2 A=.!.2r20 3=.!.2 (5)2 0 3= 252 0 6 0.24 radian 0=-= 25 1t 1t ' r = 2 inches; 0=30°=30·-= 180 -6 radIan rc3 1.047 m. 2 A=-r12 20=-21 (2)2 (-rc6 ) =-:::! r = 2 feet; 0= rc3 radians 1t 21t 2.094 feet =rO = 2·-= 3 -:::! 3 2rc :::!2.094 fe = A =.!.2 r20 .!.2 (2 )2 (rc)= 3 3 1t -71tradIans ' r 12 yards; 0=70°= 70·-= 80 18 1
93.
97.
t
99.
t
d
rm
=
=
101. 91.
t
r =5 cm; = 20 seconds; 0=3.!. radian O)=!!...= (120/3) =3.!.._201 =�60 radian/sec v='::=rO = 5·(201/3) =�3 .�20 =�12 em/sec = 26 inches; r =13 inches; v=35 12 in. 1 35 mi ' 5280 ft '--'--v=-hr ft 60 min =36,960 in.lmin v 36,960 in.lmin 0)=-= r 13 in. :::! 2843.08 radians/min 2843.08 rad 1 rev min 21trad :::!452.5 rev/min r0==3960 miles 35°9'- 29°57' =5°12' =5.2° =5.2·� 180 :::!0.09076 radian =rO =3960·0.09076 :::!359 miles r =3429.5 miles 0)=1 rev/day = 27t radians/day =�12 radianslhr v=rO)=3429.5·�:::! 12 898 mileslhr t
S
89.
1t 1tradIan' r=4m, 0=45°=45·-=180 4 A =.!.2 r20=.!.2 (4)2 (rc)4 = 2rc:::! 6.28 m2
rIn=156 minutes, inches 1 90°=-1t. 0=-6015 rev =-·360°= 2 radIans 4 =rO =6·�2 =31t:::! 9.42 inches In 25 minutes, 51t. 5 0=-6025 rev=-·3600=1500=12 6 radIans 57t 57t :::! 15.71.mches =rO =6 . -= 6 s
hr
milhr
s
103.
s
369
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exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
105.
r= UJ= = 2n
x105 miles 2.39 3 days 1 rev/27. radiansl27.3 days 12·27.3 radianslhr =rUJ= (2.39x 105)._n_ "" 2292 miles/hr 327.6 = 2 inches; r2 = 8 inches; UJI =3 rev/min =67r radians/min Find UJ2:
115.
s
7t
_ __ _
v
107.
s
'i
C
= 1.5n
radians/min = l.5n revmm 2n ./ =-3 rev/nun. 4 r = 4 feet; UJ= 10 rev/min = 20n radians/min v=rUJ = 4 · 20n ft = 80n min 80re ft 1 mi 60 min ft hr "" 2.86 milhr5280 = 8.5 feet; r = 4.25 feet; V= 9.55 V 9.55 UJ=-= r 4.25 ft 9.55 mi 5280 ft 1 rev hr 4.25 ft nu 60 min 2n "" 31.47 rev/min inis the24 hours. one fullin 24rotation makestraveled earth The hours distance The the At the equator earth. miles. circumference Therefore, circumference ofis the2n(3960 ) must travel to keep person a linearthevelocity theup with sun is: 2n(3960 ) v=!.- = 24 "" 1037 mileslhr
117.
--
119.
I
113.
hr
7r
'i
of the is r andangletheislength radius ofbya circle also central the subtended arctheIf themeasure r, then of the angle is radian. Also, 180 rad·lan = degrees. per traveled the distance speedandmeasures Linear the measures speed angular time, unit In other time.traveled per unit angle in a central change distance describes speed linear words, andthe of a circle, on thetheedgeturning a pointspeedlocated byangular of rate describes circle itself. Answers will vary. I
----
.--_.
--- =
VI =v2 'iUJI =r2UJ2 'iUJI = r2UJ2 r2UJI r2UJI 'i UJ2 =-
rni!hr
rni!hr
7r'
•
min = -_._--.---
d
25 . 500= 12,500 "" 3979 miles . = 2 trr = 2 12,500 25,000 nules. miles, 397925,000 of Earth is approximately The theradiuscircumference approximately is and miles. rotates at UJI rev/min , so VI = 'iUJI' r2 rotates at UJ2 rev/min, so v2 =r2UJ2 the belt speedweofhave Since theis linear that:connecting the the same, pulleys r=
12n 8
111.
=rB
500 =r . .!!. 25...
VI = v2 'iUJI =r2UJ2 2(6n) = 8UJ2 UJ2 =-
109.
betweenSinceAlexandria thatto bethe =distance We know theis 7.20, miles. Syene and 500 Alexandria in rays Sun's the of measure of Earth the center atSyene angle formed the centralAlexandria also must and between 7.20. Converting to radians, we have be 7.20= 7.20. � 1800=.!!.25... radian . Therefore,
I
121.
._--.--
I
-
7r
123 -125.
t
370
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Right Triangle Trigonometry
15.
Section 7.2 1.
3.
7. 9. 11.
13.
c2 =a2+b2 =62+102 =36+100=136 c =56 = 2134 complementary True True opposite 5; adjacent 12; hypotenuse=? (hypotenuse)2 =52 + 122 =169 hypotenuse =.J169 =13 hyp sinB = opp hyp =2-13 cscB= opp =�5 adj =.!2 secB= hyp =� cosB= hyp adj 12 13 adj .!2 tanB = opp adj =2-12 cotB= opp = 5 opposite 2; adjacent 3; hypotenuse ? (hypotenuse)2 = 22 +32 =13 hypotenuse =Jl3 _2_ _2_ . Jl3 2J13 sinB = opp hyp = Jl3 = Jl3 Jl3 = 13 adj = _3_ =_3_. Jl3 =3J13 cosB = hyp Jl3 Jl3 Jl3 13 tanB= opp adj =�3 hyp = .J13 cscB = opp 2 hyp Jl3 secB=-=-adj 3 adj =l cotB= opp 2 =
=
=
=
17.
=
=?
adjacent 2; hypotenuse 4; opposite (opposite)2 + 22 =42 (opposite)2 = 16-4 =12 opposite=.J12= 2J3 2J3 = J3 sinB = opp = hyp 4 2 adj =� =.!.. cosB = hyp 4 2 2J3 tanB = opp adj = 2 =J3 hyp =_4_ =_4_ . J3 = 2J3 cscB = opp 2J3 2J3 J3 3 secB = hyp adj =.i2 = 2 adj =_2_ =_2_ . J3 = J3 cotB = opp 2J3 2J3 J3 3 opposite J2; adjacent 1; hypotenuse=? (hypotenuse)2 =(J2f + 12 =3 hypotenuse =J3 J3= 16 B= = J3J2 =J3J2 . J3 cosB= adj =_J31 =_J31 . J3J3 = J3 = oadjpp= J21 = J2 cscB= opp= J2J3 = J3J2. J2J2= 162 secB= adj = J31 =J3 cotB= oadjpp=_J21_=_J21_. J2J2= J22 sin
=
=
=
=
°PP hyp
3
hyp
3
tan e
hyp
hyp
371
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
19.
23.
opposite 1 ; hypotenuse = J5 ; adjacent 12 + (adjacent)2 = (J5 (adjacent)2 = 5 - 1 = 4 adjacent = J4 = 2 _1_ = _1_ .J5 = J5 sinB = hopp = J5 J5 J5 5 yp adj = � = � .J5 = 2 J5 cos () = hyp J5 J5 J5 5 ! tan B = opp adj = 2 J5 = J5 csc () = hyp = opp 1 J5 sec B = hyp = adj -2 adj = � = 2 cot B = opp 1 =
t
=?
1 _ _1_ 2.J5 3J5 secB = _ cosB = J5 = J5 J5 = 5 3 1_ = _1_ = _5_J5 = 5J5 = J5 cot () = _ tan () 2J5 2J5 J5 10 2 5 25.
sin () = .fi2 corresponds to the right triangle: b=v'2
� � =2
a
(.fit
Using the Pythagorean Theorem: a2 + = 22 a2 = 4 - 2 = 2 a = .fi So the triangle is:
cos () = J3 2 1 B in �=__1 =__1 . J3=J3 s "2 = = tanB co B J3 =�2 . J3 J3 J3 J3 3 s 2 1 1 =1·2 = 2 csc () = -= sin () 1 2 1 =1 =2 = -2 . J3 2J3 -sec ()=-= cos () J3 J3 J3 J3 3 2 1_ = _1_ = � = �. J3 = 3J3 = J3 cot () = _ tan B J3 J3 J3 J3 3 3 J5 sin () = � '' cos () = 3 3 2 sinB =.l=?:. . �=�=� . .Js=2.Js tanB=cosB .Js 3.Js .Js .Js.Js 5 3 1 = -1 = -3 csc B = -sin () 2 2 3
b=�
� a
=.J2
.fi
adj = cos B = hyp 2 .fi opp tan B = -= adj .fi = 1 yp = � = � . sec B = hadj hyp = � = � . csc B = opp .fi adj - = 1 cot () = -= opp
.fi=.fi .fi .fi.fi .fi=.fi .fi.fi .fi .fi
372
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Right Triangle Trigonometry
27.
cosf) = -31 Using the Pythagorean Identities: sin2f) + cos2 f) = 1 =1
.J5 = .J5 = csc f) = hyp opp 1 .J5 secf) = hyp adj = 2 adj = � = 2 cotf) = opp 1
Sin2B+GJ
sin 2 f) + .!. 9
=1 sin2f)=�
31.
9
= fI 2.J2 '19 3 (Note: sin f) must be positive since f) is acute . ) sinf) = 2f = 2.J2 . � = 2.J2 tanf) = cos f) t 3 1 sin f)
=
1 =1 = -3 = 3 ..Ji= -3..Ji cscf) = -sinf) 2f 2..Ji 2..Ji..Ji 4 1 = -1 = 1 ·3 = 3 sec f) = -cos f) t 1 ..Ji ..Ji 1 =-1 = -.- = cot f) = -tanf) 2..Ji 2..Ji..Ji 4 tanf) = .!.2 corresponds to the right triangle: -_.-
29.
b= l
=
�
-_.-
33.
a=2
Using the Pythagorean Theorem: c2 = f + 22 = 5 c = .J5 So, the triangle is: b= l
t:s:.f5= a=2
secf) = 3 Using the Pythagorean Identities: tan2 f) + 1 = sec2 f) tan2 f) + 1 = 32 tan2 f) = 32 - 1 = 8 tanf) = .J8 = 2..Ji (Note: tanf) must be positive since f) is acute.) 1 1 cos f) = -sec f) = 3 sin f) tan f) = -cos f) , so 2J2 sm. f) = (tanf) )( cos f))=2"11�2·-31 = 3 1 = -1 = -3 -3 .J2 . - = 3J2 cscf) = -sinf) 2f 2.J2 2..Ji.J2 -4 1 = -1 = 1 ..Ji= ..Ji cot f) = -tanf) 2..Ji 2..Ji..Ji -4 tanf)=..Ji Using the Pythagorean Identities: sec2 f) = tan2 f) + 1 sec2 f) = (..Jir + 1 = 3 sec f) = J3 (Note: secf) must be positive since f) is acute . ) I -I = I J3 =J3 cos f) = -secf) = J3 J3 J3 3 sinf) so tan f) = --, cosf) J3 = J6 . = (tan f)) ( cosf)) = "11�2·smf) 3 -3 1_ = � = 2- = 2-. J6 = 3J6 = J6 cscf) = _ sinf) .Jf J6 J6 J6 6 2 1 1 = 1 .J2 =.J2 cot f) = --2 tanf) = .J2 ..Ji .J2 _.-
e
opp = _1 = _I_ . .J5 = .J5 sin e = hyp .J5 .J5 .J5 5 adj = l:...- = l:...- . .J5 = 2 .J5 cos e = hyp .J5 .J5.J5 5
_.-
373
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
35.
�=2
csce= 2 corresponds to the right triangle: b= l
a
43.
8
Using the Pythagorean Theorem: a2 +e = 22 a2 + 1 = 4 a2 = 4 - 1 = 3 a = .J3 So the triangle is: b= l
39.
41.
cos 1 0° = sin(9 0° - 10°) = sin 80° = 1 sin 8 0° sin 80° sin 8 0° using the identity cos e = sin ( 90° - e)
1 - cos2 20° - cos2 70° = 1-cos2 200 -sin2(9 00 -700 ) = 1 - cos2 20°- sin2(200 ) = 1 - ( cos2 20°+ sin2(200» ) =1- 1 =0 using the identities cos e = sin ( 90° - e) and sin 2 e + cos2 e = 1 . sin(90° cos 70° - 70°) � -'--cos 20° 49 . tan 20 0 - --- = tan 200 cos 20° sin 20° = tan 200 - --cos 20° = tan 20° - tan 20° =0 using the identities cos e = sin ( 900 - e) and sin e tan e = -cos e .
47.
�=2 =.J3 a
37.
45.
sin38° - cos52° = sin38°- sin(900 -52°) = sin 38° - sin38° =0 using the identity cos e = sin ( 90° - e)
8
.!. sin e = °PP hyp = 2 adj .J3 cose= hyp -= 2 _1 = _1 . .J3 = .J33 tane= °PP = adj .J3 .J3.J3 l:...- = l:...- . .J3= 2.J3 sec e = hyp = adj .J3 .J3.J3 3 cot e = 0adj = .J3 = .J3 PP 1 sin2 20° + cos2 20°= 1 , using the identity sin2 e + cos2 e = 1
--
51.
_ = 1, using the sin80°csc800= sin 80°. sin180° 1 · csc e = -1· d entity sine sin 50° = tan500 - tan 500= 0 , usmg . the tan 50° ---cos 50° sine 1·d entity · tan e = -cos e __
--
sin 35° ) sec55o . cOS 350 tan 35° . sec55° . cos35° = ( cos 35° = sin 35°·sec55° = sin 35° ·csc (900-55° ) = sin35°·csc35° 1 = sin 350 . _ sin_ 35° =1 sin e , . the l·dentitles .. tan e = -usmg cose 1 sec e = csc (9 0° - e) , and csce = -sm.e.
374
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Right Triangle Trigonometry
53.
55.
cos35° ·sin 55°+ cos 55°·sin35° = cos35°·cos(900-55°) + sin(900-55°) . sin35° = cos 35°· cos 35° + sin 35° ·sin35° = cos2 35°+ sin2 35° =1 using the identities sin8 = cos ( 90° -8 ) , cos8 = sin ( 90° -8 ) , and sin2 8 + cos2 8 = 1 . Given: sin 30° = .!.2 a.
b.
c.
61.
a.
i
cos 60° = sin ( 90° - 60° ) = sin 30° = cos2 30° = 1 - sin2 30° = 1 - 2 43 _ = .!.1 = 2 csc 2:6 = csc30° = sin130° 2
(i)
__
---
d. 57.
f.
g.
Given: tan8 = 4 sec28 = I + tan28 = 1 + 42 = 1 + 16 = 17 1 =1 b. cot8 = tan8 4
h.
a.
--
c.
d.
63.
( )
cot � -8 = tan8 = 4
csc2 8 = 1 + coe 8 1 17 1 = 1 +-1 = 1 + -== 1 + tan28 16 16 42
65.
--
59.
Given: esc8 = 4 1 sin8 = _ csc8_ = 4.!. a.
b. c.
d.
Given: sin 38° "" 0.62 cos 38° "" ? sin2 38° + cos2 38° = 1 cos2 38° = 1 - sin2 38° cos38° = .J1 - sin2 38° "" �1 - ( 0 . 62 ) 2 "" 0.78 sin38° "" 0.62 "" 0 . 79 tan380 = cos38° b. 0.78 5 cos 38° "" 0.78 5 "" 1 .2 7 cot380 = sin c. 38° 0 . 62 1 1 d. sec380 = __ ",, __ "" 1.27 cos 38° 0 . 78 5 1 1 . 61 csc38° = sin 138° ""--,,,, e. 0 . 62
sin 52° =cos ( 90° - 52° ) = cos38° ",, 0 .7 8 cos 52° = sin ( 90° - 52° ) sin 38° "" 0.62 tan 52° cot ( 90° - 52° ) = cot38° "" 1 .27 =
=
Given: sin8 = 0 . 3 sin8 + cos � -8 = sin8 + sin8 = 0 . 3 + 0.3 = 0 . 6
( )
The equation sin8 = cos ( 28 + 30° ) will be true when 8 = 90° - ( 28 + 30° ) 8 = 60° - 28 38= 60° 8 = 20°
coe 8 = esc 2 8 - 1 42 - 1 = 16 - 1 = 15 sec(90° -8) = esc8= 4 1 =16 sec 2 8 = 1 + tan 2 8 = 1 + coe1 8 1 + 15 15 =
-- =
375
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
67.
a.
b. c.
l S00 + -= SOO S + S = 10 mmutes . T = -300 100
f.
SOO l-S00 = S+ l S = 20 mmutes . T = -+ 100 100 SOO SOO so X = -tan O = -, tan 0 x SOO �n. 0 = distance in sand , so SOO . d·Istance In. sand = -sin O -x distance in sand T ( 0) = l S00 100 300 + S O SOO l S00 ---=:ta:=.On:..: O'- sin 0 300 + 100 S_ + _S_ = S__ 3tan O sin O 1_ + _ 1 _ = S 1__ 3 tan 0 sin 0 S OO 1 . tan 0 = -lS00 = -3 , so we can consIder the triangle:
20
00 �======l90 ° o
Use the MINIMUM feature: 20
00
_ _
)
(
d.
g.
I�
69.
a.
3
S_ + _S_ T= S__ 3 tan O sinO = S - Sl + -lS ;-
() 3"
e.
S_ + _S_ with the Let 1'; = S _ _ 3 tanx sinx calculator in DEGREE mode.
b.
"iniMU,""
X=70.S2B77S _ V=9. 71�O�SZ o
•
900
The time is least when the angle is approximately 70.5". The value ofx for this angIe IS. x = tan SOO 70.S3° "" 177 feet . The least time is approximately 9.7 minutes. Answers will vary. Z2 =X2 + R2 Z = �X2+ R2 = � 4002+ 6002 = � S20,000 = 200.J1j "" 721.1 ohms The impedance is about 721 . 1 ohms. sinq$ = XZ = 400 2.J1j 200.J1j 13 cosq$ = -R = 600 3.J1j Z 200.J1j 13 X 400 = -2 tanq$ = -R = 600 3 cscq$ = X � = 200.J1j = .J1j 2 400 .J1j 200.J1j secq$ = ZR = 600 = 3 600 = -3 cotq$ = X -R = 400 2 ----;:=
M
= S - S + SM "" I S.8 minutes 1000 feet along the paved path leaves an additional SOO feet in the direction of the path, so the angle of the path across the sand is 45". S + _S_ T = S - 3 tan4So sin4So S S S+10 = S-+ = S3 J2 3 · 1 J2 2 "" 1 0.4 minutes 376
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Right Triangle Trigonometry
71.
Since IOAI = IocI =1 , �OAC is isosceles. Thus,LOAC LOAC = LOCA.. Now LOCA LAOC = 1 800 LOAC LOCA (1800- B) = 1800 LOAC+LOCA = 2( LOAC ) = B
a.
+
+
+
+
B
+
LOAC=�2
C DI J CD I = CDI sinB = II OC I 1 1 cosB = IOlODCIi = IO1DI = O1 DI sinO tan�= IIACDIDI IAOIIC+DIloDI=l+ICloDIDI = l+cosO h = x·-xh = xtanB -h = (I-x) tan(nB) h (1 - x) I-x xtanB = (1 - x)tan(nO) xtanB = tan(nB)-xtan (nB) xtanB xtan(nB)= tan(nB) x( tanB tan(nO))= (nB) tan(nB)x = ------'--',tanB tan--(nO) Area �OAC =-21 OC 1 I ·I AC I = LI2 OCI .I ACI 1 . =-cosasma 2 1 . =-smacosa 2 Area �OCB = 211 oc1·1 BC I = .!.2 'IOB 12 . o1Ol cBI 1 .Ol1 BCBII = .!.2 IOB 1 2 cosfJsinfJ = .!.2 IOB 12 sinfJcosfJ
b.
c.
73.
d.
2
2
·
+
+
2
,8)
tan
+
75.
+
e.
2
=
Area �OAB = 21 1 BD1 ·O1 AI = �I BDI·l = L2 IOBI·IOl BDBII = .!.2 IOBI sin(a fJ) Ol C i cosa = 01 AI Jocl . ol BI =IOBI cosfJ Ol C i 1 Ol Ci Ol BI Area �OAB = Area �OA C Area �OCB .!.I DB Isin(a+,8) =.!.sinacosa +.!.I DB 12 sin,8cos,8 cosa . --sm( cos,8 a +,8) . . +--cos2a- sm,8cos,8 =smacosa 2 cos ,8 cos,8 . cosa . sin(a+ ,8)=--smacosa +--sm,8cos,8 cosa cos,8 sinea+ =sinacos,8 + cosasin,8 sina . = --·cosa sma = tancosaa cosa = cosfJcosa = cosfJtan,8 sinfJ = cosfJ·_cos,8 = sinfJ sin2a+ cos2a=1 sin2a + tan2,8= sin2,8 sin2a+--=1 cos2,8 . sm2a =1 sm. 2a+ I-sin 2a (1-sin2a) (sin2a+ I�i�i��a)=(1)(1-sin2a)
c.
77.
a.
I
b.
377
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
a
a
sin2 -sin4a+sin2 =1-sin2 a sin4a-3sin2a+l=0 Using the quadratic formula: 3±J5 sm. 2 -2 J5 ± . �3-SIna= 2 . �3-J5 But �3-+J5 2->1 . So, sma= -2- ' Consider the right triangle: a=
79.
7.
f(60°) =sin 60° =
9.
f
11.
13.
15. a
If () is an acute angle in this triangle, then: a a>0, b 0 and c>O. So cos () = ->0 . c 2 2 Also, since2 a +2 b2 =c , we know that:
17.
>
0
19.
Thus, 0 1 . So we now know that 0
21.
>
>
81.
23.
S ection 7.3
27. 3.
5.
True sin45° = .J22 csc45° = .J2 .J2 sec45° = .J2 cos45° =tan45° =12 cot45° =1
29.
�
(6�0) =f(300) =sin300 =�
[f(600)J =(sin600t = [ �J 4-3 2f( 60°) = 2sin60° =2· -13 2 = -13 f(60°) sin 60° -13 1 -13 22 -13 2 =--2 = -=_ 2 .-=:2 4 4cos450-2sin450=4. .J22 _ 2'.J22 = 2.J2- .J2 =.J2 1 6 tan45°-8cos60°=6·1-8·-= 2 6-4=2 2-13 sec-+2csc-= ,,2 +2·-4 3 3 =.J2+ 43-13=3.J2+34-13 ,ec' � -4= [ 2n -4= I: -4=�-4=-� "
"
r;;
1-00" 300 -00" 600 =1-[ �J -GJ =1-�-� 44 =0 Set the calculator to degree mode: sin 28°"" 0.47 . �ln�L��694715628
378
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.3: Computing the Values of Trigonometric Functions of Acute Angles
31.
33.
Set the calculator to degree mode: tan 21°", 0.38 .
45.
Set the calculator to degree mode: sec41°= cos141° ",1 .33 .
47.
�an(21�383864035
am'��093362496
---
/COS�:H5012993
35.
Use the formula R = 2V0 2 singOcos O with g = 32.2ftlsec2; () = 45° ; Vo = 100 ftlsec : 2(100 l sin45°·cos 45° ",3 10.56 feet R = 32.2 sin2 0 with Use the formula H = V02 2g g = 32.2ftlsec2 ; () = 45° ; Vo = 100 ftlsec : 1002 sin2 45° ",77.64 feet H = 2(32.2)
Set the calculator to radian mode: sin�10'" 0.31 . ��,...,.,....,-----,
sln()[���§0169944
37.
Set the calculator to radian mode: tan 0.3 ", 0.31 .
Set the calculator to radian mode: tan 511: 12 ",3.73 . _ --i ,,-:::;: <= """"� an �;)��H1050808
49.
Use the formula R = 2V0 2 singOcos O with g = 9.8 m/sec2 ; () = 25° ; Vo = 500 m/sec : 2(500 l sin 25°·cos 25° ",19,541.95 R = g
m
39.
sin2 0 with Use the formula H = V02 2g g = 9.8 m/sec2; () = 25° ; Vo = 500 mlsec : 5002 sin2 25° ",2278.14 m H = 2(9.8)
Set the calculator to radian mode: 1 sec12 = cos- ", 1.04. 12 11:
--
11:
51. 41.
43.
Set the calculator to radian mode: sin 1 ", 0.84 . �ln(1�8414709848
Use the formula t = ± 1 g sin2a O with Ocos g = 32 ftlsec2 and a = 10 feet : 2(1 0) t = ± 1 ___..:.... _'--_ ",1.20 seconds 32 sin 30°· cos 30° -----
a.
Set the calculator to degree mode: sin 1°", 0.02 . �ln�1:0174524064
b.
t
__2 ..:.... (1_0,-) --",1 . 12 seconds = ± 1_ 32 sin 45°· cos 45°
c.
t
__2 ..:.... (1_0,-) --",1.20 seconds = ± 1_ 32 sin 60°· cos 60°
379
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exist. No portion of this material may be reproduced, in any fonn or by any means, without pennission in writing from the publisher.
Chapter 7: Trigonometric Functions
53.
a.
We label the diagram as follows: _____
I I
3 !ilL
c.
____ _
1 mil
1
Note that tan () = ..!..x and sin () = .!.s , so 1 . Also, note that 1 x = -tan () and s = -sin () . = rate·tIme . , so tIme . = distance dIstance rate Then, on sand time on sand distance rate on sand 2 2s 2 ( fne ) = -3 3 3 sin () on road . on road = distance and tIme rate on road = 8 -82x = 1 - �4 1_ = 1 - tan(} = 1 4 + time4 tanon ()road So, total time = time on sand
d.
--
e.
-----
_
b.
_
( 4 tan1_() ) 2 1 = 1 + -3 sin () - -4 tan () 1 2 T(300) = 1 + -3 sin 30° 4 tan 30° 2 _+ 1 T((}) = _ 3 sin ()
2 T(45°)=I+ --3sin45° 4 tan 45° 2 1 = 1 + -.3 _1_ --"" 4·1 1 .69hr J2 Sally is on the paved road for 1 - 4 tan 45° = 0.75 hr. 2 T(600) = 1 + --3 sin 60° 4 tan 60° 2./3 1./3 = 1 +----3' 2 4. = 1 + � --1- "" 1.63hr Sally is 1on the3./3paved4./3road for 1 - 4 tan 60° "" 0.8 6 hr. T(900) = I + 3 sin2 90° 4 tan 90° But tan 90° isformul undefined, so wepath.can't use theHowever, function a for this d be 2 Themilestotalin the sand andthe8distance miles onwoul the road. time would be: �3 + 1 = �3 "" 1 .67 hours. The pathkingwoul1 dmilbeetoinleave the straight first houseto the wal the sand road. Then andandwalwalk 8kmiles on the road. Finally, 1 mile in the sand to the second house. 1 =1 = 4. Thus, the tan(} = -41 , so x = -tan(} 114 Pythagorean Theorem yields:
_ _
_ __ _
tum
tum
f.
2.!. -1 = 1 + -_1 3. 2 4 . ./3 ./3 = 1 + -43 -",, 1 .9hr 4 Sally is 1on the paved road for 1 - 4 tan 30° "" 0.57 hr.
s2 = x2 + 12 s = ,Jx2 + 1 = ,J4 2 + 1 = 07 Total time = time on sand + time on road 2.J0 8-2·4 T = 2s3 + 8-2x = 8 3 + 8
8-8 = 207 ::::: 2.75 = 207 + 3 8 3
hrs The path woulsandd bedirectly to leaveto thethe bridge. first houseThenand wal k in the cross theandbridge (approximatel y 0directl milesyonto tthhee road), t h en wal k i n the sand second house. 380
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.3: Computing the Values of Trigonometric Functions of Acute Angles
g.
1 Let =1+ 3si2n x - 4tanx Y.I
--
57.
--
Case 1: 8=25°, a=5
c /1s �
4
\
b
�
sin(25°)=� 5 ;::,_5_ ;::,11.83 in. = sin(25°) 0.4226 Case 2: 8=25°, b=5 e
Use the MINIMUM feature:
e
4
0°
55.
HiniMUM
�=67.975676
o
_
Y=1.61B0165.
The tiismapproxi e is leastmately when 1.62 8;::,hour. 67.98° . The least time Sally's time on the paved road is 1 1 1- -4tan8;::,1- 4tan67.98° ;::,0.90 hour. =8, 8=35°
�a e
c/1a �
90°
b
sin(35°) = �8 a=8sin(35°) ;::,8(0.5736) ;::,4.59 in.
5 ;::, 5 ;::,5.52 . e= cos(25°) 0.9063 There aredtwobe adjacent possibleorcasesopposite becausethethegivengiven siangl de ecoul . tan(35°) JAcl 100 I AC I =100tan(35°);::,100(0.7002);::,70.02 feet Let x the height ofthe Eiffel Tower. tan(85.361°)= 80 x=80tan(85.361°);::,80(12.3239);::,985.91 feet Let x the distance to the base of the plateau. ---
59.
cos(35°)=-8b b=8cos(35°) ;::,8(0.8192) ;::,6.55 in.
61.
m.
=
�
63.
=
/1someters � x
tan(20°)=-50x 50 137.3 7 meters x= tan(5020°) ;::,--;::, 0.3640
381
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exist. No portion of this material may be reproduced, in any fonn or by any means, without pennission in writing from the publisher.
Chapter 7: Trigonometric Functions
65.
We construct the figure below: 500
tan (32°) =x
71.
67.
=
789 ft
500
tan(23°}=y 73.
Y
Distance + 500 = tan500(32°) + --:---:tan(23°) "" 1978.09 feet Let h represent the height of Lincoln's face. =x
Let h the height of the monument. h tan (35.1°) =789 h = 789tan(35.1°}"" 789(0.7028)"" 554 . 52 Lethighwayy, andaround thethe three bay (seesegments figure).of the x,
ft
z =
·S 1 : �I 1
1400
I
-I
1..o::._cL_____ cL�0:: a b 3mi
=
I
130· y
The length of the highway + + sin(40°) = .!. = sin(140°)"" 1 .5557 ' x
z
x
x
69.
b tan(32°) =800 b = 800tan(32°)"" 499.90 tan(35o)= b+h 800 b +h = 800tan(35°)"" 560.17 Thus, Lincoln h = (bthe+ h)height -b = of560.17 -499.'s face 90""is:60.27 feet Let the length of the guy wire. x =
nu
= sin /50°)"" 1.3054 mi tan(40°) =-a1 a= tan /40°)""1.1918mi tan(50°) = b1 b = tan (50°)"" 0.8391 a+y+b=3 y =3-a-b "" of3 -1.1918 -0.8391is about: = 0.9691 The l e ngth the highway 1.5557 + 0.9691 + 1. 3054"" 3.83 miles . z
10ft
mi
190 ft
190 sm. (690) = = sin190(69°)"" 203.52 ft.
mi
x
x
382
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.3: Computing the Values of Trigonometric Functions of Acute Angles
75.
Addi some lines to the thediagram and labeling specialngpoints, we obtain following: 18
I
I
B I
I
- -,II - - I I
---
, ,
.., , , ,
, I ,
,
----
I E I I I I
- -
----
- -_ -_ -_ -_-_- -_-_ -_ -_ _- - i_ A--.JI -_-_-_L--.J ....:. _ ....:. Lf · _·· ······· 1 -j 1 3 =
MBC ,
79.
o
a
I I I
_ : If we let x length of side we see that, in tana=lx . Also, in tana=� 5-x . Therefore, we have 1.8 3 5-x x15-3x=1. 15 =4.8x 8x x=�4.8 =3.125 ft 1+3.125=4.125 ft The pl a yer shoul d hit theuppertop lcushion that is 4.125 feet from eft comer.at a point sinO 0 sinO -0 0.5 0.4 794 0.9589 0.4 0.3894 0.9 735 0.2 0.1987 0.9933 0.1 0.0998 0.9983 0.01 0.0 100 1.0000 0.001 0.0010 1.0000 0.0001 0.0001 1.0000 0.00001 0.00001 1.0000 sinO approaches 1 as 0 approaches O. -o We rearrange product as folltheows:order of the terms in this tan 1° ·tan2° ·tan3° · . . ·tan89° =( tan 1° . tan89° ) . ( tan2° . tan88° ) . . . . ( tan 44° . tan 46° ) . (tan 45° ) Now each set ofangles. parentheses contains ausing pair of compl e mentary For example, cofunction properties, we have: 1 .5
77.
5
( tan 1° . tan 89° ) =( tan 1° . tan ( 90° -1° )) =( tan1° . cot1° )
BC,
MDC ,
81.
1 ) =(tanlo . tanlo =1 ( tan 2° . tan 88° ) =( tan 2° · tan (90° -2° )) =( tan 2° . cot 2° ) 1 ) =( tan2o . _ tan 2° =1 and so on. This result holds for each pair in our product. ce we knowas: that1·1·1·tan. .45°·1 ==1 .1 , our product can beSinrewritten Therefore, tan 1° . tan 2° . tan 3° . . . . tan 89° = 1. We can rearrange product as follows:the order of the terms in this cosIO ·cos2° · . . ·cos45° ·csc46° · . .·csc89° =( cos 1° . csc89° )'( cos 2° . csc88° ) . . . . ( cos 44° . csc 46° ) . ( cos 45° ) Now each set ofangles. parentheses contains ausipaingr of compl e mentary For example, cofunction properties, we have: ( cos 1° . csc 89° ) =( cos 1° . csc (90° - ) ) =( cos 1° . sec1° ) 1 =( COSlo . _ cos 1° )=1 ( cos 2° · csc88° ) =( cos 2° · csc (90° -2° )) =( cos2° . sec2° ) 1 =( COS20 . _ cos2° )=1 and on. holds for each pair in our product. Since Thissoresult J2 we know that cos 45° =2 , our product can be J2 J2 rewrI. tten as 1· 1 ·1 .. · 1· -= 2 -. 2 Thus' cosIO ·cos2° · . . ·cos45° . csc46° . . . . csc890 = J22 Answers will vary. r
83 - 85.
383
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
Section 7.4 1. 3. 5. 7.
9.
11.
15.
tangent, cotangent 240°-180°=60° True 600°-360°=240°; 240°-180°= 60° -"2 and -32" (-3,4): a -3, b = 4 r =�a2 + b2 =�r-(_-3----:)-2 +-4 2 =�9 + 16 =55 =
(-3,-3) : a=-3, b=-3 r =�a2 +b2 = �(-3l +( _3) 2 =.J18 =3.fi y 5
4
x
=
5
.fi . br 3-3.fi .fi -sm(}=-=----= 2 .fi a ----3 .fi = .fi cos () = -= r 3.fi .fi 2 b -3 1 tan(}=-a =-= cot ()=-ba =--3-3 = 1 -3 sec ()=�a = 3.fi -3 = - .fi csc ()=�b = 3-.fi3 = -.fi ( J32 '�2 J : a = J32 ' b = �2 r = �a2 + b2 = (� J 2 +(�J = �% + � = J1 = 1 - -
4
x
-4
. =-b =-4 -3 =--3 sm() cos () = -= rb ra -3 4 4 3 tan(}=-a =-==--3 3 cot(} =-=b 4 4 r sec(}=-=-=-a -3 3 csc(} =-=b 4 (2, -3): a= 2, b=-3 r=�a2 +b2 =�r-2-2 +-(-_-3)-2 =�4+9=Ji3 a
5
5
13.
5
5
r
17.
5
5
y
-I
y 5
-I
-5
b 2 2 c o s (} = :: = J31 2 = J3 c ot e = :: = J3 1 2 = J3 1 2 b 1/ 2 b 11 2 1 J3 J3 tan () a --= J31 2 --=J3 J3 3 sec e = -r = 2 J3 = 2J3 a J31 2 J3 J3 3 csc() = - = -1 = 1 r
Ji3 =_ 3Ji3 sin(}=!:r =2 Ji3 .JG 13 Ji3 = 2 Ji3 cos ()=!!:.r . = _Ji32_ Ji3 13 b -3 3 a -= 2 --2 tan(} =-=-=-a 2 2 cot () = -= b -3 3 r sec(}=-=a Ji32 csc () = b = Ji3 -3 = _ Ji33
r
= - =
1
= --
--
--
�
384
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.4: Trigonometric Functions of General Angles
19.
( .fi2 ' _ .fi2 ) = .fi2 ' b=_ .fi2 :
r
a
29.
= (� J + ( -� J =�i + i =Ji =1
cos 3!n = (� + 3!n ) = (� + n ) =cos (� + 4 . 2n ) =cos-n4 COS
COS
y
8
.fi
2
31.
J2 -Z-
J2 Z- J2 2 1 J2
33.
a sm()=-=--=--J22 cosB=-=-=J2 tan B=�= '/J2 = .
b r
a
1
2
r
1
tan 21n = tan(O+ 21n) =tanO=0 Sincecossin()< 0 0forforpoipointsntsininquadrants quadrants andand and the angle () lies in quadrant Sinceandsintan<()<0 0forforpoipointsnints iquadrants andand n quadrants the angle () lies in quadrant Sincecotcos<() 0 0forforpoipointsntsininquadrants quadrants andand and the angle lies in quadrant Sincetansec <0 0forforpoipointsntsininquadrants quadrants andand and the angle () lies in quadrant () -30° is in quadrant so the reference angle is =30° . B
>
II
II.
35.
B
IV,
37.
2
J2 =_J2 _� cscB=�=_l_= J22 J2 J2
39.
b
41.
43.
B
=
=
II
IV.
B
>
B
IV.
B
>
a
II,
III,
III
IV,
� = � �=.fi secB=�= a '/ 2 '/ 2 '/ 2
I
III.
II
I
I
II
IV,
IV,
III,
III,
IV,
II,
() 120° is in quadrant so the reference angle is =180° - 120° =60° . a
385
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exist. No portion of this material may be reproduced , in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
8=210° is in quadrant III, so the reference angle is a=210° -180° =30° . 8=-57r4 . m. qua ant III, so the reference angle 57r 7r . a=--7r=-. 4 4 8=-87r3 . m. quadrant II. Note that 87r =-, 27r so the reference angle is --27r 3 3 27r -7r . a=7r--= 3 3 8=-135° is in quadrant III. Note that -135° + 360° =225° , so the reference angle is a=225° -180° =45°. 53 8=--27r3 . m. quadrant III. Note that 27r 27r =-47r , so the reference angle . --+ 3 3 47r -7r . a=--7r= 3 3 8=440° is in quadrant I. Note that 440° -360° =80° , so the reference angle is =80° . 157r . .mquadrant IV. Note at 8=4 157r =-77r , so the reference angle . ---27r 4 77r 4 7r a=27r--=4 4. sin 150° =sin 30° =.!.,2 since 8=150° has reference angle =30° in quadrant II. cos315° =cos45° =J22 , since 8=315° has reference angle a=45° in quadrant IV. sin 510° =sin 30° ==.!.2 , since ()=510° has reference angel a 30° in quadrant II.
45.
47
a
dr
IS
·
67.
a
IS
49
69.
a
IS
·
71
a
IS
75.
a
77.
55.
a
·
_
_
r;;
79.
th
IS
.
a
IS
57
r;;
73.
51.
·
cos ( -45° ) =sin 45° =J22 , since 8=-45° has reference angel =45° in quadrant IV. sec 240° =-sec 60° =-2, since 8= 240° has reference angle =60° in quadrant III. cot330° =-cot30° =-.[3 , since 8=330° has reference angle =30° in quadrant IV. . 8=-37r4 has . 37r4 . 7r4 J22 smce . sm-=sln-=-, reference angle = 7r4 in quadrant II. 77r cot -7r =,,3 , smce ()=-77r has cot-= 6 6 6 reference angle = 7r6 in quadrant III. 137r 7r J2 ' smce 137r cos--=-cos-=-4 7r 2 . 8=--4 has 4 reference angle = 4 in quadrant III. sin ( - 2; )= sin ; = � , since ()= 2; has reference angle a= 7r3 in quadrant III. 147r -tan -7r =-,,3 , smce 8=-147r has tan--= 3 3 3 reference angle = 7r3 in quadrant II. csc ( -315° )=csc45°=J2 , since 8=-315° has reference angle a=45° in quadrant I. sin(8n)=sin(0+8n) =sin(O)=0 tan(7n) =tan( 7r + 6n) = tan(7r)=0 sec(-3n)=sec(7r -4n) =sec(7r)=-1
65.
_
.
a
IS
81.
83.
59.
a
85.
87.
61.
63.
==
386
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exist. No portion of this material may be reproduced, in any fonn or by any means, without pennission in writing from the publisher.
Section 7.4: Trigonometric Functions of General Angles
89.
y
sin e = g , e in quadrant II 13 Since e is in quadrant II, sin e > ° and csc e > ° , while cos e < 0 , sec e < 0 , tan e < ° , and cot e 0 . If a is the reference angle for e , then 12 . . =sma 13 Now draw the appropriate triangle and use the Pythagorean Theorem to fmd the values of the other trigonometric functions of a .
4
<
4
93.
tan a 125
= -
seca 135
= -
13 csca = cota 152 12 Finally, assign the appropriate signs to the values of the other trigonometric functions of e . 5 12 sec e = -13 tan e = -cos e = -13 5 5 = -
13 csc e = 12
91.
sm. a = -35
tan a = -43
seca =-45
5 csce = --35
4 4 cot 6/ = -
4
csca =-35 cota = -43 Finally, assign the appropriate signs to the values of the other trigonometric functions of e . sIn. e = --3 sece =--5 tane = -3
x
5 cosa = 13
x
3
sin e = 2. , 900< e < 1 800 , so e in quadrant II 13 Since e is in quadrant II, cos e < 0, sec e< 0, tan e < ° and cot e < 0, while sin e > ° and csc e > 0. If a is the reference angle for e, then sin a = 2. . 13 Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of a . y
5 cot e = -12
cos e = --45 ' e in quadrant ill Since e is in quadrant ill, cos e < 0, sec e< 0, sin e < ° and csc e< 0, while tan e > ° and cot e > 0. If a is the reference angle for e , then cos a = i5 . Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of a .
12 cos a = 13
5 tana = 12
13 sec e = -12
12 cot e = -5
13 csca = 5
13 12 seca = cot a = 12 5 Finally, assign the appropriate signs to the values of the other trigonometric functions of e . 13 5 csc e = 12 tan e = -cos e = -5 12 13 387
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
95.
cos 0=--,13 1800 < 0< 2700 (quadrant III) Since 0 is in quadrant III, cos 0<0, sec 0<0, sin 0< and csc 0<0, while tan 0 > and cotO> 0. If a is the reference angle for 0 , then cosa=.!.3 . Now draw theTheorem appropritoatefmdtriathenglevalues and useof thethe Pythagorean other trigonometric functions of a . °
97.
sinO=%, tan 0< (quadrant II) Since 0 is in quadrant II, cos 0<0, sec 0<0, tan 0< and cot 0<0, while sin 0 > and cscO > 0. If a is the reference angle for 0 , then �3 . Now draw theTheorem appropritoatefmdtriathenglevalues and useof tthehe Pythagorean other trigonometric functions of a . °
°
°
°
i
s na
y
=
2
2
x
3 ../5 3../5 cosa=-../53 seca=-·-=-../5 ../5 5 2 ../5 2../5 cota=-../5 tana=-'-=-2 ../5 ../5 5 csca=-32 Finally, the approprifuncti ate signs of the otherassigntrigonometric ons ofto the0 . values 3../5 cosO=--../53 secO=--2../5 ../5 cotO=-tanO=--5 2 3 cscO=-2
3 Ji = 3Ji . 2Ji = csca sma=-3 2Ji Ji 1 Ji Ji tana= 2Ji1 =2Ji cota=----=2Ji Ji 3 seca=-=3 Fiofntheally,other1assigntrigonometric the approprifunctions ate signs ofto the0 . values 3Ji . 2Ji3 cscO=--smO=--Ji tanO=2Ji cotO=secO=-3 -4 4
5
4
4
388
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.4 : Trigonometric Functions of General Angles
99.
sec 0 = 2, sin 0 < (quadrant N) Since 0 is in quadrant N, ' sin 0 < 0, csc 0 < 0, tan 0 < and cot 0 < 0, while cos 0 > and secO > 0. IfNowa draw is the reference angle for 0 , then seca = 2 . theTheorem appropriate triangl evalues and useof thethe Pythagorean to find the other trigonometric functions of a . °
°
y 4
°
y
sma=-3 cosa =-4 cota =-4 csca =-3 seca =-4 Finally, the appropriate signs ofto O.the values of the otherassigntrigonometric functions smO = --3 cosO = --4 cotO 4 cscO= - -3 secO = --4 tan 0 = -'3'1 sinO> (quadrant II) Since 0 is in quadrant II, cos 0 < 0, sec 0 < 0, tan 0 < and cotO < 0, while sinO> and cscO> 0. IfNowa draw is thethereference angletriangl for 0 e, then tanthe= 1 / 3 . appropriate and use Pythagorean Theorem trigonometric functionstooffinda .the values of the other ·
4
101.
III ,
103.
3
5
5
= -
5
5
.J3 = .J3 .J3 cosa =-1 . =tan a = sma 1 2 2 2 .J3 2.J3 1 .J3 -J3 cota=-'-= csca=-·-=-J3 .J3 3 .J3 .J3 3 Final l y , assign the appropriate signs to of the other trigonometric functions of O.the values .!. tan 0 = -.J3 sin 0 = _ .J3 2 cos 0 = 2 2.J3 cotO= --.J3 cscO= --3 3 tanO=�,4 sinO and cot 0 > 0. If a is the reference angle for 0 , then tana = �4 . Now draw theTheorem appropriate triangl evalues and useof thethe Pythagorean to fmd the other trigonometric functions of a. °
5
5
·
3
5
5
°
°
°
a
y
°
1 J10 J10 sma=--·--=-J10 J10 10 csc = .JlO1 =.JlO J10 = 3J1O seca =-3 ._.JlO cosa =-_ J10 J10 10 3 cota = -=1 3 Finally, the appropriate signs ofto O.the values of the otherassigntrigonometric functions cscO=vlO cos f} = - 3Fa sm f} = Fa sec f} = FlO cote = -3 3 a
·
--
3
·
10
-
r.;;
10
--
- --
389
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
105.
csc 0 = -2, tan 0 > 0 => 0 in quadrant III Since 0 is in quadrant III, cos 0 < 0, sec 0< 0, sin 0< 0 and csc 0 < 0, while tan 0 > 0 and cot O > O. If a is the reference angle for 0 , then csc a = 2 . Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of a .
113.
y
115. x
J3 . = -1 sma cos a = 2 2 J3 = -J3 = J3 seca = 2 ._ 2J3 tana = -1 · 3 J3 J3 J3 J3 3 cot a = J31 = J3 Finally, assign the appropriate signs to the values of the other trigonometric functions of 0 . sinO = .!..2 cos O = _ J32 J3 2J3 tan B = secO = 3 3 cot B = J3 sin 40° +sinI300+sin 2200+sin310° = sin 40° +sin( 40° +90°) + sin( 40°+ 1 80°) + sin( 40° +270°) = sin 40° + sin 40° - sin 40° -sin 40° =0 Since f (B) = sin B = 0.2 is positive, 0 must lie either in quadrant I or II. Therefore, B + tr must lie either in quadrant III or IV. Thus, f ( B + 7t) = sin( 0 + 7t) = -0.2
117.
1 = -1 = 5 G·Iven sm. B = -51 , then csc B = -sin O 1 5 Since csc 0 > 0 , B must lie in quadrant I or II. This means that csc ( 0 + tr) must lie in quadrant III or IV with the same reference angle as B . Since cosecant is negative in quadrants III and IV, we have csc(B + tr) = -5 . sin 1° + sin 2° + sin 3° +... + sin 357° + sin 358° + sin 359° = sin 1° +sin 2° + sin3° + . . . + sin(360° _3°) + sin(360° - 2°) + sin(360° _1°) = sin 1° +sin 2° +sin3° + . . . +sin( _3°) + sin( -2°) + sin( -10) . = sin 1° +sin 2° +sin3° + . . -sin 3° -sin 2° -sin 1°
a.
�
b.
--
109.
111.
.
�
-
107.
322.J2 [sm( 2( 60°)) - cos( 2( 600)) -1 = --:n] 32.J2 (0.866 - (-0.5) - 1) 16.6 ft . Let 1'; = 322.J2 32 [sm(2x) - cos {2x) - I] R
c.
Since F ( 0) = tan 0 = 3 is positive, B must lie either in quadrant I or III. Therefore, B + 7t must also lie either in quadrant I or III. Thus, F (O+ 7t) = tan (B + tr) = 3 .
119.
--
Using the MAXIMUM feature, we find: 20
Answers will vary.
390
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.5: Unit Circle Approach: Properties of the Trigonometric Functions
Section 7.5
3.
5.
7.
. = -2 smt 3
even
1 = (1)(Js) = � Js = 2; 3 csc t = l = 1 (% ) = % 3 sec t = Js = ( Js ) = Js Js = 3;
All real number, except odd multiples of 7r2 -0.2, 0.2
. = --21 smt
cost
tant =
.J3 2
= -
3
1 tant = ] = ( - �) (�) = - � � = - � 2 csct = I.!. = { - f = -2 _ 2 ) sect = � = ( �) = � � = 2� 2 .J3 cot! = �.!. = ( � ) ( -f) =-.J3 2
11.
J5 cott = 15.
17.
fi cos t = -2
tant = _2_ = 1
cott =_fi2_ = 1
fi
fi
2
csc t =
fi
2
19.
� = ( -1) = -1 � = -fi
2
sect =
For the point (3, -4) , x = 3, y = -4, r = �x2 + y2 = �9 + 16 = J2s = 5 SIn. () = --45 cos () = -35 tan () = --43 sec () = -35
cot () = --43
For the point (-2, 3), x=-2 , y=3, r = �x2 + y2 =�4 + 9 = 03 sin () = _3_ 03 = 3 03 csc () = .j13 3 13 03 03 .j13 cos () = __2_ 03 = _ 203 13 sec () = - 2 03 03 tan () = --23 cot () =--23
-
_ fi = b p = _ fi2 ' _ fi2 '. a = _ fi 2' 2
. = -fi smt 2
I3 = (�)(%) = �
csc () = --45
)
(
2
J5 cos t = 3
� = { - 1) - � � = -fi =
For the point (- I, - I), x = - I, y = - I, r = �x2 + y2 = .J1+1 = fi sin () = -=!. fi = _ fi2 csc () = fi -1 =- fi fi fi cos () = -=!. fi = _ fi2 sec () = fi -1 = -fi fi fi -1 = 1 -=1 cot () = -1 tan () = -1 1
-
2
-
391
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
49.
25.
27.
29.
31.
33.
csc4500= csc(3600+ 90°) = csc900= 1 cot 390° = cot(180° + 1 80° + 30° ) = cot 30° = J3 337t = cos 7t + 87t cos 4 4" = COs � + 4 ' 27t = cos -7t4 .fi 2 tan(217t) = tan( O + 217t) tan ( 0) = 0
( ) ( )
55.
57.
59.
=
( ) ( )
177t = sec 7t + 47t sec 4 4" sec � + 2 · 27t = sec-47t = .fi J3 tan 197t 6 = tan 2:6 + 37t = tan 2:6 = 3 =
35.
53.
( )
61.
65.
67.
tan
( )
sec -2:6 = sec2:6 = 2 J3 3 sin ( -7t) + cos (57t) = - sin ( 7t) + cos ( 7t + 47t) = 0 + COS7r = -1 sec( -7t) + csc
(-%) = sec7t - csc %
= -1 - 1 = -2 sin _ 9; _ tan _ 9; 97t . 97t tan= -Sln-+ 4 4 = -sin � + 27t + tan � + 27t . 7t + tan -7t = -Slll4 4 2 - .fi .fi + 1 or -= -2 2 '
( ) ( )
( ) ( )
63.
45.
tan(-7t) = - tan7t = 0
(-�) = -tan � = -1 69.
The domain of the sine function is the set of all real numbers. I ((}) = tan () is not defined for numbers that are odd multiples of 2:2 . I( (}) = sec ()
is not defined for numbers that are odd multiples of 2:2 . The range of the sine function is the set of all real numbers between -1 and 1 , inclusive. That is, the interval [ -1, 1 ] . The range of the tangent function is the set of all real numbers. That is, (-00 , 00 ) .
392
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.5: Unit Circle Approach: Properties of the Trigonometric Functions
71.
73.
75.
77.
79.
81.
83.
85.
87.
The range of the secant function is the set of all real number greater than or equal to 1 and all real numbers less than or equal to -1 . That is, the interval (-ex), -1 ] or [ 1 ,ex) ) . The sine function is odd because sine-(}) = -sin () . Its graph is symmetric with respect to the origin. The tangent function is odd because tan(-(}) = -tan () . Its graph is symmetric with respect to the origin. The secant function is even because sec( -(}) = sec () . Its graph is symmetric with respect to the y-axis. If sin () = 0.3 , then sin () + sin ( () + 21t) + sin ( () + 41t) = 0.3 + 0.3 + 0.3 = 0.9 If tan () = 3 , then tan () + tan ( () + 1t) + tan ( () + 21t) =3+3+3 =9 fe -a) = -f (a) = --31 b. f (a) + f (a + 21t) + f (a + 41t) = f (a) + f (a) + f (a) = -31 + -3I + -3I =1 fe -a) = -f (a) = -2 b. f (a) + f (a + 1t) + f (a + 21t) = f (a) + f (a) + f (a) = 2+2+2 =6 f e -a) = f (a) = - 4 b. f (a) + f (a + 21t) + f (a + 41t) = f (a) + f (a) + f (a) = -4 + (-4) + (-4) = -12
89.
When t = 1 , the coordinate on the unit circle is approximately (0 . 5, 0.8) . Thus, 1 "" 1.3 sinl "" 0.8 cscl "" 0.8 1 = 2.0 sec l ",, cosl "" 0.5 0.5 0.5 ",, O . 6 0.8 tan I "" 0.5 = 1.6 cot l "" 0.8 on RADIAN mode: Set the calculator S u /s � a.
-
ln � 8414709848 os ( l ) . 5403023059 �,- an ( l ) 1 . 557407725 b.
ln� �88395 106 l/cos ( l ) 1 . 8508 1 5718 l/t an ( l ) . 6420926159
When t = 5.1 , the coordinate on the unit circle is approximately (0.4, -0.9) . Thus, 1 "" -1.1 sin 5.1 "" -0.9 csc5.l "" -0.9 I = 2.5 cos 5 . 1 "" 0.4 sec5.l "" 0.4 -0 9 tan 5. 1 "" -' 0.4- "" -2.3 cotS.l "" � -0.9 "" -0.4 Set the calculator on RADIAN mode: --
JSlm � : §158146823 Il /s ln�r : ��012977 pos(5. 1) l/cos (5. 1) 2. 645658426 . 3779777427 (5. 1) 1/t an -2. 449389416 an- (5.1) . 4082650123
a.
91.
Let P = (x, y) be the point on the unit circle that corresponds to an angle t. Consider the equation tan t = 2:.x = a . Then y = ax . Now x2 + l 1 , 1 and so x2 + a2 x2 = 1 . Thus, x = ± � l +a2 y = ± � ; that is, for any real number a , l + a2 there is a point P = (x, y) on the unit circle for which tan t = a . In other words, -ex) < tan t < ex) , and the range of the tangent function is the set of all real numbers. =
a.
a.
393
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
93.
Suppose there is a number p, 0 < p < 21t, for which sine 0 + p) sin 0 for all O . If 0 0 , then sin (0 + p) sin p sin 0 0 ; so that p n . If 0 = � then sin % + p sin . 3n - 1 sm. "2n 1, But p n . Thus, sm. 2 =
( ) ( %) ( ) ()
=
=
=
5.
=
=
-1
3.
=
=
=
=
7.
=
9.
or = 1 . This is impossible. The smallest positive number p for which sine 0 + p) = sin 0 for all 0 is therefore p 2n .
1cos_O since cos 0 has period =
95.
f ( 0) = sec 0 =
_
1 1.
:
3 ,. 27r6 = 7r3 False The graph of y = sin x crosses the y-axis at the point (0, 0), so the y-intercept is O . The graph of y = sin x is increasing for n <X< n 2 2 The largest value of y = sin x is 1 . sin x = 0 when x = 0, n, 2 n . - -
2n , so does f (O) = sec O. 97. If P = (a, b ) is the point on the unit circle corresponding to 0 , then Q = (-a, -b) is the point on the unit circle corresponding to 0 + n . -b b Thus, tan( 0 + n) = = - = tan 0 . Ifthere -a a exists a number p, 0 < p < Jr, for which tan(0 + p) = tan 0 for all 0 , then if 0 = 0 tan(p) = tan( O ) = O . But this means thatp is a multiple of n . Since no multiple of n exists in the interval (0, n) , this is impossible. Therefore, the fundamental period of f ( 0) tan 0 is Jr. sin O - O sin O = tan 0 . Since L 99. Slope ofM = cosO - O = -cosO is parallel to M, the slope of L = tan 0 . 1 01 - 103. Answers will vary.
13.
15.
1 7.
,
19.
- .
3n -n ' . = l when x = -smx 2 ' 2' 3n . = - 1 when x = --n smx ' 2 2 y = 2 sin x This is in the form y = A sin«(i}x) where A = 2 and = Thus, the amplitude is I A I = 1 2 1 = 2 2n = 2n = 2 n . and the peno. d ' T = 1
1.
(i)
IS
=
21.
(i)
y = -4cos(2x) This is in the form y A cos( (i}x) where A -4 and = 2 . Thus, the amplitude is I A I = 1 -4 1 = 4 and the period is 2n = 2n n . T= 2 =
(i)
=
Section 7.6 1.
I ; !!:..2 + 2Jrk , k is any integer
=
(i)
y = 3x2 Using the graph of y = x2 , vertically stretch the graph by a factor of 3.
23.
y = 6sin(nx) This is in the form y = A sine (i}x) where A = 6 and Jr . Thus, the amplitude is I A I = 1 6 1 6 2n = 2n = 2 . and the peno. d ' T = n (i)
=
=
IS
(i)
x
-5
394
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7. 6: Graphs o f the Sine and Cosine Functions
-
3 " , and 2" 0, "2 , " , 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y coordinates of the five key points for y 4 cos x , we multiply the y-coordinates of the five key points for y = cos x by A 4 . The five key points are (2", 4) (", -4) , We plot these five points and fill in the graph of the curve . We then extend the graph in either direction to obtain the graph shown below .
This is in the form y = A cos( OJx) where A = - -2I and OJ = �2 . Thus, the amplitude is 1A1= T
=
1- i 1 i and the period is =
=
(0,4), (;,0).
27t 27t 4 7t - -;- - 1 - 3 ' _
_
_
2
This is in the form y = A sin(OJx) where A and OJ 1A1= T
29. 31. 33. 35.
=
=
2" . Thus, the amplitude is 3
1 -% 1
=
27t = 27t
OJ
2 Jf
=
(11T. 4)
5
3
- -
x
% and the period is =
3
.
3
45.
F A H C
41. 43.
Comparing y -4 sin x to y = A sin ( ) , we find A = -4 and OJ = I . Therefore, the amplitude is 1 -41 = 4 and the period is 2"I 2" . Because the amplitude is 4, the graph of y = -4sinx will lie between -4 and 4 on the y-axis. Because the period is 2" , one cycle will begin at x = ° and end at x = 2" . We divide the interval into four subintervals, each of length 2"4 = !!... by OJX
=
=
37. J 39.
e;,o).
A B Comparing y = 4 cos x to Y = A cos ( OJx) , we find A = 4 and OJ = I Therefore, the amplitude is 1 41 = 4 and the period is 2"I = 2" . Because the amplitude is 4, the graph of y = 4 cos x will lie between -4 and 4 on the y-axis. Because the period is 2" , one cycle will begin at x = ° and end at x = 2" . We divide the interval 27r] into four subintervals, each of length 2"4 = "2 by finding the following values:
[0,2,,] 2 3"
0,
finding the following values: "2 , " , 2 , 2" These values of x determine the x-coordinates of the five key points on the graph. To obtain the y coordinates of the five key points for y = -4 sin x , we multiply the y-coordinates of the five key points for y sin x by A -4 . The five key points are e; ,4 . (2", 0) ,-4 , We plot these five points and fill in the graph of the curve. We then extend the graph in either
.
=
(0,0), (; ) (",0),
[0,
=
)
395
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
direction to obtain the graph shown below. , y 5 C�, 4)
49.
Since sine is an odd function, we can plot the equivalent form y = - sin ( 2x ) . Comparing y = -sin(2x) to y = A Sin(wx) , we find A = -1 and w = 2 . Therefore, the amplitude is 1 - 11 = 1 and the period is 22Jr = Because the amplitude is 1 , the graph of y = - sin (2x) will lie between -1 and 1 on the y-axis. Because the period is Jr , one cycle will begin at x = ° and end at x = Jr . We divide the interval [0, ] into four subintervals, each of length 4 by finding the following values: Jr .
47.
(_3;. -4) -5 (� , -4) Comparing y cos( 4x) to A Cos( wx) , we find A 1 and w 4 . Therefore, the amplitude is III 1 and the period is 24Jr Jr2 . Because the amplitude is 1, the graph of y = cos ( 4x) will lie between -1 and 1 on the y-axis. Because the period is 2 , one cycle will begin at x ° and =
=
Jr
Y =
=
=
Jr
=
Jr
=
0, -Jr4 , -Jr2 , -34Jr , and
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y coordinates of the five key points for y sin (2x) , we multiply the y-coordinates of the five key points for y = sin x by A = -1 . The five key points are (0, 0) , (: ' -1), ( ; , 0), e: ,1), ( Jr, 0) We plot these five points and fill in the graph of the curve . We then extend the graph in either direction to obtain the graph shown below. y 2
]
=
end at x = ; . We divide the interval [0, ; into four subintervals, each of length 4/ 2 = Jr8 by finding the following values: 3 Jr ' °, 8 Jr4 ' -8Jr and -Jr2 These values of x determine the x-coordinates of the five key points on the graph. The five key points are Jr
'
-
C;r, l) e:, l)
(0,1), ( ; ,0), (:,-1), e;,o) , (;,1) (-� . l ) (_3; . 0) 11"
(0, 1 )
(i · O)
(11", 0) 211"
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. y 2
Jr
X
-2
(� , 1 )
x
-2/
(-i , 0) (-� . - 1) -2
396
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publi sher.
Section 7. 6: Graphs o f the Sine and Cosine Functions
51.
( )
Comparing y = 2 sin � x to y = A Sin(mx ) , we find A = 2 and m = -21 . Therefore, the amplitude is 121 = 2 and the period is 127r1 2 = 47r . Because the amplitude is 2, the graph of y = 2 sin ( x will lie between -2 and 2 on the y-axis. Because the period is 47r , one cycle will begin at x = ° and end at x = 47r . We divide the interval [0, 47r] into four subintervals, each 47r = 7r by finding the following of length 4 values: 0, 7r , 27r , 37r , and 47r These values of x determine the x-coordinates of the five key points on the graph. To obtain the y coordinates of the five key points for y = 2 sin x ) , we multiply the y-coordinates of the five key points for y = sin x by A = 2 . The five key points are (0,0) , (7r, 2) , (27r,0) , (37r,-2) , (47r,0) We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
53.
I 1
�)
Because the amplitude is !2 , the graph of
y = - !2 cos (2x) will lie between -!2 and !2 on the y-axis . Because the period is 7r , one cycle will begin at x = ° and end at x = 7r . We divide the interval [0, 7r] into four subintervals, each of length 7r4 by finding the following values: 37r , and 7r 0, -7r4 , -7r2 , 4 These values of x determine the x-coordinates of the five key points on the graph . To obtain the y coordinates of the five key points for y = -!2 cos (2x) , we multiply the y-coordinates of the five key points for y = cos x by A = - !2 .The five key points are
(�
( - 3'IT , 2)
y
3
Comparing y = -!2 cos (2x ) to Y = A cos (mx ) , we find A = - -21 and m = 2 . Therefore, the 27r = 7r . . 1 1 amp11tu · de IS - - = - and the peno . d I' S 2 2 2
(o,-�) , ( : ,0) , (;,�) , e:,o) , (7r'-�)
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
('IT. 2)
y 2
-3
(311", -2)
- 21T -2
397
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
55.
We begin by considering y = 2 sin x . Comparing y = 2sinx to y = A Sin(mx) , we find A = 2 and m = 1 . Therefore, the amplitude is 121 = 2 and the period is 27(1 = 27( . Because the amplitude is 2, the graph of y = 2 sin x will lie between -2 and 2 on the y-axis. Because the period is 27( , one cycle will begin at x = ° and end at x = 27( . We divide the interval [0, 27( ] 27( = 7( by into four subintervals, each of length ""4 "2 finding the following values: 37( , and 27( 0, -7(2 , 7( , 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y coordinates of the five key points for y = 2 sin x + 3 , we multiply the y-coordinates of the five key points for y = sin x by A = 2 and then add 3 units. Thus, the graph of y = 2 sin x + 3 will lie between 1 and 5 on the y axis. The five key points are (0,3) , (7(,3) , e;,l), (27(,3) We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
57.
We begin by considering y = 5 cos ( 7(x) . Comparing y = 5 cos(7(x) to Y = A cos(mx) , we find A = 5 and m = 7( . Therefore, the amplitude is 1 51 = 5 and the period is 27(7( = 2 . Because the amplitude is 5, the graph of y = 5 cos ( 7( x) will lie between and 5 on the y-axis. Because the period is 2 , one cycle will begin at x = ° and end at x = 2 . We divide the interval [0,2] into four subintervals, each of length -42 = 21 by finding the following values: 0, 21 , 1, -23 , and 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y coordinates of the five key points for y = 5 cos (7(x ) -3 , we multiply the y-coordinates of the five key points for y = cos x by A = 5 and then subtract 3 units. Thus, the graph of y = cos ( 7(x ) - 3 will lie between -8 and 2 on the y-axis. The five key points are (0, 2) , -3 , (1, - 8 ) , -3 , (2, 2) We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
-5
-
-
5
(;,5),
(% ' )
(� , )
y
-4
398
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. exist. No portion of this material may be reproduced, in any form or by any means, without permIssIon m wntmg from the pubhsher.
Section 7. 6: Graphs of the Sine and Cosine Functions
59.
( )
61.
We begin by considering y = -6sin � x .
(� )
Comparing y = -6 Sin x to y = A sin(wx) , 7r
we find A = -6 and w = -3 . Therefore, the amplitude is 1 -61 = 6 and the period is 7r27r/3 6 . Because the amplitude is 6, the graph of y = 6sin � x will lie between -6 and 6 on the y-axis. Because the period is 6, one cycle will begin at x = ° and end at x = 6 . We divide the interval [0,6] into four subintervals, each of length i4 l2 by finding the following values: 0, -23 , 3, -92 , and 6 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y coordinates of the five key points for y = -6sin � x + 4 , we multiply the y coordinates of the five key points for y = sin x by A = -6 and then add 4 units. Thus, the graph of y = -6sin � x + 4 will lie between -2 and lO on the y-axis. The five key points are (0,4) , % ,-2 (3,4) , (6,4) We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
y = 5 - 3 sin ( 2x ) = -3 sin ( 2x ) + 5 We begin by considering y = -3 sin (2x ) . Comparing y = -3sin( 2x) to Y = A sin( wx) , we find A = -3 and w = 2 . Therefore, the amplitude is 1 -31 = 3 and the period is 227! 7! . Because the amplitude is 3, the graph of y = -3 sin ( 2x) will lie between -3 and 3 on the y-axis. Because the period is one cycle will begin at x = ° and end at x = We divide the interval [0, ] into four subintervals, each of length 4 by finding the following values: 37r , and 0, -4 , -2 , 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y coordinates of the five key points for y = -3sin (2x) + 5 , we multiply the ycoordinates of the five key points for y = sin by A = -3 and then add 5 units. Thus, the graph of y = -3sin(2x) + 5 will lie between 2 and 8 on the y-axis . The five key points are 3� ,8 , (7!,5) (0,5) , : , 2 , We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. =
=
( )
7r ,
7! .
7r
7r
7r
=
7r
7r
x
( )
( ) (;,5), ( )
( ) ( } (�,1O),
y
10
(7t ,5) (6, 4) 8
-7t
7t X
x
399
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
63.
Since sine is an odd function, we can plot the 21i ) . .5 (3x eqUlva ent orm y = -3sm 21i ) to 5 . (3x Companng. y =-3sm y = sin(OJx) , we find = - 23 and OJ = 21i3 . Therefore, the amplitude is \ - %\ =% and the period is � 21i / 3 =3 . Because the amplitude is 21i x) will lie ( 2.,3 the graph of y = _2.sin 3 3 5 5 . Because the d -3 on the y-aXIS. between --an 3 period is 3 , one cycle will begin at x = and end at x =3 . We divide the interval [0, 3 ] into four subintervals, each of length � by fmding the following values: 9 and 3 3 -, 3 -, 0, -, 4 2 4 Thesefivevalues of x determine the x-coordinates ofy thecoordinates key ofpoints on the graph. To obtain the the five key points for 2 y =-%sin ( ; x) , we multiply the y coordinates of the five key points for y sin x by =_2.3 .The five key points are .
1
65.
fi
A
A
A
A
We begin by considering y =-%cos ( : x) . Comparing y = -% cos ( : x) to y = cos(OJx) , we find = --23 and OJ =-1i4 . Therefore, the amplitude is \ - %\ =% and the period is 21i =8 . Because the amplitude is � , the graph 2 1i /4 of y = -�cos 2 (1i4 x) will lie between -�2 and �2 on the y-axis. Because the period is 8, one cycle will begin at x = and end at x =8 . We divide the interval [0, 8] into four subintervals, 8 2 by finding the following each of length 4-= values: 0 , 2, 4, 6, and 8 These of x determine the x-coordinates ofy thecoordinates fivevalues key ofpoints on the graph. To obtain the the five key points for y=-�cos 2 (1i4 x) +.!.,2 we multiply theycoordinates of the five key points for y cos x by -�2 and then add .!.2 unit. Thus, the graph of y = -�cos 2 (1i4 x) +.!.2 will lie between 1 and 2 on the y-axis. The five key points are ) ) ( 0 , - 1 ) , (2,� . (4,2), (6,� . ( 8, -1 ) We plot these five points andthefillgraphin thein either graph of thedirection curve. We then extend to obtain the graph shown below. (2, +)(4, 2) (-4, 2) °
°
4
=
A
=
A
=
-
,0 ,0 ) . (%' % ). ( 3 ) (�,-%). (% We plot these five points and fill in the graph of ( 0, 0 ) ,
the curve.toWeobtain thentheextend graphbelow. in either direction graphtheshown y 2
(- 8, - 1)
-2
-2 (0, - 1 )
(8, - 1 )
400
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.6: Graphs of the Sine and Cosine Functions
6 7.
I A I = 3,' T = n ,' m = 2Tn = 2nn = 2 y = ±3sin(2x)
69.
2n = 2 2n = n I A I = 3; T = 2; m = T y = ±3sin(nx) The graph is a cosine graph with amplitude 5 and period 8. Find m : 8 = 2nm 8m = 2n 2n = -n m=8 4
71.
79.
(� )
81.
( )
The equation is: y = 5 cos � x .
73.
83.
(� )
77.
The graph is a sine graph with amplitude 3 and period 4 . Find m : 4 = 2n m 4m = 2n 2n = -n m=4 2 The equation is: y = 3 sin �
( x) .
The graph is a reflected cosine graph with amplitude 3 and period 4n . Find m : 4n = 2nm 4nm = 2n 2n = -1 m=4n 2 The equation is: y = -3COS X .
75.
The graph is a reflected cosine graph, shifted up 1 unit, with amplitude 1 and period � . 2 3 2n Find m : 2 m 3m = 4n 4n m=3 The equation is: y = - cos n x + 1 .
The graph is a sine graph with amplitude �4 and period 1 . Find m : 1 = 2nm m = 2n The equation is: y = �sin 4 ( 2nx ) .
85.
The graph is a reflected sine graph with amplitude 1 and period 41l"3 . Find m : 4n3 2n m 4nm = 6n 6n = -3 m=4n 2
The graph is a reflected cosine graph with amplitude 4 and period 21l"3 . 2n = 2n Find m : 3 m 2nm = 6n 6n = 3 m=2n The equation is: y = -4 cos ( 3x ) . sin ( 1l" I 2 ) - sin ( 0) 1l" 1 2 - 0 1l" I 2 1-0 2 1l" 1 2 1l" The average rate of change is 3.1l" . f ( 1l" I 2 ) - f ( 0)
_
(% )
The equation is: y = - sin x . 401
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exist. No portion of this material may be reproduced , in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
87.
(� �) (� )
91.
. - Sin 0 f (" / 2) - f ( O ) Sin ,, / 2 ,, / 2 - 0 sin ( / 4) - sin ( 0 ) .
_
(J
0
g)(x) -2( cosx) -2cosx =
=
"
_
x
J2
2 J2 2 J2 ,, / 2 2 " " The average rate of change is J2 " .
= __ = _ 0 _ = -
89.
(g f) ( x) cos( -2x) =
0
(J g)(x) = sin (4x) 0
x
x
(g
0
f )(x) = 4(sinx) 4sinx =
93.
I (t) = 220 sin(60n t), t � 0
2n � Period: T 2nOJ 60n 30 Amplitude: 1 A 1 1 220 1 = 220 =
=
=
=
x
I
402
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.6: Graphs of the Sine and Cosine Functions
95.
V(t) = 220sin(1207tt) Amplitude: I A I = 1 220 I = 220 Period: T = 27tcv = � 1207t = � 60
99.
a.
a.
b, e.
b.
c.
d.
97.
a.
27t . Physical potential: cv = 23 ' 2 7t = -7t ; EmotlOna . 1 potentia. 1 : cv = 28 14
27t Intellectual potential: cv = 33 Graphing: 1 10
0
V = IR 220sin(1207tt) = 101 22 sin(1207tt) = I � l(t) = 22sin(120m) Amplitude: I A I = 1 22 1 = 22 _1 Period: T = 27tcv = � = 1207t 60
0
33
# 1 , #2, #3
( �; t) + 50 p(t) = 50sin e : t ) + 50 p(t) 50sin (�; t ) + 50
#1 : p(t) = 50sin #2 : c.
J p(t) = [V (t) R (Vo sin (27t/ )t t V0 2 sin2 (27t/ )t R R = _VR0_2 sin2 (27tft ) The graph is the reflected cosine graph translated up a distance equivalent to the amplitude. The period is _2/1_ , so cv = 47t / . Va2 . . IS. 1 V0 2 = The amphtude 2 R 2R The equation is: V02 p(t) = V2R0 2 cos( 47t/ )t + _ 2R V 2 0 =_ 2R (l - cos( 47t/ )t) Comparing the formulas: sin2 (27tft ) = .!.2 ( 1 - cos (47tft ))
d.
#3 : No.
=
1 10
_
b.
7 3 0 5 �=====�=�� o
_ . -
_
c.
101.
_
Physical potential peaks at 15 days after the 20th birthday, with minimums at the 3rd and 26th days. Emotional potential is 50% at the 17th day, with a maximum at the 10th day and a minimum at the 24th day. Intellectual potential starts fairly high, drops to a minimum at the 13th day, and rises to a maximum at the 29th day. Y = lsin xl , - 2,, '5, x '5, 2!r x
-I
1 03 - 105. Answers will
vary.
403
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exist. No portion of this material may be reproduced, i n any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
S ection 7.7 1.
21.
x = 4 souveniers
3.
origin; x = odd multiples of 2:2
5.
Y
7.
9.
11.
13.
1 5.
1 7.
( )
y = tan � x ; The graph of y = tan x is . honzontally compressed by a factor of 2
7r
= cosx The y-intercept of y = tan x is 0 . The y-intercept of y = sec x is 1 . secx = 1 when x = -27t, 0, 27t; secx = -1 when x = -7t, 7t y = sec x has vertical asymptotes when 37t - 7t 7t 37t x = -T ' 2' 2' T ·
23.
.
(±x) ;
y = cot The graph of y = cot x is horizontally stretched by a factor of 4. y
Y
= tanx has vertical asymptotes when 37t - -7t -7t 37t . x = -' ' ' 2 2 2 2
5
x
y = 3 tan x ; The graph of y = tan x is stretched vertically by a factor of 3.
111 6 1
IY
x
25.
Y
= 2 sec x ; The graph of y = sec x is stretched vertically by a factor of2. (0 , 2) y
1 9.
y = 4 cot x ; The graph of y = cot x is stretched vertically by a factor of 4.
I 'IT i1 2 (- 'IT, - )� it ¥, 4
y
16
1 'IT 'IT -x2 i, l\t, ( , ) I
I
I
x
404
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7. 7: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
27.
y = -3 cscx ; The graph of y = cscx is vertically stretched by a factor of 3 and reflected about the x-axis.
2
, ,
,
,"
,
, x
,
,
Ao:""-r----+--
,
,
(T '
, I
,
29.
'
:U' e�, 3)
, ,
, , ,
,
,
3) ' (--, - n 3 1T
,
- 1T
'
(� )
y = 4sec x ; The graph of y = sec x is horizontally stretched by a factor of 2 and vertically stretched by a factor of 4. y
,'
-
33.
3)
di �
( - 37T, 2 ) -
4
:(0, 1
47T
7T
-8
35.
( )
y = sec 2; X + 2 ; The graph of y = secx is horizontally. compressed by a factor of � 2n and shifted up 2 units .
l ,
, , , , , (0, 4) : , ,
( )
y = tan � x + 1 ; The graph of y = tan x is horizontally stretched by a factor of 4 and shifted up 1 unit.
y
, 31T (2 1T, - 4) , X
, ,
, ,
I
31.
-4
y = -2 csc ( nx ) ; The graph of y = cscx is horizontally compressed by a factor of � , vertically stretched by a factor of 2, and reflected about the x-axis.
Ji 7\
y 6
(-t, 2) :
-2.0 , - 1 .0 ,
(_1.2 , - 2) '
,
, ,
I
:
, , ,
! -6
n
l\l'
, , ,
,
37.
, , , ,
y
, (+ , 2) 2.0 , x ( ) _+--r-- T, -2 :
f
� (� )
y = tan x - 2 ; The graph of y = tanx is horizontally stretched by a factor of 4, vertically compressed by a factor of .!2 , and shifted down 2 units. 8
1
(0, -8
( 1T , -t) 37T
-2)r--
1T
I
X
405
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
39.
Y = 2 CSC
(� )
(g f)( x) = 4( tan x ) = 4tanx
X -1;
The graph of y = cscx is horizontally stretched by a factor of 3, vertically stretched by a factor of 2 , and shifted down 1 unit.
0
y
J
y
6
x
{ 47.
(J o g)(x) = -2 (cot x) = -2 cotx y
f 41.
-O (�)- f ( O) = tan(Jr I 6) - tan(0) = --� Jr 1 6 Jr 1 6 �6 - O
43.
I I
x
-2
J3 0 _6 = _2J3 =_ 3 Jr Jr The average rate of change is 2J3 Jr . f
i\
2
(g o f )(x) = cot(-2x)
(�) - f ( O) _ tan(2 . JrI 6) - tan(2 . 0) �6 - O
Jr l 6
Jr l 6 Jr The average rate of change is 6J3 Jr . 45.
( J o g)(x) = tan(4x)
49.
x
a.
Consider the length of the line segment in two sections, x, the portion across the hall that is 3 feet wide and y, the portion across that hall that is 4 feet wide . Then, and SIn. f) = -y4 cos f) = -x3 3 4 x = -y = - cos f) sin f) Thus, L x + y = cos3 f) + sin4 f) = 3 sec f) + 4 csc f) =
--
--
406
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
.
Section 7. 8: Phase Shift; Sinusoidal Curve Fitting
h.
c.
Let
25
1'; =
-cos3 x + -.smx-4 . -
u
O lk=== ===;;!1 1 o 25
u
[!, ! J [ J 4 4 (�,O} ( } e;,-4} e;,o)
2!. 2
4
y
=
5.
tan x
-cot(x
y
= 2COs
(3X + %)
Amplitude: I A I = 1 2 1 = 2 Period: T = 21tOJ = 21t3
(-%) =-�
x
y=
1 1 141 4 OJ
is least when e "" 0.83 . 3 + L "" cos ( 0.83 ) sin ( 0 . 83 ) ",, 9.86 feet . Note that rounding up will result in a ladder that won ' t fit around the comer. Answers will vary. L
phase shift y = 4sin(2x - 1t) Amplitude: A = = 21t 21t Period: T = = - = 1t 2 OJ Phase Shift: .t = �2 Interval defining one cycle: + T = � , 3; Subinterval width: T 7r Key points: 3; 4 ( 7r, 0 ) , '
Use MINIMUM to find the least value:
" i n i l''Iurl')
51.
1.
3.
o X=.Bn2 729 �V=9.B6S6626 o
d.
Section 7.8
Phase Shift: .tOJ = 3 6 Interval defining one cycle:
[!,! +TJ=[-�,�] Subinterval width: 27r 3 4 -4 6
+ �)
Yes, the two functions are equivalent.
-= T
/
=-
7r
407
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publi sher.
Chapter 7: Trigonometric Functions
Key points: (-� ' 2), (0, 0) , (�'-2) (�,o), (;,2)
9.
y = 4sin(7U+2)-5 Amplitude: 1 A 1 = 1 4 1 = 4 Period: 1t 2 Phase Shift: t = -1t = _ 21t Interval defining one cycle: [�,� +T] = [-�,2-�] Subinterval width: T 2 -=-= 4 poi4 nts:2 Key (_2 ) , (!_2 2 ) ( 1 _ 2 -5 ) (%-�,-9), ( 2 - �,-5) I1J
I1J
1
7.
y = -3sin (2X+�) Amplitude: 1 A 1 = 1 -3 1 = 3 Period: T = 21t = 21t2 = 1t 1t Phase Shift: t = -22 = _ .::4 Interval defining one cycle: [ � , � + T ] = [ � , 3: ] Subinterval width: T 4 points: 4 Key (- : ,0) , (0, -3) , ( : , 0) , (;,3 ) , e: , 0)
Jr
I1J
'
-
5
Jr
'
.2
'IT
,
Jr
'
,
y
(1 _ 1. -1)'
I1J
-1
2
,
_
Jr
11.
y =3COS(7U-2)+5 Amplitude: 1 A 1 = 1 3 1 = 3 Period: 1t Phase Shift: t = 21t Interval defining one cycle: [�, � +T] = [�,2+ �] Subinterval width: T 2 Key4 poi4 nts:2 I1J
x
I1J
1
408
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7. 8: Phase Shift; Sinusoidal Curve Fitting
( � ,8J , (� + � , 5 J . ( 1 + � , 2J . (% + � ,5 J . ( 2 + � , 8J
15.
¢
13.
x
(
=
1
Assuming A is positive, we have that Y = A sin(wx - ¢) = 2 sin(2x - 1) 17.
-1
�=�
[w = t2 = _2
( ; , 8)
y
IA 1 = 2; T = 1[;
( )J = 3 sin ( 2X - % )
)
Y = -3sin - 2X + % = -3sin ( - 2X - % Amplitude: 1 A 1 = 1 3 1 = 3 Period: T = 27t = 27t2 = 7t 7t Phase Shift: [w = 2"2 = �4 Interval defining one cycle: + T = : ' 5: Subinterval width: T Jr 4 4 Key points: (:,0J. ; ,3 , (Jr, -3) ,
(
19.
W
y
( _ � , 3)
5
)
y = 2 tan(4x - Jr) Begin with the graph of y = tan x and apply the following transformations: 1) Shift right units [Y = tan (x )] Jr
[ �, � ] [ ]
( ] e:,o] ,
I A I = 3; T = 31[; [w = -3 27t = 27t = -2 [ = [ = -w=T 37t 3 w 2 3 3 ¢ = _ .!.3 . 3.3 = _ 3.9 Assuming A is positive, we have that Y = A sin(wx - ¢) = 3sin % x+ %
- Jr
2) Horizontally compress by a factor of .!.4 [ Y = tan (4x - Jr) ] 3) Vertically stretch by a factor of2 [ Y = 2 tan ( 4x ) ] - Jr
e:,oJ
( T ,OJ
( � , 3)
-5 409
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
21.
( )
Y = 3CSC 2x - : Begin with the graph of = csc x and apply the following transformations: 1)
Shift right
:
Y
units
[Y ( x-:)1 =
csc
2) Horizontally compress by a factor of ..!.2
25.
3) Vertically stretch by a factor of 3 y = 3CSC 2X - :
[
( )]
JJ I lUI ,I I I
5 (- 0; , 3)
1
I Y 1 : 1 1
1
1
1 (-*1 ' -3)
1
3
1 1
I 1 (3: , 3)
1
1
'IT
"2
I
1
1
= -sec( 27l"x + ) Begin with the graph of y = sec x and apply the following transformations: 1) Shift left 7l" units [ Y = sec( x + 7l" )] 2) Horizontally compress by a factor of _27l"1_ [ Y = sec( 2Jrx + 7l" ) ] 3) Reflect about the x-axis [Y = -sec( 27l"x + 7l" ) ] y
1 1 1
1
1
( - 1,
I
23.
{ )
y = -CO 2X + � Begin with the graph of y = cot x and apply the following transformations: 1) Shift left units = cot x +
� [Y
(
7l"
�)]
27.
2) Horizontally compress by a factor of ..!.2 3) Reflect about the x-axis
1 1 1
y
1) )1
i\ TnT, ] , (
1
3
I, 1
1 1
1
}
I
(1 , I )
x
b-, - 1 )
1
I (t) = 120sin 307t t - � t � O 27t = � Period: T = 27tOJ = 307t 15 Amplitude: 1 A I = 1 120 1 = 120 7t Phase Shift: tOJ = 307t "3 = � 90
410
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exist. No portion of this material may be reproduced, in any fonn or by any means, without pennission in writing from the publisher.
Section 7.8: Phase Shift; Sinusoidal Curve Fitting
29.
a.
y 60
50
•
40
30
•
•
•
•
••
•
•
•
•
•
20 0 I 2 3 4 5 6 7 8 9101112
b.
o
x
31.
56.0 - 24.2 = 3 1 .8 = 1 5.9 Amplitude: A 2 2 + 24.2 80.2 56.0 = 40. 1 Vertical Shift: 2 2 = 21t 1t 12 6 Phase shift (use y 24 .2, x = 1):
a.
=
-
.
.
y 80 •
50
40
30
=-
•
•
•
•
.
.
. .
.
.
•
•• •
•
.
•
.
•
0 2 01 2 3 456 7 8 9 lOll 12
(� . 1 - ¢) + 40. 1 1 5 .9 sin (� - ¢ )
b.
24.2 = 1 5 .9 sin =
.
60
=
- 1 5 .9
..
20
70
=
0)
60
c.
13
x
.4 - .5 49.9 = 24.95 Amplitude: A = 75 25 = 2 2 + 100.9 25.5 75.4 = = 50.45 Vertical Shift: 2 2 21t 1t 12 6 Phase shift (use y 25.5, x = 1): 0)=-=-
=
- � = �- ¢ 2 6 21t ¢= 3 Thus, y 1 5 .9 sin
(� x- 231t ) +40. 1 y = 1 5 .9 Sin [� ( x - 4 ) ] + 40. 1 . =
c.
- 24.95
=
-
or
-� = �- ¢ 2 6 ¢ = 21t 3 Thus, y = 24.95 Sin
y 60 50
40 x 20 �������� 0123456789101112
d.
(� ·1 - ¢) + 50.45 24.95 sin ( � ¢ )
25.5 = 24.95 sin
c.
y = 1 5 .62 sin ( 0.5 17x- 2 .096 ) + 40.377
(� x_ 231t ) + 50.45 y = 24.95 sin [ � (x - 4 ) ] + 50.45 .
or
y 80
70
plnl(e9
60
'!:I=a>l<sin(bx+c)+d a=15.61996209 b=. 517364549 c= -2.095883506 d=40.37675696
50
40
30 20
0123456789101112
x
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Chapter 7: Trigonometric Functions
d.
y = 25 .693 sin(0.476x - 1 .8 1 4) +49.854
3 5 . a.
5���:�in(bx+c)+d a=25.6934405 b=.4764311009 c=-1.813776523 d=49.85374426 80
e.
33.
a.
b.
1/\13 20
3.6333 + 12.5 = 1 6 . 1 333 hours which is at 4:08 PM. Amplitude: A = 8 .2 - (- 0.6) = � = 4.4 2 2 7.6 .2 8 0.6) +( = - = 3.8 VertlcaISh'fit: 2 2 2 rc = -47Z' rc = (1) = -12.5 6.25 25 Phase shift (use y = 8.2, x = 3 .6333):
(
1
.
Amplitude: A = 13.75 - 1 0.53 = 1.6 1 2 +10.53 3.75 1 = 1 2. 14 Vertical Shift: 2 2 rc (1) = 365 Phase shift (use y = 1 3 .75, x = 1 7 2): 1 3.75 = 1 .6 1 sin 2 rc . 1 72 - tP +12. 14 365 1 .6 1 = 1 .6 1 Sin 2 rc . 1 72 - tP 365 1 = sin 344 rc - tP 365 344 rc -tP �= 2 365 tP 1 .3900 Thus, y = 1 .6 1 sin 2 rc X- 1 .39 +12 . 14 or 365 Y= 1 .6 1 Sin 2 rc ( X- 80.75 ) +12. 14 . 365 �
( �� · 3.6333 - tP) +3.8 4.4 = 4.4 Sin (�� . 3.6333 - tP ) . ( 14.5332 rc ) 1 = sm 25 8.2 = 4.4 Sin
b.
c.
rc 40.5332 rc tP 2 25 tP 0 . 2555
(�� x - 0.2555 ) +3.8 or y = 4.4 Sin [ �� ( X- O.5083 ) ] +3.8 .
10
y 9
O
7 5
d.
3
1 2.42 hours
� �-L����-L-L��__ 140 2 80 420 x
The actual hours of sunlight on April 1 , 2005 was 12.43 hours. This is very close to the predicted amount of 12.42 hours.
x 2
Y = 4.4 Sin �
)
y
Thus, Y= 4.4 Sin
d.
(
20
�
1
)
) )
[ ] Y = 1 .6 1 Sin [ 21t ( 9 1 - 80.75 ) ] +12. 14 365 �
"' 'I'
c.
( (
( �� (1 6 . 1 333 ) - 0.2555 ) + 3.8
8 .2 feet
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Review Exercises
37.
a.
0)=
-
( (
) )
(- do)
b.
c.
Chapter 7 Review Exercises
Amplitude: A = 1 9.42 - 5 .47 = 6.975 2 19.42 + 5 .47 = 1 2.445 Vertical Shift: 2 2 7t 365 Phase shift (use y = 1 9.42, x = 1 72): 19.42 = 6.975 sin 2 7t . 1 72 - rp + 12.445 365 6.975 = 6.975 sin 2 7t ·172 - ¢ 365 344 7t 1 - sm . --'I' 365 2: = 344 7t _¢ 2 365 ¢ :d.39 Thus, y = 6.975 sin 2 7t x - 1 .39 + 12.445 365 or y = 6.975 sin 2 7t ( X- 80.75 ) + 1 2.445 . 365 Y
(
[
(
)
) ]
= 6.975 sin 2 7t ( 9 1 ) - 1 .39 + 1 2.445 365 "'" 13.67 hours y
1.
135° = 1 3 5 · � radian = 3 7t radians 4 1 80
3.
1 8° 1 8 . � radian = .2:.. radian 10 1 80
5.
3 7t = _ 3 7t . 1 80 - degrees = 1 35 ° 4 4 7t
7.
5 7t = - 5_ 1 80 degrees = - 450 ° 7t . -2 2 7t
9.
. -7t = I - -1 = -1 tan-7t - sm 4 2 2 6
=
11.
3.J2 - 4-J2 - 4 · . 45 ° - 4 tan -7t = 3·./3 = -J3 3 sm 6 2 3 2 3
13.
6 cos 3 + 2 tan
;
(-%) = 6 ( - �) + 2 ( -J3) = -3J2 - 2./3
( -%) - cot ( - 547t ) = sec % + cot 5;
=
=
2+1 3
1 5.
sec
1 7.
tan 7t + sin 7t = 0 + 0 = 0
1 9.
cos 540° - tan( - 405° ) = -1 - (-1) = -1 + 1 0
25.
sin 50° sin 50° = sin 50° = 1 cos 40° sin ( 90° - 40° ) sin 50°
=
d.
39.
The actual hours of sunlight on April 1 , 2005 was 1 3 .43 hours. This i s close to the predicted amount of 1 3 .67 hours. Answers will vary.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
29.
sin 400° · sec( -50°) = sin( 40° +360°) . sec 50° 1= sin400.cos 50° = sin 400, sin (90°1 -50°) 1 . = sm40°· sin 40° =1 () is acute,so () lies in quadrant and sin () = i5 corresponds to the right triangle: --
31.
I
b=4 �='5 a
35.
(3
Using the Pythagorean Theorem: a2 +42 = 52 a2 = 25 -16 = 9 a =..[9 = 3 So the triangle is:
b=4�'=5 aadj=33 cos() =-hy p =-5
33.
() <
() <
(3
tan () = opp adj = i3 csc () = hoppy p = �4
quadrant III. Thus,sec () = --135 . 1 =-1 = --5 cos()=-sec() -¥ 13 sin () so tan () = --, cos () . =(tan())(cos()) =-12 ( --5 ) = --12 sm() 5 13 13 1 --13 = sin()1 = -= csc ()--ft 12 = tan()1 = -¥1 = -125 cot ()-sec () = -�4 and tan 0,so () lies in quadrant II . Using the Pythagorean Identities: tan2 () = sec2 ()-1 tan2 () = ( _%)2 -1 =�;-1 =� 6 tan() = ±� = ±% Note that tan 0 ,so tan () = --43 . 1 =-= 1 --4 cos()=-sec() -% 5 sin() tan() = --, cos () so sin() = (tan())(cos() ) = _�4 ( _ i5 ) =�5 . 1 =-1 =-5 csc() =-sin() t 3 1 1 4 cot() =--=-= tan() -% --3 sin () = �13 and () lies in quadrant II . Using the Pythagorean Identities: cos2 () = 1-sin2 () 25 COS2 () = I_ (�13 )2 = 1_ 144 169 = 169 (25 = ± 135 cos() = ±V169 Note that cos () must be negative because () lies
sec () = hadjy p = �3 adj 3 cot() =--= opp 4 () < 0
tan () = �5 and sin ,so () lies in quadrant III . Using the Py thagorean Identities: sec2 () = tan2 () + 1 sec2 ()= (�5 )2 +1 = 14425 +1 = 16925 sec() = ±l2569 = ±�5 Note that sec () must be negative since () lies in
37.
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exist. No portion of this material may be reproduced , in any form or by any means, without permission in writing from the publisher.
Chapter 7 Review Exercises
39.
in quadrant II. Thus, cos () = _2-13 . sin (} = * =g (_�) =_g tan(} = cos(} --& 13 5 5 csc ()-= sin(}1 = -*1 =-1213 1 1 --13 sec () = --= 5 cos () -= --& 1 1 = --5 cot(}=--=tan(} - ¥ 1 2 sin () = _2-13 and 31l"2 () 2 1t (quadrant IV ) Using the Pythagorean Identities: cos2 1-sin2 () 144 cos2 () = 1 _ (_2-13 )2 = 1-� = 169 169 cos(} = ±l16944 = ±g13 Note that cos () must be positive because () lies m quadrant IV . Thus, cos () =-1132 tan () = cossin ()() = -*-& = _2-13 (�12 ) = _2-1 2 1 --13 1 = -= csc () = -sin(} --& 5 1 1 13 sec () = --= cos(} -* =-1 2 1 1 12 cot(}=--=tan () -� = --5 <
41.
<
<
2
.
<
--
() =
.
tan () = -31 and 1 800 () 2700 (quadrant III) U sing the Pythagorean Identities: sec2 () = tan () + 1 sec2 () = "3( 1 )2 + 1 = "91 + 1 = 910 sec() = ±VflO9 = ± J103 Note that sec () must be negative since () lies in J10 quadrant III. Thus, sec () = --3 1 = ---3 J10 = --3J1O cos ()-= sec1 () = J10 J10 J10 10 3 sin () tan(} = -cos(} , so 3J1O ) =-J10 sin(} = (tan(})(cos(}) =-31 [ --10 10 1_ = _ � =-J1O csc(} = _sin1 _() = _ J10 J10 10 1 1 cot () = --= tan () -t = 3 sec () = 3 and -31l"2 () 21t (quadrant IV ) U sing the Pythagorean Identities: tan2 () = sec2 (}-1 tan2 () = 32 -1 = -1 = tan () = ±.J8 = ±2J2 Note that tan () must be negative since () lies in quadrant IV . Thus, . tan () = -2J2 . cos(}=--sec1 () =31 sin () , so tan() = -cos () 2J2 . = (tan(})( cos(}) = -2,,�(1) sm(} 2 "3 = - 3-' 3J2 1 = 3 J2 = --e sc () -= sin(}1 = -- 2../2 2J2 2 4 J2 J2 1 -1 .cot () = --= tan(} -2J2 J2 = --4
.
43.
<
9
._-
<
8
--_.-
3
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
45.
1r
<
<
cot e = -2 and 2 e 7t (quadrant II) Using the Pythagorean Identities: csc2 e 1+ cot2 e csc2 e = 1 + ( _ 2 )2 = 1 + 4 = 5 cscethat±J5csc e must be positive because e lies Note in quadrant Thus, csc = J5 . I l J5 J5 sm. () =-csce = J5 J5 =-5 cose so cot()= -sine ' COS()= ( Cote )( sin ()) = -2 ( �) = _ 2; . 1cot() =-= 1-2 --21 tane= -1cos() =--1f = ---J55 = --J52 sece = --2 2 = 2sin(4x) The graph of = sinx is stretched vertically by afactor of 2 and compressed horiz ontally by a factor of -4I . -
49.
1r
y =
=
=
II.
= -2COS ( x+ %) The graph of cos x is shift ed 2 units to the stretched releftfl ,ected acrossvertically the x-ax is.by a factor of 2, and Y
y
e
_.-
47.
Y
51 .
= tan(x + 7t) The graph of = tan x is shifted units to the left. Y
y
y
Ii
y
1r
I
x
y
53. x
= -2 tan(3x) The graph of = tan x is stretched vertically by a factor of 2, refl ected across the x-ax is, and compressed horiz ontally by a factor of .!.3 . Y
y
y
i\ I
x
416
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Review Exercises
55.
Y
( �)
= cot x +
61.
7r The graph of y = cot x is shifted units to the 4 left.
y = 4 tan
The graph of y = tan x is stretched horizontally by a factor of 2 , shifted left "* units, and stretched vertically by a factor of 4 .
y
/ 57.
y = 4 sec ( 2x ) The graph of y = sec x is stretched vertically by a factor of 4 and compressed horizontally by a 1 factor of -. 2
(-1T,4)J!
i
I I I I
y'
4:
I I I I
i
I I I I
y
63.
= 4 cos x Amplitude = 1 4 1 = 4 ; Period = 2n
65.
Y
= - 8 sin
(% x )
7r Amplitude = 1 -8 1 = 8 ; Period = 27r = 4 2
I I I 'IT
31T
"4 I I
4 I I
I I
I I
X
67.
lr\l' (-¥-,-4)
59.
(� + : )
Y = 4 sin(3x) Amplitude: 1 A1 = 141 = 4 2rc = 2rc T = Period: 3 Phase Shift: t = .Q = 0 3 {()
y = 4 sin ( 2x + 4 ) - 2 The graph of y = sin x is shifted left 4 units,
{()
compressed horizontally by a factor of .! , 2 stretched vertically by a factor of 4, and shifted down 2 units.
Y 5
y 4
x
-5
-7
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Chapter 7: Trigonometric Functions
69.
y 2sin ( 2x - 1Z" ) Amplitude: 1 A 1 = 1 2 1 = 2 Period: 1Z" 1Z" Phase Shift: t = = 0) 2 2 =
73.
y = --2 cos ( 7tx - 6 ) 3
Amplitude: 1 A1 = Period:
1- % 1 = %
Phase Shift: t = � 0) 7t
y 3
y
x
x
-3
71. 75.
= 2 7t = 2 1t 0) 3 2 Phase Shift: t = 2:. = 21t 0) 3 3 2
Period:
T
=
41t
3
The graph is a cosine graph with amplitude 5 and period 81t. Find 0) : 81t = 2 7t =
8 7t 0) 2 1t 2 1t = -1 0) = 4
81t
The equation is: y = 5 COS
y
77.
1 x
The graph is a reflected cosine graph with amplitude 6 and period 8. Find 0) : 8 = 2 1t 0) 8 0) = 2 7t 2 7t = -1t 0) = 8
-1
(� x ) .
4
The equation is: y = - 6 COS
(� x )
.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Review Exercises
79.
81.
hypotenuse = 13 ; adjacent = 1 2 Find the opposite side: 122 + ( opposite ) 2 =13 2 ( opposite ) 2 =169-144= 25 opposite= 55 = 5 o cscB= hyp = .!l. sinB= pp = � hyp 13 opp 5 secB= hyp=.!l. cosB= adj =.!2 adj 12 hyp 13 cotB= adj = .!2 tanB= opp= � adj 12 opp 5
89.
1 80 milhr 0.25 mi = 720 rad/hr 720 rad 1 rev hr 2nrad 3 60 rev nhr "" 1 14.59 rev/hr
--_. _--
91.
Since there are two lights on opposite sides and the light is seen every 5 seconds, the beacon makes 1 revolution every 1 0 seconds: I rev 2n radians n ' m= -- . = - ra dlans/second 1 0 sec 1 rev 5
93.
Let x = the length of the lake, and let y = the distance from the edge of the lake to the point on the ground beneath the balloon (see figure).
cos
3 = = r 5
a
r - -5 cscB=-=
b
5
B b
r
secB=-=5 a
4
B= -
47l"
5
b
3
a
4
x
3
500 tan ( 65° ) = -
a 3 cotB=-= --
tanB=-=- 83.
� 'OOft
(3,--4)
r
2 = 67l" , so 5
+ 7l"
Pio
Reference angle:
67l"
a= -
5
B
500 tan ( 65° ) tan ( 250 )= 500 x+y 500 x + y = --;-----:tan ( 25° ) 500 x= --;-----:-
lies in quadrant III.
7l" - 7l"= -
5
The domain of y= sec x is { xl x is any real number except odd multiples of 1-} . The range of y= sec x is {YI y � - l or y � I}.
87.
r=
2 feet, B= 3 0° or B = i n n = rB= 2·-=- "" 1 .047 feet 6
y
x
4
85.
s
mile
v
x
sinB = !: =_ .i
2
m=-= ---r
5
-5
d= �
� 0.25 mile
y
5
1 80 milhr ;
r= = 4
( 3 ,-4 ) ; a= 3, b=-4 ; 2 r= .Ja2+b2= �3 2+ ( _4 ) = �9+16= 5s =5
-5
v=
500 500 tan ( 25° ) tan ( 6SO ) "" 1 072.25 - 233 . 1 5= 839.10 Thus, the length of the lake is approximately 839. 10 feet.
3
n -n "" 1 . 047 square feet A=-I . r 2B=-1 . (2)2 . -= 2
2
6
3
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
95.
97.
Let x = the distance across the river. x tan(25°) = 50 x = 50 tan(25°) "'" 23.32 Thus, the distance across the river is 23.32 feet.
1 0 1 . a.
Let x = the distance the boat is from shore (see figure). Note that 1 mile = 5280 feet. 1454ft
� 5280 ft
rru
(
I(t) = 220 sin 301tt + a. b. c.
�}
( ( )
(
x
1454 tan(50) = --x + 5280 x + 5280 = 1454 tan(5°) X = 1454 - 5280 tan(5°) "'" 1 6, 6 1 9.30 - 5280 = 1 1, 339.30 Thus, the boat is approximately 1 1 ,339.30 feet, 1 1, 339.30 "'" 2 . 1 5 '1es, firom shore. or 5280 99.
7 Amplitude: A = 1 4.63 - 9. 2 = 2.455 2 Vertical Shift: 14.63 + 9.72 = 12. 1 75 2 21t 0) = 365 Phase shift (use y = 1 4 . 63 , x = 172): 14.63 = 2.455 sin 21t . 1 72 -tfJ + 12. 175 365 2.455 = 2.455 sin 21t · 1 72 -tfJ 365 3441t - '""' . -1 - sm 365 2: = 3441t -tfJ 2 365 tfJ "'" 1.39 Thus, y = 2.455 Sin 21t x - 1.39 + 12. 175 , 365 or y = 2.455 sin 21t (x - 80.75) + 12. 175. 365
b.
t:::: 0 c.
21t = 1 . d=Peno 301t 1 5 The amplitude is 220. The phase shift is: 1t � t= = _2:._ = _ _I_ 0) 301t 6 301t 1 80
(
[
) )
(
)
) ]
Y = 2.455 sin 21t (9 1) - 1.39 + 12. 1 7 5 365 "'" 1 2.61 hours y 20
/
d.
220 d.
-
22
The actual hours of sunlight on April 1 , 2005 was 1 2.62 hours. This is close to the predicted amount of 12.61 hours.
0
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Test
Chapter 7 Test 1.
260° =260 · 1 degree 260 · 1 ;0 radian 2607r rad·Ian =-137r rad·Ian =-180 9 -400° =-400·1 degree =-400 . 180 radian 207r radian 4007r rad·Ian = - -=- -180 9 13° 1 3 · 1 degree =13 · 1807r rad·tan = 137r 180 rad·Ian - 7r8 radian = 7r8 . 1 radian 180 degrees =-22. 5° = 7r8 7r 927r rad·Ian =-927r . 1 rad·Ian 180 degrees =810° = 927r 7r 37r 37r 4 rad·Ian =4· 1 rad·Ian 180 degrees =135° = 37r4 7r =
2.
13.
Set the calculator to degree mode: sin 17 ° '" 0.292
14.
Set the calculator to radian mode: cos 2; '" 0.309
.29237171347
�
3.
4
·
OS�L�38§13169944
=
_
-_.
5
·
1 5.
Set the calculator to degree mode: sec 229° = cos 221 9° '" -1. 5 24
,.....,-=-=77'<,.".,..,-----,
_.
6
·
1 6.
_.
=cose: )-cose:)
11.
1 7.
=0
Set the calculator to radian mode: 1 '" 2.7 47 cot 287r-9- = tan-287r 9 To remember the sign of each trig function, we prini quadrants mari ly needI andto remember thatf) siisnf)positive is positive II, while cos intrig quadrants and The sign of the other four functions can be determined directly from si ne and sinf) 1 cosme. by owmg tanf) =-cosf) , secf) =--cosf) , cscf) -:--sm1f) , and cotf) =--:--cossmf)f) . sin cosO tan secO cscO cot in - - - -- in I
IV.
1m
.
=
37r 47r ) 7r2 (-+ . 7r2 1 947r =SI.n--tan sm--tan-4 =sin ; -tan (3: ) =1-( -1) =2
o in
o
o in
o
QI
QII
QIII
QIV
+
+
0
+
+
+
+
0
+
+
+
+
+
0
+
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
18.
1 9.
Becau se f(x ) = sin x is an odd function and since f(a ) = sina =�, then 5 f(-a ) = sin(-a ) = -sina = -�5 . si{}n =72. and{} in quadrant II. Using the Pythagorean Identities : 2 cos{}2 = 1 - sm. {}2 = 1 -"7( 5 ) = 1 - 4925 = 4924 cos{) = V{24"49 = 27-16 Note that cos{} mustbe negative becau se{} lies in quadrant II. Thus,cos{} = _ 2-16 7 . 5-16 tan{} = cossi{}n{} = t'f "7 = 5 ( - 27-16 ). -16-16 = -""}2 csc{} -= sin{}1 = -t1 =-75 1 -1 = 7 -16 = -16 7 sec{} = -= --2 cos{} - 'f 2-16 -16 12 -16 2-16 12 . cot{} -= tan{}1 =--1 = --= ---5 5-16 -16 1 ±
21.
2
--_. -
-
20.
22.
5.J6
2
<
<
cos{} = �3 and 321l" {} 2 1l" (in quadrant IV ). Using the Pythagorean Identities: sm. {}2 = 1 - cos{}2 = 1 - ("32 )2 = 1 -"94 = "95 si{}n = � = � Note that si{}n mustbe negative becau se{} lies in quadrant IV . Thus,sin{) = J53 . si{}n --= - Jf = -_J5 . -3 = -J5 tan{} = -cos{} t 3 2 2 1 -1 = -J5 = --3 .3J5 csc{} = -= sin{} - Jf J5 J5 5 1 = -1 = -3 sec{} = -cos{} t 2 2J5 cot{} -= tan{}1 --= -�1 = -_J52 _· J5J5 = --5 ±
<
<
=
±
_
tan{} = -g5 and 1l"2 {} 1l" (in quadrant II) Using the Pythagorean Identities: 169 sec{}2 = tan{}2 + 1 = ( - 51 2 )2 + 1 = 14425 + 1 2s sec{) = ±�16925 = ±�5 Note that sec{} mustbe negative since{} lies in quadrant II. Thus,sec{} = --153 1 =-1 = -5 cos{} =-sec{} -¥ 13 si{}n so tan{} = --, cos{} 12 ( -5 ) =12 .sm{} = (tan{} )(cos{} ) = -5 13 13 1 1 1 3 csc{} = -si{}n = -* = -1 2 1 -1 = -5 cot{} =--= tan{} -¥ 12 The point (2,7 ) lies in quadrant I with x = 2 and y =7. Since x2 + / = r 2,we have r =�22 +72 ../53 = . So, 7__ 7 . J53 =../53 7 53 sin{} �= r _ = J53 =_ J53 J53 The point (-5, 1 1) lies in quadrant II with = -5 and y = 11. Since x2 + y2 = r2,we have r = �( _5)2 + 112 = .J146 . So, ..J146 5M6 cos{} �= r � = .J146 � = .J146.J146 146 The point (6, -3 ) lies in quadrant IV with x = 6 and y = -3 . Since x2 + / = r 2,we have = = 3 .J5. So, = �62 + (-3/ .J45 tan{)= x= -63 = -12
23.
x
±
=_
_
24.
r
y
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Test
25.
Compari ng y 2 sin (� �) to y A sin ( - ¢ ) we see that A 2 , ..!.3 , and ¢ "6 , The graph is a si ne curve with amplitude I A I 2, period = 2" 12/"3 = 6" , and phase ShIft, = ! = 1;3 = ; , The graph of y 2 sin (� - � ) wil lie between -2 and 2 on the y-axis, One period wil begin at x = t = "2 and end at 13" . W e d'IV'd x-= 2" +-¢ = 6"+ "-2 =-I e the 2 interval [; 1�" ] into four subintervals, each of 6" 3" length -=-. 2 " l[5", 1�" ] [;The fivel[key2"poi, 7;ntsl[on7;the,5graph are (; } (27r,2) , C; ,o} (5",-2) , C�" ) plot function. these fiveThepoingraph ts andcanfillthen inthebegraph of theWe sine ex tended inboth directions. =
26.
-
=
w =
=
y
=
T
-
w
=
=
wx
=
To graph y tan ( -x+ �) +2 we will start with the graph of y tan x and use transf ormati ons to obtai n th e desi red graph. y ta n x
=-
=
2n x
-7[
=
Nex t we shif t the graph � units to the left to obtai n the graph of y tan ( x + �)
w
w
=
w
J
,
4
-7[
,2" ,0
.
Now we reflect the graph ab out the y-ax is to obtain the graph of y tan ( -x + �) . =
,0
y
y
3
\
7[
x
Lastly, we shift the graph up 2 units to obtain the graph of y = ta n ( -x + � ) +2 .
-( 'IT, - 2)
\ 7[
x
423 © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently
exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
27.
For a si nusoidal graph of the form = A si n (l1)X r/J),the ampli tu de is given by IAI, the peri od is given by 2rc l1) ,and the phase shift is givenby t. l1) Theref ore,we have A = -3 , l1) = 3,and r/J = 3 ( :) = - 3: . The equati on for the graphis = 3 in (3X + 3: ) . between the difference walk issector areaofofthethelarger of the area the and area theThe smaller shaded sector. y
29.
g
-
y
28.
-
0
.
IS
"""S"i 18
-
2
v
S
walk is givenby The area of the� = 2 R 2e-!r2 2e ' A! = � (R 2 _r 2 ) r is andradius largerThesector of thesector. whereradiusR isofthetheradius the larger smaller radius because smallerTherefore than3 feetthewide. feet longer isthe3 walk , be to is R = r + 3 , and A = � ( (r +3 / -r 2 ) = � (r 2 +6r + 9 -r 2 ) = e (6r + 9 ) "2 The shaded sector has an arc length of 25 feet and a central angle of 50° = �; radians . The rad· f thi s sector r = -eS = 25 = 9rc0 eet . Thus,the area of the walk is givenby A = �( 6 (�) + 9) = 5rc36 ( 540rc + 9) =7 5 + 5rc4 ft 2 ""7 8.93 ft2 IUS
To throw the hammer 83.1 9 meters, we need 2 s = _o_ 83.1 9 m = 9.8V0mls 2 2/ 2 V0Vo2 == 815. 28.553262mlsm S are related angular speed speedto theandformula Linear = r .l1) The radius is according r = 1 90=cmr .l1)= 1. 9 m . Thus,we have 28.553 28.553 = (1. 9 )l1) radians per60 second l1) = 15.028 radians i on l1) = 15 . 028 sec . 1 minsec . 12rcrevolut radians "" 143. 5 revoluti ons per minute (rpm) Adrirpman must hammerit 83.1 theswinging Tohavethrow 9 meters, 143.5 of rate a at been upon release. Let the distance to the base of the statue. v
-
30.
•
x=
305 feet x
tan (20°) = 305 X = tan3t200 ) � 0.3640 837 . 98 feet abouty. 838 feet from the base of the The Statueshipof isLibert x
""
""
fi
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Cumulative Review
31.
Chapter 7 Cumulative Review
Let h = the height of the building and let x = the distance from the building to the first sighting.
1.
�h 50 feet
tan ( 40° ) = -h x
{-1, �}
x
3.
- ( t) )
tan ( 32° ) = h x + 50 h = (x + 50) tan ( 32° ) h=
(
tan 40°
2X2 + x - 1 = 0 (2x - 1) (x + l) = 0 x = -1 or x = -1 2
5.
+ 50 tan ( 320 )
J
h = _h_ + 50 O.6249 0.8391 h = 0.7447 h + 31.245 O.2553h = 3 1 . 245 h 1 22.39 feet The building is roughly 1 22.4 feet tall.
radius = 4, center (0,-2) Using (X_ h) 2 + (y _k ) 2 = r 2 (X _ 0) 2 + ( y - (-2) f = 4 2 x 2 + (y + 2 ) 2 = 1 6 x 2 + y 2 - 2x + 4 y - 4 0 x 2 - 2x + 1 + l + 4 y + 4 = 4 + 1 + 4 (X_1) 2 + ( y + 2 ) 2 = 9 (X _ 1) 2 + ( y + 2 ) 2 = 3 2 This equation yields a circle with radius 3 and center ( 1 ,-2). =
y 2
�
x
-6 7.
a.
y = x2 y
x
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Trigonometric Functions
y
11 . x
( )
tan " - 3 cos " + csc " = 1 - 3 .[3 + 2 4 6 6 2 = 3 - 3 .[3 2
6 - 3 .[3 2 13.
x d.
y
=
The equation is: y 3 COS
2
-1
The graph is a cosine graph with amplitude 3 and period 12. Find m: 12 = 2 7t m 1 2 m= 2 7t ,. 2 7t = m= 12 6 =
lnx
1 5.
y 3
a.
( � x) .
A
polynomial function of degree 3 whose x intercepts are -2, 3, and 5 will have the form f(x) a(x + 2) (x - 3) (x - 5) , since the x intercepts correspond to the zeros of the function. Given a y-intercept of 5, we have f(O) = 5 a(O + 2) (0 - 3) (0 - 5) = 5 30 a= 5 a= 5 1 30 6 Therefore, we have the function 1 (x + 2) (x - 3) (x - 5) . f(x) = 6 =
x
-=
e.
y
= sin x y 2
(¥, I)
y
10 (-0.08,5.01) (-2,0)
f.
y
= tan x -10
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Cumulative Review
b.
A
rational function whose x-intercepts are -2, 3, and 5 and that has the line x = 2 as a vertical asymptote will have the form a (x + 2)(x-3)(x-5) , since the x1 (x) = x-2 intercepts correspond to the zeros of the numerator, and the vertical asymptote corresponds to the zero of the denominator. Given a y-intercept of we have
5, (0) 1 =5 a (0+2)(0-3)(0-5) = 5 0-2 30a = -10 a = --31 Therefore, we have the function ! (x+ 2)(x-3)(x-5) 3 x-2 1 (x) = _=<-- � (x + 2)(X2 -8x + 15) (x-2) _ � (X3 _6x2 -x+30) (x-2) x3 -6x2 -x + 30 x3 -6x2 -x + 30 6-3x -3x + 6 _ ___
I I I I I
(3.82, 1.03)
-10
427
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8 Analytic Trigonometry Section 8.1 1.
1 7.
Domain: { x i x is any real number} ; Range: {YI-I::; y ::; 1}
3.
[ 3 , (0)
5.
1', J3 2
7.
x = sin y
9.
7r
1 9.
5
11.
True
1 3.
sin- 1 0
-
.
-I
sin() =
.
-
21.
We are finding the angle (), :::. ::; ()::; :::. , 2 2 whose sine equals 1 sin() = -1, -
-
()
= _
J2 , 2
()= :::. 4 J2 = -7t sm 2 4
sin-I ( -1 )
.
2:
2 7t sln ( - 1 ) = -2 .
J2 sm 2 We are finding the angle (), -2: ::; () ::; 2: , whose 2 2 J2 . equa I s sme T'
We are finding the angle (), - 2: ::; B ::; 2: , whose 2 2 sine equals O. sin() = 0, - 7t ::; ()::; 7t 2 2 ()= o 1 sin- 0 = 0 1 5.
tan -10 We are finding the angle (), _2: < () < 2:, whose 2 2 tangent equals O. 7t < () < 7t tan() = 0, -2 2 ()= o tan- 1 0 = 0
_
2: ::;, () ::; 2:
2
2
-I
tan- I J3
We are finding the angle (), - 2: < () < :::. , whose 2 2 tangent equals J3 . 7t 7t r:; tan B = ,,3, -- < () < 2 2 ()= 2: 3 tan- I J3 2: 3 =
-I
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Section B.1: The Inverse Sine, Cosine, and Tangent Functions
23.
COS - I
( �)
39.
-
We are finding the angle
-J3 . equa I s 2 cosme -
-
COS-I
( �) _
sin-I 0. 1 '" 0. 1 0
27.
tan - I 5 '" 1. 3 7
;
41.
35.
cos-I .fi '" 1.08
37.
COS-I cos
'"
( ( ;)) ( ( 9;)) _
3
sin-I sin
=_
;
3 .
follows the form of the
( ( ))
f)
angle
9"
9" is not 8
[ ; , ; ]. We need to fmd an in the interval [ -; , ;] for which
in the interval
-
9" is in quadrant 8 8 so sine is negative. The reference angle of 9" is 8
I
sin
=
sin f). The angle
III
"8 and we want to be in quadrant so sine will still be negative. Thus, we have " IS. m . the mterva . 98" = sm. ( -") - . S·mce - sm' I 8 8 [-; ;] , we can apply the equation above and " get sm ( sm 9" ) = sm ( sm ( ") 8 ) = -8 ' S f)
O�-'� -�. 1:�26395 Fo�-1(,f<2)/3) 1.079913649 lsin-1(,f<3)/5) .3537416059
( 4; ) follows the form of the equation ( ( )}
Since
2 2
cannot use the fonnula directly since
an� ] 8 ��063771 an-1< -3) lsin-1<-1.249045772 -0.12) -.1202898824
FI i x = cos-l cos x = x . Since
( tan ( x )) = x .
-I
equation I-I (i x = sin-I sin x = x , but we
.i973955598 o�-1<7/8) .5053605103 in-1<1/8) .1253278312
sin-I (- 0. 1 2 )'" - 0. 12
( ( )}
.
( ))
�ml� � i M1674212 o�·1<0.6) .927295218 rt-an-1(5) 1.373400767
33.
3
( ))
tan-I tan
tan-I ( O A ) - 0.3 8
follows the form of the
the equation directly and get
6
31.
-
8
57t
8
3
3" . m the mterva ' 1[""] - - , - , we can app I y
- - IS
0 � f) � 7t
COS-I 7... '" 0.5 1
-
equation I-I (i x = tan
=5
25.
29.
0 � f) � 7t , whose
'
-J3 ' cos f) = 2 f) =
f) ,
( ( ; ))
tan-I tan
IV
,
4; is
.
in the interval [0," ] ' we can apply the equation
( 4; ) = 4; .
-I
.
.
-I
.
-
directly and get COS-I cos
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8: Analytic Trigonometry
43 .
tan-I [ tan ( 4;)) follows the form of the equation rl {!(x)) = tan-I (tan(x)) = but we cannot use the formula directly since 41l"5 not in the interv al [- � , %]. We need to find an angle () in the interv al [-� , �] for which tan (4; J = tan () . The angle 4; is in quadrant so tangent is negativ e . The ref erence angle of 41l"5 is 1l"5 and we want () to be in quadrant so tangent wil still be negativ e. Thus, we hav e tan (541l" I) = tan (-"51l" I) ' S·mce -"51l" . m. the interval [-� , �], we can apply the equation ab ove and get tan-I [ tan (;4 J) = tan-I [ tan (- � J) = - � .
51. tan ( tan-I 1l") follows the form of the equation I (J -I (x)) = tan ( tan-I(x)) = x . Since is a real number, we can apply the equation directly and get tan ( tan-I 1l" ) = 1l" . 53. I( x ) = 5 sin x + 2 y = 5sinx+2 x = 5sin y +2 5sin y = x- 2 x- 2 sm. y =-5 X - =I-I ( ) y = sm. -I-5 The domain of I( x) is the set of all real numbers, or ( ) in interv al notation. To find the domain of I-I(x) we note that the argument of the inverse sine function is x -25 and that it must lie in the interv al [-1,1]. That x-2 $:1 -1$:-5 -5$:x- 2$:5 -3$:x$:7 The domain of I-I(x) is {xl-3$:x$:7}, or [-3,7] in interv al notation . 55.I(x) = -2cos( 3x) y = -2cos( 3x) x = -2 cos( 3y ) cos( 3y ) =-�2 3y = cos-I (-1) = "31 cos-I (-"2x I) =I-I ( x) The domain of I( x) is the set of all real numbers, or ( ) in interv al notation. To find the domain of I-I (x) we note that the 1l"
x,
IS
2
II
IV
-00,00
IS
45.
IS,
sin (sin -I ±J follows the form of the equation I (J-I (x)) = sin (sin-I (x)) = x . Since ± is in the interval [-1, 1] we can apply the equation directly and get sin ( sin-I ±J = ± . tan ( tan-I )4 follows the form of the equation I (J-I (x)) = tan ( tan-I (x)) = x . Since 4 is a real number, we can apply the equation directly and get tan(tan-1 4) = 4 . Since there is no angle () such that cos () = 1.2 , the quantity cos-I 1.2 is not defmed. Thus, cos( cos-I 1. 2 ) is not defined. '
47.
49.
x
Y
-00,00
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8.1: The Inverse Sine, Cosine, and Tangent Functions
The domain of I ( x ) is the set of all real numbers, or ( ) in interval notation. To find the domain of I-I ( x ) we note that the
argument of the inverse cosine function is -x 2 and that it must lie in the interval [ -1,1 ] . That is, -1 � - '::' � 1 2 2 �x �-2 -2 � x � 2 The domain of rl ( x ) is { x 1-2 � x � 2 } , or [-2, 2 ] in interval notation.
-00,00
argument of the inverse sine function is
3
and
that it must lie in the interval [ -1,1 ] . That is, - 1 � �� 1 3 -3 � x � 3 The domain of rI ( x ) is { x l-3 � x � 3 } , or [ -3, 3 ] in interval notation.
57. / ( x ) = - tan ( x + l ) - 3 y = - tan ( x + 1 ) - 3 x = - tan (Y+ l ) - 3 tan ( y + 1 ) -x - 3 y + l = tan-I ( -x - 3 ) y = - 1 + tan-I ( -x - 3 ) = - 1 - tan-I ( x + 3 ) = rl ( x ) (note here we used the fact that y = tan-I x is an odd function). The domain of I ( x ) is all real numbers such ( 2k + 1 ) 7Z' 1 where k is an integer. To that x "* 2 find the domain of I-I ( x ) we note that the argument of the inverse tangent function can be any real number . Thus, the domain of I-I ( x ) is all real numbers, or ( ) in interval notation.
61.
=
4 sin- 1 x 7Z' . - I X = -7Z' sm 4 =
h .x;= sin -7Z' = 4 2
The solution set is 63 .
-00,00
6
sin ( 2y + l ) = � 3 2 y + 1 = sm -X 3 -I
(� ) - 1 y = -1 sm-I ( -x ) - -1 = I-I ( x ) 2 3 2
{�}.
3 COS - I ( 2x ) 2 7Z' 2 7Z' COS- I ( 2x ) = 3 2 7Z' 2x = cos 3 1 2x = -2 X = --1 4 =
The solution set is
59. l ( x ) = 3 sin ( 2x + l ) y = 3 sin ( 2x + l ) x = 3 sin ( 2y + l )
.
�
{-±}.
5. 3 tan- I x = 7Z' tan- I x = 7Z' 3 x = tan 7Z' = J3 3 The solution set is {J3} .
2Y= Sin-I .
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in wri ting from the publisher.
Chapter 8: Analytic Trigonometry
67.
75.
4cos-1x - 2 n- = 2 cos-1x 2cos-1x - 2 n- = 0 2cos-1x = 2 nCOS- I = nx = cos n- = -1 The solution set is {-I} .
Let point C represent the center of Earth, and arrange the figure so that segment CQ lies along the x-axis (see figure). y
X
69.
[ [ [
I
Note that e = 29°45 ' = 29.75° . a.
D
=
24· 1 -
cos-l(tan(23. 5.I�o)tan(2 9.7---, 5.I�o)) ..c. , ::..:. -,--_ :,,,: �_":",-_,:,:::,--
_
7t
"" 13.92 hours or 13 hours, 55 minutes b.
D
=
24·
cos-l(tan(0.I�o)tan(29.75.I�o))
- ---'---'-----'--'----'-'-
I
7t
"" 12 hours c.
D
=
24· 1 -
l
COS-I (tan(22.8. �o)tan(29.75.I�o)) .: .:c..:.-"--'.::.::. ...:. ----'�-'-l -'---
_
_
7t
__
"" 13.85 hours or 13 hours, 51 minutes 71.
(
Note that e = 2 1 ° 1 8 ' = 2 1 .3° . a.
D
=
24· 1
-
· I�o)) COS-I (tan(23. 5.I�o) 21.3--" tan( .:...:...!... --'.::.::. -'------'--'---...:.
-
(
7t
-
"" 13.30 hours or 13 hours, 18 minutes b.
D=
cos-l(tan(0.I�o)tan(21.3 .I�o)) 24· 1 - ---'---'-----'--'---'-"7t
(
"" 12 hours c.
D
=
l
7t
"" 13.26 hours or 13 hours, 15 minutes 73 . a.
D
=
24 ·
1
. 5 . I�o ) tan ( 0 .I;o) ) cos - l (tan (2 3-
- ---'--'--
----''---'----"-"-
7t
"" 12 hours b.
D
= 24·
cos-l(tan(O. I �o ) tan(O.I�o ) )
": --=":":..:....!... 1- ---'--":"'-":"::":"'---'-7t
"" 12 hours c.
D
=
7t
the latitude of Cadillac Mountain, the effective radius of the earth is 27 1 0 miles. If point D(x, y) represents the peak of Cadillac Mountain, then the length of segment PD is 1 530 ft . 1 mile "" 0.29 mile . Thus, point 5280 feet D(x, y) = (2700,y) lies on a circle with radius r = 27 10.29 miles. Thus, cos e = � = 271 0 r 27 1 0.29 e = cos-1 27 1 0 "" 0.01463 radians 27 1 0.29 Finally, = re = 271 0(0.0 1463) "" 39.65 miles , 2 n-(27 1 0) = 39.65 , so and 24 t 24(39.65) . t "" 0.0558 9 hours "" 3 . 35 nunutes 2 n-(27 1 0) Therefore, a person atop Cadillac Mountain will see the first rays of sunlight about 3 .35 minutes sooner than a person standing below at sea level.
l
s
(
)
=
l
]
cos-l(tan(22- . 8 .I;o ) tan ( 0 . I;o )).-".. --'---'-:.:..:. 24· 1 - ----''---'---'-'-'-'---
"" 12 hours d.
]
At
l
cos-l(tan(22.8.I�o)tan(21.3 .I�o)) 24· 1 - ----''----'-----'-----'----'-'-
( ( (
l
I
77 .
= 0 ; b = .J3; The area is: tan- I b - tan- I = tan-l .J3 - tan- I 0 . = -n- - 0 = -n- square umts 3 3
a.
a
b.
a
a
= - .J3 ; b = 1 ; The area is: 3
( �l
tan- I b - tan- I = tan-1l - tan-1 a
]
5 n- square umts . =12
There are approximately 1 2 hours of daylight every day at the equator. 432
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8.2: The Inverse Trigonometric Functions [Continued]
79.
Here we have = 4 1 °50 ' , /31 = -87°3 1 ' , a2 = 2 1 0 1 8 ' and /32 = -1 57°5 1 ' . Converting minutes to degrees gives = ( 4 1 t ) 0 , /31 = ( -87 �6 ) 0 , a2 = 2 1 .3° , and /32 -157.85° . Substituting these values, and 3960 , into our equation gives 425 1 miles. The distance from Chicago to Honolulu is about 425 1 miles. (remember that S and W angles are negative) a
,
a
l
=
r
Find the angle B, 0 :::; B whose cosine .J3 . equals -2 :::;
l
d
=
�
7t ,
S ection 8.2 1.
Domain: {xl x odd multiples of � }, Range: {yl y :::; - l ory I} 15 2 =-2 - 15 -5"*
13. sec(cos-I �) Find the angle B, 0 :::; B :::; whose cosine equals -.21 cos B = -21 , 0 :::; B B=!!:.3 sec( COS-I �) = sec � = 2 csc(tan-' l) Find the angle B, - � B �, whose tangent equals 1 . tanB 1,
:2:
3.
7t,
:::;
7.
True
7t
1 5.
7t
7t
Find the angle B, --2 :::; B :::; -,2 whose sine J2 . equals 2 sinB = J22 , B = !!:.4 cos (sin-I J2) = cos!!:.4 = J22 2
<
<
=
csc( tan-I I) = csc: = J2
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exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.
Chapter 8: Analytic Trigonometry
1 7.
sin [ tan-I(-I)] Find the angle (), -% () %, whose tangent equals -1 . tan() = -1, () = -!!:.4 . .Ji sm. [ tan -I (-1) ] = sm. ( - 4"7t ) = - 2 <
23.
<
!!:..
Find the angle (), --7t2 :-:; ():-:; -,7t2 whose sme. equals --21 . SIn. () = --21 ' () = -!!:.6 . sec [sin -I ( _�)] = sec ( _�) = 2� 21.
sin-I [Sin ( _ 7; )] = sin-I � Find the angle (), --7t2 :-:; ():-:; -,7t2 whose sme. equals -.21 SI.n () =-,21 () = 6 sm. _I [sm. ( -67 1C )] = "67t tan (sin-I �) Let () = sin-I .!.3 . Since sin () =.!.3 and () :-:; !!:..2 , () is in quadrant and we let 2y = 1 and r=3. Solve for x: x2 + 1 = 9 x2 = 8 x = ±J8 = ±2.Ji Since () is in quadrant x = 2.Ji . tan ( sin-l �) = tan() = � = 2:n · � = � sec ( tan-I �) Let () = tan-I .!.2 . Since tan () =.!.2 and () 2 () is in quadrant and we let 2 x = 2 and y = l . Solve for 2r: 2 + 1 2 =r r2 = 5 r = ..[5 () is in quadrant sec ( tan-1 '!'2 ) = sec() =!...x = ..[52
2 5.
- !!:.. :-:;
COS-I ( cos 5; ) = COS-I ( _�) Find the angle () :-:; 7t, whose cosine .Ji . equals -2 .Ji :-:; ():-:; 7t cos () = - 2' 3 7t () = 4 COS-I ( cos 5; ) = 3;
I,
I,
(), 0 :-:;
27.
0
- !!:.. <
< !!:.. ,
I,
I.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8.2: The Inverse Trigonometric Functions [Continued]
29.
c+m ' [ - '?) 1 J2 and J2) ' S mce ' sm () = - 3 Let () = sm [ -3 -�2 .$; (}.$; �2 , () is in quadrant and we let = - and r = 3 . Solve J2for x: x2 + 2 = 9 x2 = 7 x = ±.[i Since () is in quadrant x = .[i . J7 . J2 = $4 cot [sin [ _ J2 )] = cot () = .:: = -J2 J2 sin [ tan-I(-3) ] Let (}= tan-I(-3) . Since tan(}= -3 and -�2 < () < �2 , () is in quadrant and we let x = 1 and = -3 . Solve for r: 1 + 9 = r2 r2 = 10 r = ±M Since () is in quadrant r = M . sin [ tan-I (-3) ] = sin(} = � = 20. � = 3� sec [ sin-I 2f) 2J5 and 2J5 . S mce ' sm. () = -Let () =sm. -I --�2 .$; (}.$; �2 , () is in quadrant and we let = 2J5 and = . Solve for x: x2 + 20 = 25 x2 = x = ±J5 Since () is in quadrant x = J5 . sec [ sm. 2J5 ) = sec () = -:;r = J5 = -I
.
.
37.
,
IV,
Y
31.
3
_
y
6
39.
2
,
-
IV,
y
41.
_
5
y
r
2J3 sec_ 1 -3 We are finding the angle (), O .$; () � 2.n ' 2J3 . whose secant equals -3 2J3 , O .$; (}.$; n, ()"# � sec () = -2 3 () =� 2J3 = -n sec_ 1 -3 , () "#
6
5
6
I,
5
1,
n
IV,
33.
< n,
6
IV,
-I
WecoelareJ3finding the angle () 0 < () whose cotangent equals J3 . cot(}= J3, O < (} < n () = � coel J3 = � csc-I(-I) We are finding the angle (), -�2 -< () -< �2 ' () "# 0 whose cosecant equals -1 . csc(} = -�2 <- (} -< �2 ' (}"#O (} = -�2 csc-I(-I) = -�2
5
I,
-1
5-
5
(; 5 "J
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8: Analytic Trigonometry
43.
13 ) cot ( -3 We are finding the angle B, 0 < B < 1t, whose 13 . cotangent equals -3 13 O < B <1t cotB = --, 3 21t B= 3 coel ( �) = 231t -I
We seek the angle B, --:rr2 :-:;; B :-:;; -,:rr2 whose sine equals ..!.3 Now sin B = ..!.3 so B lies in quadrant The calculator yields sin-I ( -�J "" -0.34 , which is an angle in quadrant so csc-I ( -3 ) "" -0.34 . _
is I n' I :
sec-I 4 = COS-I .!..4 We seek the angle B, 0 :-:;; B:-:;; :rr, whose cosine equals .!..4 Now cos B = .!..4 , so B lies in quadrant The calculator yields COS-I .!..4 "" 1 .32 which is an angle in quadrant so sec-I ( 4 ) "" 1 .32 . 1.
51 .
,
IV,
]���3 69095
coel (-Fs) = tan-I ( - Js) We seek the angle B, 0 :-:;; B :-:;; :rr, whose tangent equals Fs1 . Now tanB = Fs1 ' so l'les . quadrant The calculator yields tan-I ( - Js) "" -0.42 , which is an angle in ce B. iTherefore, s in quadrant quadrant B "" -0.42 +:rrSi""n2.72 coel (-Fs) "" 2 .72 . -
,
-
t��8116072
IV .
4
7
. cot - 1 2 = tan -I -21
We seek the angle B, 0 :-:;; B:-:;; :rr, whose tangent equals -.2I Now tan = -,2I so lles' . quadrant The calculator yields an -I .!..2 "" 0.46 , which is an angle in quadrant so coel ( 2 ) "" 0.46 . �an'l( l L1 (7
L1 (7
L1 (7
m
II .
I,
os'l e
-
IV .
_
45.
.
II,
m
1.
I,
We seek the angle B, --:rr2 :-:;; B :-:;; -:rr2 , B * O , whose sine equals - �3 . Now sin B = -�3 , so lies in quadrant The calculator yields sin-I ( -%J "" -0.73 , which is an angle in quadrant so csc-I (-%J "" -0.73 .
���3 6476 09
IV .
IS 1 n" :
IV,
)2�+276562
436
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
e
Section 8.2: The Inverse Trigonometric Functions (Continued]
(), 0 -::;.
63.
-::;. n,
We are finding the angle () whose tangent equals --23 . Now tan () = --,23 so () lies in quadrant The calculator yields tan-I ( -%) 9 which is an angle in quadrant Since () is in quadrant () + 2.55 . Thus, coCI ( -%) 2.55 . II.
"" 0 5 -
.
,
IV.
"" 0 59 -
57.
.
n ""
II,
""
n n - - < () < - ,
Let ()= tan- u so that tan() = u , 2 u Then, 1 1 cos ( tan- I u ) = cos () = -sec () = " sec 1
-OC) <
65.
< oc) .
67.
2
r--:;-:. 2 ()
n
-OC) <
< oc) .
- n -::;.
-::;. n
y
59.
61.
_ n -::;. () -::;. n ,
Let ()= sin-1 u so that sin() = u , 2 -1 -::;. u -::;. 1 . Then, sin () tan ( sm. - I u ) = tan () =-cos() sin () sin () �COS 2 () �1-sin2 () u
2
n
Let () = csc- 1 u so that csc() =u , --2 -::;' ()-::;' -2 , l u i ;:: 1 . Then, n () = cot () sm. () cos ( csc- I u ) = cos () = cos () . -sism() .cot() � �CSC2 ()-1 csc () csc () csc () � u Let () u= coC I Then, so that cot () = u , 0 () , 1 _ =.!. tan (CoC1 u ) = tan () = _ cot () u l g ( r C�) J = cos ( sin - I ��) ' sm. () = -12 and Let () = sm. -I -1213 . S mce 13 quadrant and we let 2= 12()and2 , =()13is. inSolve for x: x2 +122 = 132 x2 + 144 = 169 x2 25 x = ±.J25 = ±5 Since () is in quadrant x = 5 . x 5 . _1 12 1 12 g ( f- ( 13 )J = cos ( sm 13 ) = cos () = -; 13 <
U
I,
r
=
I,
=
O -::;' () -::;' n
Let () = sec- 1 u so that sec() = u , and � , l u i ;:: 1 . Then, sin ( sec - I u) = sin () = �sin 2 () = �1-cos2 () 2 ()-1 = �1- sec21 () = �sec �sec2 () () ,to
�
lui
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in wri ting from the publisher.
Chapter 8: Analytic Trigonometry
Let B = sin-I -
;r
;r
(-�). Since sin B -� and
Let B = COS-I
=
(-±). Since cos B = -± and
o :=; B :=; B is in quadrant II, and we let x = - 1 and r = 4 . Solve for y: ( 1) 2 + y 2 = 4 2 1 + y2 1 6 / = 15 y = ±M Since B is in quadrant II, y = M .
_
:=; B :=; , B is in quadrant IV, and we let 2 2 y = -3 and r = 5 . Solve for x: x 2 + ( 3) 2 = 5 2 x 2 + 9 = 25 x2 = 1 6 x = ±J16 = ±4 Since B is in quadrant IV, x = 4 . = tan sin-I h rl
_
;r ,
=
( ( - �)) ( (-�))
73 .
g
(h- ' c;)) (
M M == tan B = Z = x -1 79. a.
;)
= cos tan-I 1
Let e = tan-I g . Since tan e g and 5 5
e = coC'
=
;r
:=; e :=; , e is in quadrant I, and we let 2 2 x = 5 and y = 1 2 . Solve for r: r 2 = 5 2 + 12 2 r 2 = 25 + 144 = 1 69 r = ±M9 = ±13 Now, r must be positive, so r = 1 3 . 12 = cos tan _ 1 "5 _ 1 "5 1 2 = cos e -;-x = 13 5 g h -
;r
( ( )) (
)
Since the diameter of the base is 45 feet, we 45 have r = - = 22.5 feet. Thus, 2
b.
(2:�5) = 3 1 .89° .
e = coC' !:. h r r = h cot e � cot e = h Here we have e = 3 1 .89° and h = 17 feet. Thus, = 17 cot (3 1 .89° ) = 27.32 feet and the diameter is 2 (27.32) = 54.64 feet.
-
r
=
c.
13 = = cos -1 2 6 ;r
r . From part (b), we get h = -cot e 122 = 6 1 feet. The radius is 2 r h = __ = 6 1 "" 37.96 feet. cot e 22.5 / 1 4 Thus, the height i s 37.96 feet.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8. 3: Trigonometric Identities
8 1 . a.
2x cot () = ---:2y + gt 2 () = cot
-l
11.
[ 2y 2x+ ) gt
2
The artillery shell begins at the origin and lands at the coordinates ( 6 1 75, 2450 ) . Thus, () = cot
-l
[
2 · 6 1 75 2 . 2450 + 32.2 ( 2.27 ) 2
1
13.
::::: coel ( 2.437858 ) ::::: 22.30 The artilleryman used an angle of elevation of 22 .3 0 • b.
v
t
sec () = _0_ x x sec () ( 6 1 75 ) sec ( 22.3 0 ) = = 2.27 t = 2940.23 ft/sec Vo
---\
. y = sec x = cos -1 x -\
7t
-5
85.
o
(
5
Answers will vary.
S ection 8.3 1.
True
3.
identity; conditional
5.
0
7.
9.
----
sin () + cos () + cos () - sin () cos () sin () 2 sin () + sin () cos () + cos ()( cos () - sin ()) sin () cos () sin 2 () + sin () cos () + cos 2 () - cos () sin () sin () cos () sin 2 () + cos2 () + sin () cos () - cos () sin () sin () cos () sin () cos ()
1 5. 83
_
cos () l + sin () cos ()(I + sin ()) I - sin () 1 + sin () - 1 - sin2 () cos ()(1 + sin ()) cos 2 () l + sin () cos ()
1 7.
1 9.
False
(sin () + cos ())( sin () + cos ()) - 1 sin () cos () sin 2 () + 2 sin () cos () + cos2 () - 1 sin () cos () 2 sin () + cos 2 () + 2 sin () cos () - 1 sin () cos () 1 + 2 sin () cos () - 1 sin () cos () 2 sin () cos () sin () cos () =2
_
,,'"-
() + 1)(sin () + I) 3 sin 2 () + 4 sin () + 1 (3sin - --�-sin 2 () + 2 sin () + 1 (sin () + 1)( sin () + 1) 3 sin () + 1 sin () + 1 csc () . cos ()
=
-'-:-
1 . cos () = cot () cos () = -sin () sin ()
--
sin () 1 1 tan () · csc () = -- · _- = - cos () sin () cos ()
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8: Analytic Trigonometry
23.
2 5.
( ( (
)
sin B + -.cos -B cos B(tan B + cot B) = cos B -cos B sm B 2 2 = cos B Sin B + cos B cos Bsin B 1 = COS B cos Bsin B 1 sin B = csc B
)
J
(sec B - 1)(sec B + 1) = sec 2 B - 1 = tan 2 B
29.
(sec B + tan B)(sec B - tan B) = sec 2 B - tan 2 B = 1
33.
35.
sin u 1 - - sec u - tan u = -cos u cos u = 1 - Sin u . 1 + s�n u l + sm u cos u l - sin 2 u cos u(1 + sin u) cos 2 cos u(1 + sin u) cos u l + sin u
)(
(
)
U
1 - cos 2 u tan u cot u - cos 2 u = tan u . -tan u = 1 - cos 2 u . 2u = sm
27.
31.
37.
39.
41.
cos 2 B(I + tan 2 B) = cos 2 B · sec 2 B 1 = cos 2 B · -cos 2 B =1 (sin B + cos B) 2 + (sin B - cos B) 2 = sin 2 B + 2 sin B cos B + cos 2 B + sin 2 B - 2 sin B cos B + cos 2 B = 2 sin 2 B + 2 cos 2 B = 2(sin 2 B + cos 2 B) = 2·1 =2
3 sin 2 B + 4 cos 2 B = 3 sin 2 B + 3 cos 2 B + cos 2 B = 3(sin 2 B + cos 2 B) + cos 2 B = 3 . 1 + cos 2 B = 3 + cos 2 B 2B 2 1 cos B = 1 - 1 - sin 1 + sin B 1 + sin B (1 = 1 _ - sin B)(1 + sin B) 1 + sin B = 1 - ( I - sin B ) = 1 - 1 + sin B = sin B 1 l + tan v 1 + cot v 1 - tan v 1 1_ cot v 1 + _I_ cot v cot v 1 1 cot v cot v cot v + 1 cot v - l --
43.
( (
sec4 B - sec 2 B = sec 2 B(sec 2 B - 1) = (tan 2 B + 1) tan 2 B = tan4 B + tan 2 B
45.
_ _
_ _
) )
sec B sin B cos B sin B + = + 1 csc B cos B cos B sin B sin B + sin B = -cos B cos B = tan B + tan B = 2 tan B
--
--
--
--
-
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8.3: Trigonometric Identities
1 1 + sin O 1 + csc O 1 I - sin O 1 - -csc O csc O + 1 csc O csc O - 1 csc O csc O + 1 csc O csc O csc O - 1 csc O + 1 csc O - 1
53.
--
47.
49.
51.
(sec 0 - tan 0) 2 = sec 2 0 - 2 sec O tan 0 + tan 2 0 2 = 12 - 2 . 1 . sin 0 + sin 2 0 cos 0 cos 0 cos 0 cos 0 1 - 2 sin O + sin 2 0 cos 2 0 (1 - sin 0)(1 - sin 0) 1 - sin 2 0 (1 - sin 0)(1 - sin 0) (1 - sin 0)(1 + sin 0) I - sin O 1 + sin O ---
I - sin v + --COS y (I - sin V) 2 + cos 2 V cos y I - sin v cos v(l - sin v) 1 - 2 sin v + sin 2 v + cos 2 v cos v(l - sin v) 1 - 2 sin v + 1 cos v(l - sin v) 2 - 2 sin v cos v(1 - sin v) 2(1 - sin v) cos v(l - sin v) 2 COS y = 2 sec v
55.
__
_-
--
cos 0 + --sin O 1 - tan O 1 - cot O cos O + sin 0 sin 0 11 - cos 0 cos O sin O cos O = + sin O cos O - sin O sin O - cos O sin O cos O sin 2 0 cos 2 0 + ---cos O - sin O sin 0 - cos 0 cos 2 0 - sin 2 0 cos O - sin O (cos 0 - sin O)(cos 0 + sin 0) cos O - sin O = sin O + cos O -----
sin O sin O . sin O sin O - cos O sin O - cos O 1 sin O
57.
1 - cos O sin O 1 - cot O
Ll
tan l7 + cos O = sin O + cos O l + sin O cos O l + sin O sin 0(1 + sin 0) + cos 2 0 cos 0(1 + sin 0) sin O + sin 2 0 + cos 2 0 cos 0(1 + sin 0) sin O + 1 cos 0(1 + sin 0) 1 cos O = sec O --
---
441
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8: Analytic Trigonometry
59.
61.
tan 0 + sec 0 - 1 tan O - sec 0 + 1 tan O + (sec O - 1) tan 0 + (sec 0 - 1) tan 0 - (sec O - 1) tan 0 + (sec 0 - 1) tan 2 0 + 2 tan O(sec 0 - 1) + sec 2 0 - 2 sec 0 + 1 tan 2 O - (sec 2 0 - 2 sec O + 1) sec 2 0 - 1 + 2 tan O(sec 0 - 1) + sec 2 0 - 2 sec O + 1 sec 2 0 - 1 - sec 2 O + 2 sec O - 1 2 sec 2 0 - 2 sec O + 2 tan O(sec 0 - 1) 2 sec O - 2 2 sec O(sec 0 - 1) + 2 tan O(sec 0 - 1) 2 sec O - 2 2(sec 0 - 1)(sec 0 + tan 0) 2(sec O - l) = tan O + sec O
-
65.
--
67.
sin O cos O tan 0 - cot 0 cos O sin O sin O + - cos 0 tan O + cot O -cos O sin O sin 2 O - cos 2 0 cos O sin 0 sin2 0 + cos 2 0 cos Osin 0 sin 2 O - cos 2 0
--
1 - tan 2 0 + 1 = 1 - tan 2 0 + 1 + tan 2 0 l + tan 2 0 l + tan 2 0 2 sec 2 0 = 2 . _12_ sec 0 2 cos 2 0 =
69.
sin u cos u tan u - cot u + cos u - sin u 1= + sin u + cos u 1 tan u + cot u cos u sin u sin 2 u - cos 2 u cas u sin u + 1 sin 2 u + cos 2 cos u sin u .sm 2 u - cos 2 u +1 1 = sin 2 u - cos 2 u + l sin 2 u + (1 - cos 2 u) = sin 2 u + sin 2 u = 2 sin 2 u --
63 .
1 + sin O sec O + tan O -cos O cos O cot O + cos O cos O + cos O sin O 1 + sin O cos O cos 0 + cos O sin 0 sin O sin O 1 + sin O cos 0 cos O(l + sin 0) sin O 1 cos O cos O = tan O sec O
--
-
71.
U
sec O - csc O cos O sin O 1 1 sec O csc 0 cos O sin O sin O - cos O cos O sin 0 1 cos o sin 0 sin 0 - cos 0 cos O sin O cos O sin 0 = sin O - cos O sec O - cos O = 1 - cos O cos O l - cos 2 0 cos O sin 2 0 cos O . = Sln o ' sin O cos O = sin O tan O --
L1
=
--
442
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8.3: Trigonometric Identities
73 .
1 + sin 8 + 1 - sin 8 1 - sin 8 1 + sin 8 (1 - sin 8)(1 + sin 8) 2 1 - sin 2 8
--- + ---
79.
2
75.
77.
(
cos 2 8 = 2 sec 2 8
)(
sec 8 = sec 8 . 1 + sin 8 1 - sin 8 1 - sin 8 1 + sin 8 sec 8(1 + sin 8) 1 - sin 2 8 sec 8(1 + sin 8) cos 2 8 l + sin 8 = -_ . _-cos 8 cos 2 8 1 + sin 8 cos 3 8
)
sin 8 + cos 8 sin 8 - cos 8 cos 8 sin 8 sin 8 cos 8 sin cos 8 = -- + -- - --8 + -cos 8 cos 8 sin 8 sin 8 = sin 8 + 1 - 1 + cos 8 cos 8 sin 8 sin 2 8 + cos 2 8 cos 8sin 8 1 cos 8 sin 8 = sec 8 csc 8 sin 8 + cos 8 (sin 8 + cos 8)(sin 2 8 - sin 8 cos 8 + cos 2 8) sin 8 + cos 8 = sin 2 8 + cos 2 8 - sin 8 cos 8 = 1 - sin 8 cos 8 cos 2 8 - sin 2 8 2 1 - sin 2 8 cos 8 2 cos 8 - sin 2 8 cos 2 8 - sin 2 8 cos 2 8 . cos 2 = ( cos 2 8 - sm 2 8 ) · 2 8 2 cos 8 - sin 8
(sec v - tan v) 2 + 1 csc v(sec v - tan v) sec 2 v - 2 sec v tan v + tan 2 v + 1 csc v(sec v - tan v) sec 2 v - 2 sec v tan v + sec 2 v csc v(sec v - tan v) 2 sec 2 v - 2 sec v tan v csc v(sec v - tan v) 2 sec v(sec v - tan v) csc v(sec v - tan v) 2 sec v csc v 2 . _1_ COS y 1 sin v = 2 . _1_ . sin v COS y 1 = 2 . sin v cos y = 2 tan v
[ 2 cos 2 8 - (sin 2 8 + cos 2 8) r (cos 2 8 - sin 2 8)( cos 2 8 + sin 2 8) (cos 2 8 - sin 2 8) 2 2 (cos 8 - sin 2 8)(cos 2 8 + sin 2 8) cos 2 8 - sin 2 8 cos 2 8 + sin 2 8 = cos 2 8 - sin 2 8 = 1 - sin 2 8 - sin 2 8 = 1 - 2 sin 2 8
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8: Analytic Trigonometry
87.
1 + sin B + cos B 1 + sin B -cosB (1 + sin B) + cos B (1 + sin B) + cos B (1 + sin B) -cos B (1 + sin B) + cosB 1 + 2sin B + sin 2 B+ 2cosB(1 + sin B) + cos2 B 1 + 2sinB + sin 2 B -cos2 B sin B) s n e) 1 + 2 sin B + sin 2 B -(1-sin 2 e) 2 + 2sin B + 2cosB(1 + sin B) 2(1 si2sinB+2si n B) + 2 cosnB(I2B+ sin B) 2 sin B(l sin B) 2(1 + sin B)(1 + cos B) 2sin B(1 sin B) 1 + cose sin B (asinB+bcosB/ +(acose-bsine)2 = a2 sin2 B+ 2absinBcosB+b2 cos2 B + a2 cos2 B -2absin Bcos B + b2 sin2 B = a2(sin2 B+cos2 B)+b2(sin2 B+cos2 B) = a2 +b2 tan a + tanfJ tan a +tanfJ 1 1 cota + cot fJ --+- tana tanfJ tan a + tanfJ tanfJ + tan a tan a tanfJ = (tan a +tan fJ) . ( tantanaa+tantanfJfJ ) = tanatanfJ (sin a + cos fJ)2 + (cos fJ + sin a )(cos fJ -sin a) = sin2 a + 2sin a cos fJ + cos2 fJ + cos2 fJ -sin2 a = 2sinacosfJ+2cos2 fJ = 2 cos fJ(sin a + cos fJ) ln l secB I = ln l _case1_ 1 = ln l cosB r = -ln l cosB I ln I 1+cosel+ln I I-cosel = (1 1 + cos B 1 . 1 1-cos B J ) = ln l 1-cos2 BI = ln l sin2 BI = 2ln l sinB I 1 + 2 sin B + sin 2 B + 2 cos B(1 +
+
99.
X
+ (1 - i 2
X
+
+
89.
91 .
93.
95 .
97.
I(x) = sinx · tanx smx . · - = smx cosx sm. 2 x cosx 1-cos2 cosx cos2 cosx cosx = secx-cosx = g(x) _ cosB 1 (19) = I -sine cosB l+sinB (1 n B)( 1 + sin e) cos e· cos 19 -_ -si cosB(l+sinB) (1+sinB) . cosB 1-sin2 e-cos2 B cos B(l + sin B) 1-(sin2 e+cos2 e) cos B(l + sinB) 1-1 cos B(l + sin e) cos 19(1 + Sine) =0 = g(B) 1 - (---;- -1 ) 1200seCe ( 2SeC2 B-1 ) = 1200cos 19 cos 19 1_ [_2_ _ cos2 B ) = 1200 _ cos B cos2 B cos2 B e--'-) 1200---'-( 1+1-cos2 1 [ 2-COS2 B ) = = 1200 --:--: cos 19 cos 2 19 cos3 19 1200 ( 1+sin2 e) cos3 B Answers will vary.
101.
o
1 03 .
1 05. - 1 07.
In
444
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8. 4: Sum and Difference Formulas
S ection 8.4 1.
�( 5 _ 2 )2 + ( 1 - ( _3))2
b.
7.
J2 1 J2
3
4
2 2
9 - 6.[3 + 3 1 2 - 6.[3 9-3 6 = 2 - .[3
1 -! = ! 2 2
-
False
1 7.
= 19.
. -1t 1t ' cos -1t - sm . -1t · sm = COS 4 4 3 3 1 J2 .J3 J2 2 2 2 2 = J2 - .J6 )
=
-�( J2 + .J6 )
( -l�l
�
00.(_"-) 12
1t
�
co, 1
(� _ �) 12
12
7t
.
7t
.
7t
1
I
.J2 . .J2 . 1 + _ .[3 - .J2 + J6 - _ 2
2
2
2
4 .J2 - J6 4 .J2 + J6 . .J2 - J6 4.J2 - 4J6 4.J2 - 4J6 -4 2- 6
cos I 65° = cos ( 1 20° + 45° ) = cos I 20° · cos 45° - sin I 20° · sin 45° 2
"c
-�(.J6 + J2 )
cos - ' cos - + Sln - ' Sln 3 3 4 4
�(
2 2
[ )
.
)
�(
)
(
21t 1 51t + . 1 71t = Sln . sm 12 12 12 51t . sm 5 1t . COS -1t + COS . -1t = SIn. 4 6 4 6 = _ J2 .J3 + _ J2 . ! 2 2 2 2
. -1t . cos -1t + cos -1t . sm . -1t = sm 4 6 4 6 J2 . -1 .J3 + J2 . _ =2 2 2 2 = .J6 + J2
13.
tan 1 5° = tan( 45° - 30° ) tan 45° - tan 30° 1 + tan 45° · tan 30°
1 - .J3 3 3 - .J3 3 - .J3 "":::'3 = 3 ' 3 3 + .J3 · 3 - .[3 = .J3 1+1·-
= .J3 2 + 4 2 = .J9 + 1 6 = .fi5 = 5
3 . a.
5.
15.
= J6 - .J2
2 21.
sin 20° · cos 1 0° + cos 20° · sin 1 0° = sin(20° + 10° ) = sin 30° 2
23.
cos 70° · cos 20° - sin 70° · sin 20° = cos( 7 0° + 20° ) = cos 90° =0
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8: Analytic Trigonometry
a.
= tan 45° =1 b•
( ) = sin -% ) ( = -1
61t . = sm - 12
c.
( �� ) = cos - ) (1t �1
= cos -
d.
= cos - = 3 2
31.
33.
y
(x, 3)
y
3
x
(2../5, Y)
x
Y
sin fJ = -
�,
Y
[ )
tan a = --4 -1t < a < 1t 3' 2 1t 1 cos fJ = - , 0 < fJ < 2 2 (-3, 4) y
y
( l , y)
x
r 2 = ( _ 3) 2 + 4 2 = 25 r=5 -3 3 . a = -4 , Sln cos a = - = - 5 5 5 e + y2 = 22 , Y > 0 y 2 = 4 - 1 = 3, Y > 0 Y = .j3 .j3 tan fJ = = .j3 1
<0
y 2 = 25 - 20 = 5, y = -$
[ )
y
x2 + 3 2 = 5 2 , x > O x2 = 25 - 9 = 1 6, x > O x=4 4 tan a = -3 cos a = - , 4 5
( 2$ f + y 2 = 5 2 ,
( )
� - ( - �) = % = 2 1 + (� () �) !
. a = -3 , 0 < a < -1t sm 2 5 $ 2 - -1t < fJ < 0 cos fJ = -5 ' 2 y
sine a + fJ) = sin a cos fJ + cos a sin fJ $ $ = � . 2 +i . _ 5 5 5 5 6$ - 4 $ 2 $ 25 25 cos(a + fJ) = cos a cos fJ - sin a sin fJ 2$ _ � . _ $ =i. 5 5 5 5 8$ + 3 $ 1 1 $ 25 25 sine a - fJ) = sin a cos fJ - cos a sin fJ 2$ _ i . _ $ =L 5 5 5 5 2$ 1 0$ = -6$ + 4$ = -5 25 25 tan a - tan fJ tan( a - fJ) = ----'--1 + tan a · tan fJ
<0
tan fJ =
;Js = - � 446
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8. 4: Sum and Difference Formulas
a.
sine a + fJ) = sin a cos fJ + cos a sin fJ = (�}( ) + ( -�} [ � ) �
35.
4 - 313 10
b.
c.
(-�} (�)-(�){�)
-3-413 10 sinea -fJ) = sin a cos fJ -cos a sin fJ
x2 + 5 2 1 3 2 , < 0 x2 = 1 69 - 25 = 144, x < O x = -12 12 5 -12 = -tana cosa = 12 13 13 ' r2 = ( _ 1)2 + 13 2 = 4 =
X
=-
a -tanfJ tan(a _ fJ) = l+tantana tanfJ _ i3 -13 1 + ( - � } 13 -4-313 3 3-413 3 =
y
x
4 + 313 10
d.
=
y
cos(a + fJ) = cos a cos fJ -sin a sin fJ =
SIn. a -153 , --3n2 < a < -n tanfJ = -13, '!!:.2 . < fJ < n (- I, ,[3 ) (x, 5)
r=2
sinfJ = 132 ' cosfJ = --12 = --21 a.
[ ��-4:ff) { �::�)
b.
-48 - 2513 -39 2513 + 48 39
sinea + fJ) = sin a cos fJ + cos a sin fJ -5 - 1 213 or 5 + 1213 26 26 cos(a + fJ) = cos a cos fJ -sin a sin fJ 5 =( _ - ��}( �) C3 ){ �) 1 2 - 513 26
c.
sinea -fJ) = sin a cos fJ -cos a sin fJ -5 + 1213 26
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
x
Chapter 8: Analytic Trigonometry
d.
tan( a - fJ)
tan a --- tan fJ-'= -1 + tan a tan fJ -� - ( - v!3 ) 1 + ( - 5 ) - ( -v!3) 12 -5 + 1 2.[3 = 12 v!3
( 4) = -
tan (} +
d.
tan () + tan !!'-
1i
-1 + 2 J2 1 +1 - -J2 2 .J2 - 2 1 1 _ _ __ . 1 2J2 + 1 2 .J2 2J2
( ) = ( ��::} ( ��=:) _
12 + 5 12 -5 + 1 2v!3 . 1 2 - 5 v!3 1 2 - 5 v!3 1 2 + 5 v!3 + 1 69 v!3 69
)( )
=( - 240 37.
sin () a.
b.
c.
= .!., 3
() in
(
sin () +
8-4.fi + 1 8-1 9 - 4.fi 7
39.
quadrant II
cos (} = -.JI - sin2 (}
= -�I- (�y = -�I- i = -J% 2.fi 3
()
( )=
cos (} -
1i
3
1i
()
x2 y 2 4 ,
a lies in quadrant I . Since + = 14 2 . Now, is on the circle, so + 12 =
= = (x, l) x2 4 x2 = 4- e x = .J4-12 = v!3 r
Thus sin a = 1::. � and cos a = � = v!3 . 2 · 2 fJ lies in quadrant IV . Since + 1, ,
r
-6 ) = sm. . cos -6 + cos . sm-6 = (l) ( � ) + ( - 2� ) (�) 1i
4 4
1 tan () . tan !!.-
=
= Ji = 1 . Now,
()
' 1i
r
2
r
x2 y2 =
(�, y) is on the circle, so
y2 = 1 y2 =1- (l) 2 y = -�l - (�y = -J% = - 2� -2.fi = --2 .fi and Thus sin fJ = 1::. = 3 1 x l = -i . Thus, cos fJ = -= 1 f( a + fJ ) = sin ( a + fJ ) = sin a · cos fJ + cos a · sin fJ "31 +
v!3 - 2 .J2 -2 .fi + v!3 6 6
cos (} . cos 1i + sin (} . sin 1i 3 3
,
= ( - 2� ) (� ) + (i) (� )
r
-2 .fi + v!3 6
_3_
r
1.
3
2 ..[6 -61 - -6
1 - 2 ..[6 6
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8.4: Sum and Difference Formulas
41.
From the solution to Problem 3 9, we have J3 , sin fJ = -2J2 , and sm a = -1 , cos a = 2 2 3 1 cos fJ = - 0 Thus, 3 g (a - fJ) = cos (a - fJ) = cos a 0 cos fJ + sin a 0 sin fJ
--
0
o
45.
47.
sin (rc - B) = sin rc o cos B - cos rc o sin B = 0 0 cos B - ( -1)sin B = sin B
49.
sin (rc + B) = sin rc o cos B + cos rc o sin e = 0 0 cos e + ( -1) sin B = -sin e
6 43.
From the solution to Problem 39, we have -2 J2 and J3 , sin fJ = -sm a = -1 , cos a = 2 2 3 ' cos fJ -1 0 Thus, 3 1 "2 a sin = = 1 = J3 and tan a = cos a J3 J3 3 2 2J2 tan fJ = sin fJ = 3 = -2J2 Finally, cos fJ .!. 3 tan a + tan fJ h (a + fJ) = tan (a + fJ) = I - tan a tan fJ
51.
0
o
( J
sin !:.. + B = sin !:.. 0 cos B + cos !:.. o sin B 2 2 2 = 1 0 cos B + 0 0 sin B = cos B
=
-----
tan ( rc - e) = tan rc - tan e 1 + tan rc 0 tan B O - tan B 1 + 0 0 tan e - tan B = - tan e
53.
0
55.
!f- + (-2J2 ) 1 - � ( -2 J2 )
57.
J3 - 2J2 3 3 16 3 1+ 2 3 J3 - 6J2 0 3 - 216 3 + 2 16 3 _ 2 16 3 J3 - 6 J2 - I SJ2 + 24J3 9 - 616 + 616 - 24 27J3 - 24J2 sJ2 - 9J3 5 -15
( )
sin 3rc + e = sin 3rc 0 cos e + cos 3rc 0 sin e 2 2 2 = -l o cos B + 0 0 sin B = - cos e sin( a + fJ) + sin( a - fJ ) = sin a cos fJ + cos a sin fJ + sin a cos fJ - cos a sin fJ = 2 sin a cos fJ sin(a + fJ) sin a cos fJ + cos a sin fJ sin a cos fJ sin a cos fJ = sin a cos fJ + cos a sin fJ sin a cos fJ sin a cos fJ 1 + cot a tan fJ =
59.
cos(a + fJ) cos a cos fJ - sin a sin fJ cos a cos fJ cos a cos fJ cos a cos fJ sin a sin fJ cos a cos fJ cos a cos fJ = 1 - tan a tan fJ
449 © 2008 Pearson Education, Inc o , Upper Saddle River, NJo All rights reservedo Thi s material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publi sher.
Chapter 8: Analytic Trigonometry
61.
sinea + fJ) sin a cos fJ + cos a sin fJ . fJ sine a - fJ) sin a cos fJ - cos a sm sin a cos fJ + cos a sin fJ cos a cos fJ sin a cos fJ - cos a sin fJ cos a cos fJ sin a cos fJ + ---� cos a sin fJ ----'-cos a cos fJ cos a cos fJ sin a cos fJ cos a sin fJ cos a cos fJ cos a cos fJ tan a + tan fJ tan a - tan fJ
67.
69.
71. 63.
+ fJ) cos ( acot(a + fJ) = --'':"""':'" sin(a + fJ) cos a cos fJ - sin a sin fJ sin a cos fJ + cos a sin fJ cos a cos fJ - sin a sin fJ sin a sin fJ sin a cos fJ + cos a sin fJ sin a sin fJ cos a cos fJ sin a sin fJ sin a sin fJ sin a sin fJ sin a cos fJ + cos a sin fJ ----'sin a sin fJ sin a sin fJ cot a cot fJ - 1 cot fJ + cot a
73.
) ( ) ( )
(
Jr + 2" Jr . . . - I 1 + cos - 1 0 = sm 6 sm sm "2 . 2 Jr = sm 3 2
[
%
( �)]
sin sin- I - cos- I -
( S)
. . . 1 3 I 4 Let a = smS and fJ = cos - - . a Is m
�
sec( a + fJ)
sine B + k1t) = sin B · cos k1t + cos B · sin kn = (sin B)(- l)k + (cos B)(O) = ( _ 1)k sin B, k any integer
Jj
quadrant I; fJ is in quadrant II. Then sin a = -3 , 5 4 Jr Jr O S a S - , md coo fJ = -- , - S fJ S Jr . 2 5 2
-
65.
sin(a - fJ)sin(a + fJ) = (sin a cos fJ - cos a sin fJ)( sin a cos fJ + cos a sin fJ) = sin2 a cos2 fJ - cos2 a sin2 fJ = sin2 a(1 - sin2 fJ) - (l - sin2 a)sin2 fJ = sin2 a - sin2 a sin2 fJ - sin2 fJ + sin2 a sin2 fJ = sin2 a - sin2 fJ
cos a = ,Jl - sin 2 a 2 = 1= 1-
cos(a + fJ) 1 cos a cos fJ - sin a sin fJ 1 sin a sin fJ cos a cos fJ - sin a sin fJ sin a sin fJ
� G) � ; M � � � (-�J � �� & % 5
=
=
sin fJ = 1 - cos 2 fJ = 1-
.[
= 1-
=
=
( )]
3 . . ( -I 4 sm sm S - cos - S = sm a - fJ) = sin a cos fJ - cos a sin fJ
sin a sin fJ cos a cos fJ sin a sin fJ sin a sin fJ sin a sin fJ csc a csc fJ cot a cot fJ - 1
-I
12 12 25 25 24 25 450
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8.4: Sum and Difference Formulas
75.
cos ( tan-I �+cos3 I 2.13 ) Let a = tan- I �3 and f3 = COS- I 2.13 . a IS m quadrant f3 is in quadrant Then tan a = �3 a ' and cos f3 = -135 ' 0 :$ f3 :$ - . seca =.Jl + tan2 a = I + (�J = �1+ 1: = J¥ = % cosa =-35 sin a = .Jl-cos2 a = 1 _ (%)2 = �1- ;5 = Ns = � sin f3 = �1-cos2 f3 = I - C53 )2 = �1- 1�9 = ��:; = �� _I -5 ) _ cos ( tan- -+cos 3 13 = cos( a + f3) = cosa cos f3 -sin a sinf3 = (%}C53 ) - (�}C� ) 33 15 65 65 65 I;
o<
<
ff
2
-
I.
77.
5 _ I -3 ) cos ( sm. _ I --tan 13 Let a = sin-t 2.13 and f3 = tan- I �4 . a quadrant f3 is in quadrant Then sin a = 2.13 :$ a :$ , and tan f3 = 3 , 0 f3 COS a = .J1 -sin 2 a = I - C53 J = �1- 1�9 = ���� = �� sec f3 = �1 + tan2 f3 r.g (25 5 1+ l4)3 )2 = V 1+ 16 = V16 = �� =4 cosf3 =-5 sin f3 = �1-cos2 f3 = 1 _ (�)2 = �1- �� = Hi = % 5 _3 cos [sm. _ I --tan 13 I -] cos(a -f3) = cos a cos f3 + sina sin f3 5 . -3 = -113 . -+_ 5 13 5 = -65 + -6515 63 65 4
IS III
I.
I;
,
o
ff
2
�
1t
2
4
-
-
<
<
1t
2
-
.
�
� �
4
�
1 4
4
=
2 4
48
48
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
,
Chapter 8: Analytic Trigonometry
79.
(
5+6)
. -I 3 tan sm
7r
81.
5 . a is in quadrant
Let a = sin-I �
7r
sin a = -3 , 0 � a � 2
5
cos a = .Jl - sin2 a
� (%f
= 1-
--
I.
Let a = sm - and fJ = cos -I I ; a . , 5 .
Then
�
= 1-
IS ill 7r
4 5
cos a = .Jl - sin2 a 2 = 1_ = 1-
;5 = J¥s = �
� (�
�
)
�� = & = %
4 5 4 4 5 5 . _ 1 54 - cos-I I ) tan ( sm . - I 54 ) + tan ( cos -I 1 ) tan ( sm ) 4 . 1 - tan ( sm I 5 tan ( cos - 1 1 ) -4 + 0 4 4 = _3__ = 2. =
sin a = - = - ' -5 = tan a = -cos a 3 3 3
7r
=
4
7r .
5 4 4 4 5 ( ,(. n-1 ,3 ) )"n 6 ( � n t o tan Sin -I 5 + 6 ) 1 -tan sm. _1 "53 . tan "6 3 -13 -+43 3 -13 1--·4 3 9 + -13 12 - 12-3-13 12 9 + -13 12+3-13 12 -3-13 . 12 + 3-13 108+75-13+36 144 - 27 144+ 75-13 1 17 48+25-13 39 7r
-I
quadrant I. Then sin a = - , 0 � a � - , and 2 cos fJ = 1 , 0 � fJ � So, fJ = cos-I I = O .
.
3 sin a = -5 _- _3 . _ _- -3 tan = 5 cos a a
( . _1 4 + )
tan sm 5 cos-I I
7r
.
_
1 - -4 · 0 3
=
83.
3
cos ( cos - I u + sin-I v ) Let a = cos-I u and fJ = sin-I v . Then cos a = u, 0 � a � , and 7t
7t
.
7t
sm fJ = v, - - � fJ � 2 2 -1 � u � l , -1 � v � 1 sin a = .Jl - cos2 a = .Jl - u2 cos fJ = �I - sin2 fJ = .Jl - v2 cos ( cos-I sin-I v ) = cos(a fJ) = cos a cos fJ - sin a sin fJ = u .Jl - v2 - v.Jl - u2 U
+
+
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8. 4: Sum and Difference Formulas
85.
sin ( tan-I u - sin-I v) Let a = tan-I u and fJ = sin-I v . Then 1t
sin fJ = �1 - cos2 fJ = J1 - v2 J tan fJ = sin fJ = l - v2 v cos fJ
1t
tan a = u , - - < a < - ' and 2
1t
2
tan ( sin -I u - cos v ) = tan( a - fJ) _ tan a - tan fJ l + tan a tan fJ u - J1 - v2 -r7 -vJ l + u_ . 1 - v2 J1 - u2 uv - �.j}:7 v� v� + u � vJ1 - u2 uv - �.j}:7 vJ1 - u2 + u .j}:7
1t
-I
sm' fJ = v, - - :S; fJ :S; - . 2 2 -oo < u < oo , -l :S; v :S; l sec a = Jtan 2 a + 1 = � 1 cos a = � u2 + 1
_ _
V
cos fJ = �1 - sin2 fJ = � sin a = J1 - cos2 a = 1- 1 u2 + 1 u2 + 1 - 1 u2 + 1
�
=
u2 + 1 Jh
89.
U
(� a ) , cos (� - a ) = cos fJ . v 0, then O :s; a � , so that (�- a ) and fJ both lie in the interval [0, � ]. v < 0, then -� :s; a < 0 , so that (�- a ) and fJ both lie in the interval (�, 1t] . Either way, cos ( � - a ) = cos fJ implies � - a = fJ , or =
Ju2 + 1
sin a = cos
sin ( tan-I u - sin-I v ) = sin(a - fJ) = sin a cos fJ - cos a sin fJ = u .�_ 1 .v � � J u 1 - v2 - v Ju2 + 1 87.
If
=
1t
a + fJ = !!:... . Thus, sin-1 v + cos-1 v = !!:... .
1t
sm a = u, - - :s; a :S; - , and 2 2 cos fJ = v, :s; fJ :s;
0
If
-
:s;
�
tan ( sin-I u - cos-I v ) Let a sin-I u and fJ = COS-I v . Then .
Let a = sin-I v and fJ = COS-I v . Then sin a v = cos fJ , and since
2
2
1t .
- l :s; u :S; l , - l :S; v :S; l
cos a = J1 - sin2 a = J1 - u2 sin a = u tan a = -cos a J1 - u2 453 © 2008 Pearson Education , Inc. , Upper Saddle River, NJ. All rights reserved . This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8: Analytic Trigonometry
91.
Let a = tan-I (;) and fJ = tan-I v . Because ; must be defined, v 0 and so a, fJ O . Then 1 ' tan a = -v1 = -tanfJ = cot fJ , and Slllce tan = cot (� - ) , cot (� - a ) = cot fJ . Because v > 0 , 0 a �2 and so (�-a) 2 and fJ both lie in the interval (0, � ) . Then cot (� -a ) = cot fJ implies % - a = fJ or a = --fJ 2 . Thus, tan_ 1 (�1 ) = "2 -tan-I ' f v > 0 . 7:
a
93.
7:
a
<
<
95.
'It
'It
97.
a.
V, l
tan(tan-I l+tan-1 2)+tan(tan-1 3) tan(tan-I l+tan-1 2+tan-I 3) = tan ( (tan l+tan -I 2)+tan -I 3 ) = I-tan ( tan-I l+tan-I 2 ) tan ( tan-I 3 ) -
I
3 +3 -3+3 0 �+3 1-1·2 -1l-.. =-0 =-= 1+9 10 tan ( tan-I l)+tan( tan-I 2) . 3 1-� 13 1- 1-tan ( tan-I 1) tan ( tan-I 2 ) ·3 1-1·2 -1 From the definition of the inverse tangent function we know 0 tan-I I ,.2 , 0 tan-I 2 ,.2 , and tan -13 ,.2 . Thus, 0 tan -I 1 + tan -I 2 + tan -I 3 32,. . On the interval ( 0, 32,. ) , tan () = 0 if and only if () = ,. . Therefore, from part (a), tan-I I + tan-I 2 + tan-I 3 = ,. . Note that () = (}2 - (}I . <
h.
o<
99.
sin(sin-I v+cos-I v) = sin(sin-I v)cos( cos-I v) + cos( sin-I v)sin ( cos-I v) = v . v+ �1-v2 �1-v2 = v2 + I-v2 =1 I(x + h) -I(x) h sin(x + h) -sinx h sin xcos h + cosxsin h -sinx h cosxsinh -sin x + sinxcos h h cos xsin h -sin x(l-cos h) h sin h . 1 -cos h = cOSX'---Sl h h ll X'---
<
<
<
<
<
<
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in writing from the publi sher.
Section 8.5: Double-angIe and Half-angle Formulas
101.
If tana = x+l and tan,B = x-l, then 2 cot(a - fJ) = 2·--:----:tan (a - fJ) 2 1 + tan a tan fJ =_ 2(12(tan1 ++ (tanax+-atanI)tan(x-fJfJ)I)) - x+ I -(x-l) _1)) = 2(1+(X2 x+l-x+1 2X2 2 If formula (7) is used, we obtain tan (i - () ) = tan 2 -tan. () However, this is I + tan - tan () 2 impossible because tan 2:2 is undefined. Using formulas (3a) and (3b), we obtain sin (� - () ) ) tanh:-o) = cos ( --() 2 cos () sin () = cot ()
x2 + 32 = 52, x > 0 x2 = 25 - 9 = 16, x > 0 x=4 So , cos() 45 sin(2()) = 2 sin ()cos () = 2 . �5 . �5 = 2524 cos(2()) = cos2 ()-sin2 ()
I
tan fJ
tan a
1 03 .
2:
11:
= -
a.
b.
c.
3. s.
7.
= l�� = Ji = � = J.o:a = �
.
11:
9.
"
Section 8.5 1.
. ()2 l-COS() 2
SIn - =
()
tan()= -,43 11: < () < -2 . Thus' -2 < -2 < -4 ' which means f!...2 lies in quadrant II. x = -3, y = -4 r2 = (_3)2 +(_4)2 = 9 +16 = 25 r =5 sm. () = --45 ' cos () = --35 Sin(2()) = 2sin()COS() = 2 { - :}( - %) = �; cos(2()) = cos2 () -sin 2 () �
11:
�
a.
sin2 (), 2cos2 () , 2sin2 () sin () False sin() =�,5 O < () < 2:2 . Thus, < f!...2 < 2:4 , which means f!...2 lies in quadrant y =3, r = 5
b.
o
c.
1.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8: Analytic Trigonometry
d.
11.
d.
7t 7t -<-<-,
0 J6 7t cosO= --, Thus, 4 2 2 3 2 which means !!..2 lies in quadrant = -J6, = 3 (-J6 f + l = 32 l =9-6 =3 y = ..[3 - < O < 7t .
I.
x
a.
b.
13.
r
� 'in�2 = ��
<
>
<
7t
<
I.
x
r
a.
3
b.
6 3 -3 --9 9 9 3 c.
.
o<
sin(20) = 2 sin 0 cos 0 2J18 9 9 cos(20) = cos2 0 -sin2 0
sec 0 = 3, sm 0 0 , so 0 0 -2 . Thus, !!.. 2: , which means !!.. lies in quadrant 2 2 4 cos 0 = -31 , = 1 , = 3 . 12 + l = 32 l = 9-1 =8 y = ..J8 = 2J2 sinO = 2J23 2J2 · -1 =-4..[i . = 2·-. = 2smOcosO sm(20) 3 3 9 cos(20) = cos2 0 -sin2 0
c.
�= 1 +2�l
d.
=tt 3+ J6 =� 6
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8.5: Double-angle and Half-angle Formulas
15.
cot B = -2, sec B < O , so 21t- < B <1t . Thus, 1t < -B < 1t hich means -B IIes 4 2 2 2 " quadrant I . = -2, = 1 r2 = (_2)2 + 12 = 4 + 1 = 5 = ..f5 -2 = _ 2..f5 = , cos B sin B = � ..f5 ..f5 . 5 ,,5 5 sin(2 B) = 2 sin Bcos B 2 =_ =_ =2 �� :
- -
,w
17.
,W
m
y
x
x
r
=
a.
tan B = -3, sin B < O , so 31t 2 < B <21t. Thus, 31t < -B < 1t h'ICh means -B I"Ies quadrant II . 2 4 2 = 1, = -3 r2 = 12 + ( _3)2 = 1 + 9 = 10 r =M sin B = ;Fa - 3 , cos B = 10 = a.
{ �)( ;)
cos(2 B) = cos2 B - sin2 B
b.
15 = -3 20 5 = 25 - -25 = 25 5 c.
p
sin�2 = - 2COSB = �
d.
fI:fJ( )
tt'
c.
sin(2B) = 2 sin Bcos B =2 _ 3 3 6 10 5 cos(2 B) = cos2 B -sin2 B _ _3
{ : } (!)
( ! ) ( :)' '
P
sin�2 = - 2COS B =
gtM
-10
2 -M _ 10 2 -M = lO 20 = 2! IO-.Jlo 5 +.Jlo + COS B cos�2 = _ 2 = _ 10 2 lO +.Jlo 10
t �
( )
F
d.
�
© 2008 Pearson
!
90 = -10 -80 = - -4 =100 100 100 5
= 5 +210..f5 1 + - 2-..f5 5 + COS B � cos 2 = 2 = 2 -2..f5 _ 5 2 = 5 -210..f5
p
:
�
2..f5 5 2
�
y
=
_
b.
m
p
�
gI
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing fro m the publisher.
Chapter 8: Analytic Trigonometry
19.
21.
sin 22 . 5 ° = sin ( 4�O) = l -C�S450 = �l -2� = �2 -4J2 = M2 2 1t 1t 7 7 cos 1 1t 4 tans7 = tan ( 2 ) = - I HO'.77 l 1 - J2 2 1+ J22 '2 '[2 - J2).[22 -J2J2) 2+J2 = _�(2 - �) = _ [ 2 :) = -(J2- 1) =1-J2 cos165° = cos (-33200- ) 1 +cos330° 2 =-�l +-2.J3-2 - �2+4.J3 = J2:J3 2 .-1...
25.
�
\
151t )
cos ( � 151t4 l+cos2 �l+; �2+4J2 2 �2+J2 2 ) . [J2;J2] = [�2+J2 �2+J2 2-J2 = ( 2 J2;J2 2+ J2 ] .[2 -J2 ) 2(2 -J2)J2;J2 2 = (2-J2)�2+J2
_2_
'
23.
sec-151tg = cos-151t g
27.
__
'm( - i) = ,"t� ) 1r- -C- O- s (;- - �- -;- ) 2 J22 _ -J2 f27z =-R -2 - J¥ 4 = 2 _
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458 Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they cu;"ently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 8.5: Double-angle and Half-angle Formulas
29.
8 lies in quadrant Since x2 + = , = 15. Now, the point 2) is on the circle, so +22 = 5 = -2 .J a =- 22 =-.Ji8 li=-1 is negative5because es in quadrant 215 and Thus, sm8 = - = 2 = -cos8 =-=-15-1 =--15 . Thus, /(28) = sin(28) = 2sin8cos8 =2 { 2� ){_ � ) = _ �� =_ � Note: Since 8 lies in quadrant !!..2 must lie in quadrant Therefore, cos!!.2 . is positive. From the solution to Problem 29, we have cos 8 = _ 15 . Thus, (%) = cos % = l + �os8 a2
(a,
5
a2
y2
II.
5
b
a r
r
r; ,, 5
5
-.Ji
5,
r
II.)
a
_
=-
35.
II,
l
III.
b
r
1.
2
5
g
�
5
a
l
II.
b
5
�
(a,
5
a2
(a
II.)
.
8 lies in quadrant Since x2 + = =15. Now, the point 2) is on the circle, so + 22 = = _ 22 .J = is negative5-22 because2= 8 li=-1 es in quadrant Thus, tan8 =-=-= -2. -1 h(28) = tan(28) 2tan8 I-tan28 2(-2) -_ -4 -4 =-4 - 1-(-2l 1-4 -3 3 a lies in quadrant Since x2 + = 1 , =.Ji = 1. Now, the point (-�, ) is on the circle, so (-"41 ) + 2 =1 b2 =1_ ( _ �)2 � ( �J =-JH =- � is negative because1 a lies in quadrant Thus cos a = -= -= 1 --41 and F15 F15 Thus, sma =- =--1 =---. 4 ( 2a ) = cos ( 2a ) cos 2 a - sin 2 a a2
2
(a
31.
33.
r
pp
b
b=- I- -
(b
rt � -15
III.)
a
,
= 10 = �1O(510-15) 5
.
g
-4"
r
b
-4
r
=
15 = --1416 =-=--16 16
7 8
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459 All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8: Analytic Trigonometry
37.
Note: Since a lies in quadrant III, a2 must lie in quadrant II. Therefore, sin �2 is positiv e. From
41.
sin4 0 = ( sin2 O r
= -CO� (20)
C
= ±[ 1 - 2cos(20) + cos2 (20)] =.!.4 - .!.2 cos (20) + .!.4 cos2 (20)
the solution to Problem 3S, we have cos a =-.!.4 . Thus,
f
(� ) =sin � = l c;s -
J
=.!.4 .!.2 cos( 20) + .!.4 ( 1 + cos2 (40) )
a
_
1 = 41 --cos(20) + -81 + -81 cos( 40) 2 =g3 -21 cos(20) + g1 cos( 40) -
43. 39.
From the solution to Problem 3S, we have . = -J15 and cos a = --1 . Thus sma ' 4 4 a = tan a = 1 - cos a h 2 sin a 2
()
--
cos(30) = cos(20 + 0) = cos(20)cos 0 - sin(20)sin 0 = ( 2cos2 0 - 1 ) cos 0 - 2sin Ocos Osin 0 = 2cos3 0 - cos O - 2sin2 Ocos O = 2cos3 0 - cos 0 - 2 ( I - cos2 O ) cos 0 = 2 cos3 O - cos 0 2cos 0 + 2cos3 0 = 4cos3 0-3cos O We use the result of problem 42 to help solv e this problem: -
45.
S 4
sin(50) sine40+ 0) sin(40)cosO+ cos(40)sin 0 =cosO( 4sin 0-8sin3 0)cosO+ cos(2(20) )sin cos2 O(4sinO -8sin3 0)+ (1-2sin2 (20))sinO (1-sin2 0)(4sin 0-8sin3 0) + sinO (1-2(2sinOcos0f) 4sin0-12sin3 0+8sin5 0 + sin O ( 8sin2 Bcos2 B) =4sin0-12sin3 0 +8sin5 0 + sin 0 -8sin3 0(1-sin2 0) 5sin0-12sin3 0 +8sin5 0-8sin3 0+8sin5 0 16sin5 0-20sin3 0+ 5sin 0 =
=
J15
0
4 S -
=
J15 s .J15 =-J15 J15 sJ15
=
=
1-
IS
J15 3
=
=
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460 Pearson Education, Inc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exi st. No portion of thi s material may be reproduced, in any fonn or by any mean s, without pennission in writin g from the publi sher.
Section 8.5: Double-angle and Half-angle Formulas
47.
cos4 B -sin4 B ( cos 2 B+ sin 2 B )( cos 2 B -sin 2 B) 1 · cos ( 2B ) cos (2B )
57.
=
,ee'
(%)
�
eot '
(�)
�
=
=
49.
cot( 2B) tan(12B)
59.
2 tan B I - tan 2 B 1 I - tan 2 B 1 coe_B 2 2 tan B cot B coe B - 1 cot 2 B cot 2 B - 1 cot B 2 coe B 2 cot B cot 2 B - l 2 cot B sec( 2B) cos(1 2B) 2 cos 2 B - l 1 1 2 2 - sec 2 B _ sec2 B 1 sec 2 B sec 2 B 2 - sec 2 B cos 2 (2 u) - sin 2 (2 u) cos [ 2( 2u) ] cos( 4u) = --
_ _
51.
= --
_
53.
55.
_
=
61.
=
cos( 2B) cos 2 B - sin 2 B 1 + sin( 2B) 1+ 2sin Bcos B ( cos B - sin B)(cos 0 + sin 0) cos 2 B+ sin 2 B + 2 sin Bcos O ( cos B -sin B)( cos B+ sin 0) ( cos B + sin B)( cos 0+ sin 0) cos B - sin B sin B cos O + sin B sin B cos B - sin O cos B+ sin O cos B sin O sin B sin B cos B sin 0 -sin B + -sin B cot B - l cot B + l
© 2008 Pearson
eo,'
(%)
2 1+ cos O 1 +cosO 2
(�)
l - cosv 1+ cosv 1+ cosv l - cos v 1 1 +-secv 1 1 -secv secv + 1 secv secv -l secv sec v+ 1 secv secv secv -l secv + 1 sec v - I l - cos B I - tan 2 -2B l+ cos B B 0 l + tan 2 -2 1 + l1 -+ cos cos O 1+ cos 0 - ( 1 - cos 0) 1+ cos O 1+ cos B+ 1 - cos 0 1 + cos O 2cos O 1+ cos 0 2 1+ cos 0 2 cos O l+ cos B l+ cos O 2 cos B sin( 30) - --cos( 3B) sin ( 30 ) cos 0 - cos ( 30 ) sin 0 -sin Ocos O sin B cos B sin( 30 - B) sin Bcos B sin 20 sin OcosO 2sin Ocos 0 sin Ocos O 2 tan '
=
63.
=
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exist. No portion of thi s material may be reproduced, in any form or by any mean s , without permi ssion in wri ting from the publisher.
Chapter 8: Analytic Trigonometry
65.
tan(30)= tan( 2 0+0) aanOn2 0- + tan u 2tI-t--=t a a n(20)+t nO = I -tan(20)tanO _- 2tanO . tanO 2 tan 0I-tan2 + tanO -0 tan30 I-tan2 0 I - tan2I 0ta-n22tan2 0 0 I-tan2 0 3 taI-tnOa-tn2-an30 0 1 -3 tan2 0 t a n30 tanO 3 I-3t-an2 0
73.
LI
1
II.
7r
=
�'(lnl l -COS(20)I-ln 2) =�'ln l l-C�S20 I m (l l -CO�(2B) I"') I n(1 sin2 (1121 ) = lnl sinO I . TJ3 . SI.ll( 2SI'll-1"21 )=SI.ll( 2 '6)=sm3= cos ( 2 sin -I% )= 1 - 2 sin 2 ( sin-I %) =1-2(%) 2=1- �� 25
9
-7
�
=
69
=-
-
67.
7r
tan [ 2coS-1(-�)] Let a=COS-I ( -�) . a l ies in quadrant Then cosa=--,35 -::;a::;1t . 2 seca= - -35 tan a - .Jsec2 a -1 = �(-%J - 1= _�2: -1= -f! � tan [ 2cos -I(--35 )]= tan2a = I-t2taanan2 a _1-2...o.((-_.. ��-"')J_ _- 1- �1: . 9 -24 -24 -9-1624 =- --= =-Sin ( 2cos-1 �) Let a= cos -1-45 . a ..IS quadrant . Then cos a=-45 , ::;a::;-.1t2 sin a=.J l -cos2a =�I - (�J =� 1- ��=&=% SI.ll( 2 cos_1"54)=sm. 2a 45 -2425 = 2SI.lla cos a= 2· 35 -= Sin2 (.!.c2 os-1-35)= I-COS (2cos-I�5 ) 2 1- �5. _-"5-= 2 2 -5
75.
7
0
71.
7
_.
77.
__
© 2008 Pearson
I
III
7r
1
462 Educati on , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright l aws as they currently
exi st. No portion of this material may be reprod uced, in any form or by any mean s, without permi ssion in wri ting from the publisher.
Section 8.5: Double-angle and Half-angle Formulas
79.
( %) Let a = tan (%) sec 2 tan-I
-I
83. a
.
a.
is in quadrant I.
.
1t
v
�m+ 1 = �:6+ 1 = � = % )
�[
C
)]
v
cos a = -45 sec 2 tan-I �4 = sec ( 2a ) = cos 2a 1 1 2 2cos a -I
(
v
v
3 Then tan a = -, 4 0 < a < -2 . seca = Jtan 2 a+ 1 =
R(O) = ;16-J2 cos O(sin O - cos O) ;-J2 (cos Osm O - cos 2 0) = -16 ;-J2 1 . 2cos 2 0) = --·-(2cos OsmO16 2 = V;3 sin 20 - 2 + c�S20 = ;32-J2 [ sin ( 20 ) - I - cos ( 20 )] = ;32-J2 [ sin ( 20 ) - cos ( 20 ) -I ] 32 2 -J2 [ sm(2x) . - cos(2x) ] Let r; = -32 v
� )
b.
= -32 -1 = 7 = -7 25 25 81.
a.
�w
D = _-=csc O 2- cot O = 2D ( csc O - cot O ) cos 0 1 - cos 0 1 csc 0 cot 0 = _ sin 0 sin () = sin 0 = tan-()2 Therefore, = 2 D tan -()2 .
c.
-
W
_
_
20
_
450
W
b.
Here we have D = 15 and = 6.5 . 6.5 = 2 ( 15 ) tan "2o 13 tan-o2 = 60 o 13 -2 = tan 60 W
85.
_\
�
© 2008 Pearson Education , Inc . ,
1\
"GXir."tUM
\
X = 6 7.5 _ V = 1 B.7�51 6 6 >
o
90°
R has the largest value when 0 = 67 . 5° . Let b represent the base of the triangle . b/2 . () = cos -o2 = -hs sm2 s () ' 0 b = 2ssmh = s cos2 2 1 = -2 b · h = - 2s sin s cos . () () = s 2 sm-cos2 2 = .!..2 s 2 sin O A
0 = 2 tan .!2. 60 24.45° -\
U sing the MAXIMUM feature on the calculator:
� ( �)( �)
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exist. No portion of thi s material may be reproduced, in any form or by any means , without permi ssion in wri ting from the publisher.
Chapter 8: Analytic Trigonometry
87.
89.
91.
2sicosOnO ._cos201 . =-_ . = 2smOcosO sm(20) sinO 2 . cosO 1cos2 0 2tsec2anO0 2 tan 0 44 1+tan20 4(2 taannO)20) 4+(2t 4x 4+X2 L2 sin2 x+C =- Lcos(2x) 4 C = - L4 cos(2x) - L2 sin2 x = - ± .(cos(2x) + 2 sin 2 x) =- ± .(1 - 2sin2 x+2sin2 x) = _ L4 (1) = -.!.4 If = tan ( � ) , then
© 2008
93.
l-cos(2x)-'f(x) = sin 2 x = ---'-2 of = cos x , compress Sthoriartziontng awill ythbythea graph factandor ofshri2,nreflect across thae x axis, shi ft 1 uni t up, k vert i c ally by factor of2 . y
y
x
95.
-1� = 1 - cos-12 sin ;4 =sin (1t] 2 � 2
2
�-.!.(J6 8 8-2(v'6 +./2) �8-2(v'6 +./2) 16 4
+ .fi)
=
cos�24 = cos (�2 J = 1+cos� 2 12 ll.
z
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exi st. No portion of this material may be reproduced , in any form or by any means , without permi s s i on in wri ting from the publi sher.
Section 8.5: Double-angle and Half-angle Formulas
97.
sin3 () + sin3 «() + 120°) + sin3 «() + 240°) = sin3 () + (sin () cos (120°) + cos () sin (120°) y + (sin ()cos (240°) + cos ()sin( 240°) y ./3 ' cOS () 3 + -21 ' sm. () - ./3 ' cOS () 3 = sm. 3 () + -21 ' sm. () +T T = sin3 ()+ i ' ( -sin3 () + 3./3sin 2 ()cos ()-9sin ()cos 2 ()+3./3 cos3 () )
)
J (
(
- i ( sin3 () + 3./3 sin 2 ()cos()+9sin ()cos 2 () + 3./3 cos3 ()) 3./3 cos 3 () 3./3 · sm. 2 () cos () - -9 · sm. () cos 2 () +-_· = sm. 3 () - -·1 sm. 3 () + -8
8
8
8
3./3 cos3 () 3./3 sm. 2 ()cos ()- -9 · sm. ()cos 2 () --_· --81 · sm. 3 () ---· 8 8 8 = % . sin3 ()- � ,sin ()cos 2 () = % {sin3 ()-3sin ()(I-sin 2 ())] = % ,(sin3 ()-3sin () + 3 sin3 ()) =
99.
%.(4sin 3 ()-3sin ()) = - % ,sin (3())
(from Example 2)
Answers will vary.
© 2008
465 Pearson Education, Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently
exist. No portion of thi s materi al may be reproduced, in any form or by any means, without permi ssion in wri ting from the publisher.
Chapter 8: Analytic Trigonometry
Section 8.6 1.
3.
sin( 4B) sin(2B) = .!.2 [cos( 4B - 2B) - cos( 4B + 2B)] = "21 [ cos(2B) - cos( 6B)J sin( 4B) cos(2B) = .!.2 [sin( 4B + 2B) + sin( 4B - 2B)]
�
= [Sin( 6B) + sin(2B)J
5.
7.
9.
1 cos(3B)cos(5B) = -[cos(3B -5B) + cos(3B + 5B)] 2 = "21 [ cos( -2B) + cos(88)J = "21 [ cos(28) + cos(88)J sin 8 sin(28) = .!.2 [cos( 8 - 28) - cos( B + 28)] = "21 [ cos( -8) - cos(38)J = "21 [ cos 8 - cos(38)J
13.
1 5.
21.
[ e ) e )]
3e cos 8 = 1 sm 8 + 8 + sm. 8 - 8 sm. 2 "2 "2 2 "2 2 "2 = [sin(28) + sin 8J .
�
11.
1 9.
(
) (
sin( 48) - sin(28) = 2 sin 48 ; 28 cos 48 ; 28 = 2sin 8cos(38)
(
)
23.
) e )
cos(28) + cos( 48) = 2 cos 28 +2 48 cos 8-48 2 = 2cos(38)cos(-8) = 2cos(38)cos 8
( } (¥)
sin 8 + sin(38) 2sin ¥ os 2sin(28) 2sin(28) 2 sin(28) cos( -8) 2sin(28) = cos( -8) = cos 8 48 ; 28 cos 48 ; 28 sin(48) + sin(28) 2 sin cos(48) + cos(28) cos( 48) + cos(28) 2 sin(38) cos 8 2 cos(38) cos 8 sin(38) cos(38) = tan(38)
(
) (
( ) ( ) ( }( )
8 + 38 sm-. 8-38 . -sm 2 cos 8 cos(38) 2 2 sin 8 + sin(38) 2sin B �38 os 8 �38 sin(28) sin( -8) = -22 sin(28) cos( -8) -( -sin 8) cos 8 = tan 8
( ) ( )
8 -38 8 +38 cos sm sm. 8 + sm. (38) = 2 · 2 2 = 2sin(28)cos(-8) = 2sin(28)cos 8
© 2008
466 Pearson Education, Inc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyri ght laws as they currently
exist. No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in wri ting from the publisher.
)
Section 8.6: Product-to-Sum and Sum-to-Product Formulas
25.
sin B [sin B+ sin( 3B) ] B+B -3B 3B cos = smB 2sm 2 2 = sin B [ 2 sin( 2B) cos( -B) ] = cos B [ 2 sin( 2B) sin B ] = cos B 2 cos B - COS( 3B) ] = cos B [ cos B - cos( 3B) ]
. [ ( ) ( )] .
[ .�[
27.
29.
31.
]
00'
sin( 4B)+ sin( 8B) cos( 4B)+ cos( 8B) 2 sm. 4B +2 8B cos 4B -2 8B 2 cos 4B+2 8B cos 4B -2 8B 2 sin( 6B) cos( -2B) 2 cos( 6B) cos( -2B) sin( 6B) cos( 6B) = tan( 6B) sin( 4B)+ sin( 8B) sin( 4B) - sin( 8B) 2 sm. 4B+2 8B cos 4B -2 8B -2 sm. 4B -2 8B cos 4B+2 8B 2 sin( 6B) cos( - 2B) 2 sin( -2B) cos( 6B) sin( 6B) cos( 2B) -sin( 2B) cos( 6B) = -tan( 6B) cot( 2B) tan( 6B) tan( 2B)
( (
) ( ) (
) )
( (
) ( ) (
) )
© 2008 Pearson Education, Inc . , Upper Saddle River,
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( l ( ){ ) (a P) (a P) sin a+ sinp 2 sin ; cos ; cosa + cosp 2COS ( a;p ) cos ( a ;p ) sm. ( a+2 P ) cos ( a ;p ) a+p ) = tan ( 2-
a -p . a+p sin a+ sinp 2 sm -2- cos -2a +p a -p sina - sinp 2 sm. cos 22 . a+p a -p "" a �-li a �-Ii sm cos 2 2 = tan a ; P co a ; P
33.
--
467 NJ. All rights reserved . This material is protected under all copyright laws as they currently
exist. No portion of this material may be reprod uced, in any form or by any means, without permission in writing from the publisher.
Chapter 8: Analytic Trigonometry
35.
1 + cos(28) + cos(48) + cos( 68) = cos 0 + cos( 68) + cos(28) + cos( 48) 2 = 2 COS 0+268 cOS 0-268 + 2COS 8: 48 cOS = 2 cos(38) cos(-38) 2 cos(38) cos(-8) = 2 cos 2 (38) + 2 cos(38) cos 8 = 2 cos(38) [cos(38) + cos 8] = 2 cos (38) 2COS 38; 8 COS 38; 8 = 2 cos(38) [2 cos(28) cos 8] = 4 cos 8 cos(28) cos(38) = sin [2.7r( 852)t] + sin [2.7r( 1209)t] = 2 sm. 27r(852)t +2 27r(1209)t cos 27r(852)t -227r(1209)t = 2sin(2061m) cos(-357m) = 2sin(2061m) cos(357m)
( ) ( ) ( +
) ( 28;48)
[ ( ) ( )]
37.
a.
y
b.
Because I sin 81 :-:; 1 and I cos 81 :-:; 1 for all 8 , it follows that I Sin(2061m)\ :-:; 1 and \ cos(357 m)\ :-:; 1 for all values oft. Thus, = 2sin(2061m)cos(357.7rt):-:; 2 · 1 · 1 2 . That is, the maximum value ofy is 2 . Let J-; 2sin(2061.7rx) cos(357.7rx) .
c.
) (
(
)
y
=
=
2
-2
39.
Iu = Ixcos28 + Iysin28 - 2Ixy si n 8cos 8 = Ix COS 8 + 1 + Iy - c�s 28 _ Ixy2 sin 8cos 8 Iy cos 28 I xy sin 28 2
( � ) C
)
Iv ix sin2 8+Iy cos2 8+ 2Ixy sin 8cos 8 COS 28 + I + I 2· I + sm 8 cos 8 x I - COS28 2 2 Iy + Ixy sm. 28 Ix _ Ix cos 28 + Iy cos 28 + 2 2 2 2 Ix+Iy - Ix-Iy cos 28 + I sin 28 2 2 =
1 =
= ---
)
) (
(
Y
---
© 2008 Pearson Education ,
xy
xy
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exist. No portion of thi s materi al may be reproduced, in any form or by any means, without permi ssion in writing from the publi sher.
Section 8.6: Product-to-Sum and Sum-to-Product Formulas
41.
sin(2a) + sin(2jJ) + sin(2r) 2sin ( 2a � 2jJ ) cos ( 2a; 2jJ ) + sin(2r) 2siinn ((a + jJ)cos(a -jJ) + 2sin rcos r ) o a + 2sin rcos r 2 sin r cos(a -jJ) + 2 sin r cos r 2sin r[cos(a - jJ) + cos r] 2 sin r [2 cos ( a - � + r ) cos ( a - � -r )] 4sin rcos (" -22jJ ) cos ( 2a2-") 4sin rcos (% jJ ) cos (a �) 4sin rsin jJsin a 4sin a sin jJsin r the sum fonnulas for sin (a + and sin(a an d solve for sin a cosjJ =
=
=2
n- r
s
=
c s( - jJ)
= =
=
=
=
-
-
=
jJ)
43. Add
- jJ)
sin(a + jJ) = sin a cosjJ + cosasinjJ sin(a - jJ) = sin a cosjJ - cosasinjJ sin(a + jJ) + sin(a - jJ) = 2sin a cosjJ
2
:
sin a cos jJ = .!. [ sin(a + jJ) + sin(a jJ)]
45.
-
2 cos ( : ) cos ( ; ) 2 . � [ cos ( : � ) cos ( : + � p)] cos ( 2: ) + cos ( 2; ) cos jJ + cos Thus, cos a + cos jJ 2 cos ( a �jJ ) cos ( a;jJ ) . 2sin f}+3 2 2sin f} =-1 3 -5 6 + SI.n f} 2 6 3 f} 6 + 2kn or f} 6 + 2 IS any I n teger 4 2 The soluti on set i s {%}. On f} 2n, the so utlOn set IS {6' (5} . " 5" 4cos2 6' 6 cos2 f} .!.4 False cosf} ±-2 f} 3 + or f} 23 + is any integer On�,the2 in terval5; } f} 2 , the soluti on set i s { ;�, . a
jJ
jJ
a
=
a
p
_
a
p
+
a
p
=
a
=
a
=
Section 8.7 1.
x
=-x
4x =
7.
=
1
1
x =- =-
=
=--
7n
=
-
0�
3. 5.
·
kn, k· .
.
7n lIn
f) = 1
9.
=
= !!..
1
kn
,
2008 Pearson Educati on , Inc . ,
I
<
=
I&:i
lIn
-
n
n
=
0
�
kn, k
<
n
469 Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in wri ting from the publi sher.
Chapter 8: Analytic Trigonometry
11.
2sin2 0 - 1 = a 2sin2 0 = 1 sin2 0 = ..!.2
19.
=-
� �
sino= ± = ± 3 " + k" , k is any integer O = -"4 +k " or O = 4 On the interval 0 ::; 0 21t , the solution set is 31t 51t 71t . 4' 4 ' 4 ' 4
{2:
1 3.
}
{) U
<
21.
1 5.
23.
cos ( 20 ) = --21 4" + 2k " 2 " + 2k " or 20 = 20 = 3 3 2 " . 0 = 3 + k" or u = -3" + k k' any mteger On the interval 0 ::; 0 2n , the solution set is 21t 4n 51t . 3' 3 ' 3 ' 3
}
{�
1 7.
",
<
25.
IS
<
<
__
27.
9
{
9
'
' 9
9
}
.
=
=
30 = -2 sec2 30 2 " + 2k'"'" or 30 = 4 " + 2k " 2 3 2 3 4 4k 0 = " + 3" or k is any integer On the interval 0::; 0 2n , the solution set is 41t 81t 16n <
{
}
{
}.
( )
{ }.
COS 20 - % = -1 20 --n2 = n + 2k7t 3n + 2k 7t 20 = 2 31t . u = -+ 4 k n, k 'IS any mteger On 0 ::; 0 2n , the solution set is 3: , 7: . {)
<
© 2008 Pearson
{7; �1t } .
On 0 ::; 0 2n , the solution set is 341t , 741t . 4sec O + 6 = -2 4sec O = -8 sec O = -2 2n + 2kn or 0 = 41t + 2kn, k is any integer o= 3 3 On 0 ::; 0 2n , the solution set is 231t , �1t 3 12 cos 0 + 2 = -1 312 cos 0 = -3 1 -12 cos O = --= 2 12 3n 5n o 4 + 2kn or 0 4 + 2k1t, k is any integer On 0 ::; 0 2n , the solution set is 341t , 5: <
<
{)
tan O + 1 = a tan O = -l 31t + kn , k IS' any mteger . =4 {) u
/l
}
{) U
<
sin ( 30) = -1 31t + 2k 7t 30 = 2 2k1t , k 'Is any mteger 1t . u = -2 + -3 On the interval 0 ::; 0 21t , the solution set is 7; , 1 �1t .
{%,
2sin O + 1 = O 2sin O = -1 sin O ..!.2 71t + 2k 1t or = 1 I1t + 2k 1t , k'IS any mteger . =6 6 On 0 ::; 0 21t , the solution set is , 1
{
}
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in writing from the publi sher.
Section 8.7: Trigonometric Equations (I)
29.
( )
tan % + � = 1 -2 + -1t3 = -1t4 + k1t 1t + -2 = -12 k1t = --1t6 + 2k7t , k is any integer 1 I1t · On 0 0 21t , th e so I utlOn set . 6
o
o o
�
3 1.
33.
35.
37.
<
IS
39.
-
{I
{ }.
4 1.
sm. 0 = -21
{olo = � + 2k1r or 0 = 5; + 2k7r} , k is any integer. Six solutions are 131t 171t 251t 291t o -_ 6"1t ' 651t ' 6 ' 6 ' 6 ' 6-
43.
tan O = J33 0 0 = 5; + k1t , k is any integer Six solutions are 291t- 351t 51t 1 I1t 171t 231t 0 -- 6' ' 6'6' 6 6 ' -6- '
{1
--
45.
}
4 7.
}
{olo
}
2
IS
},
1
mteger
49.
0 = cos-1 -± <::; 1 .82 () <::; 1.82 or () <::; 21r - 1.82 <::;4046 . The solution set is {1.82, 4A6} . 5 tan () + 9 = 0 5 tan O = -9 tan () = _ 2.5
(-�)
IS
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ.
4
-
( )
cos (20) = --21 21t 2k 1t or 2 0 = 41t + 2k1t, k ' any mteger . 2O = 3 3 = � + k1t or 0 = 231t + k1t , k is any integer 21t 41t 51t 71t 81t . are 0 = -1t SIX. solutIOns 3 ' 3 ' -3 ' -3 ' -3 ' -3 . +
-
<::;
cos O = 0 0 0 = % + 2k 7t or 0= 3; + 2k1t , k is any integer 91t 31t 51t 71t 1 11t . are 0 = -1t SIX. solutIOns 2' 2 ' 2 ' 2 ' 2 ' 2 .
{1
J3 . 02 = -sln2 () = 41t + 2 k 1t or -() = 51t + 2k 1t, k ' any . 3 2 3 0 0 = 8; + 4k1t or 0 = �1t + 4k7t k is any integer. Six solutions are 81t 101t 201t 221t 321t 341t () -_ 3' -3- ' 3- ' -3- ' -3- ' 3 sin () = 004 0 = sin-1 (004) <::; 0041 0<::;0041 or 0<::;1r- OAI <::;2.73 . The solution set is {OA1, 2.73} . tan () = 5 0 = tan-1 (5)<::;1 .37 ()<::; 1 .37 or ()<::;1r + 1 .37<::;4 .51 . The solution set is {1.37, 4 . 51} . cos () = -0.9 () = cos-1 (-0.9) <::; 2.69 ()<::;2.69 or () <::; 21r - 2.69 3.59 . The solution set is {2.69, 3 . 59} . sec () = cos () = --41
() = tan-1 <::; -1 . 064 ()<::;-1.064 + 1r or () <::; -l.064 21r <::; 5 . 22 <::; 2.08 The solution set is {2.08, 5.22} . +
471 All rights reserved . Thi s material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means , without permi ssion in writing from the publi sher.
Chapter 8: Analytic Trigonometry
51.
3sin B - 2 = 0 3sin B = 2 sm. B = -23
c.
(%)
B
= sin-1 � 0.73 B � 0.73 or B � 1Z" - 0.73 � 2.41 . The solution set is {0.73, 2.41} . 53.
X
d.
55. a.
b.
-
+
any integer
+
}
>
}
n < x < 5n or 13n < x < 17n . or"6 6 6 6
IS
57.
I (x) = tan x I(x) = -4 4 tanx = -4 tan x = -1 x x = - : + k1Z" ' k is any integer I(x) < -4 4 tan x < -4 tan x < -1 Graphing Yl = tan x and Y2 = -I on the interval -; , ; , we see that Y1 < Y2 for - ; < x < - : or - ; , - : . 4
a.
I ( x) = 0 3sinx = 0 sin x = 0 x = 0 2k1Z" or x = 1Z" 2k1Z", k is any integer On the interval [-21Z", 4n], the x-intercepts of the graph of/are -2n, -n, 0, n, 2n, 3n, 4n . I ( x ) = 3 sin x +
-
{l
� �
sinx = ± = ± 21Z" k 1Z" , k . any lllteger . x = 1Z"3 k1Z" or x = 3 On the interval [0, 2n ] , the x-intercepts of the 5n . 4n 2n graph of/are -n3 ' ' ' 3 3 3 +
=
{
I(x) = 0 2 4sin x - 3 = 0 4sin 2 x = 3 sm. 2 x = -43 -+
I(x) = % 3 sm. x = -32 sm. x = -I2 1Z" 2k1Z" or x = 51Z" 2k1Z", k is 6 6 On the interval [-21Z",4n], the solution set is lIn - 7n n 5n 13n 17n . -"6 ' 6'"6'6'6'6 From the graph in part (b) and the results of part (c), the solutions of I ( x ) 2.2 on the interval [-21Z",4n] is x _ l �n < x <_ 7;
+
b.
}
{i
( ) 6
x
(
�)
1t
2
)
� 2
-6
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ.
472 All rights reserved. Thi s material is protected under all copyright l aws as they currently
exist. No portion of thi s material may be reproduced, in any form or by any mean s , without permi s s i on in wri ting from the publi sher.
Section 8.7: Trigonometric Equations (/)
59. a, d.
b.
f (x) = 3sin(2x)+2 ; g (x) = 22 y
6
x
-2 b.
c.
c.
f (x) = g (x) 3 sin ( 2x ) +2 = 22 3sin(2x) = �2 sin(2x) = �2 5lr 1r 2x = -+2klr or 2x=-+2klr 6 6 5lr x =.!!. 12 ' 12 .-+klr or x=-+klr k is any integer On [0, 7t ] , the solution set is 12 57t 12 . From the graph in part (a) and the results of part (b), the solution of f(x) g(x) on 5 7t . 7t [0, 7t] 1S 127t < x < 57t or -, 12 12 12 X
63.
-
6
g(x)
=
2
} ( ) cos x
{
{l } ( ) h(t) = 125 Sin ( 0.157t- ; )+125 Solve h(t) = 125 sin ( 0.157t- % )+125 = 125 on the interval [0,40]. 125sin ( 0 .157t - %)+125 = 125 125sin ( 0 .157t- % ) = 0 sin ( 0 .157! %) = 0 0. 157t - 1r2 = klr, k is any integer 0. 157t = klr + lr2 , k is any integer klr+!!.. t = 0.1572_, k is any integer O+!!.. For k = 0, t = 0. 1572 _ 10 seconds . lr+-1r2 For k = 1, t = 0.157_ 30 seconds . 2lr +!!.. For k = 2, t = 0.157_2 50 seconds . So during the first 40 seconds, an individual on the Ferris Wheel is exactly 125 feet above the ground when t 10 seconds and again when t 30 seconds .
+ 3
__
__
f(x)
=
-4
cos x
�
�
__
-5
}.
-
f (x) = -4cos x ; g(x) = 2cosx+3 y
-
a.
>
-
-
>
{2, }
.{I
61. a, d.
f (x) = g (x) -4 cos x = 2 cos x +3 -6cosx = 3 cos = -63 = - 21 4lr 2lr x = -+2klr or x = -+2klr, 3 3 k is any integer On [0, 2lr], the solution set is 2; , 4; From the graph in part (a) and the results of part (b), the solution of f (x) g (x) on [0,2lr] is x 2; < x < 4; or 2; , 4; -
__
�
�
�
© 2008 Pearson Education , Inc . ,
473 Upper S addle River, NJ. All rights reserved. Thi s material is p rotected under all copyright laws as they currently
exist. No portion of thi s material may be reproduced, in any form or by any mean s , wi thout permission in writing from the publi sher.
Chapter 8: Analytic Trigonometry
b.
(
�)
So during the first 40 seconds, an individual on the Ferris Wheel is more than 125 feet above the ground for times between about 10 and 30 seconds. That is, on the interval 10 x 30 , or (10,30) .
Solve h(t)=125Sin 0.157t- + 125=250 on the interval [0,80]. 125sin 0.157t- % + 125 = 250
(
)
<
( �) sin ( 0. 157t �) = 1
65.
125 Sin 0.157t - = 125 0. 157t - Jr2 = Jr2 + 2k Jr ' k is any integer 0. 157t= Jr + 2k Jr, k is any integer 2k Jr , k'1S any mteger . t= Jr0.+157 Jr For k = 0, t= -0.157 '" 20 seconds . Jr + 2Jr '" 60 seconds . For k = 1, t=--0.157 +4 Jr '" 100 seconds . Jr For k = 2, t= --0.157 So during the first 80 seconds, an individual on the Ferris Wheel is exactly 250 feet above the ground when t '" 20 seconds and again when t '" 60 seconds . Solve h(t) = 125sin 0. 157t + 125 > 125 on the interval [0,40] . l25sin 0.157t- + 125 > 125
�)
(
(
-
b.
( ) sin-1 ( - % ) + 21rk x = -"-�
0.65 3.94+2 k 5.94 + 21rk Jr x "' ---"' ---or x 0.65 0.65 k is any integer 5.94+0 + 0 or x "'--For k = 0 , x '" 3.94 0.65 0.65 '" 6.06 min '" 8.44 min 5.94 + 2Jr For k = 1 , x '" 3.940.65+ 2 Jr or x "' ---0.65 '" 15.72 min '" 18. 1 1 min For k = 2 , 3.94+4 Jr or x "'---5.94 + 4 Jr x "'---0.65 0.65 '" 25.39 min '" 27.78 min So during the first 20 minutes in the holding pattern, the plane is exactly miles from the airport when x '" 6.06 minutes , x '" 8.44 minutes , x '" 15.72 minutes , and x '" 18. 1 1 minutes . --
�)
( -�) > 0 sin ( 0.157t �) > 0 Graphing = sin ( 0. 157x - �) and 125sin 0. 157t
-
=0 on the interval [0,40] , we see that > for 10 x 30 . 1.5 Y1
<
<
d(x) = 70sin(0.65x) + 150 d (O) = 70sin ( 0.65 (0))+150 = 70sin(0)+150 = 150 miles Solve d (x) = 70sin(0.65x) + 150= 100 on the interval [0, 20] . 70sin(0.65x) + 150 = 100 70sin(0.65x) = -50 sin(0.65x) =_2.7 0.65x = sin-1 - % + 21rk
a.
-
c.
<
Y1
Y2
Y2
---
1 00
-1.5 © 2008
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in wri ting from the publi sher.
Section 8.7: Trigonometric Equations (I)
c.
Solve d ( x ) = 70sin ( 0.65x ) + 150>100 on the interval [ 0,20 ] . 70sin ( 0.65x ) + 150> 100 70sin ( 0.65x ) > -50 sin ( 0.65x ) >_ 2.7 Graphing YI = sin ( 0.65x ) and Y2 = _ 2.7 on the interval [ 0,20 ] , we see that Y1 > Y2 for 0 < x < 6.06 , 8.44 < x < 15.72 , and 18.1 1 < x < 20 .
69.
1.5
-1.5
d.
So during the first 20 minutes in the holding pattern,the plane is more than 100 miles from the airport before 6.06 minutes, between 8.44 and 15.72 minutes, and after 18. 1 1 minutes. No, the plane is never within 70 miles of the airport while in the holding pattern . The minimum value of sin ( 0.65x ) is -1 . Thus, the least distance that the plane is from the airport is 70 ( -1 ) 150 = 80 miles.
71.
sm
73.
+
67.
----� �
2
sm
If (J is the original angle of incidence and ¢ is (J = n The the angle of refraction, then sin sm ¢ 2 angle of incidence of the emerging beam is also ¢ , and the index of refraction is � . Thus,(J is n2 the angle of refraction of the emerging beam. The two beams are parallel since the original angle of incidence and the angle of refraction of the emerging beam are equal. •
sin40° = 1 .33 sin(J 2 1.33 sin(J 2 = sin 40° s 2 _ sin1.3340° 0 4833 (J 2 = sin- I ( 0.4833 ) "" 28.90° L1 • m u
Calculate the index of refraction for each: 2i sin(J1 (J 2 (J I v 2 = sin(J 2 sin 10° "" 1.2477 8° 10° sin8° sin 20° "" 1.2798 20° 15°30' = 15S sin15S 30° "" 1.3066 30° 22°30' = 22S sinsin22S sin 40° ",, 1 .3259 40° 29°0' = 29° sin 29° sin 50° "" 1.3356 50° 35°0' = 35° sin35° 60° "" 1.3335 60° 40°30' = 40S sinsin40S sin 70° ",, 1.3175 70° 45°30' = 45S sin45S sin 80° "" 1 . 2856 80° 50° 0' = 50° sin 50° Yes, these data values agree with Snell's Law. The results vary from about 1.25 to 1.34. Calculate the index of refraction: ()I = 400 ' ()2 = 260 ,. sin. (J()I = sin. 40° 260 "" 1.47
•
Mediulll I
Tbick
75.
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ.
Answers will vary.
475 All rights reserved. Thi s material i s protected under all copyright laws as they currently
exist. No portion of this material may be reproduced , in any form or by any mean s, without permi ssion i n writing from the publi sher.
Chapter 8: Analytic Trigonometry
Section 8.8 1.
9.
4x2 - x - 5 = 0 (4x - 5)(x + 1) = 0 4x - 5 = 0 or x + 1=0 x = -45 or x = -1
{
{ %} .
The solution set is -I,
11.
(2x - 1) 2 - 3(2x -1) - 4 = 0 [(2x - 1) + 1][(2x - 1) - 4]= 0 2x(2x - 5) = 0 2x = 0 or 2x - 5 = 0 x = O or x=-25 The solution set is 0, -25 ' 5. 2cos 2 0 + cos O=0 cos O(2 cos O + 1) = 0 cos O= 0 or 2cos O + 1 = 0 2cos O =-1 3n o= 2' 2 cos O =--21 0= 2n3 ' 4n3 Tr 2Tr , 4Tr 3Tr The so1utIon · set IS. -, 2 3 3' 2 . 7. 2sin 2 0-sin O -1=0 (2sin 0+l)(sin 0-1) = 0 or sin O - 1 = 0 2 sin 0+1 = 0 2sin O =-1 sin O = 1 smu =--21 0= 7n6 ' l 6In 7Tr 1 I Tr . The solution set is -Tr2 ' 6' 6
3.
{ }
{
13.
}
15.
11
{
}
{
�
.
(tan O - 1)(secO-1) = 0 tan O - 1 = 0 or sec O - 1 = 0 tan O=1 sec O = 1 5n o=� ' 4 4 The solution set is 0, Tr4 ' 54Tr . sin 2 0 - cos 2 0 = 1 + cos 0 ( 1 - cos 2 0) - cos 2 0 = 1 + cos 0 1 - 2 cos 2 0 = 1 + cos 0 2 cos 2 0 + cos 0 = 0 ( cos 0) ( 2 cos 0 + 1) = 0 cos O=O or 2 cos 0 + 1 = 0 0 = Tr2 ' 3Tr2 cos O = --21 0 = 2Tr3 ' 4Tr3 Tr 2Tr , 4Tr 3Tr . set IS. -, The so 1utIon · ' 2 3 3 2
}
}
sin 2 0 = 6(cos O+1) 1 - cos 2 0 = 6 cos 0+6 cos 2 0 + 6cos O + 5 = 0 ( cos 0 + 5) ( cos 0 + 1) = 0 cos 0 + 5 = 0 or cos 0 + 1 = 0 cos O =-5 cos O = -1 (not possible) O = Tr The solution set is {Tr} . cos(20) + 6sin 2 0 = 4 1 - 2 sin 2 0 + 6 sin 2 0 = 4 4sin 2 0 = 3 sin 2 0 = �4 sin O= .J32 Tr 2Tr 4Tr 5Tr o±
{
_
3" ' 3 '
3 '3
}
Tr 2Tr , 4Tr , 5Tr . The so 1utIon · set IS. -, 3 3 3 3 © 2008 Pearson Education, Inc. , Upper Saddle River, NJ.
476 All rights reserved. Thi s material is protected under all copyright l aws as they currently
exist. No portion of this materi al may be reproduced, in any form or by any means , without permi ssion in writing from the publi sher.
Section 8.8: Trigonometric Equations (1/)
17.
cos B=sin B sin B cos B = 1 tan B= 1 B=�4 ' 4
Sn
The solution set is 19.
2 5.
{ " 54" } · "4'
n
tan B= 2sin B sin B ' --= cos B 2 Sln u sin B= 2sin Bcos B 0= 2sinBcos B - sin B 0= sin B(2 cos B - l) or sin B = 0 2cos B - l = 0 B= 0, n cos B=-21 B= �3 ' 5n3 The solution set is �, ", 5; . /l
21.
sin B= csc B 1 Slll u=-sin B sin 2 B= 1 sin B= ±l B= �2 ' 3n2 .
23.
2 7.
}
,
cos( 4B) - cos( 6B) = 0 . (4B-6B)= 0 -2 sm (4B+6B) sm 2 2 - 2sin(5B)sin(- B) = 0 2 sin(5B)sinB= 0 sin(SB) = 0 or sin B= 0 5B = 0 + 2kn or 5B = n + 2kn or B=!:5 + 2kn B= 2kn 5 5 B=0 + 2kn or B= n + 2kn On the interval 0:5: B 2 " , the solution set is {o'::' 2" 3" 4" " 6" 7" 87r 97r .
/l
--
--
<
'5' 5'5 ' 5 ' '5' 5'5'5
{" }
3" · 2 cos(2B) = cos B 2 cos 2 B - 1 cos B 2cos 2 B - cos B - l = 0 (2cos B + 1)(cos B - 1) = 0 2cos B + l = 0 or cos B - 1 = 0 cos B= 1 cos B= --21 B=O 2n 4n B= 3 ' 3 2 " 4" ' The solution set is 0, 3' 3
The solution set is
<
{
}
{o,
sin(2B) + sin(4B) = 0 sin(2B) + 2 sin(2B) cos(2B) = 0 sin(2B) (1 + 2 cos(2B») = 0 sin(2B) = 0 or 1 + 2 cos(2B) = 0 cos(2B) =--21 2B = 0 + 2kn or 2B = n + 2kn or B=kn B =-n2 +kn 4 + 2kn 2n + 2kn or 2B=2B =3 3 2n n B=-3 +kn B=-3 + kn On the interval 0:5: B 2 " , the solution set is " " 2,, " 4 " 3 " S " 0, 3' 2' 3' 3' 2' ""3 .
2'
2 9.
=
{
© 2008 Pearson Education , Inc . ,
}
.
l + sin B = 2cos 2 B l + sin B= 2(1 - sin 2 B) 1 + sin B = 2 - 2 sin 2 B 2 sin 2 B + sin B - l = 0 (2sin B - l)(sin B + l) = 0 or sinB+ 1 = 0 2 sin B - 1 = 0 sinB =-1 Sln. u =-21 B= 3n2 5n B = !!:.6 ' 6 The solution set is 6" 56" ' 32" . /l
{"
}
}
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in wri ting from the publi sher.
Chapter 8: Analytic Trigonometry
31.
33.
35.
2sin 2 0 - 5 sin O + 3 = 0 (2 sin 0 - 3) (sin 0 + 1) = 0 2sin O - 3 = 0 or sin O - 1 = 0 sin 0 = �2 (not possible)
{�}.
43.
�
3 - sin O = cos(20) 3 - sin 0 = 1 - 2 sin 2 0 2sin 2 0 - sin O + 2 = 0 This equation is quadratic in sin 0 . The discriminant is b 2 ' = 1 - 16 = -15 0 . The equation has no real solutions. sec2 0 + tan 0 = 0 tan 2 0 + 1 + tan O = 0 This equation is quadratic in tan 0 . The discriminant is b2 ' = 1 - = -3 0 . The equation has no real solutions. - 4ac
© 2008 Pearson
'
AI
:
tan(20) + 2 sin 0 = 0 sin(20) + 2 sm' 0 = 0 --'---'cos(20) sin 20 + 2sin Ocos 20 = 0 cos 20 2sin Ocos 0 + 2sin 0(2cos 2 0 - 1) = 0 2 sin 0 ( cos 0 + 2 cos 2 0 - 1 ) = 0 2sin O ( 2cos 2 0 + cos 0 - 1 ) = 0 2sin 0(2 cos 0 - 1)(cos 0 + 1) = 0 2cos O - l = 0 or 2 sin O = 0 or 1 sin O 0 cos O = -2 0 = 0, 1t 51t o=� 3' 3 cos O + l = 0 cos O = -1 0 = 1t The solution set is �, ", 5; . =
<
4
AI
{;, 7:}.
tan 2 0 = �2 sec O sec 2 0 - 1 = �2 sec O 2sec 2 0 - 2 = 3sec O 2sec 2 0 -3sec O - 2 = 0 (2sec 0 + 1)(secO - 2) = 0 2sec O + 1 = 0 or sec O - 2 = 0 sec O = 2 sec O = --21 51t o= (not possible) 3' 3 5 " The solution set is "3' 3 '
- 4ac
39.
sin 0 - Jj cos 0 = 1 Divide each side by 2: 1. O - Jj cos O = -1 -sm 2 2 2 Rewrite in the difference of two angles form Jj , and Al = -1t usmg cos = -21 , sm = 2 3 sin Ocos ¢ - cos Osin¢ = �2 sin(O - ¢) = �2 O - ¢ = -1t6 or O - �3 = �6 O = �2 The solution set is .
The solution set is 3(1 - cos 0) = sin 2 0 3 - 3 cos O = l - cos 2 0 cos 2 0 - 3 cos 0 + 2 = 0 (cos 0 -1) ( cos 0 - 2) = 0 cos 0 - 1 = 0 or cos 0 - 2 = 0 cos 0 = 1 cos 0 = 2 0=0 (not possible) The solution set is {o} .
{" }
37.
41.
{O,
}
<
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in wri ting from the publisher.
Section 8.8: Trigonometric Equations (1/)
45.
sin () + cos () = .Ji Divide each side by .Ji : .1 sin B + � cos () = 1 Rewrite in the sum of two angles form using cos ¢ = .Ji1 ' sm ¢ = .Ji1 ' and ¢ = '4 : sin Bcos ¢ + cos Bsin ¢ = 1 sin«() + ¢) = 1 () + ¢ = -2
51.
X
=
TI
.
TI
53.
() + -4 = -2 B = ::"4 The solution set is TI
47.
j (x) = 0 4cos2 x - I = 0 cos2 X = -41 cos x =
±
TI
{:}.
\
J{ = � ±
-10
27r 47r 57r The zeros on 0 � x < 27r are -7r3 ' 3 ' 3 ' -3 . 49.
55.
j (x) = 0 sin ( 2x ) - sin x = 0 2sinxcosx - sin x = 0 sinx ( 2cos x - l ) = 0 sinx = 0 or 2 cos x - I = 0 x = 0, 7r cosx = -21 57r x = -7r3 ' 3 7r 57r The zeros on 0 � x < 27r are 0,-,7r,3 3
© 2008 Pearson Education,
j ( x) = 0 sin 2 x + 2 cos x + 2 = 0 l - cos2 x + 2cos x + 2 = 0 - cos 2 + 2 cos x + 3 = 0 cos2 x - 2cos x - 3 0 ( cos x -3 ) ( cos x + 1 ) = 0 cos x - 3 = 0 or cosx + 1 = 0 cos x = -1 cos x = 3 X = 7r (not possible) The only zero on 0 � x < 27r is 7r . x + 5 cos x = 0 Find the zeros (x-intercepts) of 1'; = x + 5 cos x :
X "" -1.31, 1 .98,3.84 22x -17 sin x = 3 Find the intersection of 1'; = 22x -17 sin x and Y2 = 3 : 5
-11:
{
I�t�r5�ctio K=.S2096017
x ",, 0.52
J
tl
11:
V =3
-5
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exist. No portion of this material may be reproduced , in any form or by any means, without permi ssion in writing from the publi sher.
Chapter 8: Analytic Trigonometry
57.
sin x + cos x = x Find the intersection of >'; = sin x + cos x and Y2 = x :
I�hr��cti�� X=1.�SB7�B2
65.
a.
Y=1.2SB72B2
-3
b.
3
- 7tI�-,-*-+-I-----� 7t
-7tI �-.o.-I,,--t-*"'-� I 7t 2�r� X=1.02169
-3
X ""
61.
-1 .02, 1.02
y=o
-3
c.
d.
/
\,,-
63.
o
( :) (�) cos e:) = 0 or cos (�)
<
=
0
38 = 900 . 38 = 2700 . -8 = 90° · -8 = 270° ' ' 2 ' 2 ' ' 2 2 8=60° 8=1 80° 8 180° 8 = 540° On the interval 0° 8 90° , the solution is 60". A(600 ) = 16sin(600 )[cos(600 ) + 1] = 16 · +1 = 12J3 in 2 "" 20.78 in 2 Graph >'; = 16sin x(cos x + 1) and use the MAXIMUM feature: <
=
� (� )
o
6sin x - ex = 2, x > 0 Find the intersection of >'; = 6 sin x - eX and Y2 = 2 : 6
<
25
x "" 0, 2.15
<
<
<
x 2 - 2 sin(2x) = 3x Find the intersection of >'; = x2 - 2 sin (2x) and Y2 = 3x : 12
<
<
1 . 26 x 2 - 2 cos x = 0 Find the zeros (x-intercepts) of >'; = x 2 - 2 cos x : X ""
59.
cos(28) + cos 8 = 0 , 0° 8 90° 2 cos 2 8 -1 + cos 8 = 0 2 cos 2 8 + cos 8 - 1 = 0 (2cos 8 - 1)(cos 8 + 1) = 0 2cos 8- 1 = 0 or cos 8+ 1 = 0 cos 8 = -1 cos 8 = -21 8 = 180° 8 = 60°, 300° On the interval 0° 8 90° , the solution is 60". cos(28) + cos 8 = 0, 0° 8 90° 2 COs 3 cOs = 0
/'
" � X i fVl U fVl X=6Q
o
Y=20.7B�61
�
90
The maximum area is approximately 20.78 in.2 when the angle is 60".
6
/
\
\
I�t�r��cti�� X=.7621S�OB
-6
X ""
2, Y=�
o
F I�tUNcti�h X=1.3�B7B�1
2.
Y=2
-6
0.76, 1 .35
© 2008 Pearson Educati on, Inc . , Upper Saddle River, NJ.
480 All rights reserved. Thi s material is protected under all copyright laws as they currently
exi st. No portion of thi s material may be reproduced, in any form or by any means, without permission in wri ting from the publisher.
Chapter 8 Review Exercises
67.
lnt�rs�ct;.n X=2.028757B
-12
69.
C h apter 8 Review Exercises
Find the first two positive intersection points of >-; = -x and Y2 = tan x .
l r. t � r s t c t i . n X=�.9131BO�
y= -2.02875B
-12
1.
y= -�.9131B
The first two positive solutions are x "" 2.03 and x "" 4.91 . (34.W sin(2B)-'107 = ----'-9. 8 sin(2B) = 107(9.8) (34.W "" 0.8659 2B "" sin- I (0.8659) 2B "" 60° or 120° B "" 30° or 60° Notice that the answers to part (a) add up to 90° . The maximum distance will occur when the angle of elevation is 90° -;- 2 = 45° : 2 45°)J "" 123.6 R ( 45°) = (34.8) sin[2( 9.8 The maximum distance is 123.6 meters. 2 sin(2x) Let I = (34.8)9.8
3.
d.
�
"2 ' 1t
whose sine
1t
1t
1t
5.
-I [ ../3)
cos - 2 Find the angle B, 0 � B � whose cosine equals -2 .
../3 ../32 ' 0 � B � cos B = -B = 51t6 ../32 ) = -51t6 . Thus , cos [ --
Y.
1t ,
1t
\ 90
_I
XM i n=0 XMax=90 Xsc l = 1 0 Y M i n=0 YMax= 1 25 Ysc l =25 Xres= l
7.
sec- l .J2 Find the angle B, 0 � B � whose secant equals .J2 . sec B = .J2, 0 B B = �4 Thus, sec- l .J2 = �4 . 1t,
�
IMtyuo:tion X=60.009053 _ V=iO?
© 2008 Pearson
B
tan- I I Find the angle B, - "2 < B < "2 ' whose tangent equals 1 . tan B = 1, --2 < B < -2 B = �4 Thus, tan _ I ( 1 ) = -4 . 1t
oV� o
1t
1t
b.
1 25
- "2 �
Thus, sin- I (I) = �2 .
a.
c.
sin- I I Find the angle B, equals 1 . sin B = 1, B = �2
� 1t
481
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in wri ting from the publisher.
Chapter 8: Analytic Trigonometry
9.
1 �" is in 1 5" IS. -" . quadrant so the reference angle of -1 5") = cos- . mce Thus, we h ave cos (-in the interval [0, ,, ] we can apply the equation above and get cos - I ( cos ( 1 5")) = cos - I ( cos ?"") = ?"" . C� )
( ( ))
sin- I sin 3; follows the fonn of the equation rl (t (x) ) = sin - I (sin (x) ) = x . Since 3" IS. m. the mterva ' I --,8 2 2 , we can appIy the equation directly and get sm. - I sm -8 = -8 .
cos ,, = cos (} . The angle I
[ ,,]
1 1.
( . e")) 3" tan - I ( tan ( 2; )) follows the fonn of the
s·
" 7
.
- 15
-7
1 5.
sin( sin-I 0.9) follows the fonn of the equation f (rl (x)) = sin (Sin - I ( ) ) Since 0.9 is in the interval we can apply the equation directly and get sin( sin - I 0.9) = 0.9 .
[- 1, 1 ],
[-; ,%]. We need to find an angle () in the interval [ - ; , ;] for which tan ( 2; ) = tan () . The angle 2; is in quadrant in the interval
1 7.
x
= x .
cos ( cos - I ( -0.3 )) follows the fonn of the equation f (rl ( x) ) = cos ( cos (x)) = x . Since -0.3 is in the interval we can apply the equation directly and get cos( COS- I (-0.3)) = -0 . 3 .
[- 1, 1 ], -I
so tangent is negative. The reference angle of 2-" IS. -" and we want () to be m' quadrant IV 3 3 so tangent will still be negative. Thus, we have 2 " = tan -3 ' S'mce -3" IS. m. the tan 3
II
( ) ( ") interval [- ; , ; ] , we can apply the equation
1 3.
7
7
'
equation rl (t(x)) = tan- I (tan(x)) = x but we cannot use the fonnula directly since 23" is not
above and get tan- I tan 2;
"
7
"
7
1 9.
( ( )) tan- I ( tan ( - � )) = - � . cos - I ( cos ( 1 �" follows the fonn of the ))
21.
=
-1 .6 ,
Since there is no angle () such that cos e = the quantity COS-I ( - 1.6 ) is not defined. Thus, cos ( cos - I ( -1.6 ) ) is not defined.
( ) . ( -21 ) = -"6
2n = sm sm cos 3" .
-1
-1
1t
equation rl (J(x)) COS- I (cos(x)) x , but ' Iy smce we cannot use the fionnuIa d!fect -- IS ' not in the interval J . We need to fmd an angle () in the interval [0, ,, for which =
[0,,,
=
1 5" . 7
]
© 2008 Pearson Education , Inc . , Upper Saddle River,
482 NJ. All rights reserved. This material is protected under all copyright l aws as they currently
exist. No portion of this material may be reproduced, in any form or by any means , without permi ssion in writing from the publisher.
Chapter 8 Review Exercises
S·mce sm = 45 - 7t2 - - -7t2 ' Iet = - 4 and r = Solve for x: x 2 1 6 = 25
. 7t whose sme Find the angle f}, - -7t2 ::; f} ::; -, 2 .J3 equals - 2 . sin f} = .J32 ' - -7t2 ::; f} ::; -7t2 f} = - !!:.3 J3 = - 7t ' So, sm. - I -2 3
.
5 .
33.
(
y
(J
Thus, sin tan - i4 ) = sin = 1.r = i5 .
© 2008
I
Y
I(x) 2 sin(3x) y = 2sin(3x) x = 2sin(3y) � sin (3y) 2 3y = sin - I =
.
- 1 ::; � ::; 1
2 -2 ::; x ::; 2 The domain of rl ( x ) is { x l -2 ::; x ::; 2 } , or [ -2, 2 ] in interval notation.
Since cot = %, 0 is in quadrant I. Let x = 3 and = 4 . Solve for r: 9 16 = r2 r 2 = 25 7! ,
x2 = 9 x = ±3 is in quadrant IV, x = 3 .
The domain of I ( x ) is the set of all real numbers, or (-00, 00) in interval notation. To find the domain of I - I ( x ) we note that the argument of the inverse sine function is �2 and that it must lie in the interval [-1, 1] . That is,
sin coel %)
< f) <
+
1
IS
[ �) ( �) �
(J
y
(1) _I X -I y = -sm 3 ( -2 ) = I ( x )
=
(
<
=
Find the angle f}, - 7t2 f} -7t2 , hose tangent . J3 3 J3 tan f} = -, 3 f} = !!:.6 J3 = -7t . So , tan _ I 3 6 sec = 2 . Thus, sec tan- I 29.
Ll {7
I
-
W
-<
[ . ( )]
[ ) Th." ,+m-' [ � ) l = tan Hl = -A <
-- '
Since f} - 4 = --4 Thus , tan sm- - -45 = tan (J = -x =3 3
_
- <
Ll {7
+
r=5
(J
483
Pearson Education , Inc . , Upper S addle River, NJ. All rights reserved. Th i s material is protected under all copyri ght laws as they currently
exist. No portion of thi s materi al may be reprod uced, in any form or by any mean s , without permission in wri ting from the publisher.
Chapter 8: Analytic Trigonometry
45.
35. /(x) = -cos x + 3 y = -cos x + 3 x = -cosy + 3 x - 3 = -cosy 3 - x = cosy y = (3 - x) = rl (x) The domain of I ( ) is the set of all real numbers, or (-00, 00) in interval notation. To find the domain of I- I ( ) we note that the argument of the inverse cosine function is 3 -x and that it must lie in the interval [-1, 1] . That is, -1 ::; 3 -x ::; 1 -4 ::; -x ::; -2 4�x�2 2 ::; x ::; 4 The domain of rl (x) is {x I 2 ::; x ::; 4 } , or [ 2,4] in interval notation. COS
47.
-I
5cos 2 0 + 3 sin 2 0 = 2cos 2 0+3cos 2 0+3sin 2 0 = 2cos 2 0 + 3( cos 2 0 + sin 2 0) = 2cos 2 0 + 3 · 1 = 3 + 2cos 2 0 I - cos O sin O (1 - O) 2 + sin 2 0 sin O
COS
+ ---
I - cos O
sin 0(1 - cos 0) 1 - 2 cos O + cos 2 0 + sin 2 0 sin 0(1 - cos 0) 1 - 2cos O + 1 sin 0(1 - cos 0) 2 - 2cos O sin 0(1 - cos 0)
X
X
37.
39.
2(l - cosO)
=
49.
Let O = sin- I u so that sin O = u , _ 7r2 ::; 0 ::; 7r2 , -1 ::; u ::; 1 . Then, cos ( sin -I u ) = cos 0 = �cos 2 0 = �I - sin2 0 = �I - u2
51.
Let O = csc- u so that csc O = u , --7r2 ::; 0 ::; -7r2 and O :/. 0 , lui � 1 . Then, 1 = -1 sm. ( csc -I u ) = sm. u = -csc O u 1
1 2 tanOcot O - sin 2 0 = tan O . tan-0 - sin 0 = I - sin 2 0 cos 2 0 sin 2 0(1 + coe 0) = sin 2 O · csc 2 0 1 = Sln2 u '-sin 2 0 =1 •
© 2008
csc O sin 0 sin O 1 + csc O 1 + _1_ sin O sin O sin O+ 1 1 I - sin O 1 + sin 0 1 - sin 0 I - sin O I - sin 2 0 I - sin O cos 2 0 l _ - sin O csc O - sin O = _ sin O l - sin 2 0 sin O cos 2 0 sin O cos O = cos O·- sin O = cos Ocot O 1
53.
=
43.
cos O cos O . cos O cos O - sin O . cos O - sin O I cos O 1 - sin O cos O I I - tan O
11
41.
sin 0(1 - cos 0) 2 sin O 2 csc O
11
484
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exist No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.
Chapter 8 Review Exercises
55.
I - sin B = cos B(1 - sm. B) --sec B B = cos B(I - sm. B) · 11 ++ sin sin B cos B ( 1 - sin 2 B ) 1 + sin B cos B ( cos 2 B ) 1 + sin B cos3 B 1 + sin B B sin B cot B - tan B = cos sin B cos B cos 2 B - sin 2 () sinBcos B 1 - sin 2 B -sin 2 B sin B cos B 1 - 2 sin 2 B sin Bcos B cos(a + 13) cos a cos 13 - sin a sin 13 cos a sin 13 cos a sin 13 cos a cos 13 sin a sin 13 cos a sin 13 cos a sin 13 cosf3 sin a sin f3 cosa = cot 13 - tan a cos(a - 13) cos a cos 13 + sin a sin 13 cos a cos 13 cos a cos 13 cos a cos sin a sin 13 = cos a cos 1313 + ---� cos a cos 13 = 1 + tan a tan 13 (1 + cos B) tan-B2 = (1 + cos B) · l +sincosB B = sm. B cos. -B cos ( 2()) 2 cot B cot ( 2B ) = 2 · sm B sm ( 2B) 2 cos B ( cos 2 B -sin 2 B ) sin B ( 2 sin () cos B ) cos 2 B - sin 2 B sin 2 B cos 2 B sin2 B sin 2 B sin 2 B = coe B - 1
67.
---
57.
59.
61.
63.
65.
© 2008 Pearson
. ( 2 e 4e ( 2e - 4e ) sin(2B) + sin (4B) 2sm -2- )cos -2cos(2B) + cos( 4B) 2cos Ce : 4e )cos ce : 4e ) 2sin(3B)cos(-B) 2cos(3B)cos( B) _ sin(3B) - cos(3B) tan (3B) +
69.
=
-
-- - -
'
1 - 8sin 2 Bcos 2 B = 1 - 2 ( 2sin Bcos B ) 2 = 1 - 2sin 2 ( 2B ) = cos ( 2 . 2B ) = cos ( 4B)
=
71.
73.
cos(2B) -- cos( ---'-----'-4B) ----"-- - tan tan(3 11) cos(2B)--'-+cos( 4B) sin(3B) sine-B) tan B tan(3B) = -2 2cos(3B) cos( - B) 2 sin(3()) sin B tan B tan(3B) 2 cos(3B) cos B = tan(3B) tan B - tan B tan(3B) =0 sin 165° sin ( 120° + 45° ) = sin 120° ·cos 45° + cos 120° . sin 45° + L1 u
u
=
=[ �){V;) ( -�){ V;) J6 .J2
75.
cos �; = cos
e; + �;)
= cos-7t4 · cos-7t6 - sm-47t · sm-7t6 .J2 J3 .J2 1 2 2 2 2
.
.
.
J6 J2
485
Education , Inc . , Upper Saddle River, Nl. All rights reserved. This material is protected under all cop yri ght laws as they currently
exist. No portion of thi s material may be reproduced, in any form or by any means , without permission in wri ting from the publisher.
Chapter 8: Analytic Trigonometry
77.
cos 80° cos 20° + sin 80° sin 20° coscos (80° 20°) 12 60° l - cos2:4 .
c.
=
.
=
sinea - fJ) sin a cos fJ -cos a sin 5fJ =(�}( - ��) _(%}C3) - 4865-15 63 tan(a + fJ) 1t-antana +attaannfJfJ i + ( -�)5 1 - (i)- (- 1 2) 11 149 .!..!12 �14 5633 2 54 -35 2425 ' sm. (2 a) 2 smacosa cos(2fJ) cos2 fJ - sin2 fJ -( �� J - C53 J 144169 16925 119169 =
65
d.
= ------.:...
=
e.
81 .
sm. a -54 0 < a < -;n2 cosa -35 tan a 34 cosfJ - -1213 tanfJ 125 n4 fJ2 n2 0< a2- <-n4' -<-<sine a + fJ) sin a cos fJ + cos a sin 5fJ = (�} ( - ��) +(�}C3) - 48+15 336565 cos(a + fJ) cos a cos fJ -sin a sin5fJ = (�} ( - ��) _ (�}C 3) -36-65 20 6556 =
=
a.
b.
f.
= -,
=
,
= --
=
.
=
=
=
. _ .
= -
=
=
,
,
l1.
g.
,
=
h.
=
=
=
© 2008 Pearson Educati on , inc . , Upper Saddle River, NJ.
cos-a2 �l +cosa 2 �l:� � � 5s = 2� =
=
=
486 All rights reserved. Thi s material is protected under all copyright l aws as they currently
exist. No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in writing from the publi sher.
Chapter 8 Review Exercises
83.
3n cos fJ = 12 3n < fJ < 2n sm. a = - -35 ' n < a < -" 13 ' 2 2' 3 4 5 tanfJ = --5 cosa = --5 ' tan a = -4 ' sm. fJ = -13 ' 12 ' n- < -a < 3n 3n < -fJ < n 2 2 4' 4 2 sin( a + fJ) = sin a cos fJ + cos a sin fJ = + - - 153 -36 + 20 65 16 65 cos( a + fJ) = cos a cos fJ - sin a sin fJ = - 153 -48 - 1 5 65 63 65 sin( a - fJ) = sin a cos fJ - cos a sin fJ - 153 = -36 - 20 65 56 65 tan a + tanfJ . tan( a + fJ) = -----'1 - tan a tan fJ 3+ - 5 -4 1 2 = .1... = .!. " � = � 1 - � - 152 �� 3 21 63 sin(2a) = 2sin a cos a =2 24 25
a.
b.
c.
f.
( �}C�) ( :}( )
g.
h.
( : }C�)-(-�}( )
(-�}C�)-( -�}( )
85.
---';-_"'-
-
l
� � � = Ju � =
3 n < a < -; 3n tan fJ = -, 12 0 < fJ < -n tan a = -, 4 2 2 5 .sm a = - -3 cos a = --4 sm. fJ = 12 cos fJ = -5 13 ' 5' 13 ' 5' n- < -a < 3n 0 < -fJ < -n 2 2 4' 2 4 sin( a + fJ) = sin a cos fJ + cos a sin fJ = _ 53 + 15 48 65 65 63 65 cos( a + fJ) cos a cos fJ - sin a sin fJ = _ 53 - 20 + 36 = -65 65 16 65
a.
_( ) ( ) { - �}(-�)
© 2008 Pearson Education, Inc . , Upper Saddle River, NJ.
C�) ( Y
-
d
c.
cos(2fJ) = cos 2 fJ - sin 2 fJ = 2 - - 153 144 - 25 169 169 1 19 169 . fJ = - cosfJ Sln2 2 1= 2:� = =
b.
( �}C ) (-�}C�)
=
( �}C ) ( �}C�)
487 All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 8: Analytic Trigonometry
c.
d.
e.
f.
g.
sinea - = sin a cos 5 - cos a sin = ( _�}C3 ) - ( -�}C�) 15 4865 =--+65 6533 t1 antaan+atantan tan(a + = -------'- 3 12 -+4 5 l - ( % )C:) 63 2 = � = �� ( _ %) = _ �� 5 sin(2a) = 2sin acosa = 2 (-�)(-�) = �� cos(2f3) = cos2 f3 - sin2 = C53 J -C�J =-126-� --�:-: =- -��-: 13)
13
13
87.
13
13
cos a2 = _�I+cosa 2
� -� �-H�-��-Jo�-�
© 2008 Pearson Education , inc . ,
�
a.
13)
13
13
b.
13)
13
13
c.
13)
d.
h.
13
13
13 13
13)
1t2 <0; secf3= 3,-
-
13
fJ
-
488 Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently
exist. No portion of thi s material may be reprod uced, in any form or by any means , without permi ssion in writing from the publisher.
Chapter 8 Review Exercises
e.
f.
( � )(�) = - �
sin(2a) = 2sm a cos a = 2 cos(2fJ) = cos2 fJ - sin2 fJ = - -2
(�J ( � J
d.
-91 - -98 = - -79
+
+ --+-
tan a tan----'-fJtan( a fJ) = --I - tan a tanfJ 2J5 J5 5 2 . J5 1 _ 2J5 5 2 4J5 10 1-1 9J5 =� . Undefined o ' sin(2a) = 2sm a cosa =4 =2
+5J5
g.
e.
( : ) ( �) : _
f.
89.
3n ' cosfJ = --2 n < fJ < 3n . = --2 n < a < sma 2 3' 3' 2' J5 2J5 sm. fJ = -J5 tana = -cosa = -3' 3' 5 ' J5 n < a < 3n n < fJ < 3n tanfJ = T ' '2 2" 4 ' '2 2" 4 sin(a fJ) sin a cos fJ cos a sin fJ
a.
+
g.
+ (4- %5) ( -%) + ( - �)( -�) = -9 + -9
c.
© 2008 Pearson Education,
-49 95 -9
0-R) �
= V� = �t = �6� = m 6 h.
=1 cos(a fJ) = cos a cos fJ - sin a sin fJ = % 2J5 2J5 9 9 =0 sin( a - fJ) = sin a cos fJ - cos a sin fJ = 4 5 9 9 1 9
+
+H -(-�)' - - - . fJ = l - cos fJ Sln2 2
cos(2fJ) = cos2 fJ - sin2 fJ
=
=
b.
_
(- �)(-%)-(- ) (-�)
� [-�l
+ o o, a = _ � = _ l 2
��
= -tt = -H �6(3-J5} =
( -%)(- : ) ( -�)(-�)
2
6 6
J6J37s
489 Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently
exist. No portion of thi s materi al may be reproduced, in any form or by any means, without permission in writing from the publ i sher.
Chapter 8: Analytic Trigonometry
91.
(
)
cos sm S3 - cos "21 et a = sm -53 and 13 = cos 21 a . m. quadrant I; 13 is in quadrant I . Then sin a = �5 , 1 0 � 13 � -n o � a � -n , and cos 13 = -, 2 2 2 cos a = .Jl - sin 2 a = = 1 - :5 = =
L
.
_I
_I
.
-I
-I
-
.
tan a - tan 13 1 + tan a tan 13 J3 3 3 4
IS
.
FG) ) J¥s � sin 13 = �1 - cos2 13 = Jl - (�J = )I - ± = l = � (
-4J3 - 9 12 1 - 3J3 12 -9 - 4J3 · 12+3J3 12 - 3J3 12 +3J3 -144 - 75 J3 1 17 -48 - 25J3 39 48 + 25J3 39
�)
cos sin- I �- cos-J = cos ( a - 13) = cos a cos 13 + sina sin 13 J3 = -45 . -21 + _35 . 2 4 3J3 4 + 3J3 =+-= 10 10 10 93.
[ ( �) �] ( )
---
tan sin-J - - tan-J Let a = sm. -J -"21 and 13 = tan -J 4"3 . a m quadrant IV; 13 is in quadrant I. Then, . = --21 ' 0 � a � -2 ' and tan 13 = -43 ' sma 0 < 13 < -n2 · cos a = .J'1- si-n:-2-a
( J
Let a = cos-J -� a is in quadrant II. Then cos a = - -35 , -2 � a � n . sin a = .Jl - cos 2 = 1 - :5 = = 1-
IS
7r
� (-�J � � � [ ( �)] = = (�)( -�) = �: a
n
J ( �J
=
sin 2 cos- J
) l=�
= 1 - - = 1 -± = J3 tana = --1 = -J3 3
© 2008 Pearson Educati on , Inc . ,
sin 2a
-
2sin a cos a
=
2
-
490 Upper Saddle River, N J . All rights reserved. This material is protected under all copyri ght l aws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in wri ting from the publi sher.
Chapter 8 Review Exercises
97.
cos B = -21 5n + 2kn , k is any B = -n3 + 2kn or B = 3 integer n 5n . On 0 ::; B 2n , the solution set is "3' 3
{
<
99.
}
2cos B + ..fi = 0 2 cos B = -..fi ..fi cos B = -2 3n 5n + 2 k n , k IS· any Integer . B = -4 + 2kn or B = 4 3n 5 n . On 0 ::; B 2n , the solution set is 4' 4
{
<
101.
1 07.
<
1 09.
}
sin(2B) + 1 = 0 sin(2B) = -1 3n + 2kn 2B = 2 3n B -4 + kn, k is any integer On the interval 0 ::; B 2n , the solution set is
1 03.
<
1 1 1.
tan ( 2B ) = 0 2B 0 + kn kn where k IS· any Integer . B = -, 2 On the interval 0 ::; B 2n , the solution set is , n, 3; .
1 05.
<
}
<
see 2 B = 4 sec B = ±2 cos B = ±-21 n 2n +kn, B = -+kn or B = 3 3 where k is any integer On the interval 0 ::; e 2n , the solution set is ; , 2; , 4; , 5; .
{
© 2008 Pearson
}
sin(2B) -eos B - 2sin B + 1 = 0 2 sin B cos B - cos B - 2 sin B + 1 = 0 cos B(2sin B - 1) - 1(2sin B - 1) = 0 (2sin B - 1)(cos B - 1) 0 or cos B = 1 sin. B = -21 B=O B = n6 ' 5n6 On 0 ::; B 2n , the solution set is =
=
{o, �
}
{o,
<
=
r:, 7:}
sin B = tan B sin B SIn. B = -COS B sin B cos B = sin B sin Bcos B - sin B = 0 sin B(cos B - 1) = 0 cos B - 1 = 0 or sin B = 0 cos B = 1 B=0 B=n On the interval 0 ::; B 2n , the solution set is {O, n } . sin B + sin(2B) = 0 sin O + 2sin Beos B = 0 sin B(l + 2 cos B) = 0 or sin B = 0 1 + 2cos B = 0 B = 0, n cos B = --21 B = 2n3 ' 4n3 On 0 ::; B 2n , the solution set is 2; ,n, 4; .
1 13.
{o, � , 5; } .
2 sin 2 B - 3sin B + l = 0 (2 sin B - 1)(sin B - 1) = 0 2sin B - 1 = 0 or sin B - 1 = 0 sin e 1 SIn. B = -21 B = n6 ' 5n6 On 0 ::; B 2n , the solution set is {!!.. , !!.. , 57r} . =
<
<
6
2
6
491 Educati on , Inc . , Upper Saddle River, NJ. All rights reserved. Th i s material is protected under all copyright laws as they currently
exist. No portion of thi s material may be reproduced, in any form or by any means , without permission in wri ting from the publi sher.
Chapter 8: Analytic Trigonometry
1 1 5.
37t B - c/J = -7t4 or B - c/J = 4 3 B - !!:.4 = !!:.4 or B _!!:.4 = 47t B = !!:.2 or B = 7t On 0 ::; B 2 Jr , the solution set is {%,Jr} .
4 sin 2 B = 1 + 4 cos B 4 ( 1 - cos 2 B ) = 1 + 4 cos B 4 - 4 cos 2 B = 1 + 4 cos B 4cos 2 B + 4 cos B - 3 = 0 (2 cos B - 1)(2cos B + 3) = 0 2cos B - 1 = 0 or 2 cos B + 3 = 0 cos B = --23 cos B = -2I (not possible) B = Jr3 ' 53Jr On 0 ::; B 2 Jr , the solution set is {�, 5;} .
<
121.
sin- I (O.7) � O.78
iS m1( :
7753974966
<
1 1 7.
sin (2B) = .fi cos B 2 sin B cos B = .fi cos B 2 sin Bcos B - .fi cos B = 0 cos B ( 2 sin B - .fi ) = 0 cos B = 0 or 2 sin B - .fi = 0 sin B = .fi2 B = Jr4 ' 34Jr On 0 ::; B 2 Jr , the solution set is {!!.. , !!.. , <
1 1 9.
4
2
1 23.
tan- I (-2) � -1 . 1 1 a n-' � i� f 07 1 48718
1 2 5.
3
3 7T , 7T 4 2
}.
tD
sec- I (3) = cosWe seek the angle B, 0 ::; B ::; Jr, whose cosine equals �3 . Now cos B = �3 , so B lies in
quadrant I. The calculator yields COS- I �3 � 1.23 , which is an angle in quadrant I, so sec- I ( 3) � 1 .23 .
sin B - cos B = I Divide each side by .fi : _1_ sin B 1_ cos B = _1_ .fi .fi .fi Rewrite in the difference of two angles form where cos c/J = � , sin c/J = � , and c/J = !!:.4 : ,,2 ,,2 sin Bcos c/J - cos Bsin c/J = � sin( B - c/J) = .fi2
�OS-I(
r��09594 1 7
_ _
1 2 7.
2x = 5 cosx Find the intersection of 1'; = 2x and � = 5cos x :
x � 1.11 The solution set is {1. 1 1} . © 2008 Pearson Education , Inc . , Upper Saddle River, NJ.
492 All rights reserved. This material i s protected under all copyright laws as they currently
exist. No portion of thi s material may be reprod uced, in any form or by any means , without permi ssion in wri ting from the publi sher.
Chapter 8 Review Exercises
1 29.
Using a difference formula: sin 15° = sine 45° - 30°) = sine45°) cos(300) -cos(45°) sin(300)
2sin x + 3cosx = 4x Find the intersection of 1'; = 2 sin x + 3 cos x and Y2 = 4x :
..fi ../3 ..fi
= 16 I�tors�ctio� X=.B666�SBl
V=3.�66s�n
-6
131.
X=�.2191071
4 ./24 = 16-4 ./2 = �4 (16 - J2) _
Verifying equality: ( 16 - ..fi ) = 16 � ..fi ..fi . ../3 - ./2
�
x '" 0.87 . The solution set is {0.87} . sinx = lux Find the intersection of 1'; = sin x and Y2 = lu x :
Int�rs�ctio�
1
2 2
2 2
4
./2 ( ../3 - 1 ) 4
V=.79710�93
-2
x '" 2.22 The solution set is {2.22} . 133.
-3sin-1 x = sm x = --3 7r
•
2 = �2-4../3
7r
_I
2 · 2 ( - ../3 ) 16
( -� ) = - � The solution set is {-�} . x = sin
135.
fi=J3
( )
Using a half-angle formula: 300 . ° = sm. 2 sm15 = - C�S 300
l
�
�
2
I= 2 = 2 - J3 = fi=J3 2 Note: since 15° lies in quadrant I, we have sinl5° O .
J4
>
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exist. No portion of this material may be reproduced, in any form or by any mean s, without permi ssion in writing from the publi sher.
Chapter 8: Analytic Trigonometry
Ch apter 8 Test 1.
Let () = sec - I
4.
(�) . We seek the angle (), such
1(1 -1 (x)) = tan ( tan- I x) = x . Since the domain of the inverse tangent is all real numbers, we can directly apply this equation to get tan tan- I � = � .
that 0 � () � re and () "* 2're ' whose secant equals � . The only value in the restricted range with a secant of � is � . Thus, sec - I 2.
Let () = sin - I
( )
(�) = � .
5.
(-,,?) . We seek the angle () , such
3.
(
.
t l
)
sin - I sin 1 �re follows the form of the equation rl (I( x)) = sin- I ( sin (x)) = but because re re , we cannot l Ire . . ' I --,-- IS not m the mterva 2 2 5 directly use the equation. We need to fmd an angle () in the interval re , re for which sin l Ire = sin () . The angle S 2 2 l Ire is in quadrant I. The reference angle of S l Ire = sm. -re . S mce l Ire . re . -re IS. m. ' - IS - and sm S S S
[ ] x ,
6.
[_ ]
"lV ,
_
( %) sec [ COS- I ( -%)] = sec ()
Let () = COS- I
-
.
- 4' = --43 7. sin-I ( 0.382 ) :::; 0.392 radian 3
[-; , ; ] , we can apply the equation . l lre-) = 5re ' above and get sm. ( smS
the interval
-I
© 2008 Pearson Educati on , Inc . , Upper Saddle
y
y
cos ()
S
S
cot ( csc- I .JiO ) Since csc- I () = !... = 10 re2 -< () -< re2 ' let = J10 and = 1 . Solve for x: x2 + e = (J1O f x 2 + I = 10 x2 = 9 x=3 () is in quadrant I. Thus, cot ( csc -I J1O ) = cot () = � = T = 3 . r
. equals -� The that --re2 <- () <- -re2 ' whose sme 2 ' only value in the restricted range with a sine of re . ThUS, sm. - I -2 re Ji = -4' - 2 IS. -4' Ji
( �) follows the form
tan tan- I
. 39 1 959453 1
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exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publi sher.
Chapter 8 Test
8.
sec -I 1.4 COS- I ( _1.41_ ) 0.775 radian
13
�
=
sin B cos B . tan B + cot B = -cos B +- sin B sin2 B cos2 B sin B cos B + ---sin B cos Bsin 2 B + cos 2 B sin Bcos B 1 sin Bcos B 2 2 sin Bcos B 2 sin(2B) 2 csc(2B) =
9.
tan -I 3 1.249 radians �
a n·'� r� 2491345772
1 0.
coC I 5 tan -I =
(t)
�
=
0.197 radian
an ·'� � 1�h955598
1 1.
1 2.
14.
csc B + cot B sec B + tan B csc B + cot B csc B - cot B sec B + tan B csc B - cot B csc2 B - cot2 B (secB + tan B)( csc B - cot B) 1 (sec B + tan B)( csc B - cot B) sec B - tan B 1 · (secB + tan B)(cscB - cot B) sec B- tan B sec B - tan B ( sec2 B - tan2 B ) ( csc B - cot B) sec B - tan B csc B - cot B sin B Sill. B tan B + cos B · B . -cos B + cos B sin2 B + -- cos2 B = -cos B cos B sin2 B + cos2 B cos B 1 cos B sec B =
sin(a + ,8) tan a + tan,8 sin a cos fJ + cos a sin fJ sin a + - sinfJ -cos a cos fJ sin a cos fJ + cos a sin fJ sin a cos fJ + ---� cos a sin fJ ---� cos a cos fJ cos a cos fJ sina cos fJ + cosa sinfJ - sin a cos fJ + cosa sinfJ cos a cosfJ f3 + f3 _
s i n a cos
I
cos
a
sin
sin
cos a cos f3
a
cos f3 + cos a sin
cosa cos fJ sin(3B) sin( B + 2B) sin Bcos( 2B) + cos Bsin( 2B) sin B ( cos2 B - sin2 B ) + cos B· 2sin Bcos B sin Bcos2 B - sin3 B 2sin Bcos2 B 3sin Bcos2 B - sin3 B 3sin B ( 1 - sin2 B ) - sin3 B 3sin B - 3sin3 B -sin3 B 3sin B - 4sin3 B
f3
=
1 5.
=
=
=
=
+
=
S ill
=
=
=
=
© 2008 Pearson Educati on , [nc . ,
495
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permi s s i on in wri ting from the publi sher.
Chapter 8: Analytic Trigonometry
1 6.
1 7.
sin B cos B tan B - cot B cos B sin B sin B cos B tan B + cot B -cos B + - sin B sin2 B- cos2 B sin Bcos B sin 2 B + cos 2 B sin Bcos B sin2 B - cos2 B sin2 B + cos2 B -cos( 2B) 1 = - ( 2 cos2 B - 1 ) = 1 - 2 cos2 B cosI5° = cos(45° - 300) = cos 45° cos 30° + sin 45° sin 30° ./2 . _ J3 + _ ./2 . -1 =_ 2 2 2 2 = ( J3 + 1 ) J6 : ./2 or ±( J6 + ./2 )
1 9.
( )
�
18.
2 0.
.
<
(
1 - COS cos-I � 2
(
<
)
)
tan 2sin -1 161 Let B = sm. -I O6 . Then sin B = 161 and B lies in quadrant Since sin B = .lr = � 1 1 , let = 6 and = 1 1 , and solve for x: x2 + 62 = 1 12 x2 = 85 x = -J85 _ = 6-J85 tan B = .lx = _6 -J85 85 tan(2B) = 1 -2 tanB tan2 B 2 1.
tan 75° = tan ( 45° + 30°) tan 45° + tan 30° 1 - tan45°tan30° 1 + J33 1 - 1 . J33 3 + J3 3 - J3 = 3 + J3 ' 3 + J3 3 -J3 3 + J3 9 + 6J3 + 3 32 - 3 12 + 6J3 6 = 2 + J3
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ.
)
(
sm. Z1 cos- 1 s-3 Let B = cos-I i5 Since 0 B 7r2 (from the range of cos - I x ), sin � B = J¥
y
r
(�J 1 - ( 6�J
12-J85 � 1 - 36 85 12-J85 85 = 85 . 49 12-J85 49
--
496 All rights reserved. Thi s material i s protected under all copyright laws as they currently
exist. No portion of thi s material may be reproduced, in any form or by any means , without permission in writing from the publisher.
Chapter 8 Test
21.
cos ( sm. - 1 32 + tan "23 -I
)
23.
Let a = sm -23 and fJ = tan - 1 -23 . Then sin = 23 and tan fJ = �2 , and both a and fJ . Y 2 1 Ie · m , quadrant I . S'mce sm a = -I = - 1 e t 3' fj 2 = 2 and fj = 3 . Solve for + 22 = 32 2 +4 =9 2 =5 = .J5 .J5 . Thus, cos a = fj = 3 Since tan fJ = Yx2 = �2 , let x2 = 2 and Y2 = 3 . 2 Solve for x2 : 2 2 + 3 2 = r22 4 + 9 = r/ r/ = 13 r2 = JG Thus, sinfJ = Y2r = b . 2 ",13 Therefore, cos ( + fJ) = cos a cos f3 - sin a sin f3 .J5 ' 03 2 - 32 ' 03 3 =3 2.J5 - 6 3JG 203 (.J5 - 3 39 Let a = 75° , fJ = 15° . Since sin a cos fJ = sin( a + fJ) + sin( a - fJ) ] ' sin 75° cos 15° = sin (90°) + sin (60°) ] = 1+ .
-I
a
YI
XI :
XI
24.
cos 65° cos 20° + sin 65° sin 20° = cos ( 65° - 20°) = cos( 45°) .J2
•
2
X I
XI
25.
XI
---L X
{
. set IS. 3lr ' 3 2lr ' 3 4lr ' 3 5lr 5lr ' The solutIOn f) = 3
a
26.
)
22.
4sin 2 f) - 3 = 0 4sin 2 f) = 3 sm. 2 f) = '43 sin f) = -+ v'32 On the interval [0, 2lr ) , the sine function takes 2lr . The lr or f) = v'3 when f) = on a value of 2 3 3 4lr and . takes on a value of - 2 v'3 when f) = 3 sme
(; )
}
.
-3 cos - f) = tan f) -3 sin f) = tan f) sin f) ' O = -cos f) + 3 sm f)
(� )
0 = sin f) co f) + 3 1 sin f) = 0 or -cos f) + 3 = 0 cos f) = --31 On the interval [0, 2lr) , the sine function takes on a value of 0 when f) = 0 or f) = lr . The cosine function takes on a value of in the second and third quadrants when f) = lr - cos - I � and f) = lr + COS- I . That is f) ",, 1 .91 1 and f) "" 4.373 . The solution set is {O, 1.91 1, lr, 4.373} .
�[
�[ �[ '7] = ± ( 2 + v'3 )
-
2 + v'3 4
�
�
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Chapter 8: Analytic Trigonometry
27.
The second equation has no solution since -1 :$ sin 61 :$ 1 for all values of 61 . Therefore, we only need to find values of 61 between 0 and 2 7r such that sin 61 = . These will occur in the first and second quadrants. Thus, B = sin-I "" 0.253 and B = 7r - Sin-I "" 2.889 . The solution set is { 0.253, 2.889 } .
cos 2 61 + 2 sin 61 cos 61 - sin 2 61 = 0 ( cos 2 61 - sin 2 B ) + 2 sin 61 cos 61 = 0 cos ( 2B ) + sin ( 2B ) = 0 sin ( 2B ) = - cos ( 2B ) tan ( 261 ) = -1 The tangent function takes on the value -1 when its argument is 3: + k7r . Thus, we need 3 7r + k7r 2B = 4 7r 61 = 37r 8 +k 2
�
�
30.
61 = ; ( 3 + 4k ) On the interval [ 0, 27r ) , the solution set is 3 7r 7 7r 1 17r I 57r . 8' 8 ' 8 ' 8
}
{
28.
�
Using the average grade, we approximate the situation by considering the start of the route as the center of a circle of radius r = 1 5 (the length of the route). Then the end of the route is located at the point P(x, y) on the circle x 2 + y 2 = 1 5 2 = 225 . We have that tan 61 = �x = 0.079
61 = tan-I 0.079 "" 4.50 sin B = �r = L 15 y = 1 5 sin B = 1 5 sin 4S "" 1 . 1 8 The change in elevation during the time trial was about 1 . 1 8 kilometers.
sin ( B + I ) = cos B sin 61 cos 1 + cos 61 sin 1 = cos 61 sin 61 cos 1 + cos 61 sin 1 cos 61 cos B cos B tan 61 cos 1 + sin 1 = 1 tan 61 cos 1 = 1 - sin 1 1 - sin 1 tan u = -- cos l Therefore, 61 = tan -l I -cossinl l "" 0.285 or 61 = 7r + tan- l I - S in I "" 3.427 cos l The solution set is { 0.285, 3 . 427 } . /l
(
29.
(
)
)
4sin2 B + 7 sin B = 2 4 sin 2 61 + 7 sin 61 - 2 0 Let u = sin 61 . Then, 4u2 + 7u - 2 = 0 ( 4u - I )( u + 2 ) = 0 4u 1 = 0 or u + 2 = 0 4u = 1 u = -2 1 U =4 Substituting back in terms of 61 , we have sin 61 = or sin 61 = -2 =
-
�
© 2008
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exist. No portion of thi s material may be reprod uced , in any form or by any mean s , without permi ssion i n writing from the publi sher.
Chapter 8 Cumulative Review
Ch apter 8 Cumulative Review 1.
5.
-1 ±
3x 2 + x = ° -b "';"'---b24-a-c x = ---'-2a _- - l ± �e - 4(3)(- 1) 2 (3)
x
y = -2 -5 7.
3x + y 2 = 9 x-intercept: 3x + 0 2 = 9 ; (3, 0) 3x = 9 x=3 y-intercepts: 3 (0) + y 2 = 9 (0, -3) , (0, 3) y2 = 9 y = ±3 Tests for symmetry: x-axis: Replace y with -y : 3x + ( -y) 2 = 9 3x + y 2 = 9 Since we obtain the original equation, the graph is symmetric with respect to the x-axis. y-axis: Replace x with -x : 3 ( -x) + y 2 = 9 -3x + y 2 = 9 Since we do not obtain the original equation, the graph is not symmetric with respect to the y-axis. Origin: Replace x with -x and y with -y : 3 (_x) + (_y) 2 = 9 -3x + y 2 = 9 Since we do not obtain the original equation, the graph is not symmetric with respect to the origin.
© 2008 Pearson Educati on , Inc . ,
= 3ex - 2 Using the graph of y = eX , stretch vertically by a factor of 3, and shift down 2 units. y
-1±.J1+i2 6 -1±.JU 6 . set I. S {-I-.JU -1 + .JU} . The solutIOn ' 6 6 3.
Y
a.
y x3 =
5
x
Inverse function: y � y
=
5
-5
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exist. No portion of thi s material may be reproduced, i n any form or by any means , without permi ssion in writing from the publi sher.
Chapter 8: Analytic Trigonometry
b.
Y=
-5
eX
(
Inverse function: y = sin -I x y
y
- 1 , ;)
5
( I. I)
2 5
,)
-2
x
(+ !:l2 )
-5
2
-2
Inverse function: y = In x
y .,.
y
5
1
x
f - l (x )
=
sin- l "
I!-- (¥, 1)
�
-5 -5
y
y
=
eX
d.
y
= cos x , y
0 ::; x ::;
rc
3 x -1 -I
-4 c.
Inverse function: y = CO S -I Y
Y = sm x , - -2 -< x <- -2 .
x
(It,- I )
rc
rc
y
2
(I' I) 2
-1 -1
x
( 1 , 0)
3
\
\
,t f - 1 (x )
=
(0. 1 )
-1
2008 Pearson
x
y , 3
-2
©
X
y =
cos x
cos -1 x
('IT, - 1 )
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exist. No portion of thi s material may be reproduced, in any form or by any means, without permi ssion in writing from the publi sher.
Chapter 8 Cumulative Review
9.
:5,
r
r
r
e
=
:5,
r:
11.
Since the remainder is 0, x - ( -1) x 1 is a factor. The other factor is the quotient: 2X4 - 3x3 _ x 2 + 3x - 1 . Use synthetic division with 1 on the quotient:
cos (tan-I 2) Let B = tan-I 2 . Then tan B = � = T ' 7t 7t -"2 B "2 ' Let x = 1 and y = 2 . Solve for = + y 2 = e + 22 =5 = /5 is in quadrant cos(tan-1 2) = cosB = � = Js = Js . 1s = 1 2
x
2
1)2 - 3 - 1 3 - 1 2 -1 -2 2 -1 -2 1 0
2
2
r
Since the remainder is 0, x - 1 is a factor. The other factor is the quotient:
1.
2x3 - x2 - 2x + 1 .
Step 4: Factoring: 2x3 - x2 - 2x + 1 = x2 ( 2x - 1) - 1 ( 2x - 1) = ( 2x - 1) ( x2 - 1 ) = ( 2x - 1) ( x - I) ( x + 1 ) Therefore, I(x) = (2x - 1)(x - l) 2 (x + l) 2
I(x) = 2xs _ x4 - 4x3 + 2X2 + 2x - 1 Step 1 : I(x) has at most 5 real zeros. Step 2: Descartes' Rule of Signs There are 3 sign changes in I ( x) so there are 3 or 1 positive real zeros. I( -x) = -2x5 _ X4 + 4x3 + 2X2 - 2x - l There are 2 sign changes in I ( -x) so there are 2 or 0 negative real zeros.
a.
( �} X - l)2 (X + l)2
= 2 X-
Step 3: Possible rational zeros:
b.
p = ±I; q = ±1, ± 2; E.q = ±1, ±..!.2
Using the Bounds on Zeros Theorem: I(x) = 2(xS - 0.5X4 - 2x3 + X2 + x - 0.5) a4 = -0.5, a3 = -2, a2 = 1, al = 1, ao = -0.5 Max { 1, 1 -0.5 1 + 1 1 1 + 1 1 1 + 1 -2 1 + 1 -0.5 1 } = Max {I, 5} = 5
c.
1 + Max { 1 - 0.5 1 , 1 1 1 , 1 1 1 , 1 -2 1 , 1 -0.5 1 } =1+2 =3 The smaller of the two numbers is 3. Thus, every zero of f must lie between -3 and 3. Use synthetic division with -1 : -1)2 - 1 - 4 2 2 - 1 -2 3 1 -3 2 -3 -1 3 -1 0 © 2008
+
d.
The real zeros are -1 and 1 (both with multiplicity 2) and � (multiplicity 1) . x-intercepts: 1, �2 , -1 y-intercept: -1 The intercepts are (0, -1) , (1, 0) , (�, 0) , and (-1,0) I resembles the graph of y = 2xs for large I xl ·
Let 1'; = 2xs _ x4 - 4x3 + 2x2 + 2x - l 2
-2
2
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exist. No portion of thi s materi al may be reproduced, in any form or by any mean s, without permi ssion in wri ting from the publi sher.
Chapter 8: Analytic Trigonometry
e.
f.
Four turning points exist. Use the MAXIMUM and MINIMUM features to locate local maxima at (-1, 0) , (0.69,0.10) and local minima at (1, 0) , (-0.29, -1 .33) . To graph by hand, we determine some additional information about the intervals between the x intercepts: Interval (-00,-1) (-1,0.5) (0.5,1) (1,00) Test 2 ° 0.7 number -2 Value -45 "" 0. 1 27 -1 ofl Below Above Above Location Below x-axis x-axis x-axis x-aXI S ( -2, -45) (0, -1) (0.7,0.1) (2, 27) Point I is above the x-axis for (0.5, 1) and (1,00 ) , and below the x-axis for (-00, -1) and (-1,0.5) . Y 2
(0.69 , 0. 1 0)
x
-2 g.
( - 0.29, - 1.33)
I is increasing on (-00, -1) , (-0.29,0.69) , and (1,00) . I is decreasing on (-1,-0.29) and (0.69,1) .
502
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Chapter 9 Applications of Trigonometric Functions
Section 9.1
3.
13.
5. 7.
(�) "" 26.6°
True True
9. b
=
COS
cos A = � c ± (10°) = c
4 ",, _4_ "" 4.06 (10°) 0.9848 B = 90° - A = 90° _ 1 0° = 80° c=
5, B = 20° tan B = �
COS
a
tan (200) = -a5 a=
1 5.
5 ",, _5_ "" 1 3.74 tan (20° ) 0.3640
. B = -b S In c c=
= 5, A = 25° cot A = �a cot (25°) = Sb
a
b=
sin (200) = �c
5 cot ( 25°) "" 5 · (2.1445) "" 10.72
csc A = � a
5 "" 5 ,,,, 14.62 sin (20°) 0.3420
esc (25°)
--
a = 6, B = 40° tan B = �a tan ( 40°) = 6b b = 6 tan (400) ""
=
�
c = 5 esc (25°) "" 5 . (2.3662) "" 1 1 .83
A = 900 - B = 90° - 20° = 70° 11.
A = 1 0° tan A = Z;a
tan (10°) = � a = 4 tan (100) "" 4 · (0. 1 763) "" 0.71
tan O = 21 ' 0 < 0 < 90° 0 = tan- I
b = 4,
1 7.
c = 9, B = 20° . B = -b S In c sin (200) = b = 9 sin (200) "" 9 · (0.3420) "" 3.08
%
6 · (0.8391) "" 5.03
cos B = � c cos (40°) = i
cos E = � c cos (200) = �9 a = 9 cos (200) "" 9 · (0.9397) "" 8.46
c
6 "" _6_ ,,,, 7.83 cos (40°) 0.7660 A 900 - B = 90° - 40° = 50° c=
=
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503 River, NJ. All rights reserved. Thi s material i s protected under all copyright laws as they currently
exist. No portion of thi s material may b e reproduced, in any form or by any mean s , without permi ssion in writing from the publi sher.
Chapter 9: Applications of Trigonometric Functions
19.
a = 5, b = 3 2 c = a 2 + b2
tan (20°) = 30 30 "'" 82.42 feet = 30 "'" -0.3640 tan ( 20° ) The truck is traveling at 82.42 ft/sec, or 82.42 ft . 1 mile . 3600 sec "", 56.2 milhr . sec 5280 ft hr
b.
= 5 2 + 3 2 = 25 + 9 = 34 = 54 "'" 5.83 tan A = -a = -5 b 3 A = tan-I "'" 59.0°
x
x
c
(n
B = 90° - A "", 90° - 59.0° = 3 1 .0° 21.
a = 2,
c
��
2
c
2
ticket. Now, tan-I
29.
40'
(�)
"'"
23.6°
tan A = 300 50 = 6 A = tan-I 6 "", 80.5° The angle of elevation of the sun is about 80.5° .
27.
a.
J3 0 re a o
31.
s feet
x
(
<::>vt
E
c
Let y the height of the embankment. =
f) = 36°2 ' "'" 36.033° sin (36.0330) = -.L 5 1 .8 y = 5 1 .8 sin (36 . 033°) "'" 30.5 meters The embankment is about 30.5 meters high.
x
© 2008 Pearson Educati on, Inc . , Upper Saddle
B
y
Let represent the distance the truck traveled in the 1 second time interval. 30 tan (1 5°) = = 0 "'" 0 3 0 "'" 1 1 1 .96 feet tan 1 5° ) . 2 67 9 The truck is traveling at 1 1 1 .96 ft/sec, or 1 1 1 .96 ft . 1 mile . 3600 sec "", 76.3 miIhr . sec 5280 ft hr x
1 8 . 8° , so a ticket
tan f) = _1_ = 2 0.5 f) = 63.4° L.DAC = 40° + 63.4° = 103.4° L.EAC = 1 03 .4° - 90° = 13.4° Now, 90° -13.4° = 76.6° The control tower should use a bearing of S76.6'E.
B = 90° - A "'" 90° - 23.6° = 66 .4° The two angles measure about 23 . 6° and 66.4° . 25.
.
c
A = sin-I
"'"
Begin by finding angle f) = L.BAC : (see figure) D
B = 90° - A "", 90° - 23.6° = 66.4° = 5, a = 2 sm. A = 52
(��)
is issued if f) < 1 8 .8° .
(�)
c
23.
60 milhr or more. 60 mi . 528� ft . I hr = 88 ftlsec. hr nu 3600 sec If tan f) < , the trooper should issue a
=5
= a2 + b2 b 2 = - a 2 = 5 2 - 2 2 = 25 - 4 = 2 1 b = m "", 4.58 sin A = E. = � 5 A = tan-I "'" 23.6° c
A ticket is issued for traveling at a speed of
c.
504 River, NJ. All rights reserved. Thi s material i s protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion i n writing from the pub l i sher.
Section 9. 1 : Applications Involving Right Triangles
33.
Let x the distance between the buildings.
39.
=
1 45 1
Let f) the central angle formed by the top of the lighthouse, the center of the Earth and the point P on the Earth's surface where the line of sight from the top of the lighthouse is tangent to 362 '1es. the Earth. Note also that 362 fieet = 5280 =
!TIl
x
tan f) = 145 1 x x = 145 1 ",, 7984 tan (10.3°) The two buildings are about 7984 feet apart. 35.
The height of the beam above the wall is 46 - 20 = 26 feet. 26 = 2.6 tan f) = 10
Not to scale
(
f) = tan-I 2.6 "" 69.0° The pitch of the roof is about 69.0° . 37.
line segment drawn from the vertex of the angle through the centers of the circles will bisect f) (see figure). Thus, the angle of the two right triangles formed is A
=
�.
-I
Let x the length of the segment from the vertex of the angle to the smaller circle. Then the hypotenuse of the smaller right triangle is x + 2 , and the hypotenuse of the larger right triangle is x + 2 + 2 + 4 = x + 8 . Since the two right triangles are similar, we have that: x + 2 = -x+8 2 4 4(x + 2) = 2(x + 8) 4x + 8 = 2x + 1 6 2x = 8 x=4 Thus, the hypotenuse of the smaller triangle is 4 + 2 = 6 . Now, sin = = , so we have =
--
that:
% = sin-I G)
(%)
()
f) = 2 · sm. _1 '31
)
"" 0.337 15° f) = CO S -I 3960 +3960 362 / 5280 Verify the airplane information: Let fJ the central angle formed by the plane, the center of the Earth and the point P. 3960 fJ -- cos 1 . 77 169° 3960 + 1 0, 000/ 5280 Note that tan f) = dl and 3690 dl = 3960 tan f) d2 = 3960 tan fJ So, dl + d2 = 3960 tan f) + 3960 tan fJ "" 3960 tan(0.337 1 5°) + 3960 tan(1 .77 169°) "" 146 miles To express this distance in nautical miles, we express the total angle f) + fJ in minutes. That is, f) + fJ "" ( 0.337 1 5° + 1 .77 1 69° ) . 60 "" 126.5 nautical miles. Therefore, a plane flying at an altitude of 1 0,000 feet can see the lighthouse 120 miles away.
(
)
� �
"" 38 . 9 0
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505 All rights reserved. Thi s material i s protected under all copyright laws as they currently
exist. No portion of thi s material may be reprod uced, i n any form or by any means , wi thout permission in wri ting from the publi sher.
Chapter 9: Applications of Trigonometric Functions
Verify the ship information:
7.
p
True
9. c = 5,
B = 45° , C = 95° = A 1 800 -B - r = 1 80° - 45° - 95° = 40° sin A sin C a
c
sin 40° sin 95° 5 a 5 sin 40° "" a= sin 95° 3 . 23 Not to
sin B = -sin C -b c
scale
Let the central angle formed by 40 nautical . rrules then. = 40 = 2 ° 60 "3 cos(a - tJ) = 3960 3960 + x cos ((2 / 3)O - 0.337 150) = 3960 3960 + x (3960 + x ) cos ((2 I 3)° - 0.337 15°) = 3960 3960 3960 + x = cos ((2 I 3)° - 0.337 15°) 3960 - 3960 x= cos ((2 / 3)0 - 0.337 15°) "" 0.06549 miles a
=
sin 45° sin 95° b 5 5 sin 45° "" b= sin 95° 3 . 55
a
1 1.
sin A sin B
( !)
1 3.
Section 9.2
3.
For similar triangles, corresponding pairs of sides occur in the same ratio. Therefore, we can write: 15 x 5 or x = 2 3 2 The rrussmg Iength ' 15 2
..
S.
c
b
b = 7, A = 40° , B = 45° C = 1 80° - A - B = 1 80° - 40° - 45° = 95°
sin A = -sin B -b a
sin 40° sin 45° 7 a 7 in 40° ",, 6.36 a= � sm 45° sin C sin B c
IS - .
b
sin 95° sin 45° 7 c 7 in 95° c= � sm 45° ",, 9.86
sin A sin B sin C
-a- = -b- = -c-
© 2008 Pearson Education , Inc . , Upper S a d d l e River, NJ.
b
sin 85° sin 45° c 3 3 sin 85° "" 4 . 23 c= sin 45°
"" 346 feet Therefore, a ship that is 346 feet above sea level can see the lighthouse from a distance of 40 nautical miles.
sin A cos B - cos A sin B
a
sin 50° sin 45° a 3 3 sin 50° . a= sin 45° ",, 3 25 sin C = -sin B --
"" (0.06549 mi) 5280
1.
b = 3, A = 50° , C = 85° B = 1 800 -A - C = 1 80° - 50° - 85° = 45°
S06 All rights reserved. This material i s protected under all copyright l aws as they currently
exist. No portion of thi s material may be reproduced, in any form or by any means , without permi ssion in writing from the publi sher.
Section 9.2: The Law of Sines
15.
b = 2, B = 40°, C = 1 00° A = 1 800 -B - C = 1 80° - 40° - 1 00° = 40°
21.
=
sin A sin C a c sin 1 1 0° sin 30° a 3 l l O° "" 3sin a= 5. 64 sin 30° sin C sin B c b si n 30° sin 40° 3 b 3sin 40° "" 3. 86 b= sin 30°
sin A sin B = a b sin 40° sin 40° a 2 2sin 40° = 2 a = sin 40° sin C sin B c b sin 1 00° sin 40° 2 c 2sin 1 00° "" 3 06 c= . sin 40° --
17.
--
A = 40°, B = 20°, a = 2 C = 1 80° - A - B = 1 80° - 40° - 20° = 1 20°
23.
A = 40°, B = 40°, c 2 C = 1 80° - A - B = 1 80° - 40° - 40° 100° =
=
sin A sin C a c sin 40° sin 1 00° a 2 2sin 40° "" 1 . 3 1 a = sin 1 00° sin B sin C b c sin 40° sin 1 00° = b 2 40° "" 1 . 3 1 2sin b= sin 1 00°
sin B sin A =-a b sin 20° sin 40° =--2 b 2� in 20° "" 1 . 06 b= sm 40° sin C sin A a c sin 40° sin 1 20° --= c 2 2sin 1 20° "" 2. 69 c sin 40° --
---
19.
A = 1 1 0°, C = 30° , c = 3 B = 1 80° - A - C = 1 80° - 1 1 0° - 30° 40°
B = 70°, C = 1 0°, b = 5 A = 1 800 -B - C = 1 80° - 70° - 1 0°= 1 00° sin B sin A = -a b sin 1 00° sin 70° a 5 l 00° "" 5sin a= 5. 24 sin 700 sin C sin B = c b sin l 0° sin 70° c 5 1 0° "" sin 5 c= 0. 92 sin 7 00 --
--
--
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Chapter 9: Applications of Trigonometric Functions
25.
= 3, b = 2, A = 50° sin B sin A a b ( 50° ) .. sin B = sin _--0-----:. 2 3 2sin ( 50° ) � 0.5 1 07 sin B = 3 B = sin-I ( 0.5 1 07 ) B = 30.7° or B = 149.3° The second value is dis carded because A + B > 1 80° . C = 1 800 -A - B = 1 80° - 50° - 30.7° = 99.3° sin C sin A c a sin 99.3° sin 50° c 3 3s�n 99.3° c= � 3 . 86 sm 50° One triangle: B � 30.7° , C � 99.3° , c � 3 .86 a
29.
__
27.
= 4, b = 5, A = 60° sin B sin A b a sin B sin 60° 4 5 5 Si �60° "" 1 .0825 sin B a
=
There is no angle B f or which sin B > 1 . Theref ore, there is no triangle with the given measurements. 31.
b = 5, c = 3, B = 1 00° sin B sin C b c sin 1 00° sin C 3 5 1 00° sin 3 sin C = � 0.5909 5 C sin-I ( 0.5909 ) C = 36.2° or C = 1 43 .8° The second value is dis carded because B + C > 1 80° . A = 1 80° - B - C = 1 80° - 1 00° - 36.2° = 43.8° sin B sin A b a sin 1 00° sin 43 .8° 5 a 5sin 43 .8° "" 3 a= .5 1 sin 1 00° One triangle: A � 43 .8° , C � 36.2° , a � 3 . 5 1
b 4, c = 6, B = 20° sin B sin C b c sin 20° sin C 4 6 . 6sin 20° "" 0 5 1 30 sm C = . 4 C = sin-I ( 0.5 1 30 ) C I = 30.9° or C = 1 49. 1 ° 2 F or both values, B + C < 1 80° . Theref ore, there are two triangles . A I = 1 800 - B - C I = 1 80° - 20° - 30.9° = 1 29. 1° sin A I sin B = -b al sin 20° sin 1 29. 1° 4 al 4sin 1 29. 1° � 9 . 08 aI = sin 20° A = 1 80° - B - C = 1 80° - 20° - 149. 1° = 1 0.9° 2 2 A sin sin B = 2 b sin 20° sin 1 0.9° a2 4 4sin 1 0.9° � 2.2 1 a2 = sin 20° Two triangles : A I � 1 29. 1° , C I � 30.9° , al � 9.08 or A ",, 1 O · 9° , C � 1 49. 1° , a2",,2.21 2 2 =
--
=
--
--
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Section 9.2: The Law of Sines
33.
a
= 2, c = 1, C = 100° sinC sinA c
35.
37. a.
a
sin 100° sin A 2 1 sin A = 2 sin l000""1.9696 There is no angle A for which sin A > 1. Thus, there is no triangle with the given measurements. = 2, c = 1, C = 25° sinA sinC
c
A
a
a
C = 1 80°- 60°- 55°= 65° sin 55° sin 65° 150 1 5 �sin 55° ""135.58 miles = sm65° sin 60° sin 65° b 1 50 150sin b = sin 65°60° ""143. 33 miles Station Able is about 143.33 miles from the ship, and Station Baker is about 135.58 miles from the ship. 135.6 ""0.68 hours t =-= r 200 0.68 hr . 60 rr;!n ""41 minutes a
c
a
sin A sin 25° 2 sin A = 2sinI 25° ""0 . 8452 A = sin -I (0.8452 ) AI = 57.7° or A = 122.3° For both values, 2A + C < 1 80°. Therefore, there are two triangles. b.
sinBI = sinC bl sin 25° 1 sin97.3° "" 2. 35 bI = I sin 25°
--
Find C ; then use the Law of Sines (see figure): B
--
C
39.
a
--
LQPR = 1 80°-25°= 155° LPQR = 1 80°- 155°-15°= 10° Let c represent the distance from P to Q.
sin 1 5° sin 10° c 1000 1000sin = sin 1 0°1 5° ""1490. 48 feet c
sinB2 = -sinC -b2 c sin 32.7° sin 25° b2 Two triangles: AI ""57.7°, BI ""97.3°, bl "" 2.35 or A2 ""122.3°, B2 ""32.7°, b2 "" 1.28
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Chapter 9: Applications of Trigonometric Functions
41.
Let h the height of the plane and x the distance from Q to the plane (see figure). =
45.
=
R
Note that LK OS = 57.7° - 29.6° = 28.lo (See figure) 79.40 �-, K _________
I I I
29.6'
s
� I I I
sin 40° sin 105° 1000 x sin 40° x = 1000 sin l05° ""665.46 feet
o
. 350 = h = h -; 665.46 h = (665 .46 ) sin 35°""381 .69 feet The plane is about 381.69 feet high. Let h the height of the tree, and let x = the distance from the first position to the center of the tree (see figure). SIll
43.
=
Using the Law of Sines twice yields two equations relating x and h. Equation 1 : sinh30° = sinx60° 60° x = hsin sin 30° sin 20° = --sin 70° Equation 2: --h x + 40 70° - 40 x = hsin sin 20° Set the two equations equal to each other and solve for h. hsin 60° = hsin70° - 40 sin 30° sin 20° sin sin 60° h sin 30° _ sin 700 20° = -40 -40 h= sin 60° sin 70° sin 30° sin 20° ""39.4 feet The height of the tree is about 39.4 feet.
--- -----
(
47.
From the diagram we find that LKS O = 1 800-79.4° - 62.1° = 38.5° and LOKS = 180° -28.1 ° -38S = 1 13.4° . We can use the Law of Sines to find the distance between Oklahoma City and Kansas City, as well as the distance between Kansas City and St. Louis. sin 38.5° sinl13.4° OK 461.1 461.1sin38.5° OK = ""3 12.8 sin 1 13.4° sin 28.1 ° sin 1 13.4 ° 46l.l KS KS = 46l.lsin 28.1 ° "" 236.6 sin 1 13.4° Therefore, the total distance using the connecting flight is 3 1 2.8 + 236.6 = 549.4 miles. Using the connecting flight, Adam would receive 549.4 -46l. l = 88.3 more frequent flyer miles. Let h the perpendicular distance from R to PQ, (see figure). =
R
---
)
sinR = sin 60° -123 184.5 sin 60° "" 0.5774 sinR = 1231 84.5 R = sin-1 (0.5774 ) "" 35.3° LRPQ = 1 80°- 60°- 35.3°""84.7° h sin 84.r = - 1 8 4.5 h = 184.5 sin 84.7°"" 1 83.7 1 feet 510
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Section 9.2: Th e Law of Sines
49.
A = 1 80°- 140°= 40°; B = 1 80°- 13 5°= 45°; C = 1 80°-40°- 45°= 95° AI I I I
sin 95° sin 42.5° DE 0.125 sin95° DE 0.125 sin 42.5° 0. 1 84 miles The approximate length of the highway is AD + DE + BE = 1.295 + 0.184 + 1. 165 2.64 mi. Determine other angles in the figure: =
1400
�
�
2mi:
I I I I
51.
B 1350 sin 40° sin 95° 2 BC 2sin 40° 1.290 mi BC = sin 95° sin 45° sin95° 2 AC 2sin 45° 1.420 mi AC = sin 95° BE = 1.290 - 0.125 = 1. 1 65 mi AD = 1.420 - 0. 125 = 1.295 mi For the isosceles triangle, LCDE LCED 1 80°-2 95° = 42. 50 �
---
=
Using the Law of Sines: sine 65 + 40)° sin 25° L 88 88sin 25° 38 5' hes L = sin 105° The awning is about 38.5 inches long.
---
�
=
53.
�
.
me
=
Let h height of the pyramid, and let x distance from the edge of the pyramid to the point beneath the tip of the pyramid (see figure). =
=
"
"
,,
"
, , "" " "
"
,,,
"
h
"
" , " ,.:'�olo_ .... �6:3.7:
100ft
'
loon
x
Using the Law of Sines twice yields two equations relating x and y: sin 46. 27° sin(90° -46. 27°) Equation 1 : h x+ 100 (x + 1 00) sin 46. 27° = h sin 43.73° xsin 46.27° + 1 OOsin 46 .27° = hsin 43.73° -1 OOsin 46.27° x hsin 43.73° sin 46.27° sin(90° 40.3°) sin 40. 3° Equation 2: h x + 200 ( x + 200 ) sin 40. 3°= hsin49.7° xsin 40.3°+200sin 40. 3°= hsin 49. 7° 200sin40.3° x = hsin 49.7°sin 40. 3°
--
= --
-------
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Chapter 9: Applications of Trigonometric Functions
Set the two equations equal to each other and solve for h. h sin 43 .73° - 1 00 sin 46.27° h sin 49.7° - 200 sin 40.3° sin 40.3° sin 46.27° h sin 43.73° · sin 40.3° - 1 00 sin 46.27° · sin 40.3°= h sin 49.7° · sin 46.27° - 200 sin 40.3° · sin 46.27° h sin 43 .73° · sin 40.3° -h sin 49.7° · sin 46.27°= 1 00 sin 46.27° · sin 40.3° - 200 sin 40.3° . sin 46.2r 100sin 46.27° · sin 40.3° - 200 sin 40.3°·sin 46.27° h= sin 43 .73° · sin 40.3° - sin 49.7° · sin 46.27° ", 449.36 f eet The current height of the py ram id is about 449.36 f eet. 55.
Using the Law of Sines:
57.
M "Cury B
x
Eartl
-- - ----15° ___-
-
-
--
-
---
57,910,000 -
C
Sun
r
149,600,000
sin B 57, 9 1 0, 000 149, 600, 000 . =149,..:.---.:.--600, 000 · sinI5° sm B 57, 9 1 0, 000 14, 960 · sin 1 5° 579 1 50 B = sin -I 14,960, sinI ", 4 1 .96° 579 1 or B", 1 3 8.04° C", 1 800 - 4 1 .96° - 1 5°= 1 23 .04° or C", 1 80° - 1 38 .04° - 1 5°= 26 .96° sin C sin 1 5° 57, 9 1 0, 000 x 57, 9 1 0, 000 . sin -Cx = -'----'--sin 1 5° 57, 9 1 0, 000 · sin 1 23.04° sin 1 5° "' 1 87, 564, 95 1 .5 km or 57, 9 1 0, 000 · sin 26 .96° x= sin 1 5° "' 1 0 1, 439, 834.5 km So the possible distances betw een Earth and Mercury are approx im ately 1 0 1 ,440,000 km and 1 87,600,000 km.
(
Since there are 36 equally spaced cars, each car 360° is separated by --= 1 0° . The angle between 36 the radius and a line segm ent connecting . . 1 70° consecutIve cars IS --= 8 5° ( see fiIgure) . If 2 we let = the radius of the wheel, we get sin 1 0° sin 85° r 22 22 sin 85° = ", 1 26 sin 1 0° The length of the diam eter of the wheel is approx im ately d = 2r = 2 ( 1 26) = 252 feet.
r
)
59.
b sin A sin B sin A - sin B a -b a -=--- = -- ---=--,--c c c sin C sin C sin C
(¥) cos (�) sin ( 2 · �) 2 sin ( ¥ ) cos ( �) 2 sin ( � ) cos ( � ) sin ( ¥) cos ( % - � ) sin ( � ) cos ( �) . A - B) . (C) sm ( sm sin ( � ) cos ( � ) sin ( ¥) sin a (A - B) ] cos ( lc) cos ( �)
2 sin
-2-
"2
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Section 9.3: Th e La w of Cosines
61.
a-b a - b = c_ a+b a+b c
Section 9.3
_
[
Sin � (A -B) cos
cos
(� c)
1.
]
3. 5.
[� (A -B) ]
(� )
sin c
[�
sin (A -B)
]
7. 9.
( )
sin � c
c
Cosines Cosines False a = 2, c = 4, A = 45° b 2 = a 2 + c 2 - 2ac cos B b 2 = 2 2 + 4 2 - 2 . 2 . 4 cos 45°
b
=
�20 - 8J2
�
2.95
a 2 = b 2 + c 2 - 2bc cos A 2bc cos A = b 2 + c 2 - a 2 2 2 _ a2 cos A= b + c 2bC 2 +4 2 2 2 20. 7025 cos A= 2.95 2 ( 2.95 )( 4 ) 23 .6 � 28.70 A = cos-1 20.7025 23 .6
1t
63 - 65.
�( X2 - Xl )2 + ( Y2 -Yl )2
J2 = 20 - 1 6 ·2 = 20 - 8 J2
(� ) cos [� (A -B)] = tan [1 (A-B) }an (1 c ) = tan [1 (A -B) ] tan [1 ( - (A + B) ) ] = tan [1 (A-B) }an [�_( A ;B ) ] A B = tan [� (A -B) ] cot ( ; ) tan [1 (A -B) ] tan [ � (A +B) ] cos
d=
_
(
11.
Answers will vary.
)
a= 2, b= 3 , C = 95° c 2 = a 2 + b 2 - 2ab cos C c 2 = 2 2 +3 2 - 2 · 2 ·3 cos 95°= 13 - 1 2 cos 95° c = .J13 - 1 2 cos 95° � 3 .75 a 2 = b 2 + c 2 - 2bc cos A b2 + c2 _ a 2 cos A = 2b C 2- 2 cos A= 32 + 3 .75 2 1 9.0625 2(3 )(3 .75 ) 22.5 A = cos-1 1 9.0625 � 3 2 . 10 22.5
(
)
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Chapter 9: Applications of Trigonometric Functions
13.
a =6, b = 5, c = 8 a2 = b2 + c2 - 2bc cos A C2 -a2 52 + 82 - 62 = 53 cos A = b2 +2bc 2(5)(8) 80 A
=
COS
-I
(��)
::::
19.
48.5°
b2 = a2 + c2 - 2ac cos B c2 _ b2 cosB= a2 +2ac 82 _ 52 75 cosB = 62 2(+ 6)(8) 96 75 )::::38. 60 B = cos _1 96
(
21.
C = 1 80° -A-B::::1 80° -48.5° -38.6°=92.9° a = 9, b = 6, c = 4 a2 =b2 + c2 - 2bccosA c2 _ a2 62 + 42 _ 92 29 cos A = b2 +2bc 48 2(6)(4) A=
COS-
I
(-��):::: 127.2°
b2 =a2 + c2 -2ac cos B c2 _ b2 92 + 42 _ 62 61 cosB = a2 +2ac 2(9)(4) 72 B= cos -1 � 72 ::::3 2. 10 C = 180° -A -B::::1 80°-127.2° -32. 1°=20.7° a = 3, b = 4, C = 40° c2 = a2 +b2 - 2ab cos C c2 = 32 + 42 - 2·3· 4 cos 400= 25 - 24 cos 40° c =,J2 5 - 24cos 40°:::: 2.57 a2 =b2 + c2 - 2bccosA c2 _ a2 42 + 2.572 _ 32 13.6049 cos A = b2 +2bc 20.56 2( 4)(2.57) A = cos 13.6049 20.56 ::::48 . 60 B = 1800 -A - C 1 80°-48.6° -40°=91.4°
23.
( )
17.
I -
(
I
(
)
-I
(
)
I COS-
(
COS
----
15.
b = 1, c = 3, A = 80° a2 = b2 + c2 - 2bc cos A a2 = 12 + 32 -2·1·3cos 800 = 10- 6 cos80° a = .J 10 - 6cos 800::::2.99 b2 =a2 + c2 - 2ac cos B c2 - b2 2.992 + 32 _ 1 2 16.9401 cos B = a2 +2ac 17.94 2(2.99)(3) 1 6.9401 ::::19.2° B= 17.94 C = 1 80° -A -B::::180° -80°-19.2°= 80.8° a = 3, c=2, B = 1 10° b2 = a2 + c2 -2accosB b2 = 32 +22 -2 . 3 . 2 cos l loo = 13 - 12cos 1 10° b = .J 13 - 12cos 1 10°::::4.14 c2 = a2 + b2 - 2ab cos C b2 _ c2 32 + 4.142 _ 22 22. 1396 cosC = a2 +2ab 24.84 2(3)( 4. 14) 1396 ::::27.00 C = cos 22.24.84 A = 1800 -B -C::::1 80°-1 10°-27.0°= 43.0° a = 2, b = 2, C = 50° c2 = a2 + b2 - 2abcos C c2 = 22 + 22 - 2·2·2 cos 500 = 8 - 8cos 50° c =.J8 - 8 cos 50°:::: 1.69 a2 = b2 + c2 - 2bccosA c2 - a2 22 + 1.692 - 22 2.8561 cos A = b2 +2bc 6,76 2(2)(1.69) 2.8561 :::: 65.0° A= 6.76 B =180° -A C::::1 80° -65.0° -500 = 65.00
)
-
)
,.,
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Section 9.3: Th e La w of Cosines 25.
a = 12, b = 13, c =5 a 2 = b 2 + c 2 - 2bc cos A 2 c2 _ a 2 1 32 + 5 2 _ 1 22 cos A = b +2bc 2(13)(5)
C }'" 67.4°
A cos-I 5300 =
2
50 130
=
2(10)( 5)
C = 1 80° - A B ""1 80° -97.9° -52.4°
33.
c
=
35.
-
( ) 2
2
B = cos-I C
(:�) ""62.2°
= 29.7° Find the third side of the triangle using the Law of Cosines: a = 1 50, b = 35, C = 1 10° c 2 = a 2 + b 2 - 2ab cos C = 1 50 2 + 35 2 -2·150·35 cosllO° = 23, 725 - 10, 500 cos 1 1 0° =,)23, 725 - 10, 500cos 1 1O° ""1 65 The ball is approximately 165 yards from the center of the green. After 1 0 hours the ship will have traveled 150 nautical miles along its altered course. Use the Law of Cosines to find the distance from Barbados on the new course. a = 600, b = 150, C = 20° c 2 = a 2 + b 2 - 2ab cos C = 6002 + 1 50 2 - 2· 600· 150cos 20° =382, 500 - 1 80, 000 cos 20° c = ,)382, 500 - 1 80, 000 cos 20° ""461.9 nautical miles Use the Law of Cosines to find the angle a. opposite the side of 60 0: _ a2 b 2 + c 2;cos A = --::-: -2bc 2 + 46 1 .9 2 - 600 2 124, 148.39 cos A = 1 502(150)( 461.9) 138,570 124, 148.39 ""153.60 A = cos-I _ 138, 570 The captain needs to turn the ship thr ough an angle of 1 80 0 - 153.60 = 26.40 -
B cos-I 0.5 = 60° C = 180° A -B "" 1 80° -60° -60° = 60° a = 5, b = 8, c = 9 a 2 = b 2 + c 2 - 2bccos A 2 c 2 _ a 2 8 2 + 9 2 _ 5 2 120 cos A = b +2bc 2(8)(9) 144 A = cos -I 120 144 "" 33 . 60 2
61 100
1 02 + 5 2 _ 82
a 2 + c2 _ b2 cos B = --2:--- ac
a +c _b cos B = --2:--- ac
11 80
(-!�) ""97.9°
2
A = cos-I 0.5 =60°
29.
a = lO, b = 8, c = 5 a 2 = b 2 + c 2 - 2bc cos A 2 c 2 - a 2 8 2 + 5 2 _ 1 02 cos A -_ b +2bc 2(8)(5) A cos-I
a +c _b -- cos B -2 ac B = cos -I 0 = 90° C = 1 800 -A -B "" 1 80 ° -67.4° -90°= 22.6° a = 2, b = 2, c =2 a 2 =b 2 + c 2 - 2bc cos A =
27.
2
31.
42 90
= 180° - A B ""1 80° -33.6° - 62.2° = 84.2° -
(
)
•
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Chapter 9: Applications of Trigonometric Functions
b.
37. a.
miles 30. 8 hours are = 461.9 15nautical 1mots required for the second leg of the trip. (The total time for the trip will be about 40.8 hours . ) Find x in the figure: t
x2 = 5002 + 1002 - 2(500)(100)cos80° = 260, 000 - 1 00, 000 cos 80° = .J260, 000-100, 000 cos 80° ",492.6 ft The guy wire needs to be about 492.6 feet long. b. Use the Pythagorean Theorem to find the value ofy: y2 = 1002 + 2502 = 72, 500 y = 269.3 feet The guy wire needs to be about 269.3 feet long. 41. Find x by using the Law of Cosines:
"'
X
,------..,. 2nd
Home "-''''--:�--� I sl
x2 = 60.52 + 902 - 2(60.5)(90) cos45° = 1 1, 760.25 - 10, 980 = 1 1, 760.25 - 5445 ..fi = �1 l, 760.25-5445 ..fi ",63.7 feet It is about 63.7 feet from the pitching rubber to first base. Use the Pythagorean Theorem to find y in the figure: 902 + 902 = (60.5 + y)2 16,200 = (60.5 + y)2 60.5 + y = .J 1 6, 200 ",127.3 y",66.8 feet It is about 66.8 feet from the pitching rubber to second base . Find B in the figure by using the Law of Cosines: + 63.72 - 902 382.06 cos B = 60.52 7707.7 2(60.5)(63.7) 382.06 B = cos (_ 7707 . 7 ",92.80 The pitcher needs to turn through an angle of about 92.8° to face first base. Find x by using the Law of Cosines:
( '?)
X
b.
c.
-\
39. a.
x2 = 4002 + 902 - 2( 400)(90) cos ( 45°) = 168, 100-36,000..fi x = �1 68,100- 36, 000.J2 ",342.33 feet It is approximately 342.33 feet from dead center to third base . 43. Use the Law of Cosines:
)
c2 = a2 + b2 - 2abcosC = 102 + 102 -2(1O)(10)cos(45°) = 100 + 100-200
[�l
= 200 -loofi = 100 ( 2 - fi) C = �100 ( 2 - fi ) = 10�2 - fi ",7.65 The footings should be approximately 7.65 feet apart. 516 © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 9.4: Area of a Triangle
45.
Use the Law of Cosines to find the length of side d: d2 = r2 +r2 -2· r· r· cos () = 2r2 -2r2 cos () = 2r2 (1-cos ()) = 4r2 C - �os (} ) = 2rl- �os () = 2rsin (�) If s = r(} is the length of the arc subtended by (), then d < s , and we have 2rsin (�) < r(} or 2 sin ( �2 ) < () Given sin () = 2 sin �2 cos �2 and cos�2 1 , we have sin(} 2sin�2 < () . Therefore, sin () < () for any angle ()> 0 .
S ection 9.4
False a = 2, c =4, B = 45° = �acsinB = �(2)(4)sin45°"" 2. 8 3 7. a = 2, b = 3, C = 95° = �abSinC = �(2)(3)sin95°"" 2.99 9. a = 6, b = 5, c = 8 s = 21 (a+b+c) =21 (6+5+8) = 219 = .Js(s -a)(s -b)(s -c) = C;)(i)(�)(%) = �3��1 "" 14. 9 8 11. a = 9, b = 6, c = 4 s = 21 (a+b+c) = 21 (9+6+4) = 219 = .Js(s -a)(s -b)(s -c) = C;)( �)(i)C21 ) = �1 ��3 "" 9. 5 6 13. a = 3, b = 4, C = 40° = � absin C = �(3)(4)Sin 40°"" 3. 8 6 15. b = 1, c = 3, A = 80° = �bcsin A = �(I)(3)sin800"" 1.48 17. a = 3, c= 2, B = 110° = �acsin B = �(3)(2)sin 1 10°"" 2. 82 19. a= 12, b = 13, c = 5 s = 21 (a+b+c) = 21 (12 +13+5) = 15 = .Js(s -a)(s -b)(s -c) = �(15)(3)(2)(10) = .)900 = 30 3. 5.
K
K
.
�
47.
�
K
. C l -cosc sm-= 2 2
K
_(a2 -2ab+b2 _c2) 4ab
49
c)( a -b -c) = "/-(a -b+4ab (a -b+c)(b+c-a) 4ab (2s -2b)(2s -2a) 4ab 4(s-b)(s-a) 4ab (s-a)(s-b) ab - 51. Answers will vary.
K
K
K
K
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Chapter 9: Applications of Trigonometric Functions
Find the area of thec =lo75t using Heron's Formula: a= 100, b= 50, 225 s = "21 (a+b+c) = 21 (100+50+75) = 2 = �s(s-a)(s-b)(s-c) ( 2�5 )( 2� )( 1 �5 )( 7� ) 52,734,375 16 ",, 1 815.4 6 Cost = ($3)(1815.4 6) = $5446. 3 8 37. Divide home plate into a rectangle and a triangle.
35.
a = 2, b = 2, c = 2 1 =3 s = 21 (a+b+c) = 2(2+2+2) = �s(s-a)(s-b)(s-c) = �(3)(1)(1)(1) = .fj "" 1. 73 23. a = 5, b = 8, c = 9 s = 21 (a+b+c) = 21 (5 +8+9) = 11 = �s(s-a)(s-b)(s-c) = �(11)( 6)(3)(2) = .J396 "" 19.90 sin -A = sinB . we know 25. From the Law of Smes a b- . asinB Solvmg for b, so we have that b = sm-A- ' Thus, sin B )sm. C = 21 ab sm. C = 21 a ( asinA a2 sin BsinC 2sin A 27. A= 40°, B = 20°, a = 2 C = 180° A-B = 180° - 40° - 20° = 120° 22 sin 20° . sin 120° "" 0.92 = a2 sinBsinC 2sin40° 2sinA 29. B= 70°, C= 10°, b=5 A= 180°- B -C = 180°- 70°- 10°= 100° n AsinC 52 sin 100°· sin 1 0° "" 2. 2 7 = b2 si2sinB 2sin 70° 31. = 110°, C = 30°, c = 3 B = 180°- A-C = 180°- 110°- 30°= 40° AsinB 32 sin1 �Oo. sin40° ",,5 .44 = c2 sin2sinC 2sm30° 33. Area of a sector = 21 r2 (} where () is in radians. () = 70 ' 1801t = 1871t A = L2 82 . 71t18 = 1121t9 A = � . 8·8 sin 70° = 32 sin 70° = 11:1t -32sin700 "" 9. 03 As
21.
K
K
K
.
.
17"
= = (17)(8. 5) = 144. 5 in2 U sing Heron's formula we get = �s(s-a)(s - b)(s -c) 1 12 +17) = 20.5 1 +b+c) = -(12+ s = -(a 2 Thus,2 =� = �(20. 5 )(8. 5 )(8. 5)(3. 5 ) = .J5183. 9 375 "" 72. 0 sq. in. So, + = = 144. 5 +72. 0 = 216. 5 in2 The area of home plate is about 216. 5 in2 . ARectangle
------
ATriangle
K A
ATotal
K
Sector
Triangle
egment
lw
Ani,ogle
-
K
8.5"
8.5"
K
(20.5)(20.5 -12)(20.5 -12)(20.5 17) -
ARectangle
ATriangle
ft2
fe
fe
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Section 9.4: Area of a Triangle
The sector.area is the sum of the area of a triangle and a = �r . r sine1t - B) = �r2 sine1t - B) = 2"1 r2 B As = � r2 sin ( 1t - B) +�r2 B As = �r2 (sin(1t - B) +B) = � r2 ( sin 1tCOSB -cos 1tsin B+B) = �r2 ( 0 . cos B -(-1) sin B + B) = � r2 (B sin B) 41. Use Heron's formula: a = 87 , b = 190 , = 173 =-21 ( a + b+ ) =-21 ( 87 +190+173) = 225 = � ( )( - b )( ) = �225 ( 225 -87 )( 225 -190)( 225 -173) = �225 ( 138 )( 35 ) ( 52 ) = �56,511,000 "" 7517.4 The feet ofbuilding groundcovers area. approximately 7 517. 4 square Area /).OAC = 2"1 1oc1·1ACI 43. = L2 Io1c1.1A1CI . = 2'1 cos asma . a = 2'1 smacos Area /).OCB = 2'1 1 OCI·IBCI I 2 .1oOBcI1.1BC = 1.2 ,1OB1l lOBI = �IOB12 cos fJsin fJ = �IOBI2 sin fJcos fJ
39.
c.
ArriangJe ec!or
haded region
d.
e.
+
I OB Isin(a+ fJ)
cs o a . sm( cs o
c
c
s
K
s
s -a
s
Area /).OAB = 2"11 BD1·1 OAI = 2"1IBDI'1 = 21 " 1 OBI·II BOBDII = �IOB I sin(a + fJ) lIOC i cos A = OA I = ' oc1.1 OBI =IOBI cos BlOCI 1 IocI lOBI Area /).OAB = Area /).OAC+Area /).OCB � � � 12 -fJ a + =
sn i a cs o a+ I OB
sn i fJ c o s fJ
. cs o 2a . fJ) =sma c osa+--2-smfJ c osfJ cs o fJ
cs o a . . c osfJ . sm( a+ fJ) =--sma c s o a+- sm fJ c osfJ cs o a cs o fJ
s -c
a = a 45. The grazing area must be considered in sections. Region AJ represents three-fourth of a circle with radius 1 00 feet. Thus, AJ = � 1t ( I00l = 75001t "" 23, 5 61.94 ft2 Angles are needed to find regions � and A3 : (see the figure) sine
a.
+ fJ)
sin
cs o fJ +c s o asin fJ
�c B
10 ft
A
In MBC, LCBA = 45°, AB = 10, AC =90. Find LBCA: sin LCBA sinLBCA 90 10 sin 45° sinLBCA 90 10 sinLBCA = IOsin45° 90 "" 0. 0786 LBCA = sin -J COS�450 ) "" 4.51° LBAC = 180°-45°- 4. 5 1°= 130.49°
b.
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Chapter 9: Applications of Trigonometric Functions
LDAC = 130.49°- 90°= 40.49° A3 = i (l O)(90)sin40.49 ", 292.19 fe The angle for the sector � is 90°- 40.49°= 49. 5 1° . A2 = i ( 90 ) 2 ( 49. 5 1 . 1 ;0 ) '" 3499. 66 fe Since the cow can go in either direction around the bam, both � and � must be doubled. Thus, the total grazing area is: 23,561.94+2(3499. 66) +2(292.19) ", 31,146 47. Perimeter: = a+b+c = 9+10+17 = 36 Area: s =-21 (a+b+c) =-21 ( 9+10+17) = 18 = �s( s -a)( s-b)( s-c) = �18( 18 -9 )( 18 -10)( 18 -17) = �18( 9 )(8 )( 1 ) ..11 296 = 36 Siequal, nce thetheperimeter and area are numerically given triangle is a perfect triangle. Perimeter: = a+b+c = 6+25 +29 = 60 Area: s =-21 (a+b+c) =-21 ( 6+25 +29) = 30 = �s( s-a)( s-b)( s-c) = �30 (30 -6)( 30 -25 )( 30 -29 ) = �30( 24 )( 5 )( 1 ) = ..13 600 = 60 Since thetheperimeter and area are numerically equal, given triangle is a perfect triangle.
49.
0
K
.
_
51.
P
=
53.
P
fi
_
__ .
K
b.
K
--
fe
a.
We know = i ah and = i abSin C , whic h means h = bsinC . From the Law of Sines, we asinB sinB know sinA a = -b- ' so b = SIn-A- . There ore, asisinn AB )Slll. C - a sin B sin C . h - (-sin A . -A ·Slll. -B C·Slll C 2 2 cosC 2_ cot 2 = = ----;===r=== Slll-2 c· ac C
C ab r ac (s-a)(s-b) c ab(s-a)(s-b)(s-c)2 r abc2 (s-a)(s-b) 2 = �. s-c = �r �(S-C) c2 r c s-c r = Area MOQ+ Area MOR + Area !1QOR 1 +-rb 1 +-ra 1 = -rc 21 2 2 =-r2 ( a+b+c) = rs Now, = �s(s-a)(s -b)(s-c) , so rs �s(s-a)(s-b)(s-c) -b)(s-c) r = --'-�s(s--a)(s -----s K
K
K
=
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Section 9.S: Simple Harmonic Motion; Damped Motion; Combining Wa ves
S ection 9.5 1. 3. 5. 7. 9. 11. 13.
21.
d(t) = cos ( 2t) e
-tlTC
y
= 2:r4 = :r2 simple harmonic ; damped d = - Scos ( ttt) d = -6cos ( 2t) d = - Ssin (7tt) d = -6sin ( 2t) d = Ssin(3t) Simple harmonic S meters 27t seconds 3 � oscillatio n/second 27t d = 6cos(7tt) Simple harmonic 6 meters 2 seconds .!. oscillatio n/second 2 d= -3Sin (k t) Simple harmonic 3 meters 47t seconds 1 47t oscillation/second d = 6+2cos(27tt) Simple harmonic 2 meters 1 second oscillatio n/second lsi s· '
2'1l'
23.
d( t)
-tI21t =e
X
cost
y
a.
b. c.
d.
15.
25.
j ( x ) = x+cosx y 2'7T
a.
b. c.
d.
17.
-'IT
a.
27.
b.
y
c.
d.
19.
j ( x ) = x-sin x
_ _
a.
b. c.
d.
-21T'
I
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Chapter 9: Applications of Trigonometric Functions
29.
j ( x) = sinx+cosx
37. a.
y 2
d
d
(0. 8)2 = - e- 0.St/2 (10) cos [ (-23n)2 - --4(10)2 J = _5 e-0.St/2 0 cos [ 49n2 - 0.40064 ] t
5
t
b. -2
31.
j ( x ) = sinx+sin ( 2x)
39. a.
y
b.
c.
y
=
d
d
(0. 7) 2 = _10 e-0.7t/2 (2 5) cos [ ( 2 n)2 - 4(25) 2 0.49 ] = _10 e-0.7t/50 cos [ 425n2 - 2500
d. t
5
t
J e.
b. 41. a.
o IE+-t"+-+-I-+'-t-'I-++
35. a.
d
d
20
sin(2x)
-2
33. a.
Damped motionfactor with aofbob0. 7 ofkg/sec. mass 20 kg and a damping 20 meters leftward
b.
The displacement of theis bob of the second oscillation aboutat 18.the3start 3 meters. The displacementt/ of0 the bob approaches zero, since e-{J·7 4 0 as Damped motionfactor with ofa bob0 . 6 ofkg/sec. mass 40 kg and a damping 30 meters leftward �
t � 00 .
c.
(0.6)2 - - 18 e-0.6 t/2 (3 0) cos [(-2n)2 - --4(30/ J 0. 3 6 ] = _18 e- 0.6 t/60 cos [ �4 3600
o 11+-+-+++4+..Ir--l...... 1
t
_
-20
t
d.
b.
e.
The displacement of theis bob of the second oscillation aboutat28the.4start 7 meters. The displacement6t/Sof the bob approaches zero, since e-{J· O 0 as �
t
� 00 .
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Section 9.5: Simple Harmonic Motion; Da mped Motion; Combining Wa ves
43. a.
b.
Damped motion with a bob of mass 1 5 kg and a damping factor of 0.9 kg/sec. 1 5 meters leftward
-15
d.
The displacement of the bob at the start of the second oscillation is about 1 2.53 meters.
c.
The displacement of the bob approaches zero, since e-O·9t130 � 0 as t � 00 .
O�320 f\ 1.25
The maximum displacement is the amplitude so we have a = 0.80 . The frequency is given by (1)
f = - = 520 . Therefore,
(1)
Irlt�rstctio� X=.�SlS216S
= 1 0401Z" and the
I I
f = - = 440 . Therefore,
OJ = 8801Z" and the 21Z" movement of the tuning fork is described by the equation d = 0.0 1 sin ( 880m ) .
rs�o
r\
[\
�
51.
L"'-
1. � ."---. .L
I�t. � �=2.1BBSSQ6
-1.25
;c(
.
25
"-
I / V=.�
V=.�
3.2 0
��
.....
�
I !
"il'lirvlu X=2.966�S23
-1.25
�
"'-
V= -3699522
Let r; = sin ( 21Z" ( 852 ) x ) + sin ( 21Z" ( 1209 ) x ) .
-1
On the interval O:=;t :=; 3 , the graph of V touches the graph of y = e-t1 3 when t = 0, 2 . The graph of V touches the graph of y = _ e-t1 3 when t = 1, 3 .
53.
3.2
'"
2.5
b.
3.2
'--.
1 25
rs�
V
1
'I \
Irlt'-':rs ol'l X=1.75�9n� -
1 . 25
o -"-
"-
-1.25
-1.25
(1)
�
IMt � X=.666�B6�2 v= -.�
V=.�
-1.25
The maximum displacement is the amplitude so we have a = 0.01. The frequency is given by
49. a.
1 25
\ \
21Z" motion of the diaphragm is described by the equation d = 0.80 cos ( 1 040m ) .
47.
We need to solve the inequality -0.4< e-tl 3 cos ( m )< 0.4 on the interval o:=; t :=; 3 . To do so, we consider the graphs of y = -0.4, y=e-tI 3 cos ( 1Z" t ) , andy = O.4 . On the interval O:=;t :=;3 , we can use the INTERSECT feature on a calculator to determine that y = e -t1 3 cos ( m ) intersects y = 0.4 when t '" 0.35, t '" 1 .75 , and t '" 2. 1 9, y = e -t1 3 cos ( 1Z"t ) intersects y = -0.4 when t '" 0.67 and t '" 1 .29 and the graph shows that -0.4< e-tl 3 cos ( 1Z"t )< 0.4 when t = 3 . Therefore, the voltage V is between -0.4 and 0.4 on the intervals 0.35 < t< 0.67 , 1 .29
15
c.
45.
c.
CBL Experiments
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�
3.2
Chapter 9: Applications of Trigonometric Functions
55.
sin x . Let >-; = -x
Chapter 9 Review Exercises 1.
= 10, B = 20° sl.n B = -b sin 200= �10 b = 10sin 200", 3. 4 2 cosB = c!!.. cos 200= !!...10 . a = lOcos 20°", 9. 4 0 A = 900-B = 90°-20°= 70° b = 2, c = 5 c2 = a 2 + b2 a 2 = c 2 _ b 2 = 52 - 22 = 25 -4 = 21 a = 51 '" 4. 5 8 sin B = c!: = �5 B = sin -I (�) '" 23. 6° A = 90° - B '" 90° -23.6° = 66.4° A= 50°, B = 30°, a = 1 C = 180°- A-B = 180°- 50°- 30°= 100° sin A sin B b a sin 50° sin 30° 1 b b = 1 �in 30° '" 0. 6 5 sm50° sin C sin A c a sin 100° sin 50° c c = 1s�n100° '" 1. 2 9 sm50°
c
c
smx approaches 1. As x approaches 0, -x 57. = C;} in x Y
3.
y
= C12 } in X
o1
-0.06 Y
= ( :3 } in x
5.
0.05
-0.015
Possible observation: As x gets larger, the graph of = (:n } in x gets closer to = 0 . y
y
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Chapter 9 Review Exercises
7.
A = 1 00° , c = 2, a = 5 sin C -sin A -= c a sin C sin 1 00° 2 5 .SIn = 2 sin l00° "" 0 . 3939 C 5 . I 2 sin l00 0 C = SIn 5 C "" 23.2° or C "" 1 56.8° The value 1 5 6.8° is discarded because A +C>1 80° . Thus, a ",, 23.2° . B = 1 80° - A -C "" 1 80° - 1 00° - 23.2° = 56.8° _
sin B b sin 56.8° b
9.
(
13.
A = cos-II 0° No triangle exists with an angle of 0'. =
)
15.
sin A a sin 1 00° 5
a = 3, c = 1, C = 1 1 0° sin C sin A c a sin 1 1 0° sin A 1 3 3 sin 1 1 0° "" sin A 2.8 1 9 1 1 No angle A exists for which sin A > 1 . Thus, there is no triangle with the given measurements.
17.
)
1 3.3 824 13.92
a = 5, b = 3, A = 80° sin B sin A a b sin B sin 80° 3 5 3 sin 80° "" sin B = 0 . 5909 5 0 . B = SIn-I 3 sin 80 5 B "" 36.2° or B "" 1 43 . 8° The value 1 43.8° is discarded because A +B > 1 80° . Thus, B "" 36.2° . C = 1 80° - A - B "" 1 80° - 80° - 36.2° = 63 .8°
(
a = 3, c = l, B = 1 00° b 2= a 2+c 2- 2ac cos B = 3 2+ e - 2·3 · 1 cos 1 00° = 1 O - 6 cos l 00° b = .J 1 O - 6 cos 1 00° ",, 3 .32
)
sin C c sin 63.8° c
sin A a sin 80° 5 5 sin 63 .8° "" 4 56 . c= sin 80°
a 2= b 2+c 2- 2bc cos A 2+c 2_ a 2 3 .32 2+1 2- 3 2 3.0224 cos A = b 2bc 6.64 2(3.32)(1) 3 .0224 A = cos -I 6.M "" 62 . 90 ----
(
a = 1, b = 3, C = 40° c 2= a 2+b 2- 2ab cos C = f +3 2-2·1 · 3 cos 40° = 1 0 - 6 cos 40° c = .J 1 O - 6 cos 40° "" 2.32 a 2= b 2+c 2- 2bc cos A 2 2 2 2+2.32 2- e cos A = b +c - a 3 2 bc 2(3)(2 .32) 1 3 .3824 "" 1 6 00 . A = cos -I 1 3 .9 2
(
=
11.
a = 2, b = 3, c = 1 a 2= b 2+c 2- 2bc cos A
)
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Chapter 9: Applications of Trigonometric Functions
19.
a = l b =.!.2' c =i3 a2 = b2 + c2 - 2bc cos A
---
27 37
C
( )
A = cos 27 37 ""39 . 60 -I
23.
c2 _ b2 cos B = a2 +2ac
91 96
----
( )
B = cos -1 91 96 ""1 8 . 60 21.
a = 3, A = 10°, b = 4 sinB = -sinA -a b sinB sin 10° 3 4 4sin10° sinB = 3 ""0 . 23 15 B = sin 4 sin3 100 BI ""13.4° or B ""1 66.6° For both values, A2 + B < 1 80° . Therefore, there are two triangles. C1 = 1 80°- A-BI ",, 1 80°- 10°-13.4°""156.6° sinA = sinC1 -a c1 sin 10° sin 1 56.6° 3 c1 si 3 � 1 56.6° "" 6.86 c = smlO0 .
-I
(
sin A = sinC2 c2 a sin 10° = sin 3.4° -c2 3 3 sin3.4° 2 = sin 10° ""1.02 Two triangles: BI "" 13.4°, C1 ""156.6°, c1 ""6.86 or B2 ""1 66.6°, C2 ""3.4°, c2 "" 1.02 c = 5, b = 4 , A = 70° a2 = b2 + c2 - 2bc cos A a2 = 42 + 52 -2·4·5 cos 700 = 41 - 40 cos 70° a = .J41 - 40 cos 70° ""5.23 c2 = a2 + b2 - 2abcosC b2 _ c2 cosC = a2 +2ab 5.232 + 42 - 52 18.3529 41.48 2(5.23)(4) C = cos 1 8.3529 41 .48 ""64.00 B = 180°- A - C 1 80°- 70°- 64.0°""46.0° -I
(
)
""
25.
)
27.
29.
--
a = 2, b = 3, C = 40° K = � absin C = � (2)(3)Sin 400"" 1 .93 b = 4, c = 10, A = 70° K = � bcsin A = � (4)( I O) sin 70°"" 1 8.79 a = 4, b = 3, c = 5 s = '21 (a + b + c) = '21 (4 + 3 + 5) = 6 K = �s(s -a)(s -b)(s ) = �( 6)(2)(3)(1) = 56 = 6 -c
31.
1
a = 4, b = 2, c = 5 s = '21 (a + b + c) = '21 (4 + 2 + 5) = 5.5 K = �s(s -a)(s -b)(s - c) = �(5.5)(1 .5)(3.5)(0.5) = ..114.4375 ""3.80
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 9 Review Exercises
33.
35.
A = 50° , B = 30° , a = 1 C = 180° - A -B = 180° - 50° - 30° = 100° a2 sinBsinC K = 2sinA
sin 60° sin 55° 3 BC 3 in � 60° � 3.17 mi BC = sm55° sin 65° sin 55° 3 AC in 3 � 65° � 3.32 mi AC = sm55°
Let B = the inclination (grade) of the trail. The "rise" of the trail is 4100 - 5000 = -900 feet (see figure).
4100 ��900n ft
BE = 3.17 - 0.25 = 2.92 mi AD = 3.32 - 0.25 = 3.07 mi For the isosceles triangle, LCDE = LCED = 180°; 55° = 62.so
900 sinB = 4100 900 �12. 70 B = sin-I 4100 The trail is inclined about 12.7° from the lake to the hotel. Let h = the height of the helicopter, x = the distance from observer A to the helicopter, and r = LAHB (see figure).
( )
37.
H
A r
41.
100ft = 180°- 40°- 25°= 1 1 5° B
sin 40° sin115° x 100 100sin 40° � 70. 92 leet x= sin l15°
c
�
39.
sin55° sin 62.so DE 0.25 55° 0 DE = ?sm5 sin 62.5° � 0.23 miles The length of the highway is 2.92 + 3.07 + 0.23 = 6.22 miles. Find the third side of the triangle using the Law of Cosines: a = 50, b = 60, C = 80° c2 = a2 + b2 - 2ab cos r = 502 + 602 - 2·50·60 cos 80° = 6100 - 6000 cos 80° = .J6100 -6000 cos 80° � 71.12 The houses are approximately 71.12 feet apart. Construct a diagonal. Find the area of the first triangle and the length of the diagonal:
43.
h_ sin 250 = !!:...x �_ 70.92 h� 70.92 sin25°� 29.97 feet The helicopter is about 29.97 feet high. a = 180°-120°= 60°; f3 = 180°- 115°= 65°; r = 180°- 600 - 65°= 55°
KI = � . 50 ·100· sin 400�1606.969 fe d2 = 502 +1002 - 2·50·100cos40° = 12, 500 - 10, 000 cos 40° d = .J12, 500 - 10, 000 cos 40° � 69.566914 feet
©
2008
527 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 9: Application s of Trigonometric Function s
Use the Law of Sines to find sin sin 1 00° 20 69. 5 66914 sm - 69.20sinl00° 5 66914 ' 20sinl000 ) ", 16.447° B = SIn -I ( 69. 566914 180°- 100° -16.447°= 63. 5 53° Find the area of the second triangle: K2 = i·20 . 69. 5 66914 . sin63. 5 53°'" 622. 8 65 ft 2 The cost of the parcel is approximately 100(1606.969 +622. 865) = $222,983.40 . Find angle and subtract from 80· to obtain B :
B
.
B
_
A ",
45.
A MB
6. 5 = '83 cos a = 5'2 3 a = COS- I (�) ", 67. 98° f3 = 180°- 67. 9 8°= 112. 0 2° z = 6. 5 tan 67. 9 8°", 16.07 in = 2. 5 tan 67.98°", 6.18 in The arc length on the top of the larger pulley is: 6 . 5 ( 112. 02· 1;0 ) ", 12. 7 1 inches. The arc length on the top of the smaller pulley is 2. 5 ( 112. 02 · 1 ;0 ) ", 4. 89 inches. The of tangency is z+ distance 16. 07between +6.18 = the22. 2poi5 nintsches. The 2(12.length 7 1 4.of89the 22.belt2 5)is about: 79. 7 inches. d= -5 COS(�t) d = 2cos(4t) Simple harmonic 2 feet '2 seconds 3. oscillation/second y
e.
80·
Y '"
A
+
49.
B
tan = 4010 = 4 = tan- I 4 ", 76. 0° 80°- 76. 0°", 4. 0° The bearing is S4.0oE . Letin thex figure the hypotenuse of the24 larger triangle -x is right below. Then the hypotenuseareofsimilar. the smaller triangle. The two triangles LA MB
51.
LA MB
=
a.
e ",
47.
+
b. c.
=
d.
53.
7t
7t
d = -3sin[�t] Simple harmonic 3 feet 4 seconds i oscillation/second
a.
b.
6. 5 = -x 2. 5 24-x 2. 5x = 6. 5 (24 -x) 2. 5x = 156 -6. 5 x 9x = 156 x = -523 24-x =-203
c.
d.
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d = _ 1 3 e-o. 6S f I 2( 2S ) cos
55. a.
d = _ 1 3e-o.6 S f I SO COS
(
[() �
2
_
2
( 0 . 65 ) 2 4(25 ) 2
� - 0.4225 t 4 2500
b.
t
Chapter 9 Test
]
Chapter 9 Test 1.
J
Let a = the angle formed by the ground and the ladder.
iJ' 05 ft
Ot+--t-+-t-T-'-+-t-+--t-H
. 10.5 Then smA 12 10.5 "" 61 . 00 A SIn. - I -12 The angle formed by the ladder and ground is about 61.0° . Let A = the angle of depression from the balloon to the airport. = --
=
Damped motion with a bob of mass 30 kg and a damping factor of 0.5 kg/sec. 20 meters leftward
57. a.
b.
2.
c.
W�l600ft ,
t � 00 .
Y
=
tan A =
CO
(}
y
�=�
26400 44 A = tan - l 4 = 1 .30
�
2 S( 2x ) + sin �
AP
Note that 5 miles = 26,400 feet .
It approaches zero, since e -0.5 f I 60 0 as
e.
5 mi
MSFC
The displacement of the bob at the start of the second oscillation is about 19.5 1 meters.
d.
59.
( )
3.
(; )
The angle of depression from the balloon to the airport is about 1.3° . Use the law of cosines to find a: a2 b2 + c2 -2bc cos A (17)2 + (19)2 -2(17)(19) cos 52° "" 289 + 361-646(0.616) 252.064 a ../252.064 "" 1 5.88 Use the law of sines to find B . =
O :S; x :S; 2n"
=
=
3
=
x
-V y = 2
.
a x
sin A
= SIn 2
1 5.88 sin 52°
cos (2x)
sin B
b
sin B 17 sin B
= � (sin 52° ) "" 0.8436 1 5 .88
Since b is not the longest side of the triangle, we know that B < 90" . Therefore, B sin- I (0.8436) "" 57.5° C 1 80° -A - B 1 80°-52° - 57.5° 70.5° =
=
©
2008
=
=
529 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material i s protected under all copyright laws a s they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 9: Applications of Trigonometric Functions
4.
Use the Law of Sines to find b : b a
6.
sin A sinbB 12 sin 41 ° sin 22° b = 1 2 · sin 22° � 6.85 sin 41 °
C = 1 80° - A - B = 1 800 - 4 1 ° - 22° = 1 1 7° Use the Law of Sines to find c : c a
sin A sin e 12
c
sin 41 ° sin 1 1 7° c=
5.
1 2 · si n 1 1 7 °
sin 41 °
� 1 6.30
Use the law of cosines to find A . a 2 = b 2 + c 2 - 2bc cos A 8 2 = (5) 2 + (1 0) 2 - 2(5)(1 O) cos A 64 = 25 + 1 00 - 1 00 cos A 1 00 cos a = 6 1 cos A = 0. 6 1 A = cos-I (0.61) � 52.4 1 ° Use the law of sines to find B . a b sin A sin B 8 5 sin 52.4 l o sin B
7.
8.
sin B = �( sin 52.4 1 ° ) 8 sin B � 0.495 Since b is not the longest side of the triangle, we have that B < 90° . Therefore B = sin -I (0.495) � 29.67° C = 1 800 - A - B = 1 80° - 52.4 1 ° - 29.67° = 97.92°
A = 5 5° , C = 20° , a = 4 B = 1 80° - A - C = 1 80° - 55° - 20° = 105° Use the law of sines to find b. sin A -sin B -= b a sin 55° sin 1 05° 4 b 4 s�n l 05° � 4.72 b= sm 55° Use the law of sines to find c. sin C sin A a c sin 20° sin 55° 4 c 4 �in 20° � 1 .67 c= sm 55° a = 3, b = 7, A = 40° Use the law of sines to find B sin B -sin A -= a b sin B sin 40° 7 3 7 . sin 40° 1 4998 sm B = � . 3 There is no angle B for which sin B > 1 . Therefore, there is no triangle with the given measurements. a = 8, b = 4, C = 70° c 2 = a 2 + b 2 - 2ab cos C c 2 = 8 2 + 4 2 - 2 . 8 . 4 cos 70° = 80 - 64 cos 70° c = ..!80 - 64 cos 70° � 7.62 a 2 = b 2 + c 2 - 2bc cos A b 2 + c 2 _ a 2 4 2 + 7.62 2 - 8 2 cos A = --,--2bc 2(4)(7.62) 1 0.0644 � 80.5° A = c os - \ 60.96
(
)
1 0.0644 60.96
B = 1 80° -A - C � 1 80° -80S -70° = 29S 9.
a = 8, b = 4, C = 70° K
� = � (8)(4) Sin 70° � 1 5 .04 square units
= ab sin C
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Chapter 9 Test
10.
a = 8, b = 5, c = lO s = -21 ( a + b + c ) = -21 ( 8 + 5 + 10 ) = 1 1.5 K = �s(s - a)(s - b)(s - c) = �1 1 . 5(1 1 .5 - 8)(1 1 . 5 - 5)(1 1 .5 - 1 0) �1 1.5(3.5)( 6.5)(1.5) = ...)3 92.4375 :::; 19.81 square units
13.
:::;
14.
We can find the area of the shaded region by subtracting the area of the triangle from the area of the semicircle. Since triangle ABC is a right triangle, we can use the Pythagorean theorem to find the length of the third side.
Then, sin A
sin 40° OB AB sin 70° sin 40° 5 AB 40° 3 . 420 sin 5 AB = sin 70° Now, AB is the diameter of the semicircle, so the 3 . 420 ra d'lUS I.S 2- = 1 . 710 . 1 2 = -1 7l' ( 1.710 ) 2 :::; 4.593 sq. umts ASemicircle = -7l' 2 r 2 ATriangle = .!.2 absin(LO)
a2 + b2 = c2 a2 + 62 = 82 a2 = 64 - 36 = 28 a = 58 = 2J7
:::;
The area of the triangle is
A = � bh = � ( 2J7 ) ( 6 ) = 6J7 square cm .
.
The area of the semicircle is A
=
.!. JlT 2 .!. 7l' ( 4 ) 2 87l' square cm .
2
=
2
=
Therefore, the area of the shaded region is 87l' - 6J7 :::; 9.26 square centimeters . 12.
= .!.2 (5)(5)(sin 40°) :::; 8.035 sq. units ATotal = Asemicircle + ATriangle :::; 4.593 + 8.035 :::; 12.63 sq. units
Begin by adding a diagonal to the diagram. 5
�pper .
= (4.2)2 + (3.5)2 - 2(4.2)(3.5) cos 32° :::; 17.64 + 12.25 -29.4(0.848) 4.9588 c = "/4.9588 2 . 2 3 Madison will have to swim about 2.23 miles. Since f.. OAB is isosceles, we know that LA = LB = 1 80°2- 40° = 700 =
=
11.
Use the law of cosines to find c: c2 = a2 + b2 -2ab cos C
15.
= � (5)(I I)(Sin 72°) :::; 26.15 sq. units
By the law of cosines,
d2 = (5)2 + (1 1)2 - 2(5)(1 1)(cos 720) = 25 + 121 - 1 10(0.309) = 1 12.008 d = "/1 12.008 :::; 10.58
Using Heron's formula:
s = 5x + 6x2 + 7x = 1 8x = 9x 2 K = �9x(9x - 5x)(9x - 6x)(9x - 7x) = "/9x · 4x·3x · 2x = �216x4 = ( 6F6 ) x2
Thus,
(6F6 )x2 = 54F6 x2 = 9 x=3 The sides are 1 5, 1 8, and 21.
Using Heron's formula for the lower triangle,
s = 7 + 8 +2 10.58 = 12.79 Alower . = �12.79(5.79)(4.79)(2.21) = "/783.9293 :::; 28.00 sq. units Total Area = 26.15 + 28.00 = 54.15 sq. units
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exist. No portion of this material may be reproduced, in any form or by any means, wi thout permission in writing from the publisher.
Chapter 9: Applications of Trigonometric Functions
16.
Since we ignore all resistive forces, this is simple harmonic motion. Since the rest position ( t = 0) is the vertical position (d = 0) , the equation will have the form d = a sin(wt) .
Interval Test Number g (x) Pos./Neg. 6 Positive -2 -oo < x < - 1 -4 Negative -1 < x < 4 0 6 4 < x < oo 5 Positive The domain of f ( x ) = �x 2 - 3x - 4 is { x l x � -1 or x � 4 } . 5.
Now, the period is 6 seconds, so 21r 6=
[ ( ;)]
Y = -2 COS ( 2X - 1r ) = -2 COS 2 X Amplitude: I A I = 1 - 2 1 = 2 21£ T=Period: = 1r 2 1r Phase Shift: t =
W
21r w=6 1r w = radians/sec 3 From the diagram we see
y 3
W
2
= sin 42° 5 a = 5 ( sin 42° )
!::
x
o ( �t ) t 346 . sin ( � ) .
Thus, d = 5 ( sin 42 ) sin
d
,
d.
or
-3
7. a.
y
Chapter 9 Cumulative Review 1.
3x 2 + 1 = 4x 3x 2 - 4x + 1 = 0 ( 3x - 1 ) ( x - 1 ) = O 1 x = - or x = 1 3 The solution set is
3.
0
b.
{l, 1} .
= eX , 0 � x � 4
60
4
0
y
= sin x , O :<:;: x :<:;: 4
1 .5
0
f ( x ) �x 2 - 3x - 4 f will be defined provided g ( x ) = x 2 - 3x - 4 � 0 . x 2 - 3x - 4 � 0 ( x - 4 )( x + 1 ) � 0 x = 4, x = - 1 are the zeros.
4
=
- 1 .5
c.
Y = eX s in x , O :<:;: x :<:;: 4 0
10
-50
�
\
4
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Chapter 9 Cumulative Review d.
9.
a
y = 2x + sin x , 0 � x � 4 8
11.
R is in lowest terms. The x-intercepts are the zeros of p(x) : 2X2 - 7x - 4 = 0 ( 2x + 1 ) ( x - 4) = 0 or x - 4 = 0 2x + 1 = 0 1 x=4 X = -2 The y-intercept is 2 . OZ - 7 . 0 - 4 - 4 4 = = R(O) = OZ + 2 . 0 - 1 5 - 1 5 15 ·
= 20, c = 1 5 , C = 40° sin C sin A c
2x2 - 7x - 4 (2x + l)(x - 4) = XZ + 2x - 1 5 (x - 3)(x + 5) p(x) = 2x z - 7x - 4; q(x) = xz + 2x - 1 5; n = 2; m = 2 Domain: { x l x 7: - 5, x 7: 3 } R(x) =
a
sin 40° sin A 20 15 20 sin 400 . sm A = 15 20 sin 400 A = Sln - 1 15
---. ( )
A1 ",, 59.00 or A ",, 1 2 1 .00 2 For both values, A + C < 1 80 0 . Therefore, there are two triangles.
2(-x)z - 7(-x) - 4 2x z + 7x - 4 which = ( _ x) z + 2 (-x) - 1 5 XZ - 2x - I 5 is neither R(x) nor -R(x) , so there is no symmetry. The vertical asymptotes are the zeros of q(x) : XZ + 2x - 1 5 = 0 (x + 5) (x - 3) = 0 x + 5 = 0 or x - 3 = 0 x=3 x = -5 Since n = m , the line y = 2 is the horizontal asymptote. R(-x) =
--
sin C sin B1 -= c b1 sin 40° sin 8 1 . 0° 15 b1 1 5 sin 8 1 .00 "" 23.05 b1 "" sin 40°
R(x) intersects y = 2 at
sin B2 sin C --= c b 2 sin 40° sin 1 9.00 15 b z 1 9 .00 "" 7 sin 5 1 .60 b "" z sin 40°
----=-2x2x2 +-2x7x--1 45 = 2
(�� , 2) . since:
---
2X2 - 7 x - 4 = 2 ( x 2 + 2 x - 1 5 )
2x2 - 7x - 4 = 2x2 + 4x - 30 - l lx = -26 26 X=11 Graphing utility:
Two triangles: A1 "" 59.0° , B1 "" 8 1 .0° , b1 "" 23 .05 or A ",, 1 2 1 .0° , B2 ",, 1 9.0° , b ",, 7.60 . z z
8
: /·
.::.�.. 1 8 - I 0 1���"'7f'::o...� ...... ---/ .
-4
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Chapter 9: Applications of Trigonometric Functions
Graph by hand: Interval
( -00, -5)
Test
Value
number
off
'" 1 2.22
-6
(-5, -0.5) -1
-0.3 125
( -0.5,3)
°
'" 0.27
(3, 4)
3.5
", -0.94
(4, 00)
5
0.55
)1
Location
Point
Above
(-6, 1 2 . 22)
x-axis Below x-axis Above x-axis Below x-axis Above x-axis
(-1, -0.3 125) (0, 0.27) (3.5, -0.94) (5, 0.55)
y 8
:(01 .1. )
- - - -
15 I - T
x
-8 13.
log 3 ( X + 8) + log 3 X = 2 log 3 [(x + 8)( x)] = 2
(x + 8)(x) 32 x2 + 8x 9 x2 + 8x - 9 0 (x + 9)(x - l) 0 x -9 or x 1 x -9 is extraneous because it makes the = =
=
=
=
=
=
original logarithms undefined. The solution set is {I} .
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Chapter 1 0 Polar Coordinates; Vectors Section 1 0. 1 1.
25.
Y 2 -2
2 -2
3. 5.
( - 2, 135") •
4
27.
x
(3 , - 1 )
b r
29.
pole, polar axis
7.
(-�, -1)
9.
True
4
\j)(-2,.-7i). o
-7i
31.
11. A 13. e
17. A 19.
(3 , 90°) 90° 0
21.
a.
r >
b.
r
c.
r >
•
0
23.
-2n
L 6
() < °
()
�
- 2n o
(6. �) .
�
()
3:7t '9 1 .. .
•
•
0,
° �
33.
• (-2, 0 )
0
(5, _ �n) (-5, 53n < 0, < 2n J (5, 8; ) 0, 2n < 4n 0
15. B
a.
r >
b.
r
c.
r >
0,
(-2, 3:7t) �
()
<°
(2, - 2n)
2n ( - 2, n) , 0, 2n < 4n (2 2n)
< 0,
°�
()
�
<
()
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Chapter 1 0: Polar Coordina tes; Vectors
35.
(1 ,
43.
� 2
�)
(1, 3;) r > O, r 0, 0 2n (-1, 3;) r 0, 2n < 4n (1, 5;) - 2n � O < 0 �
<
b.
>
c.
_
0<
�
45.
0
37.
r 2n < (3, - 54 ) r 0, 0 2n (-3, 7:) (r 0, 2n < 41t (3, 1 :1t ) n 3·0= 0 = r cos = 3 cos-= 2 = r sin = 3 sin 2:2 = 3 ·1 = 3 Rectangular coordinates of the point (3, %) are (0, 3) . = r cos = - cos 0 = -2 . 1 2 = sin = -2 sin 0 = -2 . 0 = 0 Rectangular coordinates of the point (- 2, 0) are (-2, 0) . a.
>
b.
<
39.
41.
x
0
y
0
x
y
�
0, -
�
>
c.
r
0
0
0
Y
0 a.
= r cos = 6 cos 150° = 6 (-�) = -3.J3 = rsinO = 6sin 1500= 6 . !2 = 3 Rectangular coordinates of the point (6, 150°) are (-3.J3, 3) . = rcosO = -2cos 3; = -2 ( - �) = .fi = r sin = -2 sin 3n4 = -2 . .fi2 = -.fi Rectangular coordinates of the point (-2, 3;) are (.fi, -.fi) . = r cos = - 1 cos ( -�) = -1 · ± = - ± y = rsinO = -lsin(-�) = - { -� ) = � Rectangular coordinates of the point ( -1, - �) are ( _ !2 ' .J3} 2 = r cos = -2 cos(-180°) = -2 ( -1) = 2 = rsinO = -2sin(-1800) = -2 · 0 = 0 Rectangular coordinates of the point (-2,-180°) are (2, 0) . = rcosO = 7. 5 cos1100� 7. 5 (-0.3420) � -2. 5 7 = rsinO = 7. 5 sin 1100� 7. 5 (0. 9397) � 7.05 Rectangular coordinates of the point (7. 5 , 110°) are about (-2. 5 7, 7. 0 5) . = rcosO = 6. 3 cos(3. 8 ) � 6. 3(-0.7910) � -4.98 = rsinO = 6. 3 sin(3.8) � 6. 3(-0. 6 119) � -3. 8 5 Rectangular coordinates of the point (6.3,3.8 ) are about (-4. 9 8, -3. 8 5) x
n
0
0< �
2
0
47.
0
=
49.
x
y
0
x
0
x
0
Y
51.
x
Y
-
53.
x
Y
.
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Section 1 0. 1 : Polar Coordinates
55.
57.
r == �x2 +i ==.J32 + 0 2 == 19 == 3 () == tan-I (-=;) == tan- I (% ) == tan- I 0 == 0 Polar coordinates of the point (3, 0) are (3, 0) . r == �x2 +i == �(_ 1)2 +02 == Ji == 1 () == tan-I (-=;) == tan-I ( �1 ) == tan- I 0 == 0 The point lies on the negative x-axis, so () . Polar coordinates of the point (-1, 0) are ( 1, ) The point (1, -1) lies in quadrant IV. r == �x2 +i == �12 +(_ 1)2 == -12 () == tan- I (�Y ) == tan- I (T-1 ) == tan - I (- 1) -4' Polar coordinates of the point (1, -1) are ==
==
69.
.
71.
'It
73.
75. 61.
The point (13, 1) lies in quadrant I. r == �x2 +i == �(13t +12 == ..[4 == 2 () == tan- I (�Y ) == tan-I 131 == "6 Polar coordinates of the point ( 13, 1) are ( 2, �) . The point (1.3, -2.1) lies in quadrant IV. r == �x2 + i == �1. 32 + (-2 . 1)2 ==.J(;i "" 2 .47 -2 1 ) - 1 .02 () == tan-I (�Y ) == tan-I (1.3 The polar coordinates of the point ( 1. 3, -2.1 ) are ( 2. 4 7,-1. 0 2 ) . The point (S. 3 , 4.2 ) lies in quadrant I. r == �X2 + i == .JS. 32 + 4. 2 2 == �S6. 5 3 "" 9. 3 0 ()== tan- I (-=;) == tan- I (::�) "" 0.47 The polar coordinates of the point ( S. 3, 4.2 ) are ( 9. 3 0, 0. 4 7 ) . 'It
63.
.
65.
2X2 + 2y2 == 3 2 ( x2 + y2 ) == 3 2r2 = 3 r2 � or r f{ = � x2 4y (r cos ())2 = 4r sin () r2 cos 2 () -4r sin () == 0 2xy = 1 2(r cos ())(r sin ()) = 1 r2 ( 2sin()cos() ) == 1 r2 sin2() = 1 x=4 rcos() = 4 r = cos() r2 = rcos() x2 + y2 = X x2 -x+ i = 0 1 1 2 x2 -x+-+ Y 4 4 (x- �J +i = ± r2 = cos() r3 = r cos () ( r2 )3/ 2 = rcos() ( X2 + y 2 )3/ 2 = X ( X2 + y2 )3 1 2 -x = O r=2 r2 = 4 x2 + i = 4 ==
'It
'It
59.
67.
77.
""
79.
==
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Chapter 1 0: Polar Coordinates; Vectors
81.
83.
4 r = --cos O I r(l - cos 0) = 4 r - r cos O = 4 �X2 + / - x = 4 �x2 + y2 = x + 4 x2 + / = x2 + 8x + 1 6 / = 8(x + 2 ) a. For this application, west is a negative direction and north is positive. Therefore, the rectangular coordinate is (-10, 36) . b. The distance r from the origin to (-10, 36) is r = �X2 + / = �( _ 1O)2 + (36)2 = ..)1396 = 2..)349 37.36 Since the point (-10, 36) lies in quadrant II, we use O = 1 800 + tan-l ( � . Thus,
85. 87.
x = r cos O and y = r sin O Answers will vary.
S ection 1 0.2 1.
3.
5. 7.
�
9.
)
11.
(�J -10 = 1 80° + tan- l ( _ 1 8 ) 105.5° 5 The polar coordinate of the point is ( 2..)349, 1 800 + tan- I ( - 158 )J
13.
0 = 1 80° + tan- I
(-4, 6 ) (X _ (_2» 2 + (y - 5/ = 32 (X + 2 ) 2 + (Y _ 5 ) 2 = 9 Ji
2 polar equation -r False r=4 The equation is of the form r = O . It is a circle, center at the pole and radius 4. Transform to rectangular form: r=4 r2 16 x2 + / = 16 a,
�
a >
=
(37.36, 105.5° ) For this application, west is a negative direction and south is also negative. Thus, the rectangular coordinate is (-3, -35) . The distance r from the origin to (-3, -35) r = �x2 + y2 = �(-3)2 + (-35)2 = ..)1234 35. 13 Since the point (-3, -35) lies in quadrant III, we use O = 1 800 + tan - l � . Thus, �
c.
d.
IS
�
()
B-� 2
( �:) = 180° + tan- I e:) 265.1 ° The polar coordinate of the point is (..)1234, 1800 + tan-1 e:)J (35.13, 265.1 0) . O = 1800 + tan-1
�
�
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Section 1 0.2: Polar Equations and Graphs
15.
19.
f) = � 3 The equation is of the fonn f) = a . It is a line, passing through the pole at an angle of � from 3 the polar axis. Transfonn to rectangular form: f) = � 3 tan f) = tan � 3 Z = -J3
r cos f) = - 2 The equation is of the fonn r cos f) = a . It is a vertical line, 2 units to the left of the pole. Transfonn to rectangular fonn: r cos f) = - 2 x = -2
x y = -J3x
6-� 2
21.
r = 2 cos f) The equation is of the fonn r = ±2a cos f), a > O . It is a circle, passing through the pole, and center on the polar axis. Transfonn to rectangular fonn: r 2 cos f) 2 r 2r cos f) 2 x + y 2 = 2x x 2 - 2x + y 2 = 0 (x - l) 2 + y 2 = 1 center (1, 0) ; radius 1 =
17.
4
=
r sin f) = The equation is of the fonn r sin f) = b . It is a horizontal line, units above the pole. Transfonn to rectangular fonn: r sin f) = 4 y=4
4
8-� 2 tJ
..
� 2
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Chapter 1 0: Polar Coordinates; Vectors
23.
r = - 4 sin B The equation is of the form r = ±2a sin 19, a > 0 . It is a circle, passing through the pole, and center on the line
27.
pole, center on the line
and radius 1 . 2 Transform to rectangular form: r csc B = - 2 1 r · -- = - 2 sin 19 r = - 2 sin B r2 = - 2r sin B x2 + / = - 2y x2 + / + 2y = 0 x2 + (y + l)2 = 1 center (0, - 1) ; radius 1
19 = 2: .
Transform to rectangular form: 2 r = - 4 sin B r2 = - 4r sin B x2 + / = - 4y x2 + / + 4y = 0 x2 + (y + 2)2 = center (0, -2) ; radius 2
4
y
r csc B = - 2 The equation is a circle, passing through the
t
e-I
19 = 2:
e -� 2
25.
r sec B = 4 The equation is a circle, passing through the pole, center on the polar axis and radius 2. Transform to rectangular form: r sec B = 4 I r . -- = 4 cos O r = 4 cos B r2 = 4r cos B x2 + y2 = 4x x2 - 4x + y2 = 0 (x _ 2)2 + y2 = 4 center (2, 0) ; radius 2 y
e-� 2
29. E 31. F
t
e-I
33.
H
35.
D
37.
r = 2 + 2 cos B The graph will be a cardioid. Check for symmetry: Polar axis: Replace 19 by - B . The result is r = 2 + 2 cos( -B) = 2 + 2 cos 19 . The graph is symmetric with respect to the polar aXIS. The line
19 = 2: :
2
Replace
19
by 1r - 19 .
r = 2 + 2 cos( 1t - B) = 2 + 2 [ cos(1r) cos 19 + sin(1r) sin BJ = 2 + 2( - cos 19 + 0) = 2 - 2 cos B The test fails.
e-� 2
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Section 10.2: Polar Equations and Graphs
The pole: Replace r by r . -r = 2 + 2 cos O . The test fails. Due to symmetry with respect to the polar axis, assign values to 0 from 0 to 7t . 0 r = 2 + 2 cos O 4 0 7t 2 + J3 � 3.7 6 7t 3 3 7t 2 2 27t 1 3 57t 2 - J3 � 0.3 6 7t 0
Due to symmetry with respect to the line
-
=
� 2
assign values to 0 from - 2: to 2: . 2 2 0 r = 3-3sinO - 7t 6 2 - 7t 3 + 3J3 � 5.6 3 2 7t 9 2 6 0 3 7t 3 6 2 7t 3 - 3J3 � 0.4 3 2 7t 0 2 -
-
-
-
-
-
-
-
-
-
-
-
-
e -� 2
39.
0
e-� 2
= 3 - 3sin O The graph will be a cardioid . Check for symmetry: Polar axis: Replace 0 by - O . The result is r = 3 - 3 sin(-0) 3 + 3 sin 0 . The test fails . The line 0 2: : Replace 0 by 7t - 0 . 2 r = 3 - 3 sine 7t - 0) = 3 -3 [ sin (7t) cos 0 - cos ( 7t ) sin 0 = 3 -3(0 + sin O) = 3 - 3 sin O The graph is symmetric with respect to the line 0= 2: . 2 The pole: Replace r by - r . -r = 3 - 3 sin 0. The test fails. r
41.
=
=
2 + sin 0 The graph will be a lirnayon without an inner loop. Check for symmetry: Polar axis: Replace 0 by - O . The result is r 2 + sine - 0) = 2 - sin 0 . The test fails. r
=
=
The line 0 2: : Replace 0 by 7t - 0 . 2 r = 2 + sin(7t- 0) = 2 + [ sin (7t) cos 0 - cos ( ) sin 0 = 2 + (0 + sin 0) = 2 + sin 0 The graph is symmetric with respect to the line 0= 2: . 2 The pole: Replace r by r . -r = 2 + sin 0. The test fails.
]
=
7t
]
-
541
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved, This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
,
Chapter 10: Polar Coordinates; Vectors
The pole: Replace r by - r . -r = 4 - 2 cos () . The test fails. Due to symmetry with respect to the polar axis, assign values to {} from 0 to 1t . () r = 4 -2cos(} 2 0 -1t 4 - J3 ".,2.3 6 -1t 3 3 -1t 4 2 21t 5 3 51t 4+J3 "., 5.7 6 1t 6
Due to symmetry with respect to the line () = 2:, 2 1t assign values to () from --1t to - . 2 2 () r =2+sin(} 1 --1t 2 J3 . 1 --1t 2 --".,1 2 3 3 1t 2 6 2 0 5 -1t 2 6 J3 2 .9 -1t 2+-",, 2 3 1t 3 2 -
-
-
-
-
e-¥
45. 43.
r = 4 - 2 cos () The graph will be a lirnaryon without an inner loop. Check for symmetry: Polar axis: Replace () by - () . The result is r = 4 - 2 cos(-(}) = 4 -2cos () . The graph is symmetric with respect to the polar axis. The line () = 2:: Replace () by 1t - () . 2 r = 4 - 2 cos( 1t -(}) = 4 - 2 [ cos (1t ) cos () + sin ( 1t ) sin ()] = 4 - 2(-cos(}+0 ) = 4+2cos(} The test fails.
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
r = 1+ 2 sin () The graph will be a lirnaryon with an inner loop. Check for symmetry: Polar axis: Replace () by - () . The result is r = 1+ 2 sine -(}) 1 - 2 sin () . The test fails. =
The line () = 2:: Replace () by 1t - () . 2 r = 1 + 2 sin(1t -(}) = 1 + 2 [ sin ( 1t ) cos () - cos ( 1t )sin ()] = 1 + 2(0+sin (}) = 1 +2 sin(} The graph is symmetric with respect to the line (} = 2:. 2 The pole: Replace r by - r . -r = 1 +2 sin () . The test fails. 542
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 10.2: Polar Equations and Graphs
Due to symmetry with respect to the line B = � 2 assign values to B from � to � . 2 2 B r = 1 + 2 sinB -1 - n: 2 - n: l - Ji ",, - 0.7 3 - n: 0 6 1 0 n: 2 6 n: I + Ji "" 2.7 3 n: 3 2
The pole: Replace r by - r . -r = 2 -3cosB. The test fails. Due to symmetry with respect to the polar axis, assign values to B from 0 to n: . B r = 2 -3cosB -1 0 3 Ji - 0.6 n: 2 - -",, 2 6 1 n: 3 2 n: 2 2 2n: 7 3 2 5n: 2 + 3 Ji "" 4.6 2 6 5 n:
,
-
-
-
-
-
-
-
-
-
-
-
-
-
e-� 2
e -� 2
47.
-
49.
r = 2 - 3 cosB The graph will be a limayon with an inner loop. Check for symmetry: Polar axis: Replace B by -B . The result is r = 2 -3 cos( -B) = 2 - 3 cos B . The graph is symmetric with respect to the polar axis.
r = 3 cos(2B) The graph will be a rose with four petals. Check for symmetry: Polar axis: Replace B by -B . r = 3 cos(2( -B)) = 3 cos( - 2B) = 3 cos(2B) . The graph is symmetric with respect to the polar axis . The line B = �: Replace B by n: -B . 2 r = 3 cos [ 2(n: -B) ] = 3 cos(2n: - 2B) = 3 [ cos ( 2n: ) cos ( 2B) + sin ( 2n: ) sin ( 2B) = 3( cos 2B + 0) 3 cos ( 2B)
The line B = �: Replace B by n: -B . 2 r = 2 - 3 cos(n: -B) = 2 - 3 [ cos (n: ) cosB + sin (n: ) sin BJ 2 - 3(- cosB+ 0) = 2 + 3 cosB The test fails. =
]
=
543 © 2008 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10: Polar Coordinates; Vectors
The graph is symmetric with respect to the line f)= � . 2 The pole: Replace r by - r . -r = 4sin ( 5f)) . The test fails. Due to symmetry with respect to the line f) �
The graph is symmetric with respect to the line f)= � . 2 The pole: Since the graph is symmetric with respect to both the polar axis and the line f)= � , 2 it is also symmetric with respect to the pole. Due to symmetry, assign values to f) from 0 to � . 2 f) r = 3 cos ( 2f)) 3 3 rc 2 6 rc 0 4 rc -3 3 2 rc -3 2 -
=
assign values to f) from - � to � . 2 2 B rc
--
2
1t
--
3
-
rc
--
4
-
-
-
-
1t
-
6
0
r =
4sin(SB)
B 1t
-4
-
2J3 "" 3 .S
-
2.J2 "" 2. 8
-
-2
r =
,
4sin(SB)
2
6
1t
2
-2.J2 "" -2.8
4
1t
3
-2J3 "" -3.S
rc
-
2
4
0
-
e� � 2
51.
53.
r = 4sin(5f)) The graph will be a rose with five petals. Check for symmetry: Polar axis: Replace f) by - f) . r = 4sin [ 5( -f)) ] = 4 sin( -5f)) = - 4sin ( 5f)) . The test fails. The line f) = �: Replace f) by rc- f) . 2 r = 4sin [ 5(rc - f)) ] = 4 sin(5rc - 5 f)) = 4 [ sin e 5rc ) cos ( 5f)) - cos ( 5rc ) sin ( 5f))] = 4 [ O + sin ( 5f))] = 4sin ( 5f))
r 2 = 9 cos(2f)) The graph will be a lemniscate. Check for symmetry: Polar axis: Replace f) by - f) . r 2 = 9 cos(2(-f))) = 9 cos(-2f)) = 9cos(2f)) . The graph is symmetric with respect to the polar axis. The line f) = �: Replace f) by rc - f) . 2 2 r = 9 cos [ 2(rc- f)) ] = 9 cos(2rc - 2f)) = 9 [ cos ( 2rc) cos 2f)+ sin ( 2rc ) sin 2f)] = 9(cos 2f)+ O) = 9 cos ( 2f))
544 © 2008 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 10.2: Polar Equations and Graphs
The graph is symmetric with respect to the line () = ::.- . 2 The pole: Since the graph is symmetric with res pect to both the polar axis and the line () = ::.- , 2 it is also symmetric with respect to the pole. Due to symmetry, assign values to () from 0 to ::.- . 2 () r = ± 9 cos (2(}) ±3 0 3.J2 7t ±--",±2. 1 2 6 7t 0 4 7t undefined 3 7t undefined 2
() r = 2B -7t 0. 1 - -7t 0.3 2 7t - - 0.6 4 1 0 7t 1 .7 4 7t- 3.0 2 7t 8.8 37t 26.2 2 27t 77.9
�
57.
e-� 2
55.
r = 2B The graph will be a spiral. Check for symmetry: Polar axis: Replace () by - () . r = TB. The test fails.
r = 1 -cos () The graph will be a cardioid. Check for symmetry: Polar axis: Replace () by - () . The result is r = 1 - cos( -(}) = 1 - cos () . The graph is symmetric with respect to the polar axis.
The line () = ::.- : Replace () by 7t - () . 2 r = 1 - cos( - (}) = 1 - (cos ( 7t ) cos () + sin ( 7t ) sin (}) = 1 - (-cos () + 0) = 1 + cos(} The test fails. The pole: Replace r by - r. -r = 1 - cos () . The test fails. Due to symmetry, assign values to () from 0 to 7t . 7t
The line () = ::.- : Replace () by 7t - () . 2 r = 2,,-B. The test fails. The pole: Replace r by - r. -r = 2B. The test fails.
545
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10: Polar Coordinates; Vectors
()
() r = l - cos () °
°
°
-1t6 -1t -1t2 -21t3 -51t6
../3 1 -1t6 1 --",0. 2 1t-21 -1t2 1 -21t3 -32 ../3 -51t6 1 + -",1 .9 2
3
3
1t
1t
2
r = 1 - 3 cos () -2 3../3 1 .6 1 --",2 1 2 1 5 2 ../3 1 + 3 ",3.6 2 4
--
e-� 2
59.
r = I - 3 cos () The graph will be a limac,:on with an inner loop. Check for symmetry: Polar axis: Replace () by - () . The result is r = 1 - 3 cos(-() = 1 - 3 cosB . The graph is symmetric with respect to the polar axis.
61.
The line () = � : Replace B by 1t -B . 2 r = 1 - 3 cos(1t -B) = 1 - 3 [ cos ( 1t ) cosB + sin ( 1t ) sin ()] = 1 - 3(-cos () + 0) = 1 + 3 cos () The test fails. The pole: Replace r by - r . -r = 1 - 3 cos () . The test fails. Due to symmetry, assign values to () from ° to 1t .
The graph is a cardioid whose equation is of the form r = a + b cos B . The graph contains the point ( 6, 0 ) , so we have 6 = a + b cos O 6 = a + b (l) 6 = a+b The graph contains the point 3, , so we have TC
( ;)
3 = a + b cos2 3 = a + b (0) 3=a Substituting a = 3 into the first equation yields: 6 = a +b . 6 = 3+b 3=b Therefore, the graph has equation r = 3 + 3 cos () .
546 © 2008 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 10.2: Polar Equations and Graphs
63.
The graph is a limayon without inner loop whose equation is of the form r = a + b sinB , where o < b < a . The graph contains the point (4, 0) , so we have 4 = a + b sin O 4 = a + b (0) 4=a The graph contains the point 5, , so we have
-
( �)
7r
65.
B r= 2 1 - cos B 0 undefined 7t 2 "" 14.9 6 I-J3/2 7t 4 3 7t 2 2 27t 4 3 3 57t 2 "" l . 1 6 1 + J3/2 7t 1
5 = a + bsin 2 5 = a + b (1) 5 = a+b Substituting a = 4 into the second equation yields: 5 =a+b 5 = 4+b l=b Therefore, the graph has equation r = 4 + sinB . r = 2 Check for symmetry: I - cosB Polar axis: Replace B by -B . The result is 2 r= = 2 l - cos (-B) l - cosB' The graph is symmetric with respect to the polar aXIS. The line B = 2:: Replace B by 7t -B . 2 2 r = ---,------:1 - cos (7t -B) 2 1 - (cos cos B + sin sin B) 2 1 - ( - cos B + 0) 2 1 + cos B The test fails. 2 The pole: Replace r by - r . -r = --1 - cos B The test fails. Due to symmetry, assign values to B from 0 to 7t .
-
---
7r
67.
7r
r=
Check for symmetry: 3 - 2 cosB Polar axis: Replace B by -B . The result is 1 r= = 1 . The graph'IS 3 - 2 cos(-B) 3 - 2 cosB symmetric with respect to the polar axis. The line B = 2: : Replace B by 7t -B . 2 1 r = -----:---.,3 - 2 cos ( 7t -B) 1 3 - 2 ( cos cos B + sin sin B) 1 1 3 - 2 ( - cos B + 0) 3 + 2 cosB The test fails. The pole: Replace r by - r . -r = 3 - 2 cosB The test fails. Due to symmetry, assign values to B from 0 to 7t . 7r
7r
----
547
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10: Polar Coordinates; Vectors
0 0 7t -
6
7t 3 7t 2 27t 3 57t
r=
I
3 - 2cosB 1
I ----;r:; ",O. 8
3- 3
I
-
-
2
-31
-
1
-
-
-
6
1t
I
4
---3+ 3 -;r:;
e -� 2
",0.2
I
71. r = csc 0 - 2 =
1
- 2 0 < 0 < 1t sin O ' Check for symmetry: Polar axis: Replace 0 by - 0 . r = csc( - 0) - 2 = - csc 0 - 2 . The test fails .
-
5
--
The line 0 =2: : Replace 0 by 1t - 0 . 2 r = csc (1l' - 0) - 2 1 2 sin (1l' 0) 1 2 sin 1l' cos 0 - cos sin 0 1 -2 o . cos 0-1 . sin 0 1 =_ _ _ 2 sin O = csc O - 2 The graph is symmetric with respect to the line 0= 2: . 2 The pole: Replace r by - r. -r= csc 0- 2 . The test fails. -
e-� 2
1l'
69. r =0,
O?0 Check for symmetry: Polar axis: Replace 0 by - O . r = -0 . The test fails. The line 0 = 2: : Replace 0 by 1t O . r = 1t 0 . 2 The test fails. The pole: Replace r by r. -r = O . The test fails. 0 r=O -
-----
-
-
0 7t
-
6
7t 3 7t 2
�",0.5
6
-
�"'1.0 3
-
�", 1.6
1l'
37t 2
-
21l'
Due to symmetry, assign values to 0 from 0 to 2: 2 r = csc O - 2 0 undefined 0
0
1t 6 1t
0
-
4
2 1t",3.1 37t ",4.7 2 27t",6.3
-
1t 3 1t
-
-
2
12 -2",-0.6 2 ./3 2 0 8
-
3
-
",
-
.
-1
548 © 2008 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
.
Section 10.2: Polar Equations and Graphs
e-� 2
73.
75.
r = tan O, - !!..
0
by n
-
>
0.
77.
- tan O = - tan O = -tan 0 r = tan( n - 0) = tan ( n ) l + tan ( n ) tan O 1 --
79.
Convert the equation to rectangular form: r = 2a cos 0, a >° 2 r = 2a r cos O 2 x + y 2 = 2ax x 2 - 2ax + y 2 = ° (x - a) 2 + / = a 2 Circle: radius a, center at rectangular coordinates (a, 0).
81.
a.
""
-
4
J3 - 0.6 - n --",, 3 6 ° ° n ",,0 .6 3 6 n 1 4 n J3 1.7 3 -
-
J3
-
-
Convert the equation to rectangular form: r = 2a sin O, a>0 r 2 = 2a r sin O x 2 + y 2 = 2ay x 2 + / - 2ay = 0 x 2 + (y a) 2 = a 2 Circle: radius a, center at rectangular coordinates (0, a). _
The test fails. The pole: Replace r by - r . -r = tan O . The test fails. 0 r = tan 0 - n -J3 -1.7 3 n -I -
Convert the equation to rectangular form: r sin O = a y=a The graph of r sin 0 = a is a horizontal line a units above the pole if a 0, and la l units below the pole if a < ° .
""
r 2 = cos 0: r 2 = cos( n 0) r 2 = - cos O Not equivalent; test fails. ( _r ) 2 = cos(-O) r 2 = cos O New test works. -
549 © 2008 Pearson Education, Inc., Upper Saddle River, NJ.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10: Polar Coordinates; Vectors
r 2 = sin 0 : r 2 = sin( - 0) r 2 = sinO Test works . ( _ r) 2 = sin(-O) r 2 = -sinO Not equivalent; new test fails. Answers will vary.
83.
Imaginary aXiS
7t
b.
1
Real ----t--j,o;:--L.-L.---"--axis
Section 10.3
15.
1.
-4 + 3i
3.
cos Acos B - sin AsinB
5.
magnitude, modulus, argument three False
7. 9. 11.
r = � x 2 + y 2 = � 0 2 + ( _3) 2 = ,J9 = 3 tanO = 2:'.x = -30 0= 270° The polar form of = -3 i is = r ( cos 0 + i sin 0) = 3{ cos 270° + i sin 270°) z
z
3
r = � x2 + l = � = .fi tan 0 = 2:'.x = 1 0= 45° The polar form of z = 1 + i is z = r (cosO + i sinO) = .fi (cos 45° + isin 45°)
. Imaginary axis
--_ -3'----t-1r-'---�3
---
Real
axis
-3
17.
Imaginal)' axis 1
Real -----+---LJ'---axis 1
r = � x2 +l = � 4 2 + ( _ 4) 2 = F32 = 4.fi tan 0 = 2:'.x = -4 4 = -1 0= 3 1 5° The polar form of = 4 - 4i is = r ( cos 0 + i sin 0) = 4.fi ( cos 3 15° + i sin 3 1 5°) z
z
Imaginary axis 4
13.
2
) f
r = �x 2 + y 2 = ( .J3 + ( _ 1) 2 = -/4 = 2 -1 = -.J3 tanO = -yx = 3 .J3 0= 330° The polar form of = .J3 - i is z = r ( cos 0 + i sin 0) = 2 ( cos 330° + i sin 3300)
-+-""*--1.l...-----L___+_
___
Real axis
z
550
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Section 10.3: The Complex Plane; De Moivre's Theorem
19.
r = � x 2 + y 2 = � 3 2 + ( _ 4) 2 = 55 = 5 tanB= yx = -34 B::::: 306.9° The polar fonn of z = 3 - 4i is z = r ( cosB+ i sinB) = 5 ( cos 306 . 9° + i sin 306 . 9°) -
-
--�--�--r--u----��
3
2 9.
0 . 2 ( cos 1 00° + i sin 100°) ::::: 0.2 ( -0.1736 + 0 . 9848i) ::::: -0.035 + 0. 197i
31.
Imaginary axis Real
33.
aXIS
( cos 3; + i sin 3; ) = 3 (0 - l i) = -3 i
27.
(
z
i[cos (40° - 20°) + i sin(40° - 20° )] = � (cos 20° + i sin 20°)
35.
B::::: 123 . 7° The polar fonn of z - 2 + 3i is z = r (cosB+ isinB) = .J13 (cos 123 . 7° + i sin 123 . r) =
zw
Imagi.nary aXIS
z
=
3 (cos 130° + i sin 130°) 4( cos 2700+ i s in 270°)
%[ cos (130° - 270°) + i s i n ( l 300- 270°)] %( cos (-140°) + isin (-140°») %[ cos (360°-140°) + i s in (3600-1400)] = %(cos 2200+ isin 220°) =
-1
[ i � i)
=
2( cos 120° + i sin 1200) = 2 - +
(
3· 4[ cos (1300+ 270°) + isin (1300+ 270°)] = 12( cos 4000+ isi n 400°) = 12[cos (400°- 360°) + i s in (400°- 360°)] = 12( cos 40° + isin 40°)
=
Real --�-----+--�--�� axis -2 2
25.
= 3( cos 130° + i s in 130°). 4( cos 2700+ isin 270°) =
w
23.
= 2 ( cos 40° + i sin 400) . 4( cos 20° + isin 200) = 2 . 4[cos(400 + 200 ) + i sin(400 + 20°)] = 8 ( cos 60° + i sin 60° ) 2 ( cos 40° + i sin 40° ) = 4(cos 200 + i sin 200)
=
r = � X 2 + y 2 = � (_ 2) 2 + 3 2 = .J13 _l tanB= zx = � -2 = 2
3
::::: 1 . 970 + 0 . 347 i
zw
w
21.
)
2 cos l� + i sin t8 ::::: 2 (0 . 9848 + 0. 1736i)
= -I + .J3 i
) [ � -� )
4 cos 741r + i sin 741r = 4 i = 2J2 - 2J2 i
© 2008 Pearson Education, Inc., Upper Saddle River, NJ.
551 All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10: Polar Coordinates; Vectors
41.
37.
[ 4 ( cos 400 + i sin 400)r := 43 [cos (3 · 40°) + i Sin(3 . 400)] = 64 ( cos 120° + i sin 120°)
( � � i)
= 64 - +
:=-32 + 32.J3 i 43.
1t
= cos 40 + 39.
. I
' sm
1t
40
z = 2 + 2i r = ../2 2 + 2 2 = .J8 = 2 .fi tanB= �2 = l B = 45° z = 2.fi (cos 45° + i sin 45°)
= 32 (0 + l i) = 32 i 45.
f
:=27
-1 = -.J3 tanB= 3 .J3 B = 330° w = 2 ( cos 3300 + isin3300) w =2-/2 (cos 45°+ isin 45°)· 2( cos3300+ isin 3300) = 2-/2· 2[ cos ( 45°+ 330°) + i sin (45°+ 330°)] :=
:=
(� + � ) i
27 +--1 27.J3 . =2 2 47.
z
:=
[.J3 ( cos 1 0° + i sin 10°) r :=( .J3 f [ cos( 6 . 10° ) + i sine 6 . 10° ) ] :=27 (cos 60° + i sin 60°)
w=.J3 - i r = .J3 + ( _ 1) 2 = -14 = 2
�(
[2 ( cos 1� + i sin 1�)r := 25 [ cos ( 5 . 1� ) + i sin ( 5 . 1� ) ] :=32 ( cos % + i Sin % )
[� (cos �� + i sin ��)r = ( �r [ cos ( 4 . �� ) + i sin ( 4 �� ) ] 31t + 1. SIll , 3n = 2S( cos4 4) = 25 ( - V; + V; i J .
4-/2 ( cos375°+ i sin 375°) 4-/2[ cos (375°- 360°) + i sin (375°- 360°) ] 4-/2 (cos 15° + i sin 15°)
z := 2.fi ( cos 45° + i sin 45°) w 2(cos 3300+ i sin 3300) :=2 [cos ( 45° - 330°) + i sin( 45° - 330°)]
f
2S.fi + --1 25 .fi . = --2 2
:=.fi[ cos ( - 285°) + i sin( - 285°)] :=.fi[ cos( 360° - 285°) + i sin (3600 - 285°)] = .fi(cos 75° + i sin 75°) 552 © 2008 Pearson Education,
Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 10.3: The Complex Plane; De Moivre's Theorem
49.
l-i r = �12 + (_ 1)2 = .fi tanO = -1-= 17 -1 0= 4 1t 1 .=.fi( COS 47 1t + 1S..lli 47 1t ) (1-i) 5 = [.fi ( cos 74 1t + i sin 74 1t )] 5 = (.fit [ cos( 5· 74 1t ) + ism( 5· 74 1t )] 35 1t + 1S..lli 4 35 1t ) = 4.fi2 ( COS 4 = 4 .fi( - � + � i) = -4+4 i .fi-i r = �(.fit + (_ 1) 2 = ./3 -1 -.fi2 tanO =-= .fi 0"",324 .736° .fi -i "",./3(cos324.736°+ isin 324 .736°) ( .fi_ i)6 ""'[ ./3(COS324 . 7360+ ism324 . 7360)T (./3)6 [cos( 6 ·324 .736°) + isin( 6 . 324 .736°)] = 27(cos1 948.4 16°+ i sin 1 948.4 16°) "'" 27( -0.8519 + 0.5237i) ""'-23+14 . 142i
zk
Zo
51.
Z2
55.
) (
+
4-4./3i r �4 2 + ( - 4 ./3f = J64 = 8 tan0 = -44./3 ./3 0= 300° 4 -4./3i = 8( cos 300°+ ism3000) The four complex fourth roots of 4 -4./3i 8( cos 300°+ ism3000) are: =
=-
=
zk
Zo
zl
Z2
=
Z3
57.
53. l+i
[ (
) (
3600k 3000 3600k =� cos 3000 -4- +4- +isin -4-+- 4=�[cos(7So+ 90°k)+i sin (7So +90°k)J =�[cos(7So+ 90°.0)+isin (7So + 90°· O)J =� (cos7So+ isin7S0) =�[cos(7So+ 90°.1)+isin(7So+ 90°.1)J = � (cos 16S0+ isin 165°) =�[cos(7So+90°· 2)+isin(75°+90°· 2)J =� (cos2S5°+ isin 2SS0) =�[cos(7So+ 90°·3)+isin(7SO+90°· 3)J =� (cos345°+ isin34S0)
-16i r = �02 +( _ 16)2 = ../2 56 1 6 tanO = -16 -0 0= 2 70° -16 i = 16( cos 270°+ i sin2 70°) The four complex fourth roots of -16i = 16( cos 270°+ ism270°) are: =
r = � = .fi tanO = 11-= 1 0= 45° 1 +i = .fi ( cos 45°+i sin 45°) The three complex cube roots of + i = .fi ( cos 45° +i sin 45°) are: 1
553 «:I 2008 Pearson Education, Inc., Upper
)]
+
ZI
-I
[ (
4S0 360°k . 4S0 360°k . = ff2 J2 cos 3 +- - +/sm 3 +-33 = �[cos(ISO+120°k) isin(ISO+120°k)J = �[cos(lso+120°. 0)+isin(ISo+120°· O)J = � (cosIS0+ isinISO) = �[cos(ISo+120°·1) isin(ISO+120°.1)J = � (cos 13So+ isinl3S0) = �[cos(ISO+120°.2)+isin(ISO+120°.2)J = � (cos2SSo+ isin2SSo)
Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
)]
Chapter 10: Polar Coordinates; Vectors
3600-k ) +i sin (4 00 + 3600-k )] = � [ cos (400 + 4 4 1 [ cos ( 90° k ) + isin ( 90° k )J Zo = cos ( 90° . 0 ) + i sin ( 90° · 0 ) = cos 0° + i sin 0° = 1 + 0 i = 1 = cos( 90°. 1) + sin ( 90° . 1) = cos90° + i sin 90°= 0 + Ii = i Z2 = cos (90° . 2 ) + i sin ( 90°· 2 ) cos 180°+ i sin 180°= -1 + Oi = -1 z3 = cos ( 90° . 3) + i sin ( 90° . 3) = cos 270°+ isin2700= 0-li = -i The complex fourth roots of unity are: l, i, - I, -i .
[ (2700 3600 ) (2700 3600 )]
k 4r;;: . . 6 cos -+ -- +/sm zk = ,,10
(
4
4
-
4
k
zk
+ -4
= 2[cos 67.50 + 900 k) isin 67.5 0 + 900 k)]
+ ( Zo = 2 [ cos ( 67. so + 90° · 0) + i sin( 67. so + 90°· 0) J = 2 ( cos 67. so + i sin67. so ) z, = 2[ cos(67S+ 90 °· 1 ) + isin(67S+ 90° . 1 )J = 2 ( cos 157. so + i sinI 57. so ) Z2 = 2 [ cos ( 67. s o + 90°· 2 ) + isin( 67. s o + 90°· 2 )J = 2 ( cos 247. so + i sin247. so ) z3 = 2 [ cos ( 67. s o + 90°· 3 ) + isin( 67. so + 90°· 3)J = 2 ( cos337. s o + i sin337. so )
-
=
zl
i
=
59.
r = �02 + 12 = Ji = 1 tan 0 = -I o 0= 90° i = 1 ( cos 90° + i sin 90°) The five complex fifth roots of i = 1 (cos 90° + i sin 9 0° ) are: [ ( 900+-3600k-) +isin (5900 +-3600k5-)] zk = 1f1 cos 5 5 = I [cos(ISO + 72° k ) + i sin (ISO + 72° k )] Zo = 1 [cos(ISo+ 72° · O)+isin(ISo+ 72° · O)J = cos l S o + i sin l S o = 1 [cos ( 1 8° + 72° . 1 ) + i sin ( 1 8° + 72° · 1)J = cos 90°+ i sin 90° Z2 = 1 [ cos ( 1 8° + 72° · 2 ) + i sin ( 18° + 72° · 2 )J = cos I 62° + i sin l62° Z3 = 1 [ cos ( 1 8° + 72°· 3 ) + i sin ( 1 8° + 72°· 3)J = cos 234° + i sin 234° Z4 = I [ cos ( 1 8° + 72° . 4 ) + i sin ( 1 8° + 72° · 4 )J = cos 306° + i sin 306° 61. 1 = I+Oi r = �12 +02 = Ji= 1
Imagi.nary
I 1
f
-II \ \
63.
zl
/
�
-
-- ....
....
'-
"
"
,
\ \
\
\
"
"
'-
....
....
..--
-i
/
w = r ( 0+ i 0) w:f= n = * [ (� : )
/
/
/
I
I \ I
Real aXIs
Let cos sin be a complex number. If 0 , there are distinct nth roots of given by the formula: zk cos + 2 n + i sin + 2 n IZk
65.
/
/
/
:Ins
1 = * for all k
w,
(� : )}
= 1, ... , n -I
where k 0, 2,
Examining the formula for the distinct complex nth roots of the complex number
w= r(cosO+isinO), . . (-;;-0 + ---;;-)] , = \Inr[r cos(-;;-0 + -2kn- ) + zsm where k = 0, 1, 2, . .. , n -I , we see that the zk are 2n . spaced apart by an angle of n zk
tan O = -o =O 1
0= 0° I + 0 i = 1 ( cos 0° + i sin 0° )
n
2kn
The four complex fourth roots of unity are: 554 © 2008 Pearson Education, Inc., Upper
Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 10.4: Vectors
Iv I = 4 , then 1 3vl = 1 3 1 . 1 v I = 3· 4 = 12 . = (0, 0), Q = (3, 4) 25. v = (3 -O)i+ (4 - 0)j = 3i+ 4 j = (3, 2), Q = (S, 6) 27. v= (S-3)i+(6-2)j= 2i+4j 29. = (-2, -1), Q = (6,-2) v = [6-(-2)]i+[-2-(-1)]j = 8i-j = (1, 0), Q = (0, 1) 31. v = (0 -l)i+ (1 - O)j= -i+ j 33. For v= 3i-4j, I Ivl = �32 +(-4) 2 =55=s. 35. For v = i -j, I v I = �1 2 + ( _ 1) 2 = Ji. . 37. For v= -2i+3j, I vll = �( _ 2) 2 +32 =J13. 39. 2v+3w = 2(3i-Sj)+3(-2i +3j) = 6i-1 OJ -6i+ 9j = -j 41. I v-wl = I (3i-Sj)-(-2i+3j) 1 = I S i-8jl = �S 2 +(-8) 2 = ../89 43. I v I - Iw I = 11 3i -Sj 1 - 1I -2i+3jl = �32 + (_ 5)2 �r-(_-2)--:-2-+ --:-32 = ..f34- J13 v Si Si = -Si5 = .1 45. U = - = - = I Iv I I Si I -12S + 0 v 3i-4j 3i-4j 47. = U R = P i- 4 jl = �32 +(_ 4)2 3i-4j 55 3i -4j 5 4. 3. = -I--J S S
Section 10.4
23. If
5.
unit horizontal, vertical True
7.
v+w
1. 3.
P
P
P
P
v
9.
3v
3v
• • • _---_---__ ---l.. ....
11.
13.
v-w
.
v
3v+u-2w
3v
_
15. 17. 19. 21.
True False False True
= -F +E-D D-E = H+G
C
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Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10: Polar Coordinates; Vectors
I vI = 5, a = 60° v = I v I(cosai + sin aj) = 5 ( cos(600)i+ sin ( 60°) j) 1 . J3. ) = 5 (-I+-j 2 2 5 . 5 ../3. = -I+-j 2 2 57. Ilvll = 14, a = 120° v = Iv I(cos ai+ sinaj ) = 14 ( cos(1200)i + sin(120°) j) = 14 ( - � i+ � j) = -7i+ 7../3j 59. I Iv I = 25, a = 330° v = Iv I(cosai+ sinaj ) = 25 [cos(3300)i+ sin(330°) j] = 25 (� i- � j) 25../3. 25 . =--I--j 2 2 61. Let =(3,-1). Then = +v = (3,-1)+(-4,5)= (-1,4). The new coordinate will be (-1, 4).
55.
v= ai+ bj . We want Il vll = 4 and a= 2b. Il vll = �a2 + b2 = �(2b)2 + b2 = �5b2 �5b2 = 4 5b2 = 16 b2 = 165 4J5 + [i6 = -+ �= + b= -Vs J5 5 a= 2b = 2 ( ± 4�) = ± 8� 8J5. 4J5 . or V= --I--j 8J5. 4J5. v=-I+-j 5 5 5 5 53. v= 2i-j, w= xi+3j, Ilv+wll = 5 I v+wI = 1 2i-j+ xi+ 3j I = I (2+x)i+2j l = �(2 + X)2 + 22 = �X2 +4x+4+4 = �X2 +4x+8 Solve for x: �x2 +4x+8 = 5 x2 +4x+8 = 25 x2 +4x-1 7= 0 x = -4 ± �162(1)-4(1)(-1 7) -4±J84 2 -4±2m 2 = -2± m The solution set is { -2+m, -2-m}.
51. Let
u
a.
u'
u
b.
_----!._----'. �-c-
-5
63.
F
= 40 [ cos(300)i+Sin(300)j] ../3. 1 . ) - 40 (-I+-j 2 2 = 20../3 i+ 20j _
556 © 2008 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 10.4: Vectors
F] F2 F3 F. I F.I [ (175.8 ) + sin(175.8°) jJ I F.I ( -0.9973li+ 0. 0 7324j) F2 = I F2 1 [cos{3.70)i+ sin{3.70)j] ""I F2 1 1( 0 .99792i+ 0 . 0 6453j) F3 = -150j
Fj = 40[ cos (30° ) i+ sin (30°) j] = 40 [� i+ -i j) = 20v'3i+20j F2 = 60[cos(-45°)i+ sin( 45° )j] = 60 [ � i- � j) = 30.fi i- 30.fij F = Fj + F2 = 20v'3 i+ 20j+ 30.fi i-30v'2j = ( 20v'3+ 30.fi) i+ ( 20 -30v'2) j 67. Let FI be the tension on the left cable and F2 be the tension on the right cable. Let F3 represent
69. Let be the tension on the left end of the rope and be the tension on the right end of the rope. Let represent the force of the weight of the tightrope walker. = cos ° i
65.
-
""
For equilibrium, the sum of the force vectors must be zero.
F.+ F2 + F3 = -0.99731 I1F.l i+0. 0 7324 I1F.ll j +0.99792 I1F2 I1i+0 . 06453I1 F2 I1 j-150 j = (-0.99731 I1 FIII +0.99792 I1 F2 1I) i + (0. 0 7324 I1 F.I +0.06453I1 F2 1 -150) j =0 Set the i and j components equal to zero and solve: {-0.99731 I1 F)I +0.99792 I1 F2 1 = 0 0. 0 7324 1 F)1 +0 . 064531 F2 1 -150 = 0 Solve the fIrst equation for I F2 I and substitute the result into the second equatic.n to solve the system: I F2 I = �:::��� I F)I ""0.999391 F)I 0. 0 7324 1 1 F) I + 0 . 06453(0.999391 F.I ) -150 = 0 0.1377311F) 1 = 150 I F.I = 1089. 1 I F2 1 = 0 .99939(1089.1) ""1088.4 The tension in the left end of the rope is about 1 089. 1 pounds and the tension in the right end of the rope is about 1 088. 4 pounds.
the force of the weight of the box. cos ( 155° ) i +sin ( 155° )j]
F. = I FI I [ ""I F.I ( -0 .9063i+ 0.4 226j) F2 = I F2 I [cos ( 40°) i+ sin ( 40°) j] ""I F2 1 (0.7660i+0.6428j) F3 = -1000j
For equilibrium, the sum of the force vectors must be zero.
F.+ F2 + F3 = -0 .90631I F.Ii i+0 .4226 1 FI Il j +0.7660 1I F2 1I i+0 . 64281IF2 1I j-1000j = (-0.9063 1IF. I +0. 7660 1I F2 1I) i + (0.4226 1 F.I + 0 . 64281 F2 1 -1000 ) j =0 Set the i and j components equal to zero and solve: {0.-04.9063 1IFII + 0.7660 1F2 1 = 0 226 1IF.I +0. 64281IF2 1 -1000 = 0 Solve the fIrst equation for I F2 1 and substitute
the result into the second equation to solve the system:
IF211 = �:��:� I FIII "" 1 . 1832 1IF.11 0.4226 1IF.11 + 0. 6428(1.1832 1IF.ID -1000 = 0 1 .1832 1 F.11 = 1000 I F.I "" 845. 2 I F2 11 ""1 . 1832{845 . 2 ) ""1000 The tension in the left cable is about 845 . 2 pounds and the tension in the right cable is about 1000 pounds. 557 © 2008 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10: Polar Coordinates; Vectors
71.
=3000i = 2000[cos (45°) i+ sin (45°)j] = 2000 ( '7 i+ '7 j) =1000J2 i+ 1000J2j = + =3000i+ 1000J2 i+ 1000J2j =( 3000 + 1 000J2 ) i+ 1000J2j I I = ( 3000 + lOOOJ2 t + (lOOOJ2 t ""4635. 2 The monster truck must pull with a force of approximately 4635.2 pounds in order to remain unmoved. Fl
Section 10.5
F2
F
Fl
1.
5. True 7.
c.
9.
FI
F2
F3
F4
V·W
The vectors are orthogonal.
v=2i+j, w=i- 2j w= 2(1) + 1(- 2) = 2 - 2 =0 0 v·w = cos () = I villiwI � 22 +12 �12 +(_ 2)2 O_ = � =O =_ FsFs 5 90° () "" a.
F4
F3
W
b.
=-3i; =-i+4j; =4i- 2j; =-4j A vector v=a i + b j needs to be added for equilibrium. Find vector v=a i + b j : + + + + =0 -3i+ (-i+4j) + (4i- 2j) + (-4j) + (ai+ bj) =0 Oi+ - 2j+ (ai+ bj) =0 ai+ (-2+b)j=0 a = 0; -2+ b =0 b= 2 Therefore, v= 2j . F2
v=i -j, =i+ j =1(1)+ (-1)(1) =1-1 =0 0 v·w = cos () = lI vl l wl �1 2 + (_1)2 � =0 () =90° a.
F
FI
C
3. parallel
F2
73. The given forces are:
c2 =a2 + b2 - 2ab cos
V·
b.
V
c.
11.
The vectors are orthogonal.
v=.j3i-j, w=i+ j =v'3(1)+ (-1)(1) = v'3-1 v'3-1 :--= cos () = Il vlll wl �(v'3)2 + (_ 1)2 � v'3-1 =v'3-1 F6-J2 4 .J4J2 2J2 () =75° a.
b.
c.
13.
-;===-r====
V·W
V·W
-;=====
--
The vectors are neither parallel nor orthogonal.
v=3i+ 4 j, w=4i+ 3j v,w=3(4)+4(3)=1 2+1 2= 24 24 cos () = = I vil il wI �32 +42 �4 2 +32 24 24 J25J25 25 () ""16.3° a.
b.
75. Answers will vary.
c.
V·W
-;:===--===
The vectors are neither parallel nor orthogonal.
558 © 2008 Pearson Education, Inc., Upper Saddle River, NJ.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 10.5: The Dot Product
v= 4i, w= j a. V· w = 4(0) + 0(1) = 0 + 0 = 0 0 cos () = I vVli ·liwwI = -===--=== ../4 2+ 02../0 2+ 12 =� = �4 = O 4·1 () = 900 The vectors are orthogonal. 17. v = i- a j, w = 2i + 3j Two vectors are orthogonal if the dot product is : zero. Solve for a v·w = O 1(2) + (-a)(3) = 0 2-3a = 0 3a = 2 a= -23 19. v= 2i-3j, w= i-j v, = v·w)2w= 2(1)+ (-3)(-1) (i_ j) (I 1 (�12 +(-1)2 r 5, 5. =-52 (.I-J.) =-I--J 2 2 5. 5 ') = --1--) 1. 1. v2 =V-VI = (21-' 3)') - (-1--) 2 2 2 2 21. v= i-j, w= -i-2j VI = vw·w2 = 1(-1)+(-1)(-2) (-i-2j) (I 1 1) ( �( _ 1)2 +( _2) 2J 1, 2. = --51 (1+' 2)') = --1--) 5 5 . v2 =V-VI = ( ) - ( - 51,1- 52.) ) = 56 1- 53,) 23. v= 3i+ j, w= -2i-j v, = vw·w) 2w= 3(-2)+1(-1) (-2i-j) (lI l ( �(-2)2+(-1)2 r 14. 7. ' ') =-1+-) = --57( - 21-) 5 5 1, 2. 14. 7 ) =-1--) ' ' v2 =v-vl = (31+) ) - (-1+-) 5 5 5 5
15.
b.
c.
b.
W
c.
29,
I
=
W
1-).
lI Ii I =�(-0. 02)2+(-0. 0 1f = ../0.0005 "'" 0.022 The the sun'scentimeter. rays is about 0.022intensity watts perof square I A I = �(300) 2 + (400)2 = ../250,000 = 500 The area centimeter.of the solar panel is 500 square W= II . AI = I ( -0.02i-O. Olj)· (300i+ 400j)1 = 1(-0.02)(300) + (-0. 0 1)(400)1 = 1 -6+ (-4)1 = 1- 101 = 10 This means 10 watts of energy is collected. Toandcollect the maximum number ofsolar watts, A should be parallel with the panels facing the sun. Let va = the velocity of the plane in still air, vw = the velocity of the wind, and vg the velocity of the plane relative to the ground. Vg=Va+Vw va = 550 [ cos(225°) i+sin(225°)j] = 550 [ - � i- � j) = -2 75.fi i-275.fi j Vw = 80i Vg= Va+Vw = -275.fi i-275.fi j+ 80i = (80 -275.fi ) i-275.fi j
27. a.
.
'
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Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10: Polar Coordinates; Vectors
Find the angle between vwand j: cos = I vvww il'lji jI mt(1) = mt 0.9887 = -3(0)+ 20";02 + 12 20 �8.6 0 The heading of the boat needs to be about 8.60 upstream. The velocity of1 .the8 kilometers boat directlyper across the river is about 9 hour. The time to cross the river is: 0.5 = --� 1 9.8 0.025 hours or = 19.0.58 . 60 �1.5 2 minutes. 33. Split the force into the components going down the hill and perpendicular to the hill.
The speed of the plane relative to the ground is: Ilvg 1 = �(80-275./2t + ( _ 275./2)2 = �6400 -44,000./2+ 151, 250+ 151,250 .�";246,674.6 �4 96 .7 miles per hour To find the direction, find the angle between vg and a convenient vector such as due south, -j . cos = vv .(1 -_j)j I I ;1 (80 - 275./2 )( 0) + ( - 275./2) (-1) 496.7�02 + (_1) 2 275./2�0.7830 496.7 �38. 5 0 The plane is miles traveling with ainground speed of about 4 . per hour an approximate 7 96 direction of 38.50 degrees west of south. 31. Let the positive x-axis point downstream, so that the velocity of the current is v = 3i. Let vw = the velocity of the boat in the water, and Vg = the velocity of the boat relative to the land. Then vg = vw + V and vg = since the boat is going directly across the river. The speed of the boat is I vw I = 20 ; we need to find the direction. Let vw = a i+ b j , so I VW I = ";a2 +b2 20 a2 +b2 = 400 Since vg = vw + , k j= a i+ b j+ 3i = (a + 3) i+ bj a+3 = 0 a = -3 k= b a2 +b2 = 400 9+b2 = 400 b2 = 3 91 k= b= mt�1 9.8 Vw = -3i+mtj and Vg = .J3 91j
e
=
e
e
t
t
=
e
c
c
F d = F sin8°= 5300sin8°� 73 7.6 F p = F cos8°= 5300cos8°�5248.4 The force requiisrabout ed to keep thepounds. SiennaThe fromforce rolling down the hill 3 6 7. 7 perpendicular to the hill is approximately 5248.4 pounds. 35. Let va = the velocity of the plane in still air, vw = the velocity of the wind, and Vg = the velocity of the plane relative to the ground. Vg =va +vw va = 500 [ cos( 45°) i + sin( 45°) j] ./2 ./2· -_ 500 [-1+ 2 2 'J J = 250./2i+ 250./2j vw = 60 [ cos ( 120o ) i + sin( 1200) j] = 60 [ - � i+ � jJ = -30i+ 30.J3 j
k
=
Vc
©
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All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 10.5: The Dot Product
Vg =va+vw = 2S0..fi i+ 2S0..fi j -30i+ 30J3 j = (2S0..fi - 30) + (2S0..fi + 30J3) j The speed of the plane relative to the ground is: Ivg II = �( -30 + 2S0..fit + ( 2S0..fi + 30J3t "".J26 9,12 9. 1 "" S18.8 kmlhr To find the direction, find the angle between v and a convenient vector such as due north, j. cos () = ..,.--Vv-"g,,--. jj-I II g II 1 (0) + ( 2S0..fi + 30J3) (1) - ( -30 + 2S0..fi)SI8.8.J0 2 + 12 +30J3 "" 0.7816 = 2S0..fiS18.8 () "" 38.6 ° The travelingin awith a ground of of aboutplane S18.8is kmlhr direction of 38.speed 6 ° east north (N3 8.6°E) . 37. We must determine the component force going down the ramp . Fd = F sin200= 2S0sin 20°"" 8S . S Timmy about 8S . S pounds of force to hold themust pianoexert in position . 39. W = F · AB W= 2, AB = 4i F = cos a i-sin a j 2 = ( cosa i-sina j) · 4i 2 = 4cosa -21 = cosa a = 60° 41. Since = 0 i + 0 j and v=a i+ b j, we have that O·v= O· a+O·b = O . 43. If v= ,i +bd = c o sai+sina j and w = a2 i+b2 j = cosfJi+sinfJ j, then cos(a - fJ) = V· = a,a2 + b,b2 = cosa cos fJ + sina sinfJ
Let u = a, i + bd and v= a2 i+ b2j. Then, (u+v) · (u -v) = [(a,+ a2 )i+ (b,+ b2)j] [ (a,-a2 )i+ (b, -b2 ) j] = (a,+ a2 )( a,- a2 ) + (b, + b2 )( b, -b2 ) = a, 2 - a2 2 + b, 2 - b22 = ( a,2 + b/ ) -( a/ + b/ ) = I ul12 - 1vI 2 Since the vectors have the same magnitudes and (u +v)( u -v) = 0 , the vectors are orthogonal. Because the vectors u and v are radii of the circle, we know they have the same magnitude. Since u +v and u -v are sides ofsincethe weangleknowinscribed in arethe semi circle and that they orthogonal part (a), this angle must be a right angle.by 47. (IIw I v + IlvIw) '( 1 w Iv - l lvIw) = (IIw10 2 V· v- Iwil i vI V·w + 1 wi l i vIlvw , - (II V 102 = (11 1 ) 2 V·v- (II V 1 ) 2w·w = (II 1 )2 (II V 102 - (I V1 )2 (I I 1 )2 =0 Therefore, the vectors are orthogonal. 49. (II u +v Ilf - (II u -v Ilf = (u +v) · (u +v)-(u -v) · (u -v) = (u·u + u · v+ V· u + V·v) -(u·u -u · v-v· u + V· v) = 2(u ·v)+2(u ·v) = 4(u ·v)
45.
i
g
a.
b.
w · w
W
W
W
0
a
W
©
561 2008 Pearson Education,
Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10: Polar Coordinates; Vectors
Polar coordinates of the point (-3, 3) are (-3.[i, -%) or ( 3.[i, 3;) . 9. The point (0, - 2) lies on the negative y-axis. r = �x2 +y2 = �02 + (_ 2)2 = 2
Ch apter 10 Review Exercises
3J3 ' y = 3S .lli -1t =-3 x = 3cos-61t = -2' 6 2 Rectangular coordinates of ( 3, i) are ( 3J32 ' l). 2
- 2 is undefined, so B = - 21t ; - 21t +7r =27r ' Polar coordinates of the point (0, - 2) are o
The point (3, 4) lies in quadrant I . r = �x2 + / = .J32 +42 = 5 B = tan-I (�) = tan-I (�)� 0.93 tan- I (�) + 7r � 4 . 0 7 Polar coordinates of the point (3, 4) are (5, 0.93) or (-5, 4 . 0 7) . r = 2sinB 13. r 2 = 2rsinB x2 + y2 = 2y x2 + y2 -2y = 0 x2 + / -2y+l= 1 x2 +(Y _ l)2 = e The graph is a circle with radius 1 and center (0,1) in rectangular coordinates.
11.
x = -2cos 431t = l'' Y = - 2sin 431t = J3 Rectangular coordinates of ( -2, � 1t ) are (1, J3) . 5.
a.
(-3, - �) ---..
0 .....,... ..
b.
x = -3 cos( -�) 0; y = -3sin ( -�) 3 Rectangular coordinates of (-3,-%) are (0, 3) . 7. The point (-3, 3) lies in quadrant II. r = �x2 +y 2 = �(_ 3)2 + 32 = 3.[i B = tan-I (�) = tan-I ( �3) = tan-I (-I) = - % =
=
Ytn
9 = '2
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All rights reserved. This material i s protected under all copyright laws a s they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10 Review Exercises
15.
a.
b.
r=5 r 2 = 25 x2 + y 2 = 5 2
19.
The a circle with radius and centegraph r at theis pole. 5
r = 4cosO The graph will be a circle with radius 2 and center (2, 0) . Check for symmetry: B by -B . r 4 cos( -0) = 4 cos 0 . The aXIS. graph is symmetric with respect to the polar The line 0 = � : Replace 0 by 7t 0 . r = 4 cos( 7t 0) = 4( cos cos 0 + sirl sin0) = 4(-cosB+0) = -4cosO The test fails. The pole: Replace r by - r . -r 4 cos O . The test fails. Due to symmetry with respect to the polar axis, assign values to 0 from 0 to 7t . 0 r = 4 cosO 0 4 Polar axis : Re place
The result is
=
Yt o=I
-
-
7r
17.
=
cos 0 + 3r sin0 = 6
a.
r
b.
y = - 3 x+ 2 The graph is a line withy-intercept (0, 2) and slope 3I in rectangular coordinates.
=
x + 3y 6 3y = -x + 6 1
7t 2.J3 "" 3.5 6 7t 2 3 7t 2 27t -2 3 57t - 2.J3 "" -3.5 6 -
-
Ye t=
7r
-
�
-
0
-
-
7r
-4
e
=
31T 2
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Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10: Polar Coordinates; Vectors
21.
r = 3 - 3sin O The graph will be a cardioid. Check for symmetry: Polar axis: Replace () by - O . The result is r = 3 - 3sin( - 0) = 3 + 3 sin 0 . The test fails.
23.
The line () = � : Replace () by () . r = 3 - 3sin(7t - (}) = 3 - 3( sin cos () - cos sin (}) = 3 - 3(0 + sin 0) = 3 - 3 sin O The graph is symmetric with respect to the line O = !!:..2 ' 7t -
Jr
-r =
assign values to - -7t2 - -7t3 7t -6 0 7t 6 7t 3 7t 2
Jr
3 - 3 sin 0 .
Due to symmetry with respect to the line 0 = 0
i:
The line 0 = Replace 0 by 7t - 0 . r = 4 - cos( 7t - 0) = 4 - (cos cos 0 + sin sin 0) = 4 - (- cos O + 0) = 4 + cos O The test fails. The pole: Replace r by - r . -r = 4 -cos 0 The test fails. Due to symmetry with respect to the polar axis, assign values to 0 from 0 to 7t . 0 r = 4 - cos O 3 0 7t 4 -13 � 3 . 1 6 2 7t7 3 2 7t4 2 9 27t 3 2 57t 4 + 13 � 4.9 2 6 5 Jr
Jr
The pole: Replace r by - r . The test fails.
i
r = 4 - cos O The graph will be a limayon without inner loop. Check for symmetry: Polar axis: Replace 0 by - O . The result is r = 4 - cos( - 0) = 4 - cos 0 . The graph is symmetric with respect to the polar axis.
Jr
.
'
0
from - !!:..2 to !!:..2 . r = 3 - 3 sin O 6 3 + 313 2 � 5.6 9 2 3 3 2 3 - 313 2 � 0.4 0
564 © 2008 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10 Review Exercises
r = �x2 + y 2 = �(_1)2 +(_1)2 = .J2 tanB =Lx = -1-1 = 1 B = 225° The polar form of = -1-i is = r( cosB+isinB) = .J2 (cos225°+ isin225°) . 27. r = �x 2 + y 2 = �4 2 +( _ 3)2 = .J25 = 5 tanB =Lx = _ i4 B ", 323. 1° The polar form of = 4 -3 i is = r(cosB+isinB) = 5( cos 323.1°+ isin323. 1°) . 29. 2(COS1500+isin1500) = 2 ( -.J} + l i ) = --13 +i
33.
25.
0.1(cos3500+isin3500) '" O. l(0.9848-0.l736i) '" 0.10-0.02i 0.06
z
z
0.02
- 0.06
z
35.
z
2
- ..
-1
-' 2-
= (cos 80°+ i sin 80°)( cos 50°+ i sin 50°) = 1 · 1 [cos(80°+ 50°) + i sin(80°+ 50°)] = cos 130°+ isin130° cos 80°+ i sin 80° cos 50°+ i sin 50° = i [cos(80°- 50°) + i sin(80°- 50°)] = cos 30°+ i sin 30°
zw
z
w
Imaginary axis
-;! �----'t--' --"---'--------o-----,
Imagi n a ry '1 xis
Real axis
6(cos 21t + isin 21t ) 6 (cosO + isin 0) =6 91t . ' 91t ) . ( COS-+1Sl\l= .( 1t . ' 1t ) 2 COS S +1Sl\l S =
-2
=
31.
z
w
3
5
5
Imaginary axis 3
Real -+----:---t-...l-.Jl� axis -1
�
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Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10: Polar Coordinates; Vectors
39.
z W = 5 ( cos l00 + i sin l00) . ( cos 355° + i sin355°) = 5 · 1 [ cos(1 0° + 355° ) + i sin(10° + 355°) ] = 5 ( cos365° + i sin 365° ) = 5 ( cos 5° +isin 5°) 5 ( cos 1 0° + i sin 1 0°) W cos 355° + i sin 355° = -i[ cos(10° - 355° ) + i sin(10° - 355°) ] = 5 [ cos(-345°) + i sin(-345° ) ] = 5 ( cos 1 5° +i sin 1 5° )
47.
r
i 3 e "" 5 3 . 1 3 °
tan e
[3 ( cos 200 + i sin 200 )T 33 [ cos(3 . 20° ) + i sin(3 . 20° ) ] = 27 ( cos 60° + i sin 60° )
49.
(� + � i ) ../3
27 + 27 1. =2 2 --
43.
[ (
.J2 cos 5 1t + i sin 5 1t 8 8
[ (4 ·
5 1t + i sin 8 . 2 51t + I. SlD 51t cos2
= (.J2f cos
=4
.
(
= 4(0 + l i) = 4i 45.
)
)r
)
Zk
(4 . 5;)]
--
) ( ]
Zl
-%
�e + ( -../3t = 2 tan 8 = -../3 = -../3
Z2
=
1 8 = 300° 1 - i = 2 ( cos 300° + isin 300° ) i = [2( cos 300° + isin 3000 ) r = 26 [cos ( 6 . 300° ) + i sin ( 6 . 300° ) J = 64 · ( cos 1 800° + i sin 1 800° ) = 64 · ( cos 0° +i sin 0° ) = 64 + 0 i = 64
--
�
= 3 [ cos ( 120° . 2 ) + i sin ( 120° . 2 )J 3 i = 3 ( cos 240° + i sin 2400 ) =
-%- �
51.
../3 (1-../3 r
© 2008 Pearson Education,
[ (
Zo
1 - ../3 i r
27 + O i = .J27 2 + 0 2 = 27 tan 8 = ..Q.. 27 = 0 8 = 0° 27 + Oi = 27 ( cos Oo +isin OO) The three complex cube roots of 27 + Oi = 27 ( cos oo +isin Oo ) are: 0 0 360 0 . k 00 36 0 0 .k + 3 + i sin "3 + 3 = :iffi . cos "3 = 3 [ cos ( 120° k ) + i sin ( 1 20° ·k )J = 3 [ cos ( 120° · 0) + i sin ( 120° . O)J = 3 ( cos 0° + i sin 0°) = 3 = 3 [ cos ( 120° . 1 ) + isin ( 120° . 1 ) = 3 ( cos 120° + i sin l200) = + 3 i r
=
= 27
=
3 + 4i = 5 [ cos ( 53. 13° ) + i sin ( 53. 13°)J (3 + 4it = ( 54 [cos ( 4 . 53. 1 3° ) + i sin ( 4 . 53. 13°) Jf = 625 [ cos ( 212 .52° ) + i sin ( 212.52° )J "" 625 ( -0.8432 - 0.5376i) = - 527 - 336i
Z
41.
3 + 4i = .J3 2 + 4 2 = 5
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
)]
Chapter 10 Review Exercises
The angle between v and i equals 60° . Thus, . . y x x cos60° = I vvl · li ili = ( xi+3· 1j) i 3·1 = -3 We also conclude that v lies in Quadrant I. cos 60° = "3 -1 = -x 2 3 x =-23 x2 + y 2 = 9 (%J + y2 = 9 y 2 = 9 - (%J = 9 - � = 364-9 = 247 (27 = +- 3/3 y = -V4 2 Since v lies in Quadrant I, y = 3f . So, v = X·I+. y . .J = "23.I + -3/32- .J . 71. v = -2i + j, w = 4i -3j v.w = -2(4) + 1(-3) = -8 -3 = -11 cos () = I vv·1 w1wI -11 2 _ 2 �( ) +12 �4 2 + (_ 3)2 -11 = -11 J5·5 5../5 16 () 9.7° 73. v= i- 3j, w= -i+ j v·w = 1(-1)+ (-3)(1) = -1-3 = -4 cos () = I vv·1 w1wI -4 �e + (_3)2 �(_1)2 +e -4 -2 = --2J5MF2 J5 5 () 153.4 °
53.
x
= (1, -2), Q = (3, -6) v = (3-1)i+( -6- (-2))j = 2i-4j Ilvl = �2 2 + (_4)2 = hO = 2J5 57. P = (0, -2), Q = (-1, 1) v = (-1-0)i+(I- (-2))j = -i+ 3j I vll = �(_1)2 + 32 = M 59. v+w = (-2i + j )+(4i-3j) = 2i-2 j 61. 4v-3w = 4( -2i + j)-3(4i - 3j) = -8i+4j-12i + 9j = -20 i+1 3j 63. I vl = I -2i + jl = �(-2) 2 +e = J5 65. I v � + I w I = 11 - 2i + j I + 1 4 i - 3j I = �(_2)2 +12 + �42 + (_ 3)2 = J5 5 7.24 v -2i+ j = -2i + j 67. U =- = I v� 1 -2i+ jl �(_2)2 +12 -2i+ j = -../5 . 2J5.I + -J = -5 5 J5 69. Let v = x . i + y . j, I v I = 3 , with the angle between v and i equal to 60° . I vl = 3 �X2 + y2 = 3 x2 + y 2 = 9
55. P
+
+
r:::;
r:::;
r:::;
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All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10: Polar Coordinates; Vectors
75.
v = 2i + 3j , w = -4i - 6j v · w = (2)(-4) + (3) (-6) = - 8 - 1 8 = -26 cos = vvil ·liww II () II -26 .J2 2 + 3 2 �( _4) 2 ( _ 6) 2 -26 = -26 = -26 = -1 J13J52 .J676 26 () = cos- I ( -1 ) = 1 80° Thus, the vectors are parallel. v = 3i - 4j , w = -3i + 4j V · w = (3)(-3) + (-4)(4) = -9 - 1 6 = -25 cos = I vil ·liw () I v w II -25 �3 2 + (_4) 2 �( _3) 2 + 4 2 -25 = -25 = - 1 J25J25 25 () = cos- I ( -1) = 180° Thus, the vectors are parallel. v = 3i - 2j , w = 4i + 6j v . w = (3) ( 4) + ( -2) (6) = 12 - 12 = 0 Thus, the vectors are orthogonal. v = 2i + j , w = -4i + 3j The decomposition of v into 2 vectors V I and v 2 so that v I is parallel to w and v 2 is perpendicular to w is given by: v I = v · w2 W w1 and v 2 = V - V I v . w -_ (2i + j) . (-4i + 3j) ( -41. + 3J. ) V I _- -Il w ll 2 �( _4)2 + 3 2 2 = (2)( -4)25+ (1)(3) (-4i + 3j) = --51 ( -4 1' + 3 J' ) = -45 I. --53 J.
83.
c
79.
81.
(
©
g
w
c '
w
g
c
w
Since the river is 1 mile wide, it takes the swimmer about 0.2 hour to cross the river . The swimmer will end up (0.2)(2) 0.4 miles downstream. Let F, the tension on the left cable, F2 the tension on the right cable, and FJ the force of the weight of the box. F, = I F, II [ cos(1400) i + sin (140 ) j ] ,., II F, -0.766044i + 0.6427 88 j) F2 = II F2 11[ cos (300) i + sin ( 300 ) j] ,., II F2 11 ( 0. 8 66025i + 0.5 j) F3 = - 2000j For equilibrium, the sum of the force vectors must be zero. Fl + F2 + F3 = - 0.766044 1 FJ Ii i + 0.6427 88 11 FJ I l j + 0. 8 66025 1I F2 1I i + 0.5 1I F2 1I j - 2000j = (-0.766044 11 Fl II + 0. 8 66025 11 F2 11 ) i + (0.6427 88 1I FJ I + 0.5 1I F2 11 - 2000 ) j =0 Set the i and j components equal to zero and solve: - 0.766044 1I FJ II + 0.866025 1I F2 11 = 0 0.6427 8811 FJ II + 0.5 11 F2 11 - 2000 = 0 Solve the first equation for IIF2 11 and substitute the result into the second equation to solve the system: F2 1 II FJ I ,., 0. 88 4552 1 FJ 0.6427 8811 FJ I + 0.5 (0.8 84552 11 Fl II ) - 2000 = 0 =
85.
=
=
I(
)
V 2 = V - V I = 2 1. + J. - 54 1. - 53 J. 3. = 2I. + J. - -45 1. + -J 5 6 . . 8 = -I 5 +-J 5
=
w
-lI
(
w
g
=
+
77.
Let the positive x-axis point downstream, so that the velocity of the current is v = 2i . Let v the velocity of the swimmer in the water and v the velocity of the swimmer relative to the land. Then v = v + V The speed of the swimmer is II v II = 5 , and the heading is directly across the river, so v = 5 j . Then v = v + V = 5 j + 2i = 2i + 5 j
°
=
{
)
I ,., �:�:���;
I
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Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1 0
= =
1 .085064 1 F\ II 2000 II F\ II 1 843.2 1 1b I F2 1 0.884552(1 843.2 1) 1630.41 1b The tension in the left cable is about 1 843.21 pounds and the tension in the right cable is about 1630.41 pounds. ""
AB W
5.
=
=7 �X2 + y 2 = 7 ( �x 2 + y 2 t = 72 x2 y 2 = Thus the equation is a circle with center (0, 0) and radius 7. r
+
49
=
=
20i F . AB
= = [% i 5� j). 20i = (%}20) + [ 5�} 0 ) = 50 ft-lb +
e
6. Ch apter 10 Test 7t
1 - 3.
=
3 7t 2
= =
tan g 3 sin g 3 cos g r sin g 3 r cos g l. 3 or Thus the equation is a straight line with and b O.
=
2
x=
7t
6
y = 3x
=
4.
est
In =
3
x = 2 and y = 2"-J3--- --=r = �x 2 + y 2 = �(2) 2 + ( 2 J33 ) 2 = J16 = 4 and tan g =l. = 2J3 = J3 x 2 Since (x, y) is in quadrant !, we have g = 3" .
The point (2, 2J3) in rectangular coordinates is the point ( 4, 1-) in polar coordinates.
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yright laws as they
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
c urr
ntly
Chapter 10: Polar Coordinates; Vectors
7.
r sin 2 8 + 8 sin 8 = r r 2 sin 2 8 + 8r sin 8 = r 2 y 2 + 8y = x 2 + y 2 8y = x 2 or 4 ( 2 ) y = x2 The graph is a parabola with vertex (0, 0) and focus (0, 2) in re c tangular coordinat s Yt
e
9.
.
e =� 2
x
i+li+l-H�
e = 1t
e =0
e=
8.
31t 2
r 2 cos 8 = 5 Polar axis: Replace 8 with -8 : r 2 cos( -8) = 5 r 2 cos 8 = 5 Since the resulting equation is the same as the original, the graph is symmetrical with respect to the polar axis. The line 8 = 1 : Replace 8 with :r - 8 : r 2 cos(:r - 8) = 5 r 2 (cos :r cos 8 + sin :r sin 8) = 5 r 2 (-cos 8) = 5 _ r 2 cos 8 = 5 Since the resulting equation is not the same as the original, the graph may or may not be symmetrical with respect to the line 8 = :r . 2 The pole: Replacing r with -r : ( _r) 2 cos 8 = 5 r 2 cos 8 = 5 Since the resulting equation is the same as the original, the graph is symmetrical with respect to the pole. Note: Since we have now established symmetry about the pole and the polar axis, it can be shown that the graph must also be symmetric about the line 8 = 1 .
1 0.
11.
r = 5 sin 8 cos 2 8 The polar axis: Replace 8 with -8 : r = 5 sine -8) cos 2 (-8) r = 5(-sin 8) cos 2 8 r = -5 sin 8 cos 2 8 Since the resulting equation is the not same as the original, the graph may or may not be symmetrical with respect to the polar axis . The line 8 = � : Replace 8 with :r - 8 : r = 5 sin(:r - 8) cos 2 (:r - 8) r = 5( sin :r cos 8 - cos :r sin 8)( - cos 8) 2 r = 5(0 · cos 8 - (-1) . sin 8)( cos 2 8) r = 5 sin 8 cos 2 8 Since the resulting equation is the same as the original, the graph is symmetrical with respect to the line 8 = � . The pole: Replacing r with -r : -r = 5 sin 8cos 2 8 r = -5 sin 8cos 2 8 Since the resulting equation is not the same as the original, the graph may or may not be symmetrical with respect to the pole. = (2( cos 85° i sin 850)][ 3 ( cos 22° i sin 220)] = 2 · 3[cos(85° + 22°) + i sin(85° + 22°)] = 6(cos 1 07° + i sin 107°) zw
+
+
= 3 ( cos 22° + isin 22° ) 2 ( cos 85° + i sin 85° ) = � [cos(22° - 85°) + i sin(22° - 85°)] 2 = � [cos( -63°) + i sine -63°)] 2 Since -63 ° has the same terminal side as 297° , we can write = � (cos 297° + i sin 297°) 2 5 = [3(cos 22° + isin 22o)] = 3 5 [cos(5 · 22°) + i sin(5 · 22°)] = 243(cos 1 10° + i sin 1 10°) w z
w z
12.
w
5
570
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 10 Test
13.
1 8.
Let w = -8 + 8.J3 i ; then \ wi = �( 8)2 + (8.J3)2 = ..)64 + 192 = ..)256 = 16 so we can write w in polar form as W = 16 - � + � i = 1 6(cos I 200 + i sin I 200) Using De Moivre ' s Theorem, the three distinct 3 rd roots of w are = .ift6[ cos ( 1 2r r ) + i sin ( 1 2r + r )] = 2ifi [cos( 40° + 1200k) + i sin(40° + 1200k)] where k = 0, 1, and 2 . So we have = 2ifi[cos( 40° + 120° . 0) + i sin( 40° + 120° . 0)] = 2ifi ( cos 40° + i sin 40°) "" 1.93 + 1.62i = 2ifi [cos( 40° + 120° · 1) + i sin(40° + 120° . 1)] = 2ifi (cos 160° + sin 160°) "" -2.37 + 0 . 86i Z2 = 2ifi [cos(40° + 120° . 2) + i sin( 40° + 120° . 2)] = 2ifi ( cos 280° + i sin 280°) "" 0.44 - 2.4 8i
From Exercise 17, we can write v = 10(cos315°i + sin315°j)
_
[
i
�{; ,+ �})�sv
)
+ 36
Zk
k
36
1 9.
k
<1
=
I
i
\\
2V2( c o s 1 60· + i sin 160·)
2 · Imagin a ry "xis <0
=
2 \Y2( cos 40· + i sin
400)
= ( 8J2 -3J2 , 2J2 - 7J2 ) = ( 5J2,-5J2 )
V
1 5.
I l v il =
16.
1 7.
�(5J2 t + (-5J2 t = ..)100 = 10 U =� = � ( 5J2 ' -5J2) = J22 ' J22 Il v ll 10
(
= (4, 6 )+ 2(-3,-6) -(-8,4) = (4 + 2(-3) -( -8),6 + 2(-6) -4) = (4 - 6+8, 6 -12 -4) = (6,- 10)
If Bij is the angle between vi and V j , then cos Bij = IvVi ' Vj ' Thus, I J lh ll 8_ + 6(-6) = -48 = _ _-165 cos �1 2 = 4(-3) 6-165 .J52 . J45 -8) + 6(4) = -8 = - 1 cos (),1 3 = 4(.J52 8-165 -165 . J80 4(10) + 6(15) 130 = 1 cos f)1 4 = .J52 . ..)325 = 130 +(-6)(4) = � = O cos f)2 3 = (-3)(-8) J45 . J80 60 (-6)15 = -120 = _ _8_ cos f)24 (-3)10+ J45 . ..)325 15-165 -165 (4)15 = � = _ _1_ cos f)34 = (-8)10+ J80 . ..)325 20-165 -165 The vectors vi and V j are parallel if cos Bij = 1 and are orthogonal if cos Bij = O . Thus, vectors v 1 and v 4 are parallel and v 2 and v are 3 orthogonal. The angle between vectors V I and v 2 is 8 ) 172 87° cos -1 [ IIvVIdl '' lVIv22 11 cos -1 ( - -165 "" . . =
Real axis
14.
3
j
20 - 2 1 .
Zo
Z
V I + 2V 2 - V
SV
_
22.
)
""
)
From Exercise 16, we can write v = IIv ll ' u = 10 = 10(cos315°, sin315°) = 1O(cos315°i +sin315° j) Thus, the angle between v and i is 315° .
('7, -'7)
571
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Chapter 10: Polar Coordinates; Vectors
23. We first calculate the angle 16
a.
C hapter 10 C u mulative Review
In( ex2-9) = In(l) x2 - 9 = 0 (x + 3) (x - 3) = 0 = -3 or = 3 The solution set is {-3, 3} .
x
F3
x
=
3. The circle with center point (h, k) CO, 1) and radius 3 has equation: (x - h/ + (y _ k) 2 = (X _ 0)2 + (y - l/ = 3 2 x2 =9
U sing the right triangle in the
sketch, we 1 6 conclude tan = - = 2 so that 8 = tan-1 2 � 63.435° . Thus the three force vectors in the problem can be written as F I = IIF111 (cos I 16.565° j + sin I 16.565° j) = II F111 (-0.4472 li + 0.89443j) F 2 = II F2 11 (cos 63.435° j + sin 63.435° j) = II F2 11 (0.4472li + 0.89443j) F3 = -1200j Since the system is in equilibrium, we need F I + F 2 + F3 = 0 j + 0 j . This means [II F 1 11 ( -0.4472 1) + II F 2 1 (0.4472 1) + 0] = 0 and [IIF1 11 (0.89443) + II F 2 11 CO.89443 ) - 1200] = 0 . The first equation gives II F I II = IIF 2 11 ; if we call this common value II F II and substitute into the second equation we get 2 · II F I · (0.89443) = 1200 1200 II FII = 2(0.89443) � 670.82 Thus, the cable must be able to endure a tension of approximately 670.82 lbs.
r
a
a
=
r2
+(y_l)2 Y
-4
(0, 4)
(0, -2)
5. x 2 + i = 2X4 Test for symmetry: x-axis: Replace y by -y : x2 = 2X4 - i = 2X4 which is not equivalent to x 2 + i = 2X4 . y-axis: Replace x by -x : (-x/ + i = 2 ( _X)4 + i = 2X4 which is equivalent to x 2 + i = 2X4
+( _y) 3 x2 x2
x -x
:
Origin: Replace by and y by -y +( = 2 ( -x t = 2X4 which is not equivalent to x 2 + i = 2X4 . Therefore, the graph is symmetric with respect to the y-axis.
( _X)2 _y) 3 x2 _ y 3
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572
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Chapter 10 Cumulative Review
7.
y
=
=
=
I sin x I sin x, -sm. x, sin x, -sm x,
{ {
when when when when
sin x � 0 sin x < 0 O � x � Jl' Jl' < x < 2Jl'
Y 2
- 1T
'IT
'IT
2
1T
2'
-1
x
2
We are finding the angle . sme equa Is --1 2 . () 1 SIn 2'
() ,
- � -< () -< �
2'
2
whose
= --
11.
Graphing 2 and () Jl' using polar coordinates: 3 2 yields a circle, centered at (0, 0) , with radius = 2. () Jl' yields a line passing through the point (0, 0) , 3 forming an angle of () = Jl' with the positive x-axis. 3 t r =
=
r =
=
' .\
/' = 2 x
0: 0
573
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1 1 Analytic Geometry 17.
Section 11.1
Not applicable Section 1 1.2
3.
5. 7. 9. 11.
(X + 4) 2 = 9 x + 4 = ±3 x + 4 = 3 => x = -1 or x + 4 = -3 => x = -7 The solution set is {- 7, -I}
1 9.
(c); the graph has vertex (h, k) = (O, O) and opens to the left. Therefore, the equation of the graph has the form y2 = -4ax . The graph passes through the point (-1, -2) so we have (-2f = -4a ( -1) 4 = 4a l=a Thus, the equation of the graph is l = -4x .
The focus is (4, 0) and the vertex is (0, 0). Both lie on the horizontal line y O . a 4 and since (4, 0) is to the right of (O, 0), the parabola opens to the right. The equation of the parabola is: l = 4ax l = 4 ·4·x l = 16x Letting x = 4, we find l = 64 or y = ±8 . The points (4, 8) and (4, -8) define the latus rectum. =
3, up paraboloid of revolution True
(b); the graph has a vertex (h, k) = (O, O) and opens up. Therefore, the equation of the graph has the form x 2 = 4ay The graph passes through the point (2, 1) so we have (2) 2 = 4a (l) 4 = 4a 1=a Thus, the equation of the graph is x 2 = 4 y .
D: x = -4
.
13.
(e); the graph has vertex (h, k) = (I, I) and opens to the right . Therefore, the equation of the graph has the form (y _ 1) 2 = 4a ( x - I) .
15.
(h); the graph has vertex (h, k) = ( -1, -1) and opens down. Therefore, the equation of the graph has the form (x + 1)2 = -4a (y + 1) .
=
y
x
574
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Section 1 1 .2: The Parabola
21.
The focus is (0, -3) and the vertex is (0, 0). Both lie on the vertical line x = 0 . a 3 and since (0, -3) is below (0, 0), the parabola opens down. The equation of the parabola is: x 2 = -4ay x 2 = -4 · 3 · Y x 2 = -12y Letting y = -3, we find x 2 = 36 or x = ±6 . The points (-6, -3) and (6, -3) define the latus rectum.
25.
=
-� and the vertex is (0, 0) . The focus is ( 0, � ) . a = � and since ( 0, �) is
The directrix is y =
above (0, 0), the parabola opens up. The equation of the parabola is: x 2 = 4ay 1 y => x 2 = 2y x2 = 4 · T
�, we find x2 = 1 or x = 1 . The points ( 1, �) and ( -1, �) defme the latus
Letting y =
y
10
±
rectum.
D: y = 3 x
F=
23.
x
(0, -3)
-5
The focus is (-2, 0) and the directrix is x = 2 . The vertex is (0, 0). a 2 and since (-2, 0) is to the left of (0, 0), the parabola opens to the left. The equation of the parabola is: y 2 = -4ax / = -4 · 2 · x / = -8x Letting x = - 2, we find y 2 = 16 or y = ±4 . The points (-2, 4) and (-2, -4) define the latus rectum.
(-l, � )
=
V=
(0, 0)
-5
27.
y
Vertex: (0, 0). Since the axis of symmetry is vertical, the parabola opens up or down. Since (2, 3) is above (0, 0), the parabola opens up. The equation has the form x 2 = 4ay . Substitute the coordinates of (2, 3) into the equation to find a : 2 2 = 4a · 3 => 4 = 1 2a => a = 1.3
�
5
The equation of the parabola is: x 2 = y . The D: x = 2
( �) .
�, we find x 2 = � or x = ± � . The points (�, �) and
focus is 0,
x
Letting y =
-5
575
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Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
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Chapter 1 1 : Analytic Geometry
(-�,�) define the latus rectum.
31.
y
x
v=
29.
(0, 0)
-5
The vertex is (2, -3) and the focus is (2, -5). Both lie on the vertical line x = 2 . a = 1-5 - ( -3)1 = 2 and since (2, -5) is below (2, -3), the parabola opens down. The equation of the parabola is: (X _ h) 2 = -4a (y - k) (X _ 2) 2 = -4(2) (y - ( -3)) (X _ 2) 2 = -8(y + 3) Letting y = -5 , we find (x - 2) 2 = 16 x - 2 = ±4 => x = -2 or x = 6 The points (-2, -5) and (6, -5) define the latus rectum.
D: x = -2
10
Y 5
4 3
2 x
V = ( - I , - 2)
33.
y 10 - 10
The vertex is (-1 , -2) and the focus is (0, -2). Both lie on the horizontal line y = -2 . a = 1 -1 - 01 = 1 and since (0, -2) is to the right of (-1 , -2), the parabola opens to the right. The equation of the parabola is: ( Y _ k) 2 = 4a (x - h) (y - ( -2)t = 4(1)(x - ( -1) ) (y + 2) 2 = 4(x + l) Letting x = ° , we find (y + 2) 2 = 4 y + 2 = ±2 => y = -4 or y = ° The points (0, -4) and (0, 0) define the latus rectum.
X
The directrix is y = 2 and the focus is (-3, 4). This is a vertical case, so the vertex is (-3, 3). a = 1 and since (-3, 4) is above y = 2 , the parabola opens up. The equation of the parabola is: (x - h) 2 = 4a(y - k) (X - (-3)) 2 = 4 · 1 · (y - 3) (x + 3) 2 = 4(y - 3) Letting y = 4 , we find (x + 3) 2 = 4 or x + 3 = ±2 . So, x = -1 0r x = -5 . The points (-1 , 4) and (-5, 4) define the latus rectum. y
F
=
(2, - 5) -2
576
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Section 1 1.2: The Parabola
35.
The directrix is x = 1 and the focus is (-3, -2). This is a horizontal case, so the vertex is (-1, -2). a 2 and since (-3, -2) is to the left of x = 1 , the parabola opens to the left. The equation of the parabola is: (y_k)2 = -4a(x- h) (y _(_2»2 = -4 . 2 . (x- (-l» (y + 2)2 = -8(x + 1) Letting x = -3 , we find (y + 2)2 = 16 or y +2 = ±4 . So, y = 2 or y = -6 . The points (-3, 2) and (-3, -6) defme the latus rectum.
39.
=
Y
D:
x =
The equation y2 = -16x is in the form / = -4ax where -4a = -16 or a = 4 . Thus, we have: Vertex: (0, 0) Focus: (-4, 0) Directrix: x = 4 y
x
1
41.
The equation (y - 2)2 = 8(x + I) is in the form (y _k)2 = 4a(x - h) where 4a = 8 0r a = 2 , h = -1, and k = 2 . Thus, we have: Vertex: (-1, 2); Focus: (1, 2); Directrix: x = -3 y
D: x = -3
37.
=
The equation x2 4y is in the form x2 = 4ay where 4a = 4 or a = 1 . Thus, we have: Vertex: (0, 0) Focus: (0, 1) Directrix: y -1 =
y
43. x
D: y = - 1 -3
The equation (x -3)2 = -(y + 1) is in the form (X _ h)2 = -4a(y - k) where -4a = -l or a = '41 ' h = 3, and k = -1 . Thus, we have: Vertex: (3, -1); Focus: (3,- �) ; Directrix: y i =
-
y
5 v=
(3 , -1)
577
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1 1 : Analytic Geometry
45.
The equation (y + 3) 2 = 8(x - 2) is in the form (y - k) 2 = 4a(x - h) where 4a = 8 or a = 2 , h = 2, and k = -3 . Thus, we have: Vertex: (2, -3); Focus: (4, -3) Directrix: x = ° y D: x =
49.
Complete the square to put in standard form: x 2 + 8x = 4y - 8 x 2 + 8x + 1 6 = 4y - 8 + 1 6 (X + 4) 2 = 4(y + 2) The equation is in the form (x _ h) 2 = 4a(y - k) where 4a = 4 or 1 , h = - 4, and k = - 2 Thus, we have: Vertex: (-4, 2) ; Focus: (-4, -1) Directrix: y = -3
.
a =
0 x
-
y
x
47.
Complete the square to put in standard form: / - 4y + 4x + 4 = ° / - 4y + 4 = -4x ( Y _ 2) 2 = -4x The equation is in the form (y _ k) 2 = -4a(x - h) where -4a -4 or a = 1 , h = 0, and k = 2 . Thus, we have: Vertex: (0, 2); Focus: (-1, 2) Directrix: x = 1
D: y = -3
-5
51.
=
Y
D:
x =
�""""--r:....;---..;.t-� V = (-4, -2)
Complete the square to put in standard form: / + 2y - x = 0 y 2 + 2y + l = x + 1 (y + l) 2 = x + l The equation is in the form (y - k) 2 = 4a(x - h) Where 4a = 1 or = -41 , h = -1, and k = -1 . thus, we have: Vertex: (�1 , - 1 ) ; Focus: - , -1 a
1
(� )
D. Irec · tr·IX: D:
X = -�
x = - 4"5 y
2 x
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578 All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 1 1 .2: The Parabola
53 .
Complete the square to put in standard form: x 2 - 4x = y + 4 x 2 - 4x + 4 = y + 4 + 4 (X _ 2) 2 = y + 8 The equation is in the form (x - hi = 4a(y - k) where 4a = l or a = 4"1 ' h = 2, and k = -8 . Thus, we have: Vertex: (2, -8); Focus: 2, - 341
63.
a :
( )
. . y = - 33 D lrectnx: 4
��
65.
-2
D: .'v = - TI 4
55.
57.
59.
61.
(y _ l) 2 = c(x - O) (y - l) 2 = cx (2 _ 1) 2 = c(l) => 1 = (y _ l) 2 = x
Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 = 4ay . Since the parabola is 10 feet across and 4 feet deep, the points (5, 4) and (-5, 4) are on the parabola. Substitute and solve for 25 5 2 = 4a(4) => 25 = 16a => a = 16 a is the distance from the vertex to the focus. Thus, the receiver (located at the focus) is = 1 . 5625 feet, or 1 8.75 inches from the base of the dish, along the axis of the parabola. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 = 4ay Since the parabola is 4 inches across and I inch deep, the points (2, 1 ) and (-2, 1) are on the parabola. Substitute and solve for a : 2 2 = 4a(1) => 4 = 4a => a = 1 a is the distance from the vertex to the focus. Thus, the bulb (located at the focus) should be 1 inch from the vertex. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 = cy . The point (300, 80) is a point on the parabola. Solve for c and find the equation: 3002 = c(80) => c = 1 125 x 2 = 1 1 25y
67. C
(y _ l) 2 = c(x - 2) (0 - 1) 2 = c(I - 2) 1 = -c => c = -1 (y _ 1) 2 = - (x - 2)
,
Iy
t. . . . . . . . . . =....1 ..;.; !
(X - 0) 2 = c(y - l) x 2 = c(y - l) 2 2 = c(2 - l) 4=c x 2 = 4 ( Y - l)
<", o, O«<= ," ( I S0.h)
Since the height of the cable 1 50 feet from the center is to be found, the point (150, h) is a point on the parabola. Solve for h: 1 50 2 = 1 125h 22, 500 = 1 125h 20 = h The height of the cable 1 50 feet from the center is 20 feet.
(y _ 0) 2 = c(x - (- 2)) / = c(x + 2) 1 2 = c(O + 2) => 1 = 2c => c = "21 Y 2 = "21 (x + 2) 579
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.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 1 1 : Analytic Geometry
69.
point (30, y) is a point on the parabola. Solve for y:
Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 = 4ay . a is the distance from the vertex to the focus (where the source is located), so a 2. Since the opening is 5 feet across, there is a point (2.5, y) on the parabola. Solve for y: x 2 = 8 Y
302 = -144y 900 = -144y -6.25 = Y
=
The height of the bridge 30 feet from the center is 25 - 6.25 = 1 8.75 feet. To find the height of the bridge, 50 feet from the center, the point (50, y) is a point on the parabola. Solve for y:
2.52 = 8y 6.25 = 8y y = 0.78125 feet
50 2 = -144y => 2500 = -144y => y = -17.36 The height of the bridge 50 feet from the center is about 25 - 17.36 7.64 feet.
The depth of the searchlight should be 0.78 125 feet. 71.
73 .
=
75. a.
Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 = 4ay . Since the parabola is 20 feet across and 6 feet deep, the points (10, 6) and (-1 0, 6) are on the parabola. Substitute and solve for a : 102 = 4a(6) => 100 = 24a => a � 4 . 17 feet The heat will be concentrated about 4. 17 feet from the base, along the axis of symmetry.
Imagine placing the Arch along the x-axis with the peak along the y-axis. Since the Arch is 625 feet high and is 598 feet wide at its base, we would have the points ( -299, ° ) , (0, 625) , and (299, 0 ) The equation of the parabola would have the form y = ax 2 + c . U sing the point (0, 625 ) we have .
625 = a (Ol + c 625 = c
y = ax2 + 625 . Next, using the point (299, 0 ) we get 0 = a (299)2 + 625 -625 = ( 299) 2 a 625 a = ---(299 ) 2
The model then becomes
Set up the problem so that the vertex of the parabola is at (0, 0) and it opens down. Then the equation of the parabola has the form: x 2 = cy . The point (60, -25) is a point on the parabola. Solve for c and fmd the equation:
602 = c( - 25) => c = -144 x 2 = -144y
Thus, the equation of the parabola with the same given dimensions is
625 x2 + 625 . y = --(299)2
(60.-23)
zIs i
b.
To find the height of the bridge 10 feet from the center the point ( 1 0, y) is a point on the parabola. Solve for y:
102 = -144y 100 = -144y -0.69 � Y
The height of the bridge 1 0 feet from the center is about 25 - 0.69 = 24.3 1 feet. To fmd the height of the bridge 30 feet from the center the
c.
625 2 + 6 25 , we get y = ---( 299) 2 Width (ft) x Height (ft), model 63. 12 567 283.5 225.67 239 478 459.2 1 54 308
U smg .
X
No; the heights computed by using the model do not fit the actual heights.
580
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Section 1 1.3: The Ellipse
77.
Cy2 + Dx = 0 C 0, D 0 Cy2 = -Dx=:> y2 = - DC X equation of a parabola with vertex at axis of symmetry is the x-axis. (0,This0)isandthewhose The focus is ( -:c' 0 ) . The directrix is x = :c . The parabola opens to the right if -� 0 and to the left if - � O . C/ + Dx + Ey + F = 0 C 0 If D 0 , then: Cy 2 + Ey = -Dx-F E E2 ) = -Dx-F+E2 C ( 2 +-y+-C 4C2 4C ( y+ E )2 = 1 (-Dx-F+ E2 ) 2C C 4C 2 = -D(x+ F _ L) ) (y+ !i.. 2C C D 4CD (y+ 2�r = -g( x- E24��CF ) This is the equation of a parabola whose . E2 -4CF - E ) ' and whose vertex ( 4CD ' 2C axis of symmetry is parallel to the x-axis. If D = 0 , then Cy2 + Ey + F = 0 =:> y = -E±�E2C2 -4CF If E 2 -4CF = 0 , then y = - � is a single 2 horizontal line. If D = 0 , then Cy2 + Ey + F = 0 y = --E±--'-:-�2CE-:-2 -4CF -If E 2 - 4CF 0 , then E 2 -4CF and y = _E+ �2C y = -E- �2CE2 -4CF are two horizontal *
>
79.
d.
*
Y
=:>
- 4 CF
<
Section 1 1.3 1.
<
3.
*
*
a.
D = 0 , then Cy2 + E + F = 0 y = --E±--'--�2CE-2 -If E 2 -4CF 0, then there is no real solution. The graph contains no points. If
= �(4 _ 2)2 + ( _2_( _5) )2 = �2 2 +32 = JU x-intercepts: 0 2 1 6 -4x 2 4x2 = 16 x2 = 4 x = ±2 (-2,0),(2,0) y-intercepts: y 2 = 16-4(0) 2 y2 = 16 = ±4 (0,-4),(0,4) The intercepts are (-2, 0), (2, 0) , (0, -4) , and (0,4) . left 1; down 4 d
=
�
Y
Y
5.
1S
b.
7.
ellipse
9.
(0,-5) and (0,5)
�
11.
True
13.
(c); the major axis is along the x-axis and th vertices are at and
15.
(b); the major axis is along the y-axis and th vertices are at and
---'-:-=----
(-4,0) (4,0) .
(0, -2) (0,2) .
c.
=:>
>
lines.
581
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Chapter 1 1 : Analytic Geometry
1 7.
x2 + / =1 25 4
y 5
The center of the ellipse is at the origin. a = 5, b = 2 . The vertices are (5, 0) and (-5, 0). Find the value of c :
x
c 2 = a 2 - b 2 = 25 - 4 = 2 1 � c = 51 The foci are (51, 0) and (-51, 0) .
-5 ( 0, -4)
y
5
(0, 2)
23.
(-5, 0) -M�������� -5 (- 'fil , 0) (0, -2)
4/ + x 2 = 8
Divide by 8 to put in standard form:
4y 2 + x 2 = -8 -8 8 8
y2 = 1 x2 + 8 2
�
The center of the ellipse is at the origin.
-5
19.
( 0, 2.J3) and ( 0, - 2.J3) .
The foci are
a = J8 = 2Ji, b = .J2 .
x2 + / =1 9 25
The vertices are
(2Ji, 0 ) and (
-
2Ji, 0) .
Find
the value of c :
c 2 = a 2 - b2 = 8 - 2 = 6 c = J6 The foci are ( J6, 0) and ( - J6, 0) .
The center of the ellipse is at the origin. a = 5, b = 3 . The vertices are 5) and (0, -5) . Find the value of c:
(0,
c 2 = a 2 _ b 2 = 25 - 9 = 16 c=4 The foci are (0, 4) and (0, -4).
y
3
(0, 1/2)
x
(- 3, 0) -s
x
(0, - V'i)
25. 21.
( vo. 0)
-3
4x 2 + / = 16 Divide by 1 6 to put in standard form: 4x 2 + 16 � x2 + / =/ =1 16 16 16 4 16
x2 + / = 1 6
�
2
2
x y =1 -+
16 16 This is a circle whose center is at radius 4. The focus of the ellipse is
(0, 0) and (0, 0) and the vertices are (-4, 0) , (4, 0) , ( 0, -4) , ( 0,4) . =
y
The center of the ellipse i s at the origin.
5 (0, 4)
a = 4, b = 2 .
The vertices are (0, 4) and (0, -4). Find the value of c:
(- 4, 0) -5
c 2 = a 2 _ b 2 = 1 6 - 4 = 12 c = m = 2.J3
(4, 0)x 5
-5 (0 , -4)
582
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Section 1 1.3: The Ellipse
27.
Center: (0, 0); Focus: (3, 0); Vertex: (5, 0); Major axis is the x-axis; a = 5; c = 3 . Find b: 2 b = a 2 - c 2 = 25 - 9 = 1 6
b=4
x2
-
Write the equation:
25
)'
y2
+-
16
33.
Focus: ( -4, 0 ) ; Vertices: (-5, 0) and (5 , 0) ; Center: (0, 0) ; Maj or axis is the x-axis. a 2= 5 ; c = 4 . Find b: b = a 2 - c2 = 25 - 1 6 9 � b = 3 2 x / Write the equation: - + - = 1 25 9 =
=1
y 5
29.
Center: (0, 0); Focus: (0, -4); Vertex: (0, 5); Major axis is the y-axis; a = 5; c = 4 . Find b: 2 b = a 2 - c 2 = 25 - 1 6 = 9
b=3
x
-
Write the equation: )'
2
9
y2
+-
25
(0 , -3)
-5
35.
=1
(0, 5)
(3, 0)
(-
5
x
y 5
4 13
(0, Vf3)
(0, -5)
31.
Foci: (0, ±3); x-intercepts are ±2. Center: (0, 0); Major axis is the y-axis; c = 3; b = 2 . Find a : 2 a 2 = b 2 + c = 4 + 9 = 1 3 � a = J13 2 2 x y Write the equation: - + - = 1
x
Foci: (±2, 0); Length of major axis is 6. Center: (0, 0); Major axis is the x-axis; a = 3; c = 2 . Find b: 2 b = a2 - c2 = 9 - 4 = 5 � b = 2 x / Write the equation: - + - = 1 9 5
J5
37.
Center: (0, 0); Vertex: (0, 4 ) ; b = 1 ; Major axis is the y-axis; a = 4; b 1 . 2 x / Write the equation: - + - = 1 =
y
5
16
1
x
x
-5
583
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Chapter 1 1 : Analytic Geometry
39.
41.
c 2 = a 2 _ b 2 = 1 6 - 4 = 12 � c = .Ji2 = 2J3
Center: (-1, 1) Major axis: parallel to x-axis Length of major axis: 4 = 2a � a = 2 Length of minor axis: 2 = 2b � b = 1
Thus, we have: Center: (-5, 4) Foci: (-5 - 2J3,4) ,
(x + l) 2 + (y _ l) 2 = 1 4 1 Center: (1, ° )
Vertices:
(-5 + 2J3, 4) (-9, 4), (-1 , 4) y 7
(-5 + 2 '13, 4)
Major axis: parallel to y-axis Length of major axis: 4 = 2a � a = 2 Length of minor axis: 2 = 2b � b = 1
x
(x - l) 2 + L = 1 1 4
43.
-3
(X - 3) 2 + (y + l) 2 = 1 is in the 4 9 (X _ h) 2 + ( Y _ k) 2 = 1 (major axis parallel form b2 a2 to the y-axis) where a = 3, b = 2, h = 3, and k = -1 . Solving for c: c 2 = a 2 - b 2 = 9 - 4 = 5 � c = .J5 .
47.
The equatIOn
Thus, we have: Center: (3, -1) Foci: (3, -1 + .J5), Vertices:
Complete the square to put the equation in standard form:
x 2 + 4x + 4y 2 - 8y + 4 = 0 (x 2 + 4x + 4) + 4(y 2 - 2y + l) = -4 + 4 + 4 (x + 2) 2 + 4(y - l) 2 = 4 (X + 2) 2 4(y - l) 2 4 -'-----'- + 4 4 4 (X + 2) 2 + (y _ l) 2 = 1 1 4 _
(3, -1 - .J5)
----"'_-c-
The equation is in the form
(3, 2), (3, -4)
(x _ h) 2 ( y _ k) 2 -'--a--:2-'- + b 2 = 1 (maj or axis parallel to the x-axis) where a = 2, b = 1, h = - 2, and k = 1 . Solving for c: c 2 = a 2 - b 2 = 4 - 1 = 3 � c = J3
y 5
Thus, we have: Center: (-2 , 1) Foci: (- 2 - J3, 1) , Vertices:
45.
Divide by form:
16 to put the equation in standard
(- 2 + J3, 1) (-4 , 1), (0, 1)
(-2 - 1/3, 1) (-2, 2)
(x + 5) 2 + 4(y - 4) 2 = 16 (x +-5i--'- + 4(y- 4) 2 16 -'----"' 16 -'-- 16 16 (X + 5) 2 + (y _ 4) 2 = 1 16 4
(-2, 0)
y 3
(-2 + 1/3, 1 ) x
-2
The equation is in the form
(x _ h) 2 + (y _ k) 2 = 1 (major axis parallel to the -'-a-2-'-- b 2 x-axis) where a = 4, b = 2, h = -5, and k = 4 . Solving for c : 584
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Section 1 1 .3: The Ellipse
49.
Complete the square to put the equation in standard form:
y-axis) where Solving for c:
2X2 + 3y 2 - 8x + 6y + 5 = ° 2(x 2 - 4x) + 3(/ + 2y) = -5 2(x 2 - 4x + 4) + 3(y 2 + 2y + 1) = - 5 + 8 + 3 2(x - 2) 2 + 3(y + l) 2 = 6 2(x- 2)-'--2 + 3(y + 1) 2 6 ----'6 6 6 2 2 (X _ 2) + (y + l) = 1 3 2
a = 3, b = 2, h = 1,
and k = - 2 .
c 2 = a 2 _ b 2 = 9 - 4 = 5 -+ c = .J5
Thus, we have: Center:
( 1, -2 )
(1, -2 + .J5) , (1, -2 - .J5) Vertices: (1, 1) , (1, -5)
Foci:
�--'--
y
5
The equation is in the form
(X _ h) 2 ( y _ k) 2 -'--a-:2-'- + b 2 = 1
(maj or axis parallel to the
x-axis) where
a = 13, b = J2, h = 2, and k = - l . Solving for c : c 2 = a 2 - b 2 = 3 - 2 = 1 -+ c = 1 Thus, we have: Center: (2, -1) Foci: ( 1 , -1), (3, -1) Vertices: (2 - 13, - 1) ,
53.
Complete the square to put the equation in standard form:
4x 2 + / + 4y = 0 4X 2 + / + 4y + 4 = 4 4x 2 + (y + 2) 2 = 4 4x 2 + (y + 2) 2 4 4 4 4 � + (y + 2) 2 = 1 4 1
(2 + 13 , - 1)
y 5
-=----'--
The equation is in the form
(x _ h) 2 + (y _ k) 2 = 1 (major axis parallel to the -'--b-:2-'-: a2 y-axis) where a = 2, b = 1, h = 0, and k = -2 .
-5 51.
Solving for c:
Complete the square to put the equation in standard form:
c 2 = a 2 - b 2 = 4 - 1 = 3 -+ c = 13
9 x 2 + 4/ - 1 8x + 1 6y - 1 1 = ° 9(x 2 - 2x) + 4(/ + 4y) = 1 1 9( x 2 - 2x + 1) + 4(/ + 4 y + 4) = 1 1 + 9 + 1 6 9(x - l) 2 + 4(y + 2) 2 = 36 _ -'-1) 2 + 4(y + 2) 2 36 9(x---'36 36 36 (x - l) 2 + (y + 2) 2 = 1 9 4
Thus, we have: Center: (0,-2) Foci: (0, - 2 + 13),
Vertices: (0, 0), (0, -4) x
-2
-=--'--
(0, -2 -
The equation is in the form
(X _ h) 2 + (y _ k) 2 = 1 -'-b 2-'-- a 2
(0, - 2 - 13)
"V3) -s
(maj or axis parallel to the
(0 , -4)
585
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Chapter 1 1 : Analytic Geometry
55.
Center: (2, -2); Vertex: (7, -2); Focus: Major axis parallel to the x-axis; a = 5; Find b:
(4, -2); c=2 .
61.
(2, -2 + Vfl)
x
-3 (-3, -2) (0, -2) -7
57.
(1, 3);
(x _ 1) 2 + (y - 2) 2 = 1 b2 a2 Since the point (1, 3) is on the curve: o 1 =1 + a2 b2 � = 1 � b2 = 1 � b = 1 b2 Find a : a 2 = b 2 + c 2 = 1 + 9 = 10 � a = .J1Q (x - ll + (y - 2) 2 = 1 Write the equation: 1 10
b 2 = a 2 - c 2 = 25 - 4 = 21 � b = 51 . (X _ 2) 2 + (y + 2) 2 = 1 . the equation: Wnte 21 25 y
(1, 2); Focus: (4, 2); contains the point Major axis parallel to the x-axis; c = 3 . The equation has the form: Center:
(2, -2 - V21)
Vertices: (4, 3), (4, 9); Focus: (4, 8); Center: (4, 6); Major axis parallel to the y-axis; a = 3; c = 2 . Find b:
y
5
b 2 = a 2 _ c 2 = 9 - 4 = 5 � b = .J5 (X _ 4) 2 + (y _ 6) 2 = 1 Write the equation: 9 5
(-2, 2) (1 -
Vfb,
•
2) -2
63.
( 1 , 3)
�-+-___�
(4, 2)
( 1 , 2)
(1
+
x
VlO, 2)
-1
Center: ( 1 , 2); Vertex: (4, 2); contains the point ( 1 , 3); Major axis parallel to the x-axis;
a=3.
-3 59.
7
-1
The equation has the form:
x
Foci: (5, 1), (-1 , 1) ; Length of the major axis = 8; Center: (2, 1); Major axis parallel t o the x-axis; a = 4; c = 3 Find b:
(x - l) 2 + (y _ 2) 2 = 1 b2 a2
Since the point ( 1 , 3) is on the curve:
-o + -12 = 1 9 b � = 1 � b2 = 1 � b = 1 b2 Solve for c: c 2 = a 2 - b 2 = 9 - 1 = 8 . Thus, c = ±2.[i . (x _ 1) 2 + (y - 2)2 = 1 Write the equation: 1 9
.
b 2 = a 2 - c 2 = 16 - 9 = 7 � b = .J7 (x - 2) 2 + (y _ 1) 2 = 1 Write the equation: 7 16
y 5
(1
(2, 1
-
-
( 1 , 3) 2 V'i., 2) �..j---�
V7)
-5
-2
(1
+
2 V2, 2)
(4, 2)
(1, 1) -1
4
586
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Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
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Section 1 1.3: The Ellipse 65.
Rewrite the equation:
71.
y = ..}1 6 - 4x 2 / = 16 - 4x 2 , y � O 4x 2 + / = 1 6, y�O x2 + y2 = 1 y�O 4 16
.
b 2 = a 2 _ c 2 = 2500 - 625 = 1 875 b = .J1 875 "" 43 . 3 The ceiling will be 43.3 feet high in the center.
y
5
(0, 4)
73 .
-1
67.
x
-8 (O, �)
69.
The center of the ellipse is (0, 0). The length of the major axis is 20, so a = 10 . The length of half the minor axis is 6, so b = 6 . The ellipse is situated with its maj or axis on the x-axis. The equation is:
2 + ,L2 = 1 . _x_ 3600 625
10 2 + - = 1 -3600 625 L = I - 100 3600 625 y 2 = 625 . 3500 3600 y "" 24.65 feet The height 30 feet from the center: 30 2 + / =1 -3600 625 L = I - 900 3600 625 y 2 = 625 . 2700 3600 y "" 2 1 .65 feet The height 50 feet from the center: 50 2 + / =1 -3600 625 L = I - 2500 625 3600 y 2 = 625 . 1 100 3600 y "" 1 3.82 feet
y
(2, 0)
The equation is:
The height / 1 0 feet from the center:
y = - ..}64 - 1 6x 2 / = 64 - 1 6x 2 , y ::; 0 1 6x 2 + y 2 = 64, y ::; O x2 + / =1 4 64 ' 2
Place the semi-elliptical arch so that the x-axis coincides with the water and the y-axis passes through the center of the arch. Since the bridge has a span of 120 feet, the length of the major axis is 120, or 2a = 1 20 or a = 60 . The maximum height of the bridge is 25 feet, so
b = 25 .
Rewrite the equation:
(-2, 0)
Assume that the half ellipse formed by the gallery is centered at (0, 0) . Since the hall is 100 feet long, 2a = 100 or a = 50 . The distance from the center to the foci is 25 feet, so c = 25 Find the height of the gallery which is b :
x2 + / =1. 100 36
587
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Chapter 1 1 : Analytic Geometry
75. If the x-axis is placed along the 100 foot length and the y-axis is placed along the 50 foot length, the equation for the ellipse is: Find y when x = 40:
The equation of the orbit is:
y2 x2 =1 ---�2 + 93 x l06 92.99 x 106 2
x2 i -+- = 1 . 50 2 25 2
(
)
We can simplify the equation by letting our units for x and y be millions of miles. The equation then becomes: x2 i
40 2 + y2 = 1 252 = I _ 1 600 625 2500 9 y 2 = 625 · 25 y = 15 feet
5 02
) (
L
+ =1 8649 8646.75
--
81.
To get the width of the ellipse at x = 40 , we need to double the y value. Thus, the width 10 feet from a vertex is 30 feet.
The mean distance is 507 million - 23.2 million = 483.8 million miles. The perihelion is 483.8 million - 23.2 million = 460.6 million miles. Since a = 483.8 x l06 and c = 23.2 x 106 , we can find b:
t (
(
b 2 = a 2 _ c2 = 483.8 x 10 6 - 23.2 x l06 = 2.335242 x 1017 b = 483.2 x l 06
77. Because of the pitch of the roof, the major axis will run parallel to the direction of the pitch and the minor axis will run perpendicular to the direction of the pitch. The length of the major axis can be determined from the pitch by using the Pythagorean Theorem. The length of the minor axis is 8 inches (the diameter of the pipe).
t
The equation of the orbit of Jupiter is: 2 x -------::-2
y2
+
(483.8 x l 06) (483.2xl06)2
=1
We can simplify the equation by letting our units for x and y be millions of miles. The equation then becomes: 2
x ----
2
Y 234,062. 44 + 233,524.2 = 1
2(4)
=
83. a.
8
The length of the major axis is
�( 8l + ( 10)2 = .J164 = 2.J4i inches.
79. Since the mean distance is 93 million miles, a = 93 million. The length of the major axis is 1 86 million. The perihelion is 1 86 million - 94.5 million = 9 1 .5 million miles. The distance from the center of the ellipse to the sun (focus) is 93 million - 9 1 .5 million = 1.5 million miles. Therefore, c = 1 .5 million. Find b:
(
t (
b 2 = a 2 _ c2 = 93 x l06 - 1 .5 X 106 = 8 . 64675 x 1015 = 8646.75 X 101 2 b = 92.99 X 106
Put the equation in standard ellipse form:
AX 2 + Cy 2 + F = 0 AX 2 + Cy 2 = -F AX 2 Cy 2 -- + -- = 1 -F -F 2 x Y2 = 1 --- + (-F I A) (-F I C) where A :t= 0, C :t= 0, F :t= 0 , and -F I A and -F I C are positive. If A :t= C , then - :t= - . So, this is the equation of an ellipse with center at (0, 0). If A C , the equation becomes: Ax 2 + Ai = _ F � x 2 + y 2 ,;, -F A
� �
t
h.
=
This is the equation of a circle with center at
(0, 0) and radius of
�-: .
588 ©
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Section 1 1.4: The Hyperbola
85.
Answers will vary.
Write the equation: y=
Section 1 1.4 1.
3.
d
y
�
V2 =
x-intercepts:
F2
(0,
0 2 = 9 + 4x 2 4x 2 = -9 x 2 = 94 (no real solution)
1 9.
right 5 units; down 4 units
7.
hyperbola
11.
True; hyperbolas always have two asymptotes.
13.
(b) ; the hyperbola opens to the left and right, and has vertices at (± 1, 0) . Thus, the graph has an
equation of the form
VI
21.
x2 - Y 2 = 1 . b2
x
-2 '12)
= (0, 6)
= (0, --4)
Foci: (-5, 0), (5, 0); Vertex: (3, 0) Center: (0, 0); Transverse axis is the x-axis;
a = 3; c = 5 .
Find the value of b:
b2 = c2 - a2 = 25 - 9 = 16 => b = 4 x2 - 2 = 1 . Write the equation: 9 �6
2 y2 4 -b =1.
�
(3 , 0)
Center: (0, 0); Focus: (0, -6); Vertex: (0, 4) Transverse axis is the y-axis; a = 4; c = 6 . Find the value of b:
F2
(a); the hyperbola opens to the left and right, and has vertices at (±2, 0) . Thus, the graph has an equation of the form
=
b2 = c2 _ a2 = 36 - 1 6 = 20 b = J2Q = 2J5 2 2 Write the equation: �6 - �O =1 .
y 2 = 9 + 4 (0) 2 y2 = 9 Y = ±3 -? (0, -3) ,( 0, 3) The intercepts are (0, -3) and (0, 3) .
5.
( 1 , 0) 5
y-intercepts:
1 7.
2 V2 x
= ( _2 _ 3) 2 + ( 1 -( _4))2 = �( -5l + (5) 2 = ../25 + 25 = Fo = 5J2
--
1 5.
-
4-2 - �2 = 1 .
2
y = _ A:. x 3
Center: (0, 0); Focus: (3, 0); Vertex: ( 1 , 0); Transverse axis is the x-axis; a = 1; c = 3 . Find the value of b: b2 = c2 - a 2 = 9 -1 = 8
,
y
,
b = J8 = 2J2
589
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Chapter 1 1 : Analytic Geometry
23.
Vertices: (0, -6), (0, 6); asymptote: y 2 x ; Center: (0, 0); Transverse axis is the y-axis; a = 6 . Find the value of b using the slope of the a = b6 = 2 => 2b = 6 => b = 3 asymptote: b =
27.
2
Write the equation:
2
25
-
the center of the hyperbola is at (0, 0). a = 5, b = 3 . The vertices are (5, 0) and
( -5, 0) .
Find the value of c: 2 c 2 = a + b 2 = 25 + 9 34 => c = J34 The foci are ( $4, 0) and ( -$4, 0) .
Find the value of c: c 2 = a 2 + b 2 = 3 6 + 9 = 45
c = 3..[5
x l - 9 =1
=
2
The transverse axis is the x-axis. The asymptotes
�6 - x9 = 1 . F2 1
3
3
=
(0, 3� )
V =
0'
are y = S' x; y = - S' x . y
2x
10
x
-10
25.
Foci: (-4, 0), (4, 0); asymptote: y = - x ; Center: (0, 0); Transverse axis is the x-axis; c = 4 . Using the slope of the asymptote: - � = - 1 => - b = -a => b = a
a
.
Find the value of b: b 2 = c 2 a 2 => a 2 + b 2 = c 2 b 2 + b 2 1 6 => 2b 2 = 1 6 => b 2 = _
=
b = J8 = 2-fi
a
= J8 = 2-fi
Write the equation:
29.
g
( = 4)
4x 2 - y 2 = 1 6 Divide both sides by 1 6 to put in standard form: y 2 16 x 2 y 2 4x 2 - = - => -- = 1 16 16 16 4 16 The center of the hyperbola is at (0, 0). a = 2, b 4 . The vertices are (2, 0) and (-2, 0). Find the value of c: c 2 = a 2 + b 2 = 4 + 16 = 20 =
C
c = J2Q = 2..[5 The foci are (2..[5,0) and (- 2..[5, 0) .
( a = b) 2 2
Xg � = 1 . -
The transverse axis is the x-axis. The asymptotes are y = 2x; y = - 2x .
�90
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Section 11.4:
31.
-9x2 9 Divide both sides by 9 to put in standard form: / 9x2= 9 � / x 2= 9 center 9 of9 the hyperbola 9 1 is at (0, 0). The a 3, b 1 . The ces areof(0,c: 3) and (0, -3). Findverttheivalue c2 a2 b 2 9 1 10 c .JlO The foci are (O,.JlO) and (0, -.JlO) . /
35.
=
---
=
=
-
---
=
+
=
=
+
=
+
=
=
=
+
The the verti hyperbola a 1,center b 1of. The ces areis(at1, (0,0) 0).and (-1, 0). Find the value of c: c2 a2 b 2 1 1 2 c J2 The foci are (J2, 0) and ( J2 0). The verse axiares iys thex;x-axi The trans asymptotes y -xs. . The equation is: x2- / 1 . The of the hyperbola is at (0, 0). a 6,center b 3 The vertices are (0, -6) and (0, 6) . Find the value of c: 2 c2 a2 b 36 9 4 5 c +t5 3.J5 The foci are (0, -3.J5 ) and (0, 3-/5) . The verseesaxisare iys the2x;y-axis.-2x . The The trans asymptot equation is: 362 92 1 . (7, -1 ); Vertex: (6, - 1) ; Center:verse(4, -axis1 ); iFocus: Trans s parallel to the x-axis; a 2; c 3 Find the value of b: b 2 c2-a2 9 - 4 5 b -/5 Write the equation: ( x -44 ) 2 ( y 5 1)2 1 . =
1
-
=
=
37.
=
=
=
=
=
+
.
=
+
=
=
Y
=
L_�
39.
_x2 25 Divide both sides by 25 to put in standard form: / x2 1 25 The25 center of the hyperbola is at (0, 0). a 5, b 5. Thevertices are ( 0,5) and (0, -5). Find the value of c: c2 a2 b2 25 25 50 c go 5.J2 The foci are (0, 5.J2) and (0, -5.J2) The ransverse axisare iys thex;y-axis The tasymptotes y -. x . /
=
--- =
=
=
=
=
.
=
=
+
+
=
,
=
=
33.
The Hyperbola
=
=
=
.
=
=
�
=
+
=
=
=
.
y
=
=
x
59 1
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exist. No portion of this material may be reproduced, in any fonn or by any means, without pennission in writing from the publisher.
Chapter 11:
41.
43.
Analytic Geometry
Center: (-3, (-3,-2); -4); Focus: (-3, -8); Vertex: Trans a=2;verse c=4.axis is parallel to they-axis; Find thevalue ofb: b2=c2_a2=16-4=12 b=J0. =2..J3 2=1. Write the equation: (y :4)2 (X+3) 12
45.
=>
Foci : (3,(5,7),7);(7,Trans 7); Vertex: (6, 7); Center: x-axis; a=1; c=2.verse axis is parallel to the Find the value ofb: b2=c2-a2=4-1=3 b=..J3 7 Write the equation: (X -5)2 1 (y -3 )2=1. Y 12
V1=(4,7)
\5, 7 +{3) \
/ I.
I
Y
/
47.
- 7=-I3(x - 5)
V2=(6.7) epz = (7,7) (5.7)
Y
(1, -1)a=; 2 . Vertices:verse(-1,axis-1),is parallel (3, -1); toCenter: Trans the x-axis; Asymptote: y+1='23 (x -1) Using the slope of the asymptote, fmd thevalue ofb: b b2 -23 b=3 -=-= aFind thevalue of c: c2=a2+b2=4+9=13 c=JU . (x-l4 )2 (y+91)2=1 . Wnte. the equatIon:
(y +3)2 (x _2)2 =1 4 9 The a=2,center b=3.of the hyperbola is at(2, -3). The Findvertices thevalueareof(0c,: -3) and(4, -3). c2=a2+b2 4+9 13 c=JU Foci: ( 2-JU, -3) and ( 2+JU,-3) . Transverse axis: y=-3, parallel to x-axis. Asymptotes: y+3=2'3 (x-2); y+3=-2'3 (x-2) =
x
=
=>
y+ 3
-7 -,J3(x - 5) =
=
�(x - 2)
x
592
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Section 11.4:
49.
(y_ 2)2-4(x+2)2=4 Divide both sides by 4 to put in standard form: (y -2)2 (x+1 2)2= 1 · 4 The b= 1of. the hyperbola is at (-2, 2). a=2,center Thevertices of c: are (-2, 4) and (-2, 0 ). Find the value c2=a2+b2=4 +1 =5 c=.J5 Foci: ( -2, 2-.J5 ) and (-2, 2+.J5) . Transverse axis: x=-2 , parallel to the s. y -2= 2(x+2); y 2=-2(x+2) y-axi Asymptotes:
y 4
y+2=x+l x
�
y
-
2
=
-2(x +2) \
y 1Y
-
-
2
=
y+ 2 53. .
2(x + 2)
=
-(x + 1)
Complete the squares to put in standard form: x 2-l-2x-2y-1 =0 (x 2-2x+1)-(l+2y+1)=1+1-1 (x_1)2_ (y+l)2=1 The the hyperbola center vertices areis(1,(0,-1)-1). and a= 1, b= 1of. The (2, -1). Find the value of c: c2=a2+b2= 1 +1 =2 c=.J2 Foci: ( 1-.J2 , -1 ) and ( 1+.J2 , -1 ) . Transverse axis: y=-1 , parallel to x-axis. Asymptotes: y +1 = x-I; y +1 = -1) . -(x - 1) �
x \
\
51.
The Hyperbola
Y
(x+1)2_(y+2)2=4 Divide both sides by 4 to put in standard form: (x+1)4 2 (y+2)2 =1 . 4 -2). (-1, s i hyperbola the of center The a=2, b=2. The c: -2) and ( 1 , -2). the valueareof(-3, Findvertices c2=a2+b2=4+4=8 c=Fs =2Ji Foci: ( -1-2.J2 , -2) and ( -1+2.J2 , -2) Transverse axis: y=-2 , parallel to the x-axis. Asymptotes: y +2= x+ 1; y+2=-(x+1)
+
1
=
-
Y 2
(x
-4
55.
Complete the squares to put in standard form: l-4x2-4y-8x-4 =0 (l-4y+4)-4(X2+2x+1)=4+4-4 (y_2)2-4(x+1)2=4 2 _ (x+1)2=1 (y -2) 4 1 is (-1, 2). hyperbola the of center The a=2, b= 1 . The the valueareof(-1,c: 4) and (-1, 0). Findvertices c2=a2+b2=4+1 =5 c=.J5 Foci: ( -1 , 2-.J5 ) and ( -1 , 2+.J5) . �
593
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 11:
Analytic Geometry
Trans verse axis:y-2=2(x+I); x=-1 , parallel to they-axis. y-2=-2(x+I). Asymptotes: I
,Y
-2
��V2
=
=
F2 =
59.
2(x + 1)
(-1, 2 +VS) (-1, 4)
Complete the squares to put in standard form: y2_4x2-16x-2y-19= 0 (/ -2y+I) -4(x2+4x + 4)=19+1-16 (y_I)2-4(x+2)2=4 (y_41)2 (x+12)2=1 The a= center b=1.of the hyperbola is (-2, I). The valuvert e of ices are (-2, 3) and (-2, -I). Find the =a2+b2=4+1=5 =../5 Foci: ( -2, 1-../5) and ( -2, 1+../5) . Transverse axis: x=-2, parallel to they-axis. Asymptotes: y-l=2(x+2); y-l=-2(x+2) . '{5) y 2,
x
c:
c
57.
Complete the squares to put in standard form: 4x2-/ -24x-4y+16= 0 4(x2-6x+9)_(y2+4y+4)=-16+36-4 4(x-3)2_ (y+2)2=16 2 (y+2)2=1 (X-3) 16 is (3, -2). 4 The=2,center of the hyperbola b=4. The verti value of ces are (1, -2) and (5, -2). Find the c2=a2+b2=4+16=20 =50 =2../5 Foci: (3-2../5, -2) and (3+2../5, -2) . Transverse axis: y=-2 , parallel to x-axis. Asymptotes: y+2=2(x -3); y+2=-2(x-3)
2
c
1
=
-2(x
+
2)
\
F2
\
=
(-2, Y
1
+
Y
1
=
2(x
+
2)
a
x
c:
c
y +2
F[
=
=
-2(x - 3)
VI
=
Y
3
7
-3
(3 - 2 J5. -2)
•
61.
(I, -2)
x
eF2 =
(3 +2 ",,r;:;5._. -2 )
Rewrite the equation: y=�16+4x2 /=16+4x2, y�O /-4x2=16, y�O 2 16/-4x =1, y�O
VI (5, -2) =
x
594
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 11.4:
63.
Rewrite the equation: y =_ �r_-25- -+- x2y2=_ 25+ x2, 0 x2_ y2= 25, x252 -/25 =1, Y ::s;
y::s;o
The Hyperbola
The equation the hyperbola location of theoffueworks displathaty is describes the /--x2 -----"-: 00=1 27, 575, 9 2 550 Siindincevidual the fIreworks displx=ay 5280 is dueandnorthsolvofe tthhee at A, we let equation for 52802 - 27, 575,y2 900=1 -5502 2 =-91. 16 27, 5� 5, 900 / = 2, 513, 819, 044 = 50, 138 Therefore, t h e fIreworks diduesplanorth y wasof50,th138e feet (approxi m ately 9 . 5 miles) person at A. Tothe deter mineofthetheheight, we fIrstusedneedto generat to obtaein equation hyperbola the hyperboloid. acingthetheequation center ofofthethe hyperbola at the Plorigin, 2 / =1 . hyperbola will have the form ';--The 200 center diameter is 200 feetaso web2have a=-= 100. We also know that the base 2 diameter is 400 feet. Since the center of the hyperbola is at the origin, the points (200, -360) and (-200, -360) must be on the graph of our hyperbolaTherefore, (recall the center is 360 feet above ground). (200)2 (-360)2=1 -(100)2 b2 4- 360b22=1 3602 3=b2 b2=43, 200 b=../43, 200=120J'j The equation of the hyperbola is x2 / -----=1 10, 000 43, 200 300 150 . At the top of the tower we have x=-= 2 y.
y
10
x
Y
65.
Fiplarstce,notesuchthatthatallthpoints where a burstwould couldbetathkee e t i m e difference sahyperbola me as thatwithforAtheandfIrstB asburst,the would formwiath a f o ci . Start diagram: (x,y), ........
B
............ .. ......
...........
14
67.
N
............ I 1
2 miles
:
,
�I
A
Assume a coordinate system with the x-axis containing BA and the origin at the midpoint of BA. The ordered pair ( y) represents the location of thefeet fIreworks. We know thatperson soundat point travelAs atis 1100 per second, so the 1100 feetat poiclosernt B.toSince the fIreworks displayofthan the person the difference the distance from (x, y) to A and from (x, y) to B is the constant 1100, the point (x, y) lies on a hyperbola are atA andB. The hyperbola whose has thefociequation / -ax22 --=1 b22a=1100, so a= 550. Because the where dista5nce60 fbetween theeachtwoperson peopleis atis a2 focus milesof the (10, e et ) and hyperbola, 2c=10, 560we have =5280 b2=c2_a2= 52802 -5502= 27, 575, 900 x,
c
595
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 11:
Analytic Geometry
1502 ---=1 y2 10, 000 43, 200 2 .25 y-=1 43, 200 / =54000 y",232.4 The height of the 232.4+360= 592.tower 4 feetis. approximately Since deflected angle, thetheparticles asymptotesarewill be y=at±xa 45°. Sithencehyperbola, the vertexweisknow 10 cmthfrom at a=the10.center The of slope of the asymptotes is given by ±!!..a . Therefore, we ha v e b -=1 b -=1 O aUsing =>the origin 10 as=>b=l the center ofpartithecle path hyperbola, the equation of the would be x2 / x�o ---=1 100 100 Assume -ax22 --b2/=1. If",thae eccentri co.ity When is closeb tois close 1, thento 0, the and b", hyperbola is veryarenarrow, the asymptotes close tobecause O . the slopes of If tishmuch e eccentrilargercitythanis verya andlarge,b isthen very large. The resul t i s a hyperbol a that i s very wide,large. because the slopes of the asymptotes arevery For ':;'a2 -.;..b2=1 , the opposite is true. When the ricity isthecloseslopes to 1, ofthethehyperbola isvery wicleccent dosee because asymptotes are to O. When the eccentricity i s very large, narrowlarge.because the slopes ofthethehyperbola asymptotesisveryarevery --
69.
73 .
a.
b.
71.
x2 y2=1 (a=2, b=1) This s a hyperbola with horizontal transverse axis, icentered at (0, 0) and has asymptotes: y=±21 x x42 (a=1, b=2) y2 --=1 This is a hyperbola withvertical transverse axis,1 centered at (0, 0) and has asymptotes: y=±2x . Siasymptotes, nce the twotheyhyperbolas have the same areyconjugates. 4-
3
2
x 2 -y =1
4
75.
C
C
-3
Put the equation in standard hyperbola form: AX2+c/ + F = 0 A· C 0, F 0 Ax2+C /=-F A-FX2 + Cy-F2=1 2 +( y�)2 =l ( x�) _ _ Sithinsceis -Fa hyperbola / A and -Fwith/ Ccenter have(0,opposite signs, 0). The ransvverse x-axis ifif --FF / A> The ttrans erse axis axis isis thethe y-axis A o0 .. <
*
/
596
<
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 11.5:
Section 11.5 1. 3.
5. 7.
21.
sinA cosB +cosA sinB l -�OSB A-C cot(2B)=Bx2+2xy+3y2-2x+4y+l0= 0 A=1, B=2, and C=3 ; B2-4AC=22-4(1)(3)=-8 2 Since B ellipse. -4AC 0, the equation defines an True x2+4x+y+3=O A=1 andC=O ; AC=(I)(O)=O . Since AC= 0, the equation defmes a parabola. 6x2+3y2-12x+6y= 0 A=6 andC=3 ; AC=(6)(3)=18. Since Aellipse. C> 0 andA C, the equation defmes an 3x2-2y2+6x +4= 0 A=3 andC=-2; AC=(3)(-2)=-6. Since AC 0, the equation defines a hyperbola. 2i x2- + x= 0 A=-1 and C=2; AC=(-1)(2)=-2. Since AC 0 , the equation defines a hyperbola. x2+i -8x+4y= 0 A=1 and C= 1; AC=(1)(1)=1. Since AC> 0 andA= C , the equation defmes a circle.
11.
13.
X
-
23.
7t
,
,
.
7t
7t
.
,
7t
,
7t
,
7t
,
,
--
,
,
Y
25.
<
19.
X
,
<
1 7.
- =-
_
'#
1 5.
x2+4xy+y2-3= 0 A=1, B=4, and C=1; 1-1 0= 0 = cot(2B)= A-C B 4 4 !!:. 2B= 2=::;. B=!!:.4 = cos--y 4 sm-4 =.J22 x' _ .J22 y' =.J22 (x' y') .J2 .J2 +-y cos-=-x y=x sm-+y 4 4 2 2 .J2 ('x +y') =2 5x2+6xy +5 y2-8= 0 A=5, B=6, and C=5; 5 -5 0 cot(2B)= A-C =-=-=O B 6 6 2B=!!:.2=::;. B=!!:.4 .J2 .J2 sm-=-x x=x cos--y 4 4 2 2 =.J22 (x' _ y') .J2 .J2 y=x sm-+ y cos-=-x +-y 2 2 4 4 .J2 ('x +y') =2 13x2-6J3 xy + 7i -16= 0 A=13, B=-6J3 , and C= 7; 13-7 6_ =_ 13 cot(2B)= A-C B =-6 J3=_ -6 J3 3 2 B=2B=-=::;. 3 3 13 , 1 , --y x x cos--y sm-=-x 3 3 2 2 =�( x' -J3y' ) J3, +-1, = x sm -+ y cos -= -x 3 3 2 2 =� ( J3x' + y' ) --
<
9.
Rotation of Axes; General Form of a Conic
.
7t
,
7t
,
Y
,
,
-
-
y
,
,
1t
7t
=
.
.
1t
,
1t
,
.
1t
1t
597
y
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 11:
27.
Analytic Geometry
4x2-4xy+y2-s.J5 x-16.J5 y = 0 25x2-36xy+40y2-12J13 x-sJ13 y= 0 A=4, B=-4, and C= 1 ; A=25, B=-36, and C=40; A-C 4-1 =--3 cos2() =--3 25-40 5 cos2()=-5 cot(2()) =--= cot(2()) = AB-C=--=-' B -4 4' 5 -36 12' 13 I-/3 =H; =.k =2�; 2.[5 'inO= � � = 2 = = 'in =l -H) 2 Vs.J5 5 ' 5 I+ I+ � 1 3 =V13f9 =_JG3_ =3JG fL _1 =.J5 co,O= co, =� H) = 2 13 2 Vs J5 5 3JG , ---y 2JG , , cos() -y, sm. () =--x 2J5 , J5 ,--y x = x x=x,cos() -y,.sm () =-x 13 13 5 5 JG (3x' -2y' ) = J5( x, -2Y' ) -13 5 2JG ,+--y 3JG , ,sm . ()+y ,cos() =--x 2J5 ,+-y J5 , y=x y=x, sm. ()+y, cos() =-x 13 13 5 5 JG(2'x+ 3 Y' ) = J5( 2x,+y' ) =13 5 x2+4xy+y2-3=0 ; ()=45° (-; (X' - Y' »),+4(-; (X' - Y' ») (-; (X' +y' )H� (x' + Y' )J -3=0 � ( X,2 1 2x) + y'2)+1 2( X,2 y,2)+� ( X,21 +2x'y' y,21) -3=0 x _2X,2 x'y' + -2y,2+2 ,2 2y,2+2_X,2+ x'y' + 2y,2=3 3X,2 y,2 3 x,2 y,2 ---=1 1 3 ' Hyperbola; center at the origin, trans v erse axi s i s the x -axi s , vertices (±1 , 0). 29.
-
'
--
o
o
31.
y
_
/
_
+
_
X
_
_
_
=
598
© 2008 Pearson Education, Inc., Upper Saddle River, Nl. All rights reserved . This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 11.5:
33.
35.
Rotation of Axes; General Form of a Conic
5x2+6xy+ 5y2-8= 0; () =45 ° (J2 )(J2 ) (J2 , , 2 -8=0 (J2 5 T(x' - Y')) 2+6 T(x' - Y') T{x'+ Y') +5 T{x+y) J %( X,2_ 2x'y' + y,2) +3 ( X,2_ y,2) +%( X,2+ 2x'y' + y,2) -8= 0 �2 X' 2_ 5x'y' +�2y,2+ 3X,2_ 3y,2+�2 X' 2+ 5x'y' +�2y,2=8 8X,2+ 2y,2=8 X, 2 y, 2 -+-=1 Ellipse; center at the origin, major axis is the y' -axis, vertices(0,1 ±2)4. 13 x 2-6J3 xy+ 7i -16= 0 ; () =60 ° 13
G
(X' - Jly')
J (� -6Jl
(X' - Jl y')
)(�
) G
, , Jlx' + y ) + 7 (Jlx' + y ) (
)
2
-16
x -3 =
0
3Jl r:: ,2 13 ,2 r:3: " + ,2 16 " -,,3y r:: ,2) + 73x,2 + 2"jxy r:: , , ,,3x - 2xy 4( x - 2,,3xy + 3y'2) - -- ( ( Y ) ; 2 9,2 21 ,2 7Jl " 13 39 9 7 ,2 16 13Jl " r::3 " +-y +-x +-xy +-y -x,2 ---xy +-y,2 --x,2 + 3 "jxy =
=
4
2
4
2
2
4
2
4
4X,2 + 16y,2 16 X,2 ,2 =
-
37.
Y +-=1
Ellipse; center at the origin, major axis is the x' -axis, vertices ( ±2, 0). 1 4x2-4xy+i -8.J5x-16.J5 y= 0; () "" 63. 4 ° 4 ( �( x' -2Y'))' -4 [ � (x' -2Y'))[� (2x' +y')H �(2x' + Y')), -8.J5[ � ( x' -2Y' ))-16 .J5[ � (2 x' + y,))= 0 �( X' 2_ 4x'y' + 4y'2) _�( 2X,2-3x'y' -2y'2) +�( 4x' 2+ 4x'y' + y,2) -8x' +16y' -32x' -16y' = 0 4" +-y1 ,2- 40x' = 0 16 ,2--x8 ,2+-12"xy +-y8 ,2+-4 x ,2+-xy 16" +-y -x45 '2 --xy 5 5 5 5 5 5 5 5, 5y 2-40x' = 0 y,2=8x' Parabola; vertex at the origin, axis of symmetry is the x axis, focus at (2, 0). 4
I
599
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any fonn or by any means, without pennission in writing from the publisher.
x
Chapter 11:
39.
Analytic Geometry
25x2-36xy+40 /-12.Ji3x-S.Ji3y=0 ; fhd3.7° 25( � (3x' -2y')J -36(� (3x' -2Y'))( � (2x'+3y'))+40(� (2x'+ 3Y'))' -12.Ji3( � (3x' -2Y')) -S.Ji3(� (2x'+3Y')] = 0 �: ( 9X, 2-12x'y'+ 4y, 2) -�� ( 6X, 2+ 5x'y' -6y, 2)+�� ( 4X, 2+12x'y'+ 9y, 2) -36x'+ 24y'-16x'-24y'= 0 SO " 216 , 2 225 300" 100 216 I , , , 2 2 2 -x --xy +-y --x --xy +-y 13 13 13 13 16013 4S0"13 360 , 2+-xy , 2- 52x '= 0 +-x +-y 13 13 13 13x, 2+ 52y, 2-52x' = 0 x, 2-4x'+ 4y, 2= 0 (x'-2) 2+4y, 2=4 2-+ y, 2 1 -'---(x'-- -2) -"--4 at (2,1 0), major axis is the x' -axis, vertices(4, 0) and(0, 0). Ellipse; center -
=
y
y'
5
x 41.
16x2+ 24xy+9 /-130x+90y= 0 A-C=16-9 7 7 A=16, B=24, andC=9·, cot(2(})=-B -24 =-24=>cos(2(})=-25 1 -1+ - =�16 =4=>(}�36.90 -f-fsin(}=� =&9 =s3 ; cos (}=� S 3 ,=-1 (4x ' -3Y )' x= x, cos() -y, sm. () =-x45 , --y 5 5 y=x' sin(}+ Y 'cos (}=�5 x'+ i5 y' =.!.5(. 3x'+ 4y') 600
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Section 11.5:
Rotation of Axes; General Form of a Conic
16(� ( 4x' -3y')J +24(� ( 4x'- 3Y'))(� ( 3x'+4y'))+9G ( 3x'+4y')J -130(� ( 4x'- 3y'))+90G ( 3X '+4Y ') )= 0 �� (16x,2- 24x )/ + 9y,2)+�: ( 12x,2 +7 x'y '-12y,2) +:5( 9X,2+24x'y '+16y'2)-104x'+78y'+ 54x'+ny'= 0 288 ,2 288 168" --y 384" 144 ,2+-x 256 ,2--xy -x 25 25 +-y 25 8125 ,2+-xy 25 25 216" 144 ,2- 50x '+150y=' 0 +-x 25 25 ,2 25 ,2+-xy+-y 25x -50x'+2150y'= 0 x, - 2x2'=-6 Y ' (x'- 1) =-6y' +1 (X ' -1)2=-6( Y ' -�) Parabola; vertex (1, �) , focus (1, -�) ; axis of symmetry parallel to the y' axis. y 5
x
43. 45. 47. 49. 51.
A=l, B=3, C=-2 B2-4AC=32- 4(1)(- 2)=17>0 ; hyperbola A=1, B -7, C 3 B2- 4AC=(_7)2-4(1)(3) 37> 0 ; hyperbola A=9, B 12, C=4 B2-4AC=122-4(9)(4)= 0 ; parabola A=lO , B=-12, C=4 B2- 4AC=(-12)2_4(10)(4)=-16<0 ; ellipse A=3, B=-2, C=1 B2-4AC=(_2)2-4(3)(1)=-8<0 ; ellipse =
=
=
=
601
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 11:
53.
55.
Analytic Geometry
A' =Acos28+Bsin8cos8+Csin28 B' =B(cos28-sin28)+2(C-A)( sin8cos8) C' =Asin28-Bsin8cos8+ Ccos28 D' = Dcos8+Esin8 E' =-Dsin8+Ecos8 F ' =F B'2=[B(cos28-sin28)+2(C-A)sin8cos8f =[B cos28-(A-C)sin28f =B2 cos228-2B(A -C) sin28cos28+(A-C)2 sin228 4A'c'=4[A cos28+B sin8cos8+Csin28][A sin28-Bsin8cos8+Ccos28] =4[AC+C�S2B) + B( sin22B) +C( l-C�S2B)J[Ae-C�S2B) B(sin22B) +C( 1+C�S2B)] =[A(1 +cos28) B(sin28)+C(l-cos28)][A(I-cos28)-B(sin28)+C(l+ cos28)] =[(A+C)+Bsin28+(A -C)cos28][(A+C)-(Bsin28+(A -C)cos28)] =(A+C)2 -[B sin28+(A -C) cos28]2 =(A+C)2_[B2 sin228+2B(A-C)sin28cos28+(A-C)2 cos228] B,2_4A'C' =B2 cos228-2B(A-C)sin28cos28+(A _C)2 sin228 -(A+ci +B2 sin228+2B(A -C)sin28cos28+(A-C)2 cos228 =B2(COS22B+sin228)+(A-C)2(COS228+sin228)-(A+C)2 =B2+(A_ C)2 -(A+C)2=B2+(A2 -2AC+C2)-(A2+2AC+C2) =B2 -4AC d2=( Y. 2- YI)2+(X2 - XI)2. =(X2, SInuLJ+ Y2, cosuLJ - X,ISInB- YI' COSuLJ)2+(X'2COSu-Y2, SInuLJ-X,I COS u+ YI, SInu11 )2 =((X� -x;)sin8+(y� - Y;) cos8) 2+(( x� -x;)cos8-( Y; -y;)sin8) 2 LJ 2(X'2-X'I)( Y'2- YI, ) s.m ucosu+ LJ LJ (Y2' -YI, )2COS2u =(X'2 - X,I )2SIn. 2u+ +(x; _ X;)2COS28-2(x� -x;)(Y; -y;)sin8cos8+(y; -y;)2 sin28 =(x; -X;)2( sin28+cos28) +( Y; -y;)2( cos28+sin28) =[(x; _X;)2+( Y; -y ;)2J(Sin28+cos28) =(X'2 - X,I )2+( Y'2 - YI, )2 Answers willvary. -
+
57.
11
.
Ll
.
11
59.
602
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Section 11.6:
Section 11.6 1. 3. 5. 7.
9.
cosB; sinB � , ellipse, parallel, 4, below 2 True e 1; 1 ; parabola; directrix is perpendicul ar to. the polar axis and 1 unit to the right of the pole 4 4 2-3sinB 2 l-�2sinB 3 4 ep 2, e "2; p 3 Hyperbola; directrix is parallel to the polar axis and �3 units below the pole. 3 3 4-2cosB 3 4 '. 1 1--cosB 21 3 3 ep "4 e "2; p "2 Ellipse; directrix is perpendicular to the polar axis and �2 units to the left of the pole. r
r
=
(�, 0).
'
=
=
Directrix
y 2
----'-_--'-_-+-�_+_--'-�x Polar axis
-2 1 5.
=
r = ----
=
P
=
1
r= ----
=
11+, cose B1 , 1 Parabola; r to theex ispolar axis unitdirectrix to the rightis perpendicul of the pole;avert r = ---
ep =
P=
=
11.
13.
Polar Equations of Conics
8 4+3sinB 8
r = ----
2 1+�4 sinB
Ellipse; directrix is parallel to the polar axis 83 units above the pole; vertices are 3 (!7'2:)2 and (8' 2rc ) . Also: "21 (8+78) 732 so the center . at (8 327'321r ) ( 247'321r )' and 247 274 so that the second focus is at ( 247 + 247'321r ) ( 487'321r ) a =
=
_
=
IS
c=
=
_
0
=
=
(�'I)
Polar x
Axis
603
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exist. No portion of this material may be reproduced, in any fonn or by any means, without pennission in writing from the publisher.
Chapter 11:
17.
Analytic Geometry
the second focus . at (38 +38'2Tr) = ( 316'2Tr) .
9 = r ---3-6cosB 9 = 3 r=3( 1-2cos B) 1 - 2cosB 3 ep = 3 , e = 2 , p = 2 Hyperbola; directrix is perpendicular to the polar axis �2 units to the left of the pole; vertices are (- 3, 0) and (1,1t ) . Also: = i( 3-1)=1 so the center is at (1+1, Tr )=( 2, Tr ) [or (-2, 0)], and c=2- 0=2 so that the second focus is at (2+2, Tr )=(4, Tr ) [or (- 4, 0)].
IS
x Polar ---+----it-:--:-:-l�axis
(4, Jt)
a
Directrix
21.
Directrix )I
6 r(3-2sinB)=6=> r=3-2sinB r= 6 1- 2sin 3( 1- % sinB) % B 2' ep = 2 , e =- p = 3 3 is parallel to the polar axis 3 Ellipse; directrix units below the pole; vertices are ( 6 , �)2 and (�'5 31t). 2 Also: ="21 ( 6+56) =5"18 so the center . at ( 6 .!i5' 2Tr) =(g5' 2Tr)' and c=g5 - 0= g5 so that the second focus is at c:+1:';) =( 245 , ;) . ----;----7
(1, "iT)
19.
x
Polar axis
8 r=--2-sinB8 4 r = ---;,...-----;:2(1-�sinB} 1-�sinB 1' ep = 4 , e = - p = 8 2rix is parallel to the polar axis 8 Ellipse; di r ect units below the pole; verti c es are (8 , �2) and (�3' 31t2). Also: = i( 8+�) =1: so the center is at (8 !i3' 2Tr)=(�3' 2Tr)' and =�3 - 0=�3 so that
a
IS
_
y
(6, �)
a
_
(2, Jt)
C
(2, 0)
31t 5' 2
(.§ )
x
Polar axis
Directrix
604
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Section 11.6:
23.
(�) (_) ( )( )
6 6 6sec O cos O r= ---2sec O - 1 2 1 - 1 2 - cos O cos O cos O COS O 6 = 6 cos 0 2 - cos 0 = 2 - cos 0 6 3 1 1 - -cos 2 O ep= 3 , e= 1 ' p=6
27.
29.
Ellipse; directrix is perpendicular to the polar axis 6 units to the left of the pole; vertices are ( 6 , 0 ) and ( 2 , ) Also : a = � ( 6 + 2 ) = 4 so the center is at ( 6 - 4, 0 ) = ( 2 , 0 ) , and = 2 - 0 =2 so that the second focus is at ( 2 + 2 , 0 ) =( 4 , 0 ) .
c
.
Directrix
31. x
Pol.ar --t-(-2,-Jt-)..--+----...(-6,�O )axls
33. 25.
----
8 r= 4 + 3sin O 4r + 3r sin 0 = 8 4r=8 - 3rsin O 16r2 = (8 -3rsin 0) 2 16(x 2 + i) = (8 _3y) 2 1 6x 2 + 1 6i = 64 - 48 Y + 9 i 16x 2 + + 48Y - 64 = 0
7y2
2
1t
Polar Equations of Conics
r= --1 + cos 0 r+ rcos O = 1 r = l - rcos O r2 =(1 - r cos 0) 2 x2 + i =(l - x) 2 x2 + i = 1 - 2x + x 2 i + 2x - 1 = O
----
r= 3 - 69cos O 3 r - 6 rcos 0 = 9 3 r = 9 + 6 r cos 0 r = 3 + 2rcos O r2 = (3 + 2 r cos 0) 2 x 2 + y 2 = (3 + 2 X ) 2 x 2 + y2 = 9 + 12x + 4x2 3x2 - y 2 + 12x + 9 = 0
---
8 r= 2 -sin O 2r - rsin O = 8 2r = 8 + rsin O 4r2 = (8 + rsin 0)2 4(x 2 + i) =(8 + y) 2 4x 2 + 4i = 64 + 16y + i 4x 2 + 3y 2 - 1 6y - 64 = 0 r(3 - 2 sin 0) = 6 3 r- 2 rsin O = 6 3 r = 6 + 2 rsin 0 9 r2 = (6 + 2 rsin O) 2 9(x 2 + i) = (6 + 2y)2 9x 2 + 9 i = 36 + 24y + 4 i 9x 2 + 5y 2 - 24y - 36 = 0
605
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 11:
35.
Analytic Geometry
6secB r= 2secB-l 6 r= 2-cosB 2r-rcosB=6 2r=6+rcosB 4r2=(6 +rcosB)2 4(x2 + /)=(6+x)2 4x2 +4y2=36+12x+x2 3x2+4y2-12x-36=O r=l+esinB e=1; p= 1 1 r=--1+sinB ep r= l- ecosB e= 4'5; p=3 12 12 5 5-4cosB r=_-=4 l--cosB 5 = l- esinB e=6; p=2 12 r=---1-6sinB Consider the following diagram: ----
---
37.
39.
r
45.
43.
( r, 0)
p= ( r,
0)
Polar ---P-o-e I -O-f""--'�-r--""""+--- Axis
(Focus f)
p
Directrix D
d(F, P) = e· d(D, P) d(D, P) = +rsinB r= e( p+ sinB) r= ep+er sinB r- er sin B= ep r(l-esin B) = ep r= l- esinB P
r
_-,ep,--_
_-'ep'---_
P=
ep
d(D, P)
_-'ep'-----_
r
B)
e
__
41.
d(F, P) = e· d(D, P) d(D, P) = p-rcosB r= e( p-r cosB) r= ep-er cosB +er cosB= ep r(r+ cos = r=_---'ep'1+ecosB-_ Consider the following diagram:
Section 11.7 1. 3.
Directrix D
5.
21< = 7r 4 2 ellipse False; for example: x=2cost, =3 sint defme the same curve as in problem 3. 131 3; =
Y
Polar
---P-o-le-O--l<-"';'--'---"-4QI---+(Focus f)
Axis
-- p--.
606
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 11.7:
7.
x = 3t + 2 , y = t + 1 , 0 � t � 4 x = 3(y- l) + 2 x = 3y- 3 + 2 x = 3y- l x - 3y+ l = 0
13.
Plane Curves and Parametric Equations
x = 3t 2 , Y = t + 1 , - ex) < t < x = 3( y- l) 2
ex)
y 5
y
x
-5
x
9.
15.
x = t + 2 , y = Ji , t�O y = .Jx - 2 y
5
x = 2et , y = 1 + et , t �0 y= I +-x2 2y = 2 + x y 5 4 3 2 1
-5 11.
1
x = t 2 + 4 , y = t 2 - 4 , - ex)
17.
2
3
4
5
x
x = Ji, y = t 3/2 , t�O y= ( 2 ) 3/2 y = x3 X
y 10
x
-10 -3
607
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Chapter 11:
19.
Analytic Geometry
x = 2 cost , y = 3sint , 0:S;t:S;27t -x = cost 2::. = sint 3 2 1 + f = cos2 t + sin2 t = 1 x2 i -+ 4 9 =1
23.
(J(J
x = sec t , Y = tan t , O:S; t :s; 2:4 sec2 t = 1 + tan 2 t x2 = 1 + i x2 - i = 1 1 :s; x :s; J2, O:S; y :s; 1
y 2
y
(..[2, 1) x x
25. 21.
x = 2 cost , y = 3sint , -7t:S;t:S;O -x = cost 2 1 + f = cos 2 t + sin 2 t = 1 x2 + i = 1 y:S;O 4 9
(J(J
x = sin 2 t , Y = cos 2 t , O:S; t :s; 27t sin 2 t + cos 2 t = 1 x+y = l
-1
y 3
-1
x
-3
27.
x = t , y = 4t - 1
x = t + 1 , y = 4t +3
29.
x = t - 1, Y = t 2 -2t +2
31.
x = t , y = t2 + 1 x = t , Y = t3
33.
x = t 3/2 , y = t
35.
x = t + 2 , y = t ; O:S; t :s; 5
37.
x = 3cos t , y =2 sin t ; 0:S;t:S;27t
39.
Since the motion begins at (2, 0), we want x 2 andy 0 when t O. For the motion to be clockwise, we must have x positive and y negative initially. x = 2 cos ( w t ) , y = -3sin ( w t ) 27t =2=>w =7t w x = 2 cos (7tt ) , Y = -3sin ( ttt ) , O:S; t :s; 2
x = iff , y = t x = Ji, y = iff
=
=
=
608
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Plane Curves and Parametric Equations
Section 11.7:
41.
Since the motion begins at (0, 3), we want x 0 andy 3 when t O. For the motion to be clockwise, we need x positive andy positive initially. x =2sin ( wt ) , y = 3cos ( wt ) 27t = 1 � w = 21t x =2sin ( 21tt ) , y = 3 cos ( 2nt ) , O � t � 1 =
=
43.
y
=
C1
y
(--4, 16)
(0,0)
x
-16
(4, 16)
C1
45.
x = t sin t , Y = t cos t
47.
x = 4sin t - 2 sin(2t)
7
x
-16
C2
y
16 (1, 1)
(-1,1)
C2
x
-7.5
7.5
-16
C3
49.
y
x
a.
b.
-16
Use equations (1): x = ( SOcos 90o ) t = 0 Y = - �( 32 ) t 2 + ( SOsin 90 o ) t + 6 2 = -16t2 + SOt + 6 The ball is in the air until y = O . Solve: -16t2 + SOt + 6 -= -0 - - - -SO ± �r-SO-::- 2 - 4( -1 6)( 6) t = ---'-----'-'-'2( -16) -SO ± .J2s84 -32 "" -0.12 or 3.24 The ball is in the air for about 3.24 seconds. (The negative solution is extraneous.)
609
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Chapter 11:
c.
d.
Analytic Geometry
The maximum height occurs at the vertex of the quadratic function. t = -� = - � = 1.5625 seconds 2a 2(-16) Evaluate the function to find the maximum height: -16 (1 .56 25) 2 +50(1 .56 25) + 6 = 45.06 25 The maximum height is 45.0625 feet. We use = 3 so that the line is not on top of they-axis. 50
c.
10
1=10
x
53.
a.
=
c.
d.
Bill: = 5(t - 5) Y2 = 5 Bill will catch the train if Xl = t 2 = 5(t - 5) P = 5t - 25 t 2 - 5 t + 25 = 0 Since b 2 - 4ac = ( _ 5) 2 - 4(1)(25) = 25 - 100 = -75 < 0 the equation has no real solution. Thus, Bill will not catch the train. x2
x2
o
50
100
Train x
Use equations (1 ): = ( 145cos 20o ) t y = - !2 ( 32 ) t2 + ( 145 sin 20o ) t + 5 The ball is in the air until y = O . Solve: -16t 2 + ( 145sin 200 ) t + 5 = 0 ---- ---::"-- - -- --145 sin 20° ±�r( 1 4 5 s in 2 0o )2 _ 4 ( - 1 6 ) ( 5 ) t = --------��------�-------2( -16) "" -0.1 0 or 3.20 The ball is in the air for about 3.20 seconds. (The negative solution is extraneous.) Find the horizontal displacement: X = ( 145cos 20o )( 3.20 ) "" 436 feet x
Let YI =1 be the train's path and Y2 = 5 be Bill's path. a. Train: Using the hint, Xl -1 (2)t 2 = t 2 2 Yl = 1
b.
Bill
5
b.
51.
y
•
e.
The maximum height occurs at the vertex of the quadratic function. b = 145 sin 20° "" 1.55 seconds t = -2(-16) 2a Evaluate the function to find the maximum height: -16 ( 1 .55 )2 + (1 45 sin 200 ) (1.55) + 5 "" 43.43 The maximum height is about 43.43 feet. 25 0
o�
--......." 440
-5 0
610
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Section 11.7:
55.
a.
b.
Use equations (1): x = (40cos 45°)t = 20J2t Y = - �(9.8)t 2 +(40sin 45° )t + 300 2 = -4.9t 2 + 20J2t + 300
57.
a.
Plane Curves and Parametric Equations
At t = 0 , the Paseo is 5 miles from the intersection (at (0, 0)) traveling east (along the x-axis) at 40 mph. Thus, x = 40t - 5 , y = 0 , describes the position of the Paseo as a function of time. The Bonneville, at t 0 , is 4 miles from the intersection traveling north (along they-axis) at 30 mph. Thus, x = 0 , y = 30t - 4 , describes the position of the Bonneville as a function of time. =
The ball is in the air until y = O . Solve: -4.9t 2 + 20J2t + 300 = 0 - 20J2 ±�( 20J2 r - 4(-4.9)(300) t 2( - 4.9) "" -5.45 or 1 1 .23 The ball is in the air for about 1 1 .23 seconds. (The negative solution is extraneous. ) Find the horizontal displacement: x = ( 20J2 ) (1 1 .23) "" 3 17.6 meters = ------'--'-----'--------
c.
d.
b.
The maximum height occurs at the vertex of the quadratic function. 20Ji "" 2.89 seconds t = -2a 2(-4.9) Evaluate the function to find the maximum height:
b
(0,0)
P
=
Let d represent the distance between the cars. Use the Pythagorean Theorem to find the distance: d = �( 40t - 5) 2 + (30t -4)2 .
c.
Xl'lin=0 Xl'lax=.2 Xscl=.l Yl'lin=-2 Yl'lax=7 Yscl=l Xres=l
- 4.9(2.89)2 + 20.)2 (2.89)+ 300 3 40.8 meters
=
e.
d.
"i�il"\ul"\ --' �=.12B00002 "--______
V=.2
The minimum distance between the cars is 0.2 miles and occurs at 0. 128 hours.
e.
y 5 1 = 0.125
1=0
Bonneville t =0.25
1=0.25
"-..: "-.,.
1=0.125
5
Paseo x
1 =0
611
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Chapter 11:
59.
a.
b.
Analytic Geometry
We start with the parametric equations cos and h sin -1. We are given that 0 ftlsec2 , and h 3 feet. This gives us
x=(vo O)t Y= 2gt2+(vo O)t+ . =45° , g=32 = .J2 x=(vo . cos45°)t=-vot 2 Y=-1.2(32)t2+(vo sin45°)t+3 .J22 =-16t2+-vot+3 where Vo is the velocity at which the ball
�
61.
leaves the bat. Letting mileslhr have
Vo= 90 =132 ftlsec , we .J22 +3 Y=-16t2+-(132)t =-16t2+66Ji t+3 The height is maximized when 33.J2 t= 2(66Ji -16) = 16 2. 9168sec Y=-16(2. 9168)2 +66Ji (2. 9168) + 3 139. 1 The maximum height of the ball is about 139. 1 feet. From part (b), the maximum height is reached after approximately 2. 9168 seconds. x=(vo cos O) t =[132. �}2. 9168) 272. 25 The ball will reach its maximum height when it is approximately 272. 25 feet from home plate.
63 .
1.
/ -16x a=4 =
This is a parabola. Vertex: (0, 0) Focus: (-4, 0) Directrix:
�
3.
�
d.
Answers will vary.
Chapter 11 Review Exercises
�
c.
Y=-16(3. 3213)2+66Ji (3. 3213)+3 136.5 Since the left field fence is 37 feet high, the ball will clear the Green Monster by 136.5-37=99.5 feet. x=(X2- XI) t+ XI , Y=(Y2- Yl) t+Yl> -00
The ball will reach the left field fence when feet. cos O ) t
x= 310 x=(vo =66Ji t 310=66Ji t 31� 3. 3213 sec t=66,,2 Thus, it will take about 3. 3213 seconds to
X=4 X2 Y2=1 -25 This is a hyperbola. a= 5, b=1. Find the value of c: c2=a2+b2=25+1=26 c=£6 Center: (0, 0) Vertices: (5, 0), (-5, 0) Foci: (£6,0), (-£6,0) 1 Asymptotes: Y= 15 Y =--x 5 -
�
x;
reach the left field fence.
612
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Chapter 11 Review Exercises
5.
7.
9.
11.
x2 / +25 16 = 1 This is an ellipse. a = 5, b = 4 . Find the value of c: c2 = a 2 -b 2 = 25 - 1 6 = 9 c=3 Center: (0, 0) Vertices: (0, 5), (0, -5) (0, 3), (0, -3) Foci: x2 + 4y = 4 This is a parabola. Write in standard form: x2 = -4y + 4 x 2 = -4(y - 1) a =1 Vertex: (0, 1) Directrix: y = 2 Focus: (0, 0)
x 2 - 4x = 2y This is a parabola. Write in standard form: x 2 - 4x + 4 = 2y + 4 (x - 2) 2 = 2(y + 2) a = -21 Vertex: (2, -2) Focus: 2, D"rrectnx: y = --52 y2 - 4y - 4x2 + 8x = 4 This is a hyperbola. Write in standard form: (/ - 4y + 4) - 4(X 2 - 2x + 1) = 4 + 4 - 4 (y _ 2) 2 - 4(x - 1) 2 = 4 (y - 2) 2 (x _ 1) 2 = 1 4 1 a = 2, b = 1 . Find the value of c: c 2 = a2 + b2 = 4 + 1 = 5 c = .J5 Center: (1, 2) Vertices: (1, 0), (1, 4) Foci: ( 1, 2 - .J5 ) , ( 1, 2 + .J5 ) Asymptotes: y - 2 = 2(x - 1); y - 2 = - 2(x -1)
( -i)
13.
4x 2 - / = 8 This is a hyperbola. Write in standard form: / x 2 -2 8 =1 a = J2, b = J8 = 2 J2 . Find the value of c: c 2 = a 2 + b 2 = 2 + 8 = 10 c = .JlO Center: (0, 0) Vertices: ( -J2, 0 ) , ( J2, 0 ) Foci: ( -.JlO, O ) , ( .JlO, O ) Asymptotes: y = 2x; y = -2x
15.
4x2 + 9/ - 1 6x - 1 8y = 1 1 This is an ellipse. Write in standard form: 4x2 + 9/ - 1 6x - 1 8y = 1 1 4(x2 - 4x + 4) + 9(y2 - 2y + 1) = 1 1 + 16 + 9 4(x - 2) 2 + 9(y - 1) 2 = 36 (x - 2) 2 + (y _ 1) 2 = 1 4 9 a = 3, b 2 . Find the value of c: c2 = a2 - b 2 = 9 - 4 = 5 � c = .J5 Center: (2, 1); Vertices: (-1, 1), (5, 1) Foci: ( 2 - .J5, 1 ) , ( 2 + .J5, 1 ) =
613
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 11:
17.
19.
21.
Analytic Geometry 23.
4x 2 - 16x + 16y + 32 = 0 This is a parabola. Write in standard form: 4(x2 - 4x + 4) = -16y - 32 + 16 4(x - 2) 2 = -16(y + l) (x - 2i = -4(y + l) a=1 Vertex: (2, -1); Focus: (2, - 2) ; Directrix: y = °
Hyperbola: Center: (0, 0); Focus: (0, 4); Vertex: (0, -2); Transverse axis is the y-axis; a = 2; c = 4 . Find b: b 2 = c 2 2 = 16 - 4 = 12 b = .J12 = 2.J3 x2 = 1 Write the equation: y24 -12 _
a
-
y
9x 2 + 4i - 1 8x + 8y = 23 This is an ellipse. Write in standard form: 9(x 2 - 2x + 1) + 4(i + 2y + 1) = 23 + 9 + 4 9(x - l) 2 + 4(y + l) 2 = 36 (x _ 1) 2 + (y + 1) 2 = 1 4 9 a = 3, b = 2 . Find the value of c: c2 = a 2 -b 2 = 9 - 4 = 5 c = J'S Center: (1, -1) Vertices: (1, -4), (1, 2) Foci: ( 1, - I - J'S ) , ( I , - I + J'S ) Parabola: The focus is (-2, 0) and the directrix is x = 2 . The vertex is (0, 0). a 2 and since (-2, 0) is to the left of (0, 0), the parabola opens to the left. The equation of the parabola is: y2 = -4ax y2 = -4 . 2 . x i = -8x
Fl
25.
=
(0, -4)
Ellipse: Foci: (-3, 0), (3, 0); Vertex: (4, 0); Center: (0, 0); Major axis is the x-axis; a = 4; c = 3 . Find b: b 2 = a 2 - c 2 = 16 -9 7 b = J7 x2 + y2 = 1 Write the equation: 16 7 =
-
y
s
(o, V7)
=
Fl
=
(-3, 0)
(o, -I/7 )
-s
x
614
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Chapter 1 1 Review Exercises
27.
Parabola: The focus is (2, -4) and the vertex is (2, -3). Both lie on the vertical line x = 2 . a 1 and since (2, -4) is below (2, -3), the parabola opens down. The equation of the parabola is: (x - h/ = -4a(y - k) (x _ 2) 2 = -4 . 1 . (y - (-3)) (X _ 2) 2 = - 4(y + 3)
_ 5) 2 = 1 Write the equation: (X +164) 2 + (y 25
=
y
2
VI
x 33.
= (- 4, 0)
x -2
Hyperbola: Center: (-1 , 2); a = 3; c = 4 ; Transverse axis parallel to the x-axis; Find b: 2 = c 2 - a 2 = 16 - 9 = 7 � = J7 Wnte · the equatIOn: (x + 1) 2 (y -7 2)2 --1 9
b
b
.
29.
Hyperbola: Center: (-2, -3); Focus: (-4, -3); Vertex: (-3, -3); Transverse axis is parallel to the x-axis; a = 1; c = 2 . Find b: b 2 = c 2 - a 2 = 4 -1 3 = .fj Write the equation: ( +1 2) 2 (y + 3i = 1 3 Y + 3 Y3(x 2) Y + 3 \f3(x + 2) y =
b
x
=
3
/
/
I
=
Y
+
35.
(-2, -3 + V3 ) x
/
/
(x + 1 ) - 2 - 1/7 3
-8
Hyperbola: Vertices: (0, 1), (6, 1); Asymptote: 3y + 2x - 9 = a ; Center: (3, 1); Transverse axis is parallel to the x-axis; a = 3 ; The slope of the asymptote is -�3 ; Find b: - 2 � -3b = - 6 � = = a 3 3 Write the equation: (X -93) 2 (y _4 l) 2 = 1
-b -b
31.
Ellipse: Foci: (-4, 2), (-4 , 8); Vertex: (-4, 10); Center: (-4, 5); Major axis is parallel to the y axis; a = 5; c = 3 . Find b: b 2 = a 2 _ c 2 = 25 - 9 = 16 � = 4
y - 1 = - 3�x - 3) , "VI = (0, 1)
b
b=2
Y
"-
" y - l = 3;1.(X - 3) " /
5
"-
"
-5
615
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Chapter 11:
Analytic Geometry
43. 4x2 + 10xy + 4i - 9 = 0 37. i + 4x + 3y - 8 = 0 A = 4, B = lO, C = 4 A = 0 and C = 1; AC = (0)(1) = O . Since A C = 0 , the equation defines a parabola. B 2 - 4AC = 10 2 - 4(4)(4) = 36 > 0 Hyperbola 39. x 2 + 2i + 4x - 8y + 2 = 0 45. x2 - 2xy + 3 y 2 + 2x + 4 Y - 1 = 0 A = 1 and C = 2; AC = (I)(2) = 2 . Since A = 1, B = - 2, C = 3 A C > 0 and A "* C , the equation defines an ellipse. B 2 - 4AC = ( _ 2) 2 - 4(1)(3) = -8 < 0 Ellipse 41. 9x2 -12xy + 4i + 8x + 12y = 0 A = 9, B = -12, C = 4 B 2 -4AC = (-12) 2 - 4(9)(4) = 0 Parabola 47. 2x2 + 5 xy + 2i --9 = 0 2 2 - 2 = 0 -)0 2 B = - -)0 B = A = 2" B = 5 and C = 2', cot(2 B ) = A B- C = -5 2 4 J2 '- -y' J2 = -(x' J2 - y') X = x'cos B -y 'sm. B = -x 2 2 2 J2 -y' J2 = -(x'+ J2 y') y = x'sm. B + y'cos B = -x'+ 2 2 2 2 (x'- y') + 5 (x'- y , » (x'+ y , ) + 2 (x '+ y') + 0
--
J
(":
7t
7t
( ": )(": ) (": ( x'2 _ 2x'y'+ y'2 ) + % ( X '2 ,_ y ,2 ) + ( X'2 + 2x'y'+ y ,2 )-� = 0
J
y ,2 = 1 x ,2 - _9 X ,2 _ _1 y ,2 = _9 -)o 9 X ,2 _ y , 2 = 9 -)0 2 2 2 1 9 Hyperbola; center at (0, 0), transverse axis is the x'-axis, vertices at (x', y ') = (± 1, 0) ; foci at (x', y ') = ( ±.JiQ, 0 ) ; asymptotes: y' = ±3x' . y
x
616
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Chapter 11 Review Exercises
49.
6x2 + 4 xy + 9y2 _ 20 = 0 A = 6" B = 4 and C = 9', cot ( 2{}) = A - C = 6 - 9 = --3 � cos ( 2{}) = - -3 B 4 4 5 sm o = = g = 2 v'5 o os O = = [ = v'5 -> MH' 2 fs 5 ' 2 v'S 5 2 J5 y' = J5 '- -J5 ( x'- 2y ' ) x = x'cos {} - y 'sm. {} = -x 5 5 5 . 2 J5 J5 J5 y = x'sm {} + y'cos {} = x '+ -y' = - ( 2x '+ y' ) 5 5 5 ' 6 ( x'- 2 Y ' l + 4 ( x'- 2Y ' l ( 2x'+ Y ' l + 9 ( 2X '+ Y ' l - 20 = 0
-- H- %))
1H %))
-( � ) ( � )( � ) (� ) � ( x'2 _ 4x' y'+ 4y ,2 ) + � ( 2X ,2 _ 3x'y'- 2y'2 ) + � (4X ,2 + 4x'y'+ y ,2 ) - 20 = 0 ,
24 24 y ,2 + -8 x ,2 _12 x'y'- _8 y ,2 + _ 36 X ,2 + 36 x'y'+ 9 y , 2 = 20 _6 x ,2 _ -x'y'+ _ 5 5 5 5 5 5 5 5 5 , , X 2 2 IOx ,2 + 5y ,2 = 20 � _ + L = 1 2 4 Ellipse; center at the origin, major axis is the y'-axis, vertices at ( x " y ' ) = ( 0, ±2 ) ; foci at ( x " y ' ) = ( 0, ±.J2 ) . _
y
x
51.
4x2 -12xy + 9y 2 + 12 x + 8 y = 0 A - -C = 4 - 9 = 125 � cos ( 2{} ) = 5 A = 4, B = -12, and C = 9; cot ( 2{} ) = B -12 13 5 1 - 153 1 + 1 3 = (9 = JJi3 -> O d 3.7 ' sm o = 2 = f4 = 2.J13 . oos O = 2 V 13 13 V 13 13 ' 3J13 '- --y 2J13 ' = J13 ( 3x'- 2y' ) x = x'cos {) - y'sm. {} = --x 13 13 13 3J13 J13 ( 2x'+ 3y' ) y = x'sm. {} + y'cos {} = 2 J13 13 13 x '+ 13 y ' = -
1(
1(
)
)
-- --
617
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 11: Analytic
4
Geometry
(� (3x '- 2y ') J - 12 (� (3x '- 2Y '))(� ( 2x '+ 3Y ')) + 9 (� ( 2x '+ 3y ')J + 1 2 ( � ( 3x '- 2Y ' ) ) + 8 ( � ( 2X '+ 3Y ' ) ) 0 =
(
) 13 (
) 13 (
� 9x '2 - 1 2x ' y '+ 4y , 2 _ g 6x ,2 + 5x ' y '- 6y ,2 + � 4x '2 + 1 2x ' y '+ 9 y , 2 13
)
24..JU 16..JU 24.JG 3 6.JG y' 0 x '+ y '+ x '13 13 13 13 3 6 , 4 8 , , 1 6 , 72 , 60 , , 72 , 3 6 , 1 08 , , 8 1 , 3 '=0 1ux -x 2 - - x y + -y 2 - - x 2 - - x y + -y 2 + - x 2 + - x y + - y 2 + 4v/';; 13 13 13 13 13 13 13 13 13 4 .JG , , /,;; x' 1 3y 2 + 4v 1 3x ' = 0 � y 2 = 13 +
=
--
Parabola; vertex at the origin, focus at ( x ', y ')
=
y
( - 1] , 0) .
x
53 .
4
55.
r = ---
ep
I - cos O
=
4,
e
=
1,
p
=
ep
( 2, n ) .
l�X
2 - sin O
4
Parabola; directrix is perpendicular to the polar axis 4 units to the left of the pole; vertex is
=
3,
e
=
2"1
p
=
6
Ellipse; directrix is parallel to the polar axis 6 units below the pole; vertices are 6, and 2, 3; . Center at ( 2, ; } other
( %) ( ) focus at ( 4, ; )
y
_(-12_ ,_ 3t)-+---+
3
6
r = ---
-iO'x Polar Axis
y
__ _ _
(3, n; )
�
-
__----:::+_-__tIf_:__---=.�
Polar axis
Di rectrix 618
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Chapter 11 Review Exercises
57.
63.
2 S r = ---4 + S cos O 1 + 2cos O ep = 2 , e = 2 , P = 1 Hyperbola; directrix is perpendicular to the polar axis I unit to the right of the pole; vertices are 0 and ( - 2, 1t) . Center at (�, 0 ; other
(�, )
focus at
( -�' 7r) y
y
2
)
.
Directrix
-2
x Polar Axis
-+-"''++-+---�
---
59.
x = 4t - 2 , y = 1 - t, - 00 < t < 00
65.
4 r = --I - cos O r - r cos O = 4 r = 4 + r cos O r 2 = (4 + r cos O) 2 X2 + y 2 = (4 + X) 2 x2 + y 2 = 16 + Sx + x 2 - Sx -16 = 0
x = 4(1 - y) - 2 x = 4 - 4y - 2 x + 4y = 2 x = 3sin t , y = 4 cos t + 2, O :s; t :s; 21t y-2 -x = sm. t -' 4 = cos t 3 sin 2 t + cos 2 t = I � + (y _ 2) 2 = 1 9
16
Y
8
(0, 6)
/
61.
S r = ---4 + S cos O 4r + Sr cos O = S 4r = S - Sr cos 0 = 2 - 2r cos O r 2 = (2 - 2r cos O) 2 x 2 + y 2 = (2 _ 2X) 2 x 2 + = 4 - Sx + 4x 2 3x 2 - - Sx + 4 = 0
x
r
/
y 2
/
-2
( 1 , 0)
2
x
-2
tan 2 t + 1 = sec2 t � Y + 1 = x 619
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Chapter 11:
Analytic Geometry
69. Answers will vary. One example: y = -2x + 4 x = t , y = -2t + 4 x = 42 t y = t
=
-
-- '
71.
�4 = cos ( m t ) ; 2::.3 = sin ( m t ) 2n = 4 � 4m = 2n � m = 2n = -n m 4 2 :. � = cos ; t f = sin ; t �
73.
( } ( ) x = 4 cos ( ; t } y = 3 sin ( ; t ) ' 0
�
t�4 77.
=
79.
For the hyperbola: Foci: (-3, 0), (3, 0); Vertices: ( -..[5, 0 ) , ( ..[5, 0 ) ; Center: (0, 0) a = ..[5; c = 3; b2 = c2 - a 2 = 9 - 5 4 � b = 2
=
=
75.
=
)
-
Write the equation in standard form: 2 2 =l 4x2 + 9/ = 36 � �9 + L 4 The center of the ellipse is (0, 0). The major axis is the x-axis. a = 3; b = 2; c 2 = a 2 _ b 2 = 9 - 4 = 5 � c = ..[5. For the ellipse: Vertices: (-3, 0), (3, 0); Foci: ( -..[5, 0 ) , ( ..[5, 0 )
x 2 -/= The equation of the hyperbola is: 5 4 Let (x, y) be any point in the collection of points. The distance from -----(x, y) to (3, 0) = �'-(x _ 3) 2 + y 2 . The distance from (x, y) to the line x 1 6 is x - 1 6 · 3 3 Relating the distances, we have:
1 1 ( ) (
16 �(x - 3) 2 + y 2 4"3 x - "3 16 2 (x - 3) 2 + y 2 = -9 x - 3 16 32 x + 256 x2 - 6x + 9 + y 2 = 169 x 2 - "3 T 2 2 2 16x - 96x + 144 + 16y = 9x - 96x + 256 7x 2 + 16/ = 1 12 7x2 + 16/ = 1 1 12 1 12 x2 + / 16 7 = 1 The set of points is an ellipse. Locate the parabola so that the vertex is at (0, 0) and opens up. It then has the equation: x2 = 4ay . Since the light source is located at the focus and is 1 foot from the base, a 1 . Thus, x2 = 4y . The width of the opening is 2, so the point (1, y) is located on the parabola. Solve for y: 1 2 = 4y � 1 = 4y � Y = 0.25 feet The mirror is 0.25 feet, or 3 inches, deep. Place the semi-elliptical arch so that the x-axis coincides with the water and the y-axis passes through the center of the arch. Since the bridge has a span of 60 feet, the length of the major axis is 60, or 2a 60 or = 30 . The maximum height of the bridge is 20 feet, so b = 20 . The X2 Y2 . IS: . equatIon 900 + 400 = 1 . The height 5 feet from the center: 52 + / 900 400 = 1 2 l � L 400 900 875 � y '" 19.72 feet y 2 = 400 . 900 The height 1 0 feet from the center:
I
=
I 1
a
-
620
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Chapter 11 Review Exercises
640, 000 360,000 = 1 y2
83.
B
1
B.
x
First note that all points where an explosion could take place, such that the time difference would be the same as that for the fIrst detonation, would form a hyperbola with A and as the foci. Start with a diagram: N
9 360, 000 1 6 = 202, 500 y = 450 The second explosion should be set off 450 feet due north of point a. Use equations (1) in section 9.7. = ( 80cos ( 35° )) t Y = - � (32)t 2 + ( 80sin ( 35° )) t + 6 2 The ball is in the air until y = O . Solve: b. -16 t2 + ( 80sin ( 35° )) t + 6 = 0
l
l
81 .
- 80 sin
A
B
... ... D,
... 1 ( I OOO, y)
(0 , 0) (a, O) 2000
feet
: I
( 1 000, 0)
B
c.
±�(
80 sin
(35°)) 2 - 4( - 1 6)(6)
2(-16) '- 45 .89 ± v'24 8 9 .s-4 -32 "" -0.13 or 2.99 The ball is in the air for about 2.99 seconds. (The negative solution is extraneous.) The maximum height occurs at the vertex of the quadratic function. -80sin ( 35° ) "" 1.43 seconds t= 2(-16) 2a Evaluate the function to fInd the maximum height: -16(1 .43 ) 2 + ( 80 sin (35°) ) ( 1 .43) + 6 '" 3 8 . 9 ft
-b=
Since A and are the foci, we have 2c = 2000 c = 1000 Since is on the transverse axis and is on the hyperbola, then it must be a vertex of the hyperbola. Since it is 200 feet from we have a = 800 . Finally, b2 = c2 - a2 = 1000 2 - 800 2 = 360, 000 Thus, the equation of the hyperbola is Y2 640, 000 360, 000 = 1 The point ( 1000, y ) needs to lie on the graph of the hyperbola. Thus, we have
D(
(35°)
t = ----------�------�------ --
D,
... ... ... ... ... ... (- 1 000, 0)... ...
y2
( 1000 ) 2
10 2 + y2 = 1 900 400 L = I _ 100 400 900 800 Y 2 = 400 · - � Y "" 1 8.86 feet 900 The height 20 feet from the center: 202 -+900 400 = 1 L = I - 400 400 900 500 y 2 = 400 . 900 � y "" 14.91 feet
B,
d.
Find the horizontal displacement: = ( 80cos ( 35° )) (2.99) "" 196 feet (or roughly 65.3 yards) x
e.
-so
621
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Chapter 11:
Analytic Geometry
3. 2x 2 + 3y 2 + 4x - 6y = 13 Rewrite the equation by completing the square in x andy. 2X 2 + 3/ + 4x - 6y = 13 2x 2 + 4x + 3/ - 6y = 13 2 ( x2 + 2x) + 3 ( y 2 - 2y ) = 13 2 ( X 2 + 2x + 1 ) + 3 ( / - 2y + 1 ) = 13 + 2 + 3 2(x + 1) 2 + 3 (y _ 1)2 = 18 (x _..- ( _ 1))2 ( Y _ 1)2 ...-'. --'-9--'-'-- + 6 = 1 This is the equation of an ellipse with center at (-1, I) and major axis parallel to the x-axis. Since a 2 = 9 and b 2 = 6 , we have a = 3 , b = J6 , and c 2 = a 2 _ b 2 = 9 - 6 = 3 , or c = ../3 . The foci are (h ± c,k) = ( -1 ± ../3, 1 ) or ( -1 - ../3, 1 ) and ( -1 + ../3,1 ) . The vertices are at (h ± a, k) = ( - I ± 3, I) , or (-4, 1) and (2,1) .
Chapter 11 Test 1.
(x + 1)2 y2 = 1 9 4 Rewriting the equation as _ 0)'--2 = 1 , we see that this is the (x _ (_1))2 00.:(y -=-' 32 22 equation of a hyperbola in the form (x _ h) 2 ( Y _-=-' k)-2 = 1 . Therefore, we have b2 a2 h = -1 , k = 0 , a = 2 , and b = 3 . Since a 2 = 4 and b 2 = 9 , we get c 2 = a 2 + b 2 = 4 + 9 = 13 , or c = 03 . The center is at (-1, 0) and the transverse axis is the x-axis. The vertices are at (h ± a, k) = (-1 ± 2, 0) , or (-3, 0) and (1, 0) . The foci are at (h ± c, k) = ( -1 ± 03, 0 ) , or ( -1 - 03,0 ) and ( -1 + 03, 0 ) . The asymptotes are y - O = ± "23 (x - (-l)) , or y = - "23 (x + 1) and y = �(x 2 + 1) .
-'---
2.
4.
8y = (x - I) 2 _ 4 Rewriting gives (x - l) 2 = 8y + 4 (x _ I) 2 = 8 y - -
( ( i)) (x - l/ = 4(2 { y - ( -i))
This is the equation of a parabola in the form (x - h / = 4a (y - k ) . Therefore, the axis of symmetry is parallel to the y-axis and we have (h, k) = 1, - and a = 2 . The vertex is at
( �) ( h, k) = ( 1, - � ) , the axis of symmetry is x = 1 , the focus is at (h, k + a) = ( 1, -i + 2 ) = ( 1, %) ,
The vertex (-1,3) and the focus (-1, 4.5) both lie on the vertical line x = -1 (the axis of symmetry). The distance a from the vertex to the focus is a = 1.5 . Because the focus lies above the vertex, we know the parabola opens upward. As a result, the form of the equation is (x _ h) 2 = 4a (y - k) where (h, k) = (-1, 3) and a = 1 .5 . Therefore, the equation is (x + 1)2 = 4(1.5)(y - 3) (x + 1)2 = 6( Y - 3) The points (h ± 2a, k) , that is (-4, 4.5) and (2, 4.5) , define the lattice rectum; the line y = 1 .5 is the directrix.
and the directrix is given by the line y = k -a , or y = --52 .
622
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 11 Test
5.
The center is ( h, k ) = (0, 0) so h = 0 and k = 0 . Since the center, focus, and vertex all lie on the line x = 0 , the major axis is the y-axis. The distance from the center (0, 0) to a focus (0, 3 ) is c = 3 . The distance from the center (0, 0) to a vertex (0, -4) is a = 4 . Then,
where h = 2 , k = 2 , and a = 2 . This gives us (Y _ 2)2 (x _-;;-'2) 2 1 4 b2 Since the graph contains the point (x, y) = ( 2 + M, , we can use this point to determine the value for b. (2 + M - 2r "":"""
5)
b2 = a2 _ c2 = 42 _ 32 = 1 6 - 9 = 7
....:.... -;:-----'-- = 1 b2
The form of the equation is (x - h/...!. + ( y _ k )2 = 1 �....,.b2
� _ .!Q
a2 h = 0 , k = 0 , a = 4 , and b = J7 . Thus,
4
where we get x2 y2 -+- = 1
1 10
Therefore, the equation becomes ( Y _ 2) 2 "":"""-(x _ 2)-'-2 = 1 4 8 Since c 2 = a 2 + b 2 = 4 + 8 = 1 2 , the distance from the center to either focus is c = 2.fi . Therefore, the foci are c = 2.fi units above and below the center. The foci are Fi = ( 2 2 + 2.fi) and F2 = ( 2, 2 - 2.fi ) . The asymptotes are given by the lines y - k = ±!:( x - h ) . Therefore, the asymptotes b are 2 - 2) y - 2 = ±-(x 2Ji Ji - 2) + 2 y = ±-(x 2
=
,
= (0. 4)
Vl
-5 6.
=
b2 = 8 b = 2 Ji
To graph the equation, we use the center ( h, k ) = (0, 0) to locate the vertices. The major axis is the y-axis, so the vertices are a 4 units above and below the center. Therefore, the vertices are � = (0, 4) and V2 = (0, -4) . Since c = 3 and the major axis is the y-axis, the foci are 3 units above and below the center. Therefore, the foci are Fi = ( 0, 3 ) and F2 = (0, -3) . Finally, we use the value b = J7 to fmd the two points left and right of the center: ( -J7, 0 ) and ( J7, 0 ) . y 5
b2 5
4" = [;2
16
7
- =
=
(2, 4)
The center ( h, k ) = (2, 2) and vertex (2, 4) both lie on the line x = 2 , the transverse axis is parallel to the y-axis. The distance from the center (2, 2) to the vertex (2, 4) is a = 2 , so the other vertex must be (2, 0) . The form of the equation is
(y _ k )2 ( X _ h )2 = 1 a2
b2
623
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Chapter 11:
7.
Analytic Geometry
(-7 , 24 )
2X 2 + 5xy + 3y 2 + 3x - 7 = 0 is in the form AX 2 + Bxy + C/ + Dx + £y + F = 0 where A = 2 , B = 5 , and C = 3 . B 2 - 4A C 5 2 - 4 ( 2 ) ( 3 ) = 25 - 24 =1 2 Since B - 4AC > 0 , the equation defines a
24
=
7
hyperbola. 8.
We have cot ( 28 ) = - 14 so it follows that cos ( 28 ) = 15
3x 2 - .xy + 2 / + 3 y + 1 = 0 is in the form Ax 2 + Bxy + C/ + Dx + Ey + F = 0 where A = 3 , B = -1 , and C = 2 . B 2 - 4A C = ( _1 ) 2 - 4 ( 3 )( 2 ) = 1 - 24 = -23 Since B 2 - 4A C < 0 , the equation defines an
-
,in 0
()
� ( ;) 1- -
�
2
5 = fl6 � V 25 5 �
( ;5 ) = {9 = �
1+ -
V25
2
5
8 = cos- 1 % ", 53. 130 With these values, the rotation formulas are 3 4 3 4 x = - x ,- - y , and y = - x '+ - y ' 5 5 5 5 Substituting these values into the original equation and simplifying, we obtain
x2 - 6xy + 9 / + 2x - 3y - 2 = 0 is in the form AX 2 + Bxy + C/ + Dx + £y + F = 0 where A = 1 , B = -6 , and C = 9 . B 2 - 4A C = ( _6 ) 2 - 4 ( 1 ) ( 9 ) = 36 - 36 =0 Since B 2 - 4A C = 0 , the equation defmes a
4 1x 2 - 24xy + 34/ - 25 = 0 4 3 4 2 3 4 3 4 1 S" x ,- S" y , - 24 S" x , - S" y , S" x , + S" y ,
(
)(
)2 ( +34 S" x '+ S" ) = 25 (4 3
parabola. 10.
1 - co, ( 20 ) = 2
�
l + co, ( 20 ) = ,", 0 = 2
ellipse. 9.
.
4 lx 2 - 24.xy + 34y 2 - 25 = 0 Substituting x = x ' cos 8 - y ' sin 8 and y = x ' sin 8 + y ' cos 8 transforms the equation
Y
'
)
Multiply both sides by 25 and expand to obtain
into one that represents a rotation through an angle 8 . To eliminate the x ' y ' term in the new equation, we need cot ( 28) = A B- C . That is, we need to solve 4 1 - 34 cot ( 2 ll ) = ---24
(
)
(
2 2 2 2 41 9X , - 24x ' y ' + 16y , - 24 1 2x , - 7x ' Y '- 1 2 y ,
+34 ( l 6x ,2 + 24x ' y '+ 9y ,2 ) = 625
,
62 5 x 2 + 1 2 5 0y , 2 = 625 X ,2 + 2y ,2 = 1 x,2 y ,2 -+- = 1
u
cot ( 28 ) = - � 24 Since cot ( 28 ) < 0 , we may choose 90° < 28 < 1 80° , or 45° < 8 < 90° .
1
1
2
.J2
Thus: Ji. = 1 and b = II = 2 V"2 This is the equation of an ellipse with center at ( 0, 0 ) in the x ' y '- plane . The vertices are at a =
624
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)
Chapter 11 Test
�
c
-
center
c=-
plane
-2rcosB = 3 = 2rcosB+3 = (2r cos B + 3)2 x2 + y2 =(2x+3)2 x2 +l =4x2 +12x+9 3x2 + 12x -l =-9 3(X2 +4x)-l =-9 3(X2 +4x+4)-y2 = -9+12 3(x+2)2 _ y2 =3 (X+2)2 --= y23 1 (1) {X=3t-2 y = l - Ji (2) t x Y 0 x=3(0)-2=-2 y=I-JQ=1 1 x=3(1)-2=1 y=I-J! =0 4 x=3(4)-2=10 y=I-v'4=-1 9 x = 3 (9) -2 = 25 y = I-J9 =-2 r
(-1,0) and (1,0) in the x'y'- plane. J22 12 21 2 =a2 - b2 =1--=The foci are located at (± � ,0] in the x' y '- plane. In summary: x'y' xy(0,0) (0,0) (±1,0) (�,�}( -�,-�) ( 2J25 ' 3J210 ] ' (O,± '7) _2J2 3J2 ( 5 ' _ 10 ) ( 3J210 ' 2J25 ) ' (± '7,0) _ 3J2 2J2 ( 10 ' _ 5 ) The graph is given below. plane
vertices
minor axis intercepts
r r2
�----':.....
12.
foci
(x,y) (-2,1) (1,0) (10,-1) (25,-2)
y :)
-5
11.
Towe find thtoe elrectangul athr eequati oblneforfrtomhe curve, need i m i n ate vari a the equations. Wex =can3t-2start by solving equation ( 1) for 3t = x+2 x+2 t =-3 ng this result for into equation Substi t uti gives T-3- ' -2 :S;x:S;25 y=I- V/x+2
3 = l-ecosB 1-2cosB Therefore, = 2 and = �2 Since I , this is the equation of a hyperbola. 3 ( I -2 cos B) = 3I-2cosB r-2rcosB = 3 Since x = rcosB and x2 + y2 = we get r
ep
e
r=
p
t
t.
e>
.
. ( 10 , - 1 )
---
r
t
r2 ,
625
(2)
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Chapter 11: 13.
Analytic Geometry
We can draw the parabola used to form the reflector on a rectangular coordinate system so that the vertex of the parabola is at the origin and its focus is on the positive y-axis. y
-2
1.
4 ft
2
(-2, 1 . 5)
Chapter 11 Cumulative Review
(2, 1 .5 )
o
x
2
-2
The form of the equation of the parabola is x 2 = 4ay and its focus is at (0, a) . Since the point (2, 1 .5) is on the graph, we have
3.
2 2 = 4a (1 .5) 4 = 6a a - 13
The microphone should be located i feet (or 8 inches) from the base of the reflector, along its axis of symmetry. y
(-2, 1 . 5)
=
(2, 1 .5 )
x
F
- (0 �
6 - x � x2 0 � X2 + x - 6 x2 + x - 6 � O (x + 3)(x - 2) � 0 f(x) x 2 + x - 6 x = - 3, x 2 are the zeros of f . Interval (-00 , -3) ( -3, 2) Test Value -4 0 Value off 6 -6 =
4 ft
2
f (x + h) - f (x) h -3 ( X +h) 2 + 5 ( X + h) - 2 - ( -3x 2 + 5x - 2 ) h -3 ( X2 + 2xh + h2 ) + 5x + 5h - 2 + 3x2 - 5x + 2 h -3x 2 - 6xh - 3h 2 + 5x + 5h - 2 + 3x 2 - 5x + 2 h -6xh - 3h 2 + 5h = -6x - 3h + 5 h
5.
3
, 1. )
(2 , 00 ) 3 6
Conclusion Positive Negative Positive The solution set is { x 1 - 3 � x � 2 } , or [-3, 2] . f (x) log4 (x - 2) a. f (x) = log4 (x - 2) = 2 =
x - 2 = 42 x-2 = 16 x = 18
b.
The solution set is { 1 8} . f (x) = log4 (x - 2) � 2 x - 2 � 4 2 and x - 2 > 0 x - 2 � 1 6 and x > 2 � 18 and x > 2 x
2 < x � 18 (2, 1 8]
626
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Chapter 11 Review Exercises 7.
sin ( 28 ) = 0.5
28 = !!:. + 2krr. => 6 511" + 2krr. => or 28 = 6 where k is any integer.
9.
8 = .!!:.... + k7t 12 5rr. 8 = - + krr. 12
Using rectangular coordinates, the circle with center point (0, 4) and radius 4 has the equation: (X _ h ) 2 + ( y _ k ) 2 = r 2 (X _ O)2 + ( y _ 4 )2 = 4 2 X2 + / - 8y + 1 6 = 1 6 X 2 + / - 8y = 0
Converting to polar coordinates : r 2 - 8r sin 8 = 0 r 2 = 8r sin 8 r = 8 sin 8 y
9
.x
-5
11.
8 < 90° rr. rr. krr. 28 = - + krr. => 8 = - + - where k is any 4 8 2 ' integer. On the interval 0° < 8 < 90° , the solution is 8 = � = 22.5° .
cot ( 28) = 1, where
0°
<
627
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Chapter 1 2 Systems o f Equations and Inequalities Section 12 .1 1.
3. 5. 7.
13.
3x + 4 = 8 - x 4x = 4 x=1 The solution set is {I} . inconsistent False 2X - y = 5 5x + 2y = 8 Substituting the values of the variables: 2(2) - (-1) = 4 + 1 = 5 5(2) + 2(-1) = 10 - 2 = 8 Each equation is satisfied, so x = 2, y = -1 , or (2, -1) , is a solution of the system of equations.
{
15.
p x - 4Y = 4 1� X -3Y = _ �
j
Substituting the values of the variables: 3(2) - 4 � = 6 - 2 = 4
() �(2) 2 - 3 ( �2 ) = 1 - �2 = -�2
1 7.
Each equation is satisfied, so x = 2, y = ! or ( 2, !) , is a solution of the system of equations. '
11.
3x + 3y + 3z = 4 x - y - z=0 2y - 3z = -8 Substituting the values of the variables: 3(1) + 3(-1) + 2(2) = 3 - 3 + 4 = 4 1 - (-1) - 2 = 1 + 1 - 2 = 0 2(-1) - 3(2) = -2 - 6 = -8 Each equation is satisfied, so x = 1, y = -1, = 2 , or (1, - 1 , 2) , is a solution of the system of equations. 3X + 3 y + 2z = 4 x - 3y + z = 10 5x - 2y - 3z = 8 Substituting the values of the variables: 3(2) + 3 ( -2) + 2(2) = 6 - 6 + 4 = 4 2 - 3 ( -2) + 2 = 2 + 6 + 2 = 10 5(2) - 2(-2)- 3(2) = 10 + 4 - 6 = 8 Each equation is satisfied, so x = 2 , y = z = 2 , or (2, -2, 2) is a solution of the system of equations. x+y = 8 x-y= 4 Solve the first equation for y, substitute into the second equation and solve: y = 8-x x-y = 4 x - (8 - x) = 4 x-8+x = 4 2x = 12 x=6 Since x = 6, y = 8 - 6 = 2 . The solution of the system is x = 6, y = 2 or using ordered pairs (6, 2) . Z
{
9.
{ {
{ 1
-2,
{
{
{x-y = 3 1
"2 x + y = 3 Substituting the values of the variables, we obtain: =2+1 =3 li��) : : Each equation is satisfied, so x = 4, Y = , or 1
(4, 1) , is a solution of the system of equations.
628
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 12.1:
1 9.
{
5X - y = 1 3 2x + 3y = 1 2
Systems of Linear Equations: Substitution and Elimination 25.
3
Multiply each side of the first equation by and add the equations to eliminate y:
{
= 51 x=3
Substitute and solve for 5(3) - y = 1 3 15- y = 13 -y = - 2 y=2
= {4x+2y= 3 y
y: 27.
The solution of the system is x = 3, y = 2 or using ordered pairs 21.
{
(3, 2).
3X = 24 x + 2y = 0
x
Solve the first equation for and substitute into the second equation:
{
X+2
;:
8 0
8 + 2y = 0 2y = - 8 y = -4
{
29.
3X - 6Y = 2 5x +4y = 1
3, 2
Multiply each side of the first equation by and each side of the second equation by then add to eliminate
{
y:
6x -1 2y = 4 1 5x + 1 2y = 3 21x
=7 1 X=3
Substitute and solve for 3 ( 1 / 3 ) - 6y = 2
y:
1 y = -6
using ordered pairs (�, -�) .
4x+2(l-2x)=3 4x+2-4x=3 0=1 This equation is false, so the system is inconsistent. X- y = 0 { 23x+2y =7 Solve the first equation for y, substitute into the second equation and solve: 2X { y=3x+2y= 7 3x+2(2x)=7 3x+4x7x=7 =7 x=1 Since x=1, y = 2(1) = 2 The solution of the system is x=1, y 2 or using ordered pairs (1, 2). x+2y==84 { 2x+4y Solve the first equation for x, substitute into the second equation and solve: X = 4-2y { 2x+4y=8 2(4-2y)+4y = 8 8-4y+4y=8 0= 0 These equations are dependent. The solution of the system is either x = 4 -2 , where y is any real 4-x , where x is any real number. number or y=-2 Using ordered pairs, we write the solution as {(x, y )l x = 4 -2y, y is any real number} or as {(X,y)1 y = 4; X , x is any real number} . Y
1 - 6y = 2 -6y = 1
The solution of the system is
1 - 2x
=
The solution of the system is x = 8, y = - 4 or using ordered pairs (8, -4) 23.
y,
Solve the first equation for substitute into the second equation and solve:
1 5X - 3Y = 39 2x + 3y = 1 2 1 7x
2X + y = 1 { 4x+2y= 3
x = J. , y = _J. 6 3
or
629
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Chapter 12:
31.
Systems of Equations and Inequalities
{ 2x - 3y = -1
10x + y = 1 1 Multiply each side of the first equation by -S, and add the equations to eliminate x: -lOx + IS y = S 10x + y = 1 1 16y = 16 y=1 Substitute and solve for x: 2x - 3(1) = -1 2x -3 = -1 2x = 2 x=1 The solution of the system is x = 1, y = 1 or using ordered pairs ( l , 1). 3
35.
{
33.
{I �
� x+ ! y = 3
_ 4 x 3 y = _1 Multiply each side of the first equation by -6 and each side of the second equation by 12, then add to eliminate x: -3X - 2Y = -18 3x - 8y = -12 - 10y = -30 y= 3
{
Substitute and solve for x: 1 1 -x +-(3) =3 2 3 -1 x + l = 3 2
r: : : : � Solve the second equation for x, substitute into the first equation and solve: {2X + 3Y � 6
37.
x = y + -2
( �)
2 Y + + 3Y = 6 2y + l + 3y = 6 Sy = S y=l Since y = 1, x = 1 + �2 = �2 . The solution of the = system is x = �, 2 y 1 or using ordered pairs
1 -x 2 =2 x=4 The solution of the system is x = 4, y = 3 or using ordered pairs (4, 3). 3x - Sy 3 l Sx + Sy = 21 Add the equations to eliminate y: 3x - Sy = 3 ISx + Sy = 21 1 8x = 24 X= -43 Substitute and solve for y: 3 ( 4/3 ) - Sy = 3 4 - Sy = 3 -Sy = -l y = -1 S
{ {
=
The solution of the system is x = � , y = !S or 3 using ordered pairs � .
(�, )
630
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39.
{
Section 12.1: Systems of Linear Equations: Substitution and Elimination
!+! = 8 x y
43.
� - 2. = O x y
1 Rewn. te Iettm' g u = -1 , v = -: y x u+ v = 8 3u- 5v = O Solve the first equation for u, substitute into the second equation and solve: u=8-v 3u - 5v = 0
{
{
41.
x - 2y + 3z = 7 2x+ y + z = 4 -3x+ 2y - 2z = -10 Multiply each side of the first equation by -2 and add to the second equation to eliminate x; and multiply each side of the first equation by 3 and add to the third equation to eliminate x: -2x+ 4y - 6z = -14 2x+ y + z = 4 5y - 5z = - 1 0 3x - 6y + 9z = 2 1 -3x + 2y - 2z = - 1 0 - 4y + 7z = 1 1 Multiply each side of the first result by � and 5 add to the second result to eliminate y: 4y - 4z = -8 -4y + 7z = 1 1 3z = 3 z=1 Substituting and solving for the other variables: x - 2(-I) + 3(1) = 7 y-l = -2 x+ 2+3 = 7 y = -1 x= 2 The solution is x = 2, Y = -1, z = 1 or using ordered triplets (2, -1, 1) .
3(8 - v) - 5v = O 24 - 3v - 5v = 0 -8v = -24 v=3 Since v = 3, u = 8 - 3 = 5 . Thus, x = ! = ..!.. , u 5 y = ..!.. = ..!... T he solution of the system is v 3 . 1 -1 . -, x =-1 , y = -1 or usm. g ordered paIrs 5 3 3 5
{
1
( )
x- y = 6 2x- 3z = 1 6 2y + z = 4 Multiply each side of the first equation by -2 and add to the second equation to eliminate x: -2x+ 2y = -12 2x - 3z = 16 2y - 3z = 4 Multiply each side of the result by -1 and add to the original third equation to eliminate y: -2y + 3z = - 4 2y + z = 4 4z = 0 z=O Substituting and solving for the other variables: 2y + O = 4 2x- 3(O) = 1 6 2y = 4 2x = 1 6 x=8 y=2 The solution is x = 8, y = 2, z = 0 or using ordered triplets (8, 2, 0).
45.
{
x- y -z = l 2x+3y + z = 2 3x+ 2y = 0 Add the first and second equations to eliminate z: x- y -z = 1 2x+ 3y +z = 2 3x+ 2y = 3 Multiply each side of the result by -I and add to the original third equation to eliminate y: -3x - 2y = -3 3x+2y = 0 0 = -3 This equation is false, so the system is inconsistent.
63 1
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
47.
{
-y +4z = 12 y -4 z = 7 0 = 19 This result is false, so the system is inconsistent.
x- y- z = 1 -x + 2y - 3z = -4 3x - 2y - 7z = 0 Add the first and second equations to eliminate x; multiply the first equation by -3 and add to the third equation to eliminate x: x- y- z = 1 -x + 2y - 3z = -4 y -4z = - 3
51.
x+ y- z = 6 3x - 2y + z = -5 x + 3y - 2z = 14 Add the first and second equations to eliminate z; multiply the second equation by 2 and add to the third equation to eliminate x+ y- z = 6 3x - 2y + z = -5 4x- y = 1 6x - 4y + 2z = -10 x + 3y - 2z = 14 7x - y 4 Multiply each side of the first result by - 1 and add to the second result to eliminate y: -4x + y = -1 7x - y = 4 3x = 3 x=1 Substituting and solving for the other variables: 4 (1) - y = 1 3(1) - 2(3) +z = -5 -y = -3 3 - 6 +z = -5 y=3 z = -2 The solution is x = I, y = 3, z = - 2 or using ordered triplets (1, 3, -2) . z:
-3x + 3y + 3z = -3 3x - 2y - 7z = 0 y -4z = -3 Multiply each side of the first result by -I and add to the second resul t to eliminate y: -y +4z = 3 y -4 z = -3 0= 0 The system is dependent. If z is any real number, then y =4z - 3 . Sol ving for x in terms ofz in the first equation: x - (4z - 3) -z = 1 x -4z + 3 -z = 1 x - 5z + 3 = 1 x = 5z - 2 The solution is { (x, y,z) 1 x = 5z - 2, y = 4z - 3 , z is any real number} . 49.
{
{
2x - 2y + 3z = 6 4x - 3y + 2z = 0 - 2x + 3 y - 7z = 1 Multiply the first equation by -2 and add to the second equation to eliminate x; add the first and third equations to eliminate x: -4x + 4y - 6z = -12 4 x - 3y + 2z = 0 y -4z = -12 2x - 2y + 3z = 6 - 2x + 3 y - 7z = 1 y -4z = 7 Multiply each side of the first result by -1 and add to the second resul t to eliminate y:
53.
{
x + 2y - z = -3 2x -4 y + z = -7 - 2x + 2y - 3z = 4 A dd the first and second equations to eliminate z; multiply the second equation by 3 and add to the third equation to eliminate z: x + 2y - z = - 3 2x - 4y + z = - 7 3x - 2y = - 1 0 6x - 12y + 3z = - 2 1 - 2x + 2y - 3z = 4 4 x - I Oy = - 1 7 Multiply each side of the first result by -5 and add to the second result to eliminate y:
632
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Section 12.1: Systems of Linear Equations: Substitution and Elimination
Solve by substitution: 5x + 45 = 3(x + 30) 2x = 45 x = 22.5 So, 22.5 pounds of cashew s should be used in the mixture.
-1 5x + lOy= 5 0 4x - lOy= - 1 7 - 1 1x
= 33 X= -3
Substituting and solving for the other variables: 3 (-3) - 2 y= - 1 0 -9- 2 y= - 1 0 - 2 y= - 1 1 y= 2 -3 + 2
(�J
61.
-Z= -3
-3 + 1 - z= -3 -z= - 1 z= l
The solution is ordered triplets 55.
57.
x = -3, y= .!., z= 1 2
( �, J . -3,
1
63.
Let x the number of $ 25-design. Let y the number of $4 5-design. Then x + y the total number of sets of dishes. 25x + 45y the cost of the dishes. Setting up the equations and solving by substitution: x + y = 200 25x + 45y = 7400 Solve the first equation for y, the solve by substitution: y = 200 - x 25x + 45(200 - x) = 7400 25x + 9000 - 45x = 7400 - 20x = -1 600 x = 80 y = 200 - 80 = 1 20 Thus, 80 sets of the $25 dishes and 120 sets of the $45 dishes should be ordered. =
=
=
=
=
{
y= 2 (l8) + 1 y= 3 6 + 1 y= 3 7
In 2005 there were 1 8 commercial launches and 37 noncommercial launches. 59.
=
Multiply each side of the first equation by .!., 3 multiply each side of the second equation by .!., 4 and add the result to eliminate y x + y = 200 x - Y = 1 50 2x = 350 x = 1 75 1 75 + y = 200 y = 25 The airspeed of the plane is 175 mph, and the wind speed is 25 mph.
or using
Let x the number of commercial launches and y the number of noncommercial launches. Then: x + y = 55 and y = 2x + l Solve by substit ution: x + (2x + 1) = 55 3x= 54 x= 1 8
=
{
Let I be the length of the rectangle and w be the width of the rectangle. Then: 1 = 2w and 21 + 2w = 90 Solve by substit ution: 2(2w) + 2w = 90 4w+ 2w = 90 6w = 90 w = 1 5 feet 1 = 2(1 5) = 30 feet The fl oor is 1 5 feet by 30 feet.
=
Let x the plane' s airspeed and y the wind speed. Rate Time Distance 600 W ith W ind x + y 3 600 x-y 4 Against (x + y)(3) = 600 (x - y)(4) = 600
Let x the number of pounds of cashews. Let y is the number of pounds in the mixture. The value of the cashews is 5x . The value of the peanuts is 1 .50(30) = 45. The value of the mixture is 3 y . Then x + 30 = y represents the amount of mixture. 5x + 45 = 3 y represents the value of the mixture. =
=
633
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
65.
Let x = the cost per package of bacon. Let y = the cost of a carton of eggs. Set up a system of equations for the problem: 3X + 2y = 1 3.45 2x + 3y = 1 1 .45 Multiply each side of the first equation by 3 and each side of the second equation by -2 and solve by elimination: 9x + 6y = 40.35 -4x - 6y = -22.90 5x 1 7.45 x = 3 .49 Substitute and solve for y: 3(3.49) + 2y = 1 3 .45 10.47 + 2y = 1 3 .45 2y = 2.98 y = 1 .49 A package of bacon costs $3.49 and a carton of eggs cost $ 1 .49. The refund for 2 packages of bacon and 2 cartons of eggs will be 2($3049) + 2($ 1 .49) =$9.96.
69.
{
67.
y = ax 2 + bx + c At (-1 , 4) the equation becomes: 4 = a(- 1) 2 + b(-l) + c 4 = a-b+c At (2, 3) the equation becomes: 3 = a(2) 2 + b(2) + c 3 = 4a + 2b + c At (0, 1 ) the equation becomes: l = a(0) 2 + b(0) + c l=c The system of equat ions is: a- b+c = 4 4a + 2b + c = 3 c=l Substitute c = 1 into the first and second equations and simplify : a- b+1 = 4 4a + 2b + 1 = 3 a- b = 3 4a+2b = 2 a = b+3 Solve the first result for a, substitute into the second result and solve: 4(b + 3) + 2b = 2 4b + 1 2 + 2b = 2 6b = - 1 0 b = - -5 3 a = --5 + 3 = -4 3 3 The solution is a = .i , b = -�, c = 1 . The 3 3 5 . 4 . IS y = -x 2 --x + 1 . equatIOn 3 3
{
Let x the # ofmg of compound 1 . Let y the # of mg of compound 2. Setting up the equations and solving by substitution: 0.2X + OAy = 40 vitamin C 0.3x + 0.2y = 30 vitamin D Multiplying each equation by 1 0 yields 2X + 4Y = 400 6x + 4y = 600 Subtracting the bottom equation from the top equation yields 2x + 4y - ( 6x + 4 y ) = 400 - 600 2x - 6x = -200 -4x = -200 X = 50 2 ( 50 ) + 4y = 400 1 00 + 4y = 400 4y = 300 y = 300 = 75 4 So 50 mg of compound 1 should be mixed with 75 mg of compound 2. =
=
{
{
71.
{0.06Y - 5000r = 240
0.06Y + 6000r = 900 Multiply the first equation by - 1 , the add the result to the second equation to eliminate Y. -0.06Y + 5000r = -240 0.06Y + 6000r = 900 1 1 000r = 660 r = 0.06 Substit ute this result into the first equation to find y.
634
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Section 12.1: Systems of Linear Equations: Substitution and Elimination
If only half of the orchestra seats are sold, the revenue i s $ 14,600. SO, 50 X + 35 Y + 25Z = 14, 600 . Thus, we have the followi ng system: x + Y + Z = 500 50x + 35y + 25z = 1 7, 1 00 25x + 35y + 25z = 1 4, 600
0.06Y - 5000(0.06) = 240 0.06Y - 300 = 240 0.06Y = 540 Y = 9000 The eq ui lib ri um level ofi ncome and i nterest rates i s $ 9000 mi lli on and 6%. 73.
{
{
12 = 11+ 13 5 - 311- 512 = 0 1 0 - 512 - 713 = 0 Substit ute the expressi on for 12 i nto the second and thi rd equati ons and si mplify: 5 - 311- 5(11+ 13) = 0 -S11- 513 = -5 1 0 - 5(11+ 13) - 713 = 0 -511- 1 213 = - 1 0 Multi ply both si des o f the first result by 5 and multi ply b oth si des of the second result by -S to eli mi nate II: -4011- 2513 = -25 4011+ 9613 = SO 7 113 = 55 13 - � 71 Sub sti tuti ng and solvi ng for the other vari ab les: = -5 -S11- 5 275 = -5 -SI1- 71
Multi ply each si de of the first equati on by -25 and add to the second equati on to elimi nate z; multi ply each si de of the thi rd equati on by- l and add to the second equati on to eli mi nate z : -25x - 25y - 25z = - 1 2, 500 50x + 35y + 25z = 1 7, 100 = 4600 25x +l0y 50x + 35y + 25z = 1 7, 1 00 -25x - 3 5y - 25z = -14, 600 25x = 2500 x = 1 00 Substi tuti ng and solvi ng for the other vari ables: 25(1 00) + lOy = 4600 1 00 + 210 + z = 500 2500 + 10y= 4600 3 1 0 +z = 500 I Oy= 2 1 00 z = 190 y= 2 1 0 There are 1 00 orchestra seats, 2 1 0 mai n seats, and 1 90 b alcony seats.
( ��)
77.
SO -S11= -71 I 1= .!.Q 71 5 5 = 65 I 2 = .!.Q + 71 71 7 1 1 0 , I = 65 , I3 = 55 The so1utt· On I S I1= 71 71 2 71
()
75.
.
(� )
Let x = the number of servi ngs of chi cken. Let y = the number of servi ngs of com . Let = the number of servi ngs of 2% mi lk. P rotei n equati on: 30x + 3 y + 9z = 66 Carbohydrate equati on: 35x + 1 6y+ 1 3z = 94.5 Calci um equati on: 200x + lOy+ 300z 910 Multi ply each si de of the first equati on by -16 and multi ply each si de of the second equati on by 3 and add them to eli mi nate y; multi ply eac h si de of the second equati on by -5 and multi ply eac h si de of the thi rd equati on by S and add to eli mi nate y : -4S0x - 4Sy- 144z = - 1 056 1 05x + 4Sy + 39z = 2S3.5 - 375x - 1 05z = - 772.5 z
=
-
Let x the number o f orchestra seats. Let y the number of mai n seats. Let z = the numb er ofb alcony seats. Si nce the total number of seats i s 500, x + y+z = 500 . Si nce the total revenue i s $ 17, 1 00 i f all seats are sold, 50x + 35y + 25z = 1 7 , 1 00 . =
=
635
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
79.
-175x-80y- 65z = -472.5 1600x+80y+2400z= 7280 1425x + 2335z = 6807.5 Mul tiippllyy each each sisiddee ofof thethe second first resulresult byt by19 5andto mul t el-7125x-1995z iminate x: = -14,677.5 7125x+ 11, 675z = 34,037. 5 9680z = 19,360 z=2 Substi tuting and=sol-772. ving5for the other variables: -375x-l05(2) -375x-210 = -772. 5 -375x = -562. 5 x = 1. 5 30(1.5) + 3y + 9(2) = 66 45+3y+18= 66 3y =3 y= 1 The shoulandd serve2 servi1. 5nservi 1 servidientigtiofancom, gs ofn2%gs ofmichilk.cken, LetLet xy thethe pripriccee ofof 11 hamburger. order of fries. Let z the price of 1 drink. We{8X+can6y+const6zruct= 26.10 the system lOx + 6y + 8z = 31.60 A system involving only 2 equatio ns that contain 3 or more unknowns cannot be solved uniquely. Multiply the first equation by -.!.2 and the second equation by .!.,2 then add to eliminate y: -4x-3y-3z = -13. 05 5x+3y+4z = 15.80 x +z= 2. 75 x= 2. 7 5-z Substi tute and+3y+4z solve for= 15.y in80terms ofz: 5 ( 2.75-z ) 13. 7 5+3y-z = 15. 80 3y = z+2. 05 1 41 y=-z+3 60
Solutions of the system are: x = 2. 75 -z, 1 41 y= -z+-. 3 60 Since wevalareuesgiofvenz thatthatgi0.v6e0two-deci z 0.9m0al, -weplace choose val0.7u5es ofx y 1.andy 00 . with 1. 7 5 x 2.25 and The possible values of x, y, and z are shown in the table. z y x 2.13 0. 89 0.62 2.10 0. 90 0.65 2. 07 .091 0. 68 2. 04 . 092 0. 7 1 2. 0 1 .093 0.74 1. 98 . 094 0.77 1.9 5 0. 9 5 0.80 1.92 0. 9 6 0.83 1.89 0. 9 7 0.86 1.86 0. 98 0.89 LetLet xy Beth's t i m e worki n g al o ne. Bill's time working alone. Let z Edie's time working alone. We can use the following tables to organize our work: Beth Bill Edie Hours to do job x y z Part ofjob done 1 1 -1 in 1 hour x y z In 10 hours they complete 1 entire job, so 1 1) = 1 10 (1-x +-+yz 1 1 -x1 +-+1 -= z 10 Bill Edie Hours to do job y z Part of job done 1 -1 in 1 hour y z �
=
81.
=
�
�
�
�
�
=
=
=
=
-
-
Y
-
636
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Section 12.2: Systems of Linear Equations: Matrices
15 hours�) th ey complete 1 entire j ob , so ( 15 � + = l . 1 1 1 -+-=15 B eth Bi ll Edi e H ours to do j ob x y P art of j ob done -1 1 1 i n 1 hour x -y W ith all 3 working for 4 hours and B eth and B ill working for an additional 8 hours, they complete 1 entire j ob, so 4 ( .!.x + .!.y + .!.) +8 ( .!.x + .!.y ) = 1 g + g + � =l x y W e have the system 1 1 1 -x1 +-+-= Y 10 1 1 -1 -+-= g + gy + � = l15 x y Sub tract the second equation from the first equation: -x1 +-+Y1 -1 =101 -y1 + -1 = -151 1 x x=3030 Substitute x 30 into the thi rd equation: g + g + � =l 30 Y -12y + -4 = -35
-12 + --12 = --12 -15 Y -12Y + -4 35 8 3 =4015 P lugging 40 to find y: -12y + -4 =53 4 3 -12y +-= 40 -5 12 1 y 2 y=24 W orki ng alone, it would take B eth 30 hours, Bill 24 hours, and Edie 40 hours to complete the j ob.
In
y
z
z
z
z
z
z
z =
z
z
z
z
83.Answers wi ll vary.
z
z
Section 12.2
z
1.
matrix
3. True 5.
z
W riting the augmented matri x for the system of equations:
{4;:��:� � [� -� I:]
z
7.
=
z
{2X+3Y -6 = 0 4x-6y+2 = 0 W ri te the systemi n standard form and then write the augmented matrix for the system of equati ons: { 2X+3Y = 6 � [2 3 1 6] 4x-6y -2 4 -6 -2 =
z
9.
N ow consid er the system consisting of the last result and the second origi nal equati on. Multiply the second original equation b y and add it to the last result to elimi nate
W riting the augmented matrix for the system of equati ons:
{0.O lx-0.03Y =0.06 �[0.0 1 -0.03 1 0. 06] 0.13x+ 0.l0y= 0.20 0.13 0.10 0.20
y: -12
637
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Chapter 12: Systems of Equations and Inequalities
11.
13.
{
b.
Wri tinogns:the augmented matrix for the system of equati x- y+ z 10 -1 1
1 20 '�l
x+ y+2z= 2 [i Writinogns:the augmented matrix for the system of equati x+ y-z�2 {5x+3 3x-2yy-z=1=2 � l: -23 0 ;] Writinogns:the augmented matrix for the system of equati [� -1 �1 � � 0 12X-3:�:�:: _: � -3 4 0 5 1 x +4y = 5 4x-5y+z=0 4 0 [ 1 -3 -2 � { x-3y=-2 2 -5\ 5 ] 2x-5y=5 =-2fj2+ 1 -3 -2 [� =�1� ]�[_2(1)+2 2(-3)-5I -2(-2)+5 1 � [� �31�2] x-3y+4z { 4 3 -53 64 66] � -53xx-5+3yy+4z +6z===66 =-3fj + [i, 6 :] �[ : + 3 4 -3(�+6l -6 { �l �
�
3 x+ 3 y
=
5
3
a.
a.
-3
-3
+r2
-5 3 -4 -6 4
) 5
(
1
�
b.
-3 1
� 3( [1 � � [20 =�
1) -3
-3 4 -5 3 4
3( ) 3 -3(-3 ) - 5
R2
-3
r2
-
4
- 2 2 4 :J x-3 y+2z=-4=-6 -3-5 23 -6-4] � { 2x-5y+3z -6 4 6 -3x-6y+4z=6 =-2fj 2 -6] [i 6, 2 � [-2 (l-3� +2 -2 -3-6 - -2 2 + -2 (-; )-4 ] [ 0 2 -6] -3 -6 4 3
3
R2
6 L]
1
r2
-3
[i,[ 6 �l � [� : 6 -3 4 -5 3 4
-3 -5
-5
R2
r3
4 -3 -5 3 5(1) -5 5( -3) +3 5(4) + 4 '(3
-1
17.
=5fj + 1
1 -1
1 5.
R3
8
-1
6
�]
2
3(-3)-6 3(2)+4 � -=!]2
-15 10
4
( ) 3 4
-6
-3
-5
3
1
-3(4)+6
-5
-5
-3 4
4
3
4
638
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23.
[�-4 =�1 �4 =�]6 {2:��:::;: =� -4x+y+4z=6
a.
R\
=-2r2 +
�
1
4
6
-2 (- 5 )
-
3
-2(6)+1 6 4
1
-4
-
�
5 1- 3
-4
Consi stent; (5, -1) .
or
3
:
= =2 -
x4, {(XI, X2, X3, X4) XI 1, x2 x3 3 - 2x4, X4 is any real number}
6
I
=
Consi stent;
I
=
XI 2 - 4x4 X2= 3 - x3 -3x4 X3' X4 are any real numbers
or
I
=2
- 4x4, X2 {(XI, X2'X3, X4) XI x3 and x4 are any real numbers}
35.
1 l;� =
=
3 - X3 - 3x4,
:2
XI + X4= -2 x2 + 2x4 X3 - X4
-0 0=0
Consi stent;
2 : � _-2::
X3 X4 x4 is any real number
or usi ng ordered pai rs
{:0=3:�
or 37.
{
{(XI, x2, X3, x4) I XI
=
x3
=- 2
- X4, X2
x4, x4 is any real number}
=2
- 2x4,
X + Y= 8 x - y= 4
W ri te the augmented matri x:
[� -�I!J�[� _�I_!J(R2=-1i+r2) �[� � I�] (R2 =-tr2) �[� �I�J (RI=-r2+1i)
x = 6, y = 2 (6,2).
Consi stent;
The soluti on i s pai rs
i s any real number
or real number}
=
1;� �
-2(--22)-21
Inconsi stent 29.
2
=
4+1
{X+2Z =-1 y-4z0=0=-2 {Xy==-1-2z -2+4z z{(x,y,z)lx=-1-2z,y=-2+4z,
:� : �
X2 + x3 + X4
- X4 X3 3 - 2x4 X4 is any real number
-5 1
5,
J l
Consi stent;
-2-2] 6 [ 4+2 6 6-2-2 ] 4 6 ->[!4 =� ! �] {:: �1 x= y=-1,
b.
27.
31.
'i
[� =� � =�l [-2(2)+5 -5 � 2 �5 ->[: T �l -4
25.
Section 12.2: Systems of Linear Equations: Matrices
zi s any
or usi ng ordered
639
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
39.
{2X-4Y=-2
3x+2y= 3
45.
I-�]�[� -� I-�] (Rl==-3t�) + �[� -� I-�] (R2 � r2) �[� -� 1-�] (R2= r2) �[ 1 01 !] (Rl= 2r2+�)
-�
o
t
The so Iutl· On IS·
(�'i). 41. { X+ 2Y=4 2x+4y=8
x=-,12 y=-43 4"
o
o
.
[� �I:]�[� �I�] (R2=-2�+r2)
o
This is a dependent system.
is a ny rea l number is a ny rea l number}
o
I
_ 1.
I -I
The solution is
8
o
-1-
or
t
o
W rite the a ugmented ma trix:
1.
or -, 1
W rite the a ugmented ma trix: 1 -1
W rite the a ugmented ma trix:
43.
I
1
The so Iuti· on IS·
47.
The solution is I or
-'3
'3
. or usm g ordered pairs
x+2y=4 x=4-2y x=4 -2y,y {(x,y) x= 4-2y,y 12X+3Y= 6 x- y=-2 2[ 3 1 6]�[1 t 3] (Rl=t�) t 1 -1 t �[10 -f 3] (R2=-�+r2) �[� 11�] (R2=-ir2) �[� � 11] (Rl=-tr2+�) x=%, y=1 (%, )
3[15 -55 1 213]�[151 -15 1 211] (Rl=trl) � [ 1 -301 1 61] (R2=-15r1 +r2) 5 [ 1 � 15]1 (R2=30r2) �[01 01154] (Rl= 1 r2+�) x=-,43 y=-51 ( 43 -5 ). {2xx- y-3z==166 2y+ z= 4 [2 0 -30 166] 2 14 [� 01 -12 -30 46] (R2=-2�+r2) 2 14 r� 01 -11 -t0 261 (R2= r2) 2 14 [� 01 01 -t 21 (Rl= r2+r1 0 4 0 R3 =-2r2 +r3) [� 01 01 _-t1. 21 (R3=tr3) 0 0 1 [� 01 01 00 28] [Rl==trr33++r21 0 1 0x= 8,yR2= 2, = 0 (8,2,0).
W rite the a ugmented ma trix:
W rite the a ugmented ma trix:
[�
3x-5y= 3 {15x+5y=21
8
o
o
1 .
The solution is
640
t
Z
1j
or
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
49.
1
l
Section 12.2: Systems of Linear Equations: Matrices
x-2y+ 3z = 7 2x+ y+ z = 4 -3x+2y-2z = -IO
53.
-x+ y+ z = -I -x+2y-3z = -4 3x-2y-7 z = 0 W rit e t he a ugment ed mat rix:
[=: � -� ��] [ : -� =� ] [ : �� -�] [� � �! ��] l l 3 -2 -7
0
1 -4 3 -2 -7 0
� -
1 � 0
0
�
1
2x-2y-2z = 2 51. 2x+ 3y+ z = 2 3x+2y = 0 W rite the a ugment ed ma trix: 3 2
�
�
o
o
55.
5 3 -3 -
1 -4 -3 0
�
3 -7
�
0 0 -3 There is no solut ion. The system is inconsist ent .
z
I
-2 3 -7 1
1 -1 t 3 0 -4 -1 2 _4 7 0 0
(R3 = -r2 + r3)
z
2x-2Y+ 3 z = 6 4x- 3 y+ 2z = 0 -2x+ 3y-7 z =1 W rite the a ugmented mat rix:
�
(R, �t,,)
0
1 [ ! �� ��] [ � =� ! �1 [ 1( [� =1 ��1 ( -2
0 0
�
-
0 The ma trix i n the la st step rep resents the system X - 5z = -2 X = 5Z-2 y-4z = -30 r, equiva lent ly, y=4z -3 0 =0 0 =0 The solution is x= 5z -2, y= 4z -3 , is a ny rea l number or {(x,y,) z I x= 5z -2,y= 4z - 3, is a ny rea l number} .
The solution is x= 2, y= -1, z = 1 or (2,-1,1) .
[� -� -� �] [i -� -1 �] [� -� -� �] [� � � �]
o
(Rl = -Ij )
o
(Rl=trl) R2 =-41j+r2 R3 21j + r3 =
Rl r2 + Ij R3 -r2 + r3 =
]
]
0 0 -9 1 There is no solut ion. The syst em is inconsist ent . =
641
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Chapter 12: Systems of Equations and Inequalities
57.
l
X+Y- Z= 6 3x-2y+ z=-5 x+3y-2z= 14
[� -� -: -�] ->[i -� �� 1] ->[i 0 �� �l
W rite the augmented matri x: 3 -2
1
14
-2
[�o 0 =! �j [I0 0 ¥1.] o0 0 [�o 0 � �l I, 2
�
�
�
1
t-s -t -1 1 -2 -2
The solution is x== 59.
!
The solution is X=-3,y=�,Z=1 or (R, --h)
6 1.
(-3,�,1).
3X+ y- z=j
2x- y+ z=1 4x+2y
8
=3
= (R3 b)
y== 3, Z= -2,
or
(1, 3,-2) .
x+2y- z=-3 2x-4y+ z==-7 -2x+2y- 3z==4
[ � -� -: =�l [[�oI -� -I-� -�=�]l 0 o [0I 0 j
W rite the augmented matrix: -2
�
�
�
2 -3
4
6 -5 -2
2 1 -t "8 6 -5 -2
o0
(R2 =-tr2)
--;\- -¥ -t "8I II II -4 -4
642
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
63.
1
{
Section 12.2: Systems of Linear Equations: Matrices
x+ y+ z+ W= 4 2x- y+ z = 0 3x+ 2Y+ z - W= 6 x-2y-2z +2w = -1 W rite the a ugmented ma trix: 1
65.
[� -� � �1 � [ -� -� -:1 and [� -:1 (Interchange) [� : 1 [f � -� � ��1 [f ; -j �� ��1 [f ; ! -� �1 [ � ; ! -1 ;1 [f ; P �1
X + 2Y + z = l 2x - y + 2z = 2 3x + y + 3z = 3
3 2 1 -I 6 1 -2 -2 2 -1
�
o
o
�
o o
�
o
o
-I
-2 -3 -3
-I
-4
-6
-5
- I -2 - 4 -3 - I -2 - 8 -3 -3 -5
1 2
4
-2 - 8 -3 -3 1 -5 -3
-I
r2
( R2
{
r3
=
The ma trix in the la st step rep resents t he system X + 2Y + Z = 1 - 5y = 0 0=0 Substitute a nd solve: -5y = 0 x + 2(0) +z = 1 y=o z = l-x The solution is y 0, Z 1 - x, x is a ny rea l number or { (x, y,z) I y = 0, z = I - x, x is a ny rea l number} .
-r2 )
=
4
( R, �h)
67.
4
=
{
x-y+ z= 5 3x + 2y - 2z = ° W rite the a ugmented ma trix: l � 2 �1
[� � 1 �] �[�
4
( R, �h)
4
1 ]
0 2 1 -1 -3 The ma trix in the la st step rep resents the syst em X=2 or, equiva lently, x = 2 y =z-3 y -z = -3 Thus, the solution is x = 2 , y z - 3 , z is a ny rea l number or { (x, y,z)lx = 2, y = z -3, is a ny rea l number} . °
4
The solution is x= I , Y= 2, z= 0 , W =1 or ( 1 , 2, 0, 1 ).
{
{
=
643
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1 r � �1 =� �01
Chapter 12: Systems of Equations and Inequalities
2X + 3 Y - z = 3 x - y -z = O 69. -x + y +z = O x + y + 3z = 5 W rit e t he a ugment ed mat rix:
Y-7
=���jl
� [� I
n -1
----'- [� 1 � o 10 [� o1� -8 11 � oo 01.!2.0 rj � ----,
o 2 o 0 =�0 �0 -S -7
7 4 5
C:) = -7
13 x =9 1 3 7 1 9 or Th us, t h e so 1utlOll · IS· = -, Y = - , z = 9 3 7 19 9'18'18 .
18 18
(1 ) { 4x + y + z - w = 4 x- y + 2z + 3w = 3 Writ e t he a ugment ed mat rix: 1 [; -1 2 � 1;] (int ercha nge) -1 2 �1 1:] a nd r2 X
�1 ( int erCha nge ) -11 11 31 05 a nd r2 � �1 �1 o 0 00 o 2 45 (int ercha nge)
� rr� �
X-S
Y = 18
1 3 5
-1 1 1 1
C:) = 7 7
71.
1j
1j
{
The mat rix in t he la st st ep rep resent s t he syst em x - y + 2z + 3w = 3 5y - 7z - 1 3w = -S The second equat ion yields 5y - 7z - l 3w = -S 5y = 7z + 1 3w - S 13 S 7 + -w-y = -z 5 5 5 The fl rst equat ion yields x - y + 2z + 3w = 3 x = 3 + y - 2z - 3w Subst itut ing for y: x = 3 + - S + 7 z + 1 3 w - 2z - 3w S S S 3 x = --z - -2 w+ -7 5 5 5 3 - -2 w+ -7 , · · x = --z Thus,t h e so lutlOnlS 5 5 5 1 3 --S , z a nd wa re a ny rea l numbers or y = -7 + 5 5 5 , x, y , Z) x = - � z - � w + .?' y = .? z + � w- !' 5 5 5 5 5 5
r) a nd r4
( R2 = -2r) + r2 )
-7 -7 IS 9
(
-7 -7 -7 18
The mat rix in t he la st st ep rep resent s t he syst em X - SZ = -7 y - 7z = -7 19 z =IS Subst it ut e a nd solve:
{(W
z
)
i
W
}
z a nd w a re a ny rea l numbers .
644
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Section 12.2: Systems of Linear Equations: Matrices
(I nt ercha nge)
73. Ea ch oft he point s must sat isfy t he equat ion 2 y = ax + bx + c . (1, 2) : 2 = a + b +c (-2, -7): - 7 = 4a - 2b + c (2 3) - 3 = 4a + 2b + c Set up a mat rix a nd solve:
r3 a nd fj
:
,-
[� -� -�l 4
2
-3
o o o
The solut ion is a = - 2, b = 1, c = 3 ; so t he equat ion is y = _2x 2 + + 3 .
o o
X
75. Ea ch oft he point s must sat isfy t he equat ion f(x) = ax 3 + bx 2 + cx + d. f(-3) = -12 : -27a + 9b - 3c + d = - 1 1 2 f(-I) = -2 : -a + b- c + d = -2 f(l) = 4 : a + b+c+d= 4 8a + 4b + 2c + d = 1 3 f(2) = 1 3 : Set up a mat rix a nd solve: -1 2
I
The solut ion is a = 3, b = -4, c = 0, d = 5 ; so t he equat ion is f(x) = 3x3 - 4x2 + 5 .
[-�� : =� ��1 1 1 8 4
o 0 "3I .!±3 0 -"35 - "3 0 0 0 -5 - 25 o 0 "3I .!±3 0 -1. _1. 0 3 3
4 13
645
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as
they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
== =
Condition on investment equa tion: z = O.5x x - 2z = 0 Set up a ma trix a nd solve:
77. Let x the number of servings of sa lmon stea k. Let y the number of servings of ba ked eggs. Let z the number of servings ofa corn squa sh. P rotein equa tion: 30x + 15 Y + 3z = 78 Ca rbohydra te equa tion: 20x + 2y + 25z = 59 V ita min A equa tion: 2x + 20y + 32z = 75
0.07
o
1
10,000 1 0.01 0.02 80 -3 -10,000 -1
(I ntercha nge) r3
1 10,000 0.08 680 0 -2
1 0,000 2
1 -1
a nd r 1
-
3
-� �� � � � � -� �� �
o
o
]
-1 - 2000
o
]
8000 -10,000
]
1 2000
o
0
4000
1 0
o
4000
1 2000
]
]
Ca rletta should invest $4000 in Trea sury bills, $4000 in Trea sury bonds,a nd $2000 in corpora te bonds.
== =
81. Let x the number of D elta s produced. Let y the number of Beta s produced. Let z the number of Sigma s produced. Pa inting equa tion: l Ox + 1 6 y + 8z = 240 D rying equa tion: 3x + 5 y + 2z = 69 P olishing equa tion: 2x + 3 y + z = 41 Set up a ma trix a nd solve:
Substitute z = 1 a nd solve: -1 98y - 295(l) = - 69 1 -198y = -396 y=2 x + 10(2) + 16(1) = 37.5 x + 36 = 37.5 x = 1 .5 The dietitia n should serve 1 .5 servings of sa lmon stea k, 2 servings of ba ked eggs, a nd 1 serving of a corn squa sh.
] [ [� � � ��l [� � � �:O] ->[i 10 16 3 5 3
2
-t
2
== =
79. Let x the a mount invested in Trea sury bills. Let y the a mount invested in Trea sury bonds. Let z the a mount invested in corpora te bonds. Tota l investment equa tion: x + y + z = 10, 000 Annua l income equa tion: 0.06x + O.07y + 0.08z = 680
-t
646
o
3
8 240 2 69 1
41
1 41
-3
-25
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
Section 12.2: Systems of Linear Equations: Matrices
--)
1 0 0 0 * o I 0 0 2 '6 o 0 I 0 23 o 0 0 1 �
16 44 -, -, 12 =2, I) =23 The so Iuti· on IS· I, =23
= =
{-
3.
83. Rewrite the system to set up the ma trix a nd solve: 212 = 4 4 + 8 - 212 = 0 8 = 514 + 1, � I, + 514 = 8 I, + 313 = 4 4 = 313 + I, � -� -� = O �+� = �
z
{ [�I � � �:] (Interchange) [� j � -� -�� �]�] and --) [� � -I� �] � 0 0
3
-I
o 0 o 0
3 -5 -4 -6 -8
--)[�� �� � �� �:] ] �r� ; � - 6 -8 3 -5 -4
-I
0 -2
5 8 o 2 6 8 I
]
1
I 20,000 I I 0.07 0.09 0.11 2000 1 20,000 --) 7 9 11 200,000 (R2 =100'2 ) 20,000 ----, o 2 1460,000 (R2 ='2 -7/i)
[I I 1 ] -'-[I 1 ] --)[� �I��:���] (R2=t'2 ) [ 1 - 11 - 1 0, 000] (R[ = - '2) ---,
0 4 0
[--) � � � o 0 o 0
[
-I
'2
28 23
85. Let x = the a mount invested in Trea sury bills. Let y the a mount invested in corp ora te bonds. Let z the a mount invested in j unk bonds. Tota l investment equa tion: x+y+ =20,000 Annua l income equa tion: 0.07x+ 0.09y+ O.llz=2000 Set up a ma trix a nd solve:
The compa ny should p roduce 8 D elta s,S Beta s, a nd 10 Sigma s.
I
14
Ii
----'0.
a
'i o 1 2 30, 000 The ma trix in the la st step rep resents the -z = -10' 000 system y + 2z = 30, 000 Therefore the solution is = -10, 000 + z , y = 30, 000 -2z , z is a ny rea l nu mber. P ossible investment stra tegies:
{X
x
(InterChange) ') and'4
Amount Invested At
7% 0 1 000 2000 3000 4000 5000
-2
;�
9% 1 0,000 8000 6000 4000 2000 0
1 1% 1 0,000 1 1 ,000 12,000 1 3,000 14,000 1 5,00 0
647
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Chapter 12: Systems of Equations and Inequalities b.
1
Tota l invest ment equa tion: x + y + z = 25, 000 Annua l income equa tion: 0.07x + 0.09y + 0. l lz = 2000 Set u p a ma trix a nd solve: 1 1 1 25, 000 0.07 0.09 0.11 2000 1 1 I 25, 000 = 1 00 7 9 1 1 200, 000 1 2 5, 000 --"- 1 = - 71j ) 0 2 4 25, 000 1 2 5, 000 --"- 1 = 2 12, 500 0 - 1 1 2, 500 = 1j 2 1 2, 500 The ma trix in the la st step rep resents the x - z = 12, 500 system y + 2z = 1 2, 500 T hu s, the solu tion i s x = z + 1 2, 500 , y = -2z + 12, 500 , z is a ny rea l nu mber. P ossible invest ment stra tegies:
[ �[ [ [
] 1 ] (R 2 r2) 1 ] (R 2 r2 1 ] (R 2 tr2) 1 ] (R( -r2) 1
----r
----r
{
{
=
Amount Invested At
7% 30,000
c.
9% 12,500 8500 4500 0
9% 0
1 1% 0
This will yield ($30,000)(0.07) $2100, which is more tha n the requ ired income. =
d.
87.
Amount Invested At
7% 12,500 14,500 1 6,500 1 8,750
] (R2 = tr2) 1 ] (R( -r2)
1 3 0, 000 2 -5000 o -1 3 5, 000 = 1j I 2 -5000 T he ma trix in the la st step represents the x - z = 35, 000 system y + 2z = -5000 Thu s, the solu tion is x = z + 35, 000 , y = -2z - 5000 , z is a ny rea l nu mber. H owever, y a nd z ca nn ot be nega tive. From y = -2z - 5000 , we mu st ha ve y z = O. O ne p ossible invest ment stra tegy
=
Let x the a mou nt of su pplement 1 . Let y the a mou nt of su pp lement 2. Let z the a mou nt of su pplement 3. 0.20X + 0.40 y + 0.30z = 40 V ita min C 0.30x + 0.20y + 0.50z = 30 V ita min D Mu ltiplying ea ch equa tion by 1 0 yields 2X + 4 Y + 3z = 400 3x + 2y + 5z = 300 Set up a mat rix a nd solve:
{ {
1 1% 0 2000 4000 6250
Answers will va ry. =
=
Tota l investment equa tion: x + y + z = 30, 000 Annua l income equa tion: 0.07x + 0.09y + 0. llz = 2000 Set up a ma trix a nd solve: I 1 I 3 0, 000 0.07 0.09 0. 1 1 2000 1 1 1 30, 000 = I 00r2 7 9 1 1 200, 000 1 30, 000 --"- 1 = r2 - 71j ) o 2 4 - 10, 000
[
�[ [ -r
1
1
1
] ] (R 2 ] (R(
)
648
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Section 12.3: Systems of Linear Equations: Determinants
{
{
The ma trix in the la st step rep resents the system X + i Z = 50 y - i z = 75
15. x + y = 8 x-y = 4 D = I 1 =-1 - 1 =-2 1 -I D = 8 1 =-8 -4=-12
Therefore the solution is x = 50 - 2. 4 Y = 75 + � Z , z is a ny rea l number. 8 P ossible combina tions: Supp lement 1 Supp lement 2 Supp lement 3 Omg 50mg 75mg 8mg 36mg 76mg 16mg 77mg 22mg 78mg 8mg 24mg Z ,
x
y
17.
Section 12.3
{2x5X +- 3yy == 11 32 D = I � -� 1 = 15+2 = 17 D = 1 13 -1 1 = 39+12 = 51 12 3 D = 1 5 13 1 = 60 - 26 = 34 y 2 12 x
1. determina nts 3. Fa lse
I! � 1 = 3(2) - 4(1) = 6 - 4 = 2 7. I_� � 1 = 6(3) - (-1)(4) = 18+4 = 22 9. I-! -� 1 = -3(2) - 4(-1) = -6+4 = -2 11. J � �l �JI�� j�41: � H : ��I
Find the solutions by Cra mer's Rule: Dy 34 Dx = 5 1 = 3 y =x ==-= 2 D 17 D 17 The solution i s (3, 2).
5.
19.
�
�
= 3[( -1) ( -2) -2 (5)]-4[1 (-2) - 1 (5)]
=3 ( -8) -4 ( -7)+2 (3) =-24 +28+6 =10 4 6
1 1
D = 1 8 =4 -8=-4 1 4 Find the solutions by Cra mer's Rule: Dy -4 Dx = -= -12 6 y = =-= 2 x=_ D -2 D -2 The solution is (6, 2).
89 - 91. Answers will var y.
13.
I 1 1 4 -1 1
+2[1 (2) - 1 ( -1)]
=
Find the solutions by Cra mer's Rule: Dy -24 Dx = 48 = 8 y = x == = -4 D 6 D 6 The solution is (8, -4) . -
1 1 1 1 1 1 -3 1
-I 2 -I ° =4 -1 -3 -3 4
{3Xx + 2y == 240 D =I � � 1 = 6-0 = 6 Dx = 1 2� � 1 = 48 - 0 = 48 Dy 1 3 24 1 = 0 - 24 = - 24 1 0
° ( 1) 6 ° +2 6 -1 4 __ 1 4
=4[ -1 (4) -0 (-3)] + 1[6 (4) - 1 (0) ] +2[6 (-3) - 1 ( -1)] = 4( -4) + 1 (24) +2 (-1 7) =-16 +24 - 34 =-26
649
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
{
3X -6Y = 24 21. 5x+4y = 12 D = - = 12 - (-30) = 42
29.
I� �1 Dx = 1 24 -6 1 = 96 - (-72) = 168 12 4 D = 1 3 24 1 = 36 - 120 = -84 y 5 12
1 1 1 D, = I � : 1 = 1 -6= -5
Find the solutions by Cra mer's Rule: 15 D y -5 D 3 y=-=-=l -"2 =X=-x =-D -5 2 D -5 The solution is 1 .
3X - 2Y = 4 23. 6x -4y = 0 D = 3 -2 = -12- (-12) = 0 6 -4 Since D = 0, Cra mer's Rule does not app ly. -4Y = -2 25. 2X 3x+2y = 3 D = - = 4+12 = 16
(%, )
1 1
{
r::1 :� �
D= 2 3 =- 2 - 3=-5 1 -1 6 3 =_6_�= _� Dx = .1 -1 2 2 2
Find the solutions by Cra mer's Rule: D -84 Dx =168 =4 y=--..L x=_ =-- = -2 D 42 D 42 The solution is (4,-2) .
{
3
31 .
I � �1 Dx = 1 -2 -4 1 = -4+12 = 8 3 2 2 D = 1 -2 1 = 6+6 = 12 y 3 3
{ 3x - 5y= 3 1 5x + 5y= 2 1 D= \ 3 -5 \ = 1 5- (-75)=90 15 5 Dx = 1 3 -5 \ = 1 5 - (-105)=120 21 5 y = \ 3 3 \ =63 - 45= 1 8 1 5 21 D
Find the solutions by C ra mer's Rule: D x= Dx = 120 = 4 y= y = 1 8 ="51 Ii 90 "3 1) 90 The solution is
Find the solutions by Cra mer's Rule: 8 = 1 Y = Dy = 12 = 3 x = D; = 16 "2 1) 16 4 The solution is .
(�, %)
2x - 3y = -1 27. { lOx+ lOy = 5 D = 1 2 -3 1 = 20- (-30) = 50 10 10 D = 1 -1 -3 1 = -10 - (-15) = 5 5 10 D = 1 2 -1 1 =10- (-10) = 20 y 10 5
33.
x
(�, �)
{
x + y - z= 6 3x - 2y + z =-5 x + 3y - 2z = 14 -1 D= 3 - 2 3 -2 =1 - 2 1 _ 1 3 1 + ( _ 1) 3 3 -2 1 -2 1 = 1(4 - 3) - 1(-6 - 1) - 1(9+2) =1+7 - 1 1 =-3
1
Find the solutions by Cra mer's Rule: D 20 2 Dx = 5 = -1 x=D 50 10 y =Dy = 50 = -5 The solution is 10 5
(J..., 3.).
.
1 1
1 1
650
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 12.3: Systems of Linear Equations: Determinants
Dx
Dy
6 -1 = -5 - 2 1 14 3 - 2 =6 - 2 1 _ 1 -5 1 + ( _ 1) -5 -2 3 -2 14 -2 14 3 =6(4 - 3) - 1(10 - 1 4) - 1(-1 5 + 28) =6+ 4 - 1 3 =-3 6 -1 = 3 -5 1 1 14 -2 1 + (_1) 3 = 1 -5 14 - 2 1 -2 1 14 = 1(1 0 - 14) -6(-6 - 1) - 1(42 + 5) -4 + 42 - 47 =-9 1 6 = 3 - 2 -5 1 3 14
=
Dz
1
1 1
/
1 / _ 6/ 3 /
1 1
Dx = -7-34 -422 -3-11 =-3 1 - � _�1 -2 1 -� _�I + (- I)I -� - �I ==-30-34-2 -3(12 -2) -2(21-4) -1(-14 + 16) =-66 -3 -1 Dy = 2 -7 1 -2 4 -3 =1 1 -74 -31 1 _(_3)1 -22 -31 1+(_1) 1 - 22 -47 1 = 1(21-4)+ 3(-6 + 2) -1(8 -14) = 1 7-12+6 =11 2 -3 Dz = -22 -42 -74 = I I - � -: 1 - 21 _ � -:1 +(-3) 1_� -�I ==-2+12+12 1(-16+14) -2(8 -14) -3(4 -8) =22 Find the solutions by Cra mer's Rule: Dy 1 1 1 Dx = --66 = -3 y -=-=x=_ = D 22 2 D 2222 D z = z = =1 D 22
1
/ -5 /
/ � �� / - 1 / � �� / +6/ � -� /
=I -
= 1(-28 + 1 5) - 1(42 + 5) +6(9 + 2) = -13 - 47 + 66
=6 Find the solutions by Cra mer's Rule: -3 -9 =-=1 y= -L=-=3 -3 -3 6 -2 =-= -3 The solution is (1, 3, -2) .
x=_DDx z = _DDz
35.
{
D D
{
( �)
The solution is -3, , 1 .
37.
x + 2y - z=-3 2x - 4y z = -7 -2x + 2y - 3z= 4 -1 -4
+ D= -22 22 -31 =1 1 - � _�1 -2 1 _ � _�I +(- I) I _ � ==10+8+4 1(12 -2) -2(-6 + 2 ) -1 ( 4 -8) =22
x - 2y + 3z= 1 3x + y - 2z = 0 2x - 4y +6z = 2
1 2 3 D = 3 - 1 -2 2 -4 6 = 1 1� - �1-(- 2) 1� - �1 + 3 1� �l = 1(6 -8) + 2( 18 4) + 3(-12 -2 ) =-2 + 44-42 =0 Since D = 0 , Cra mer's Rule does not app ly. +
651
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{
Chapter 12: Systems of Equations and Inequalities
39.
x + 2y - z=O 2x -4y + z = 0 -2x + 2y - 3z = 0 1 2 -1 D= 2 -4 - 2 2 -3 =1 -4 1 _ 2 2 1 + (_ 1) 2 -4 2 -3 -2 -3 -2 2 = 1(12 - 2) - 2(-6+ 2) - 1(4- 8) = 10+ 8 + 4 = 22 o 2 -1 Dx = 0 -4 1 = 0 [By Theorem (12)] o 2 -3 0 -1 D = 2 0 1 =0 [By Theorem (12)] -2 0 -3 I 2 0 Dz = 2 -4 o =0 [By Theorem (12)] -2 2 0 Find the solutions by Cramer's Rule: Dy 0 Dx =-= 0 O -=O y=-= x= D 22 D 22 z = Dz = � = O D 22 The solution is (0, 0, 0).
\
\ \
\ \
43.
Solve for x: =5
I: � 1
3x -4x = 5 -x= 5 x=-5
\
45.
Solve for x: x 1 4 3 2 =2 -1 2 5 x
l � � 1 - l l _� � 1 + l l _� � 1=2
x ( 1 5 -4) - ( 20+ 2) + ( 8 + 3 ) = 2 l lx - 22 + 1 1 = 2 l lx = 1 3 13 x= II
y
41.
47.
x 2 3 1 x 0 =7 -2 6 x
{
1 1
1 1
l � x_( -�2x1 -)2-1 �2(-_2�) \++33(11�-6x�)1==77
- 2x2 + 4 + 3 - 1 8x = 7 _ 2X 2 - 1 8x = 0 - 2x ( x + 9 )= 0 x =0 or x=-9
x - 2y + 3z=0 3x + y - 2z= 0 2x -4y +6z = 0 1 -2 3 -2 D= 3 2 -4 6 = 1 1 - 2 _ (_ 2) 3 - 2 + 3 3 1 -4 6 2 6 2 -4 = 1(6 - 8) + 2(1 8 +4) + 3(-12 - 2) = -2+44-42 =0 Since D= 0 , Cramer's Rule does not apply.
1
Solve for x:
x y z 49.
u
v
w
=4
2 3 By Theorem ( 1 1), the value of a determinant changes sign if any two rows are interchanged. 1 2 3 Thus, =-4 . x y z
1
U
v
w
652
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Section 12.3: Systems of Linear Equations: Determinants
51.
xY z Let \ u v wi = 4 . 2 3 x -3 -6 -9 1 = -3 1 1 x
u
Y
v
55. Y
z 2 3 v
z
w
u =-3 - I l u
x
(
)
2 2
v
z w
2 3
i
[Theorem ( l l )]
= 3(4) =12
53.
z
�1
=4
3 2x 2y 2z u - I v - 2 w- 3 2 3 [Theorem (14)] z = 21 x Y u-I v-2 w-3 x Y z = 2(-1) \ 2 3 [Theorem (1 1)] u- I v - 2 w - 3 x Y z = 2(-I)( - I) \ U - I v - 2 w - 3 1 [Theorem (1 1)] 2 3 xYz [Theorem (1 5)] = 2(-1)(- 1) I u v w (R2 -r3 +r2 ) I 2 3
[Theorem ( 14)]
w
Y
xY Let I u v
xY z Let \ u v w \ = 4 2 3 2 3 x-3 y-6 z - 9 2u 2v 2 w 3 2 = 21 x -3 y- 6 z - 9 [Theorem (14)] u v w x-3 y-6 z-9 2 = 2(-1) 1 3 [Theorem (11)] u v w x-3 y-6 z-9 = 2(-1)(-1) \ u w [Theorem (1 1)] v 3 2 xYz [Theorem (1 5)] = 2(-1)(-1) u v w (Rl = -3r3 +rl ) I 23
=
= 2(-1)(-1)(4) =8
57.
x Y xl Yl I I = 0 X2 Y2 X I Y2Yl II I _ y I X2Xl II I + I I X2xl Y2YI I = 0 X(YI - Y 2 ) - Y (XI - X2 ) + (xI Y2 - X2Yl ) = 0 X(YI - Y 2 ) + Y (X2 - XI ) X2YI - XI Y 2 Y (X2 - Xl) = X2YI - X1 Y 2 + X(Y2 - Yi ) Y (X2 - Xl) - YI (X2 - Xl ) = X2YI - XI Y2 + X(Y2 - YI ) - YI (X2 - XI ) (X2 - XI ) (y - Yl ) = X (Y2 - Yl ) + X2YI - X1 Y 2 - Y1 X 2 + Y1 XI (X2 - XI )(Y-Yl ) = (Y2 - Yl )X - (Y2 - YI )XI (X2 - XI )(y - YI ) (Y2 -YI )(X - XI ) (Y - YI ) = (YX22 -Yl - Xl ) (X-Xl )
Expa nding t he det ermina nt :
=
= 2(-1)(-1)(4) =8
=
f
\
This is t he 2-point form oft he equat ion fora line.
653
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws
as
they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
59.
ExpandiXng D,x wexobtain D = -21 YII Y22 Y33 1 1 1
LetA =(xl, YI ),B = (x2 , Y2 ) , and C= (x3 , Y3 ) represent cesonofoura trtriiangle. Forn the first siquadrant. mplicity,theSeewevertipositi a ngle i be the poifromnt closest toaxithes, andy-axiB sbe, thefigure. be poithenpoiLett "between" nAt farthest theA andY poi n ts C. C
A (x,
= �1 ( XI (Y2 - Y3 )-x2 (YI - Y3 )+x3 (YI - Y2 ) ) = -(21 xIY2 -xIY3 -x2 YI x2 Y3 + x3� -X3 Y2 ) which is the same as the area of triangle ABC. Ifdiftheferentl vertiycthan es ofshown the triainnglethearefigure, posittheionedsigns may the absolute value of D will gibevereversed. the area ofThus,the triangle. ces of a triangle are (2, 3), (5, 2), and If(6,the5),verti then: 2 5 6 D= .!.2 3 2 5 1 = � ( 2 1 � � 1 - 5 1 � � 1 +6 1 � � I ) = � [2(2 -5) -5(3 - 5) + 6(3 - 2)] = � [2(-3) - 5(-2) + 6(1)] = �1 [-6+ 10+ 6] The=5area of the triangle is 151 = 5 square units. If a = 0, then b 0 and e 0 since { by = s . ad -be 0 , and the system is ex+dy=t The solution of the system is Y = :.-,b t -dy t - d(t) = tb - sd . . Cramer' s X = --= e e be Rule, we get D = I � � I = -be , D = I st db I = sd -tb ' Dy = I � ; 1 = 0 -se = -se , so +
D (x" O)
E (x" O)
Let D = (xl ' O), E = (x2 ' 0) , and F= (x3 ' 0). Wethe areas find theof tarea ngle ABCandbyBEFC subtracti rapezoiof tridsaADEB fromng the area of trapezoid ADFC. Note: AD = YI , BE = Y2 , CF = Y3 ' DF = x3 -XI ' DE = x2 -xI , and EF = X3 - x2 . Thus, the areas of the three trapezoids Iare as follows: = -(X -X + 2 3 I )( YI Y3 ) I -X = -(X + 2 2 I )( YI Y2 ) ' and I -X + = -(X 2 3 2 )( Y2 Y3 ) ' The area= of our triangle ABC is I -X )( + ) = -(2I X3 -XI )( YI + Y3 ) --(X 2 2 I YI Y2 - -(2I x3 -X2 )( Y2 + Y3 ) I I I 1 - ��� I = ��� + ��� - ��� - ��� --x21 2 Y2 + -2I xIYI + -2I xI Y2 --x2I 3Y2 --x21 3Y3 + -2I x2 Y2 + -2I x2 Y3 I I 1 + ��Y 1 2 - ��Y 1 2 + ��� I = ��� - ��� - ��� = -(21 x3 YI -xIY3 - x2 YI + xIY2 - x3 Y2 + X2 Y3 ) KADFC KADEB
KBEFC KABC
KADFC - KADEB - KBEFC
61.
*
*
*
--
Usmg
x
654
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Section 12.3: Systems of Linear Equations: Determinants
x = DDx = ds--betb = tdbe- sd and Dy --= --bese -bs , wh' h ' the so iutl' On. Note Y = -= that these solutions agree if = O. If b = 0, then a;t 0 and d 0 since ad - be ;t 0 and the system is {axex+dy=t=s . The solution of the system is x = a y = t -dex atad- es . Usmg. Cramer' s Rule, we get D = I : � 1 = ad, Dx = I ; � 1 = sd , and DY = I ae st l =at _ es , so x= DDx = aSdd = !...a . Dy at - es h' h ' the soiutl' On. and y = -= Note that Dthese soladutions agree if e = O . If e = 0, then a;t 0 and d ;t 0 since ad - be ;t 0 , and the system is {ax+bY ==st . The solution of the system is y = d x = s -aby = sda-dtb Using Cramer' s Rule , we get D = 1 aO db 1 = ad ' Dx = 1 st db 1 = sd - tb ' and Dy = 1 a0 st 1 = at , so x = DDx = sdad- tb and Dy at t ' h' h 1S' the soiutl'On. Note Y =-=-=D ad d that these solutions agree if b = O. If d = 0, then b;t 0 and e ;t 0 since ad - be 0 , and the system is {exax+bY ==ts . The solution of the system is x = !-.e , s -bax esbe- at Usmg. Cramer ' s Rule, we y = --= IC
D
IS
x
d
;t
!.... ,
63.
-- --=
IC
IS
=
=
D x
IS
-- --
(
dy
!.... ,
-- --- .
W
=
D
,
--- , w
get D = I: �I = 0 - be = -be, D Ist bol 0 - tb -tb ' and DY = lae st l = at - es , so x = D = --betb = !-.e and y at - es = es - at , whi.ch . the Y = -= sola =0.ution.D Note-bethat thesebe solutions agree if Evaluating the determinant to show the relationship : all a12 a13 ka21 ka22 ka23 a31 a32 a33 _- all 1 kan ka23 I -a1 2 1 ka2 1 ka23 1 + a13 1 ka21 a32 a33 a31 a33 a31 = all (ka22a33 - ka23a32 ) -a1 2 (ka21 a33 -ka23a31 ) + a13 (ka21a32 -ka22a31 ) = kall (a22a33 -a23a32 ) -kaI 2 (a21 a33 -a23a31 ) + ka13 (a21 a32 - a22a31 ) = k all (a22a33 - a23a32 ) -a12 (a21 a33 -a23a31) +a13 (a21a32 -a22a31 ) ) = k ( all 1 a22a32 a23a33 1 -a12 I a21a31 a23a33 1 + a13 1 a21a31 a2a322 I) = k a21all a22a1 2 a23a13 a31 a32 a33 relEvalatiuoatinshingpthe: determinant to show the all + ka21 a12 + ka22 a13 ka23 a21 a22 a23 a31 a32 a33 = (all + ka21) 1 a22a32 a23a33 1 _ (a12 + ka22 ) 1 a31a21 + (aI3 + ka23 ) 1 a21a31 a22a32 1 = (all + ka21)(a22a33 -a23a32 ) - (aI2 + ka22 )(a21 a33 -a23a31 ) +(aI3 + ka23 )(a21 a32 -a22a31 )
65.
+
IC
;t
--- .
655
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Chapter 12: Systems of Equations and Inequalities
= all (a22a33 - a23a32 ) + ka2l (a22a33 -a23a32 ) -a12 (a2I a33 - a23a31 ) - ka22 (a21 a33 - a23a31 ) a21a32 - a22a31)+ ka23 (a21a32 - a22a31) = all +a13( (a22a33 - a23a32) + ka21a22a33 -ka2la23a32 - a1 2 (a21a33 - a23a31 ) -ka22a2l a33 + ka22a23a3l + a13 (a2Ia32 -a22a31) + ka23a2l a32 ka23a22a31 = all -(a22a33 - a23a32 ) - a12 (a21 a33 - a23a31 ) +a13(a2Ia32 -a22a31 ) = all l ::� ::: I -a1 2 I::: ::: I +a13 I :: : ::� I = a2lall a22al2 a23a13 a31 a32 a33
13.
1 5.
Section 12.4 1. 3. 5. 7.
9.
1 1.
1 7.
inverse identity False A+B = [� � -�]+[_ � � _�] 3 + 1 -S + 0] = [ 1 + 0+4 (- 2) 2 +3 6+ (-2) = [ -� : -:] 4A = 4 [0I 23 -S6] = [44.·01 44.3· 2 4(4·6-S)] = [40 128 -2024] 3A - 2B=3 [01 23 -6S] _ 2 [-24 31 -20]
=[� � = [-8 7
7 °
AC�[� � -�lU !] 0(1) + 3(2) + (-5)(3)] = [0(41(4) )++ 3(62(6) +)+(-65)(-2) (- 2) 1(1)+ 2(2)+ 6(3) = [284 -239] � -�] CA = -2 3 = 6(04(0)) ++ 2(11(1)) 6(34(3)) ++ 21((22)) 64(-(-55))+1(+ 2(6)6) - 2(0)+ 3(1) - 2(3)+ 3(2) - 2(-5) + 3(6) = [� �� 3 0 28
[
[ : �l ' [�
1
=�:l
U �l([� � -�H-� U[1 5 �}l-� : -!] = 22 :� =��l -11 22
c(A + B) �
�
1 9.
21.
AC - 312 CA - CB
=[� �
3
-:Jl
7
-9] _ [ 3 0] 23 0 3 -9] 20
- ��] - [-! � -�] - IS] 22
656
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Section 12.4: Matrix Algebra
23 .
all = 2(2)+(-2)(3) = -2 al Z = 2(1)+(-2)(-1) =4 al 3 = 2(4)+(-2)(3) = 2 al 4 = 2(6)+(-2)(2)= 8 aZ I = 1(2) + 0(3) = 2 azz = 1(1)+0(- 1) 1 aZ3 =1(4)+0(3)=4 aZ4 = 1(6) + 0(2) = 6 [ � - �][ � _: ; �] = [- � 4 � :]
31.
=
25.
=
27.
°
� [� : I-�
!l[ �l
5
1
[�[ : m: _;] 1(1)
=
2 9.
/'j
[� �l �l 1(1)+2(- 1)+3(2) 1(2) + 2(0) + 3(4) ] [0(1)+(-1)(-1)+4(2) 0(2) + (-1)(0) + 4(4) [9 146]
=
+ 0(6) + \( 8)
\(3) + 0(2)
+
1(-1)
2 (\) + 4 (6) + \ (8)
2(3) + 4 (2) + \( - I )
3(1)
3(3)
+ 6(6) + \(8)
+ 6 (2) + \( - \)
A = [� �] identity and use the matrito findx wittheh theinverse: Augment row operations [� � I � �] ) � [� � I � �] ( Interchange and rz � [ 2 -21 1 01 -31 ] (Rz =-3/'j +rz )
33.
]
A = [� �] wherea*O. identity and use Augment row operatithoensmatrito findx wittheh theinverse: [a2 a1 1 1 01] [a1 ta i t 01] ( Rl = t rl ) �1 (Rz = - a + rz ) � [� � [� �
[ � �] the identity and use row Augment find thex witinhverse: operatio1nstheto matri [� � 1 �1 0 �]1 ( Interchange) �U 1 1 1 0 0] 1 and rz � [� -1 1 1 - 2] (Rz = -2rl + rz ) �[� � 1 _� �]- ( Rz = -rz ) � [� � 1 _ � 1 �]- 1](Rl = -rz + ) Thus, [-1 2
t]
°
°
/'j
A=
--[;
�
�
A
-I
=
.
6 57
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
35.
[ �l
A = � �� - 2 -3 0 Augment theto matri h the identity and use row operations fmd thex wiintverse:
->
[-2� ��- 0� �0 0� �l1 (R, = 2 r, + r, J �l I � [� �� 2 l = R -h , ( �l i � ! [� �: - o1 0 0 2 2 l� ! =1 2 ] l� : -� -�
->
3
,
->
->
_
_
.1
2.
39.
1
->
41.
37.
Augment h the identity and use row operationstheto matri find thex wiintverse: � -� � � 3 1 2 0 0 1 - 4 -3 � - 2 -3 0 4 3 -� (R2 = - r2 ) - 2 - 1 -3 0
[� � [o� [o�
�
-1
�l
-1
�l �ll
�] - 11 ['{! -!1 5
3
-;;
2 -7
-7
7
9
7
7
1. 7
_
.£. 7
-;;
1. 7
7
->
-t
l l�
0 -3 - 2 1 1 4 3 - 1 ( R3 = t r3 ) � 00 0 I 0 I (RR2I :--3r34+r3 �+ r2 ) 0 I I Thu, 2 I {2X+ y = 8 x+y =5 Rewrite the system of equations in matrix form: A = [� �J X = [; J B = [!] Find the inverse of A and solve1 X1]= A-I B : From Problem 29, A- I = [- 1 -2 , so X = £ I B= [_: -�] [!] = [�l The solution is x = 3, y = 2 or (3, 2) . { 2X+ y = 0 x+y = 5 Rewrite the system of equations in matrix form: A = [ � � l X = [;l B = [ �] Find the inverse of A and solve X = A- I B : From Problem 29, A- I = [- 11 -21] , so X = £I B= [_ � -n[�] = [��l The solution is x = -5, y = 10 or (-5, 10) . 7 -7
7
.
7
658
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 12. 4: Matrix Algebra
43.
{2x6X ++ 2y5Y == 2 Rewrite the system of equations in matrix form: A = [� �J. X [;J. B = GJ Find the inverse of A and solve X = A- I B : From Problem 31, A- I = [- 11 _2.23] , so X=A- I B=[_ � - t] [�] = [-�l The solution is x=2,y = - 1 or (2, - 1) . { 2x6X ++ 2y5Y == 135 Rewrite the system of equations in matrix form: A= [ � �l X = [;l B = [ I �] Find the inverse of A and solve_X = A- I B : From Problem 31, A- I = [- 11 2.]32 , so X=[I B= [_: -t][I�] = [!l The solution is x = �, y = 2 or (�, 2) . X + y== -3a a*"O { 2ax+ay Rewrite the system of equations in matrix form: A = [� �l X = [ ;l B = [=:] Find the inverse of A and solve X = A- I B : From Problem 33, A- I = [- 11 -1.-; ] , so X � A-'B � [_ : - �] [=!lf�l The solution is x = -2, y = 1 or (-2, 1) . 7
49.
=
45.
47.
X + y =2a a*"O { 2ax+ay =5 Rewrite the system of equations in matrix form: A =[� �J x=[;J B= [;] Find the inverse of A and solve X = [ I B : From Problem 33, A- I = [- 1 -1.-; ] , so _ 1 [ I X =A- B = - 1 1.�][�7 ] = [1.�] . The soIutl· On · x = -a2 , y = -a3 or (-,a2 a3) X - 2y+y+z=z = 01 y - 2x - 3 = -5 Rewrite the system of equations in matrix form: X� A� : Find the inverse of A and solve X = A- I B : 3 -3 From Problem 35, A- I = - 2 2 so -4 5 - 2 3 3 X =A- I B = - 2 -2 -4 5 The solution is x = - 2, y = 3, z = 5 or (-2, 3, 5) . X =2�:;:� - 2x - 3y 2 Rewrite the system of equations in matrix form: X� B� A� : Find the inverse of A and solve A-I B : 1
a
51.
53.
a
{
1
IS
-
.
U �� n [ l B�[=!l [ -�l ' [ =�Elf�l U �� il [ l m X
=
659
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
[ - �l ' [ �mH-;j
From Problem 35, A- I = - � � so -4 5 -2 X =A-I B= - � -� -4 5 12 21 z= 1 or The so1uhon· lS· x =-,y=--,
X'
59.
9
55.
3x+ y+2z = 1 Rewrite the system of equations in matrix form: X� B� A� � � Find the inverse of A and solve X = A-I B : From Problem 37, A-I = t 7 so 2
-;;
I
-;;
t
61.
-;;
7
57.
7
-;;
-;;
[; � I � �] � [� � I -; �] � [� ! I -t �]
[ -ll m m [-:3 _1] ' 3� 4] 1. ] [ 9 ] [ t -!7' � �7: ' 34 =-,85 z = -12 or The so1utl·On lS· x=--,y -;;
Fmm Pmblem 37, I X =A- B= L 7 7 The solution is x = .!.3 , y = 1, z = �3 or (.!.,3 1, �)3 . A= [; �] Augment identity and use row operatitheonsmatrito findx withthetheinverse:
[ i
{3x+2yx+ y+ z = z=8
-� [ : ; -;], so -
y matrix on theThereleft.is noThus,waythitos obtai matrinx thehasidnoentiintverse. A = [1015 23] Augment identity and use row operatitheonsmatrito findx witheth theinverse:
G� � I � �] � [1 � � I -i �] ( R2 =- t � + r2 ) � [ol t0 -I)� 01] ( RI = I� � ) There is no way to obtain the identity matrix on
7
x+ y+ z = 2 3x+2y- z = -73 3x+ y+2z = -103 Rewrite the system of equations in matrix form:
the left; thus, there is no inverse.
[ -l} [ l [,t]
A� � � X� : B� Find the inverse of A and solve X = A- I B :
660
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63.
[A = -31 -� -�-1]
Section 12.4: Matrix A lgebra
67.
Augment identity and use row operatitheonsmatrito findx wittheh theinverse:
17� 02 [ - . 132 [ . 132 [ - . 132
r3
69.
6
6
o 3
A=
17� 01 L ei [ -. 132
Thus,
8 35
. 135 - . 13 . .. - . 132 . 131 ... 131 133 ... •
A- I :..
..13 1
.. 8 1
. 135
- . 13 1 1
.. •
•
•
l !:l ! -
132
[Ft�.. - . 13 1 . 13 1 - . 13 3 13 3
- . e ... . 133... - . e ... . 137...
J J - . 134 4. 88 E -4J :: . 137 . 86 J) ..
•
-0.02 0. 06 0.07 0.06 61 -12 1 0 - 12 7 ; B= -9 4 - 1 12 Enter the matrices into a graphing utility and use below:to solve the system. The result is shown
] []
[ [ 4 . 56661 7862 J [ -6 . 44363 1 134 J [ -24 . e7467857J J
71.
- . 132 . 81 J 13 1 83 J J
Thus, 5 7 , 07 . y,., - 6.the44,solutiz :.o. n- 24.to the07 system or (4. 5i7,s -6.44.4,-24. ) 25 61 - 12 21 18 - 12 7 7 3 - 1 -2 A=
[
4
t7�i'���7488492J [2. 4568138199 J [ 8 . 26513137321 J J
[ 0.0.0011 -0.0.0025 -0.0.0011] •
4
A- I B
"6
65.
- . 134 . 135 . 13 1 . 136
A - I :..
->
->
r�- l1 r 0. 02 - 0.04 - 0.0 1 0.0 1 1 Thus, - 0.0.0022 0.0.0051 - 0.0.0043 -0.0.0003 . A
-16
[-�[ :-�2-���5 �0� �0 0�1] �] ) � -3 --1 �1 0 0 (Interchange and � [i -� ;; ! -�] � - � �l [� � I! � r� : � � -; Jl There no waythereto obtai the left;is thus, is nonintheverse.identity matrix on [2185 -612 rl
2 1 18 = 21 1012 -1152 8
- 0.02 0. 0 1 0.03
] [] ;
x :..
B=
Thus, z :.. 8o.n27toorthe(-system 1.19, 2.is46,8.2-71.19) . , y:.. 2 .the46 ,soluti A = [: �1 B = [1:��00] AB= [: 1�] [1:1���] 1853.40 ] = [36((71.71.0000))+12+9((158.158.6600)) ] [2116. 20' Nitotalkki'tuis ttotal tion is2$1853. ion istui$2116. 0. 40, and Joe s x :..
73.
a.
b.
=
661
© 2008 Pearson Education, Inc., Upper Saddle River, N J . All rights reserved. This material i s protected under all copyright laws a s they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
75.
a.
b.
2 3 [700500 350500 400850] 2 3 500 700 The 3 by 2 matrix is: 350 500 . 400 850 The 3 by 1 matrix representing the amount of The rows of the by matrix represent stainless steel and aluminum. The columns represent l O-gallon, 5-gallon, and I -gallon. . The by matrix is:
nmtm,l i c.
d. e.
{ :]
a.
[]
Thus, r'
11,500] [700500 350500 400850] . 1 � [17,050 3 Thus, 11, 5 00 pounds of stainless steel and 17,050 pounds of aluminum were used that day. The 1 by 2 matrix representing cost is: [ 0.10 0. 0 5 ] . The total cost of the day ' s production was: 11 500] [ 0 . 10 0. 0 5 ] . [ 17,050 [ 2002.50] . The total cost of the day ' s production was $2002. 5 0. 21 1 110 11 1 The days usage of materials is:
'
77.
[ ]
K=
b.
=
=
=
al2 =
=
al 3 =
=
=
=
a23
a3 1 =
=
a32 =
=
=
a33 =
c.
fun.
(interChange ) and 'i
al l =
a22 =
Augment the matrix with the identity and use row operations to find the inverse:
->
]
20 19 14 because 47(1) + 34(- 1) + 33(0) 13 47(0) + 34(1) + 33(-1) 1 47(-1) + 34(1) + 33(1) 20 44(1) + 36(-1) + 27(0) 8 44(0) +36(1)+27(- 1) 9 =44(-1)+ 36(1)+27(1)=19 47(1)+ 41(-1) + 20(0) 6 47(0) +41(1) +20(- 1) 2 1 47(-1)+41(1)+20(1) 14 13 � M;l � A; 20�T;8�H;9�I; 19� S;6�F; 21 �U; 14� N The message: Math is a2 1 =
[ ]
[�1 1� � 0� �0 �]1 0 [� : 0 : H] [� _� ! =�
=H
r2
->
662
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 12.5: Partial Fraction Decomposition
79.
A=
�]
[:
7.
If D = ad - be � 0 , then a � 0 and d � 0 , or b � 0 and e � O . Assuming the former, then
[: � I � �] �[: ; !I � �] l O
d
-
;;
bc a
a
-;;;
-, --'-
1
0
0
0
0
ad-bc
c_ _ 1. + _b a a(ad-bc) -c ad-bc
d ad-bc -c ad-bc
-b ad-bc a ad-bc
Thus , A- l = ic
- /5
9 The proper rational expression is: x2 + 5 = 1 + _2 9_
0 1
_ .f.
->
/5
-b ad-be
x2 - 4
9
a ' (R2 -- ad-bc ,.. ) 2
/bc
a
=
.
=
x -4
11.
d
D -e
-b a
]
x _4
The rational expression x(x - l) x 2 - x is improper, so --:---(x + 4)(x - 3) x 2 + x - 12 perform the division: 1 2 2 x + X-12 X -x+ 0 x 2 + x - 12 - 2x + 1 2 The proper rational expression is: x( x - I) = 1 + -2 2x + 1 2 1 + - 2(x - -6)
)
Section 12.5
( x + 4)(x - 3)
1.
True
3.
3x + 6x3 + 3x 2 = 3x2 ( x 2 + 2x + 1 )
5.
x -4
. . 1 expreSSIOn 5x3 +2 2x - I IS. Improper, . The ratIOna x -4 so perform the division: 5x + 2x - l - 20x 22x - l The proper rational expression is: -I 5x3 + 2x - I 5x + 22x 2 2
where D = ad - be .
4
1
)
x 2 - 4 x2 + 5 x2 - 4
[ ] [� !",�" �1 �[� : ",�J [ 1 1 -+[; -+ [; _� -;l [ -fr] �[
�
. . x2 + 5 . . -- IS Improper, so The ratIOnal expreSSIOn x2 - 4 perform the division:
13.
= 3x 2 ( x + 1 ) 2
Find the partial fraction decomposition: 4 B --= -A + -x(x - I)
X (X -
The rational expression + is proper, since x -1 the degree of the numerator is less than the degree of the denominator.
(x + 4)(x 3)
x + x - 12
l)
(__) = 4
x(x - I )
x
x-I
X (X - l)
(
A x
+� x-I
4 = A(x - l) + Bx
)
Let x = 1 , then 4 = A(O) + B B=4 Let x = O , then 4 = A(-1) + B(O) A = -4 4 =-4 + 4 x(x - l) x x - I
---
--
663
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Chapter 12: Systems of Equations and Inequalities
15.
Find the partial fraction decomposition: A Bx+C 1 -+ 2 x(x +1) x x2 +1 A B +C X(X2 + 1) x(x;+I) = X (X2 + 1) x + x� +1 1 = A(x 2 + 1) + (Bx + C)x Let x = 0 , then 1 = A(0 2 +I) +(B(O)+C)(O) A=1 1= A(12 +I) +(B(l) +C)(I) Let x = I , then 1= 2A+B+C 1= 2(1) +B+C B+C = -I Let x = -I , then 1= A« 1) 2 +I) +(B(-I)+C)(-I) 1= A(I +1)+(-B +C)(-I) 1= 2A + B - C 1 = 2(1)+B - C B - C = -I Solve the system of equations: B + C = -l B - C = -1 2B = - 2 B = -1 -1 + C = -1 C=O -x 1 = -I + ---:-2 2 x(x + 1) X x + 1
J
(
-(
1 9.
)
B=
x
.!.2
Let x = -1 , then (_1) 2 = A(- 1 - 1)(-1 + 1) + B(-1 + 1) + C(- 1 - 1) 2 1 = A( -2)(0) + B(O) + C (-2 i 1 = 4C C = -1 4 Let x = 0 , then 0 2 = A(0 - 1)(0 + 1) + B(0 + 1) + C(0 - 1) 2 O = -A + B + C A = B+C A = -1 + -1 = -3 2 4 4 1. .1 1. x2 = _4_ + 2 2 + _4_ 2 (x - l) (x + l) x - I (x - l) x + l _____
Find the partial fraction decomposition: x A + B ---- = (x - l)(x - 2) x - I x - 2 Multiplying both sides by (x - 1)(x - 2) , we obtain: x = A(x - 2) + B(x - l) Let x = l, then I = A(l - 2) + B(I - I) 1 = -A A = -1 Let x = 2, then 2 = A(2 - 2) + B(2 - 1) 2=B x -1 + 2 = -(x - l)(x - 2) x - I x - 2
-- --
----
--
Let x = 1 , then 1 2 = A(l - 1)(l + I) + B(1 + 1) + C(1 - 1) 2 1 = A(0)(2) + B(2) + C(0) 2 1 = 2B
_
1 7.
Find the partial fraction decomposition: A + B + -C x2 --:;---- = 2 2 (x - I) (x + l) x - I (x - l) x + l Multiplying both sides by (x _ I)2 ( +I) , we obtain: x2 = A(x - l)(x + l) + B(x + l) + C(x - l) 2
21.
--
Find the partial fraction decomposition: 1 1 (X _ 2)(X2 +2x+4) Bx+C A +-=----2 2 x 2 (x- 2)(x +2x+4) x +2x+4 Multiplying both sides by (x - 2)(x2 +2x+4) , we obtain: 1 = A(x2 + 2x + 4) + (Bx + C)(x - 2) Let x = 2 , then 1 = A ( 22 + 2(2) + 4 ) + (B(2) + C) (2 - 2) 1 = 1 2A A =� 12
664
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws
as
they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 12.5: Partial Fraction Decomposition
Let x = 0, then 1 = A ( 02 + 2(0) + 4 ) + (B(O) + C)(O-2) 1 = 4A -2C 1 = 4 ( 1112 )- 2C -2C=�3 C =-.!.3 Let x = 21 , then 1 = A ( 1 2(1) 4) (B (I) C)(I -2) 1=7A-B-C 1 = 7 (1 1l2 ) -B+-31 B=--121 __ 3x -8 = _x -_2 - -L2xx- 134 rr -2rr( x+4) = -x-2 x + 2x+4 23. Find the partial fraction decomposition: -L
12
+
+
+
+
+
Let x = 0 , then 02 = A(O - 1)(0 + 1)2 + B(O + 1)2 + C(O _1)2 (0 + 1) + D(0-1)2 O =-A+B+C+D A-C= B+D A - C = -41 + -41 = -21 Let x = 2 , then 2 2 = A(2 - 1)(2 + 1) 2 + B(2 + 1)2 + C(2 _1)2 (2+ 1) + D(2-1/ 4=9A+9B+3C+D 9A+3C = 4 - 9B - D 9A+3C = 4 - 9 (±) - ± = % 3A+C= -21 Solve the system of equations: A-C=.!.2 3A+ C = .!.2 4A = 1 A = .!.4 �+C=.!. 4 21 C= --4
12 2 x +
+
+ ---7-'''-'--'':'''
- -- - - Mul , we obtaitinp:lying both sides by x2
A
(x _ I) 2 (x + I ) 2
x-I
x2
=
A(x - I)(x + 1) 2 +
+
B
(x _ I ) 2
B(x
+
C
x+1
+
D
(x + I ) 2
(x _ 1 ) 2 ( x + 1) 2
+ 1 )2
+ C(x _ I ) 2 (x + 1) + D(x - I ) 2
Let x = 1 , then = A(1 -1)(1 + Ii + B(l + 1)2 + C(1 - 1)2 (l + 1)+ D(l-1)2 1 =4B B= .!.4 Let x = - 1 , then (_1)2 = A(-1-1)(-1+1) 2 +B(- 1+1)2 + C(- 1-1)2 (-1 + 1)+D(-1-1)2 1=4D D=.!.4 e
25.
Find x-3 the partial fractiA on decomposi tion: B C ------: x+2 -x+l -(x+2)(x+ I) 2 :- = -(x+ 1)2 2 Multiplying both sides2by 2)(x 1) , we obtain : - 3 = A (x 1) + B(x 2)(x 1) C(x Let x = - 2 , then 2 B(-2+ 2)(-2+ 1) + C(- 2+ 2) 1 -2 -3-5== A(-2 ) A=-5A +
x
+
+
+
(x +
+
+
+
+
+ 2)
+
665
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
Let x = - 1 then 3 = A(- 1 + 1)2 + B(- 1 + 2)(-1 + 1) + C(- 1 + 2) - 1 -- 4=C C= - 4 Let x = 0 then 0 - 3 = A(O+ 1)2 + B(O+ 2)(0+ 1) + C(O+ 2) -3 = A+2B+2C -3 = -5 + 2B + 2(-4) 2B =10 B=5 x-3 --+--+--5 5 - 4 7'" (x+2)(x+l)2 x+2 x+1 (x+l)2 Find the partiAal fractiB oCx+D n decomposition: ( ) x x2 x2 + 4 Multiplying both sides by x2 (x2 + 4) we obtain : x+4 = Ax(x2 +4)+ B(x2 +4)+ (Cx+D)x2 Let x = 0 then 0+ 4 = A(0)(02 4) B(02 + 4) ( C(O) + D) (0)2 4=4B B=1 Let x = I then A(l)(12 +4)+B(12 +4)+(C(I)+D)(l)2 1+4=5=5A+5B+C+D 5=5A+5+C+D 5A+C+D= 0 Let x = - 1 then -1+4 = A(-I)« - 1)2 +4)+ B«- 1)2 +4) + (C(- I)+ D)(- 1)2 33== -5A+5-C+D 5A+5B-C+D -5A - C+D= - 2 Let x = 2 then 2 + 46=16A+8B+8C+4D = A(2)(22 + 4) + B(22 + 4) + (C(2) + D)(2)2 6 = 16A+8+8C+4D 16A+8C+4D= 2 Sol5A+C+D= ve the system 0of equations: -5A- C+D = - 2 2D=-2 D= - 1 ,
27.
x+4 2 x x2 + 4
--
X
.L .i.
29.
,
,
,
+
---,4 -'-----'-
X
- + - + -
+
5A+C-lC = 01-5A + 4(-1)4 =-2 16A +16A+8 8(1-5A)40A - = -2 - - 24A=-6 A = .!.4 C=I-5 (±) =I- % = - ± x+4 = -t + -+ 1 --'-- t-x-l�--::x2(x2 +4) x2 x2 +4 ) = + _x21 + - .Lx2( x+4 +4 Find the partial fraction decomposition: A + --::-Bx+C x2 +2x+3 ----= ,---.., x+ +2x+4 (x + 1)(x2 +2x+4) 1 x2 Multiplying both sides by (x + 1)(x2 + 2x + 4) we obtain: x2 + 2x+ 3 = A(x2 + 2x+4) + (Bx+ C)(x+l) Let x = - 1 then (_1)2 +2(-1)+3 = A«-1)2 +2(-1)+4) +(B(-I)+C)(-I+I) 2=3A A = �3 Let x = 0 then 02 + 2(0) 3 = A(02 2(0) + 4) (B(O) C)(O I) 3=4A+C 3=4 ( 2/3 ) +C C= .!.3 Let x = 1 then 12 +2(1)+3 = A(12 +2(l)+4)+(B(I)+ C)(l+ I) 6=7A+2B+2C 6 = 7 ( 2 3 ) + 2B + 2 ( 1 3 ) 2B = 6 _H._1. = 1.3 3 3 B=.L3 x2 + 2x + 3 =_3_+ 3 x + 3 (x+l)(x2 +2x+4) x+l x2 +2x+4 3 +2x+ 4 =_x+l3 _+�x2l(x+l) =
,
+
,
,
,
+
,
+
+
+
+
,
,
/
/
1.
______
.L
1
1.
_ _ _
666
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Section 12.5: Partial Fraction Decomposition
31.
Find thexpartial fractiAon decomposi tion: B (3x-2)(2x+1) = 3x-2 + 2x+1 MUltiplying both sides by (3x -2)(2x + I) we obtain: = A(2x + I) + B(3x -2) Let x = - -21 , then -� = A ( 2 ( -1/2 ) + 1 ) + B ( 3 ( -1/2) -2 ) - �2 =- 22 B B = -71 Let x = -23 , then � = A ( 2 ( 2/3 ) + 1 ) + B ( 3 ( 2/3 ) -2 ) 3 �=2 3 3 A= �7 2 =�+� (3x-2)(2x+1) 3x-2 2x+1 Find thex partial fractixon decomposi tion:B A x2 +2x-3 (x+3)(x-l) =--+-x+3 x-I Multiplying both sides by (x + 3)(x -1) , we obtain : x= A(x-1)+B(x+3) Let x = l , then 1 = A(l-1)+B(l+3) 1 =4B B =-41 Let x = -3 , then -3 = A(-3 -1) + B(-3+ 3) -3 =-4A A = �4 x =_4_+_4_ x2 +2x-3 x+3 x-I -----
--
35.
--
--::---::c-
,
,
x
=
37.
A
33.
7(0)
----
----=-___
1
Find7x+3 the partial fracti7x+3 on decomposition: x3 _2x2 -3x x(x-3)(x+1) A B+C =-+ Multiplying bothxsidesx-3by x(xx+1-3)(x + 1) , we obtain: 7x + 3 = A(x-3)(x + 1) + Bx(x+ 1) + Cx(x-3) Let x+= =0 , then I) + 1) --
I
X
Find the partial fraction decomposition: x2 + 2x + 3 = --+ Ax + B --::-Cx +--:D :(x2 +4)2 x2 +4 (x2 +4)2 Multiplying both sides by (x2 + 4)2 we obtain : x2 +2x+3 = (Ax+B)(x2 +4)+Cx+D x2 + 2x + 3 Ax3 + Bx2 + 4Ax + 4B + Cx + D x2 +2x+3 = Ax3 +Bx2 +(4A+C)x+4B+D A=O; B=l; 4A+C= 2 4B+D=3 4(0)+C = 2 4(l)+D = 3 D=-l C=2
3
--
A(O - 3)(0 + + B(O)(O
+ C(O)(O - 3)
3 = -3 A
A = -I
Let x = 3 then + + + 24 12B B=2 Let7(-1)+3 x = -1=, A(-1-3)(-1+1)+B(-1)(-1+1) then +C(-1)(-1-3) -4=4C C=-l 7x+3 -1 -1 2 +---+ -= ----,x3 _ 2x2 -3x x x-3 x+1 7(3) + 3 = A (3 - 3)(3 ,
\)
B(3)(3
\) + C(3)(3 - 3)
=
1
--
667
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
39.
41.
Perform synthetic division to find a factor: 2)1 -4 5 -2 2 -4 2 -2 0 x3 -4x2 +5x-2 = (x-2)(x2 -2x+l) = (x-2)(x-l) 2 Find the partial fraction decomposition: x3 -4x2 +5x-2 (x-2)(x-l)2 A B C =--+--+--x-2 x-I (x-l)2 Multiplying both sides by (x -2)(x _1) 2 , we obtai2 n: 2 x = A(x-l) +B(x-2)(x-l)+C(x-2) Let x = 2 , then 22 = A(2-1) 2 +B(2-2)(2-1)+C(2-2) Let4=Ax = 1 , then 12 = A(1-1)2 +B(1-2)(1-1)+C(1-2) 1=-C C=-1 Let x = 0 , then 02 = A(0-1)2 +B(0-2)(0-1)+C(0-2) 0 = A+2B-2C 0=4+2B -2(-I) -2B=6 B=-3 -1 -3 --4 --+ --= --+ --:-x3 --4x--::-x2 2-+ 5x-2 x-2 x-I (x-l)2 Find the3 partial fraction decomposition: Ex+F 3 Ax+B + Cx+D + ---x2 3 = --(x +l6) x2 +16 (x2 +16)2 (x2 +16) Multiplying both sides by (x2 + 16)3 , we obtain: +16) x3 = (Ax+B)(x2 +16)2 + (Cx+D)(x2+Ex+F x3 = (Ax + B)(x4 + 32x2 + 256) + Cx3 + Dx2 + 16Cx+ 16D+ Ex+ F
x3 = Ax5 + BX4 + 32Ax3 + 32Bx2 +3 256Ax 2 + DxEx+ + 256B + CX16D+ + 16Cx+ F X3 = Ax5 + BX4 + (32A + C)X3 + (32B + D)x2 + (256A + 16C + E)x +(256B + 16D+ A=O; B = O ; 32A+C= 1 32(0)+ C = 1 C =1 256A+16C+E = 0 32B+D =0 256(0) + 16(1) + E = 0 32(0)+D = 0 E=-16 D=O 256B+16D+F = 0 256(0)+16(0)+F = 0 F = O -16x 3 ----:(x2:-x+3-16)- ::-=3 (x2 +x16)2 + ---.,(x2 + 16) Find the4 partial fractio4n decomposiAtion: B x-3 2x+l (x-3)(2x+l) - = --+-2X2 -5x-3 ---Multiplying both sides by (x-3)(2x+ 1), we obtain: 4 = A(2x + 1) + B(x -3) Let x = - -21 , then 4 = A ( 2 ( - �) + 1 ) +B (- � -3) 4=- :"2 B B = - �7 Let x = 3 , then 4 = A(2(3) + 1) + B(3 -3) 4 = 7A A ='±7 4 ____ 2x2 -5x-3 = _x-3 + 2x+l F)
43.
�----::-
7
±
_
7_
_l
_ _
668
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Section 12. 6: Systems of Nonlinear Equations
45.
on decomposition: d the partial fracti Fin2x+3 2x+3 X4 -9x2 x2(x-3)(x+3) A B C D =-+-+--+-X x2 x-3 x+3 Multiplying both sides by x2 (x -3)(x + 3) , we obtain: 2x+ 3 = Ax(x-3)(x + 3) + B(x -3)(x+ 3) + Cx2(X+ 3) + DX2(X-3) then Let2·0+3x = =0 ,A·0(0-3)(0+3)+B(0-3)(0+3) + C . 02(0+ 3)+ D· 02(0-3) 3=-9B B =-.!.3 Let2·3 x+ =3 =3 ,Athen . 3(3 -3)(3 + 3) + B(3 -3)(3 + 3) + C . 32(3 +3) + D·32(3 -3) 9=54C C=.!.6 then x = -3=,A(-3)(-3-3)(-3+3) Let2(-3)+3 + B(-3-3)(-3 +3) +C(-3)2(-3 +3) +D(-3)2(-3-3) -3 = -54D D =�18 , then Let2·1+3x ==1 A·l(I-3)(1+3)+B(I-3)(1+3) + C . 12(1 +3) + D . f(I-3) 5 = -8A-8B+4C-2D 5 = -8A-8 ( -113 ) + 4 ( 116 ) -2(1118 ) 5 = -8A+-+83 -32 --91 -8A=�9 A=-�9 2x+3 x-3 + _x+318_ X4 -9x2 = _9X + _3x2 + _6_ _
___
.2.
_
1
1
Section 1 2.6 1.
y = 3x+ 2 graph is a line. The x-i0=ntercept: 3x+2 3x = -2 X =- -23 y-intercept: y = 3 (0 ) + 2 = 2 y
(0 , 2)
2
3.
x
y2 = x2 -1 x2 y2 = 1 x2 y2 1 ---= 12 graph is a hyperbola with center (0, 0), The at es c verti and x-axis, the along s axi transverse (-1,0) and (1,0). The asymptotes are y =-x and y=x . _
e
y
5.
{y = x2 +1 y=x+l
x
..L
y =x + l 669
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Chapter 12: Systems of Equations and Inequalities
7.
9.
(0, 1) and (I , 2) are the intersection points. Solve by substitution: x2 + 1 = x+ 1 x2 -x = 0 x(x-1) = 0 x = 0 or x = 1 y=l y = 2 Solutions: (0, 1) and ( 1 , 2) {yy=8-x = �36-x2 -""""'_( 4
-
"2, 4 + "1/2)
(4 + "2, 4
-
{y = J;
y=2-x
-5
"1/2)
-8
poi(2.5n9,ts.5.4 1) and (5.4 1, 2. 59) are the intersection Solve by substitution: �36-x2 = 8-x 36-x2 = 64-16x+x2 2x2 -16x+28= 0 x2 -8x+ 14 = 0 x = 8 ± .J64-56 2 8 ± 2J2 2 =4 ± J2 Ifx = 4+J2, y = 8- ( 4+J2) = 4-J2 Ifx = 4-J2, y = 8- ( 4-J2) = 4+J2 Solutions: (4 + J2, 4 - J2 ) and (4 - J2, 4 + J2)
11.
y ;:;:: 2 - x
(1, I ) is the intersection point. Solve by substitution: J; = 2-x x=4-4x+x2 x2 -5x+4=0 (x-4)(x-1) = 0 x = 4 or x = I y = -2 or y=l Eliminate (4, -2); we must have y � 0 . Solution: ( I , 1) {X = 2Y x= i -2y x = 2y
-----
x
-5
(0, 0) and (8, 4) are the intersection points. Solve by substitution: 2y =i -2y i -4y=0 y(y-4) = 0 y =0 or y=4 Solutions:x = 0(0,or0)xand=8 (8, 4) 670
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Section 12.6: Systems of Nonlinear Equations
13.
X2 + y2 = 4 {X2 +2x+ l = 0
1 7.
y
y
i -X = 4
-5
-5
(-2, 0) is the intersection point. Substitute 4 for x2 + y2 in the second equation. 2x+4= 0 2x=-4 x=-2 y = �4-(-2)2 = 0 Solution: (-2, 0) { y = 3x-5 15. x2 + y2 = y
Y
5
5
5
{X2 2+-x=4 l =4
(-1, 1. 7 3),on(-1,poin-1.ts.7 3), (0, 2), and (0, -2) are the intersecti Substitute x + 4 for l in the first equation: x2 +x+4= 4 x2 +x= 0 x(x+1)x == 00 or x = - 1 l =3 l =4 y = ±2 Y = ±.J3 Solutions: (0, -2), (0, 2), ( .J3) , ( - J3) { xy =4 x2 + l =8 - 1,
y = 3x - 5
1 9.
- 1,
y
x
(1, -2) and (2, 1) are the intersection points. Solve by substitution: x2 +(3x-5)2 = 5 x2 +9x2 -30x+ 25 = 5 10x2 -30x+ 20 = 0 x2 -3x+2 = 0 (x - 1)(x-2) = 0 x = 1 or x = 2 y=-2 y = l Solutions: (1, -2) and 1) (2,
671
(-2, -2) and (2, 2) are the intersection points. Solve by substitution: x2 + (;J = 8 x2 +�= 8 x2 4 + 16 = 8x2 X x4 -8x2 +16= 0 (x2 _4)2 = 0 Xy== 22 oror x2yx ===4--22 Solutions: (-2, -2) and (2, 2)
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
21.
{X2 + y2 = 4
25.
y = x2 -9 y
{
secondon equati on for y, substitute into theSolvfirste theequati and solve: 2X2 + y2 = 18 xy=4 => y=4x 2X2 +(�J 18 2X2 +�= x2 18 2x4 + 16 = 18x2 2x4 -18x2 +16=0 x4 -9x2 +8=0 ( x2 -8 ) ( x2 -1 ) = 0 or x2 = 1 x2 = 8 x = ±.J8= ± 2.J2 or x = ± l If x = 2.J2 : y = _2.J24_ = .J2 4 _=_.J2 Ifx =-2.J2: y =_ -2.J2 Ifx = 1 : y = -41 =4 4 Ifx=-l : y=-=-4 -1 Solutions: ( 2.J2, .J2) , ( -2.J2, -.J2), (1, 4), (-1, -4) Substiotuten andthesolfirstve:equation into the second equati { y= 2x+l 2X2 + i = 1 2x2 + ( 2x + 1 )2 = 1 2x2 + 4x2 + 4x + 1 = 1 6x2 +4x 0 2x ( 3x+2) = 0 2x= 0 or 3x+2 = 0 x=O or X =- -23 Ifx = 0: y = 2(0)+ 1 = 1 Ifx=- � : y = 2 ( - �) +I=- � +1=- � Solutions: (0" 1) (-�3 ' -.!.3 ) =
-10
23.
No solution; Inconsistent. Solve by substitution: x2 +(x2 _9)2 = 4 x2 + X4 -18x2 + 81 = 4 X4 -17 x2 + 77 = 0 17 ± �289 -4(77) 2 l7 ± r-l9 2 real solutions to this expression. There are no Inconsistent. -4 {yy ==x26x-13 27.
(3, 5) is the intersection point. Solve by substitution: x2 -4 = 6x-13 x2 -6x+9= 0 (X-3)2 = 0 x-3= 0 x=3 Solution: y(3,=5(3)2) -4 = 5
=
672
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 12.6: Systems of Nonlinear Equations
29.
Solve theequati firstoequati osolve: n for y, substitute into the second n and x+y+l = 0 � y=-x-l +6y-x=-5 {x2x2++(-x/ _1)2 + -x -x = x2 +X2 + 2x+ 1-6x-6-x = -5 2x2 -5x=0= 0 x(2x-5) x= 0 or x = � Ifx = O:5 y = -(0)-1 = -1 Ifx = -2 : y = - -25 -I = - -27 Solutions: (0,-1), (f ,- f) secondon and equatisolve: on for y, substitute into tSolhe vfirste theequati 4X2 -3xy+9/ = 15 2x+3y = 5 � y = - -32 x+-35 4x2 -3X ( - � X+ �) +9 ( - � x+ �r = 15 4x2 +2x2 -5x+4x2 -20x+25 =15 IOx2 -25x + 10 = 0 2X2 -5x+2 = 0 (2x -1)(x -2) 0 X = -21 or x = 2 Ifx = -21 : y=- �(�) + � = � 5 1 Ifx=2: y=- -23 (2)+-=3 3 Solutions: (�, �} ( 2, �) Mul each sidetoofelithemisecond add ttheiplyequations nate y: equation by 4 and {3x2X2 +-4// =31=-7 � 12x2x2 +4/ _4y2 = -7 =124 13x2 = 117 x2x==93 ± If x = 3 : 3(3)2 + / 31 � y2 = 4 � = ±2 If x=-3 : 3(-3)2 +y2 = 31 � / =4� y= ±2 Solutions: (3, 2), (3, -2), (-3, 2), (-3, -2) 6(
31.
-1)
3 5.
3x2 +5y2 = 12 Mul tisipldye each sidsecond e of thequati e first oequati onanbyd a5dandd each of the n by 3 the equations to eliminate y: 35x2 -15/ = -25 9x2 + 15y2 = 36 44x2 = 11 x2 = -4 X=± -21 Ifx = ! : 3 (�J +5/ =12 � / = � � y= ± % Ifx= - ! : <- �J +5y2 = 12 � y2 = � � y= ± % Solutions:
-5
1
{
=
33.
3 7.
(�, %} (�, -%} (-�, %} (-� , -%)
Mul sideoofns theto elisecond and taddiplytheeachequati minateequation by 2 {3x2x2 +-2xy == 1 02 6x2x2 +-2xy2xy = 104 7x2 = 14 x2 = 2 x = ±.fi Ifx =.fi: 3 ( .fir - .fi . y = 2 � - .fi . y=-4 � y=�.fi y=2.fi lfx = -.fi : 3 ( -.fit - ( -.fi ) y = 2 � .fi 'y=-4 � y=-=i.fi � y=-2.fi Solutions: ( .fi, 2.fi) , ( - .fi, -2.fi ) xy
xy:
�
=
�
�
�
=
{7X2 -3y2 +5 = 0 3x2 +5y2 = 12 {7x2 -3/ =-5
Y
673
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Chapter 12: Systems of Equations and Inequalities
39.
{ 2 2X2 + / = 2 x - 2/ + 8 = 0 {22 2 + y2 = 2
43.
X
x - 2/ = -8 Multiply each side of the first equation by 2 and add the equations to eliminate y: 4x 2 + 2/ = 4 x 2 - 2/ = -8 5x 2 = -4 x 2 = --4 5 No real solution. The system is inconsistent. 41.
1
I 3 =7 +y2 x2
Multiply each side of the second equation by 2 and add the equations to eliminate y: 5 2 2 - - = -3 x /
2 = 14 +y2 x2 6
{X4x2 2+_2/y2 == 1246
�2 = l l x
x2 = 1 x = ±1 Ifx = l : 1 =7 _3_2 + _ 4 (1) / => y = ±-1 2 Ifx = -I : 1 =7 3 +-4 ( _ 1) 2 / => y = ±-1 2 Solutions: 1 ! 1 , - ! -1 ! -1 ! '2 ' 2 ' '2 ' ' 2
Multiply each side of the second equation by 2 and add the equations to eliminate y: x 2 + 2/ = 16 8x 2 - 2/ = 48 9x 2 = 64 64 x2 = 9 x = ±-8 3 Ifx = � : 3 2 '38 + 2y 2 = 1 6 => 2y 2 = 80 "9 => y 2 = 40 => y = ± 2 M 9 3 8 Ifx = -- : 3 2 8 -'3 + 2y 2 = 16 => 2y 2 = 80 "9 40 => y = ± -2M => y 2 = 9 3 Solutions: 2M 2M 2M 3' 3 ' 3' 3 ' 3' 3 ' _ � _ 2M 3' 3
()
45.
( )
[� ) [� _ ) [_� [ )
� -3 -2 y2 = x2
)
{
( )( )( )( ) -
I +� = 6 _ x4 l �-� = 19 X4 l Multiply each side of the first equation by -2 and add the equations to eliminate x: -2 - 1 2 = -12 X4 y 4 � - � = 19 x4 y 4 - �4 = 7 y l = -2 There are no real solutions. The system is inconsistent.
674
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Section 12.6: Systems of Nonlinear Equations 47.
{X2 -3XY 2+2/ = 0 x +xy = 6
Subtract the second equation from the first to eliminate the x 2 term.
49.
-4xy+2y2 =-6 2xy-/ =3 Since y"# 0 , we can solve for x in this equation to get y2 +3 ' y "# O x= � Now substitute for x in the second equation and solve for y. x2 +xy = 6 ( Y�; 3 J + ( y�; 3 )y = 6 y4 +6y2 +9 + y22+3 = 6 4/ y4 +6/ +9+2y4 +6y2 = 24/ 3/ -12/ +9 = 0 y4 _4y2 +3 = 0 (/ - 3 )(/ -1)= 0 Thus, y = ± 13 or y = ± 1 . Ify=l: x = 2·1 = 2 Ify=-I : x= 2(-I) = -2 Ify=13: x= 13 Ify = -13 : x = -13 Solutions: (2, I), (-2,-1), (13 , 13 ), ( -13 ,-13) {/ + y + x2 -X -2 = 0 x-2 y+l+--=O y Multiply each side of the second equation by -y and add the equations to eliminate y: / + y + x2 - X -2 = 0 -/ -y -x+2 =O x2 -2x = 0 x ( x-2 ) = 0 x = 0 or x = 2
51.
Ifx =O: y2 + Y + 02 _ 0 -2 = 0 => / + Y -2 = 0 => (y+2)(y-l) = 0 => y=-2 or y=1 Ifx = 2: / + y + 22 -2 -2 = 0 => / + y = 0 => y(y + I) = 0 => y = 0 or y = -I Note: y "# 0 because of division by zero. Solutions: (0, -2), (0, I), (2, -I) Rewrite each equation in exponential form: {loglogx (4y)x y == 35 4yy == xx53 Substitute the first equation into the second and solve: 4x3 = x5 x5 _ 4x3 = 0 X3 (x2 -4) =20 x3 = 0 or x = 4 => x = 0 or x = ±2 The base of a logarithm must be positive, thus x "# 0 and x "# - 2 . Ifx = 2 : y = 23 = 8 Solution: (2, 8) Rewrite each equation in exponential form: lnx = 4lny => x = e4 1n y = e1n y' = y4 log 3 X = 2 + 210g 3 y x _- 32+2Iog, y -- 32 ' 32 1og, y -- 32 ' 3log, y' - 9y2 x = y4 2 So we have the system { x=9y Therefore we have : 9/ = / => 9/ - / = 0 => / (9 -/) = 0 /(3+ y)(3-y)- = 0 y = 0 or y = 3 or y = 3 Since In y is undefined when y ::; 0 , the only solution is y = 3 . Ify=3: X =y4 => x=34 =81 Solution: (81, 3 ) �
�
53.
675
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{ +x+x+l+/ -3Y -+2Y =O= 0
Chapter 12: Systems of Equations and Inequalities
55.
X2
59.
;
x 2 {(XX + t) + (Yy - tf == t ( +t)2 + ( -tf t
X2 +
x +
i
-
3y
+2
=
0
)'
--I-�-
(-t· t) 2
-2
(-i-. -!.) .
57.
2
_
,.\'
/ +4y+4=x-l+4 (y+2)2 = x+3 Substitute this result into the first equation. (x-l) 2 +x+3 =4 x2 -2x + 1 + x + 3 = 4 x2 -x = 0 x(x-l) = 0 x = 0 or x = 1 Ifx = 0 : (y+2) 2 = 0+3 y + 2 = ±.J3 => y -2 ± .J3 If x = 1 : (y + 2) 2 = 1 + 3 y + 2 ±2 => y = -2 ± 2 The points of intersection are: ( 0, -2 - J-j), ( 0, -2 + J-j ) , (1, -4 ) , ( 1, 0) . =
v)' x + l + 'x- = O
-2
Complete the square on the second equation.
=
x, => x =-2y { 2 x+2y=0 (x-l) +(y_l)2 5 (-2y-l) 2 +(y-l/ =5 4/ +4y+l+ / -2y+l = 5 => 5/ +2y-3 = 0 (5y-3)(y+l) = 0 y = -35 =0 . 6 or y=-l x = 65 = -1.2 or x = 2 The points of intersection are ( -%, %), (2, -1) .
y
Solve the first equation for substitute into the second equation and solve:
s
(O, .J3"- 2)
=
i + 4y - x +
5
(x -
Ii
+
(y
- Ii
=
61.
{y=2 _x-34
x,
Solve the first equation for substitute into the second equation and solve:
x -6x + / + 1 = 0 4 y=- x-3 x-3=-y4 4 x=-+3 y (; +3J -6 (; +3) +/ +1 =0 1 6 24 -+-+9---18+ 2y Y 24Y / +1=0 1 6 2 -8 = 0 -+y y2
5
x
x +
1 =0
(O, -.J3"- 2)
- -
Y
x
2y = 0
-5
676
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Section 12.6: Systems of Nonlinear Equations
y4
=
1 6 + _ 8y 2 0 / - 8/ + 1 6 = 0 (/ =0 y2 - 4 = 0 y2 = 4 y = ±2 x = -4 + 3 = 5
_4) 2
Ify = 2: Ify = - 2:
67.
YI = �12-x4; Y2 = -�12-x4; Y3 = ..J2j"-;; Y4 =-·J2lx Use INTERSECT to solve: Graph:
nT��-----' XM i n= -4 . 1 XMax=4 . 7 Xsc l = l YM i n= -3 . 1 YMax=3 . 1 Ysc l = l Xres= l
24
x = -+3 = 1 -2 The points of intersection are: ( 1 , -2), (5, 2). Y
I I I I
5
x
X=.5B009379
x=
Graph: YI = (2 / 3); Use INTERSECT to solve:
X
1\
3. 1
x =
I n t � r s t <: t i o n X=1.B120615
69.
Y2 = (-x) e 1\
YI x; Y2 = Inx
Graph: = 2 1 Use INTERSECT to solve: 3.1
1--'---.... -+.. +-....---.. -
-3. 1 0.48, y = 0.62 or (0.48, 0.62)
65.
Solution:
x= Graph: Yl =�2 _ X 2 ; Y2 4 x 3 Use INTERSECT to solve: 3. 1 -4 .
=
71.
1
Int�rs�ction )1= - 1 . 6'17705 '1'= -.B9'1113
..-/\
x
=
x
-3 . 1 - 1 .65, y = -0.89 or (-1 .65, -0.89)
-3 . 1
x = 2.35, y
=
0.85 or (2.35, 0.85)
Let and y be the two numbers. The system of equations is: -y 2 => y+2 +/ 10 Solve the first equation for substitute into the second equation and solve: (y + 2}2 + / = 1 0 y 2 + 4y + 4 + / = 10 / + 2y - 3 = 0 (y + 3) (y - l} = 0 => y = -3 or Ify = -3 : Ify = 1 : The two numbers are 1 and 3 or -1 and -3.
{X2 = x
7 1--�..o...f---+"""",*-=-� 14 . 7
Solution:
4 .7
Int�YS�(tion )I=2.�'15750B Y=.BS:26055
Int�rs�(tion )I=.'1B390757 '1'=.61637017
Solution:
V= - 1 . 0 5 0 5 7 9
= x x= =
Solutions: = 0.58, y 1 . 86; = 1.81, y = 1 .05; 1 .8 l, y -1 .05; ; O.58, y -1 .86 or (0.58, 1 .86), ( 1 .8 1 , 1 .05), ( 1 .8 1 , -1 .05), (0.58, -1 .86)
-s
63.
v= ·1.B56B03
=
x=
x,
x=-3+2 =-1 x=I+2=3
677
y=I
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
73.
Let x and y be the two numbers. The system of equations is: XY = 4 � X =
f l X2 + / = 8
-;
77.
(-;J + y2 = 8
16 + y 2 = 8 / 1 6 + / = 8/ / - 8/ + 1 6 = 0 (/ - 4t = 0 / =4 y = ±2 4 = -2 If y = 2 : x = -4 = 2 · Ify - 2 : x = -2 2 ' The two numbers are 2 and 2 or -2 and -2.
79.
=
{
�=j
a + b = 10 � a = 10 - b Solve the second equation for a , substitute into the first equation and solve: lO-b 2 3 b 3(1 0 - b) = 2b 30 - 3b = 2b 30 = 5b b=6�a=4 a + b = l O; b - a = 2 The ratio of a + b to b - a is If = 5 .
Solve the first equation for x, substitute into the second equation and solve:
75.
{
Let x and y be the two numbers. The system of equations is: X - = XY -l + - = 5 x y Solve the first equation for x, substitute into the second equation and solve: x - xy = y
�
x(1 - y) = y � x = � 1-y 1 + -1 = 5 L y l -y 1 - y + -1 = 5 -y y 2-y -- = 5 Y 2 - y = 5y 6y = 2 1 1 y = -1 � x = _3_ = 1. = _1 3 1 -1 1 2 The two numbers are ! and t .
81.
Let x the width of the rectangle. Let y the length of the rectangle. 2X + 2 Y = 1 6 xy = 1 5 Solve the first equation for y, substitute into the second equation and solve. 2x + 2y = 1 6 2y = 1 6 - 2x y=8-x x (8 - x) = 1 5 8x - x 2 = 1 5 x 2 - 8x + 1 5 = 0 (x - 5)(x - 3) = 0 x = 5 or x = 3 The dimensions of the rectangle are 3 inches by 5 inches.
{
=
=
Let x the radius of the first circle. Let y the radius of the second circle. 21t X + 21t Y = l 21t 1t X2 + n / = 20n Solve the first equation for y, substitute into the second equation and solve: 2n x + 2ny = 12n x+y = 6 y = 6-x 1tX 2 + n / = 20n x2 + / = 20 x2 + (6 - x) 2 = 20
{
=
=
678
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Section 12.6: Systems of Nonlinear Equations
83.
x 2 + 36 - 12x + x 2 = 20 � 2x 2 - 12x + 1 6 = 0 X2 - 6x + 8 = 0 � (x - 4)(x - 2) = 0 x = 4 or x = 2 y=2 y=4 The radii of the circles are 2 centimeters and 4 centimeters. The tortoise takes 9 + 3 = 1 2 minutes or 0.2 hour longer to complete the race than the hare. Let r the rate of the hare. Let t the time for the hare to complete the race. Then t + 0.2 = the time for the tortoise and r - 0.5 = the rate for the tortoise. Since the length of the race is 2 1 meters, the distance equations are: rt = 21 � r = �
85.
,
{xy
=
(� )
=
{
432 - 8x - 1 728 + 32 = 224 x 2 432x - 8x - 1728 + 32x = 224x 8x 2 - 240x + 1 728 = 0 x 2 - 30x + 2 1 6 = 0 ( x - 12 )( x - 1 8 ) = 0 x - 1 2 = 0 or x - 1 8 = 0 x = 12 x = 18 The cardboard should be 12 centimeters by 1 8 centimeters. --
1
( r - 0.5 )( t + 0.2 ) = 2 1
Solve the first equation for r, substitute into the second equation and solve: 1 0.5 t + 0.2 ) = 2 1 _
(� }
4.2 - 0.5t - 0. 1 = 2 1 21+t 4 2 l Ot 2 1 + � - 0.5t - 0. 1 = 1 0t . ( 2 1 )
(
Let x = the width of the cardboard. Let y = the length of the cardboard. The width of the box will be x - 4 , the length of the box will be y - 4 and the height is 2. The volume is V = (x - 4)(y - 4)(2) . Solve the system of equations: = 216 � y = 216 x 2(x - 4)(y - 4) = 224 Solve the first equation for y, substitute into the second equation and solve. ( 2X - 8 ) 2 6 - 4 = 224
)
87.
2 l Ot + 42 - 5t 2 - t = 2 1 01 5t 2 + t - 42 = 0 ( 5t - 14 ) ( t + 3 ) = 0 5t - 14 = 0 or t + 3 = 0 5t = 14 t = -3 14 = 2.8 t =5 t = -3 makes no sense, since time cannot be negative. Solve for r: 2 1 = 7.5 r=2.8 The average speed of the hare is 7.5 meters per hour, and the average speed for the tortoise is 7 meters per hour.
{
Find equations relating area and perimeter: X 2 + y 2 = 4500 3x + 3y + (x - y) = 300 Solve the second equation for y, substitute into the first equation and solve: 4x + 2y = 300 2y = 300 - 4x y = 1 50 - 2x x 2 + (1 50 - 2x) 2 = 4500 x 2 + 22, 500 - 600x + 4x 2 = 4500 5x 2 - 600x + 1 8, 000 = 0 x 2 - 120x + 3600 = 0 (x - 60) 2 = 0 x - 60 = 0 X = 60 y = 1 50 - 2(60) = 30 The sides of the squares are 30 feet and 60 feet.
679
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
89.
1 w:
Solve the system for and P
93.
{2/+2W= Iw=A Solve the first equation for 1 , substitute into the second equation and solve. 2/ =P-2w I=--w 2
mx
between
n
b:
=>
-w-w 2 2=A w2 --w+A 2 =O � ± -4A ± v� 1f 2' V4 w= 2 = 2 4 E2 +- �p2 2-l6A p ± �p2 -16A 4 2 2 If W= p+�p -16A then 4 1 = 2 _ p+�p42 -16A = p_�p42 -l6A 2 If w= p_�p -l6A then 4 1=-2 p_�p42 -16A p+�p42 -16A If it is required that length be greater than width, then the solution is: -l6A --.:..-42 -w p_�p42 -l6A and l = -p+�p Solve the equation: m 2 -4(2m -4) = 0 m2 -8m+16 = 0 ( m_4)2 = 0 m=4 Use the point-slope equation with slope 4 and the point (2, 4) to obtain the equation of the tangent line: y-4 = 4(x-2) y-4 = 4x-8 y = 4x-4 P
P
95.
Y
P
mx
P
=>
=>
=>
P
91.
{y = x2 + 2 y=mx+b Solve the system by substitution: x2 + 2 = + b x2 - + 2 -b = 0 Note that the tangent line passes through ( 1, 3). Find the relation 3 = m(l)+b b = 3-m m a d Substitute into the quadratic to eliminate b: x2 -mx+ 2-(3-m) = 0 x2 -mx+ (m-I) = 0 Find when the discriminant equals 0: ( _m )2 -4 (1 )( m-1) = 0 m2 -4m+4= 0 ( m_2 )2 =0 m-2 = 0 m1 = 2 b=3-m =3-2= The equation of the tangent line is y = 2x + 1 . Solve the system: {2x2 +3i = 14 = mx+b Solve the system by substitution: 2X2 + 3 ( + b )2 = 14 2X2 + 3m2x2 + 6mbx + 3b2 = 14 (3m2 + 2 ) x2 + 6mbx+3b2 -14 = 0 Note that the tangent line passes through (1, 2). Find the relation between m and b: 2 = m(l) + b b = 2 -m Substitute into the quadratic to eliminate b: (3m2 + 2)x2 + 6m(2 -m)x + 3(2 -m)2 -14 = 0 (3m2 + 2)x2 + (l2m -6m 2 )x + (3m2 -12m -2) = 0 Find when the discriminant equals 0: (l2m -6m 2 )2 -4(3m2 + 2)(3m2 -12m -2) = 0 144m2 +96m+ 16 = 0 9m 2 +6m+1= 0 (3m + 1)2 = 0 3m+1 = 0 m=--31 b = 2 -m = 2 - ( -�) = f The equation of the tangent line is y = -.!. x + 2. . 3 3 mx
P
/ 2
Solve the system:
=>
=>
680
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 12.6: Systems of Nonlinear Equations
97.
{x2- /y==3mx+b
Solve the system:
1j
x2_(mx+b ) 2 =3 x2 _ m2 x2 -2mbx -b2 = (l _ m2) x2-2mbx-b2 -3 = 0 = m(2) + b b = 1-2m m b:
Solve the system by substitution:
3
(2, b: (1-m2)x2-2m(l -2m)x -(1-2m)2- = 0 (l_ m2)x2 +(-2m+4m2)x- l+4m-4m2-3 = 0 (l_ m2)x2+(-2m+4m2)x+( -4m2 +4m-4) = 0 0: ( -2m+4m2t -4(1- m2)( -4m2+ 4m -4) = 0 4m2-16m3 + 16m4 -16m4 + 16m3 -16m+ 16 = 0 4m2-16m+ 16 = 0 m2 -4m+4=0 (m-2f =0 = 2x -3m=. 2 r2 : + r, �� r1r2=-a =-r2- ab (-r2- �}2 = � c -r22 --rab 2 --= a 0 ar/ +br2+c = 0
Note that the tangent line passes through Find the relation between and
I).
1j
Substitute into the quadratic to eliminate
3
101.
Find when the discriminant equals
The equation of the tangent line is 99.
{"
-
The solutions are:
=>
I
= -r2 ab = _[ -b±.Jb22a -4ac l - �a b"+.Jb2 -4ac 2b 2a 2a -b"+.Jb2-4ac 2a = -b+.Jb22a -4ac r2 = -b-.Jb22a -4ac . and
10010
10
Since the area of the square piece of sheet metal is square feet, the sheet's dimensions are feet by feet. Let the length of the cut.
x=
I I
10
x
:x
'
--I
10
10 -2x;
= 10 -2x; = x . x 0 10 -2x 0 x x =( ) ( ) ( ) = (10 -2x)(10 -2x)( x) = (10-2x) 2(x) 9 (1O-2x) 2(x)=9. (1O-2x) 2(x) = 9 ( 100-40x+4x2 ) x = 9 100x -40x2+ 4x3 = 9 4x3 -40x2 + 100x -9 = 0 . = 4x3 -40x2+ 1OOx -9
width
height
--
r-
,
The dimensions of the box are: length
y
I
=
--
Note that each of
these expressions must be positive. So we must have > and > => < 5, that is, o < < 5. So the volume of the box is given by V length . width . height
Solve for 1j and �
Substitute and solve:
a.
1j
In order to get a volume equal to
cubic feet,
we solve
So we need to solve the equation
Graphing
YI
on a
calculator yields the graph
681
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Chapter 12: Systems of Equations and Inequalities 80
-,
4 z.r�· H=.09��6ns
-40
10
4 ZoQyo· H=�.Z7��796
-40
True
7.
satisfied
9.
False
v=o
1 1.
80
-,
5.
10
x �0 Graph the line x = O . Use a solid line since the inequality uses 2:. Choose a test point not on the line, such as (2, 0). Since 2 2: 0 is true, shade the side of the line containing (2, 0). Y 5
v=o
80
-,UIO z.r�· H=S.6nOS91
-40
v= -ZE -11
The graph indicates that there are three real zeros on the interval [0, 6]. Using the ZERO feature of a graphing calculator, we find that the three roots shown occur at x::::: 0.093 , x::::: 4.274 and x::::: 5 . 632 . But we've already noted that we must have 0<x<5 , so the only practical values for the sides of the square base are x::::: 0.093 feet and x::::: 4.274 feet. b.
1 3.
y 5
Answers will vary.
Section 12.7 1.
-5
3x + 4 < 8 - x 4x<4
1 5.
x<1
{ x l x
x �4 Graph the line x = 4. Use a solid line since the inequality uses 2:. Choose a test point not on the line, such as (5, 0). Since 5 2: 0 is true, shade the , 0). side of the line
or
(-00,1)
false, shade the opposite side of the line from (0, 0).
The graph is a circle. Center: (0, 0) ; Radius: 3 y 5 (0,3)
(3,0)
(-3,0)
Graph the line 2x + y = 6 . Use a solid line since the inequality uses 2:. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 � 6 is
x2 + l = 9
5
-5
2x + y � 6
x
(0, -3) -5
682
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17.
x2 + l
>
1
Graph the circle x 2 + y 2
=
1.
23 .
Use a dashed line
since the inequality uses >. Choose a test point not on the circle, such as (0, 0). Since 0 2 + 0 2 >
{
Section 12.7: Systems of Inequalities
x + y� 2 2x + y;::: 4
Graph the line x + y 1
=
2. Use a solid line since
the inequality uses:'S. Choose a test point not on the line, such as (0, 0). Since ° + 0::;; 2 is true, shade the side of the line containing (0, 0). Graph the line 2x + y 4. Use a solid line since the
is false, shade the opposite side of the circle from
=
inequality uses 2:. Choose a test point not on the line, such as (0, 0). Since 2(0) + ° � 4 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
19.
y�x 2 - 1 Graph the parabola y
=
x 2 - 1 . Use a solid line
since the inequality uses �. Choose a test point not on the parabola, such as (0, 0) . Since ° � 0 2 - 1 is false, shade the opposite side of the parabola from (0, 0).
2 5.
{
2X - y �
4
3x + 2 y;::: - 6
Graph the line 2x - y
=
4.
Use a solid line since
the inequality uses:'S. Choose a test point not on the line, such as (0, 0). Since 2(0) ° �4 is true, -
shade the side of the line containing (0, 0). Graph the line 3x + 2 Y - 6 . Use a solid line since the =
inequality uses 2:. Choose a test point not on the line, such as (0, 0). Since 3(0) + 2(0) ;::: - 6 is
21.
xy;::: 4 Graph the hyperbola
xy
true, shade the side of the line containing (0, 0). The overlapping region is the solution. =
4.
Use a solid line
since the inequality uses ;::: . Choose a test point not on the hyperbola, such as (0, 0). Since ° . ° ;::: 4 is false, shade the opposite side of the hyperbola from xy;:::4
-y=4 x
y 5
683
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Chapter 12: Systems of Equations and Inequalities
27.
{
2X - 3Y �°
3 1.
3x + 2y�6
Graph the line 2x - 3 Y
=
- 2 . Use a solid line
=
2 . Use
a solid line since the inequality uses 2::. Choose a test point not on the line, such as (0, 0). Since 2(0) + ° � 2 is false, shade the opposite side of
true, shade the side of the line containing (0, 0) . . The region is the solution.
the line from (0, 0). The overlapping region is the solution.
x - 2y�6 2x - 4y � ° =
=
containing (0, 0). Graph the line 2x + y
=
the inequality uses�. Choose a test point not on the line, such as (0, 0). Since 3(0) + 2(0) �6 is
Graph the line x - 2y
2x + y � 2
since the inequality uses 2::. Choose a test point not on the line, such as (0, 0). Since 2(0) + ° � - 2 is true, shade the side of the line
true, shade the side of the line containing (0, 3). Graph the line 3x + 2y 6 . Use a solid line since
{
2X + y �- 2
Graph the line 2 x + y
0 . Use a solid line since
the inequality uses�. Choose a test point not on the line, such as (0, 3 ). Since 2(0) - 3(3) �° is
29.
{
33.
6 . Use a solid line since
the inequality uses�. Choose a test point not on the line, such as (0, 0). Since 0 - 2(0) � 6 is true,
{
2X + 3Y � 6 2x + 3y �°
Graph the line 2x + 3 y
=
6 . Use a solid line since
the inequality uses 2::. Choose a test point not on the line, such as (0, 0). Since 2(0) + 3(0) � 6 is
shade the side of the line containing (0, 0). Graph the line 2x - 4 y 0 . Use a solid line since the =
inequality uses 2::. Choose a test point not on the line, such as (0, 2). Since 2(0) - 4(2) � ° is false,
false, shade the opposite side of the line from (0, 0). Graph the line 2x + 3 y 0 . Use a solid =
line since the inequality uses�. Choose a test point not on the line, such as (0, 2). Since 2(0) + 3(2) �° is false, shade the opposite side of
shade the opposite side of the line from (0, 2). The overlapping region is the solution. y 5
the line from (0, 2). Since the regions do not overlap, the solution is an empty set. y
684
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35.
{
:s;
X2 + y2
9
39.
x+y �3
Graph the circle x2 + l 9 . Use a solid line since the inequality uses :s; . Choose a test point not on the circle, such as (0, 0). Since 0 2 + 02 :s; 9 is true, shade the same side of the =
X2 + 2
�
:s;
16
y �x - 4
Graph the circle x 2 + l 1 6 . Use a sold line since the inequality is not strict. Choose a test point not on the circle, such as (0, 0) . Since 02 + 02 :s; 1 6 is true, shade the side of the circle
=
circle as (0, 0). Graph the line x + y
{
Section 12.7: Systems of Inequalities
=
containing (0, 0) . Graph the parabola y x2 - 4 . Use a solid line since the inequality
3 . Use a solid line since
the inequality uses 2: Choose a test point not on the line, such as (0, 0) . Since 0 + 0 � 3 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
=
.
is not strict. Choose a test point not on the parabola, such as (0, 0) . Since 0 � 02 - 4 is true, shade the side of the parabola that contains (0, 0) . The overlapping region is the solution. y=x2-4
Y 5
37.
{
-5
y � x2 - 4 41.
y:S;x - 2
Graph the parabola
y
=
x 2 - 4 . Use a solid line
since the inequality uses 2: . Choose a test point not on the parabola, such as (0, 0). Since o � 02 4 is true, shade the same side of the =
-
�4
y � x2 + 1
Graph the hyperbola xy
x - 2 . Use
a solid line since the inequality uses:S; . Choose a test point not on the line, such as (0, 0). Since O:S; 0 - 2 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. y = x2
.xy
=
4 . Use a solid line
since the inequality uses �. Choose a test point not on the parabola, such as (0, 0). Since o· 0 � 4 is false, shade the opposite side of the hyperbola from (0, 0). Graph the parabola y x 2 + 1. Use a solid line since the inequality
-
parabola as (0, 0). Graph the line y
{
=
uses � . Choose a test point not on the parabola, such as (0, 0). Since 0 � 02 + 1 is false, shade the opposite side of the parabola from (0, O).The overlapping region is the solution. Y.·
4
5
-If
v = x2 +
i
..
1
-5
-5
685
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{
Chapter 12: Systems of Equations and Inequalities
4
3.
�
x �O 2X +
�:
. 45
x + 2y � 6
{
x �° x+
�:�
2x + y � 4
Graph x � 0; y � 0 . Shaded region is the first
Graph x � 0; y � 0 . Shaded region is the first
quadrant. Graph the line 2x + y = 6 . Use a solid line since the inequality uses S. Choose a test point not on the line, such as (0, 0). Since 2(0) + ° � 6 is true, shade the side of the line
quadrant. Graph the line x + y = 2 . Use a solid line since the inequality uses 2':. Choose a test point not on the line, such as (0, 0). Since ° + 02':2 is false, shade the opposite side of the line from (0, 0). Graph the line 2x + y = 4 . Use a solid line
containing (0, 0). Graph the line x + 2y = 6 . Use
since the inequality uses 2':. Choose a test point not on the line, such as (0, 0). Since 2(0) + ° � 4 is
a solid line since the inequality uses S. Choose a test point not on the line, such as (0, 0) . Since ° + 2(0) � 6 is true, shade the side of the line
false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is unbounded.
containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices:
Find the vertices: The intersection of x + y = 2 and the x-axis is
The x-axis and y-axis intersect at (0, 0). The intersection of x + 2y = 6 and the y-axis is (0, 3).
(2, 0). The intersection of 2x + y = 4 and the y
The intersection of 2x + y = 6 and the x-axis is
{
axis is (0, 4). The two corner points are (2, 0), and (0, 4).
(3, 0). To find the intersection of x + 2y = 6 and 2x + y = 6 , solve the system: x + 2y = 6 2x + y = 6
Solve the first equation for x: x = 6 - 2y . Substitute and solve: 2(6 - 2y) + y = 6 12 - 4y + Y = 6 l 2 - 3y
=
2x+
6
-3y = -6
.y=
4
x �o
y =2
y�o
x = 6 - 2(2) = 2 47.
The point of intersection is (2, 2) . The four corner points are ( 0 , 0), ( 0 , 3), (3 , 0), and (2, 2) .
x + y�2
2x + 3y � 12 3x + y � 12
y
Graph x � 0; y � 0 . Shaded region is the first quadrant. Graph the line x + y = 2 . Use a solid
line since the inequality uses 2':. Choose a test point not on the line, such as (0, 0). Since ° + °� 2 is false, shade the opposite side of the line from (O, 0). Graph the line 2x + 3y = 12 .
Use a solid line since the inequality uses S. Choose a test point not on the line, such as (0, 0). Since 2(0) + 3(0) � 12 is true, shade the side of the line containing (0, 0). Graph the line 3x + y = 12 . Use a solid line since the inequality 686
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Section 12.7: Systems of Inequalities
x� y� x+y� 2 x+y� 8 2x+ Y � 10 x� y�
3(0) + � 12
uses:S. Choose a test point not on the line, such as (0, 0). Since 0 is true, shade the
0
0
side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices: The intersection of
2). (2, 3x + = 12 3x + y = 12 { 2X+3Y = 12 3x+ y = 12
(0,
x+y = 2 x + y = 2 x 2x + 3y = 12 (4, 2x+3y = 12 and the y-axis is
The intersection of
axis is
49.
Graph
and the
�2
and
2x+3(12-3x) = 12 2x+36 -9x = 12 -7 x = -24 X = -247 y = 12-3 ( 24 ) = 12 7 2 = .!3. . (7'24 712) . 2), (2, 24 12) . (4, (7'7
Substitute and solve:
7
_
. . 0f'mtersechon The pomt
.
(0, 4),
2x+ y =
a solid line since the inequality uses:S. Choose a test point not on the line, such as (0, 0). Since 0 l O is true, shade the side of the line
2(0) + �
containing (0, 0). The overlapping region is the solution. The graph is bounded.
x+ y = 2 2). x+y 2 (2, x+y =8 8). 2x + y = x+ y = 8 2x + y = 1 { x+y = 8 2x+ y = 10 y: y = 8 -x 2x+8-x = 10 x = 2 y=8-2= 6 (2,2),6). 8), (2, (2, 6).
Find the vertices: The intersection of (0,
The intersection of
is
0) . The intersection of
axis is (0,
IS
The five comer points are (0,
x+ = 8 .
�8
7
7
Use a solid
solid line since the inequality uses:S. Choose a test point not on the line, such as (0, 0) . Since 0+0 is true, shade the side of the line containing (0, 0). Graph the line 10 . Use
, solve the system:
y: y = 12-3x
y
line since the inequality uses 2::. Choose a test point not on the line, such as (0, 0). Since 0+0 is false, shade the opposite side of the line from (0, 0). Graph the line y Use a
and the y-axis is (0, 4). The intersection of y and the x-axis is 0).
Solve the second equation for
x+ = 2.
O . Shaded region is the first
quadrant. Graph the line
0). The intersection of
To find the intersection of
0;
and the y-axis is =
and the x-axis and the y
The intersection of
10 and
the x-axis is (5, 0) . To find the intersection of and 0 , solve the system:
0),
0), and
Solve the first equation for
.
Substitute and solve:
The point of intersection is The five comer points are (0, (5, 0), and
(0,
0),
Y
.
687
2008 Pearson Education, Inc . , Upper S addle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently cxist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
51.
j 1x
X+2
;: �
57.
a.
Let x the amount invested in Treasury bills, and let y the amount invested in =
=
corporate bonds. The constraints are: x + y :<::; 50, 000 because the total investment
+2Y:<::; 1O
Graph x � 0; y � 0 . Shaded region is the first
cannot exceed $50,000. x � 3 5, 000 because the amount invested in Treasury bills must be at least $35 ,000. y :<::; 10, 000 because the amount invested in
quadrant. Graph the line x +2y = 1 . Use a solid
line since the inequality uses 2':. Choose a test point not on the line, such as (0, 0). Since ° + 2(0) � 1 is false, shade the opposite side of the
corporate bonds must not exceed $ 10,000 . x � 0, y � ° because a non-negative amount
line from (0, 0). Graph the line x +2y = 1 0 . Use a
must be invested. The system is x+y:<::; 5 0, 000 x � 3 5, 000 y:<::; 10, 000 x �O y �O
solid line since the inequality uses�. Choose a test point not on the line, such as (0, 0). Since ° +2(0) :<::; l O is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices: The intersection of x + 2y = 1 and the y-axis is b.
(0, 0.5). The intersection of x +2y = 1 and the x-axis is ( 1, 0). The intersection of x +2y = 10
Graph the system. y
. x= 35,000
80.000
and the y-axis is (0, 5). The intersection of x +2y = 10 and the x-axis is ( 10, 0). The four corner points are (0, 0.5), (0, 5), ( 1 , 0), and (l0, 0). y 16
y= 10.000
x
-20,000 -20,000
80,000
.
x + y-, . _ 50 , 000
(35.000, 0) (50.000.0)
The corner points are (35 ,000, 0), (35 ,000, 10,000), (40,000, 10,000), (50,000, 0).
x
53 .
55.
{
59.
The system of linear inequalities is: x+
a.
Let x the # of packages of the economy blend, and let y the # of packages of the =
=
;::
superior blend; The constraints are: x � 0, y � 0 because a non-negative # of
y�°
packages must be produced. 4x + Sy:<::; 75 · 16 because the total amount of
x� °
"A grade" coffee cannot exceed 75
pounds. (Note: 75 pounds = (75)( 16) ounces.) 12x +Sy:<::; 120· 16 because the total amount of
The system of linear inequalities is: x:<::; 20 y � 15
"B grade" coffee cannot exceed 120 pounds. (Note: 120 pounds ( 120)( 16) ounces.) Simplifying the inequalities, we obtain:
x + y:<::; 50
=
x - y:<::; ° x� °
4x+Sy:<::; 75· 16
12x+Sy:<::; 120· 16
x+2y:<::;7 5· 4
3x+2y:<::; 120· 4
688
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Section 12.8: Linear Programming
Section 12.8
The system is:
{::�
x + 2y:S: 300
b.
1.
objective function
3.
z = x+y
3x + 2y:S: 480
Vertex
Value of z = x + y
Graph the system.
(0, 3)
z = 0+3 = 3
(0, 6)
z = 0+ 6 = 6
(5, 6)
z = 5 + 6 = 11
y
3x+Zy=480
400
(5, 2)
z=5+2=7
(4 0)
z =4+0=4
The maximum value is 1 1 at (5, 6), and the minimum value is 3 at (0, 3). X+ Zy
-100
=
5.
300
The comer points are (0, 0), (0, 150), (90, 1 05), ( 1 60, 0). 61.
a.
Let x = the # of microwaves, and let y = the # of printers. The constraints are: x � 0, y � 0 because a non-negative # of 7.
weight cannot exceed 1600 pounds. 2x + 3 y :s: 150 because the total cargo volume cannot exceed 150 cubic feet. Note that the inequality 30x + 20y:S: 1600 can be
{
b.
:s:
Vertex
Value of z = x + l 0y
(0, 3)
z = 0 + 10(3) = 3 0
(0, 6)
z = 0 + 10(6) = 60
(5, 6)
z = 5 + 10(6) = 65
(5, 2)
z = 5 + 10(2) = 25
(4 0)
z = 4 + 10(0) = 4
The maximum value is 65 at ( 5 , 6), and the minimum value is 4 at (4, 0).
items must be shipped. 3 Ox + 20 y :s: 1600 because a total cargo
simplified: 3x + 2y
z = x + 10y
160 .
The system is: 3X + 2y:S: 160 2x + 3y:S: 150 x �O; y � °
z = 5x + 7y Vertex
Value of z = 5x + 7 y
(0, 3)
z = 5(0) + 7(3) = 2 1
(0, 6)
z = 5(0) + 7(6) = 42
(5, 6)
z = 5(5) + 7(6) = 67
(5, 2)
z = 5(5) + 7(2) = 39
(4 0)
z = 5(4) + 7(0) = 20
The maximum value is 67 at (5, 6), and the minimum value is 20 at (4, 0).
Graph the system. 9.
Maximize z = 2x + y subject to x � 0,
y � 0, x + y :s: 6, x + y � 1 . Graph the
constraints.
-20
The comer points are (0, 0), (0, 50), (36, 26),
C�O , 0) .
- .
: :x;.- y:=) :-:-:-:-::- :-:-: :-:
689
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Chapter 12: Systems of Equations and Inequalities
(0, 1), (1, 0), (0, 6), (6, 0). = 2x + y (0, 1) = 2(0)+ 1 = 1 (0, 6) = 2(0)+6 = 6 (1, 0) = 2(1)+0 = 2 (6 0) = 2(6 )+0 = 12 12 (6, 0). = 2x + 5Y x 0, y 0, x + y 2, x 5, y 3.
3x + 2y = 12, 2X+3Y == 1212 {3x+2y
Vertex
Value of z z
z
z
11.
Minimize
z
�
�
�
subject to �
constraints.
=
�
Graph the
....... ....... ... ...... .... . - -. . - .
: - : - : . :a: .
-
: : - : ti
. '
-
:-:-:-:-:-:-: :-:.
.
.
y: y = 6 -�x2
2X+3 ( 6-% X) 12 2x+18-�x2 = 12 5 --x=-6 2 12 X = -5 y = 6-�2 (.!25 ) = 6-�5 = .!25 (2.4, 2.4) . (0, (2, 0), (0, 4), (4, 0), (2.4, 2.4 ). 3x + 5y (0, 2) = 3(0)+5(2) = 10 (0, 4) = 3(0) 5(4) = 20 (2, 0) = 3(2)+ 5(0) = 6 (4, 0) = 3(4)+ 5(0) = 12 (2.4 2.4) = 3(2.4) + 5(2.4) = 19.2 20 (0, = 5x 4y x 0, y 0, x + y 2, 2x 3y 12, 3x + y
Substitute and solve:
at
12 and
solve the system:
Solve the second equation for
z
The maximum value is
2x + 3y =
To find the intersection of
The comer points are Evaluate the objective function:
.
:.: : : :-: :.:-: :
The point of intersection is
The comer points are 2), Evaluate the objective function: Vertex
(0, 2), (2, 0), (0, 3), (5, 0), (5, 3). = 2x + 5y (0, 2) = 2(0)+ 5(2) = 10 (0, 3) = 2(0)+ 5(3) = 15 (2, 0) = 2(2)+ 5(0) = 4 (5, 0) = 2(5)+ 5(0) = 10 (5 3 1 2(5) + 5(43 ) =(2,250). = 3x + 5y x 0, y 0, x + y 2, 2x + 3y 12, 3x + 2y 12.
z
The comer points are Evaluate the objective ftmction: Vertex
z
z
z
z
z
The maximum value is
z
z
1 5.
Maximize
z
�
�
�
�
+
z
�
at
4).
subject to
+
�
�
�
12. Graph
the constraints.
at
subject to
Minimize �
z=
13.
+
z
Value of z
The minimum value is
Value of z =
�
Graph
the constraints.
3x y = 12, 3y = 12 {2X+ 3x+ y = 12
To find the intersection of +
2x + 3y = 12
and
solve the system:
690
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Section 12.8: Linear Programming
Solve the second equation for y: y = 12 - 3x
e)
;
(
(!Q)
!Q
20 = 10 = 3 3 3 The point of intersection is ( 10/3 10/3 ). The comer points are (0, 10), ( 10, 0), ( 10/3 , 10/3 ). Evaluate the objective function: y = 10 - 2
Substitute and solve: 2x + 3(12 - 3x) = 12 2x + 3 6 - 9x = 12 -7x = -24 x = 24 7 4 y = 12 - 3 7 = 12 - 7 = Ii . . 24 12 . The pomt 0f'mtersectlOn IS 7'7 .
)
(274, I�). Evaluate the objective function:
_
Vertex
Value
(0, 10)
z =
(10, 0)
z
f
o
z
= 5x + 2y
5(0) + 2(10) = 20
= 5(10) + 2(0) = 50
The comer points are (0, 2), (2, 0), (0, 4), (4, 0),
Vertex
17.
The maximum value is 50 at ( 1 0, 0). 1 9.
Value of z = 5x+4y
Let x the number of downhill skis produced, and let y = the number of cross-country skis =
P
.
(0, 2)
z = 5(0) + 4(2) = 8
produced. The total profit is:
(0, 4)
z = 5(0) + 4(4) = 16
(2, 0)
z
(4, 0)
z = 5(4) + 4(0) = 20
Profit is to be maximized, so this is the objective function. The constraints are: x � 0, y � 0 A positive number of skis must be
= 5(2) + 4(0) = 10
= 70x + 50 y
2x + y � 40
produced. Manufacturing time available.
The minimum value is 8 at (0, 2) .
x + y � 32
Finishing time available.
Maximize z = 5x+2y subject to x �O,
Graph the constraints.
y �O, x + y�10, 2x + y �10, x + 2y �10 . Graph the constraints.
40
To find the intersection of 2x + y = 10 and
{
x + 2y = 1 0 , solve the system:
Solve the first equation for y: y = 32 - x .
To find the intersection of x + y = 32 and +
{
2y
2x + y = 40 , solve the system: =
IO
x+y = 32 2x+ y = 40
Substitute and solve: 2x + (32 - x) = 40
2X+ y = 10 x + 2y = 10
x=8 y = 32 - 8 = 24
Solve the first equation for y: y = 10 - 2x . Substitute and solve: x + 2(l O - 2x) = 10
The point of intersection is (8, 24). The comer points are (0, 0), (0, 32), (20, 0), (8, 24) . Evaluate the objective function:
x + 20 - 4x = 10 -3x = - 10 10 X=3 691
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Chapter 12: Systems of Equations and Inequalities
Vertex
Value of P = 70x + 5 0y
(0, 0)
P = 70(0) + 50(0) = 0
(0, 3 2)
P = 70(0) + 50(32) = 1 600
(20, 0)
P = 70(20) + 50(0) = 1 400
21.
Let x = the number of rectangular tables rented, and let y = the number of round tables rented. The cost for the tables is: C = 28x + 52y . Cost is to be minimized, so this is the objective function. The constraints are: x � 0, y � 0 A non-negative number of tables
(8 24) P = 70(8) + 50(24) = 1 760 The maximum profit is $ 1 760, when 8 downhill skis and 24 cross-country skis are produced. With the increase of the manufacturing time to 48 hours, we do the following: The constraints are: x � 0, y � 0
A positive number of skis must be
2x + y :<:; 48
produced. Manufacturing time available.
x + y :<:; 3 2
Finishing time available.
x + y :<:; 3 5
must be used. Maximum number of tables.
6x + lOy � 250
Number of guests.
Rectangular tables available. x:<:; 1 5 Graph the constraints. y x= 1 5 x + y = 35
Graph the constraints.
The comer points are (0, 25), (0, 35), ( 1 5 , 20), ( 1 5 , 1 6) . Evaluate the objective function: Vertex (0, 25)
{
To find the intersection of x + y = 3 2 and 2x + y = 48 , solve the system: x + y =3 2
23 .
x = 16 y =3 2 - 1 6 = 1 6
(0, 0)
P = 70(0) + 5 0(0) = 0
(0, 32)
P = 70(0) + 5 0(32) = 1 600
(24, 0)
P = 70(24) + 5 0(0) = 1 680
C = 28(0) + 5 2(25) = 1 3 00
(0, 35)
C = 28(0) + 52(35) = 1 820
(15, 20)
C = 28(1 5) + 5 2(20) = 1 460
Let x = the amount invested in junk bonds, and let y = the amount invested in Treasury bills . The total income is: 1= 0.09x + 0.07 y . Income is to be maximized, so this is the objective function. The constraints are: x � 0, y � 0 A non-negative amount must be
The point of intersection is ( 1 6, 1 6). The comer points are (0, 0), (0, 32), (24, 0), ( 1 6, 1 6). Evaluate the objective function: Value of P = 70x + 50y
28x + 52y
.
Substitute and solve: 2x + (32 - x) = 48
Vertex
=
(15 1 6) C = 28(15) + 52(16) = 1 252 Kathleen should rent 15 rectangular tables and 1 6 round tables in order to minimize the cost. The minimum cost is $ 1 252.00.
2x + y =48
Solve the first equation for y: y = 32 - x
Value of C
x + y :<:; 20, 000 x:<:; 1 2 , 000 y � 8000
(16 1 6) P = 70( 1 6) + 5 0( 1 6) = 1 920 The maximum profit is $ 1 920, when 16 downhill skis and 1 6 cross-country skis are produced.
a.
y �x
invested. Total investment cannot exceed $20,000. Amount invested in junk bonds must not exceed $ 1 2,000. Amount invested in Treasury bills
must be at least $8000. Amount invested in Treasury bills must be equal to or greater than the amount invested in junk bonds.
692
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Section 12.8: Linear Programming Value of 1=0.09x + 0.07y
Vertex
Graph the constraints.
(12000, 8000)
1=0.09(12000) + 0.07(8000) =1640
(8000, 8000)
1=0.09(8000) + 0.07(8000) =1280 (10000, 10000) I =0.09(10000) + 0.07(10000) =1600
The maximum income is $ 1 640, when $ 1 2 ,000 is invested in junk bonds and $8000 is invested in Treasury bills. 2 5.
The comer points are (0, 20,000), (0, 8000), (8000, 8000), ( 1 0,000, 1 0,000) . Evaluate the objective function:
pork. The total cost is: C =0.75x + 0.45y . Cost is to be minimized, so this is the objective function. The constraints are: x � 0, y � 0 A positive number of pounds
Value of 1=0.09x + 0.07y
Vertex
(0, 20000)
Let x = the number of pounds of ground beef, and let y = the number of pounds of ground
I =0.09(0) + 0.07(20000)
=1400
(0, 8000)
1=0.09(0) + 0.07(8000)
x:5: 200
(8000, 8000)
1=0.09(8000) + 0.07(8000)
y �50
(10000, 10000)
1=0.09(10000) + 0.07(10000)
=560
=1280
pork must be used. 0.75x + 0.60y � 0.70(x + y) Leanness
=1600
condition (Note that the last equation will simplify to
The maximum income is $ 1 600, when $ 1 0,000 is invested in junk bonds and $ 1 0,000 is invested in Treasury bills. y�x
b.
y � ! x .) Graph the constraints. 2
:.: {: .. .
: •• : : : : : : :::: •. i<.�j"�:ilJ!jOO:.:.·:·.· . .
.
.
.
.. .... .. . . . . .. .. . .
. . .ioo:: :: .: :: :: :: .: :: ::: :: :: \2:6i (09 .
Amount invested in Treasury bills
must not exceed the amount invested in junk bonds. Graph the constraints. .
must be used. Only 200 pounds of ground beef are available. At least 50 pounds of ground
.
. 40 . . . .
: : . . .. . .. . . . . . .. 20: : .. .. . . . : . .. : . : . .
.
. . . . . . . . . . . ·x·
. : . .. . . . SO . . . . .{otc.. {S o:-::� 0: . .. . . . . . . . . . .. . . . . . . . . . . . . .. . . . . . .
.
·
·
.
·
The comer points are ( 1 00, 50), (200, 50), (200, 1 00). Evaluate the objective function: Vertex
(100, 50)
(200, 50)
The comer points are ( 1 2,000, 8000), (8000, 8000), ( 1 0,000, 1 0 ,000) . Evaluate the objective function:
(200 100)
Value
C=
C
C
= =
of C = 0.75x+ 0.45y
0.75(100) + 0.45(50) 97.50 =
0.75(200) + 0.45(50) = 172.50 0.75(200) + 0.45(100) = 195
The minimum cost is $97. 50, when 1 00 pounds of ground beef and 5 0 pounds of ground pork are used.
693
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Chapter 12: Systems of Equations and Inequalities
29. Let x
27. Let x
= the number of racing skates manufactured, and let y = the number of figure
skates manufactured. The total profit is: P = lOx + 12y . Profit is to be maximized, so
A positive number of skates must
be manufactured. 6x + 4y:'O: 120 Only 120 hours are available for
x + y�6
made. At least 6 fasteners are needed.
4x + 2y:'O:24
Only 24 hours are available.
Graph the constraints.
fabrication. Only 40 hours are available for
x + 2y :'0: 40
the number of metal fasteners, and let y
is the objective function. The constraints are: x � 2, y � 2 At least 2 of each fastener must be
this is the objective function. The constraints are: x� 0, y � 0
=
the number of plastic fasteners . The total cost is: C = 9 x + 4 y . Cost is to be minimized, so this
=
.
.
.
.
.
.
..
. . ... . . . . . . . . . ..... . . . . ..
. . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . .. . . . .
.
finishing. Graph the constraints.
::ro::"
. .......... ·x··
The comer points are (2, 4), (2, 8), (4, 2), (5, 2). Evaluate the objective function:
{
Vertex
Value of C = 9x + 4y
(2, 4)
C = 9(2) + 4(4) = 34
To find the intersection of 6x + 4y = 120 and x+2y = 40 , solve the system: x + 2y = 40
Solve the second equation for x: x = 40 - 2y
31.
Substitute and solve: 6(40 - 2y) + 4y = 120
R
- 8y = - 120 y = 15 The point of intersection is ( 1 0, 1 5). The comer points are (0, 0), (0, 20), (20, 0), ( 1 0, 1 5). Evaluate the objective function:
(0, 20) (20, 0) (1 0 1 5)
a.
Value of P = lOx + 12y
P
P
P
=
=
x l . -:'0: RatIO 0f seats. y 12 -
The constraints are: 8:'0: x:'O: 1 6 80:'O:y:'O:120
= 1 0(0) + 1 2(0) = 0 1 0(0) + 1 2(20)
= Fx + Cy is the objective function to be
maximized. The constraints are: 8:'0: x :'0: 16 Restriction on first class seats. 80:'0: y:'O: 120 Restriction on coach seats.
x = 40 - 2(1 5) = 1 0
P
Let x = the number of first class seats, and let y = the number of coach seats. Using the hint, the revenue from x first class seats and y coach seats is Fx + Cy, where F > C > O. Thus,
240 - l2y + 4y = 1 20
(0, 0)
C = 9(2) + 4(8) = 50 C = 9(4) + 4(2) = 44
C = 9(:;) + 4(2} = 5 3 is, 2) The minimum cost i s $34, when 2 metal fasteners and 4 plastic fasteners are ordered.
6X + 4Y = 120
Vertex
(2, 8) (4, 2)
240
1 2x:'O: y
= 1 0(20) + 12(0) = 200
Graph the constraints.
= 1 0(10) + 12(15) = 280 The maximum profit is $280, when 10 racing skates and 15 figure skates are produced.
694
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Chapter 12 Review Exercises 12x =y
y
Vertex
(S, SO) (S, 120) (15, 1 0) JlO SOl2 F> 0
",---...- y = 120 I-....-... . '1----.'1----+-
y =80
Since and the maximum value of R occurs at The maximum revenue occurs when the aircraft is configured with first class seats and coach seats.
x
(S, 96), (S, 120), (10, 120). R=Fx+Cy (S, 96) R=SF+96C (S, 120) R=SF+120C (10 120) R=10F+1 20C C> 0,> SFl20C96C. > 96C, SF + 120C + F > 0, 10F > SF, 10F + 120C> SF + 120C. 10 120 x -1 y S SSO��xy��16120 Sx �y
The comer points are and Evaluate the objective function: Vertex
c.
Answers will vary.
Chapter 12 Review Exercises .
1
so
Since
{ 5x2X-+ 2yy =S= 5
y = 2x-5 .
Solve the first equation for y:
so
5x 2(2x -5) = S 5x + 4x-1O = S 9x = IS y = 2(2) -5 =x 4= -52 = -1 x = 2, y=-1 (2,-1). =� {3X-4Y x-3y =-2 x: x =3y +-21 3(3Y+±)-4Y = 4 3 =4 9y + --4y 2 5 5y =-2 y =-12 X=3(�)+�= 2 x = 2, y = ± (2, �) .
Substitute and solve: +
Thus, the maximum revenue occurs when the aircraft is configured with first class seats and coach seats. -<
The constraints are:
The solution is 3.
Graph the constraints. v
120
Value of
Since
b.
R=Fx+Cy R=SF+SOC R=SF+120C R=15F+120C R=C10F +SOC > 0, 120C>(15,96C,120). 15
Value of
(15,120)
or
Solve the second equation for
Substitute into the first equation and solve:
(15, 120), (10, SO).(S, SO), (S, 120),
The comer points are and Evaluate the objective function:
The solution is
or
695
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Chapter 12: Systems of Equations and Inequalities
5.
{3x+2y-4=0 x-2y-4 = 0
Solve the first equation for
x: x = 2Y + 4
3(2y+4)+ 2y-4 == 00 6y+12+2y-4 8yy == -8 -l x = 2(-1) + 4 = 2 x = 2, y = (2, -1) . {yx == 3y+4 2x-5 x = 3(2x -5) + 4 x=6x-15+4 -5x =-11 X=-115 Y = 2 C51) -5 = - � 11 3 ) . x = 5'11 y = -53 (5'-5 {�x-x- �3y+4=0 2 2 y+ �3 = 0 3 {-3x+ 3X-9Y+12 = 0 9y-8 = 0 4=0
Substitute into the second equation and solve:
The solution is
7.
The solution is =
13.
9.
1 5.
3x-2y+3z = -16
x; -2
{ -2X-4Y+2Z = -12 2x- y+3z =-13
or
-5y+5z=-25 y-z =5 -3 x: -3x-6y+3z = -18 3x-2y+3z = -16 -8y+6z = -34 8 y: 8y-8z = 40 -8y+ 6z =-34 -2z = 6 z=-3 y-(-3) = 5 x+2(2)-(-3) = 6 y = 2 x+4+3 = 6 x = -1,y = X2,=z-1= -3 (-1, 2,-3) .
Multiply each side of the first equation by add to the third equation to eliminate
Multiply each side of the first result by to the second result to eliminate
3, 2
-3
and
and
and add
Substituting and solving for the other variables:
MUltiply each side of the first equation by and each side of the second equation by and add to eliminate
{4X+9x-6y6y-26y: == 00
{2x-X+ 2Y-y+3zZ==-136
Multiply each side of the first equation by add to the second equation to eliminate
There is no solution to the system. The system is inconsistent.
{2X3x-2y + 3y -13 = 0 =0
8
There is no solution to the system. The system of equations is inconsistent.
Multiply each side of the first equation by and each side of the second equation by -6 and add:
1 1.
or
Multiply each side of the second equation by and add to eliminate
- l or
Substitute the first equation into the second equation and solve:
· · The so1 utlOn IS
y: 3(2)-2y = 0 -2yy =-6 =3 x = 2, y = 3 (2, 3). {3X-22Y x--y3 =12 x: {-3x+3x-2y2y == -368 0 =-28
Substitute and solve for
13x -26 = 0 13xx == 226
The solution i s
696
or
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1 7.
Chapter 12 Review Exercises
{ x+2y-4z 2x -4Y +Z == 27-15
5x- 6y-2z = -3 Multiply bythe2,firandst equation byto -1eliminate and thex:second equation then add -2X+4Y-Z = 15 { 2x+ 4y -8z = 54 8y -9z = 69 Mul t ipl y the secondto equation byx: -5 and add to the-5x-l0y+20z third equation el i minate = -135 5x- 6y -2z =-3 -16y+18z =-138 both sidesresulof thet to fieliminate rst resulty:by 2 and addMultiply to the second 16y-18z = 138 -16y+18z = -138 The-16y+18z system0=0is= -138 dependent. 18z+138=16y y=-z+forSubstituting x: ( into) the second equation and solving x+2 8z+8 -4z= 27 x+-z+--4z= 27 4 4 x=-z+4 4 7 39 The so utlOn x = -z4 +-4 = -z98 +-698 , z . 7 39 ' any real number or {(x,y,z) 1 x="4z+4 9 69 .zls anyrea numb} y=-z+-, er . 8 8 = 8 3X+2Y {x+4y= -1 + [ 42]+[35 -42] [ 2+11+3 +4+(-4)5] = [43 -49] -1+5 4 4 9
9
2 5.
2 7.
29.
39
·
IS
,
y
IS
21.
A
C
1
=
2
0
1
-1
-4
=[� �] Augment the matrix identity and use row operations to finwith d thetheinverse: 6 11 [� 3 3 ° �] Interchange [� 613 °1 �] ( and r2 ) [� -61°1 -41] (R =-4 + r2 ) [� �I-� ;]-1 (R2 =- ir2) [� °11 -i t-1] (RI = -3r2 + ) Thus A-I = [ t 2 ] . A
Ij
�
2
�
1
1 9.
_
CB
22 -13
69
·
-8
=
8
7
-3
-I
=
69
9
1
0
23 .
69
8
.0] [ 6 ] [ 1 0 ] [ 6·1 6 6A= 6· -12 42 = 6(-1) 6·2 6·2 6·4 = -612 2412 AB�[ � 2�H41 1 _20] [ 2(1(44))++4(0(11)) 2(1(--33))++0(4(11)) 2(1(00))++0(4(--22))1 [-1124(4)-2+-32(1) 0]-1(-3)+2(1) -1(0)+2(-2) -2 5-4 = [31 -45] -[41 -31 02 ] 5 2 3(0) -4(-2)] [= 3(4)1(4)+5(1) -4(1) 3(-3) -4(1) 1(-3)+5(1) 1(0)+5(-2) 5(4)+2(1) 5(-3)+2(1) 5(0)+2(-2) [= 98 -132 -108]
�
�
Ij
�
,
5
1j
-� 6
3"
0
=
2+2
697
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Chapter 12: Systems of Equations and Inequalities
31.
A=
[� � �l 1 -1
3 5.
2
x witheth theinveridenti rAugment ow opera3thetionsmatri to find se: ty and use [�1 -1� 2 �0 �0 �l1 1 3 3 1 0 O� 0 -1-4 -2-1 -1-1 0 �
l [ [� -! -� --: -! �l [� ! -� : �! �l [01 0 -32 -2 3 0 1 2. .=-1 ->l� : � -t J -11 (�� :�;;,: ,J Thus , [ 1 (�
=
-
"
o
£1
7
=
-
=
�
)
� �
7
37.
[� �� I �] (interchange ) [� �16� \ �]2 and r2 [ 1 16-50 \ -5] (R2 = -3fj + 2 ) [ 1 10�l [0l O1 / ] Ii
1
o
r
o
1
t
The so utlOn x -,25 y -101 or (-,25 101 ) 5X-6Y-3Z 6 4x-7 y-2z=-3 3x+ y-7z 1 Write the augmented matrix:
{
1 ·
0
·
IS
=
=
-
.
=
=
[:3 =� ��-7 -�11
7
2. 7
-7
->[:1 �� -�1 [00 -11-2 -4-� -26-3:1
Augment identity and use row operathetionsmatri to fixnwid theth theinverse:
[-� - -� I � �] ( interchange ) � [-� -� I � �] and r2 �[ � � I �0 �]-1 1 -2 .. [0 0 1 1 4] (RI ) There is no inverse because there is no way to
�
I'j
----r
Write the augmented matrix: [1� �� \ �] �
->
�
=
�
o
->
{10x+lOy 3x- 2y 1 5
->
= - I'j
obtai singularn th.e identity on the left side. The matrix is
�
l�1
[00 � =� it] o
21 41 104 II
698
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Chapter 12 Review Exercises
60 1
9 2 39 1 3
11
- IT
3 9.
IT
41.
11
The so utlOn x = , y = -,133 z = -133 or (9 , �3 ' �3 ) . -2z= 1 2x+3y =-3 4x-3y-4z = 3 · IS
I ·
{ x-x- y+y- 5zz== 6 2x-2y+ Z = 1 Write the augmented matrix: [2� -2=� -�I �lI �[i -i �� �] =-hj �[i -i : -I:]] I [i -i The system is dependent. 0
11
9
{X
(R2
_
o
�
{zx== -1y+1
43.
o
-1
0
The solution is x = y + z = -I , y is any real number or {(x,y,z)l x = y+l, z =-I, yisany real number} . x- y- z- t = 1 2x + - z + 2t = 3 x-2y-2z-3t = 3x-4y+ z+ 5t = -3 Write -the augmented matrix: 1,
{ [�1 -2 -2 -3-� �] 3 -4 [i �-1� �:4 ��8 -6-:] ( Interchange ] 1 [� �i =j �� -6 and Y
0
: =:
1
T he so utlOn x = --2 ' y = --23 ' z = --3 or ( !2 ' 3.3 ' �4 ) . IS ·
I ·
_
_
1
�
4
_
�
0
5
-3
o
-:
1
r2
699
r3
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Chapter 12: Systems of Equations and Inequalities
51.
[R)Rl == r-3r2 +2'1+r) R4 = r2 +r4
)
D
[RRl2=-r= -r44 +rl+r2)
4 5.
47.
4 9.
R) = -r4 +r) The solution is x = 4, Y = 2, = 3, = -2 or (4, 2, 3,-2) . . 5 3 4 =3(3)-4(1) = 9-4= 11 3/ t
D
8
Dx D
- 13
D --.-1.. D
'
-8 8
+
x
y
1
-� ; : = I I � :H-� :H-� �I = 1(6-6) -4(-3 -24) + 0(-1-8) = 1(0) -4(-27) + 0(-9) = 0+ 108+ 0 = 108 2 -3 0 5 0 1 = 21 6 6 = 2(0 -6) -1(0 -2) -3(30 -0) = 2(-6) -1(-2) -3(30) = -12+2-90 =-100 2
Set up ands Rule: evaluate the determinants to use Cramer' D= I � �2 1 =1(2)-3(-2)=2+6=8 Dx = 1 : �2 1 =4(2)-4(-2)=8+8=16 Dy = 1 � : 1 =1(4)-3(4)=4-12=-8 16 2 y = = -= -1 The solution is x = = -= or (2,-1) . 3y -13 = 0 {2X 3x-2y= 0 Writ e the system is standard form: 2X+3Y= 13 {3x-2y =0 Set up ands Rule: evaluate the determinants to use Cramer' D= 1 23 -23 1 =-4-9=-13 D = 1 130 -23 1 =-26-0= -26 D = 1 23 130 1 = 0-39=-39 - 26 2 Th e so utlOn x=-=-= y= [i = _ 13 =3 or (2, 3). x+2y- z= 6 2x- y+3z= - 13 3x-2y+3z =-16 evaluate the determinants to use Set up ands Rule: Cramer' 1 2 -1 D= 2 -1 3 3 -2 3 = 1 1 =� � 1 - 21 =� � 1 +(-1) 1 � =� I = 1( -3 6) -2(-3 +6) +(-1)(-4+3) =3+6+1=10 ---L
53 .
Z
{3x+2y=4 x-2y=4
5 5.
o
1
D y
·
IS ·
-3 9
'
+
700
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Chapter 12 Review Exercises
Letx = 0, then 6 = A(0-4)+B(0) -4A = 6
6 2 -1 -13 = -1 3 -16 -2 3 = 6 1 =� � 1 -2 1 =� ! � 1 + (- 1) 1 =�! =� 1 = 6( -3 +6) - 2(-39 + 48) +(-1)(26 -16) = 18-18-10 6 -1= -10 = 2 -13 3 3 -16 3 =11 =� ! � 1 -6 1 � � 1 +(-1)1 � =�! 1 = 1( -39 + 48) -6(6 -9) +(- 1) (-32 + 39) = 9+1 18-7 2 =620 = 2 -1 -13 3 -2 -16 = 1 1 =� =:! 1 -2 1 � = : ! 1 +6 1 � =� 1 = 1 (16-26) -2(-32 + 39) + 6(-4+ 3) = -10-14-6 = -30 -10 The so utJOn = - = 10 = -1 , -30 = -3 or = � = 1020 = 2 , z = ; = 10 Dx
A =--23
61.
D y
----=---
Dz
y
D
1 ·
· IS
X
Dx D
(-1, 2,-3) .
57.
D
8.
+-
1-4 = A(l)(l-I)+B(I-1)+C(1) -3 = C C=-3 Let x = 0 , then 0-4 = A(O)(O-I) + B(O-I)+C(0)2 -4= -B Let Bx==42 then 2 -4 = A(2)(2-1) +B(2 -1) +C(2)2 -2 = 2A+B+4C 2A = -2-4-4(-3) 2A = 6 A = 3 4 -3 x-4 2x (x-l) = -+X3 -+- x 2 x-I Find the partial fraction decomposition: A Bx+C x --+ ---= ---:(x2-+9)(x+l) x+l -x2 +9 Multiply both sides by (x + 1)(x2 + 9) . x = A(x2 +9)+(Bx+ C)(x+1) Let x = -1 , then -1 = A ( ( _1 ) 2 +9 ) +( B ( -I ) +C )( -1+1 ) -1 = A(lO)+(-B+C)(O) -1 = 10A A = -J.10 .. Let x = 0 then ----=---
8
63 .
k
k.
59.
-
,
-
Let I ; � 1 = Then 1 2X2a Y 1 = 2 ( ) = 16 by Theorem (14). The of the determinant by by whenvalue the elements of a columnis multiplied are multiplied Find the partial fraction decomposition: b
6 = ---.2.3 + _32_ x(x-4) x x-4 Find the partial fraction decomposition: A B2 +-C = 2x x-4 (x-l) X x x-I2 Multiply both sides by x (x-l) x-4 = Ax(x-l)+B(x-l)+ Cx2 Let x = 1 , then 2
6 ) = X(X_4) ( A + �) X(X-4) ( x(x-4) x x-4 6= A(x-4)+Bx Letx=4, then 6= A(4-4)+B(4) 4B = 6 B = 'i2
,
701
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Chapter 12: Systems of Equations and Inequalities
o=A
( 02 + 9 ) + ( B ( 0 ) + C) ( 0 + 1 )
67.
( I�) + C
0 = 9A + C
Let
,
Multiply both sides by
then 1 = A ( 12 + 9 ) + ( B ( 1 ) + C) (1 + 1)
Let
1 = A(l O) + (B + C)(2) 1 = 1 0A + 2B + 2C
( 1�) + 2B + 2 ( :0 )
1 = 10 -
9 1 = - 1 + 2B + S
then
,
A = .!. 4 x = -1 ( _ 1)2 = A(- 1 + 1)« - 1)2 + 1) + B(- I - I )« - 1)2 + 1) + ( C (- I ) + D)( -1 - 1)(- 1 + 1 ) 1 = - 4B
Let
2B = .!. S B =� 10
65.
--
(x2 + 1)(x2 - 1) (x2 + l )(x - l )(x + 1) B + Cx + D A + -= -x - I x + 1 x2 + 1 (x - l )(x + 1)(x2 + 1) . x2 = A(x + 1)(x2 + 1) + B(x _ 1)(x2 + 1) + ( Cx + D)(x - l )(x + 1) x=1 12 = A( l + I )(12 + 1) + B(l - I ) (l 2 + 1) + (C(l) + D )(1 - 1) ( 1 + 1) 1 = 4A
0=9 -
C=� 10 x=1
Find the partial fraction decomposition:
,
then
B = -.!. 4 x=0 02 = A(0 + 1)(02 + 1) + B(0 - 1)(02 + 1) + (ceO) + D)(O - l) (O + 1) O= A-B-D
Find the partial fraction decomposition:
Let
--
Cx+ ----: D :x3 ---:- = Ax + B + ----: ---,(x2 + 4)2 x2 + 4 (x2 + 4)2 (x2 + 4)2 . x3 = (Ax + B)(x2 + 4) + Cx + D x3 = Ax3 + Bx2 + 4 Ax + 4 B + Cx + D x3 = Ax3 + Bx2 + (4A + C)x + 4B + D A = 1; B = 0 4A + C = O 4(1) + C = 0 C = -4 4B + D = 0 4(0) + D = 0 D=O x + ---:-:4x-x3 -- --::----::- --::- = -(x2 + 4)2 x2 + 4 (x2 + 4)2
Multiply both sides by
,
then
± ( ±) - D
O= - -
D = .!. 2 x=2 22 = A(2 + 1)(22 + 1) + B(2 - 1)(22 + 1) + (ce2) + D)(2 - 1)(2 + 1) 4 = I SA + S B + 6C + 3D
Let
,
then
6C = 0 C=O
702
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as they currently
exist. No portion of this material may b e reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12 Review Exercises .
69.
Solve theequation first equation second and solve:for y, substitute into the
{
75.
2X + y + 3 = 0 � y = - 2x - 3 x2 + y2 = S x2 + (- 2x - 3)2 = S � x2 + 4x2 + 12x + 9 = S Sx2 + 1 2x + 4 = 0 � ( Sx + 2)(x + 2) = 0 2 => x = x= -2
Substitute
S
71.
y=1
Solutions: (-�, _ lSI ) ' (-2, 1) . Multiply side of theto eliminate second equati and add theeachequations xy: on by 2 S
f 2xy +
y2 = 1 0 l-xy + 3l = 2
�
�
( �) � ( ) �
J2 J2 : X = -2 2 =3 Ify = - 2J2 : X = - 2 - 2 J2 = 4 3 3 Ify = J2 : x = - J2 Ify = - J2 : x = J2 Ify = 2
2x(-J2) + (- J2t = 1 0 � - 2J2x = 8 � x = - 2J2 (2J2, J2), (-2J2 , -J2)
Solutions: Substitouten andintosolve: the second equation into the first equati
(
Solutions:
{
)(
)
, ,
- 4J2 2J2 4J2 - 2J2 -J2 J2 ) ( 3 ' 3 ' 3 ' 3 ' ( J2 , -J2)
X2 + l = 6y x2 = 3y 3y + l = 6y l - 3y = 0 y(y - 3) = 0 � y = 0 y = 3 y = 0 : x2 = 3(0) � x2 = 0 � x = 0 x2 = 3(3) � x2 = 9 � x = ± 3 Y=3: (0, 0) , (-3, 3), (3, 3).
If If Solutions:
and solve:
3x2 + 4xy + sl = 8 3( _y)2 + 4(-y)y + sl = 8 3l - 4l + sl = 8 4l = 8 l = 2 => y = ± J2
2x (J2) + (J2t = 1 0 � 2J2x = 8 � x = 2J2 Ify = - J2 :
73.
or
Substitute x = -y and solve:
2xy + y2 = 1 0 - 2xy + 6l = 4
7l = 14 y2 = 2 y = ± J2
Ify = J2 :
{
x,
3x2 + 4.xy + sl = 8 x2 + 3.xy + 2l = 0 x2 + 3.xy + 2 y2 = 0 (x + 2y)( x + y) = 0 � x = - 2y x = -y x = - 2y 3x2 + 4xy + sl = 8 3(- 2y)2 + 4(- 2y)y + Sy2 = 8 1 2y2 _ 8y2 + sl = 8 9l = 8 2J2 8 Y 2 = - => y = ± -3 9
- or
11 y = --
Factor the into second substitute the equation first equati, solve on andforsolve:
or
703
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Chapter 12: Systems of Equations and Inequalities
77.
{
X2 - 3x + / + y = - 2 X2 - X + y + l = 0 y
83.
--
Multiply side of theto eliminate second equation and add theeachequations y: by -y _
79.
Graph the line - 2x + y 2 . Use a solid line since the inequality uses ::;. Choose a test point not on the line, such as (0, 0). Since - 2(0) + 0 � 2 is true, line acontaining Graphshade the linethe sixd+eyof=the2 . Use solid line (0,since0). the inequality not on the line, such asuses(0, 0).Choose Since 0a +test0 ;:::point 2 is false, shade theoverlapping opposite regi sideoofn isthethelinsolution. e from (0, 0). The =
x2 3x + y2 + Y = - 2 _ x2 + y2 - Y = 0 - 2x = -2 x=1 Ifx = l : 12 - 3(1) + / + y = - 2 / +y = o y(y + l) = 0 y = 0 y = -1 y "* 0 X
{- 2Xx ++ yy �;::: 22
2:.
_
y
or Note that because that would cause division in the original system. Solution:by( 1zero , -1) 3x + 4y � 1 2 Graph the line 3x + 4y = 1 2 . Use a solid line sinceonthetheinequality not line, suchuses as (0,� 0).. Choose Since a test point 3( 0) + 4( 0) � 1 2 is true, shade the side of the line containing (0, 0).
The unbounded.ofFind the vertices: To findgraphtheisintersection x + y 2 and -2x + y = 2 , solve the system:
{- 2xx ++ y == 22
=
Y
Solve the first equation for x: x Substitute and solve:
y
=
2
-
y
.
- 2(2 - y) + y = 2 - 4 + 2y + y = 2 3y = 6 y=2 x = 2-2=0
The The point comerofpoiinntersection t is (0, 2).is (0, 2). 81.
y � x2
Graph the parabola y = x2 • Use a solid curve since the inequality uses � . Choose a test point not on the parabola, such as (0, 1). Since 0 � 12 is false,(0, shade the opposite side of the parabola from 1).
85.
{
�;�
x+ 2x + 3y � 6 x ;::: 0; y ;::: O .
Graph Shaded region is the first quadrant. Graph the line x + y = 4 . Use a solid line since thetheinequality usesas ::;(0,. Choose a test point not on line, such 0). Since 0 + 0 � 4 is true, shade the side of the line containing (0, 0). Graph the line 2x + 3y = Use uses as::;. (0, 0). Choosea solida testlinepointsincenottheoninequality the line, such Since 2(0) + 3(0) � is true, shade the side of the line containing (0, 0).
y
5
6
x
.
6
704
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Chapter 12 Review Exercises y
89.
8
02 0
x
The overlapping region is the solution. Theandgraph isaxisbounded. Find the vertices: The x-axis y i n tersect at The intersection of 2x + y and the y-axis is 2). The intersection of 2x + y and the x-axis is 2), and The three comer points are x ;::: 2x+ ; : � x+2y ;::: 2 Graph x ;::: y ;::: Shaded region is the first quadrant. Graph the line 2x + y Use a solid line since thetheinequality usesas Choose a test point not on line, such Since + ng is true,Graph shadethethelinesidexof+ the2y line2 . Use containi atestsolidpointlinenotsinceon thetheline, inequality uses . Choose a such as Since + line from 2 is false, shade the opposite side of the =
6
(0,
3
{
=
87.
16 :S . (0, 0). =
16 (0, 0). x
�
=
° ° ;:::
(0, 0), (0,
°
0;
0.
=
8.
91.
:S. (0, 0).
2(0)
°
y2
(0, 0).
6
(3, 0). (3, 0).
16
(0, 0).
(0, 0).
3
{X2 y2
Graph the system of inequalities: + � x+y ;::: 2 Graph the circle x2 + .Use a solid line notsinceonthe2theinequality circle, suchusesas ChooseSincea test point + � is true, shade the side of the circle containing Graph the line + y 2 . Use a solid line since inequality thethe line, such asuses . Choose Since a +test poi2 nits notfalse,on shade the oppositregion e side ofis thethe solution. line from The overlapping
° :S 8 (0, 0).
�
4
=
=
x2
2
:S
(0, 0).
2(0) � (0, 0).
x2
Graph the system of inequalities: { y� xy � Graph the parabola y . Use a solid line since the inequality uses . Choose a test point not on theshade parabola, such asside(1, 2).of theSinparabol ce 2 � a ifrom s false, the opposite Graph(1,the2).hyperbola xy Use a solid line sinceonthetheinequali ty usessuch .asChoose aSitestnce point not hyperbola, (1, 2). 2 � 4 is astrue,(1, shade theoverl sameappisidengofregitheon is hyperbola 2). The the solution.
y
=
1
4.
:S
1·
x
9
overlapping region is the solution. The graphof isThe bounded. Find the vertices: The intersection x + 2y 2 and the y-axis is 1). The intersection of x + 2y 2 and the x-axis is (2, The intersection of 2x and the y-axis is The intersection of 2x + y and the x axis is (2, andThe four comer points are 1), =
=
+
(0, 8).
(4, 0). (0, 8), 0),
y
(0,
0).
=
8
=
8
(0,
(4, 0).
7 05
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Chapter 12: Systems of Equations and Inequalities
93.
Maximize z = 3x + 4y subject to x:2: 0, y:2: 0 , 3x + 2Y :2: 6 , x y � Graph the constraints. +
8.
x = 12_3( 247 ) = 12- 772 = g7 The pomt. f'mtersectton , ( 12 24 ) . The comer points are (0, 0), (0,4), (4,0), (g7 ' 247 ) . Evaluate the objective function: Vertex Value of z = 3x + 5y = 3(0) + 5(1) = 5 (0,(0, 4)1) zzz==3(0)3(1)++5(4)5(0)= 203 (1, 0) (4, 0) z = 3(4)+ 5(0) = 12 ( ' 24 Z =3 + 5 4 = 1 6 7 7 ) C;) e7 ) � The minimum value is 3 at (1, 0). 5Y= 5 { 2X+ 4x+l0y= A Multiply side of the first equation by -2 and eliminate each x: = -10 {-4X-I0Y 4x+1Oy = A 0= A-I0 If thereofareelimination to be infinitely many solutions, the result should be 0 O Therefore, . A -10 = 0 or = . 10 y = ax2 +bx+c At (0, 2the equation becomes: 1 = a(0) + b(O) + c c=1 At 0) the equation becomes: 2 0= a(1) +b(I)+c 0= a+b+c a+b+c= O At (-2, 1) the equation2 becomes: 1 = a(-2) +b(-2)+c 1 = 4a-2b+c 4a-2b+c The system =of1 equations is: {4a-2b+c a+ b+c== O1 c= 1 Substitute c = 1 into the first and second equations and simplify: .
0
IS
7 ' -:;-
1), ( 1 ,
The comerthepoints are (0,function: 3), (2, 0), (0, 8), 0) . Evaluate objective Vertex Value ofz = 3x+4y (0, 3) z = 3(0) + 4(3) = 12 (0, z = 3(0) + 4(8) = 32 (2, 0) z = 3(2)+4(0) = 6 0) = 3(8) + 4(0) = 24 The maximum value is 32 at (0,.8) Minimize z = 3x+ 5y subject to x:2: 0, y:2: 0 , x+y :2: 1 , 3x+2y�12 , x+3y � 12. Graph the constraints.
=
(8,
g
8)
(8
95.
97.
z
A
99.
=
1)
(1,
To find the intersection of 3x+ 2y = 12 and x + 3y = 12 , solve the system:
= 12 {3X+2Y x+3y=12
Solve the second equation for x: x = 12 -3Y Substitute and solve:
3(12-3y)+2y = 12 36-9y+2y=12 = -24 y= 24-7y
7
706
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Chapter 12 Review Exercises
a+b+ l = O
Multiaddplytoeachthe side the second and first ofequation to eliequati minateon by
{
4a - 2b + l = 1
a+ b = -1
4a - 2b = 0
Solve first equation secondtheequation and solve:for substitute into the a = -b - l
a,
y =5
5 + 3z = 1 1
x + 9 = 10
Thus, medium boxes, and large boxes ofsmall cookiesbox,should be purchased. Let the speed of the boat in still water, and let the speed of the river current. The distance from Chiritza to the Flotel Orellana is kilometers. Rate Time Distance trip downstream trip downstream The system of equations is: z=2 I
y=
3
x2 -
3
1 05 .
x+1 .
=
$6.00
y
x=I
5
$9.00
x = y =
{%
6x + 9 y = 6 .90(1 00) .
5/2
1 00
x-y
3
1 00
(x + y) = 1 00
3(x - y) = 1 00
Solve the first equation for Solve by substitution: 6x + 9y = 690
6,
y: y = 1 00 - x .
6x + 9(1 00 - x) = 690
1 5x + 1 5y = 600
6x + 900 - 9x = 690
1 5x - 1 5y = 5 00
-3x = - 2 1 0 x = 70 y = 1 00 - 70 = 3 0
The blend iscoffee madeandup of pounds poundsof theof the per-pound per-pound coffee. Let the number of small boxes, let the number ofof large mediuboxes. m boxes, and let the number Oatmeal raisin equation: Chocolate chip equation: Shortbread equation: 70
$6 .00$9.00-
30
y
=
z
3
5,
30x = 1 1 00 1 1 00 1 1 0 X= =3 30 0 _ 3Y = 1 00
C� )
--
1 1 0 - 3y = 1 00
=
1 0 = 3y
=
10 y=3
The speed of the boat is the speed of the current is
x + 2 y + 2z = 1 5
{
x+y
Multiply both sides sides ofofthethe second first equati on obyn by multiply both equati and add the results.
x + y = 1 00
x
2
1 00
=
x + y = 1 00
1 03 .
x + 5 + 2(2) = 1 0
3z = 6
The quadratic function is -.!. '!: Let the number of pounds of coffee that costs per pound, and let the number ofThenpounds of coffeerepresents that coststhe totalperamount pound.of in the blend. value of the blend will becoffee represented by theTheequation: Solve the system of equations:
{
-x - y - 2z = - 1 0
Substituting and solving for the other variables:
b = - '!: 3 2 1 a = - - I = -3 3
x
x + 2y + 2z = 1 5 y +3z = 1 1
4(-b - I) - 2b = 0 - 4b - 4 - 2b = 0 - 6b = 4
101.
-1
x:
x + y + 2z = 1 0
y + 3z = 1 1
1 1 0 / 3 � 36.67 km/hr ; 1 0 / 3 '" 3 . 3 3 kmIhr .
x + 2y + 2z = 1 5 x + y + 2z = 1 0 y + 3z = 1 1
707
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Chapter 12: Systems of Equations and Inequalities
1 07.
Let x the number of hours for Bruce to do the job alone, let y = the number of hours for Bryce to do theforjobMartyaloneto ,doandtheletjobz alone. the number of hours Then lIx represents the fraction of the job that Bruce does in one hour. lIy represents the fraction of the job that Bryce does in one hour. liz represents the fraction of the job that Marty doesequation in one hour. The representing working together is: Bruce and Bryce
1 09.
=
Let x theeachnumber engines produced week,ofandgasoline let y the number of diesel produced each week. The total cost is:engines C = 450x + 550y . Cost is to be minimized; thusare: , this is the objective function. The constraints 20 ::; x ::; 60 number of gasoline engines needed and capacity each week. 1 5 ::; y ::; 40 number of diesel engines needed and capacity each week. x + y 50 number of engines produced to prevent layoffs. Graph the constraints. =
=
=
: : . : . ; . : . : -: - : - : . ; ' ; ' ; ' ; ' ; ' - : ' ; ' ; ' : ' - : ' ; ' ; , " , ' , . �
1 = -3 = 0.75 -1 + -1 = -x y (4 /3) 4
The equation representing working together is: Bryce and Marty
_ The equation representing Bruce and Marty
. .
.!. + .!. = 1 _ = � = 0.625 y z (8 / 5) 6
.
working together is:
.!. + .!. = _1_ = � = 0.375 x z ( 8 / 3) 8
{y-X-II y-I
Solve the system of equations:
!
+ = 0.75 + z-I = 0.625 x-I + z-I = 0.375 u = x - I , v = y -I , u + v = 0.75 V + W = 0.625 u + w = 0.375
: : : : : : : : : : JO : : :
The comer points are (20, 30), (20, 40), (35, 1 5), (60, 1 5), (60, 40) Evaluate the objective function: Vertex Value ofC = 450x + 550y (20, 30) = 450(20) + 550(30) = 25, 500 (20, 40) C = 450(35) + 550(40) = 3 1, 000 (35, 1 5) C = 450(35) + 550(1 5) = 24, 000 (60, 1 5) C = 450(60) + 550(1 5) = 35, 250 i60, 401 C = 450i60}+ 550140) = 49, 000 The minimum cost is $24,000, when 35 gasoline engines and 1 5 diesel engines are produced. The excess capacity is 15 gasoline engines, since only 20 gasoline engines had to be delivered.
Let
C
Solve the first equation for u: u = 0.75 - v . Solve the second equation for w: w = 0.625 - v . Substitute into the third equation and solve: (0.75 - v) + (0.625 - v) = 0.375 -2v = -1 v = 0.5 u = 0.75 - 0.5 = 0.25 w = 0.625 - 0.5 = 0. 1 25
Solve for x, y, and z : x = 4, y = 2, z = 8 (reciprocals) Bruce the jobin 8inhours. 4 hours , Bryce in 2 hours , can and doMarty
708
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Chapter 12 Test 1.
{-2X + y = -7
2.
{
Chapter 12 Test
.!. X - 2Y = 1 3 5 x _ 30y = 1 8
We choosethetofirstuseequation the methodby of- 1 eli5 tomiobtai nationn theand multiply equivalent system
4x + 3y = 9
Substitution: We solve the first equation for y, obtaining y = 2x - 7 Next we substitute equation and solve this for x.result for y in the second
{-5X5x +- 30y 30Y = -1 5 = 18
replace the second equatitheonequi by thevalentsumsystem of theWe two equations to obtain
{-5X + 30Y0 == 3-1 5
4x + 3y = 9 4x + 3 (2x - 7) = 9 4x + 6x - 2 1 = 9 l Ox = 30 30 = 3 x=10
second equation is athatcontradiction anditselfhashas nonoThesolution solution. This means the system and is therefore inconsistent.
Wex = can now obtain the value for y by letting 3 in our substitution for y.
3.
y = 2x - 7 y = 2 (3) - 7 = 6 - 7 = -1
{
x - y + 2z = 5 (I) 3x + 4y - z = -2 (2) 5x + 2y + 3z = 8 (3)
We use the method of elimination and begin by eliminating the variable y from equation (2). Multiply each side of(2). equation (1) by 4 and add theour result to equation This result becomes new equation (2).
The solution of the system is x = 3 , y = - 1 or (3, -1) . Elimination: Multiply each side ofofthex infirsttheequation by 2 soare that the coefficients two equations negatives each other. The result is the equivalentofsystem
x - y + 2z = 5 3x + 4y - z = -2
{-4X + 2Y = -14
4x - 4y + 8z = 20 3x + 4y - z = -2 7x + 7z = 1 8 (2)
We now eliminate the variable y from equation (3) by multiplying each side of equation by 2 and addingourthenewresultequatito equati on (3). The result becomes o n (3). x - y + 2z = 5 2x - 2y + 4z = 10
4x + 3y = 9
We canbyreplace the ofsecond equation of thisThe system the sum the two equations. result is the equivalent system
(1)
{-4X + 2y = -14
5x + 2y + 3z = 8
5y = -5
5x + 2y + 3z = 8 + 7z = 1 8 (3) 7x
Now we solve the second equation for y. 5y = -5
Our (equivalent) system now looks like
We thisandvaluesolveforfory ix.nto the orig-2x iback-substitute nal+firsty = equation -7
Treat equations andtwo variables, as a systemandof two equations containing eliminate x variable by multiplying each side ofequation equation(3).the(2)The byresult -I and adding the result to becomes our new equation (3). 7x + 7z = 1 8 - 7x - 7z = - 1 8
{
-5 = -l y=5
-2x + ( -I) = -7 -2x = -6 -6 = 3 x=-2
The solution of the system is x = 3 , y = -l or
x - y + 2z = 5 (I) 7x + 7z = 1 8 (2) 7x + 7z = 1 8 (3) (2) (3)
7x + 7z = 1 8
7x + 7z = 1 8 0 = 0 (3)
We now have the equivalent system
(3, -1) .
709
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Chapter 12: Systems of Equations and Inequalities
{
x - y + 2z = 5 7x + 7z = 1 8 0=0
adding our the result to equation resultandbecomes new equati on The -2
(1) (2) (3)
3x + 2y - 8z = -3 6x - 3y + 1 5z = 8
This is equivalent to aSisystem ofoftwothe equati oonsns with three variables. n ce one equati contains three variables andwilonel becontains only two vari a bles, the system dependent. There infinitely manyfor solutions. We solveare equation and determine that . . SubstI. tute th'1S expreSSlOn. mto equation ( 1 ) to obtain in terms of (2)
x = -z +
(2)
7
1
z.
-5z = 0 z=O
Back-substitute solve for
_ y + 2z = 5
z=0 y. -7y + 3 lz = 1 4 -7y + 3 l (0) = 1 4 -7y = 1 4 y = -2
18 -z + - - y + 2z = 5 7 17 -y + z = 7 17 y = z-7 18 17 · x = -z + - , Y = z - - , 7 7
The so utlOn 1S is any real number or { \ ; is any real number} . 1 ·
z
{
4.
1
\
8
{
,
x=
(3)
5.
(2) 3.
We of elimination and begin byThe elicoefficients museinatitheng method theon variable from equation inother equations ( 1 ) and arethe two negatives of each so we si m ply add equations together. This result becomes our new equation x
z=0
into
, y = -2 , z = o
{
4X - 5Y + Z = 0 -2x - y + 6 = - 1 9 x + 5y - 5z = 1 0
{
O. 4x - 5y + z = 0 -2x - y + Oz -25 x + 5y - 5 z = 1 0
, -2, 0
We first terms check arethe onequations tosidmake sureequati that oalln variable the left e of the and the constants areweonputtheitriinghtwithside.a coeffici If a ent variable i s missing, of Our system can be rewritten as
3x + 2y - 8z = -3 ( 1 ) -3x - 2y + 3z = 3 (2) 6x - 3y + 1 5z = 8 (3)
(2).
x
and
x.
The solution of the original system is � or (� }
(2)
start by clearing the fraction in equationby byWemultiplying both sides of the equation 6x - 3y + 1 5z = 8
(3)
3x + 2y - 8z = -3 3x + 2(-2) - 8(0) = -3 3x - 4 = -3 3x = 1 1 X=3
3x + 2y - 8z = -3 ( 1 )
t
into equation and
Finally, back-substitute equation ( 1 ) and solve for
,z
-x - y + z = 1
z -5 .
y = -2
(X, y, Z) x = -z +
y = z-
3X + 2Y - 8Z = -3 ( 1 ) -5z = 0 (2) -7y + 3 1z = 14 (3)
We sidessolve of theequation equation byfor by dividing both
18
-
( ;) -z +
{
- 7y + 3 1z = 1 4 (3)
Our (equivalent) system now looks like
x
y x - y + 2z = 5
(3). (3). - 6x - 4y + 1 6z = 6 6x - 3y + 1 5z = 8
(2)
=
The augmented matrix is
(2) . 3x + 2y - 8z = -3 -3x - 2y + 3z = 3
[-� =� � � 1
- 5z = 0 (2)
Webynowmultiplying eliminate theeachvariable fromoequati side of equati n ( 1 ) byon
1
x
(3)
5
-5
- 5 10
710
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Chapter 12 Test
6.
The matriwithx hasthreethreeequations. rows andTherepresents aumns system three col tosystem the lehasft ofthree the vertical barWeindicate thatx, y,theand Z variables. can let these variables. The column to the rionghtthe ofridenote the vertical bar represents the constants ght side of the equations. The system is 3x + 2y + 4 z = -6 3X + 2 Y + 4Z = -6 Ix + 0Y + 8z = 2 or x + 8z = 2
{
7.
-2x + ly + 3z = -1 l
3
2
3
-1
8
3
8
4
=
- 1
5
3
2
-3
24
=
12
6
1 2.
-22
[ ] -+ [� : ;] [ 1 ] Therefore, [ =
2 _�
[ A I 12 ]
=
( R2 = -5 lj + r2 )
(R, +, )
�l] .
(R, = -tr2 + 'i )
We first form the matrix [B I 13 ] =
cannot be computed because the dimensions are mismatched. To multiply two matrices, wex to need the number of columns in the first matri bematrix. the same asx Athehasnumber of rows in the second Matri 2 columns, but matrix has 3 rows. Therefore, the operation cannot be performed. Here we are taking the product of a 2 3 matrix and a 3 x 2 matrix. Since the number of columns inrowstheinfirstthematri x ismatri the same as the number of second x (3 in both cases), the operati on can be performed and will result in a 2 x 2 matrix. AC
[AI12]=[� ! I � �] Next we use row operations to transform into reduced row echelon form. [� 1� I -� �] ° � -----" 5 4 [ I -oj 1 ] (R, 1 lj )
£'
11
�
We first form the matrix
1 -� 11, 0 -----" 0 _ _ 1 � t _% 1 0 2 -1 -----" 0 1 _ _� 1
A - 3C "
=
9.
-2x + y + 3z = - 1 1
[� =�] + [; � ] = + l �] [� =�l [-;I � [� [� �]-t �1 [� =�l -[I: �!l [ 3 -�9l
2A + C = 2
6
8.
{
1 1.
[� �l �l � � �l 2 3
0 0 ° 1
N ext we use row operations to transform [B I 13 ]
C
1 0.
x
BA
�][� �j [ ]
"[� [
�2
= 1 ' 1 + (-2) . 0 + 5 ' 3 1 ' (- 1 ) + (-2 ) ' (-4) + 5 ' 2 0·1+3·0+1·3 0 · (-1) + 3 (--4) + 1 . 2
=
©
16 17 3 -10
] 71 1
2008 Pearson Education, I nc . , Upper Saddle River, NJ.
All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
�[�
0
1
3
0
3
71
-2
-2
0
[Rl :��r3 1
1
-5
-4
3
-2
B-' =
0'
� - 7 r3 + r2
[� Tl {6X+3Y
Thus,
0
ofequatis,othen insystem actually The consists of one two variables. system isnumber dependent and therefore has an infinite of solutions. Any ordered pair satisfying the equation or is a solution to the system.
+�
-5
1 3.
1 5.
= 12
2x - Y = -2
We start by writing the augmented matrix for the system.
2
12
-1
-2
3
12
-1
-2
-? ----'-
-r
� -,-
-r
-r
----'-
2
1
3
12
-
-!-
3
1
--!-
o
- -!-
-1
0
1
3
1 2 3
0 1
=
=
x = -!- '
4
+
1 13
7
1 13
o
3
3 _ 2 -10
-?
r +�
=3
7
2
4
-2 0
1
row represents the equation . Using back-substitute into the equation (from the second row) and obtain
8X + Y = 56
z=0
] [� ; I :6
y + 3z = -2 y + 3(0) = -2 y = -2
Next we usematri rowxoperations to transform augmented into row echelon form.the
Using and the equation and obtain y = -2
G ; I : ] R2 = RI r2 � I �] matrix is now in row echelon [The� augmented -8
7
[� � � � ] �[� !I i3 �l [� � ; -6=�] ->[� � } -3��] 3 ->[�matri� x is now]in row echelon form. The last The
We start by writing the augmented matrix for the system.
6
3 12 -10
Next we usematri rowxoperations to transform augmented into row echelon form.the
7
�
x + 2y + 4z = -3 2x + 7y + 1 5z = - 1 2 4x + 7y + 1 3z = - 1 0
4
12
18
= -4x + 28 ,
[� � � � ]
-1
1
----'- 1 0
x + .!.
-2
6
-r
14.
-1
{
4
We start by writing the augmented matrix for the system.
1 ] [� we use row operations to transform the Next augmented matrix into row echelon form. [6 1 ] [6 1 ] (RR2I == r2fj ) [ 1 ] (RI = t fj ) [ 6 1-1] (R2 -6fj rJ [[ 1 ] (R2 = i r2 ) 1 ] (R2 t 2 ) The solution of the system is Y or (-!- , 3) { Y= 3
Y Y
x + .!. = 7 ,
z=0,
z 0 y + 3z = -2 =
we
we back-substitute into (from the first row)
x + 2y + 4z = -3
+
x + 2y + 4z = -3
x + 2(-2) + 4(0) = -3
The solution is
form. Because the bottom row consists entirely
x=1 x = l , y = -2 ,
z = O or
712
(1, -2, 0) .
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1 6.
Chapter 12 Test
{2X+2Y-3 =5 Z x-y+2z = 8
1 9.
3x+5y-8z = -2 We start by writing the augmented matrix for the system. �1 -; 3 5 -8 -2
[� : 1 [�3 �15 -8�3 -2�] 1 = [� � �3 :] 3 5 -8 -2 1 =[ 74 � : � =u = [� : �:� =��j 1 = [� � -:t j'l
Next we usematri rowxoperations to transform augmented into row echelon form.the
20.
(R, =*" )
The last row represents the equation 0 = -4 whichno isolution s a contradiction. has and is be Therefore, inconsistent.the system 1 7.
1-32 75 1 =(-2)( 7 )-(5)(3)=-14-15 =-29 2 6 4 0 -1 2 -4 - 2 1: �41 -(-4) 1� 1 �41 +6 1� 1 :1 = 2[4(-4)-2(0)]+4[1(-4)-(-1)(0)] + 6[1(2) -(-1)4] = 2(-16)+4(-4)+6(6) =-32-16+36 =-12 -4
1 8.
4X+3Y = -23 { 3x-5y = 19 The determinant variables is D of the coefficients of the D = 1 43 -53 1 = (4)( -5)-(3)(3) = -20-9 = -29 Since D;t; 0 , Cramer's Rule can be applied. Dx = 1��3 �51 = (-23)( -5)-(3)(19) = 58 Dy =1 ; ��31 =(4)(19)-(-23)(3)=145 x = DxD =�= -2 -29 145 =-5 y = DDy = -29 The solution of the system is x = -2 , y = -5 or (-2,-5) . 4X-3Y+2Z = 15 -2x+ y-3z = -15 5x -5y + 2z = 18 The determinant variables is D of the coefficients of the 4 -3 2 D = -2 -3 5 -5 2 = 4 1�5 -;1 - ( _3) 1 �2 -;1 +2 1-52 �51 = 4(2-15) + 3( -4+ 15) + 2(10 -5) = 4( -13)+3(11)+2(5) =-52+33+10 =-9 Since D ;t; 0 , Cramer's Rule can be applied. 15 -3 2 Dx = -15 -3 18 -5 2 -_ 15 1 -51 -32 1 -(-3) 1-1518 -32 1 + 2 1-1518 -51 1 = 15(2 -15) + 3( -30 + 54) + 2 ( 75 -18) = 15( -13)+ 3(24)+ 2( 5 7 ) =-9
{
713
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Chapter 12: Systems of Equations and Inequalities
4 15 2 Dy = -2 - 1 5 -3 5 18 2 -1 5 -3 _ - 2 -3 -2 - 1 5 +2 15 =4 18 2 5 2 5 18 = 4 ( -30 + 54) - 1 5 (-4 + 1 5) + 2 ( -36 + 75) = 4(24) - 1 5 (1 1) + 2 (39) = -9 4 -3 1 5 Dz = -2 -15 5 -5 1 8 2 5 5 _ 2 =4 +15 _ ( 3) 5 5 = 4 (1 8 - 75) + 3 ( -36 + 75) + 1 5 (10 - 5) = 4( -57) + 3 (39) + 1 5 (5) = -36 D 9 Dx = -9 = 1 , y = _ x=_ Y = - = -1 , D -9 D -9 z = Dz = -36 = 4 D -9 x = 1 , Y = -1 , z = 4 (1, -1, 4) .
1
1 1
1 1
22.
1
�
y = x+l
solSubsti ve fortutex:x + for y into the first equation and 1
2 (x + l) 2 - 3x 2 = 5 2 (x2 + 2x + l) - 3x2 = 5 2X 2 + 4x + 2 - 3x 2 = 5 _x 2 + 4x - 3 = 0 x 2 - 4x + 3 = 0 (x - l)(x - 3) = 0 x=1 x=3
orthese values into the second Back substi t u t e equati on to determine y: x=l : y = I+I = 2
1� �� 1 1� �� 1 1� � 1
x = 3 : y = 3+1 = 4
The solutions of the system are (1, 2) and (3, 4) .
23.
The solution of the system is or
21.
{2y2 -y3X2- x == 51
{
{4xX2 +- l3y ::;� 1000
Graph the circle x2 + l = 100 . Use a solid ity usesas (0,::; .0).Choose poicurvent notsin ceonththee inciequal rcle , such Sincea test ci02r cle02as::;(0,1000);isthtrue, at is ,shade insidethethesame cir clesi.de of the Graph the line 4x - 3y = Use a solid line since thethe ilninequali a test- 3(1) poin t not0 ison e, suchtyasuses(0, 1).Choose Sin ce 4(0) fal(0,se,I).shade the opposi side ofisthethe lsolineufrom tion. The overl appingteregion +
3X 2 + l = 12 l = 9x 9x l x: 3x2 + (9x) = 1 2 3x2 + 9x - 1 2 = 0 x 2 + 3x - 4 = 0 (x - l )(x + 4) = 0 x=1 x = -4
Substitute for into the first equation and solve for
2:
o.
.
�
y 12
or Back substi ute thesene values into the second equati on to tdetermi y: x = 1 : l = 9(1) 9 y = ±3 x = -4 : l = 9(-4) = -36 y = ±.J-36 =
(n ot real) The solutions of the system are (1, -3) and (1, 3) .
4x - 3y = 0
714
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Appendix: Graphing Utilities
11.
9.
XMin=-1(a XMax=20 Xsc.l=2 YMin=-1500 YMax=500 Ysc.l=100 Xres:;:l
Zt�O �=.31n09B9
Y=O
The positive x-intercepts are 0.32 and 1 2.30. 13.
Answers may vary. A possible answer follows: Xmax -Xmin = 8 - ( -4) = 1 2 We want a ratio of 3:2, so the difference between Ymax and YmiD should be 8. In order to see the point (4,8) , the Ymax value must be greater than 8. We might choose Ymax = 1 0 , which means 1 0 - Ymin = 8 , or YmiD = 2 . Since we are on the order of 1 0, we would use a scale of 1. Thus, YmiD = 2 , Ymax = 1 0 , and Y.c1 = 1 will make the point (4,8) visible and have a square screen.
XMin=-50 XfVlax=50 Xsc.l=10 YMin=-10000 YfVlax=25000 Ysc.l=5000 Xres=l
The positive X-intercepts are 1 .00 and 23.00. Section 5
Problems 1-8 assume that a ratio of 3:2 is required for a square screen, as with a TI-84 Plus. 3 -( -3) 6 3 2 -(-2) 4 2 A ratio of 3:2 results in a square screen.
3.
5.
7.
9- 0 9 3 4- (-2) 6 2 Ymax - YmiD A ratio of 3:2 results in a square screen. Xmax -Xmin
Xmax -Xmin = Ymax-Ymin
6- (-6) g
= 3 2 -(-2) 4 A ratio of 3: 1 results in a screen that is not square. =
Xmax -Xmin
9-0 9 3 4 - (-2) 6 2 A ratio of 3:2 results in a square screen. 760
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Section 3: Using a Graphing Utility to Locate Intercepts and Check for Symmetry
3.
27.
XMin=-4 XMax=l Xscl=l YMin=-4 YMax=4 Yscl=l Xres=l
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,-7), (-2,-2), and (- 1, 1). Ztr9 x= -1.707107
29.
y=o
The smaller x-intercept is roughly -1.7 1 . 5.
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,7.5), (-2,6) , and (- 1,4.5).
XMin=-5 XMax=5 Xscl=l YMin=-5 YMax=5 Yscl=l Xres=l
31.
The smaller x-intercept is roughly -0.28.
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,- 1.5), (-2,0), and (- 1, 1.5).
7.
XMin=-8 XMax=8 Xscl=l YMin=-20 YMax=50 Yscl=5 Xres=l
Section 3
1.
1JJ�����-6
\
X�lax=4 Xscl=l YMin=-5 YMax=5 Yscl=l Xres=l
\
V
"--./
The positive x-intercept is 3.00.
j
9.
"--./
Ztr9 X=-3.�1�t;1�
y=o
The smaller x-intercept is roughly -3.4 1.
XMin=-8 XMax=8 Xscl=l YMin=-50 YMax=350 Yscl=50 Xres=l
The positive X-intercept is 4.50.
759
© 2008 Pearson Education, me., Upper S addle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of thi s material may be reproduced , in any form or by any means, without permi ssion in writing from the publisher.
Appendix: Graphing Utilities c.
8
17.
10
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,-1), (-2,0) , and (-1,1).
-8 20
d.
19. 5
-5
15.
- 20
a.
Each ordered pair from the table corresponds to a point on the graph . Three points on the graph are (-3,5), (-2,4) , and (-1,3). 21.
5 -4
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,-4), (-2,-2) , and (-1,0).
8
b.
10
- 10
23.
-8
c.
8
-10
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,8), (-2,6) , and (-1,4) .
10 25.
d.
-8
20
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,11), (-2,6) , and (-1,3).
5
758
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Section 2: Using a Graphing Utility to Graph Equations c.
11.
•
"',"
"
J\.
P:' v' :,
a.
"110
-
5.
-4
-8 d.
b.
- 5. ,. ';
9.
"
. ...,.... 'uc'
"'
: 15
I' ' ( .,
-5 I
� ,
""'� . n: .
,L
:lJ r.,.'.-
,
I 10
'c 'I 5
10 -8
-4 d.
b.
-5. "in _ ;....q--.. ..
10 -8
c.
-10 V"
.
'. j'c,
13.
.. ..
5
- 20
a.
-5t�;
110
,
t '<J )' '\:'.,
15
-4
-8
d.
'
-8
c.
a.
5
l '1, �'m '
'
2U
b. - 10
5
' 10 -8
757
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently
exist. No portion of thi s material may be reproduced, in any form or by any means , without permi ssion in writing from the publisher.
Appendix: Graphing Utilities c.
5.
-
10
a.
10
4
-8
-5
3.
-4
20
d.
b. 5
5
5
10 -8
c.
4
-
8
-10
- 20
a.
5
-5
8
, j
-10
10
-4
8
b.
d.
-)()
-8 8
c.
7.
-10
d.
-5
10
���
-20
a.
to
4
-5
20
5
8
b. -1 0
5
10 -8
756
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of thi s material may be reproduced, in any form or by any means , without permi ssion in writing from the publi sher.
Appendix Graphing Utilities 1 5.
Section 1 1. 3. 5.
7.
9.
11.
13.
( -1, 4) ; Quadrant II
(3, 1) ; Quadrant I Xmin = -6 Xmax = 6 X sci = 2 Ymin = -4 Ymax = 4 Y sci = 2
17.
Xmin = - 1 0 X max = 1 1 0 X sci = 10 Ymin = - 1 0 Ymax = 160 Y sci = 1 0
fi =(1,3);P =( 5,1 5) 2 d (F; ,12 ) = �,-(5---1)-2 +- (-1 5- --3)-2 = � (4) 2 + (12) 2 = .J16 + 144 = .J160 = 4.jlo
Xmin = -6 Xmax = 6 X sci = 2 Ymin = -1 Ymax = 3 Y sci = 1
1 9.
fi =(-4,6);12 =( 4,-8)
d (fi, 12 ) = �,-(4---(--4-)) 2- +- (-_ 8- _-6)--:- 2 = �(8) 2 + (-14) 2 = .J64+ 196 = .J 260 = 2.J6s
Xmin = 3 Xmax = 9 X sci = 1 Ymin = 2 Ymax = 1 0 Y sci = 2
Section 2
Xmin = -1 1 Xmax = 5 X sci = 1 Ymin = -3 Ymax = 6 Y sci = 1
1.
4
a.
-4
Xmin = -30 X max = 50 X sci = 1 0 Ymin = -90 Ymax = 50 Y sci = 10
8
b. -\0
755
10
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Chapter 14: Counting and Probability
9.
x; 2-3
Multiply each side of the first equation by and add to the second equation to eliminate multiply each side of the first equation by and add to the third equation to eliminate
{3x+x- y-2y+3zz == -815
x:
-2x+4y- z = -27 -3x+6y-3z = -45 3x+ y - 3z = -8 7y-6z = -53 x- 2y+ z = 15 � 2x-4y+2z = 30 -2x+4y- z = -27 -2x+4y-z = -27 z=3 Substituting and solving for the other variables: z = 3:::> 7y-6 ( 3 ) = -53 7y = -35 y = -5 z = 3,y = -5:::> x-2( -5)+3 = 15 x+1 O +3 = 15:::> x = 2 The solution is x = 2, y = -5, z = 3 or (2, -5, 3) . �
11.
= 3 sin ( 2x + ll' ) = 3 sin ( 2 ( x + ; )) Amplitude: 1 A 1 = 131 = 3 21t T =-=ll' Period: 2 ll' ¢ -ll' . Phase Shift: (() = -2 = --2 Y
y
x
754
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Chapter 14 Cumulative Review
Chapter 14 Cumulative Review 1.
3x2 - 2x = -1 3x2 - 2x+l = 0 -b±Jb2 - 4ac -x = ---2a -(-2) ± �(_ 2) 2 - 4(3)(1) 2 ± v'4=12 2(3) 6 2±.,/-8 2± 2.[ii l± .[ii 3 6 6
. . {l-
The solutIOn set IS 3.
Xl"lin=-l Xl"lax=4 Xscl=l Yl"lin=-H30 Yl"lax=20 Yscl=20 Xres=l
hi 1 + hi 3 3
-- ' --
}
Step 4: From the graph we see that there are x-intercepts at -0. 2 and 3 . Using synthetic division with 3: 3)5 -9 - 7 - 3 1 - 6 1 5 1 8 33 6 2 5 6 11 o Since the remainder is 0, x- 3 is a factor. The other factor is the quotient: 5x3 +6x2 + 1 1x+ 2 . Using synthetic division with 2 on the quotient: -0.2) 5 6 1 1 2 -1 -1 -2 5 5 10 0 Since the remainder is 0, x- ( -0.2) = x+ 0.2 is a factor. The other factor is the quotient: 5x2 + 5x + 1 0 = 5 (x2 + + 2 ) .
.
y = 2 (x+1) 2 - 4 Using the graph of y = x2 , horizontally shift to the left 1 unit, vertically stretch by a factor of 2, and vertically shift down 4 units. y
x
-5
(-2, -2)
(-1. -4)
-5
5.
X
Factoring, f(x) = 5(x2 + + 2) (x - 3)(x + 0.2) The real zeros are 3 and - 0.2. The complex zeros come from solving x2 +x+2 = o. - 1± �1 2 - 4 ( 1)(2) 2 x = -b±Jb - 4ac 2 ( 1) 2a -1±v1-8 -1±H 2 2 - 1 ±.J7i 2 Therefore, over the set of complex numbers, f(x) = 5x4 -9x3 - 7x2 - 3 1x-6 has zeros
f(x) = 5x4 -9x3 - 7x2 - 3 1x -6 Step 1: f(x) has at most 4 real zeros. Step 2: Possible rational zeros: p = ± 1,± 2,±3,±6; q = ±1,± 5; 2 3 6 1 2,± -,±3,±-,±6,± P = ± 1,±-,± q 5 5 5 5 Step 3: Using the Bounds on Zeros Theorem: f(x) = 5 (x4 - 1. 8x3 -1.4x2 -6. 2x - 1 .2 ) a = -1 .8, a = - 1 .4, at = -6.2, ao = - 1 .2 2 3 Max { 1,1- 1 . 21+1- 6.21 + 1 - 1.4 1+1 - 1 .8 1} = Max {I, 1 0.6} 1 0.6
X
------
=
{_1..2 +.J72 i' _1..2 _ .J72'i _1..5' 3}.
1 + Max {I - 1 . 2 1 ,I -6.21 ,1 - 1.41 ,1 - 1 .81} = 1+6. 2 = 7.2 The smaller of the two numbers is 7.2. Thus, every zero of f lies between -7.2 and 7.2. Graphing using the bounds: (Second graph has a better window.)
7.
log (9) = log (3 2 ) = 2 3 3
753
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Chapter 14: Counting and Probability
11.
12.
310
4
26
We are choosing letters from distinct letters and digits from distinct digits. The letters and numbers are placed in order following the format LLL DDDD with repetitions being allowed. Using the Multiplication Principle, we get Note that there are only possibilities for the third letter. There are possible license plates using the new format. Let A Kiersten accepted at USC, and B Kiersten accepted at FSU. Then, we get P (A) P(B) and
=
1 5.
26· 26·23 ·10 ·10·10·10 = 155,480,000 155,23 4 80,000
6 . =C (53,5 ) ·C (42,1 )
53! 42! 5!·48! 1!·4 1! 53.52.5 1.50.49 42.4 1 ! 4 1! 5·4·3·2·1 53·52·5 1 ·50·49·42 5·4·3·2
=(
a.
=
1 3 . a.
p( win on $1 play )
1-0. 70 = 0. 3 0
Kiersten has a chance of not being admitted to FSU.
16.
Since the bottle is chosen at random, all bottles are equally likely to be selected. Thus,
5 =-5 =-1 = 0.25 8+5+4+3 20 4 There is a 25% chance that the selected bottle contains Coke. 11 0. 5 5 P( Pepsl. U IBC ) = 8+3 = -= 8+5+4+3 20 There is a 55% chance that the selected bottle contains either Pepsi or IBC
$1
1 120,526,770 "" 0. 0000000083 =
The number of elements in the sample space can be obtained by using the Multiplication Principle:
6·6·6·6·6 = 7,776 C 5,2
2
55
5·5·5 = 125 2 . 125 = 5·4·125 C ( 5, 2 ) . 125 = � = 1250 2! ·3! 2 The probability of getting exactly 2 fours on 5 rolls of a die is given by 1250 p(exactly 2 fOurs ) = -7776 "" 0.1608 .
.
1 4.
)
Consider the rolls as a sequence of slots. The number of ways to position fours in slots is ( ) . The remaining three slots can be filled with any of the five remaining numbers from the die. Repetitions are allowed so this can be done in different ways. Therefore, the total number of ways to get exactly fours is
P( Coke ) =
b.
120,526,770
)(
Since each possible combination is equally likely, the probability of winning on a play is
to at least one of the universities. b. Here we need the Complement of an event.
30%
n
53! 42! 5!·( 53-5) ! ]!. ( 42-1 ) !
= 0. 60 , = 0. 7 0 , P ( AnB ) = 0.35 . Here we need to use the Addition Rule. P(A UB) = P(A)+P(B)-P(AnB) = 0. 60+0. 7 0-0. 3 5 = 0. 9 5 Kiersten has a 95% chance of being admitted =
1
n
=
p('B) = I-P(B)
65
The number of different selections of numbers is the number of ways we can choose white balls and red ball, where the order of the white balls is not important. This requires the use of the Multiplication Principle and the combination formula. Thus, the total number of distinct ways to pick the numbers is given by (white balls ) (red ball )
Since the ages cover all possibilities and the age groups are mutually exclusive, the sum of all the probabilities must equal
0.1-0.03 8 1=0.19 0. 23+0.29 +0. 25 +1. 0. 0 1 = 0. 8 1 The given probabilities sum to 0. 8 1. This means the missing probability (for 18-20) must be 0.19. +
752
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Chapter 14 Test
33. a.
b.
c.
35. a.
b.
37.
39.
365· 364· 363····348 = 8 .6 3462 8387 x1045
P(10,6 ) =
7.
C (11, 5 ) =
P(no one has same birthday) 348 = 365· 364· 363··· 0.6 531 365 1 8 �
P(at least 2 have same birthday) = 1 - P(no one has same birthday) = 1 - 0.6 531 = 0.3469 P(unemployed) = 0.058
2.
P($1 bill) n($1 bill) = .i 9 n(S)
8.
Let S be all possible selections, so n(S) = 100 . Let D be a card that is divisible by 5, so neD) = 20. Let PN be a card that is 1 or a prime number, so n(PN) = 26 . 20 = .!. = 0.2 P(D) = neD) = n(S) 100 5 n(P P(PN) = N) = 26 = � = 0.26 n(S) 100 50
9.
From the figure: n( biology chemistry physics ) = 22 + 8 + 2 + 4 +9+ 7 +15 = 67 Therefore, n( none of the three ) = 70 -67 = 3 or
n ( biol.
and
8·7·6·5·4! 4!· 2·1
8·7 . 6 . 5 = 4 . 7 .6 . 5 2
= 840 There are 840 distinct arrangements of the letters in the word REDEEMED.
chem. ) -n ( biol. and chern. and phys. )
10.
From the figure: n( physics chemistry ) = 4 +2 + 7 +9+15 +8 = 45 or
5.
8! 4!2!l!l!
--= ----
=
= ( 8+2 ) -2 = 8
4.
Because the letters are not distinct and order matters, we use the permutation formula for non distinct objects. We have four different letters, two of which are repeated (E four times and D two times) . n!
From the figure: n( only biology and chemistry ) =
Since the order in which the colors are selected doesn't matter, this is a combination problem. We have n = 21 colors and we wish to select = 6 of them. 21! 21! C ( 21,6 ) = 6!( 21 -6 )! 6!15! 21· 20·19·1 8·17 ·16·15! 6!1 5! 21·20·19·1 8·17 ·16 6· 5· 4· 3·2·1 = 54, 264 There are 54,264 ways to choose 6 colors from the 21 available colors. r
From the figure: n( physics ) = 4 +2 + 7 +9 = 22 or
3.
11! = J..!.!.. 5!(1 1 - 5)! 5!6! 11·10·9· 8· 7 ·6! 5· 4· 3·2·1·6! 11·10·9· 8· 7 5· 4· 3·2·1 = 462
P(not unemployed) = 1 - P(unemployed) = 1 - 0.058 = 0.942
Chapter 1 4 Test 1.
10! = 10! (1 0 -6)! 4! 10·9· 8· 7·6· 5· 4! 4! =10·9· 8· 7·6· 5 = 1 51,200
6.
Since the order of the horses matters and all the horses are distinct, we use the permutation formula for distinct objects. 87 ! P( 8,2 ) = _8_ _ = 8! = · ·6! = 8 . 7 = 56 ( 8 - 2 ) ! 6! 6! There are 56 different e a t bets for 8-horse race. x c a
7! = 7 ·6· 5· 4· 3 .2 ·1 = 5040 751
an
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Chapter 14: Counting and Probability
69. a.
P(freshman or female)
=
7.
P(freshman) + P(female) - P(freshman and female) n(freshman) + n(female) - n(freshman and female)
9.
n(S)
1 8+1 5 - 8 25 33 33 P(sophomore or male)
b.
=
P(sophomore) + P( male) - P(sophomore and male) n(sophomore) + n(male) - n(sophomore and male)
71.
73 .
25 33
C(8, 3) =
1 7.
1 9.
=
.
-=
There are 2 choices of material, 3 choices of color, and 10 choices of size. The complete assortment would have: 2 · 3 · 1 0 = 60 suits. There are two possible outcomes for each game or 2 · 2 · 2 · 2·2 · 2 · 2 = 27 = 1 28 outcom for 7 games.
Since order is significant, this is a permutation. 7 P(9, 4) = _9_! _ 9! = 9 · 8 · · 6 · 5! =3024 (9 - 4)! 5! 5! ways to seat 4 people in 9 seats. =
0 .167
21.
23.
25.
Chapter 14 Review E xercises
5.
8! = � 8· 7 6 . 5! 56 ( 8 - 3)! 3! 5! 3! 5!·3·2 · 1
s
27.
3.
= n(A u C) = 20+ 5 = 25
13.
-
1.
n(neither in A nor in C)
8! 8 . 7 . 6 . 5! = 3 36 8! P(8, 3) = _ _ _ = = (8 - 3)! 5! 5!
1 5.
peat least 2 with same birthday) = 1 - P(none with same birthday) = 1 - n(different birthdays) n(S) · 363·362·361 · 360 .. · 354 = 1 - 365·364 36512 "" 1 - 0.833 = The sample space for picking 5 out of 10 numbers in a particular order contains 0 10! = 10 ! = Pl( ,5) = 30, 240 POSSI'ble (10 - 5)! 5! outcomes. One of these is the desired outcome. Thus, the probability of winning is: neE) = n(winning) peE) = n(S) n(total possible outcomes) 1 = __ "" 0.000033069 30, 240
From the figure:
11.
n(S)
15+ 1 8 - 8 33
From the figure: n(A and C) = n(An C) = 1+6 = 7
Choose 4
ers --order is significant: 8 7 · · 5·4! = 8! P(8 , 4) = __ = � = · 6 1680 (8 - 4)! 4! 4! ways a squad can be chosen. Choose 2 teams from 14-order is not significant: 14! = � = 14 . 13 . 12! =9 1 C (1 4 , 2) = ( 1 4-2)!2! 1 2! 2! 12! · 2 · [ ways to choose 2 teams. runn
_
There are 8·10·1 0·10·10 ·10· 2 1,600, 000 possible phone numbers. =
There are 24 · 9 · 1 0 · 1 0 · 10 = 2 16, 000 possible license plates.
29. Since there are repeated letters: 7''- = 7 ·6 ·5 · 4 · 3·2 · 1 = 1 260 2! · 2! 2·1·2 · 1 can be formed .
{ Dave } , {Joanne} , {Erica} , { Dave, Joanne} , {Dave, Erica} , {Joanne, Erica } , {Dave, Joanne,Erica } 0,
3 1 . a.
n(A) = 8, nCB) 12, n(An B) = 3 n(A u B) = n(A)+ nCB) - n(An B) = 8+1 2 - 3 = 1 7 =
different words
C(9, 4) · C(9, 3) · C(9, 2) = 126·84·36 = 38 1 024 committees can be formed. C(9, 4) · C(5, 3) · C(2, 2) = 1 26·1 0·1 = 1260 committees can be formed. ,
b.
6
From the figure: n(A) = 2 0+ 2 + + 1 = 29 750
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Section 14.3: Probability
neE) = n {2, 4,6, 8, 1 0} = 2- = ! n(S) 10 10 2
55.
33.
P(E) =
35.
5 5 = -1 P(white) = n(white) = n(S)· 5+1 0+8+7 3 0 6
37.
The sample space is: S {BBB,BBG, BGB, GBB,BGG,GBG,GGB,GGG} =
P(3 boys) =
39.
n(3 boys) = .!. n(S) 8
43.
45.
47. 49.
51.
53.
59.
The sample space is: S {BBBB, BBBG, BBGB, =
BGBB,GBBB, BBGG, BGBG,GBBG,BGGB, GBGB,GGBB,BGGG,GBGG, GGBG,GGGB, GGGG}
P(1 girl , 3 boys) = 41.
57.
n(l girl, 3 boys) = � = .!. 16 4 n(S)
61.
d
63 .
3)
n(sum of two ice is P(sum f two d·Ice IS. 3) = --'--------'n(S) n { 1 ,2 or 2, 1 } 2 � = n(S) 36 1 8 0
d
P(white or green) = P(white)+ P(green) n(white) + n(green) n(S) 9+8 17 9+8+3 20 P(not white) = 1 - P(white) = 1 n(white) n(S) � = 1 - = .!..!. 20 20 P(strike or one) = P(strike)+ P(one) n(strike) + (one) n(S) 3+1 -4 8 8 2 n
n(sum of two ice is7) P(sum f two d·Ice IS. 7) = ----'-------'n(S) n {1 ,6 or 2,5 or 3,4 or 4,3 or 5,2 or 6, 1 } 6 = -1 n(S) 36 6 0
P(never gambled online) = 1 - P(gambled online) = 1- 0.05 = 0.95
There are 30 households out of 1 00 with an income of $30,000 or more. peE) = neE) = n(30, 000 or more) = � � n(S) n(total households) 100 10 =
65.
peA U B) = peA)+ PC B ) - peAn B) = 0.25+ 0.45 - 0. 1 5 = 0.55
There are 40 households out of 1 00 with an income of less than $20,000. peE) = neE) = n(less than $20,000) = � = � n(S) n(total households) 100 5
67. a.
peA U B) = peA)+ PCB ) = 0.25+ 0.45 = 0.70
b.
P(AuB) = P(A)+P(B)-P(An B) 0.85 = 0.60 + PCB)-0.05 PCB) = 0.85 - 0.60+0.05 = 0.30
c.
d.
P(theft not cleared) = 1 - P( theft cleared) = 1-0. 1 3 = 0.87
e.
f.
P(does not own cat) = 1 - P(owns cat) = 1 - 0.34 = 0.66
g.
h.
P(l or 2) = P(1)+ P(2) = 0.24 + 0.33 = 0. 57 P(1 or more) = 1 - P(none) = 1- 0.05 = 0 .95 P(3 or fewer) = 1 - p( 4 or more) = 1- 0.17 = 0.83 P(3 or more) = P(3)+ P(4 or more) = 0.2 1 + 0.1 7 = 0.38 P(fewer than 2) = P(O) + = 0.05+ 0.24 = 0.29 P(fewer than 1) = P(O) = 0.05 + P(3) P(l, 2, or 3) = P(l) + = 0.24+0.33+0.2 1 = 0.78 P(2 or more) P(2) + P(3) + P(4 or more) = 0.33 + 0.2 1 + 0.1 7 = 0.7 1
P(I)
P(2)
=
749
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Chapter 14: Counting and Probability
1 3 . The sample space of tossing two fair coins and a
2 1 . The sample space is:
fair die is: S = {HH1, HH2, HH3, HH4, HH5, HH6,
{I 1 Yellow, 1 1 Red, 1 1 Green, 1 2 Yellow, 1 2 Red, 1 2 Green, 1 3 Yellow, 1 3 Red, 1 3 Green, 1 4 Yellow, 1 4 Red, 1 4 Green, 2 1 Yellow, 2 1 Red, 2 1 Green, 2 2 Yellow, 2 2 Red, 2 2 Green, 2 3 Yellow, 2 3 Red, 2 3 Green, 2 4 Yellow, 2 4 Red, 2 4 Green, 3 1 Yellow, 3 1 Red, 3 1 Green, 3 2 Yellow, 3 2 Red, 3 2 Green, 3 3 Yellow, 3 3 Red, 3 3 Green, 3 4 Yellow, 3 4 Red, 3 4 Green, 4 1 Yellow, 4 1 Red, 4 1 Green, 4 2 Yellow, 4 2 Red, 4 2 Green, 4 3 Yellow, 4 3 Red, 4 3 Green, 4 4 Yellow, 4 4 Red, 4 4 Green} There are 48 equally likely events and the
S
HTl, HT2, HT3, HT4, HT5, HT6, TH1, TH2, TH3, TH4, TH5, TH6, TTl, TT2, TT3, TT4, TT5, TT6} There are 24 equally likely outcomes and the probability of each is J..-. 24
1 5. The sample space for tossing three fair coins is:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} There are 8 equally likely outcomes and the
probability of each is
probability of each is 2.. 8 1 7. The sample space is: S
=
h.
I
1 1 1 1 1 = -+-+-+ - = 48 48 4 8 48 1 2
The probability of
23 . A , B,
getting a 2 or 4 followed by a Red is 1 1 1 P(2 Red) + P(4 Red) = -+- = - . 12 12 6
27. Let P(tails) = x, then P(heads) = 4x
x + 4x = 1
=
{I Yellow Forward, 1 Yellow Backward, 1 Red Forward, 1 Red Backward, 1 Green Forward, 1 Green Backward, 2 Yellow Forward, 2 Yellow Backward, 2 Red Forward, 2 Red Backward, 2 Green Forward, 2 Green B ackward, 3 Yellow Forward, 3 Yellow B ackward, 3 Red Forward, 3 Red Backward, 3 Green Forward, 3 Green B ackward, 4 Yellow Forward, 4 Yellow Backward, 4 Red Forward, 4 Red Backward, 4 Green Forward, 4 Green Backward} There are 24 equally likely events and the
probability of each is
5x = 1 1 X=5 P( tails) = .!. , 5 29.
I
I
24
P(heads) = � 5
P(2) = P(4) = P(6) = x P(1) = P(3) = P(5) = 2x P(l) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 2x + x + 2x + x + 2x + x = 1 9x = 1
l4 ' The probability of
1 X=9
getting a I, followed by a Red or Green, followed by a Backward is: pel Red Backward)+pel Green Backward) 24
C, F
25. B
1 9. The sample space is:
S
�. The probability of
48 getting a 2, followed by a 2 or 4, followed by a Red or Green is P( 2 2 Red) + P(2 4 Red ) + P(2 2 Green ) + P( 2 4 Green )
{I Yellow, 1 Red, 1 Green, 2 Yellow, 2 Red, 2 Green, 3 Yellow, 3 Red, 3 Green, 4 Yellow, 4 Red, 4 Green} There are 12 equally likely events and the
probability of each is
=
P(2) = P(4) = P(6) = .!. 9
I
12
P(l) = P(3) = P(5) = � 9
=-+-=-
31.
peE) =
neE) n(S)
=
n{1, 2, 3} 10
=
� 10
748
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exist. No portion of this material may be reproduced, in any form or by any means , without permi ssion in writing from the publi sher.
Section 14.3: Probability
6 1 . Choose 2 players from a group of 6 players. Thus,
47. The 1 st person can have any of 365 days, the 2nd
person can have any of the remaining 364 days. Thus, there are (365)(364) 132,860 possible ways two people can have different birthdays.
there are C(6, 2) =15 different teams possible.
=
63. a .
49. Choosing 2 boys from the 4 boys can be done
C(4,2) ways. Choosing 3 girls from the 8 girls can be done in C(8,3) ways. Thus, there are a total of: C(4, 2)· C(8, 3)=
8!
4!
b.
(4- 2)! 2! (8- 3)!3! 4! 8! 2! 2! 5!3! 4· 3 · 2! 8·7·6·5 ! 2·1· 2!
5!3!
=6·56 =336 51. This is a permutation with repetition. There are
�=90, 720 2! 2!
53. a.
b.
c.
different words.
65.
If numbers can be repeated, there are (50)(50)(50) 125,000 different lock combinations. If no number can be repeated, then there are 50·49 · 48=117, 600 different lock combinations. =
Answers will vary. Typical combination locks require two full clockwise rotations to the first number, followed by a full counter-clockwise rotation past the first number to the second number, followed by a clockwise rotation to the third number (not past the second). This is not clear from the given directions. Perhaps a better name for a combination lock would be a permutation lock since the order in which the numbers are entered matters.
Answers will vary.
C(7, 2)·C(3,1)= 21·3=63 Section 14.3
C(7, 3)· C(3, 0) = 35·1= 35 C(3, 3)· C(7, 0)=1·1=1
1 . equally likely
55. There are C(100, 22) ways to form the first
3. False; probability may equal
O.
In such cases, the corresponding event will never happen.
committee . There are 78 senators left, so there are C(78, 13) ways to form the second committee . There are C(65, 10) ways to form the third committee. There are C(55, 5) ways to form the fourth committee . There are C(50, 16) ways to form the fifth committee. There are C(34, 17) ways to form the sixth committee. There are C(17, 17) ways to form the seventh committee . The total number of committees = C(100, 22)· C(78, 13)·C(65, 10)·C(55, 5) . C(50, 16)·C(34,17)· C(17, 17)
5. Probabilities must be between 0 and 1 , inclusive.
Thus, 0, 0.01, 0.35, and 1 could be probabilities. 7. All the probabilities are between 0 and 1.
+
+
The sum of the probabilities is 0.2 0.3 + 0.1 0.4 = 1. This is a probability model. 9. All the probabilities are between 0 and 1.
+
The sum of the probabilities is 0.3 + 0.2 + 0.1 0.3 0.9. This is not a probability model.
"" 1.157x10 76
=
1 1 . The sample space is: S =
57. There are 9 choices for the first position, 8
choices for the second position, 7 for the third position, etc. There are 9·8·7·6·5 · 4· 3·2·1=9!= 362, 880 possible batting orders.
{HH, HT, TH, TT}.
Each outcome is equally likely to occur; so
peE)= neE) n(S) 1
. The probabilities are: 1
I
I
P(HH)=-, P(HT)=-, P(TH)=-, P(TT)=-. 4
59. The team must have 1 pitcher and 8 position
4
4
4
players (non-pitchers). For pitcher, choose 1 player from a group of 4 players, i.e., C(4, 1). For position players, choose 8 players from a group of 11 players, i.e., C(11, 8). Thus, the number different teams possible is C( 4,1)· C(11, 8)= 4 ·165= 660. 747
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Chapter 14: Counting and Probability
27.
Section 14.2 1.
C(5,3) =
1; 1
3 . pennutation 29. 5. True 7.
9.
6! 6! 6 . 5 . 4! = 30 P(6, 2) = _ _ _ = = (6 - 2)! 4! 4!
11.
13.
· . 6 . 5 ·4! 8 = 1680 P(8, 4) = _ !_ _ = � = 8 7 (8 - 4)! 4! 4!
1 5.
C(8, 2) =
8! = � = 8 . 7 . 6! = 28 (8 - 2)! 2! 6! 2! 6! · 2 · 1
1 7.
C(7,4) =
7! = � = 7 . 6 .5 . 4! = 35 (7 - 4)! 4! 3! 4! 4! · 3 · 2 · 1
1 9.
C(15, 15) =
21.
26! = � = 10,400,600 C(26, 13) = (26 - 1 3)! 1 3! 1 3! 1 3!
25.
{123, 124, 1 34, 234} 4 · 3! = 4 C(4,3) = 4! = (4 - 3)! 3! I! 3! 4 choices for the first letter in the code and 4 choices for the second letter in the code; there are (4)(4) 16 possible two-letter codes. =
33. There are two choices for each of three
positions; there are digit numbers.
(2)(2)(2) 8 possible three=
4 choices for the first position, 3 choices for the second position, 2 choices for the third position, and 1 choice for the fourth position. Thus there are (4)(3)(2)( 1) 24 possible ways four people can be lined up.
35. To line up the four people, there are
=
37. Since no letter can be repeated, there are 5
choices for the first letter, 4 choices for the second letter, and 3 choices for the third letter. Thus, there are (5)(4)(3) 60 possible threeletter codes. =
15! = �= �=1 (15 - 15)! 15 ! O! 15! 15! · 1
39. There are 26 possible one-letter names. There
are (26)(26) 676 possible two-letter names. There are (26)(26)(26) 1 7,576 possible threeletter names. Thus, there are 26 + 676 + 1 7,576 1 8,278 possible companies that can be listed on the New York Stock Exchange. =
=
{abe,abd,abe,aeb,aed,aee,adb,ade, ade,aeb,aee,aed,bae,bad,bae,bea, bcd,bee,bda,bde,bde,bea, bee,bed, cab,cad,cae,eba,ebd, ebe,eda,edb, ede,eea,eeb,eed,dab, dae,dae,dba, dbe,dbe,dca,deb,dee,dea,deb,dec, eab,eae,ead,eba,ebe,ebd,eea,eeb, eed, eda,edb, ede}
P(5,3) =
5! = 5 . 4 . 3! = 10 (5 - 3)! 3! 2 · 1 · 3!
3 1 . There are
4! 4! 4 . 3 . 2 . 1 = 24 P(4,4) = _ _ _ = = (4 - 4)! O! 1 P(7, 0) = _7 !_ _ = 7! = 1 (7 - O)! 7!
23.
{abc,abd,abe,aed,ace,ade,bcd,bee,bde,ede}
=
4 1 . A committee of 4 from a total of 7 students is
given by:
C(7,4) =
7! = � = 7 ·6 · 5 · 4! = 35 (7 - 4)! 4! 3! 4! 3 · 2 · 1 · 4!
35 committees are possible. 43. There are
2 possible answers for each question.
Therefore, there are i 0 = 1 024 different possible arrangements of the answers .
5! = � = 5 · 4 · 3 · 2! = 60 (5 - 3)! 2! 2!
45. There are 5 choices for the first position,
{ 1 23, 124, 132, 1 34, 142, 143, 2 1 3, 2 1 4, 23 1, 234, 24 1, 243, 3 1 2, 3 1 4,32 1, 324, 34 1,342,4 1 2, 4 1 3,421, 423, 43 1, 432} 4! 4! 4 · 3 · 2 · 1 = 24 P(4,3) = _ _ _ = = 1 (4 - 3)! 1!
4
choices for the second position, 3 choices for the third position, 2 choices for the fourth position, and 1 choice for the fifth position. Thus, there are (5)(4)(3)(2)(1) 120 possible arrangements of the books. =
746
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Chapter 14 Counting and Probability Section 14.1
A = { those who will purchase a major appliance }
25. Let
B = { those who will buy a car} n(U) = 500, n(A) = 200, n(B) = 150, n(A n B) = 25 n(A u B) = n(A) + nCB) - n(A n B)
1 . union 3. True; the union of two sets includes those elements
or
that are in one both of the sets. The intersection consists of the elements that are in sets. Thus, the intersection is a subset of the union.
both
= 200 + 150 - 25 = 325
n(purchase neither) = 5. 7.
9.
11.
n(A)+ n(B)- n(AnB)
u
B)
= 500 - 325 = 175
{a},{b},{c},{ d},{a,b},{a,c},{a, d}, {b,c},{b,d},{c,d},{a,b,c},{a,b,d}, {a,� d},{b,� d},{a,b,c, d}
= 150 - 25 = 125 27. Construct a Venn diagram: 15
n(A) = 1 5, nCB) = 20, n(A n B) = 10 n(A U B) = n(A) + nCB) - n(A n B) = 1 5 + 20 - 1 0 = 25
n(A u B) = 50, n(A nB) = 10, nCB) = 20 n(A u B) = n(A) + nCB) - n(A n B) 50 = n( A) + 20 -10 40 = n( )
1 3 . From the figure:
A
n(purchase only a car) = n(B) - n(A u B)
0,
A
n(U)- n(
and
n(A) = 15 + 3 + 5 + 2 = 25
(a) 15
(b) 15
(c) 15
(d) 25
(e) 40 29. a.
1 5. From the figure :
n(A orB) = n(AuB)
=
1 1, 597 There were 11,5971 thousand males 18 years old and older who were widowed or divorced.
1 7. From the figure:
n(A but not C) = n(A) - n(AnC) = 25 - 7 = 18 n(A and B and C) = n(
A
n
b.
B n C) = 5
2 1 . There are 5 choices of shirts and 3 choices of
ties; there are (5)(3)
=
+ n(divorced)
= 2, 643 + 8, 954
= 15 + 2 + 5 + 3 + 10 + 2 = 37
1 9. From the figure:
n(widowed or divorced) = n( widowed)
n(married, widowed or divorced) = n(married) + n( widowed ) + n (divorced) = 62, 486 + 2, 643 + 8,954 = 74,083
There were 74,083 thousand males 18 years old and older who were married, widowed, or divorced.
15 different arrangements.
23. There are 9 choices for the first digit, and 10
choices for each of the other three digits. Thus, there are (9)(10)(10)(10) 9000 possible four digit numbers.
3 1 . There are 8 choices for the DOW stocks, 1 5
=
choices for the NASDAQ stocks, and 4 choices for the global stocks. Thus, there are (8)(15)(4) 480 different portfolios. =
33. Answers will vary. 745
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 13: Sequences; Induction; the Binomial Theorem
9.
Chapter 13 Cumulative Review 1.
3.
=
2ex ex
( )
1n eX = 1n
x = 1n
(%)
(%)
�
0. 9 16
The solution set is 5.
=
Ix2 1 = 9 x2 = 9 or x2 = -9 x =±3 or x = ±3i =5 = -5 2
Center point (0, 4); passing through the pole (0,4) implies that the radius 4 using rectangular coordinates: (x _ h ) 2 + (y -k l r2 (X _ O ) 2 + (y _4 ) 2 = 4 2 x2 +y2 - 8y+16 = 16 x2 +/ - 8y = 0 converting to polar coordinates: r2 - 8rsinB = ° r2 = 8rsinB r = 8sinB
{1n(%)} .
11.
cos-I (-0.5) We are fmding the angle B, �B � cosine equals -0.5 . cosB = -0.5 �B � 27t 27t B = -=:>cos -I ( -0.5 ) = 3 3 -1[
Given a circle with center -( 1 , 2) and containing the point (3, 5), we first use the distance formula to determine the radius. r = ( 3 _ ( _I ) ) 2 + (5 _ 2) 2
-1[
�
1[,
whose
1[
= �4 2 +32 =.J16+9 = 55 =5 Therefore, the equation of the circle is given by (x _ ( _ 1) )2 +(y _ 2 )2 = 52 (x+ l) 2 + (y _ 2) 2 = 52 x2 + 2x+l+/ - 4y+ 4 = 25 x2 + / + 2x - 4y- 20 = °
7.
Center: (0, 0); Focus: (0 , 3); Vertex: (0, 4); Major axis is the y-axis; a = 4; c = 3 . Find b: b2 = a 2 _ c 2 = 16 - 9 = 7=:>b = .fi Write the equation using rectangular coordinates: x2 / -+- = 1 7 16 Parametric equations for the ellipse are: x = .fi cos ( t ) ; y = 4 sin ( t ) ; ° � t � 2 1[
1[
744
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Chapter 13 Test
13.
{3m+ 2 ) 5 =
(�)(3mt +(�)(3mt (2 )+G)(3m)3 (2)2 +C)(3m)2 (2 )3 +(!)(3m)(2t +G} 2)5
= 243m5 +5·8 1m4 ·2+10·27 m3 ·4+1O · 9 m 2 ·8+5 · 3m · 16+32 = 243m5 + 8 1 0m4 +1 080m3 + 720 m 2 +240 m+32
1 4 . First we show that the statement holds for n =
(1+1) = 1+1 = 2
1.
The equality is true for n = 1 so Condition I holds. Next we assume that true for some k, and we determine whether the formula then holds for
(1+iXl+�X 1+j) . . ( 1+;) = +I .
n
is
k+ 1. We assume that
(1+iX 1+�X l+j) ... ( 1+I) = k+ 1 . (1+iX l+�X l+±}.{ 1+-}X1+ k� 1 ) = (k+ 1)+ 1 = k+ 2 (1+iXI+�X 1+j}.. ( 1+-}X1+ k� 1 ) = [(1+iXI+�Xl+j) . . . ( 1+-})]( 1+ k� 1 )
Now we need to show that
.
We do this as follows:
(using the induction assumption)
= {k+ l ) . I+ {k+I ) · k-1- = k+l+1 +1 = k+2
Condition II also holds. Thus, formula holds true for all natural numbers.
1 6. The weights for each set form an arithmetic
1 5. The yearly values of the Durango form a
geometric sequence with first term and common ratio r 15% loss in value).
= 0.85
al = 31, 000
sequence with first term
d
a, = 100
and common
difference = 30 . If we imagine the weightlifter only performed one repetition per set, the total weight lifted in 5 sets would be the sum of the first five terms of the sequence.
(which represents a
an = 3 1, 000 . (0.85r -'
a n = a, +( n -1) d a = 1 00 +(5-1)(30) = 1 00 + 4 ( 30 ) = 220 5 Sn = i(a+an )
The nth term of the sequence represents the value of the Durango at the beginning of the nth year. Since we want to know the value after 1 0 years, we are looking for the 1 1 th term o f the sequence. That is, the value of the Durango at the beginning of the 1 1 th year. a l = a, . r"-l = 3 1, 000 · ( 0. 85)'0 = 6, 1 03. 1 1
S = �(1 00+220 ) = �(320) = 800 5 Since he performs 1 0 repetitions in each set, we multiply the sum by 1 0 to obtain the total weight
l
1 0 years, the Durango will be worth $6, 1 03. 1 1 .
After
lifted.
1 0 ( 800 ) = 8000
The weightlifter will have lifted a total of 8000 pounds after 5 sets .
743
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 13: Sequences; Induction; the Binomial Theorem
8.
-2,- 1 0,-1 8,-26,... -10 - ( -2) = -8 , - 1 8- ( - 1 0) = -8 , -26 - (-18) = -8
1 0.
The ratio of consecutive terms is constant. Therefore, the sequence is geometric with
a l = -2 . an = al + ( n - 1) d =-2 + (n- 1) (-8) = -2- 8n+ 8 = 6-8n The sum of the fIrst n terms of the sequence is
common ratio
n
n-I
HH] % [ (�J ) � [ (�J )
11.
[ J[
]
al =- 1 + 7 = 1 3 2' 2
-�
_
2n- 3 2n+ 1 2 ( n- 1) - 3 2 (n- 1)+ 1
and fIrst term
2n - 3 . 2n - 1 (2n- 3) (2n-1) 2n+ 1 2n- 5 (2n+ 1)(2n-5)
The ratio of consecutive terms is not constant. Therefore, the sequence is not geometric.
The sum of the fIrst n terms of the sequence is given by
. . 12. F or this geometrIc senes we have
Sn = % ( al + an )
and
a l = 256 . Since I r I =
converges and we get
� e -�) = � ( 27- n)
=
n-I
The difference of consecutive terms is not constant. Therefore, the sequence is not arithmetic.
The difference between consecutive terms is constant. Therefore, the sequence is arithmetic with common difference d =
-3 5
2n- 3 a n = -2n+ 1 2n - 3 2n- 5 2n - 3 2 ( n - 1) - 3 = ----a - a = --2n+ 1 2 (n- 1)+ 1 2n+ 1 2n- 1 (2n - 3) ( 2n- 1) - ( 2n - 5) (2n+ 1 ) ( 2n+ 1)( 2n - 1 ) ( 4n2 - 8n+ 3 ) - ( 4n2 - 8n - 5 ) 4n2 - 1 8 4n2 - 1 n
= -!!..+ 7 - -�+ 7 2 2 n 7 + n- 1 7 = --+ --2 2 1 2
a l = 25 .
m
� = (-2+ 6- 8n) � = (4- 8n) � = n( 2 - 4n)
a -a
and fIrst term
" ' " ' - r "25 �25 Sn " a 1 1-r 1--2 5 = 1 5 1= 25 ' 1 -
Sn = (a + an )
an =-!!..+ 7 2
r =�
The sum of the fIrst n terms of the sequence is given by
given by
•
8
lQ = 2' � = 2 .I = � ..!. = 2 25 5 1 0 5 ' 4 5 4 5
The difference between consecutive terms is constant. Therefore, the sequence is arithmetic with common difference d = -8 and fIrst term
9
.
25,1 0,4'"58' . .
7 2
r = -64 = 1 256 -4
I-�I = � < 1 ,
the series
S = � = 256 = 256 = 1 024 5 1 - r l-(-t) � 00
742
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Chapter 13 Test
69. This is an ordinary annuity with
( )( ) -[[ f ]
P= $500
and
4.
n = 4 30 = 1 20 payment periods. The . . d IS' 0 . 08 = 0 . 02 . Thus, mterest rate per peno 4 2 1+0.02 0 - 1 "" $244, 1 29.08 A = 500 0. 02
-al = -- = -0 = 0 1+8 9
2 l . an = n - 1 n+8
2 a2 = 2 - 1 = 3 2+8 1 0 2 a3 = 3 - 1 = 8 3+8 0 15 - -5 42 - 1 - a4 - -4+8 1 2 4 2 as = 5 -1 = 24 5+8 13
The first five terms of the sequence are 1 34, and 404.
Notice that the signs o f each term alternate, with the first term being negative. This implies that the general term will include a power of - 1 Also note that the numerator is always 1 more than the term number and the denominator is 4 more than the term number. Thus, each term is in the form
( t ( ::!) �) ) (k ) -1
. The last numerator is
11
1 0 terms. 2 3 --+ 4 .. +1 1 = -1 k + 1 --+5 6 7 . 14 k=1 +4
which indicates that there are 10
6.
3 The first five terms of the sequence are 0 , 10 ' 24 8 5 11 ' 4" ' and 13 '
) ))
- 2 + 3 - 4 + '+ 1 1 5 6 7 " 14
.
e -1
)
[(�r -l]+[(�)' -2]+[(�r -3]+[(�r -4]
1i � � � = 3 - 1+ 9 - 2+ 27 - 3+ 8 1 - 4 = 1 3 0 - 1 0 = _ 680 81 81 5.
a l = 4; an = 3an _ 1 + 2 a 2 = 3al + 2 = 3 ( 4 + 2 = 14 a3 = 3a2 + 2 = 3 ( 14 + 2 = 44 a4 = 3a3 + 2 = 3 ( 44 + 2 = 1 34 as = 3a4 + 2 = 3 ( 1 34 + 2 = 404
-k
=
Chapter 13 Test
2.
�[(�J ]
k
6, 12,36, 144, ... 1 2 -6 = 6 and 36 - 12 = 24
The difference between consecutive terms is not constant. Therefore, the sequence is not arithmetic . .ll =
6
2
and
36 = 3 12
The ratio of consecutive terms is not constant. Therefore, the sequence is not geometric. 7.
4, 14, 44,
a n = _ .!.. 4 n 2 I n � -2 ' 4 a n - I _' L 4 n-1 2
_
_ L 4n -I . 4 2 1 n- = 4 .4 2 1
Since the ratio of consecutive terms is constant, the sequence is geometric with common ratio
r=4
and first term
The sum of the first given by Sn
n = a I 1-r 1-r n = -2· 1 - 4 1-4 = 1 - 4n
a l = - 21 . 4 I
n
=
-2 .
terms of the sequence is
. -
�(
--
)
74 1
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 13: Sequences; Induction; the Binomial Theorem
57.
(X+2)5
=
( �}5 + ( � JX4 ' 2+G}3 . 22 +GJX2 . 23 +( :JXl . 24 +G} 25
= X5 + 5 · 2 x 4 + 10 · 4x 3 + 10 · 8x 2 + 5 ·16x + 1 · 32 = x5 + lOx4 + 40x 3 + 80x 2 + 80x + 32 59.
(2x+3)5 =
( �J c2X)5 +( � J c2X)4 ' 3+GJ c2X)3 . 32 +GJ c2X)2 . 33 +(:J c2X)1 . 34 +G} 35
4 3 2 = 32x5 + 5 · 16x · 3 + 10 · 8x · 9 + 10 · 4x · 27 + 5 · 2x · 81 + 1 . 243 2 = 32x5 + 240x 4 + nOX3 +1080x + 81 Ox+ 243
61.
n
= 9, j = 2, x = x,
a
=
( 92JX7 ·22 = � . 4X7 = 2!7!
c.
2 9·8 . 2 ·1
(�r (�r () ) n � n ( ) ()
4x 7 = 144x 7
The coefficient of x 7 is 144.
63.
n
= 7, j = 5, x = 2 x,
( 7J (2X)2 . 15
a
=1
7 .6 . 2 = � ' 4x 2 (1) = 4x = 84x 2 5 5! 2! 2 ·1 The coefficient of x 2 is 84.
,
a.
= 80, d = -3,
n
a
=
25
2
= 25 d.
(80 + 8) = 25(44) = 1100 bricks
a.
r
3 = 4
b.
() ( 2 %r
3 20 � 4
=
13 5 16
height is
0
feet .
nth
10g
log 0.025
"" 12 .82 3 log 4 The height is less than 6 inches after the 13th strike . Since this is a geometric sequence with I r I < 1 , the distance is the sum of the two
�) �) ( (
"" 8 .44 feet .
After striking the ground the
�
Distance going up: 15 15 Sup = = = 60 feet. 1-
After striking the ground the third time, the height is
�
(l-�) m
67. This is a geometric sequence with
= 20,
0.025
20
infinite geometric series - the distances going down plus the distances going up. Distance going down: 20 20 Sd,w. � � � 80 roo'
1100 bricks are needed to build the steps.
Q,
�
�
2 5 = 80 + (25 -1)(-3) = 80 - 72 = 8 bricks
b. S2 5
0.5
log ( 0.02 5
65. This is an arithmetic sequence with a
If the height is less than 6 inches or 0.5 feet, then:
time, the
The total distance traveled is 140 feet.
740
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Chapter 13 Review Exercises
= 1: 2 . 31-1 = 2 and 31 -1 = 2 ... . + 2 · -13k-1 = 3k+-1,I-1 then 2+6+18+ 2 + 6 + 18 + . +. 2. . 3k k+ 12 · 3k k = [ 2 + 6+ 18+ . + 2 3 - J + 2·3 = 3 k - 1 + 2 · 3k = 3 · 3k - 1 = 3 k + l - l Condi true. tions and are satisfied; the statement is = 1: (3 ·1-2)2 =land-·l21 (6·12 -3· 1 -1) =1 +42 + . . +(3k-2)2 = � ' k ( 6k2 -3k-l) , th12en+42 +72 + . . . +(3k-2)2 +(3(k+l) -2)2 = [12 +42 +72 + . . +(3k-2)2 J +(3k+l)2 = ± . k ( 6k2 -3k-l) +(3k+l)2 = �. [6k3 -3k2 -k + 18e + 12k + 2J = ± -[6e + 15e + 11k + 2J = � ' (k+l) [6k2 +9k+2J = ± - [6k3 +6e +9k 2 +9k+2k+2J = ± . [6k2 ( k + 1)+9k(k + 1) + 2(k +1)J = ± ' (k+l) [6k2 +12k+6-3k-3-1] = ± ' (k+l) [6(k2 +2k+l) -3(k+l) -IJ = ± ' (k+l) [6(k+l)2 -3(k+l)-IJ tCondi rue. tions and are satisfied; the statement is 5·42·1 l O ( 25 ) - -2!3!5! 5·2·41 ··3·2· 3 · 2·11 ---
51. I:
II:
n
If
·
53. I:
II:
I
II
I
II
n
If
e
55.
739
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 13: Sequences; Induction; the Binomial Theorem
31.
33.
Arial =th3,metdic= 4, an = al +(n-l)d a9 =3+(9-1)4=3+8(4)=3+32=35 Geometric 1 al = 1, r = 10 ' n = 1 1; all = al rn-I all _- 1· (-101 ) -_ ( 101 ) 1 10, 000, 000, 000 Arial =thmJ2et,icd = J2, n = 9, a = al + (n -1)d n J2 J2 J2 a9 = J2+""(9 -1) = + 8J2 = 9 12.7279 a20 =ioans:l + 19d = 96 ; Solaa7lv=+ea6dtlh+e=6dsyst31=em31 of equat aSubtl +19dract t=he96second equation from the first equat -13ddio==n-655and solve for d. aal ==31-6( 5) = 31-30 = 1 al + (n -1) d n =1+(n-l)(5) 1+5n-5 =5n-4 General formula: {an } = { 5n -4} a =a +17d =8; alSolQv=ae thl e+9d=0 s l l aal l+17d +9d ==syst8 em of equations: Subt ractiontandhe second equatd. ion from the first equat sol v e for -8d ==-81 al = -9(1) = -9 1 1-1
35.
37.
-
10
41.
3
S
43.
45.
l-r
2
a, = 21 ' r 3 Since IrI 1 , the series diverges. al = 4, r =-1 Since I rI < , the series converges. n - al _ 4 _ 4 -- 8 - � - (I - i) - (i) 32·1 + 1) = n = 1: 3· 1 = 3 and -(1 f 3+6+9+ . . +3k =-(3k2 k +1) , then + 9 + . . +. .3k+3k]+3(k+1 3=[+36+6+9+ + 3(k + 1) ) 3k2 =-(k+1)+3(k+1) = (k + l) e: 3) = 3(\+ « k + 1) Condi true. tions and are satisfied; the statement is >
47.
=2
2
1
S
=
39.
an == -9+(n al +(n-l)d - 1)(1) = -9+n-l =n-lO General formula: { an } = { n-l0} al = 3, r =-1 Since I r I < 1, th3e series 3converges. n = � = _(1 -_�) =_(�)_= 2.2 al = 2, r = --1 Since IrI < 1 , the series converges.
49. 1 :
a
II:
d
3
I
+
I
1)
+
1)
II
738
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as
they c urrently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in wtiting froni the publisher.
Chapter 13 Re view Exercises
Chapter 13 Review Exercises 1.
7.
9.
11.
13.
= _�
5+3 7 4+3 = "6'7 a5 =(_ 1)5 5+2 3+3 =- 56 ' a4 = (-1)4 4+2 2+3 = 4"5 ' a3 =(-1)3 3+2 1+3 =- "34 ' a2 =(-1)2 2+2 al =(-1)1 1+2
=
al = 2, a2 = 2 -2 = 0, a3 = 2 - = 2, a4 = 2 -2 0, a5 = 2 - = 2 4 + 2) = (4 . 1 + 2)+( 4·2 + 2)+(4·3+ 2)+(4 . 4 +2) = ( 6)+(10) +(14)+(18) = 48 �)4k 12 31 41 -131 = L:\ 3 ( -1) (-k1 ) 3, %, � , %, 136 ' ' ' Geometric 1--+---+,,·+ {all}= (n= +{ 1n++5)5}-(Arin +t5)hm=etnic+ 6 -n -5 = 1 (%)3 2 -31 =-21 r=-=-· Sn = '::2 [6+n+5]='::2 (n+ll) S. =6 [1- (�J l {en } = { 2n3 } Examine the terms of the sequence:is no2,common 16,54, di128,250, . .there is no There f f e rence; Neicommon ther. ratThere common ratio; neither. io. is no common difference or 1) {sn } =(11{ +2)3n } Geomet r i c 50(50+ ) 3825 ( 3 = k I 3 = 3k I 2 n+ 1 3 3 23 2 r = --= 23n --= 23n 12311+3-3n = 23n= 8 Sn = 8 [\�8�' ) = 8 [ =�" ) = % (8 -1) 30(3� + 1) ) -30(9) = 3 ( 0,4, 8, 12, . . Arithmetic = 4 -0= 4 1395 -270 = 1125 Sn = '::2 ( 2(0) + (n -1)4 ) ='::2 (4(n -1)) 2n(n -1) °
k=1
°
hI
k=1
21.
3
d
1 5.
={I��y ] ={I Nl ]
23.
1 7.
25.
k=1
=
k=1
k=1
1 9.
d
k=1
k=1
k=1
k =1
=
=
= ± (1- 21187 ) 1093 = L2 2186 2187 2187 "" 0.49977
=
737
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Chapter 13: Sequences; Induction; the Binomial Theorem
39.
The 1xO term in ) 1 f( �) ( X2 t-) (.!.x ) = f( �) x24-3) occurs when: 24-3j24=3j =0 j=8 The12 coeffici ent12·11· is 1 0· 9 � ( 8 ) = 8!4! = 4·3·2·1 =495 }=o
43.
J
}=o
41.
J
The X4 term in l �) ( -2)) xl O-%) ) f f ( l �) (X)I O-) (2) ( occurs when: Fx 10-�j=4 2 j =-6 2 j=4 The10 coefficient is 10 . 9 . 8 . 7 ( 4 ) ( _2)4 = � 6!4! . 16= 4·3·2·1 . 1 6=3360 )=0
J
=
)=0
J
-�
(1 .001)5 =(1+1O-3 t = (�}5 + (� }4 . 1 O-3+ G) e . ( 1 O-3 t + G ) e . (1 O-3 t + . . ... 1= 1 ++ 05(0.005.001)+ 0+.010(0. 0 00001) + 10(0 0 00000001) + . 00010+ 0 0 00000010+· · . = 1 .00501 (correct to 5 decimal places) n! = (n-1)n! !(1)! = n(n-1)! ( n-1n ) = ( n-1)! ( n-(n-1))! n-1)! = ( ( n ) = n!(nn!-n)! = n!n!O ! = n!n!. 1 = n!n! = 1 =
45.
n
n
49.
= (�} 11/ + ( � } ln-l . 1+ G} ln-2 . 12 + . . + ( :} In-n ·1/1 = ( �) + C) +· · ·+ (:) 5 +( ) + (�)(�) � (�J (�) G) (�)3 (�J + G) (�J (�J + (�) (�) (�J + G ) (�J = (� + �J = (1)5 = 1
736
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Section 13.5: The Binomial Theorem
1 9.
21.
2 5.
2 7.
29.
31.
(X-2)6 = ( �}6 + ( �}5(_ 2)+ ( �}4(_ 2)2 + (�) X\-2)3 + (!}2 (_ 2)4 + G) X(-2)5 + ( :}O (_ 2)6 . 4 +20x\-8)+15x2 · 1 6+6x· (-32)+64 == x6x6 -12x5 +6x5(-2)+15x4 + 60x4 -160x3 + 240x2 -192x + 64 (3x+1)4 = (�}3X)4 + (:}3X)3 + G}3X)2 + (;) (3X)+ (:) = 81x4 +4·27x3 +6· 9x2 +4·3x+1 = 81x4 +108x3 +54x2 +12x+1 (Fx
+ J2 t = (�)(Fx)6 + ( �) (Fxt (J2 y + G)Fxf (J2/ + ( �) (Fx t (J2)3 +(!)(Fxt(J2f +G}Fx)(J2t + (:)(J2)6
3/2 + 15· 4x + 16·2 4J2x1 / 2 + 8 = x33 + 6J2J2x5/ 22 + 15· 2x2 2 + 20·J2 2J2x = x + 6 x5/ + 30x + 40 x3/ 2 + 60x + 24J2x / + 8 (ax+by)5 = (�} ax)5 + (� }ax)4 . by + (�}ax)3 (bY/ + G}ax)2 (by)3 + (:) ax(by)4 + G}by)5 = a5x5 + 5a4x4by +10a3 x3b2/ +10a2x2b3/ +5axb4/ +b5 / n9= 9, j = 2, x = 2x, a = 3 n10= 10, j = 4, x = x, a = 3 ·8·7 9 10· . 128x\9) . 8 1x6 = ( 2) (2X)7 ·32 =� ( 4 ) X6 . 34 = � 2!7! 4!6! 4·3·2·1 . 8 1x6 9·2·18 . 128x7 . 9 Ox6 = 17, 0 1 = The coefficient ofx6 is 17,010. 7 = 41, 472x n=12, j =5, x=2x, a=-l 7 The coef fi c i e nt of x i s 41, 4 72 . 12( 5 ) (2x)7 · (-1)5 =-·128x 12!5!7! 7 (-1) n7= 7, j = 4, x = x, a = 37 ·6·5 9 · 8 � 3 . 34 x = ( ) . 7 ( -128)x = 12·11·10· 4 4!3! . 8 1x3 = 3·2· 1 . 8 1x3 = 2835x3 5· 4 · 3 ·2·1 -101 , 3 76x7 = n9= 9, j = 2, x = 3x, a = The coefficient ofx7 is -101, 376. . 2 187x7 . 4 (2 ) (3X)7 . (_ 2)2 =� 2!7! = 9·2 ·81 . 8748x7 = 314, 928x7 33.
35.
37.
-
2
735
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 13: Sequences; Induction; the Binomial Theorem
. nsum=3:of (3-2) 180°=180° whi c hi s t h e t h e angl e s ofa t r i a ngl e . Assume atoffaorconvex any intpolegerygonk, thwie tsumh siofdes tihs e(kangl-2)etsh.180° A convex pol y gon wi t h . witkhe+thanglI ksisidedessesisconsi plus satstrofiangla convex e. Thus,polthyegonsum of (k -2) .180°+ 180° = «k + I) -2) · 1 80°. Condi true. tions and are satisfied; the statement is
2 . . If 2+4+6+ +2k =k +k+2, t h en 2+4+6+"·+2k+2(k+l ) = [22 + 4 + 6 + . . + 2k] + 2k + 2 = k + k + 2 + 2k + 2 = + 2k2 + I) + (k + I) + 2 =(k+I) +(k+I) +2 n=l: 2· 1 =2andI2 +1+2=4:;t2 n = 1: a + (1-1)d = a and 1· a + d 1(1-12 ) = a f a + (a + d)k(k+ (a-1)+ 2d) + " . + [a + (k -l)d] = ka +d 2 then a + (a + d) + (a + 2d) + + [a + - l)d] + (a + kd) [a + (a + d) + (a + 2d) + . . . + [a + - l)d]] + (a + kd) =ka+d k(k2-1) +(a+ kd) = (k + I)a + d [ k(k2-1) + k ] = (k + I)a + d [ k2 - � + 2k ] k2 2+k-] =(k +l)a+d [=(k+l)a+d [ (k �l)k ] ] ) = (k+1) a+d [ (k +I)((k+1)-1 2 tCondi rue. tions and are satisfied; the statement is
33. I:
29. II:
II:
k
(e
I:
I
31. I:
II:
Section 13.5
I
. . ·
=
I
(k
II
1.
3.
(k
5.
7. 9.
II.
13.
II
1 5.
Pascal Triangle n! False', (jn ) = j! ( n-j)! (35 ) 3!2!5! 5·4·3·2 3·2·1·2·1·1 _ 5.2·14 - 10 7·6 (75 ) - 5!2!7! 7·6·5·4·3·2·1 -5·4·3·2 ·1·2· 1 2· 1 21 50·4 ! 50 9 (4509) = � = = 49 !1! 49 !·1 1 =50 1000) = 1000! 1000!O! = I = I (1000 (5523) =� 23! 32! "" 1 . 8664 X I OI 5 (4725 ) =�"" 25! 22! 1 .4834 X I OI 3 _
_
- -
1
734
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Section 13.4: Mathematical Induction
= 1: 1(1 + 1) = 2 and -·31 1(1 + 1)(1 + 2) = 2 f 1·2+2· 3 +3·4+ . . +k(k+l) =-·31 k(k +1)(k +2), then 1· 2 + 2·3 +3·4 + . . . +k(k + 1) + (k + 1)(k + 1+1) = [1 . 2 + 2 · 3 +3· 4 + . . + k(k + 1)] + (k + 1)(k + 2) = � . k(k + 1)(k + 2) + (k + 1)(k + 2) = (k + 1)(k + 2) [� k + 1] = 31 (k + 1)(k + 2)(k + 3) = 31 (k + 1)((k + 1) + 1)((k + 1) + 2) Conditions and are satisfied; the statement is true. = 1: a -b is a factor of al -bl a-b. =l 12 +1=2is divisible by2 fa -ba -bis ias faafctactoroofr ofaakk+1 -bk_ bk+, 1show. that If k2 +2k is divisible by 2 , then (k + 1) + (k + 1) k 2 + 2k + 1 + k + 1 ak+1 _ bk+1 = aa .. akk -b-a .. bk a . -b· k =(e +k)+(2k+2) = a bk +k bk b Sidivnicesibkle2 by+ k2,isthdienvis(ikb+le1)by2 +2 (kand+ 12k) is+ 2 is =a(ak -bk)+b (a-b) di v i s i b l e by 2 . are satisfied; the statement is Siis nacefactao-br ofisaa-bfact, tohrenof aa-bk -bkis aandfactaor-bof Condi t i o ns and true. k+1 _ bk+1 . a Condi tions and are satisfied; the statement is =l 12 -1+2=2is divisible by2 t r ue . f k2 -k + 2 is divisible2by 2 , then (l+a y =1+a I +1 . a (k + V -(k + 1) + 2 = k 2+ 2k + 1-k -1 + 2 Assume thatitythholeredis.s anWeinneedtegertok show for whithcath if =(k -k+2)+(2k) t h e i n equal Sidivnicesibkle2 by-k2,+ t2henis di(kvi+si1)b2le-(kby 2+and1) + 22kisis (1 + a t 1 + ka then di v i s i b l e by 2. ( 1 + a )k + 1 1 + ( k + 1 ) a . (l+a)k+1 =(I +at(l+a) tCondi rue. tions and are satisl fied; the statement is (I +ka)(I +a) = 1: If x > 1 then X = x > 1 . = +a + ka l+ka2 Assume, for some nat u ral number k, t h at i f = 1+ (k +1) a +ka2 xThen> 1 ,xtk+hen1 >x1k, for> 1 x>. 1, 1+(k+l )a 1xk+ = xk . x> 1· x = x > 1 Condi true. tions and are satisfied, the statement is k > 1) ( x tCondi rue. tions and are satisfied; the statement is
1 7. I :
II:
n
I
_ .
_ .
I
1 9. I:
n
II
25. I :
:
II:
II:
n
=
I
=
I
21. I:
II:
n
II
I
:
II
I
27. I :
�
n=l :
II:
�
I
23. I :
II:
�
II
�
n
�
I
II
t
I
II
733
© 2008 Pearson Education, Inc ., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 13: Sequences; Induction; the Binomial Theorem
1 1 . I:
II :
1 = -1 and-1 =-1 n = 1 : 1(1+1) 2 1+1 2 1 1 + -1 + ··· + 1 k , then If 1·2 + 2·3 3·4 k(k+1) k+1 1 + 1 + 1 + ... + 1 + 1 = [1.21 + 1 + 1 + .. . + 1 ] + 1 1.2 2·3 3 · 4 2·3 3·4 k(k + 1) (k + 1)(k + 2) k(k + 1) (k + 1)(k + 1 + 1) k 1 k k+2 1 =--+ k + 1 (k + 1)(k + 2) =-_.--+ k + 1 k + 2 (k + 1)(k + 2) k 2 +2k+1 (k+1)(k+1) k+1 k+1 (k+1)(k+2) (k+1)(k+ 2) k+2 (k+1)+1 --
--
----
Conditions and are satisfied; the statement is true. I
13. I:
II:
II
1 =1 n = 1 : 1 2 = 1 and-·1(1+1)(2·1+1) 6 1 12 +2 2 +3 2 + ... +e =-·k(k+1)(2k+1) , then 6 12 +22 +32 + ... +k 2 +(k+1) 2 = [12 +22 +32 + ... +k 2 J +(k+1)2 = � k(k+1)(2k+1)+(k+1) 2 = (k + 1) [� k(2k + 1) +k + 1 ] = (k + 1{� k 2 + � k +k +1] = (k + 1{� e + � k + 1 ] = � (k + 1) [ 2k 2 + 7k +6] = -61 . (k + 1)(k + 2)(2k + 3) = i · (k + 1) ((k + 1)+ 1) ( 2(k + 1)+ 1)
If
Conditions and are satisfied; the statement is true. 1 -1) = 4 n = 1 : 5 -1 = 4 and -·1(9 2 1 4+ 3 +2+ ... +(5-k) = 2 ·k(9-k) , then I
II
1 5. I :
II:
If
1 4 +3 + 2+ ··· + (5 -k) +( 5 - (k + 1)) = [4+3 + 2 + ... + (5 -k)]+ (4-k) = -k(9-k) 2 +(4 -k) = 29 k- 21 k 2 +4-k = - 21 k 2 + 27 k+4 = - 21 ' [ k 2 -7k-8] 1 1 = - -21 ·(k + 1)(k - 8) = -·(k 2 + 1)(8-k) = -2 . (k + 1)[9 -(k + 1)]
Conditions and are satisfied; the statement is true. I
II
732
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Section 13.4: Mathematical Induction
1 07. 1 09.
Answers wil vary. Answers iwially,varybut .thBote domai h increase (ogeomet r decrease) exponent n of a rliecthe sequence i s t h e set of nat u ral numbers whi of an exponent aldomai l realnnumbers . ial function is the set of
5.
I: n = 1: 3 . 1 -1 = 2 and -21 . 1 (3 . 1 + 1) = 2 II: If 2 + 5 + 8 + . . + (3k = 21 " k(3k + , th2en+ 5 + 8 + . . + (3k -1) + [3(k + 1) -1] = [2 + 5 + 8 + . . + (3k 1 ] + (3k + 2) =.!..2 k(3k + 1) +(3k + 2) = l..2 e +�k2 +3k + 2 = l..2 e + 2.2 k + 2 = .!..2 (3e + 7k + 4) = -·21 (k +1)(3k +4) = l1 ·(k +1) (3(k + 1)+ 1) tCondi rue. tions I and I are satisfied; the statement is I: n=l: 21-1 =landi-l=1 II: If + 2 +222 +. .. . +k-2k-1 1 =k2+Ik-1 , then + 2 + 2 + + 2 k-+1 2 k = [1 + 2 + 22 + ... + 2 ] + 2 = 2kk+-1_+ 2k = 2 ·2k -1 = 2 l l tCondi rue. tions I and I are satisfied; the statement is - 1)
-
Section 13.4 1.
3.
I: n = 1: 2·1 = 2 and 1(1 + 1) = 2 II: If2+4+6+···+2k=k(k+l ) ,t h en 2= +[2+4+6+ 4 + 6 + . . +. .2k+2k]+2(k+l + 2(k + 1) ) ==k(k(k +1+ 1))(k++2)2(k + 1) =(k+l) ((k+l)+I) tCondi rue. tions I and I are satisfied; the statement is I: n = 1 + 2 = 3 and -·21 1(1 + 5) = 3 1 then II: If 3+4+5+ . . . +(k+2) =-·k(k+5), 2 k + 2) + [(k + 1) + 2] 3= [34+4+5+ + 5 + . ....+ (+(k+2)]+(k+3) = -·21 k(k +5) +(k +3) =�k2 2 +�k+k+3 722 1=-k2 2 +-k+3 = � . ( k 2 + 7 k + 6) =-·(k+1 21 )(k+6) = � . (k+1) ((k+l)+5) tCondi rue. tions I and I are satisfied; the statement is
7.
1
1:
+
1
1)
)
-1
II: If 1+4+42 + . . . +4k-l = 31 · (4k -1) ,then 1+4+42 + . . +4k-1k-+41 k+I-k1 = [1+4+42 + . . +4 J +4 = 3l · ( 4k -1) +4k = 31 . 4k - 31 + 4 = � . 4k _ ± = ± (4 . 4k -1) = � . (4k +1 -1) Condi true. tions I and I are satisfied; the statement is k
73 1
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Chapter 13: Sequences; Induction; the Binomial Theorem
89.
This nis=an(12)(30) ordinary=annui y with peri= $100ods. Theand 360 tpayment mt. erest rate per peno. d · 012. 1 2 001 . Thus, [1+0.00.01tO1 -1 ] "" $349,496.4 1 A = 100 [ Thin =s(is4)an(20)ordi=narypayment annuity periwithods.=The$500intanderest rate per period is 4 = 0.02 . Thus, [ [1+00..002tO2 -1 ] "" $96, 5.9 A = 500 Thiands nis=an(12)(10) ordinary= annui t y wi t h A = $50,000 120 payment peri o ds. The mt. erest rate per perl.od · 012.06 = 000. 5 . Thus, 50,000 = [ 0.005 -1 ] = 50,000 [[1 + 00.0.05t005 20 -1 1 "" $305 . 1 0 Thial =s 1is, argeomet ric= sequence with = 2, n 64 . Find the 1-264 sum of the1-264 geometric series: S64 = 1 ( 1-2 ) = -1 = 264 _ 1 grains The common rat i o , r = 0 9 0 < 1 . The sum i s : . 1 .9 0.11 0 . S=--=-=lO 1-0 The multiplier is 10. This is an infinit1e.0geomet ric series with 3 a=4, and r=-1 . 09 4 Find th"um (1- 1 .03 f $72 .67 . 1 . 09 IS
91.
80
--
=
-
)
>
n
0.08
8
I S --
1 [1 + 0.005] 2 0
1 03 .
P �----'''----
99.
Pde<
>
""
1 1 0. 2 1
=
B:
= i9
= 1 .845 x l 0 1 9
97.
--'----"-
d
P
95.
GiFinvden:n when al = 1000, an < 0r.0=1 0: .9 100 (0.9 < 0.0 1 < (n -1 log(0.9) < log (0.00001) n -1 loglo(0g(0.00001) 9) . loglo(0g .(0000 1 ) +1 0 9 ) . OnamolithnetIlltwil hbedaylessortDecember 20, 2007, the h an $0 0 1 . . Find the sum ofn the geometric series:l l-r1-r ) = 1000 [ 1-(01-0.9.t9 ) S111 = al (-.0. 1 l ) = $9999 .92 = 1000 ( 1-(09t thethsumetm iofc serieachessequence: A:Fia n=dAri$1000, wi t h : Filnd the1000sum of=th-1,e aritnhmet= 1000ic series: SIOOO = -(1000+ 1) = 500(1001) $500,500 2 r1i9c .sequence with n = aFin1d=Thith1,es sirusm=a 2,geomet of t h e geomet r i c seri e s: SI9 =1 ( I-1-2i 9 ) = 1--1i9 -1= $524,287 results in more money. Thesequence amountwitpaih adl each day f o rms a geomet r i c = 0.0 1 and_ r = 2 . 1 1-2222 =41, 943.03 1_r22 S2 =al · -=0.0 1 · -1-r Theworkedtotaall payment l 22 days.would be $41,943.03 if you =a ·r =0 0 1 2 =20, 9 71 5 2 . . l The paymentwil onvary.theWi220dth tdayhis payment is $20,971pl.a5n,2. the Answers bulonekdayof thcane payment isalatly treduce he end tshoemioveralssingl even dramat i c payment .seNottheicamount e that wipaith doneonsithcek 220dayd youday woul d l o which is about half the total payment for the 22 days. 0 ),, 1 1 ( 0.9 ) "- 0.0000 1
.
P
88
93.
101.
P
B
1 05.
a
"
22
22 - 1
( )2 1
730
© 2008 Pearson Educati on, Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they c u rrently
exist. No portion of thi s material may be reproduced , in any form or by any means, without permi ssion in writing from the publi sher.
Section 13.3: Geometric Sequences; Geometric Series
69.
= 8, r = -21 Since�I r I < 1, th8e serie_8s converges . S = 1-r = _(1-_�) = (�_) =16 = 2, r =--41 Since�I r I < 1, the 2series converges. S = 1-r = 1- (- ) = _2_ = �5 ( �) (%) = 8 , r = 23 Since \rI > 1, the series diverges. = 5, r =41 Since I r 1 < th5e series converges . S - G; - (1- � - (�)5 - 320 ) = 21 ' r= 3 Since IrI > 1 , the series diverges. =6, r= -32 Since�I r I < 1, the 6series converges . S = 1-r = 1- ( j ) = _6(�)_ = �5 ( ) �)k;1 (-23 )k = �)k;1 . -23 . (-23 )k-1 =�)k;1 (-23 )k-1 = 2, r = 32 Since I r I < 1, t2he seri2es converges. S - l-r - 1- 2. - 6
83.
al
�
00
71 .
85.
al
75.
77.
79.
_
_
_
an
c.
_
_
00
n
+
al
00
81.
al
b.
al
00
4
al O
1,
al
-
al
al
_
=
87. a.
al
00
�
as
00
73.
Fisystndemtheofcommon ratio ofthe terms and solve the equat i o ns: x+2x --=r x+3 --=r; x+2 x + 3 x2 + 4x + 4 = x2 + 3x x x +x 2 = x+2 Thi=s $18,000, is a geometrri=c seri1.05,es win =th5. Find the 5th term:= 18000(1 .05)S-1 = 18000(1.05t = $21, 879. 1 1 Fisequence: nd the 10th term of the geometric = 2, r =100-1.9, n = 10 = 2(0.9) = 2(0.9)9 = 0.775 feet Find when -1 < 1 : 2(0.9)n I < 1 (0.9r- < 0.5 (n -1) log(0.9) < log(0.5 ) 5 ) n-l > lloog(0. g (0.9) n > 1l0og(0g(0..95 )) 1 "" 7.5 8 On the 8th swing the arc is less than 1 foot. Find the sum of t1he first 15 swings:5 SI S - 2 [ 1_(0.1-09.9) 5 ) -_ 2 ( 1-(0.0. 19Y ] = 20 (1-(0. 9Y S ) = 15. 8 8 feet Find the 2infinite2sum ofthe geometric series: S =--=-=20 1-0.9 0. 1 feet d.
00
00
a1
00
a1
_
_
3
-1 3
729
© 2008 Pearson Education, inc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of thi s material may be reproduced , in any form or by any means, without permi ssion in writing from the publisher.
Chapter 13: Sequences; Induction; the Binomial Theorem
43.
45.
47.
49.
a1 = 1, r = -1, n = 9 a9 = 1 . ( _ 1)9 - 1 = ( _ 1)8 = 1
8
a1 = 004, r = 0. 1, n = a8 = Oo4 . (O. lt l = Oo4 (O. lf = 0.00000004 14 = 2 a = a n- l a1 = 7 , r = ' n 1 r 7 a n = 7 · 2 n-1 •
1 a 1 = -3 , r = __ = -.!. ' an = a 1 r n- 1 3 -3 n-l n- 2 a n = -3 = -
( ) ( ) -�
51.
243 = a 1 ( -3 t· •
•
59.
�
8 1 9 1 . 75
Therefore,
4- 1 � � =� = = r2 a a1 r 2 - 1 r 2 1 575 = 225 r2 = 7 r = .J225 = 1 5 an = a1 · r n- l = a 1 . 1 52- 1 7 = 1 5a1 7 a1 = 15 , an = 2 · 1 5 n - 1 = 7 · 1 5 n - 2 15 •
[ ) [ )
61.
1
243 = a 1 ( -3 f 243 = -243a 1 -1 = a 1 nl an = - ( -3 ) -
53.
a1 = -1, r = 2 � � Sn = al = 1 - 2n = -I 1-2 l-r
63.
3
65.
-32767
7
Therefore 55.
•
67.
Since
1 a1 = "4 ' r = 2 � .!. � .!. = = - (1 - 2n ) S11 = a1 1-2 4 l-r =
1 a1 = 1, r = 3 I r I < 1,
the series converges.
� S = � = _1_ = _1_ = l-r 12
[ ) 4[ )
ro
± ( 211 - 1 )
( �) (�)
728
© 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any mean s, without permission in writing from the publi sher.
Section 13.3: Geometric Sequences; Geometric Series
1 5.
n+l ) n+l n ( 2 r = _ = 2(-3--3) = i/3 Thetherefore rat2(�)io tofheconsecut iveistgeomet erms isrconst a. nt, sequence i c - 21 13 ' 2 -- 2 2 /3 - 23/ 3 - 2' 4 - 24/3 ( 3n+1-1 ) 3n-n .-2nn+ r = ( 23�:n+11 ) = -3121 = 3n-(n -l) ·2"-(n+l) = 3.2-1 = �2 3
e1 -
1 7.
27.
__
3 1 -11 2 r = eJ (�J = eJ+ 3 3
e
, e3 -
e
Thetherefratoreio tofheconsecut erms isrconst sequenceiveistgeomet ic. ant, 29.
19.
1 = -i = -2 =-2 ' t2 = -22 =22 =-4 ' 344-1 334 27 33-1 32 t3 = -- = -3 = - t4 = -- =-=23 2 8 ' 2 2 16 { n+2} = (n + 1 + 2) -(n + 2) = n 3 -n -2 = 1 d
21.
23.
2 5.
-2, -8, . .. -2 = -8 = 2 -4 = r=-4 -1 -2
-
1
,
--4,
erms isrconst tTheherefratoreio tofheconsecut sequenceiviestgeomet ic. ant,
Thetherefratoreio tofheconsecut iveistgeomet erms isrconst ant, sequence i c . 3 1 -1 3° 1 3 2-1 3 1 3 t
{m}+1
9
Thetherefratoreio tofheconsecut erms isrconst sequenceiveistgeomet ic. ant,
+
Theconstdiantffe, rence consecutiisvearitethrmsmetiisc. thereforebetwtheene sequence { 4n 2 } Examine the terms of the sequence: 4, 16, 36,is64,no 100, . . difference; there is no There common common ratio; neither.
33.
35.
{3-�n } d = (3-�(n+1»)- (3-�n) =3--n23 - -23 -3+-n23 = - -23 The dif erence between consecutive terms is
37.
39.
constant, therefore the sequence is arithmetic. 1, 3, 6,is10,no .common . Neithdierfference or common ratio. There
41.
as = 2·3s -1 = 2 . 34 = 2·81 = 162 an = 2 . 3"-1 as = 5(_1)s-1 = 5(-1)4 = 5·1 = 5 an = 5 . (_1)"-1 as = 0 · ( 2"1 )S-1 = 0· (2"1 )4 = l an = O · ( � ) n - = 0 as = h . (htl =h.(hf = h·4=4.[i an =h . (h ),, -1 = (h )" a1 = 1, r = -. 21 ' n = 7 °
64
727
-.
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Chapter 13: Sequences; Induction; the Binomial Theorem
61 .
al = 35 , d = 37-35 = 2 , an = al +(n-1)d = 35+(27 -1)(2) = 35+26(2) = 87 27 27 = 1647 = 2(35+87) = 2(122) The amphitheater has 1647 seats. Thewithyearlal =y35,000, salaries dform= 1400,an arithmet= 280,000 ic sequence. d the $280, number000of. years for the aggregate salary tFionequal =!:[2a 2 l +(n-1)d] 280,000 = !:[2(35, 2 000) + (n -1)1400] 280,000 = n [35,000 + 700n -700] 280,000 = n(700n + 34,300) 280,000 = 700n 2 +34,300n 400 = n 2 +49n n 2 +49n-400 = 0 492-----4(1)(-400) -49± �n = ----'2(1) -49 ± .J4001 -49 63.2 5 2 nIt t::::a: kes7.l3about or2 n8::::years : - 56.13to have an aggregate salary ofyearsat lewiastl be$280,$319,0002.00The. aggregate salary after 8
7.
a 27
S27
63 .
9.
5.
=3
1 2
erms isrconst tTheherefratoreio tofhe1consecut sequenceiviestgeomet ic. a3nt, al = -3 (�) -%, a2 = -3 (�J a3 = -3 (�J -%, a4 = -3 (�J = 136
13.
erms isrconst tTheherefratoreio tofheconsecut sequenceiveistgeomet ic. ant, i- 1 2° -2 ci =-4 = -2 2 = 2 4' c2 = -224--1 = '22i = 2 21 ' 23-1 = '222 2 = 1, C3 = 4 24-1 23 = 2 c4 =--=4 22
Answers h increasen of (anor aridecrease) at asequence constantwiisratlthevary ,e butset. oftBothenatdomai t h met i c unralisnumbers whialllreale the domai n of a l i n ear funct i o t h e set of numbers.
_I
Section 1 3 .3
3.
n l n 3 +-
11.
±
A
3n+n 1 = = -3
l 3 2 24 S3 = 3 = 27, S4 = 3 =81
Sn
----- ::::: ----
1.
r
Thetherefore ratio tofheconsecut iviestgeomet erms isrconst ant, sequence i c . S = 31 = 3, S = 3 = 9,
Sn
65.
Falnegatse;ivthe e(ocommon rathisioresulcantsbeinposia sequence tive or of r 0, but t only Os).
= 1000 (1 + 0.�4 J'2 = $1082. 4 3
geometric annuity
726
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Section 13.2: A rithmetic Sequences
45.
a1 = 4 , d = 4. 5 -4 = 0. 5 , an = a1 +(n-l)d 100 = 4+(n-l)(0. 5 ) 96 =0. 5 (n-l) 192 = n-l 193 = n = �2 (al +an)= 1932 (4+100) = 1 �3 (104) = 1 O,036 a\ = 2(1)-5 = -3, agO = 2(80)-5 = 155 80 (-3+155) = 40(152) = 6080 2 1 1 11 a\ = 6--(1) 2 =-2 , alOo = 6- -2 (100) = -44 = 1�O C21 +(-44)) = 50 ( - 7; ) = -1925 a1 = 14, d = 16-14 = 2 , an = a\ +(n-1)d a120 = 14+(120-1)(2) = 14+ 119( 2) = 252 120 =-(14+252) = 60(266) = 15,960 2
55.
S
57.
SgO =
49.
S
SIOO
51.
=
S120
53 .
= 2 [2(25) + (30 -1)(1)] = 15(50 + 29) = 15(79) There= 1185 are 1 185 seats in the theater. Thetolmightrower colandored1 tilteileins have 20p rowtiles iThe n. the botnumber t h e t o up thwie th ta\ria=ngl20,e.decreases Thid =s-1,is anandbyarin1th=asmet20.weicmove sequence Fi n d t h e sum: = 202 [2(20)+(20-1)(-1) ] = 10(40 -19) = 10(21) There= 210are 210 lighter tiles. colandored1 titleileins have 19p row tiles. iThen the botThenumber todarker m rowdecreases t h e t o up thwie th ta\riangl19,e. dThi=s-1,is anandbyarin1t=hasmet19we.icFimove sequence nd the sum: =�[2(19)+(19-1)(-1) ] 2 =�(38-18) =�(20) = 190 2 2 There are 190 darker tiles. The Siairncecooln srepresent at the ratsethofousands 5. 5 ° F per 1000, wefeet. of feet have d = -5. 5 . The ground temperature is 67°F so we have I; = 67 -5. 5 = 61. 5 . Therefore, {Tn} = { 61. 5 + (n -1)(-5. 5 ) } = {-5. 5 n+67} or {67-5. 5 n} Afthaveer 1'sthe=parcel of ai r has ri s en 5000 feet, we 61.5 +(5-1)(-5. 5 )=39. 5 . The5000parcel feet. of air wil be 39.5°P after it has risen S30
Sn
47.
The= t25ota+l 26number of seat s i s : . . + 27 + . + (25 + 29 (1)) Thid =s1,is a\the=sum25, ofandan arin =th30met. ic sequence with Find th30e sum of the sequence:
S
Fisolnvdetthheecommon diequat fferenceions:of the terms and syst e m of (2x+1)-(x+3) = d x-2 = d �
(5x+2)-(2x+l) = d 3x+l = d 3x+l = x-2 2x = -3 X = - -23 �
59.
725
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exist. No portion of this material may be reproduced , in any form or by any means , without permi ssion in writing from the publisher.
Chapter 13: Sequences; Induction; the Binomia l Theorem
25.
27.
29.
31.
al = 2, d = "25 -2 = "21 ' an = al +(n-l)d agO = 2+(80-1)-21 = -832 ag = al +7d = 8 a20 = al +19d = 44
33.
ions by subtracting the Solfirstveeqthuate systion efromm oftheequatsecond: 12d = 36 � d = 3 al =8-7(3) = 8-21 = -13 : a rmul o f e v Recursi al = -13 an = a, _1 + 3 nth term: an = al + ( n -1) d
alg = al +17d = -9 ions by subtracting the Solfialr4stv=eeqaltuathe+13dsystion efr=mom-1ofthequat second: e 4d = -8 � d = -2 al = -1-13(-2) = -1+26 = 25 al = 25 a" = an _I -2 : a rmul o f e v Recursi nth term: an = al +(n-l)d = 25+(n-l)(-2) = 25 -2n+2 = 27 -2n
= -13+(n-l)(3) = -13+3n -3 = 3n-16 a9 = al +8d = -5 aI 5 = al +14d =31 ions by subtracting the Solfirstveeqtuathe systion efromm oftheequatsecond: 6d =36 � d = 6 al = -5-8(6) = -5-48 = -53 al = -53 a" = an_I + 6 : a rmul o f e v Recursi nth term: an = al + ( n -1) d = -53+(n -l)(6) = -53+6n-6 = 6n-59 al 5 = al +14d = 0 a40 = al +39d = -50 ions by subtracting the e systion efrmomoftheequatsecond: fiSol25drstvee=qthuat-50� d = -2 al = -14(-2) = 28 : a rmul o f e v Recursi al = 28 an = an _I -2 nth term: an = al +(n -1)d = 28+(n-l)(-2) = 28-2n+ 2 = 30-2n
39.
al = 2, d = 4 -2 = 2, an = al + (n -1)d 70 = 2 +(n-l)2 70 = 2+2n-2 70 = 2n n = 35 =-n2 ( al + an ) = -352 (2+70) =�(72) 2 = 35(36) = 1260 al = 5, d = 9-5 = 4, an = al +(n- l)d 49 =5+(n-l)4 49 = 5+4n -4 48 = 4n n = 12 Sn = "2n (al +an ) = 212 (5+49) = 6(54) =324 al = 73 , d = 78 -73 = 5, an = a1 + ( n ) d 558 = 73+(n-l)(5) 485 = 5(n -1) 97 = n-l 98 = n =!:(2 al + an) = 982 (73 + 558) = 49( 631) = 30,919
S"
41.
43.
-1
S"
724
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Section 13.2: Arithmetic Sequences
Section 13.2 1. 3.
5.
7.
arithmetic
d = sn - sn _1 = (n + 4) -(n -1 + 4) = (n + 4) -(n + 3) = w1 een consecutive terms is Theconst= din+4-n-3 fanterence bet , therefore the sequence is arithmetic SI = 1+4 = 5, s2 = 2+4 = 6, S3 = 3+4 = 7, . s4 =4+4 = 8 d = an -an_I = (2n -5)-(2(n -1)-5) =(2n-5)-(2n-2-5) = 2n -5 -2n + 7 = 2 Theconstdiantf e, rence betorewtheene seconsecut isvearitethrmsmetiisc. t h eref uence i q al = 2 ·1-5 = -3, a2 = 2 . 2 -5 = -1, a3 = 2 . 3 -5 = 1, a4 = 2 ·4 -5 = 3 d =cn -cn_1 = (6 -2n) -(6-2(n-l» = ( 6 -2n ) -( 6 -2n + 2) = 6 -2n -6 + 2n -2 = -2
13.
1 5.
=
1 7.
Theconstdiafntfe, rence betorewtheene seconsecut isvearitethrmsmetiisc. t h eref uence i q cl = 6 -2 ·1 = 4, c = 6 -2·2 = 2,
9.
= In ( 3n ) - In ( 3n-l ) = nln(3 )-(n-l)ln(3) = (ln3)(n -( n -1) = (ln3)( n -n + 1) Theconst= ln3diantffe, rence betorewtheene seconsecut isvearitethrmsmetiisc. t h eref uence i q SI = In ( 31 ) = ln(3), S2 = In ( 32 ) = 2ln(3), S3 = In (33 ) = 3ln(3), S4 = In ( 34 ) = 4 ln (3) an = al +(n-l)d = 2+(n-l)3 = 2+3n-3 = 3n-l aS I = 3·51-1 = 152 an = al +(n-l)d 5 +(n -1)(-3) = 5-3n+3 = 8-3n aS I = 8-3·51 = -145 an = al +(n-l)d = O +(n-l)-21 1 1 =-n-2 2 =!(n-l) 2 1 aS I =-(51 2 -1) = 25 an = al +(n-l)d = ..fi +'(n -1)..fi = ..fi + ..fin - ..fi = ..fin aS I = 51..fi al = 2, d = 2, an = al + (n-l)d al Oo = 2 + (100 -1)2 = 2 + 99(2) 2 + 1 9 8 = 200 al = 1, d = -2-1 = -3, an =al +(n-l)d a90 = 1+(90-1)(-3) = 1+89(-3) = 1-267 = -266
c3 = 6-2·3 = 0, c24 = 6-2·4 = -2 d = tn - tn_I = (� - ± n ) - (� - ± (n -1» ) = (� - ± n ) - (� - ± n+ ±) 1 -1 n- -1 = - -1 =-21 - -31 n--+ 2 3ween3consecut 3 ive terms is Theconstdiantf e, rence bet therefore the se1 quence is ari1 thmetic. 1 1 1 tI =-2 - -3 ·1 =-6' t2 =-2 - -3 ·2 = - -6' 1 -1 ·3 = - -1 t4 =--1 -1 ·4 = - -5 t3 =-2 3 2' 2 3 6
1 9.
1
21.
=
23 .
723
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Chapter 13: Sequences; Induction; the Binomial Theorem
81.
BI = 1.0 IBo -100 = 1. 0 1(3000) -100 = $2930 $2930
John's balance is payment. 83.
Phil' s balance is first payment.
b. c.
d.
b.
After months there are rabbits . 87.
This is the Fibonacci sequence.
after making the
k=O
k =O
' '
�. '(, 1 . �� 669296668
It will take
n
= 12
to approximate
1 ( 1. 3 ) = el .3
correct to
8
decimal places.
al = 0.4 , a2 = 0.4 + 0. 3 . 22-2 = 0.4 + 0.3 = 0. 7 , a3 = 0.4 + 0. 3 . 23-2 = 0.4 + 0. 3 ( 2 ) = 1. 0 , a4 = 0.4 +0 .3 . 264-2 = 0.4 +0. 3 ( 4 ) = 1. 6 , as =0.4 +0. 3 . 25-27 = 0.4 +0.3 ( 8 ) = 2. 8 , a6 = 0.4 +0.3 . 2 -2 = 0.4 +0. 3 ( 16) = 5. 2 , a7 = 0.4 +0. 3 . 2 -2 = 0.4 +0. 3 ( 32 ) = 10.0 , as = 0.4 + 0.3 · 2S-2 = 0.4 + 0.3 ( 64) = 19. 6 0.4, 0.7 , 1.0, 1. 6, 2. 8 , 5.2 , 10. 0, 19. 6 . 5, a5 = 2. 8 , a9 = 0.4 + 0. 3 . 29-2 = 0.4 + 0. 3 ( 128) = 38. 8 alO = 0.4 + 0.3 · i O-2 = 0.4 + 0. 3 ( 256) = 77.2 a9 , all = 0.4 +0. 3 ·il-2 = 0.4 +0. 3 ( 512 ) = 154 of2003 313 154
The first eight terms of the sequence are
and
Except for term which has no match, Bode 's formula provides excellent approximations for the mean distances of the planets from the sun.
c.
The mean distance of Ceres from the Sun is approximated by
e.
Pluto 's distance is approximated by
d.
f.
mature pairs of
1 ( 1. 3 ) = el .3 "" ± It: = 1 O!.30 + .!2.l ! + ... + 1 4.34! ",, 3 . 630170833 7 1 ( 1. 3 ) = el .3 "" ± I ;: = 1.O3!0 + 1.l!3 1 + ... + 1.7!3 "" 3. 669060828 1 ( 1. 3 ) = el .3 "" 3.669296668
uM ( se� ( 1 . 3"n/n ! , n , \3 , 1 3 » 3 . 669296667 uM ( se"l ( 1 . 3"n/n ! , n , \3 , 1 2 » 3 . 669296662 9 1 . a.
=
after making the first
BI = 1. 005Bo -534.47 = 1. 005(18,500) -534.47 = $18,058.03 $18,058.03
89. a.
al = 1, a2 1, a3 = 2, a4 = 3, as = 5, a6 = 8, a7 = 13, as = 21, an = an_I + an-2 ag = a7 + a6 = 13 + 8 = 21 7 21 1, 1, 2, 3, 5, 8, 13
85.
and that of Uranus is
as = 19. 6 .
but no term approximates Neptune ' s mean distance from the sun.
According to Bode ' s Law, the mean orbital distance
UB
will be
AU from the sun.
93 . Answers will vary.
722
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Section 13. 1: Sequences
35.
3 7.
39.
a1 = 2, a2 = 3 + 2 = 5, a3 = 3 + 5 = 8, a4 =3+8=11, as = 3+11 = 14 a1 = -2, a2 = 2+(-2) = 0, a3 =3+0=3, a4 = 4+3 = 7, as = 5+7 = 12 a1 = 5, a2 = 2·5 = 10, a3 = 2·10 = 20, a4 = 2· 20 = 40, as = 2 . 40 = 80
61.
63.
67.
1 2 -3 + . .. +-13 = I13 --k -+-+ 2 3 4 13 + 1 k + 1 1 + "' +(_1) 6 (�) = ± (_1)k (�) 1_�+� 3 9 27 3 k=O 3 hI
_
a + (a + d ) + (a + 2d) + . . · + (a + nd) = or
43.
45.
47.
49. 51.
53.
55.
57.
59.
a1 = 1, a2 = 2, a3 = 2 ·1 = 2, a4 = 2 . 2 4, as =4·2 = 8 a1 = A, a2 = A+d, a3 = (A+d)+d = A+2d, a4 = (A + 2d) + d = A + 3d, as = (A +3d)+d = A+4d a1 = .fi, a2 = f2;J2, a3 = �2 + �2 + .fi , Q4 = �2+ �2+ f2;J2 , as = �2+ �2+ �2+ �2+.fi k=1I (k +2) = 3+4+5 +6+ 7 + . .. +(n+ 2} n2 Ik=1 -k22 = 2 1 . . · + -1 Ik=O -31k = I+ -31 + -91 +-+ 27 3" 1 "' + -1 Ik=O k3 1+1 = -31 + -91 +-+ 27 311 k k=I2 (-I) lnk = ln2 - ln 3+ ln 4- .. . +(- lt lnn 1+2+3+ . . · +20 = k=1I20 k
69.
=
71.
73.
75.
II
II
9 49 25 + 2 + - + 8 + - + 1 8 + - + 32 + . . . + 2 2 2 2
1
77.
II
11 -1
79.
_
= k=1I (a+(k-1}d) II
II
= 40(5} = 200 Ik=140 5 = � 04 I40 k = 40(40+1) 2 = 20(41} = 820 20 = 5Ik+ 20 I20 3 Ik�20 (5k +3) = k�I20 (5k )+ k�I3 k� k� = 5 ( 20(2� + I} ) + 3(20} = 1050 + 60 = 1110 Ik=116 (k 2 + 4 ) = k=1I16 k 2 + k=1I16 4 = 16(16+1}(26 . 16+1} +4(16} = 1496 + 64 = 1560 k�0 2k = 2k�/k = 2 [� k - � k ] = 2 [ 60( 6� + I} _ 9(92+ 1} ] = 2[1830-45] = 3570 Ik=S20 k3 =k=1I20 k3 - k=1I4 e = [ 20(2� +I} _ [ 4(:+I} = 2102 -102 = 44,000 times
hi
J
II
k=OI (a+kd)
J
72 1
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Chapter 13 Sequences ; Induction; the Binomial Theorem Section 1 3 . 1 1.
1(2) = 2-12 =.!.2 '. 1(3) = 3 3-1 = 3.3
23 .
7. True; a sequence is a function whose domain is
the set of positive integers.
11.
13.
1 5.
1 7.
19.
10! = 10·9·8·7 ·6·5 ·4·3 ·2·1 = 3,628,800 9! = 9 .8 . 7 . 6! = 9 . 8 . 7 = 504 6! 6! 3!· 7! 3·2·1·7·6·5·4! 4! . 4! . = 3 2 ·1· 7 ·6 5 = 1,260 SI = 1, S2 = 2, S3 = 3, S4 = 4, S5 = 5 2 2 1 1 1 a1 - -1+2 - -3' a2 - -2+2 - -4 - -2 ' 4 4 2 a3 - 3+23 - 5'3 a4 - -4+2 - -6 - -3 ' 5 5 a5 =--=5+2 7 cI = (_1)1 +1 (1 2 ) = 1, c2 = (_1)2+ 1 (22 ) = _4, c3 = (-1)3+1 (32 ) = 9,c4 = ( _ 1)4+ 1 (42 ) = _16, (_1)5+1 (5 2 ) 25 2 2 = -4 =-2 S -- 31i+-1 -- 24 - -21 ' S2 =-3 2 + 41 10 5 ' S3 = 3323+ 1 = 288 = 72 ' S4 = 3 42 + 1 = 8216 = 418 ' S5 = 352+5 1 = 24432 = 618 Cs
21.
=
-
t
3. sequence
9.
1 (_1)1 t1 - (1+1)(1+2) 2·3 - - "6 ' 2 - 1 -1 (_1) 2 (2+1)(2+2) 3·4 - 12 ' -1 = - 1 = t3 = (3 +(_1)3 4· 1)(3 + 2) 5 20' (_1)4 - 1 - 1 t4 - (4+1)(4+2) 5·6 30' (_1) 5 = -1 = - 1 t5 = (5+1)(5+2) 6 .7 42 1
-
27. Each telm is a fraction with the numerator equal
to the term number and the denominator equal to one more than the term number.
n an = - n+1 1
2.
29. Each term is a fraction with the numerator equal
to and the denominator equal to a power of The power is equal to one less than the term number.
a
Ii
=-n2 1-1
-1 n -1 1.
3 1 . The terms form an alternating sequence. Ignoring
the sign, each term always contains a The sign alternates by raising to a power. Since the first term is positive, we use as the power .
all = (_1)"-1
=
I
33. The terms ( ignoring the sign) are equal to the
. ( l)n+1 . an = ( -1 )11+1 ·n
term number. The alternating sign is obtained by
usmg -
720
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Chapter 12 Cumulative Review
j.
x 2 - 2x - 4y + I = O x 2 - 2x + 1 = 4y 4y = (x _ l)
13.
2
4
(2 cos x - I)(cos x + 2) =
I.
y = 3 sin(2x)
00
I cos x = - or cos x = -2 2 Since cos x = -2 is impossible, we are left with 1 cos x = 2 Jr x = ± + 2Jrk, where k is an integer
3
The solution set is
{i ;
-2 Y
}
2 cos 2 x + 3 cos x - 2 =
-1
k.
(
2 - 2 cos 2 x = 3 cos x
1 y = - (x - l) 2
y
2 sin 2 x = 3 cos x 2 I - cos 2 x = 3 cos x
xx=±
= sm x
+ 2Jrk, k is any integer
}
2Jr . Peno d : - = Jr 2 Amplitude: 3 y 3 x
71 9
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Chapter 12: Systems of Equations and Inequalities
y = X3
c.
g.
Y
y = In x
X=O y
x x
y
d.
=
1
h.
-
x
Y�
-2 2 1�
�l (- 1 , - 1)
y = ..Jx
e.
Yt 2
�
-
x
n
=
=
1
v
1
�
/'
! 'Ii
71 -\
x
-2
y
5
+
( -'M-i' ,O.) (1
2[-
2
·
(O,-or f.
[� r [ H
The graph is an ellipse. 2 x2 � + =1
2
2
2X2 + 5y 2 = 1
i.
(o.vi-) . ' \
\
( 't ·O)
x
x 2 _ 3y 2 = 1 The graph is a hyperbola x2 y 2 =
eX
I
- - -
1
(fJ' x
y = o- -
-[ � )' = 3
1
1 y
-1
718
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Chapter 1 2 Cumulative Review
Chapter 1 2 Cumulative Review
1.
9.
x(x2X2=2x-10-xor=) =002x -1 = 0 x=The solution set is { �} . -8x-3 =2x30 -3x2 -8x -3 appears to Thehave2x3 -3x2 graph of = an x-intercept at x = 3 . Using synthetic division: -3 -8 6 9 3 3 Therefore, 2x3 -3x2 -8x -3 = 0 (x -3) (2X2 + 3x + 1) = 0 (x -3)(2x+ 1)(1 x+ 1) = 0 x = 3 or x = -- or x = -1 The solution set is {-1, -�, 3} . log3 (lxo-1g ) + log3 ++ 1) )== 3 ((x -1)(2x =32 2X2 -x-1= 9 2X25)(x + == 00 X=-5 orx=-2 Since x = makes the original logarithms undefined, the solution set is {%} . g(x) = -x4-x3+ 1 g(-x) = (-x + 1 =-x-4-+x 1 = -g(x) Thus, griics anwitoddh respect functitoonthande oriitsgigraph symmet n. i s
f(x) = 3x 2 + 1 Usihorinzgontthaelgraph of y = 3x , shi ft t h e graph ts to the riunightt. , then shift the graph vertiycallyuniupward 2
5
1
2
......---1'.:-.- - - - - x
O,
3.
5
-5
1';
-5
Domai n : Range: (1, Horizontal Asymptote: y = 6 is a line. Thex-y i=nt3xgraph e3x+6 rcept: y-y=3(0)+6 intercept: 0= 3x=-6 =6 x=-2 ( -00, (0 )
-3
11
o
2
1
y
.
a.
1
(0
)
+
2
5.
(2x
I)
2 2
(x - 1)(2x + l)
(2x -
- x - lO 2)
b.
2
)4
y
and
( 0, 2) .t
( - 2, 0)
2
2(-X)3
4
y2 = is a circle with center (0, 0) Theradix2 +ugraph s 2.
-2
7.
.Y
2
2 3
(0. -2)
717
© 2008 Pearson Education, Inc:, Upper Saddle River, N J . A U rights reserved. This material i s protected under a U copyright laws a s they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 12: Systems of Equations and Inequalities
2x-3y � 2 2(0)-3(0)� 2 �2 (0, 0) °
false
The point
2x -3 � 2 y = �3 x - �3 .
is not a solution. Thus, the
graph of the inequality
Y
includes the
28. Let
=
=
unit price for flare j eans, unit price for camisoles, and unit price for t-shirts. The given information yields a system of equations with each of the three women yielding an equation. (Megan)
Because the inequality is non-strict, the line is also part of the graph of the solution. The overlapping shaded region (that is, the shaded region in the graph below) is the solution to the system of linear inequalities. 8
Value of obj . function,
From the table, we can see that the maximum value of is and it occurs at the point
half-plane below the line
.Y
(x, y) z (0,1) z = 5(0)+8(1) = 8 (3,2) z = 5(3)+8(2) = 31 (0,8) z = 5(0)+8(8) =64 z 64, (0,8) . i c t
Comer point,
?
=
{2ii + 2c+3t+ 4t==42.905
2 2 Y = -x 3 3 --
i + 3c 2t = 62 +
(Paige) (Kara)
We can solve this system by using matrices.
y=
-
5= 4 5 ( R) = til ) 1 3 2 ��62 1 3 2 ��62 = 1 -5 �2 � ��17 1 = 2 � ;'17� = 1 � � �2 12 = ( R, = h l z=6 y -z = 2. 5 y-z = 2. 5 y-6 = 2. 5 y = 8. 5 z = 6 x + 3z = 42. 5 x+3z = 42. 5 x+3( 6) = 42. 5 x = 24. 5 $24 5 0, $8. 50, $6.. 00.
[� � �
1 2X + 4
( 4,2) (8,0) . z = 5x + 8y. z x O �:+ y � 8 x-3y � -3 2x + y = 8 x -3y = -3 y = -2x+8 -3y = -x-3 y = -31 x+1
The graph is unbounded. The comer points are and 27. The objective function is
We seek
the largest value of that can occur if and y are solutions of the system of linear inequalities
{
l [� � � l [� l [� l [ ��:l [� ! �l �:l °
°
°
°
° °
The last row represents the equation Substituting this result into
The graph of this system (the feasible points) is shown as the shaded region in the figure below. The comer points of the feasible region are and
. (from the
second row) gives
(0,1) , (3,2), (0,8) . y
Substituting first row) gives x - 3y = -3
into
Thus, flare j eans cost and t-shirts cost
2x + y = 8
(from the
camisoles cost
716
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Chapter 12 Test
24.
A+B = O C=O 6A+3B+D = 4 3C+E = 0 9A = -3
3x+7 (x +3)2 x+3 3x+7 --+ A B ----;:-= (X+3)2 x+3 (X+3)2
The denominator contains the repeated linear factor . Thus, the partial fraction decomposition takes on the form -"'7'
-
From the last equation we get
(x + 3)2 3x + 7 = A ( x + 3) + B 3x+ 7 = Ax+(3A+ B)
Substituting this value into the first equation
Clear the fractions by multiplying both sides by .
gives
The result is the identity
x
A=3.
7 = 3A+B 7 = 3(3)+B -2 = B 3x+7 =--+ 3 -2 ' x+3 (x+3)2 (x+3)2 4x2 -3 X(X2 +3t
2 6.
and
{�2x-3y:L�8
2
The inequalities
x x2 + 3 .
The denominator contains the linear factor
A = _ .l3 6 ( - t) +3 (t) +D = 4 -2+1+D = 4 D=5 4x2 -3 -3 3 x +---. 5x ---.=-+ x 2 +3) (x 2 (x2 +3)2 x(x2 +3)
into the third
Therefore, the partial fraction decomposition is 1 1
Thus, the partial fraction decomposition is
x�0 y�0
x+2y � 8 1 y � --x+4 2 (0,0) . x+2y � 8 0+2(0) � 8 �8 (0,0)
and
require that
the graph be in quadrant I .
and
the repeated irreducible quadratic factor The partial fraction decomposition takes on the form
4x2 -3 = -+ A --Bx+ C + ----::Dx+E -----;:X x2 + 3 X ( x2 + 3 t ( x2 3 t x( x2 + 3 r 4x2 - 3 = A (X2 + 3) 2 + X(X2 + 3)(Bx + C) + x(Dx + E) 4x2 -3 = (A + B)X4 + Cx3 +( 6A+3B+D)x2 +(3C+E)x+(9A)
Test the point
+
o
We clear the fractions by multiplying both sides by
E=0. B =.l3
equation gives us
Therefore, we have Substituting this result into the second equation gives
25.
From the second equation, we
Substituting
W e equate coefficients of like powers of to obtain the system
7 = 3A+B
B =!3 . C=0.
know Substituting this value into the fourth equation yields
or
{3 = A
A = -!3 .
to obtain the identity
?
false
The point
graph of the inequality
Collecting like terms yields
x + 2y � 8 y = -!2 x 4 .
is not a solution. Thus, the
half-plane above the line
includes the +
Because
the inequality is non-strict, the line is also part of the graph of the solution.
Equating coefficients, we obtain the system
2x-3y � 2 2 2 y :'O: -x-3 3 (0,0) .
Test the point 715
© 2008 Pearson Education, I nc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
