1. SOLUTION MANUAL

2. THOMAS CALCULUS TWELFTH EDITION BASED ON THE ORIGINAL WORK BY George B. Thomas, Jr. Massachusetts Institute of Technology AS REVISED BY Maurice D. Weir Naval Postgraduate School Joel Hass University of California, Davis INSTRUCTORS SOLUTIONS MANUAL SINGLE VARIABLE WILLIAM ARDIS Collin County Community College 608070 _ISM_ThomasCalc_WeirHass_ttl.qxd:harsh_569709_ttl 9/3/09 3:11 PM Page 1 3. The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson Addison-Wesley from electronic files supplied by the author. Copyright 2010, 2005, 2001 Pearson Education, Inc. Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-321-60807-9 ISBN-10: 0-321-60807-0 1 2 3 4 5 6 BB 12 11 10 09 This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permit- ted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes.All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. 608070 _ISM_ThomasCalc_WeirHass_ttl.qxd:harsh_569709_ttl 9/3/09 3:11 PM Page 2 4. PREFACE TO THE INSTRUCTOR This Instructor's Solutions Manual contains the solutions to every exercise in the 12th Edition of THOMAS' CALCULUS by Maurice Weir and Joel Hass, including the Computer Algebra System (CAS) exercises. The corresponding Student's Solutions Manual omits the solutions to the even-numbered exercises as well as the solutions to the CAS exercises (because the CAS command templates would give them all away). In addition to including the solutions to all of the new exercises in this edition of Thomas, we have carefully revised or rewritten every solution which appeared in previous solutions manuals to ensure that each solution conforms exactly to the methods, procedures and steps presented in the text is mathematically correct includes all of the steps necessary so a typical calculus student can follow the logical argument and algebra includes a graph or figure whenever called for by the exercise, or if needed to help with the explanation is formatted in an appropriate style to aid in its understanding Every CAS exercise is solved in both the MAPLE and computer algebra systems. A template showingMATHEMATICA an example of the CAS commands needed to execute the solution is provided for each exercise type. Similar exercises within the text grouping require a change only in the input function or other numerical input parameters associated with the problem (such as the interval endpoints or the number of iterations). For more information about other resources available with Thomas' Calculus, visit http://pearsonhighered.com. 5. TABLE OF CONTENTS 1 Functions 1 1.1 Functions and Their Graphs 1 1.2 Combining Functions; Shifting and Scaling Graphs 8 1.3 Trigonometric Functions 19 1.4 Graphing with Calculators and Computers 26 Practice Exercises 30 Additional and Advanced Exercises 38 2 Limits and Continuity 43 2.1 Rates of Change and Tangents to Curves 43 2.2 Limit of a Function and Limit Laws 46 2.3 The Precise Definition of a Limit 55 2.4 One-Sided Limits 63 2.5 Continuity 67 2.6 Limits Involving Infinity; Asymptotes of Graphs 73 Practice Exercises 82 Additional and Advanced Exercises 86 3 Differentiation 93 3.1 Tangents and the Derivative at a Point 93 3.2 The Derivative as a Function 99 3.3 Differentiation Rules 109 3.4 The Derivative as a Rate of Change 114 3.5 Derivatives of Trigonometric Functions 120 3.6 The Chain Rule 127 3.7 Implicit Differentiation 135 3.8 Related Rates 142 3.9 Linearizations and Differentials 146 Practice Exercises 151 Additional and Advanced Exercises 162 4 Applications of Derivatives 167 4.1 Extreme Values of Functions 167 4.2 The Mean Value Theorem 179 4.3 Monotonic Functions and the First Derivative Test 188 4.4 Concavity and Curve Sketching 196 4.5 Applied Optimization 216 4.6 Newton's Method 229 4.7 Antiderivatives 233 Practice Exercises 239 Additional and Advanced Exercises 251 5 Integration 257 5.1 Area and Estimating with Finite Sums 257 5.2 Sigma Notation and Limits of Finite Sums 262 5.3 The Definite Integral 268 5.4 The Fundamental Theorem of Calculus 282 5.5 Indefinite Integrals and the Substitution Rule 290 5.6 Substitution and Area Between Curves 296 Practice Exercises 310 Additional and Advanced Exercises 320 6. 6 Applications of Definite Integrals 327 6.1 Volumes Using Cross-Sections 327 6.2 Volumes Using Cylindrical Shells 337 6.3 Arc Lengths 347 6.4 Areas of Surfaces of Revolution 353 6.5 Work and Fluid Forces 358 6.6 Moments and Centers of Mass 365 Practice Exercises 376 Additional and Advanced Exercises 384 7 Transcendental Functions 389 7.1 Inverse Functions and Their Derivatives 389 7.2 Natural Logarithms 396 7.3 Exponential Functions 403 7.4 Exponential Change and Separable Differential Equations 414 7.5 Indeterminate Forms and L'Hopital's Rule 418^ 7.6 Inverse Trigonometric Functions 425 7.7 Hyperbolic Functions 436 7.8 Relative Rates of Growth 443 Practice Exercises 447 Additional and Advanced Exercises 458 8 Techniques of Integration 461 8.1 Integration by Parts 461 8.2 Trigonometric Integrals 471 8.3 Trigonometric Substitutions 478 8.4 Integration of Rational Functions by Partial Fractions 484 8.5 Integral Tables and Computer Algebra Systems 491 8.6 Numerical Integration 502 8.7 Improper Integrals 510 Practice Exercises 518 Additional and Advanced Exercises 528 9 First-Order Differential Equations 537 9.1 Solutions, Slope Fields and Euler's Method 537 9.2 First-Order Linear Equations 543 9.3 Applications 546 9.4 Graphical Solutions of Autonomous Equations 549 9.5 Systems of Equations and Phase Planes 557 Practice Exercises 562 Additional and Advanced Exercises 567 10 Infinite Sequences and Series 569 10.1 Sequences 569 10.2 Infinite Series 577 10.3 The Integral Test 583 10.4 Comparison Tests 590 10.5 The Ratio and Root Tests 597 10.6 Alternating Series, Absolute and Conditional Convergence 602 10.7 Power Series 608 10.8 Taylor and Maclaurin Series 617 10.9 Convergence of Taylor Series 621 10.10 The Binomial Series and Applications of Taylor Series 627 Practice Exercises 634 Additional and Advanced Exercises 642 7. TABLE OF CONTENTS 10 Infinite Sequences and Series 569 10.1 Sequences 569 10.2 Infinite Series 577 10.3 The Integral Test 583 10.4 Comparison Tests 590 10.5 The Ratio and Root Tests 597 10.6 Alternating Series, Absolute and Conditional Convergence 602 10.7 Power Series 608 10.8 Taylor and Maclaurin Series 617 10.9 Convergence of Taylor Series 621 10.10 The Binomial Series and Applications of Taylor Series 627 Practice Exercises 634 Additional and Advanced Exercises 642 11 Parametric Equations and Polar Coordinates 647 11.1 Parametrizations of Plane Curves 647 11.2 Calculus with Parametric Curves 654 11.3 Polar Coordinates 662 11.4 Graphing in Polar Coordinates 667 11.5 Areas and Lengths in Polar Coordinates 674 11.6 Conic Sections 679 11.7 Conics in Polar Coordinates 689 Practice Exercises 699 Additional and Advanced Exercises 709 12 Vectors and the Geometry of Space 715 12.1 Three-Dimensional Coordinate Systems 715 12.2 Vectors 718 12.3 The Dot Product 723 12.4 The Cross Product 728 12.5 Lines and Planes in Space 734 12.6 Cylinders and Quadric Surfaces 741 Practice Exercises 746 Additional Exercises 754 13 Vector-Valued Functions and Motion in Space 759 13.1 Curves in Space and Their Tangents 759 13.2 Integrals of Vector Functions; Projectile Motion 764 13.3 Arc Length in Space 770 13.4 Curvature and Normal Vectors of a Curve 773 13.5 Tangential and Normal Components of Acceleration 778 13.6 Velocity and Acceleration in Polar Coordinates 784 Practice Exercises 785 Additional Exercises 791 Copyright 2010 Pearson Education Inc. Publishing as Addison-Wesley. 8. 14 Partial Derivatives 795 14.1 Functions of Several Variables 795 14.2 Limits and Continuity in Higher Dimensions 804 14.3 Partial Derivatives 810 14.4 The Chain Rule 816 14.5 Directional Derivatives and Gradient Vectors 824 14.6 Tangent Planes and Differentials 829 14.7 Extreme Values and Saddle Points 836 14.8 Lagrange Multipliers 849 14.9 Taylor's Formula for Two Variables 857 14.10 Partial Derivatives with Constrained Variables 859 Practice Exercises 862 Additional Exercises 876 15 Multiple Integrals 881 15.1 Double and Iterated Integrals over Rectangles 881 15.2 Double Integrals over General Regions 882 15.3 Area by Double Integration 896 15.4 Double Integrals in Polar Form 900 15.5 Triple Integrals in Rectangular Coordinates 904 15.6 Moments and Centers of Mass 909 15.7 Triple Integrals in Cylindrical and Spherical Coordinates 914 15.8 Substitutions in Multiple Integrals 922 Practice Exercises 927 Additional Exercises 933 16 Integration in Vector Fields 939 16.1 Line Integrals 939 16.2 Vector Fields and Line Integrals; Work, Circulation, and Flux 944 16.3 Path Independence, Potential Functions, and Conservative Fields 952 16.4 Green's Theorem in the Plane 957 16.5 Surfaces and Area 963 16.6 Surface Integrals 972 16.7 Stokes's Theorem 980 16.8 The Divergence Theorem and a Unified Theory 984 Practice Exercises 989 Additional Exercises 997 Copyright 2010 Pearson Education Inc. Publishing as Addison-Wesley. 9. CHAPTER 1 FUNCTIONS 1.1 FUNCTIONS AND THEIR GRAPHS 1. domain ( ); range [1 ) 2. domain [0 ); range ( 1] c_ _ _ _ c_ 3. domain 2 ); y in range and y 5x 10 y can be any positive real number range ). c _ b ! ! _ 4. domain ( 0 3, ); y in range and y x 3x y can be any positive real number range ). c_ r _ c ! ! _ 2 5. domain ( 3 3, ); y in range and y , now if t 3 3 t , or if t 3 c_ r _ c ! ! 4 4 3 t 3 tc c 3 t y can be any nonzero real number range 0 ). c ! ! c_ r ! _4 3 tc 6. domain ( 4, 4 4, ); y in range and y , now if t t 16 , or if c_ c% r c r _ c% c ! !2 2 t 16 t 16 2 2 2c c t 4 16 t 16 , or if t t 16 y can be anyc% c c ! c ! % c ! ! 2 22 2 t 16 t 16 # "' c c2 2 nonzero real number range ). c_ c r ! _1 8 7. (a) Not the graph of a function of x since it fails the vertical line test. (b) Is the graph of a function of x since any vertical line intersects the graph at most once. 8. (a) Not the graph of a function of x since it fails the vertical line test. (b) Not the graph of a function of x since it fails the vertical line test. 9. base x; (height) x height x; area is a(x) (base)(height) (x) x x ; b # # # # # # # # # " " x 3 3 3 4 perimeter is p(x) x x x 3x. b b 10. s side length s s d s ; and area is a s a d b # # # # #" # d 2 11. Let D diagonal length of a face of the cube and the length of an edge. Then D d and j j b # # # D 2 3 d . The surface area is 6 2d and the volume is .# # # # # # $ $# j j j j j d 6d d d 3 3 33 3 # # $ 12. The coordinates of P are x x so the slope of the line joining P to the origin is m (x 0). Thus, x x x " x, x , . " " m m# 13. 2x 4y 5 y x ; L x 0 y 0 x x x x xb c b c b c b c b b c b" " " # # 5 5 5 25 4 4 4 4 16 2 2 2 2 2 2 x x c b 5 5 25 20x 20x 25 4 4 16 16 4 2 20x 20x 252 2 c b c b 14. y x 3 y 3 x; L x 4 y 0 y 3 4 y y 1 y c b c b c b c b c b 2 2 2 2 2 2 2 2 2 y 2y 1 y y y 1 c b b c b 4 2 2 4 2 Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 10. 2 Chapter 1 Functions 15. The domain is . 16. The domain is .a b a bc_ _ c_ _ 17. The domain is . 18. The domain is .a bc_ _ c_ ! 19. The domain is . 20. The domain is .a b a b a b a bc_ ! r ! _ c_ ! r ! _ 21. The domain is 5 5 3 3, 5 5, 22. The range is 2, 3 .a b a bc_ c r c c r r _ 23. Neither graph passes the vertical line test (a) (b) Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 11. Section 1.1 Functions and Their Graphs 3 24. Neither graph passes the vertical line test (a) (b) x y 1 x y y 1 x or or x y y x k k b b " c b c" c" c 25. x 0 1 2 26. x 0 1 2 y 0 1 0 y 1 0 0 27. F x 28. G x 4 x , x 1 x 2x, x 1 , x 0 x, 0 x a b a b c b 2 2 x " 29. (a) Line through and : y x; Line through and : y x 2a b a b a b a b! ! " " " " # ! c b f(x) x, 0 x 1 x 2, 1 x 2 c b (b) f(x) 2, x x 2 x x ! " ! " # # $ ! $ % 30. (a) Line through 2 and : y x 2a b a b! # ! c b Line through 2 and : m , so y x 2 xa b a b a b " & ! c c c b " c b! c " c" " " " & & c # $ $ $ $ $ f(x) x , 0 x x , x c b # # c b # & " & $ $ (b) Line through and : m , so y xa b a bc" ! ! c$ c$ c$ c $c$ c ! ! c c" Line through and : m , so y xa b a b! $ # c" c# c# b $c" c $ c% # c ! # f(x) x , x x , x c$ c $ c" ! c# b $ ! # Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 12. 4 Chapter 1 Functions 31. (a) Line through and : y xa b a bc" " ! ! c Line through and : ya b a b! " " " " Line through and : m , so y x xa b a b a b" " $ ! c c c " b " c b!c" c" " " " $ $c" # # # # # f(x) x x x x x c c" ! " ! " c b " $ " $ # # (b) Line through 2 1 and 0 0 : y xa b a bc c 1 2 Line through 0 2 and 1 0 : y 2x 2a b a b c b Line through 1 1 and 3 1 : y 1a b a b c c c f x x 2 x 0 2x 2 0 x 1 1 1 x 3 a b c c b c 1 2 32. (a) Line through and T : m , so y x 0 x a bT T T T T T T# c # # "c! # # # ! " c b c "a b f x , 0 x x , x T a b J ! c " T T T # # # (b) f x A, x A x T A T x A x T a b ! c c # T T T T # # $ # $ # 33. (a) x 0 for x [0 1) (b) x 0 for x ( 1 0] c 34. x x only when x is an integer. 35. For any real number x, n x n , where n is an integer. Now: n x n n x n. By b " b " c b " c c definition: x n and x n x n. So x x for all x .c c c c c c d 36. To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part. Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 13. Section 1.1 Functions and Their Graphs 5 37. Symmetric about the origin 38. Symmetric about the y-axis Dec: x Dec: xc_ _ c_ ! Inc: nowhere Inc: x! _ 39. Symmetric about the origin 40. Symmetric about the y-axis Dec: nowhere Dec: x! _ Inc: x Inc: xc_ ! c_ ! x! _ 41. Symmetric about the y-axis 42. No symmetry Dec: x Dec: xc_ ! c_ ! Inc: x Inc: nowhere! _ Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 14. 6 Chapter 1 Functions 43. Symmetric about the origin 44. No symmetry Dec: nowhere Dec: x! _ Inc: x Inc: nowherec_ _ 45. No symmetry 46. Symmetric about the y-axis Dec: x Dec: x! _ c_ ! Inc: nowhere Inc: x! _ 47. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even. 48. f x x and f x x f x . Thus the function is odd.a b a b a b a b c c c cc& " " "c& cx xx& && a b 49. Since f x x x f x . The function is even.a b a b a b b " c b " c# # 50. Since f x x x f x x x and f x x x f x x x the function is neither even nor b c c c b c c c a b a b a b a b a b a b# ## # odd. 51. Since g x x x, g x x x x x g x . So the function is odd.a b a b a b a b b c c c c b c$ $ $ 52. g x x x x x g x thus the function is even.a b a b a b a b b $ c " c b $ c c " c % # % # 53. g x g x . Thus the function is even.a b a b c" " c " c c"x x# # a b 54. g x ; g x g x . So the function is odd.a b a b a b c c cx x x x# #c " c" 55. h t ; h t ; h t . Since h t h t and h t h t , the function is neither even nor odd.a b a b a b a b a b a b a b c c c c" " " c " c c " " ct t t 56. Since t | t |, h t h t and the function is even.l l c c$ $ a b a b a b Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 15. Section 1.1 Functions and Their Graphs 7 57. h t 2t , h t 2t . So h t h t . h t 2t , so h t h t . The function is neither even nora b a b a b a b a b a b a b b " c c b " c c c c " c odd. 58. h t 2 t and h t 2 t 2 t . So h t h t and the function is even.a b a b a b a b l l b " c l c l b " l l b " c 59. s kt 25 k 75 k s t; 60 t t 180 " " " 3 3 3 60. K c v 12960 c 18 c 40 K 40v ; K 40 10 4000 joules # # # a b a b2 61. r 6 k 24 r ; 10 s k k 24 24 12 s 4 s s 5 62. P 14.7 k 14700 P ; 23.4 v 628.2 in k k 14700 14700 24500 v 1000 v v 39 3 63. v f(x) x 2x 22 2x x 72x x; x 7 "% c c % c b $!) ! $ # 64. (a) Let h height of the triangle. Since the triangle is isosceles, AB AB 2 AB 2 So, b # # # h 2 h B is at slope of AB The equation of AB is# # # b " " ! " c" a b y f(x) x ; x . c b " ! " (b) A x 2x y 2x x 2x x; x . c b " c b # ! "# 65. (a) Graph h because it is an even function and rises less rapidly than does Graph g. (b) Graph f because it is an odd function. (c) Graph g because it is an even function and rises more rapidly than does Graph h. 66. (a) Graph f because it is linear. (b) Graph g because it contains .a b! " (c) Graph h because it is a nonlinear odd function. 67. (a) From the graph, 1 x ( 2 0) ( )x 4 x# b c r % _ (b) 1 1 0x 4 x 4 x x# # b c c x 0: 1 0 0 0 c c x 4 x 2x 8 x 2x x (x 4)(x 2) # # c c c b# x 4 since x is positive; x 0: 1 0 0 0 c c x 4 x 2x 8 2 x 2x x (x 4)(x 2)# c c c b # x 2 since x is negative; c sign of (x 4)(x 2)c b 2 b b c c % Solution interval: ( 0) ( )c# r % _ Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 16. 8 Chapter 1 Functions 68. (a) From the graph, x ( 5) ( 1 1)3 2 x 1 x 1c b c_ c r c (b) x 1: 2Case c 3 2 x 1 x 1 x 1 3(x 1) c b c b 3x 3 2x 2 x 5. b c c Thus, x ( 5) solves the inequality. c_ c 1 x 1: 2Case c 3 2 x 1 x 1 x 1 3(x 1) c b c b 3x 3 2x 2 x 5 which is true b c c if x 1. Thus, x ( 1 1) solves the c c inequality. 1 x: 3x 3 2x 2 x 5Case b c c3 2 x 1 x 1c b which is never true if 1 x, so no solution here. In conclusion, x ( 5) ( 1 1). c_ c r c 69. A curve symmetric about the x-axis will not pass the vertical line test because the points x, y and x, y lie on the sama b a bc e vertical line. The graph of the function y f x is the x-axis, a horizontal line for which there is a single y-value, , ! !a b for any x. 70. price 40 5x, quantity 300 25x R x 40 5x 300 25x b c b ca b a ba b 71. x x h x ; cost 5 2x 10h C h 10 10h 5h 2 22 2 2 h 2 2 h 2 h 2 2b b b b a b a b 72. (a) Note that 2 mi = 10,560 ft, so there are 800 x feet of river cable at $180 per foot and 10,560 x feet of land a b# #b c cable at $100 per foot. The cost is C x 180 800 x 100 10,560 x .a b a b b b c# # (b) C $a b! " #!! !!! C $a b&!! " "(& )"# C $a b"!!! " ")' &"# C $a b"&!! " #"# !!! C $a b#!!! " #%$ ($# C $a b#&!! " #() %(* C $a b$!!! " $"% )(! Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the point P. 1.2 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS 1. D : x , D : x 1 D D : x 1. R : y , R : y 0, R : y 1, R : y 0f g f g fg f g f g fgc_ _ c_ _ b b 2. D : x 1 0 x 1, D : x 1 0 x 1. Therefore D D : x 1.f g f g fgb c c b R R : y 0, R : y 2, R : y 0f g f g fg b 3. D : x , D : x , D : x , D : x , R : y 2, R : y 1,f g f g g f f gc_ _ c_ _ c_ _ c_ _ R : 0 y 2, R : yf g _g f " # 4. D : x , D : x 0 , D : x 0, D : x 0; R : y 1, R : y 1, R : 0 y 1, R : 1 yf g f g g f f g f gc_ _ _ g f 5. (a) 2 (b) 22 (c) x 2# b (d) (x 5) 3 x 10x 22 (e) 5 (f) 2b c b b c# # (g) x 10 (h) (x 3) 3 x 6x 6b c c c b# # % # Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 17. Section 1.2 Combining Functions; Shifting and Scaling Graphs 9 6. (a) (b) 2 (c) 1c c " " c b b3 x 1 x 1 x (d) (e) 0 (f)" x 4 3 (g) x 2 (h)c " " b " b b #" b# b bx 1 x 1 x 1 x x 7. f g h x f g h x f g 4 x f 3 4 x f 12 3x 12 3x 1 13 3xa ba b a b a b a b a b a ba b a b a ba b c c c c b c 8. f g h x f g h x f g x f 2 x 1 f 2x 1 3 2x 1 4 6x 1a ba b a b a b a b a b a ba b a b a ba b c c c b b2 2 2 2 2 9. f g h x f g h x f g f fa ba b a b a ba b b " 1 1 x x 5x x 1 4x 1 4x 1 4x1 x b % b b b b " 10. f g h x f g h x f g 2 x f fa ba b a ba ba b : ; c 2 x 2 x 1 2 x 8 3x x 7 2x 2 3 c c b c c $ c c b c 2 2 2 x 2 x x x c $c c $c 11. (a) f g x (b) j g x (c) g g xa ba b a ba b a ba b (d) j j x (e) g h f x (f) h j f xa ba b a ba b a ba b 12. (a) f j x (b) g h x (c) h h xa ba b a ba b a ba b (d) f f x (e) j g f x (f) g f h xa ba b a ba b a ba b 13. g(x) f(x) (f g)(x) (a) x 7 x x 7c c (b) x 2 3x 3(x 2) 3x 6b b b (c) x x 5 x 5# # c c (d) xx x x x 1 x 1 1 x (x 1)c c c c c x x 1 x x 1 c c (e) 1 x" " cx 1 xb (f) x" " x x 14. (a) f g x g x .a ba b a b l l " l c "lx (b) f g x so g x x .a ba b a b " c " c b "g x g x x g x x x g x x g x x x xa b a b a b a b a b c" b " b " b " b " " " " " (c) Since f g x g x x , g x x .a ba b a b a b l l # (d) Since f g x f x x , f x x . (Note that the domain of the composite is .)a ba b a b l l ! _# The completed table is shown. Note that the absolute value sign in part (d) is optional. g x f x f g x x x x x x x x x a b a b a ba b l l b " l l l l " " c " l c "l c " b " # # x x x x x x 15. (a) f g 1 f 1 1 (b) g f 0 g 2 2 (c) f f 1 f 0 2a b a b a b a b a b a ba b a b a bc c c c (d) g g 2 g 0 0 (e) g f 2 g 1 1 (f) f g 1 f 1 0a b a b a b a b a b a ba b a b a b c c c 16. (a) f g 0 f 1 2 1 3, where g 0 0 1 1a b a b a b a ba b c c c c c (b) g f 3 g 1 1 1, where f 3 2 3 1a b a b a b a ba b c c c c c (c) g g 1 g 1 1 1 0, where g 1 1 1a b a b a b a ba bc c c c c Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 18. 10 Chapter 1 Functions (d) f f 2 f 0 2 0 2, where f 2 2 2 0a b a b a ba b c c (e) g f 0 g 2 2 1 1, where f 0 2 0 2a b a b a ba b c c (f) f g f 2 , where g 1 " " " " " " # # # # # # # c c c c c5 17. (a) f g x f g x 1a ba b a ba b b 1 1 x x x b g f x g f xa ba b a ba b 1 x 1 b (b) Domain f g : , 1 0, , domain g f : 1,a b a b c_ c r _ c _ (c) Range f g : 1, , range g f : 0,a b a b _ _ 18. (a) f g x f g x 1 2 x xa ba b a ba b c b g f x g f x 1 xa ba b a b k ka b c (b) Domain f g : 0, , domain g f : ,a b a b _ c_ _ (c) Range f g : 0, , range g f : , 1a b a b _ c_ 19. f g x x f g x x x g x g x 2 x x g x 2xa ba b a b a b a b a ba b a b c cg x g x 2 a b a b c g x x g x 2x g x c c c a b a b a b 2x 2x 1 x x 1c c 20. f g x x 2 f g x x 2 2 g x 4 x 2 g x g xa ba b a b a b a b a ba b a b a b b b c b 3 3 x 6 x 6 2 2 b b3 21. (a) y (x 7) (b) y (x 4) c b c c# # 22. (a) y x 3 (b) y x 5 b c# # 23. (a) Position 4 (b) Position 1 (c) Position 2 (d) Position 3 24. (a) y (x 1) 4 (b) y (x 2) 3 (c) y (x 4) 1 (d) y (x 2) c c b c b b c b c c c# # # # 25. 26. 27. 28. Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 19. Section 1.2 Combining Functions; Shifting and Scaling Graphs 11 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 20. 12 Chapter 1 Functions 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 21. Section 1.2 Combining Functions; Shifting and Scaling Graphs 13 53. 54. 55. (a) domain: [0 2]; range: [ ] (b) domain: [0 2]; range: [ 1 0] # $ c (c) domain: [0 2]; range: [0 2] (d) domain: [0 2]; range: [ 1 0] c (e) domain: [ 2 0]; range: [ 1] (f) domain: [1 3]; range: [ ]c ! ! " (g) domain: [ 2 0]; range: [ ] (h) domain: [ 1 1]; range: [ ]c ! " c ! " Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 22. 14 Chapter 1 Functions 56. (a) domain: [0 4]; range: [ 3 0] (b) domain: [ 4 0]; range: [ ] c c ! $ (c) domain: [ 4 0]; range: [ ] (d) domain: [ 4 0]; range: [ ]c ! $ c " % (e) domain: [ 4]; range: [ 3 0] (f) domain: [ 2 2]; range: [ 3 0]# c c c (g) domain: [ 5]; range: [ 3 0] (h) domain: [0 4]; range: [0 3]" c 57. y 3x 3 58. y 2x 1 x 1 c c % c# ## a b 59. y 60. y 1 1 " b b b b" " " " " * # # # $ x x xx# # ## a b 61. y x 1 62. y 3 x 1 % b b 63. y 16 x 64. y x % c c % c x # # $ # " "# # 65. y 3x 27x 66. y " c " c " c " ca b $ $ # ) $x x$ Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 23. Section 1.2 Combining Functions; Shifting and Scaling Graphs 15 67. Let y x f x and let g x x , c # b " a b a b "# h x x , i x x , anda b a b b # b" " # # "# "# j x x f . The graph ofa b a b c # b B" # "# h x is the graph of g x shifted left unit; thea b a b " # graph of i x is the graph of h x stretcheda b a b vertically by a factor of ; and the graph of# j x f x is the graph of i x reflected acrossa b a b a b the x-axis. 68. Let y f x Let g x x , " c c a b a b a bx # "# h x x , and i x xa b a b a b a b c b # c b #"# "#" # f x The graph of g x is the " c a b a bx # graph of y x reflected across the x-axis. The graph of h x is the graph of g x shifteda b a b right two units. And the graph of i x is thea b graph of h x compressed vertically by a factora b of .# 69. y f x x . Shift f x one unit right followed by a a b a b$ shift two units up to get g x x .a b a b c " b #3 70. y x f x . " c B b # c c " b c# a b a b a b a b$ $ Let g x x , h x x , i x x ,a b a b a b a b a b a b c " c " b c#$ $ $ and j x x . The graph of h x is thea b a b a b a b c c " b c# $ graph of g x shifted right one unit; the graph of i x isa b a b the graph of h x shifted down two units; and the grapha b of f x is the graph of i x reflected across the x-axis.a b a b 71. Compress the graph of f x horizontally by a factora b " x of 2 to get g x . Then shift g x vertically down 1a b a b " #x unit to get h x .a b c "" #x Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 24. 16 Chapter 1 Functions 72. Let f x and g xa b a b b " b "" # " x x# # B# # Since b " b "" " # " # B x # # , we see that the graph of f x stretched a b# "% horizontally by a factor of 1.4 and shifted up 1 unit is the graph of g x .a b 73. Reflect the graph of y f x x across the x-axis a b $ to get g x x.a b c $ 74. y f x x x c# c" # a b a b a ba b#$ #$ x x . So the graph c" # #a b a b a b#$ #$ #$ of f x is the graph of g x x compresseda b a b #$ horizontally by a factor of 2. 75. 76. 77. x y 78. x y* b #& ##& b " "' b ( ""# b "# # # # & $ % ( x xy y# # # # # # # # Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 25. Section 1.2 Combining Functions; Shifting and Scaling Graphs 17 79. x y 80. x y$ b c # $ b " b " b # % b "# ## c # # " # $ # c c" a b a bx y x y# # # # # # # #a b < a b 81. x y 82. x y$ c " b # b # ' ' b b * c &%a b a b # # $ " # # # # b " b "a b < a b x y x yc " # $ ' c c# c c $ c# # # # # $ # # # " # # 83. has its center at . Shiftinig 4 unitsx y# # "' *b " ! !a b left and 3 units up gives the center at h, k .a b a b c% $ So the equation is < a b a bx 4 4 3 y 3c c c # # # # b " . Center, C, is , and b " c% $a b a bx y 4 3 b % c $# # # # a b major axis, AB, is the segment from to .a b a bc) $ ! $ 84. The ellipse has center h, k .x y# # % #&b " ! !a b a b Shifting the ellipse 3 units right and 2 units down produces an ellipse with center at h, ka b a b $ c# and an equation . Center,a b < a bx 3 yc % #& c c## # b " C, is 3 , and AB, the segment from toa b a b c# $ $ is the major axis.a b$ c( 85. (a) (fg)( x) f( x)g( x) f(x)( g(x)) (fg)(x), oddc c c c c (b) ( x) (x), odd f f g g( x) g(x) g f( x) f(x) c cc c c Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 26. 18 Chapter 1 Functions (c) ( x) (x), odd g g( x) g(x) g f f( x) f(x) fc cc c c (d) f ( x) f( x)f( x) f(x)f(x) f (x), even# # c c c (e) g ( x) (g( x)) ( g(x)) g (x), even# # # # c c c (f) (f g)( x) f(g( x)) f( g(x)) f(g(x)) (f g)(x), even c c c (g) (g f)( x) g(f( x)) g(f(x)) (g f)(x), even c c (h) (f f)( x) f(f( x)) f(f(x)) (f f)(x), even c c (i) (g g)( x) g(g( x)) g( g(x)) g(g(x)) (g g)(x), odd c c c c c 86. Yes, f(x) 0 is both even and odd since f( x) 0 f(x) and f( x) 0 f(x). c c c 87. (a) (b) (c) (d) 88. Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 27. Section 1.3 Trigonometric Functions 19 1.3 TRIGONOMETRIC FUNCTIONS 1. (a) s r (10) 8 m (b) s r (10)(110) m ) 1 ) 4 110 55 5 180 18 9 1 1 1 1 2. radians and 225) s 10 5 5 180 r 8 4 4 1 1 1 1 3. 80 80 s (6) 8.4 in. (since the diameter 12 in. radius 6 in.)) ) 1 1 1 180 9 9 4 4 4. d 1 meter r 50 cm 0.6 rad or 0.6 34 ) s 30 180 r 50 1 5. 0 6 sin 0 0 cos 1 0 tan 0 3 0 und. cot und. und. 0 1 sec 1 und. 2 csc und. und. 2 ) 1 ) ) ) ) ) ) c c c " c c " c c" c c c# " c c " 2 3 3 4 3 2 2 3 2 3 1 1 1 # # " " " # " . sin cos tan und. 3 cot 3 3 sec und. 2 csc ) ) ) ) ) ) ) c c c " c c ! c c c " c ! c c " c # c " c c# 3 3 3 2 3 3 2 3 3 3 2 2 3 3 2 3 1 1 1 1 1 # ' % ' & # # # " " " " " # # # " " " 2 # 7. cos x , tan x 8. sin x , cos x c c 4 3 2 5 4 5 5 " 9. sin x , tan x 8 10. sin x , tan x c c c 8 3 13 5 12 12 11. sin x , cos x 12. cos x , tan x c c c " " # 5 5 3 2 3 13. 14. period period 4 1 1 15. 16. period 2 period 4 Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 28. 20 Chapter 1 Functions 17. 18. period 6 period 1 19. 20. period 2 period 2 1 1 21. 22. period 2 period 2 1 1 23. period , symmetric about the origin 24. period 1, symmetric about the origin 1 # 25. period 4, symmetric about the s-axis 26. period 4 , symmetric about the origin 1 Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 29. Section 1.3 Trigonometric Functions 21 27. (a) Cos x and sec x are positive for x in the interval , ; and cos x and sec x are negative for x in the c1 1 2 2 intervals , and , . Sec x is undefined c c3 3 2 2 2 2 1 1 1 1 when cos x is 0. The range of sec x is ( 1] [ ); the range of cos x is [ 1].c_ c r " _ c" (b) Sin x and csc x are positive for x in the intervals , and , ; and sin x and csc x are negative a bc c !3 2 1 1 1 for x in the intervals , and , . Csc x isa b c !1 1 3 2 1 undefined when sin x is 0. The range of csc x is ( 1] [1 ); the range of sin x is [ ].c_ c r _ c" " 28. Since cot x , cot x is undefined when tan x 0 " tan x and is zero when tan x is undefined. As tan x approaches zero through positive values, cot x approaches infinity. Also, cot x approaches negative infinity as tan x approaches zero through negative values. 29. D: x ; R: y 1, 0, 1 30. D: x ; R: y 1, 0, 1c_ _ c c_ _ c 31. cos x cos x cos sin x sin (cos x)(0) (sin x)( 1) sin x c c c c c c 1 1 1 # # # 32. cos x cos x cos sin x sin (cos x)(0) (sin x)(1) sin x b c c c1 1 1 # # # 33. sin x sin x cos cos x sin (sin x)(0) (cos x)(1) cos x b b b 1 1 1 # # # 34. sin x sin x cos cos x sin (sin x)(0) (cos x)( 1) cos x c c b c b c c1 1 1 # # # 35. cos (A B) cos (A ( B)) cos A cos ( B) sin A sin ( B) cos A cos B sin A ( sin B)c b c c c c c c cos A cos B sin A sin B b 36. sin (A B) sin (A ( B)) sin A cos ( B) cos A sin ( B) sin A cos B cos A ( sin B)c b c c b c b c sin A cos B cos A sin B c 37. If B A, A B 0 cos (A B) cos 0 1. Also cos (A B) cos (A A) cos A cos A sin A sin A c c c c b cos A sin A. Therefore, cos A sin A 1. b b # # # # Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 30. 22 Chapter 1 Functions 38. If B 2 , then cos (A 2 ) cos A cos 2 sin A sin 2 (cos A)(1) (sin A)(0) cos A and b c c 1 1 1 1 sin (A 2 ) sin A cos 2 cos A sin 2 (sin A)(1) (cos A)(0) sin A. The result agrees with theb b b 1 1 1 fact that the cosine and sine functions have period 2 .1 39. cos ( x) cos cos sin sin x ( 1)(cos x) (0)(sin x) cos x1 1 1b B c c c c 40. sin (2 x) sin 2 cos ( x) cos (2 ) sin ( x) (0)(cos ( x)) (1)(sin ( x)) sin x1 1 1c c b c c b c c 41. sin x sin cos ( x) cos sin ( x) ( 1)(cos x) (0)(sin ( x)) cos x 3 3 31 1 1 # # #c c b c c b c c 42. cos x cos cos x sin sin x (0)(cos x) ( 1)(sin x) sin x 3 3 31 1 1 # # #b c c c 43. sin sin sin cos cos sin7 1 4 3 4 3 4 3 4 2 2 3 6 21 1 1 1 1 1 1 # # # # # " b b b b 44. cos cos cos cos sin sin11 2 2 2 1 4 3 4 3 4 3 4 2 2 3 2 61 1 1 1 1 1 1 # # # # # " b b c c c c 45. cos cos cos cos sin sin1 1 1 1 1 1 1 12 3 4 3 4 3 4 2 23 1 3 2 2 c c c c c c " # # # # b 46. sin sin sin cos cos sin5 2 2 2 1 3 4 3 4 3 4 3 1 32 2 2 2 1 1 1 1 1 1 1 # # # # # " b c c b c b c c 47. cos 48. cos# #b b b # # # # # b b c c1 1 8 4 1 4 1 cos 1 1 cos2 2 5 1 2 3 2 10 8 1 2 3 1 1 # # # 49. sin 50. sin# # # # # # # c c cc c c b1 1 1 4 8 4 1 cos 1 1 cos2 3 3 1 2 2 2 6 1 8 3 2 1 1 # # # 51. sin sin , , ,2 3 2 4 5 4 2 3 3 3 3 3 ) ) ) 1 1 1 1 52. sin cos tan 1 tan 1 , , ,2 2 2sin cos 3 5 7 cos cos 4 4 4 4) ) ) ) ) 2 2 2 2 ) ) 1 1 1 1 ) ) 53. sin 2 cos 0 2sin cos cos 0 cos 2sin 1 0 cos 0 or 2sin 1 0 cos 0 or) ) ) ) ) ) ) ) ) )c c c c a b sin , , or , , , ,) ) ) ) " # 1 1 1 1 1 1 1 1 2 2 6 6 6 2 6 2 3 5 5 3 54. cos 2 cos 0 2cos 1 cos 0 2cos cos 1 0 cos 1 2cos 1 0) ) ) ) ) ) ) )b c b b c b c 2 2 a ba b cos 1 0 or 2cos 1 0 cos 1 or cos or , , , b c c ) ) ) ) ) 1 ) ) 1" # 1 1 1 1 3 3 3 3 5 5 55. tan (A B)b sin (A B) cos (A B) cos A cos B sin A sin B sin A cos B cos A cos Bb b c b bsin A cos B cos A sin B cos A cos B cos A cos B cos A cos B sin A sin B cos A cos B cos A cos Bc b c tan A tan B 1 tan A tan B 56. tan (A B)c sin (A B) cos (A B) cos A cos B sin A sin B sin A cos B cos A cos Bc c b c csin A cos B cos A sin B cos A cos B cos A cos B cos A cos B sin A sin B cos A cos B cos A cos Bb c b tan A tan B 1 tan A tan B 57. According to the figure in the text, we have the following: By the law of cosines, c a b 2ab cos# # # b c ) 1 1 2 cos (A B) 2 2 cos (A B). By distance formula, c (cos A cos B) (sin A sin B) b c c c c c b c# # # # # cos A 2 cos A cos B cos B sin A 2 sin A sin B sin B 2 2(cos A cos B sin A sin B). Thus c b b c b c b# # # # c 2 2 cos (A B) 2 2(cos A cos B sin A sin B) cos (A B) cos A cos B sin A sin B.# c c c b c b Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 31. Section 1.3 Trigonometric Functions 23 58. (a) cos A B cos A cos B sin A sin Ba bc b sin cos and cos sin) ) ) ) c c 1 1 # # Let A B) b sin A B cos A B cos A B cos A cos B sin A sin Ba b a b b c b c c c b c1 1 1 1 # # # # sin A cos B cos A sin B b (b) cos A B cos A cos B sin A sin Ba bc b cos A B cos A cos B sin A sin Ba b a b a ba bc c c b c cos A B cos A cos B sin A sin B cos A cos B sin A sin B b c b c b ca b a b a b a b cos A cos B sin A sin B c Because the cosine function is even and the sine functions is odd. 59. c a b 2ab cos C 2 3 2(2)(3) cos (60) 4 9 12 cos (60) 13 12 7.# # # # # " # b c b c b c c Thus, c 7 2.65. 60. c a b 2ab cos C 2 3 2(2)(3) cos (40) 13 12 cos (40). Thus, c 13 12 cos 40 1.951.# # # # # b c b c c c 61. From the figures in the text, we see that sin B . If C is an acute angle, then sin C . On the other hand, h h c b if C is obtuse (as in the figure on the right), then sin C sin ( C) . Thus, in either case, c 1 h b h b sin C c sin B ah ab sin C ac sin B. By the law of cosines, cos C and cos B . Moreover, since the sum of the a b c a c b 2ab 2ac # # # # # # b c b c interior angles of a triangle is , we have sin A sin ( (B C)) sin (B C) sin B cos C cos B sin C1 1 c b b b 2a b c c b ah bc sin A. b b c b c a bh a b c a c b h h ah c 2ab 2ac b 2abc bc # # # # # # b c b c # # # # # Combining our results we have ah ab sin C, ah ac sin B, and ah bc sin A. Dividing by abc gives .h sin A sin C sin B bc a c b law of sines 62. By the law of sines, . By Exercise 61 we know that c 7. Thus sin B 0.982.sin A sin B 3 c 3/2 3 3 2 7# 63. From the figure at the right and the law of cosines, b a 2 2(2a) cos B# # # b c a 4 4a a 2a 4. b c c b# #" # Applying the law of sines to the figure, sin A sin B a b b a. Thus, combining results, 2/2 a b 3/2 3 # a 2a 4 b a 0 a 2a 4# # # # # # " c b b c3 0 a 4a 8. From the quadratic formula and the fact that a 0, we have b c # a 1.464. c b c c # # c4 4 4(1)( 8) 4 3 4 # 64. (a) The graphs of y sin x and y x nearly coincide when x is near the origin (when the calculator is in radians mode). (b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The curves look like intersecting straight lines near the origin when the calculator is in degree mode. Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 32. 24 Chapter 1 Functions 65. A 2, B 2 , C , D 1 c c1 1 66. A , B 2, C 1, D " " # # 67. A , B 4, C 0, D c 2 1 1 " 68. A , B L, C 0, D 0 L 21 69-72. Example CAS commands: Maple f := x -> A*sin((2*Pi/B)*(x-C))+D1; A:=3; C:=0; D1:=0; f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )]; plot( f_list, x=-4*Pi..4*Pi, scaling=constrained, color=[red,blue,green,cyan], linestyle=[1,3,4,7], legend=["B=1","B=3","B=2*Pi","B=3*Pi"], title="#69 (Section 1.3)" ); Mathematica Clear[a, b, c, d, f, x] f[x_]:=a Sin[2 /b (x c)] + d1 c Plot[f[x]/.{a 3, b 1, c 0, d 0}, {x, 4 , 4 }] c 1 1 Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 33. Section 1.3 Trigonometric Functions 25 69. (a) The graph stretches horizontally. (b) The period remains the same: period B . The graph has a horizontal shift of period. l l " # 70. (a) The graph is shifted right C units. (b) The graph is shifted left C units. (c) A shift of one period will produce no apparent shift. C l l ' 71. (a) The graph shifts upwards D units for Dl l ! (b) The graph shifts down D units for Dl l ! 72. (a) The graph stretches A units. (b) For A , the graph is inverted.l l ! Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 34. 26 Chapter 1 Functions 1.4 GRAPHING WITH CALCULATORS AND COMPUTERS 1-4. The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and has little unused space. 1. d. 2. c. 3. d. 4. b. 5-30. For any display there are many appropriate display widows. The graphs given as answers in Exercises 5 30c are not unique in appearance. 5. 2 5 by 15 40 6. 4 4 by 4 4c c c c 7. 2 6 by 250 50 8. 1 5 by 5 30c c c c Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 35. Section 1.4 Graphing with Calculators and Computers 27 9. 4 4 by 5 5 10. 2 2 by 2 8c c c c 11. 2 6 by 5 4 12. 4 4 by 8 8c c c c 13. 1 6 by 1 4 14. 1 6 by 1 5c c c c 15. 3 3 by 0 10 16. 1 2 by 0 1c c Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 36. 28 Chapter 1 Functions 17. 5 1 by 5 5 18. 5 1 by 2 4c c c c 19. 4 4 by 0 3 20. 5 5 by 2 2c c c 21. by 22. byc"! "! c' ' c& & c# # 23. by 24. byc' "! c' ' c$ & c# "! Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 37. Section 1.4 Graphing with Calculators and Computers 29 25. 0 03 0 03 by 1 25 1 25 26. 0 1 0 1 by 3 3c c c c 27. 300 300 by 1 25 1 25 28. 50 50 by 0 1 0 1c c c c 29. 0 25 0 25 by 0 3 0 3 30. 0 15 0 15 by 0 02 0 05c c c c 31. x x y y y x x .# # #b # % b % c # c c # b ) The lower half is produced by graphing y x x . # c c c # b ) # 32. y x y x . The upper branch# # #c "' " " b "' is produced by graphing y x . " b "' # Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 38. 30 Chapter 1 Functions 33. 34. 35. 36. 37. 38 39. 40. CHAPTER 1 PRACTICE EXERCISES 1. The area is A r and the circumference is C r. Thus, r A . # 1 1 1# # # % #C C C 1 1 1 # 2. The surface area is S r r . The volume is V r r . Substitution into the formula for % 1 1# $ % $ % "# % $ S V 1 1 $ surface area gives S r . % %1 1# $ % #$ V 1 Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 39. Chapter 1 Practice Exercises 31 3. The coordinates of a point on the parabola are x x . The angle of inclination joining this point to the origin satisfiesa b # ) the equation tan x. Thus the point has coordinates x x tan tan .) ) ) x x # a b a b# # 4. tan h tan ft.) ) &!!rise h run &!! 5. 6. Symmetric about the origin. Symmetric about the y-axis. 7. 8. Neither Symmetric about the y-axis. 9. y x x x y x . Even.a b a b a bc c b " b " # # 10. y x x x x x x x y x . Odd.a b a b a b a b a bc c c c c c c b b c& $ & $ 11. y x cos x cos x y x . Even.a b a b a bc " c c " c 12. y x sec x tan x sec x tan x y x . Odd.a b a b a b a bc c c c csin x cos x cos x sin xa b a b c c c # # 13. y x y x . Odd.a b a bc c ca b a b a b c b" c c# c b" b" c b# c# x x x x x x x x x % $ % % $ $ 14. y x x sin x x sin x x sin x y x . Odd.a b a b a b a b a b a bc c c c c b c c c 15. y x x cos x x cos x. Neither even nor odd.a b a bc c b c c b 16. y x x cos x x cos x y x . Odd.a b a b a b a bc c c c c 17. Since f and g are odd f x f x and g x g x . c c c ca b a b a b a b (a) f g x f x g x f x g x f x g x f g x f g is evena ba b a b a b a b a b a b a b a ba b c c c c c (b) f x f x f x f x f x f x f x f x f x f x f x f is odd.3 3 3 a b a b a b a b a b a b a b a b a b a b a bc c c c c c c c c (c) f sin x f sin x f sin x f sin x is odd.a b a b a b a ba b a b a b a bc c c (d) g sec x g sec x g sec x is even.a b a b a ba b a b a bc (e) g x g x g x g is evenl c l lc l l l l l a b a b a b Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 40. 32 Chapter 1 Functions 18. Let f a x f a x and define g x f x a . Then g x f x a f a x f a x f x a g xa b a b a b a b a b a b a b a b a b a ba bc b b c c b c b b g x f x a is even. ba b a b 19. (a) The function is defined for all values of x, so the domain is .a bc_ _ (b) Since x attains all nonnegative values, the range is .l l c# _ 20. (a) Since the square root requires x , the domain is ." c ! c_ " (b) Since x attains all nonnegative values, the range is ." c c# _ 21. (a) Since the square root requires x , the domain is ."' c ! c% %# (b) For values of x in the domain, x , so x . The range is .! "' c "' ! "' c % ! %# # 22. (a) The function is defined for all values of x, so the domain is .a bc_ _ (b) Since attains all positive values, the range is .$ " _#cx a b 23. (a) The function is defined for all values of x, so the domain is .a bc_ _ (b) Since e attains all positive values, the range is .# c$ _cx a b 24. (a) The function is equivalent to y tan x, so we require x for odd integers k. The domain is given by x for # # k k1 1 # % odd integers k. (b) Since the tangent function attains all values, the range is .a bc_ _ 25. (a) The function is defined for all values of x, so the domain is .a bc_ _ (b) The sine function attains values from to , so sin x and hence sin x . Thec" " c# # $ b # c$ # $ b c " "a b a b1 1 range is 3 1 .c 26. (a) The function is defined for all values of x, so the domain is .a bc_ _ (b) The function is equivalent to y x , which attains all nonnegative values. The range is . ! _& # 27. (a) The logarithm requires x , so the domain is .c $ ! $ _a b (b) The logarithm attains all real values, so the range is .a bc_ _ 28. (a) The function is defined for all values of x, so the domain is .a bc_ _ (b) The cube root attains all real values, so the range is .a bc_ _ 29. (a) Increasing because volume increases as radius increases (b) Neither, since the greatest integer function is composed of horizontal (constant) line segments (c) Decreasing because as the height increases, the atmospheric pressure decreases. (d) Increasing because the kinetic (motion) energy increases as the particles velocity increases. 30. (a) Increasing on 2, (b) Increasing on 1, _ c _ (c) Increasing on , (d) Increasing on ,a bc_ _ _" # 31. (a) The function is defined for x , so the domain is .c% % c% % (b) The function is equivalent to y x , x , which attains values from to for x in the domain. The l l c% % ! # range is .! # Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 41. Chapter 1 Practice Exercises 33 32. (a) The function is defined for x , so the domain is .c# # c# # (b) The range is .c" " 33. First piece: Line through and . m y x xa b a b! " " ! c" c b " " c! c " c" " c ! " Second piece: Line through and . m y x x xa b a b a b" " # ! c" c c " b " c b # # c! c " c" # c " " f x x, x x, x a b " c ! " # c " # 34. First piece: Line through and 2 5 . m y xa b a b! ! 5 5 5 2 2 2 c ! c ! Second piece: Line through 2 5 and 4 . m y x 2 5 x 10 10a b a b a b ! c c c b c b c! c c c 5 5 5 5 5 5x 4 2 2 2 2 2 2 f x (Note: x 2 can be included on either piece.) x, x 2 10 , 2 x 4 a b J ! c 5 2 5x 2 35. (a) f g f g f fa ba b a b a ba b c" c" " "" " c" b # " (b) g f g f g ora ba b a ba b # # " " " # b # #& &2 " # (c) f f x f f x f x, xa ba b a ba b !" " "x x (d) g g x g g x ga ba b a ba b " " b # b# b # " b # b # x x x " b# % x 36. (a) f g f g f fa ba b a b a ba b c" c" c" b " ! # c ! #$ (b) g f f g g ga ba b a b a b a ba b # # # c # ! ! b " "$ (c) f f x f f x f x x xa ba b a b a b a ba b # c # c # c (d) g g x g g x g x xa ba b a ba b b " b " b "$ $$ 37. (a) f g x f g x f x x x, x .a ba b a ba b b # # c b # c c# # g f x f g x g x x xa ba b a b a b a ba b # c # c b # % c# # # (b) Domain of f g: (c) Range of f g: c# _ c_ # Domain of g f: Range of g f: c# # ! # 38. (a) f g x f g x f x x x.a ba b a ba b " c " c " c% g f x f g x g x xa ba b a ba b " c (b) Domain of f g: (c) Range of f g: c_ " ! _ Domain of g f: Range of g f: ! " ! " 39. y f x y f f x a b a ba b Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 42. 34 Chapter 1 Functions 40. 41. 42. The graph of f (x) f x is the same as the It does not change the graph.# " a bk k graph of f (x) to the right of the y-axis. The" graph of f (x) to the left of the y-axis is the# reflection of y f (x), x 0 across the y-axis. " 43. 44. Whenever g (x) is positive, the graph of y g (x) Whenever g (x) is positive, the graph of y g (x) g (x)" # " # " k k g (x) is the same as the graph of y g (x). is the same as the graph of y g (x). When g (x) is k k" " " " When g (x) is negative, the graph of y g (x) is negative, the graph of y g (x) is the reflection of the" # # the reflection of the graph of y g (x) across the graph of y g (x) across the x-axis. " " x-axis. Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 43. Chapter 1 Practice Exercises 35 45. 46. Whenever g (x) is positive, the graph of The graph of f (x) f x is the same as the" # " a bk k y g (x) g (x) is the same as the graph of graph of f (x) to the right of the y-axis. The # " "k k y g (x). When g (x) is negative, the graph of graph of f (x) to the left of the y-axis is the " " # y g (x) is the reflection of the graph of reflection of y f (x), x 0 across the y-axis. # " y g (x) across the x-axis. " 47. 48. The graph of f (x) f x is the same as the The graph of f (x) f x is the same as the# " # " a b a bk k k k graph of f (x) to the right of the y-axis. The graph of f (x) to the right of the y-axis. The" " graph of f (x) to the left of the y-axis is the graph of f (x) to the left of the y-axis is the# # reflection of y f (x), x 0 across the y-axis. reflection of y f (x), x 0 across the y-axis. " " 49. (a) y g x 3 (b) y g x 2 c b b ca b " # # 3 (c) y g x (d) y g x c ca b a b (e) y 5 g x (f) y g 5x a b a b 50. (a) Shift the graph of f right 5 units (b) Horizontally compress the graph of f by a factor of 4 (c) Horizontally compress the graph of f by a factor of 3 and a then reflect the graph about the y-axis (d) Horizontally compress the graph of f by a factor of 2 and then shift the graph left unit." # (e) Horizontally stretch the graph of f by a factor of 3 and then shift the graph down 4 units. (f) Vertically stretch the graph of f by a factor of 3, then reflect the graph about the x-axis, and finally shift the graph up unit." 4 51. Reflection of the grpah of y x about the x-axis followed by a horizontal compression by a factor of then a shift left 2 units.1 2 Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 44. 36 Chapter 1 Functions 52. Reflect the graph of y x about the x-axis, followed by a vertical compression of the graph by a factor of 3, then shift the graph up 1 unit. 53. Vertical compression of the graph of y by a 1 x2 factor of 2, then shift the graph up 1 unit. 54. Reflect the graph of y x about the y-axis, then 1 3 compress the graph horizontally by a factor of 5. 55. 56. period period 4 1 1 57. 58. period 2 period 4 Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 45. Chapter 1 Practice Exercises 37 59. 60. period 2 period 2 1 1 61. (a) sin B sin b 2 sin 2 3. By the theorem of Pythagoras, 1 1 3 c 3 b b 3 # # a b c a c b 4 3 1.# # # # #b c c (b) sin B sin c . Thus, a c b (2) . c c 1 3 c c sin 3 b 2 2 2 4 4 4 2 3 3 3 1 3 3 # # # # # 62. (a) sin A a c sin A (b) tan A a b tan A a a c b 63. (a) tan B a (b) sin A c b b a a a tan B c sin A 64. (a) sin A (c) sin A a a c c c c b # #c 65. Let h height of vertical pole, and let b and c denote the distances of points B and C from the base of the pole, measured along the flatground, respectively. Then, tan 50 , tan 35 , and b c 10. c h h c b Thus, h c tan 50 and h b tan 35 (c 10) tan 35 b c tan 50 (c 10) tan 35 b c (tan 50 tan 35) 10 tan 35 c c h c tan 50 10 tan 35 tan 50 tan 35c 16.98 m. 10 tan 35 tan 50 tan 50 tan 35c 66. Let h height of balloon above ground. From the figure at the right, tan 40 , tan 70 , and a b 2. Thus, b h h a b h b tan 70 h (2 a) tan 70 and h a tan 40 c (2 a) tan 70 a tan 40 a(tan 40 tan 70) c b 2 tan 70 a h a tan 40 2 tan 70 tan 40 tan 70b 1.3 km. 2 tan 70 tan 40 tan 40 tan 70b 67. (a) (b) The period appears to be 4 .1 Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 46. 38 Chapter 1 Functions (c) f(x 4 ) sin (x 4 ) cos sin (x 2 ) cos 2 sin x cosb b b b b b b1 1 1 1 x 4 x xb # # # 1 since the period of sine and cosine is 2 . Thus, f(x) has period 4 .1 1 68. (a) (b) D ( 0) ( ); R [ 1 1] c_ r ! _ c (c) f is not periodic. For suppose f has period p. Then f kp f sin 2 0 for all " " # #1 1b 1 integers k. Choose k so large that kp 0 . But then" " " # b1 1 1b (1/2 ) kp 1 f kp sin 0 which is a contradiction. Thus f has no period, as claimed. " " # # b1 1b (1/ ) kp CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES 1. There are (infinitely) many such function pairs. For example, f(x) 3x and g(x) 4x satisfy f(g(x)) f(4x) 3(4x) 12x 4(3x) g(3x) g(f(x)). 2. Yes, there are many such function pairs. For example, if g(x) (2x 3) and f(x) x , then b $ "$ (f g)(x) f(g(x)) f (2x 3) (2x 3) 2x 3. b b ba b a b$ $ "$ 3. If f is odd and defined at x, then f( x) f(x). Thus g( x) f( x) 2 f(x) 2 whereasc c c c c c c g(x) (f(x) 2) f(x) 2. Then g cannot be odd because g( x) g(x) f(x) 2 f(x) 2c c c c b c c c c c b 4 0, which is a contradiction. Also, g(x) is not even unless f(x) 0 for all x. On the other hand, if f is even, then g(x) f(x) 2 is also even: g( x) f( x) 2 f(x) 2 g(x). c c c c c 4. If g is odd and g(0) is defined, then g(0) g( 0) g(0). Therefore, 2g(0) 0 g(0) 0. c c 5. For (x y) in the 1st quadrant, x y 1 x b bk k k k x y 1 x y 1. For (x y) in the 2nd b b quadrant, x y x 1 x y x 1k k k kb b c b b y 2x 1. In the 3rd quadrant, x y x 1 b b bk k k k x y x 1 y 2x 1. In the 4th c c b c c quadrant, x y x 1 x ( y) x 1k k k kb b b c b y 1. The graph is given at the right. c Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 47. Chapter 1 Additional and Advanced Exercises 39 6. We use reasoning similar to Exercise 5. (1) 1st quadrant: y y x xb bk k k k 2y 2x y x. (2) 2nd quadrant: y y x xb bk k k k 2y x ( x) 0 y 0. b c (3) 3rd quadrant: y y x xb bk k k k y ( y) x ( x) 0 0 b c b c points in the 3rd quadrant all satisfy the equation. (4) 4th quadrant: y y x xb bk k k k y ( y) 2x 0 x. Combining b c these results we have the graph given at the right: 7. (a) sin x cos x 1 sin x 1 cos x (1 cos x)(1 cos x) (1 cos x)# # # # bb c c b c sin x 1 cos x # 1 cos x sin x sin x 1 cos x c b (b) Using the definition of the tangent function and the double angle formulas, we have tan .# # b c x 1 cos xsin cos 1 cos x # # # # # "c # "b # # x x cos 2 x cos 2 x 8. The angles labeled in the accompanying figure are# equal since both angles subtend arc CD. Similarly, the two angles labeled are equal since they both subtend! arc AB. Thus, triangles AED and BEC are similar which implies a c 2a cos b b a c c c b ) (a c)(a c) b(2a cos b) c b c) a c 2ab cos b c c# # # ) c a b 2ab cos . b c# # # ) 9. As in the proof of the law of sines of Section 1.3, Exercise 61, ah bc sin A ab sin C ac sin B the area of ABC (base)(height) ah bc sin A ab sin C ac sin B. " " " " " # # # # # 10. As in Section 1.3, Exercise 61, (Area of ABC) (base) (height) a h a b sin C# # # # # # # #" " " 4 4 4 a b cos C . By the law of cosines, c a b 2ab cos C cos C . " c b c " b c# # # # # # 4 2ab a b c a b # # # Thus, (area of ABC) a b cos C a b# # # # # #" " b c # # b c " c " c " c4 4 ab 4 4a b a b c a b a b c a b 9 # # # # # # # # # # # a b 4a b a b c 2ab a b c 2ab a b c c b c b b c c b c" "# # # # # # # # # # ## 16 16 a b c da b a ba b a b (a b) c c (a b) ((a b) c)((a b) c)(c (a b))(c (a b)) b c c c b b b c b c c c" "# # # # 16 16c d c da b a b s(s a)(s b)(s c), where s . c c c < a b c a b c a b c a b c a b cb b c b b c b b c b b # # # # # Therefore, the area of ABC equals s(s a)(s b)(s c) . c c c 11. If f is even and odd, then f( x) f(x) and f( x) f(x) f(x) f(x) for all x in the domain of f.c c c c Thus 2f(x) 0 f(x) 0. Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 48. 40 Chapter 1 Functions 12. (a) As suggested, let E(x) E( x) E(x) E is an c f(x) f( x) f( x) f( ( x)) f(x) f( x)b c c b c c b c # # # even function. Define O(x) f(x) E(x) f(x) . Then c c f(x) f( x) f(x) f( x)b c c c # # O( x) O(x) O is an odd functionc c c f( x) f( ( x)) f( x) f(x) f(x) f( x)c c c c c c c c # # # f(x) E(x) O(x) is the sum of an even and an odd function. b (b) Part (a) shows that f(x) E(x) O(x) is the sum of an even and an odd function. If also b f(x) E (x) O (x), where E is even and O is odd, then f(x) f(x) 0 E (x) O (x) b c b" " " " " "a b (E(x) O(x)). Thus, E(x) E (x) O (x) O(x) for all x in the domain of f (which is the same as thec b c c" " domain of E E and O O ). Now (E E )( x) E( x) E ( x) E(x) E (x) (since E and E arec c c c c c c c" " " " " " even) (E E )(x) E E is even. Likewise, (O O)( x) O ( x) O( x) O (x) ( O(x)) c c c c c c c c c c" " " " " (since O and O are odd) (O (x) O(x)) (O O)(x) O O is odd. Therefore, E E and" " " " " c c c c c c O O are both even and odd so they must be zero at each x in the domain of f by Exercise 11. That is," c E E and O O, so the decomposition of f found in part (a) is unique." " 13. y ax bx c a x x c a x c b b b b c b b c b# # # b b b b b a 4a 4a 2a 4a # # # # (a) If a 0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift of the vertex toward the y-axis and upward. If a 0 the graph is a parabola that opens downward. Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward. (b) If a 0 the graph is a parabola that opens upward. If also b 0, then increasing b causes a shift of the graph downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the right. If a 0 the graph is a parabola that opens downward. If b 0, increasing b shifts the graph upward to the right. If b 0, decreasing b shifts the graph upward to the left. (c) Changing c (for fixed a and b) by c shifts the graph upward c units if c 0, and downward c? ? ? ? c units if c 0.? 14. (a) If a 0, the graph rises to the right of the vertical line x b and falls to the left. If a 0, the graph c falls to the right of the line x b and rises to the left. If a 0, the graph reduces to the horizontal c line y c. As a increases, the slope at any given point x x increases in magnitude and the graph k k ! becomes steeper. As a decreases, the slope at x decreases in magnitude and the graph rises or fallsk k ! more gradually. (b) Increasing b shifts the graph to the left; decreasing b shifts it to the right. (c) Increasing c shifts the graph upward; decreasing c shifts it downward. 15. Each of the triangles pictured has the same base b v t v(1 sec). Moreover, the height of each ? triangle is the same value h. Thus (base)(height) bh" " # # A A A . In conclusion, the object sweeps " # $ out equal areas in each one second interval. 16. (a) Using the midpoint formula, the coordinates of P are . Thus the slope a 0 b 0 a bb b # # # # of OP . ? ? y x a/2 a b/2 b Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 49. Chapter 1 Additional and Advanced Exercises 41 (b) The slope of AB . The line segments AB and OP are perpendicular when the product cb 0 b 0 a a c c of their slopes is . Thus, b a a b (since both are positive). Therefore, ABc" c c b b b a a a # # # # is perpendicular to OP when a b. 17. From the figure we see that 0 and AB AD 1. From trigonometry we have the following: sin EB, ) )1 2 AB EB cos AE, tan CD, and tan . We can see that:) ) ) AE CD EB sin AB AD AE cos ) ) area AEB area sector DB area ADC AE EB AD AD CD w )" " " # # #a ba b a b a ba b2 sin cos tan sin cos " " " " " " " " # # # # # #) ) ) ) ) ) )a b a ba b2 sin cos ) ) 18. f g x f g x a cx d b acx ad b and g f x g f x c ax b d acx cb da ba b a b a b a ba b a b a ba b a b b b b b b b b b Thus f g x g f x acx ad b acx bc d ad b bc d. Note that f d ad b anda ba b a ba b a b b b b b b b b g b cb d, thus f g x g f x if f d g b .a b a ba b a ba b a b a b b Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 50. 42 Chapter 1 Functions NOTES: Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 51. CHAPTER 2 LIMITS AND CONTINUITY 2.1 RATES OF CHANGE AND TANGENTS TO CURVES 1. (a) 19 (b) 1? ? ? ? f 28 9 f 2 0 x 3 1 x 1 ( 1) f(3) f(2) f(1) f( ) c c c" c # c c # c c 2. (a) 0 (b) 2? ? ? ? g g(1) g( 1) g g(0) g( 2) x 1 ( 1) 2 x 0 ( 2) 1 1 0 4 cc c c c c c c c # c c 3. (a) (b)? ? ? 1 ? 1 h 1 1 4 h t t h h h h 0 3 3 3 c 3 4 4 3 4 4 6 6 3 1 1 1 1 1 1 1 1 1 1c c c c c c c c # # # 4. (a) (b) 0? 1 ? 1 1 ? 1 1 1 ? 1 1 1 g g( ) g(0) (2 1) (2 1) g g( ) g( ) (2 1) (2 ) t 0 0 t ( ) 2 c c c c b c c c c c " c c c c # 5. 1? ?) R 3R(2) R(0) 2 0 8 1 1 c c # # b c c " 6. 2 2 0? ?) P P(2) P(1) (8 16 10) ( ) 2 1 1 c c c b c " c % b & c 7. (a) 4 h. As h 0, 4 h 4 at P 2, 1 the slope is 4.? ? y x h h h 2 h 3 2 3 4 4h h 3 1 4h h b b a bb c c c b b c c b 2 2 2 2 a b (b) y 1 4 x 2 y 1 4x 8 y 4x 7c c c c ca b 8. (a) 2 h. As h 0, 2 h 2 at P 1, 4 the? ? y x h h h 5 1 h 5 1 5 1 2h h 4 2h h c c c c c a bc b c c c c c c c c 2 2 2 2 a b slope is 2.c (b) y 4 2 x 1 y 4 2x 2 y 2x 6c c c c c b c ba ba b 9. (a) 2 h. As h 0, 2 h 2 at? ? y x h h h 2 h 2 2 h 3 2 2 2 3 4 4h h 4 2h 3 3 2h h b b a b a b a b a bb c b c c c c b b c c c c c b 2 2 2 2 P 2, 3 the slope is 2.a bc (b) y 3 2 x 2 y 3 2x 4 y 2x 7.c c c b c ca b a b 10. (a) h 2. As h 0, h 2 2 at? ? y x h h h 1 h 4 1 h 1 4 1 1 2h h 4 4h 3 h 2h c c c a b a b a b a bb c b c c b b c c c c c 2 2 2 2 P 1, 3 the slope is 2.a bc c (b) y 3 2 x 1 y 3 2x 2 y 2x 1.c c c c b c b c ca b a ba b 11. (a) 12 4h h . As h 0, 12 4h h 12, at? ? y x h h h 2 h 2 8 12h 4h h 8 12h 4h h 2 2 b b b b a bb c b b b c b b 3 3 2 3 2 3 P 2, 8 the slope is 12.a b (b) y 8 12 x 2 y 8 12x 24 y 12x 16.c c c c ca b 12. (a) 3 3h h . As h 0, 3 3h h 3, at? ? y x h h h 2 1 h 2 1 2 1 3h 3h h 1 3h 3h h 2 2 c c c c c c c c b c c c c c c c c c ca b 3 3 2 3 2 3 P 1, 1 the slope is 3.a b c (b) y 1 3 x 1 y 1 3x 3 y 3x 4.c c c c c b c ba ba b 13. (a) 9 3h h . As h 0,? ? y x h h h 1 h 12 1 h 1 12 1 3h 3h h 12 12h 11 9h 3h h 2 c b b a b a b a b a bb c b c c " b b b c c c c c b b 3 3 2 3 2 3 9 3h h 9 at P 1, 11 the slope is 9.c b b c c c2 a b (b) y 11 9 x 1 y 11 9x 9 y 9x 2.c c c c b c b c ca b a ba b Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 52. 44 Chapter 2 Limits and Continuity 14. (a) 3h h . As h 0,? ? y x h h h 2 h 3 2 h 4 2 3 2 4 8 12h 6h h 12 12h 3h 4 0 3h h 2 b a b a b a b b c b b c c b b b b c c c b c b 3 2 23 2 3 2 2 3 3h h 0 at P 2, 0 the slope is 0.b 2 a b (b) y 0 0 x 2 y 0.c c a b 15. (a) Q Slope of PQ ? ? p t Q (10 225) 42.5 m/sec" c c 650 225 20 10 Q (14 375) 45.83 m/sec# c c 650 375 20 14 Q (16.5 475) 50.00 m/sec$ c c 650 475 20 16.5 Q (18 550) 50.00 m/sec% c c 650 550 20 18 (b) At t 20, the sportscar was traveling approximately 50 m/sec or 180 km/h. 16. (a) Q Slope of PQ ? ? p t Q (5 20) 12 m/sec" c c 80 20 10 5 Q (7 39) 13.7 m/sec# c c 80 39 10 7 Q (8.5 58) 14.7 m/sec$ c c 80 58 10 8.5 Q (9.5 72) 16 m/sec% c c 80 72 10 9.5 (b) Approximately 16 m/sec 17. (a) (b) 56 thousand dollars per year? ? p t 2004 2002 174 62 112 c c # (c) The average rate of change from 2001 to 2002 is 35 thousand dollars per year.? ? p t 20022 2001 62 27 c c The average rate of change from 2002 to 2003 is 49 thousand dollars per year.? ? p t 2003 2002 111 62 c c So, the rate at which profits were changing in 2002 is approximatley 35 49 42 thousand dollars per year." # a bb 18. (a) F(x) (x 2)/(x 2) b c x 1.2 1.1 1.01 1.001 1.0001 1 F(x) 4.0 3.4 3.04 3.004 3.0004 3c c c c c c 5.0; 4.4;? ? ? ? F F x 1.2 1 x 1.1 1 4.0 ( 3) 3.4 ( 3) c cc c c c c c c c 4.04; 4. ;? ? ? ? F F x 1.01 1 x 1.001 1 3.04 ( 3) 3.004 ( 3) c c !!%c c c c c c c c 4. ;? ? F x 1.0001 1 3. ( 3) c !!!%c !!!% c c c (b) The rate of change of F(x) at x 1 is 4. c 19. (a) 0.414213 0.449489? ? ? ? g g(2) g(1) 2 g g(1.5) g(1) x 2 1 1 x 1.5 1 0.5 1.5 c c " c c # c c c " ? ? g g(1 h) g(1) x (1 h) 1 h 1 h b c b c b c" (b) g(x) x 1 h 1.1 1.01 1.001 1.0001 1.00001 1.000001 1 h 1.04880 1.004987 1.0004998 1.0000499 1.000005 1.0000005 1 h 1 /h 0.4880 0.4987 0.4998 0.499 0.5 b b b c 0.5 (c) The rate of change of g(x) at x 1 is 0.5. Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 53. Section 2.1 Rates of Change and Tangents to Curves 45 (d) The calculator gives lim . h ! 1 h h b c" " # 20. (a) i) f(3) f(2) 3 2 1 1 6 c c c " c " " c" #3 6 ii) , T 2f(T) f(2) T T T T(T 2) T(2 T) T 2 T 2 Tc c # c # c # # c c# c # c c c c " c " " # # #T T T 2 T (b) T 2.1 2.01 2.001 2.0001 2.00001 2.000001 f(T) 0.476190 0.497512 0.499750 0.4999750 0.499997 0.499999 f(T) f(2) / T 2 0.2381 0.2488 0.2a b a bc c c c c 500 0.2500 0.2500 0.2500c c c (c) The table indicates the rate of change is 0.25 at t 2.c (d) lim T # " " c#T 4 c NOTE: Answers will vary in Exercises 21 and 22. 21. (a) 0, 1 : 15 mph 1, 2.5 : mph 2.5, 3.5 : 10 mph c c c c c c s 15 0 s 20 15 10 s 30 20 t 1 0 t 2.5 1 3 t 3.5 2.5 (b) At P , 7.5 : Since the portion of the graph from t 0 to t 1 is nearly linear, the instantaneous rate of change " # will be almost the same as the average rate of change, thus the instantaneous speed at t is 15 mi/hr. " c # c 15 7.5 1 0.5 At P 2, 20 : Since the portion of the graph from t 2 to t 2.5 is nearly linear, the instantaneous rate of change willa b be nearly the same as the average rate of change, thus v 0 mi/hr. For values of t less than 2, we have 20 20 2.5 2 c c Q Slope of PQ ? ? s t Q (1 15) 5 mi/hr" c c 15 20 1 2 Q (1.5 19) 2 mi/hr# c c 19 20 1.5 2 Q (1.9 19.9) 1 mi/hr$ c c 19.9 20 1.9 2 Thus, it appears that the instantaneous speed at t 2 is 0 mi/hr. At P 3, 22 :a b Q Slope of PQ ? ? s t Q (4 35) 13 mi/hr" c c 35 22 4 3 Q (3.5 30) 16 mi/hr# c c 30 22 3.5 3 Q (3.1 23) 10 mi/hr$ c c 23 22 3.1 3 Q Slope of PQ ? ? s t Q (2 20) 2 mi/hr" c c 20 22 2 3 Q (2.5 20) 4 mi/hr# c c 20 22 2.5 3 Q (2.9 21.6) 4 mi/hr$ c c 21.6 22 2.9 3 Thus, it appears that the instantaneous speed at t 3 is about 7 mi/hr. (c) It appears that the curve is increasing the fastest at t 3.5. Thus for P 3.5, 30 a b Q Slope of PQ ? ? s t Q (4 35) 10 mi/hr" c c 35 30 4 3.5 Q (3.75 34) 16 mi/hr# c c 34 30 3.75 3.5 Q (3.6 32) 20 mi/hr$ c c 32 30 3.6 3.5 Q Slope of PQ ? ? s t Q (3 22) 16 mi/hr" c c 22 30 3 3.5 Q (3.25 25) 20 mi/hr# c c 25 30 3.25 3.5 Q (3.4 28) 20 mi/hr$ c c 28 30 3.4 3.5 Thus, it appears that the instantaneous speed at t 3.5 is about 20 mi/hr. 22. (a) 0, 3 : 1.67 0, 5 : 2.2 7, 10 : 0.5 c c c c c c c c c A 10 15 A 3.9 15 A 0 1.4 t 3 0 day t 5 0 day t 10 7 day gal gal gal (b) At P 1, 14 :a b Q Slope of PQ ? ? A t Q (2 12.2) 1.8 gal/day" c c c12.2 14 2 1 Q (1.5 13.2) 1.6 gal/day# c c c13.2 14 1.5 1 Q (1.1 13.85) 1.5 gal/day$ c c c13.85 14 1.1 1 Q Slope of PQ ? ? A t Q (0 15) 1 gal/day" c c c15 14 0 1 Q (0.5 14.6) 1.2 gal/day# c c c14.6 14 0.5 1 Q (0.9 14.86) 1.4 gal/day$ c c c14.86 14 0.9 1 Thus, it appears that the instantaneous rate of consumption at t 1 is about 1.45 gal/day. c At P 4, 6 :a b Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 54. 46 Chapter 2 Limits and Continuity Q Slope of PQ ? ? A t Q (5 3.9) 2.1 gal/day" c c c3.9 6 5 4 Q (4.5 4.8) 2.4 gal/day# c c c4.8 6 4.5 4 Q (4.1 5.7) 3 gal/day$ c c c5.7 6 4.1 4 Q Slope of PQ ? ? A t Q (3 10) 4 gal/day" c c c10 6 3 4 Q (3.5 7.8) 3.6 gal/day# c c c7.8 6 3.5 4 Q (3.9 6.3) 3 gal/day$ c c c6.3 6 3.9 4 Thus, it appears that the instantaneous rate of consumption at t 1 is 3 gal/day. c At P 8, 1 :a b Q Slope of PQ ? ? A t Q (9 0.5) 0.5 gal/day" c c c0.5 1 9 8 Q (8.5 0.7) 0.6 gal/day# c c c0.7 1 8.5 8 Q (8.1 0.95) 0.5 gal/day$ c c c0.95 1 8.1 8 Q Slope of PQ ? ? A t Q (7 1.4) 0.6 gal/day" c c c1.4 1 7 8 Q (7.5 1.3) 0.6 gal/day# c c c1.3 1 7.5 8 Q (7.9 1.04) 0.6 gal/day$ c c c1.04 1 7.9 8 Thus, it appears that the instantaneous rate of consumption at t 1 is 0.55 gal/day. c (c) It appears that the curve (the consumption) is decreasing the fastest at t 3.5. Thus for P 3.5, 7.8 a b Q Slope of PQ ? ? A t Q (4.5 4.8) 3 gal/day" c c c4.8 7.8 4.5 3.5 Q (4 6) 3.6 gal/day# c c c6 7.8 4 3.5 Q (3.6 7.4) 4 gal/day$ c c c7.4 7.8 3.6 3.5 Q Slope of PQ ? ? s t Q (2.5 11.2) 3.4 gal/day" c c c11.2 7.8 2.5 3.5 Q (3 10) 4.4 gal/day# c c c10 7.8 3 3.5 Q (3.4 8.2) 4 gal/day$ c c c8.2 7.8 3.4 3.5 Thus, it appears that the rate of consumption at t 3.5 is about 4 gal/day. c 2.2 LIMIT OF A FUNCTION AND LIMIT LAWS 1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x 1. (b) 1 (c) 0 (d) 0.5 2. (a) 0 (b) 1c (c) Does not exist. As t approaches 0 from the left, f(t) approaches 1. As t approaches 0 from the right, f(t)c approaches 1. There is no single number L that f(t) gets arbitrarily close to as t 0. (d) 1c 3. (a) True (b) True (c) False (d) False (e) False (f) True (g) True 4. (a) False (b) False (c) True (d) True (e) True 5. lim does not exist because 1 if x 0 and 1 if x 0. As x approaches 0 from the left, x 0 x x x x x x x x x xk k k k k k c c approaches 1. As x approaches 0 from the right, approaches 1. There is no single number L that all thex x x xk k k kc function values get arbitrarily close to as x 0. 6. As x approaches 1 from the left, the values of become increasingly large and negative. As x approaches 1" cx 1 from the right, the values become increasingly large and positive. There is no one number L that all the function values get arbitrarily close to as x 1, so lim does not exist. x 1 " cx 1 Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 55. Section 2.2 Limit of a Function and Limit Laws 47 7. Nothing can be said about f(x) because the existence of a limit as x x does not depend on how the function ! is defined at x . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when! x is close enough to x . That is, the existence of a limit depends on the values of f(x) for x x , not on the! !near definition of f(x) at x itself.! 8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of the x 0 value f(0) itself. 9. No, the definition does not require that f be defined at x 1 in order for a limiting value to exist there. If f(1) is defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) 5. x 1 10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If lim f(x) exists, its value may be some number other than f(1) 5. We can conclude nothing about lim f(x), x 1 x 1 whether it exists or what its value is if it does exist, from knowing the value of f(1) alone. 11. lim (2x 5) 2( 7) 5 14 5 9 x c( b c b c b c 12. lim x 5x 2 (2) 5(2) 2 4 10 2 4 x # a bc b c c b c c b c # # 13. lim 8(t 5)(t 7) 8(6 5)(6 7) 8 t ' c c c c c 14. lim x 2x 4x 8 ( 2) 2( 2) 4( 2) 8 8 8 8 8 16 x c# a b$ # $ # c b b c c c b c b c c c b c 15. lim 16. lim 3s(2s 1) 3 2 1 2 1 x s # x 3 2 3 5 2 2 4 2 x 6 2 6 8 3 3 3 3 b b b b c c c # $ < 17. lim 3(2x 1) 3(2( 1) 1) 3( 3) 27 x c" c c c c # # # 18. lim y # y 2 y 5y 6 (2) 5( ) 6 4 10 6 0 5 2 2 4 4b b b b # b b b # b " # # 19. lim (5 y) [5 ( 3)] (8) (8) 2 16 y c$ c c c %$ %$ %$ "$ %% 20. lim (2z 8) (2(0) 8) ( 8) 2 z ! c c c c"$ "$ "$ 21. lim h ! 3 3 3 3 3h 1 1 3(0) 1 1 1 1 2 b b b b b 22. lim lim lim lim lim h 0 h 0 h 0 h 0 h 0 a b 5h 4 2 5h 4 2 5h 4 2 5h 4 4 h h 5h 4 2 5h 4 2h 5h 4 2 h 5h 4 2 5h 5b c b c b b b c b b b bb b b b 5 5 4 2 4 b 23. lim lim lim x x x & & & x 5 x 5 1 x 25 (x 5)(x 5) x 5 5 5 10 c c " " c b c b b# 24. lim lim lim x x x c$ c$ c$ x 3 x 3 1 x 4x 3 (x 3)(x 1) x 1 3 1 2 b b " " b b b b b c b# c Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 56. 48 Chapter 2 Limits and Continuity 25. lim lim lim (x 2) 7 x x x c& c& c& x 3x 0 x 5 x 5 (x 5)(x 2)# b c " b b b c c c& c # c 26. lim lim lim (x 5) 2 5 3 x x x # # # x 7x 0 x x 2 (x 5)(x 2)# c b " c # c c c c c c 27. lim lim lim t t t " " " t t 2 t 2 1 2 3 t 1 (t 1)(t 1) t 1 1 1 (t 2)(t 1)# # b c b b c c b b b # b c 28. lim lim lim t t t c" c" c" t 3t 2 t 2 1 2 1 t t 2 (t 2)(t 1) t 2 1 2 3 (t 2)(t 1)# # b b b c b c c c b c c c b b c 29. lim lim lim x x x c# c# c# c c c c b b c b2x 4 2 2 1 x 2x x (x 2) x 4 2 2(x 2) $ # # # c 30. lim lim lim y 0 y y ! ! 5y 8y y (5y 8) 5y 8 3y 16y y 3y 16 3y 16 16 8$ # # % # # # # b b b c c c c # " ca b 31. lim lim lim lim 1 x 1 x 1 x 1 x 1 1 1 x x xc c c c c1 x 1 x 1 x x 1 x 1 x 1 1 c c c 32. lim lim lim lim 2 x 0 x 1 x 1 x 1 1 1 x 1 x 1 x 1 x 1 x 1 x 1c b b b c c bb c b c b cx x x 1 x 1 x x 1 x 1 1 2x 1 2 2 c a b a b a ba b a ba b a ba b 33. lim lim lim u 1 u 1 u 1 u 4 u 1 u u 1 (u 1) u u 1 1 1 1 3 u (u 1)(u 1) u (u 1) (1 1)(1 1)% $ # # # # c " c b b c b b b b b " b c b " b b b a b a b a b 34. lim lim lim v v v # # # v 8 v 2v 4 4 4 4 12 3 v 16 (v 2) v 2v 4 (v 2)(v 2) v 4 (v 2) v 4 (4)(8) 32 8 $ # % # # # c b b b b c c b b c b b b b a b a b a b 35. lim lim lim x x x * * * x 3 x 3 x 9 6x 3 x 3 x 3 9 3 c c c c b b " " " b 36. lim lim lim lim x 2 x 4(2 2) 16 x x x x % % % % 4x x 2 x 2 x 2 x x(4 x) x 2 x 2 xc c c c c b c# b b 37. lim lim lim lim x 3 4 2 4 x 1 x 1 x 1 x 1 x 1 x 3 2 x 3 x 3 (x 1) x 3 2 (x 1) x 3 (x 3) 4 c b c b c # b b # c b b c b b # b c b b # b 38. lim lim lim x x x 1 c c c" " a bx 8 3 x 1 x 8 x 8 (x 1) x 8 (x 1) x 8 x 8# # # # # # b c b b c $ b b $ b b b $ b b b $ b c * lim lim c x 1 x 1 c c (x 1)(x 1) (x 1) x x 1 2 x 3 3 3 b c b b ) b $ c c " b ) b $ b # # 39. lim lim lim x 2 x 2 x 2 a bx 12 4 x 2 x 12 4 x 12 4 (x 2) x 12 4 (x 2) x 12 4 x 12 16# # # # # # b c c b c b b c b b c b b b c lim lim x 2 x 2 (x 2)(x 2) (x 2) x 12 4 x 2 4 x 12 4 16 4 2 c b c b b b " b b b # # 40. lim lim lim x 2 x 2 x 2 c c c x 2 x 5 3 x 2 x 5 3 x 2 x 5 3 x 5 3 x 5 3 x 5 9 b b c b b b b b b b c b b b c a b a b a b# # # # # # lim lim c x 2 x 2 c c a b x 2 x 5 3 (x 2)(x 2) x 2 4 2 x 5 3 9 3 3 b b b b c c c b b b # # Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 57. Section 2.2 Limit of a Function and Limit Laws 49 41. lim lim lim x 3 x 3 x 3 c c c 2 x 5 x 3 2 x 5 2 x 5 (x 3) 2 x 5 (x 3) 2 x 5 4 x 5c c b c c b c b b c b b c c c a b# # # # # # lim lim lim x 3 x 3 x 3 c c c 9 x 3 x 6 3 (x 3) 2 x 5 (x 3) 2 x 5 (3 x)(3 x) 2 x 5 2 4 2 c c b b c b b c c b b c b # # # # 42. lim lim lim x 4 x 4 x 4 4 x 5 x 9 4 x 5 x 9 4 x 5 x 9 5 x 9 5 x 9 25 x 9 c c b c b b c b b c b b b c b a b a b a b# # # # # # lim lim lim x 4 x 4 x 4 a b a b 4 x 5 x 9 4 x 5 x 9 16 x (4 x)(4 x) 4 x 8 4 5 x 9 5 25 5 c b b c b b c c b b b b b # # # # 43. lim 2sin x 1 2sin 0 1 0 1 1 44. lim sin x lim sin x sin 0 0 0 x 0 x 0 x 0 a b a b c c c c 2 2 2 2 45. lim sec x lim 1 46. lim tan x lim 0 x 0 x 0 x 0 x 0 1 1 1 sin x sin 0 0 cos x cos 0 1 cos x cos 0 1 47. lim x 0 1 x sin x 1 0 sin 0 1 0 0 1 3cos x 3cos 0 3 3 b b b b b b 48. lim x 1 2 cos x 0 1 2 cos 0 1 2 1 1 1 1 x 0 a ba b a ba b a ba b a ba b2 2 c c c c c c c c 49. lim x 4 cos x lim x 4 lim cos x 4 cos 0 4 1 4x x x c c c1 1 1 a b a b b b b b c b c c1 1 1 1 1 50. lim 7 sec x lim 7 sec x 7 lim sec x 7 sec 0 7 1 2 2 x 0 x 0 x 0 a b a b b b b b b 2 2 2 2 2 51. (a) quotient rule (b) difference and power rules (c) sum and constant multiple rules 52. (a) quotient rule (b) power and product rules (c) difference and constant multiple rules 53. (a) lim f(x) g(x) lim f(x) lim g(x) (5)( 2) 10x c x c x c c c (b) lim 2f(x) g(x) 2 lim f(x) lim g(x) 2(5)( 2) 20x c x c x c c c (c) lim [f(x) 3g(x)] lim f(x) 3 lim g(x) 5 3( 2) 1x c x c x c b b b c c (d) limx c f(x) f(x) g(x) lim f(x) lim g(x) 5 ( 2) 7 lim f(x) 5 5 c c c c x c x c x c 54. (a) lim [g(x) 3] lim g(x) lim 3 x x x % % % b b c$ b $ ! (b) lim xf(x) lim x lim f(x) (4)(0) 0 x x x % % % (c) lim [g(x)] lim g(x) [ 3] 9 x x % % # # # c (d) lim 3 x % g(x) f(x) 1 lim f(x) lim 1 0 1 lim g(x) 3 c c c c x x x % % % 55. (a) lim [f(x) g(x)] lim f(x) lim g(x) 7 ( 3) 4 x b x b x b b b b c (b) lim f(x) g(x) lim f(x) lim g(x) (7)( 3) 21 x b x b x b c c Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 58. 50 Chapter 2 Limits and Continuity (c) lim 4g(x) lim 4 lim g(x) (4)( 3) 12 x b x b x b c c (d) lim f(x)/g(x) lim f(x)/ lim g(x) x b x b x b c7 7 3 3c 56. (a) lim [p(x) r(x) s(x)] lim p(x) lim r(x) lim s(x) 4 0 ( 3) 1 x x x x c# c# c# c# b b b b b b c (b) lim p(x) r(x) s(x) lim p(x) lim r(x) lim s(x) (4)(0)( 3) 0 x x x x c# c# c# c# c (c) lim [ 4p(x) 5r(x)]/s(x) 4 lim p(x) 5 lim r(x) lim s(x) [ 4(4) 5(0)]/ 3 x x x x c# c# c# c# c b c b c b c "6 3 57. lim lim lim lim (2 h) 2 h h h h ! ! ! ! (1 h) 1 h(2 h) h h h 1 2h h 1b c bb b c# # # b 58. lim lim lim lim (h 4) 4 h h h h ! ! ! ! ( 2 h) ( 2) h(h 4) h h h 4 4h h 4c b c c cc b c# # # c c 59. lim lim 3 h h ! ! [3(2 h) 4] [3(2) 4] h h 3hb c c c 60. lim lim lim lim h h h h ! ! ! ! " " c c#b c# c bh 2 h 2 c c" c c c# b c c c c b c " h 2h 2h( h) h(4 2h) 4 2 ( 2 h) h c 61. lim lim lim lim lim h h h h h ! ! ! ! ! 7 h 7 h 7 h 7 7 h 7 h 7 h 7 h 7 h 7 h 7 h 7 (7 h) 7 h 7 h 7 b c b c b b b b b b b b b c " b b " #7 62. lim lim lim lim h h h h ! ! ! ! 3(0 h) 1 3(0) 1 h 3h 1 3h 1 h 3h 1 h 3h 1 1 h 3h 1 (3h 1) 3hb b c b b c " b b " b b " b b b b " b c " lim h ! 3 3 3h 1 1 b b # 63. lim 5 2x 5 2(0) 5 and lim 5 x 5 (0) 5; by the sandwich theorem, x x ! ! c c c c # # # # lim f(x) 5 x ! 64. lim 2 x 2 0 2 and lim 2 cos x 2(1) 2; by the sandwich theorem, lim g(x) 2 x x x ! ! ! a bc c # 65. (a) lim 1 1 1 and lim 1 1; by the sandwich theorem, lim 1 x x x ! ! ! c c x 0 x sin x 6 6 2 2 cos x # c (b) For x 0, y (x sin x)/(2 2 cos x) c lies between the other two graphs in the figure, and the graphs converge as x 0. 66. (a) lim lim lim 0 and lim ; by the sandwich theorem, x x x x ! ! ! ! " " " " " # # # # # # #c c c x 1 x 24 4 # # lim . x ! 1 cos x x c " ## Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 59. Section 2.2 Limit of a Function and Limit Laws 51 (b) For all x 0, the graph of f(x) (1 cos x)/x c # lies between the line y and the parabola " # y x /24, and the graphs converge as x 0. c " # # 67. (a) f(x) x /(x 3) c * ba b# x 3.1 3.01 3.001 3.0001 3.00001 3.000001 f(x) 6.1 6.01 6.001 6.0001 6.00001 6.000001 c c c c c c c c c c c c x 2.9 2.99 2.999 2.9999 2.99999 2.999999 f(x) 5.9 5.99 5.999 5.9999 5.99999 5.999999 c c c c c c c c c c c c The estimate is lim f(x) 6. x c$ c (b) (c) f(x) x 3 if x 3, and lim (x 3) 3 3 6. c c c c c cx 9 x 3 x 3 (x 3)(x 3)# c b b b c x c$ 68. (a) g(x) x / x 2 c # ca b # x 1.4 1.41 1.414 1.4142 1.41421 1.414213 g(x) 2.81421 2.82421 2.82821 2.828413 2.828423 2.828426 (b) (c) g(x) x 2 if x 2, and lim x 2 2 2 2 2. b b b x 2 x 2 x 2 x 2 x 2 # c c b c c x # 69. (a) G(x) (x 6)/ x 4x 12 b b ca b# x 5.9 5.99 5.999 5.9999 5.99999 5.999999 G(x) .126582 .1251564 .1250156 .1250015 .1250001 .1250000 c c c c c c c c c c c c x 6.1 6.01 6.001 6.0001 6.00001 6.000001 G(x) .123456 .124843 .124984 .124998 .124999 .124999 c c c c c c c c c c c c Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 60. 52 Chapter 2 Limits and Continuity (b) (c) G(x) if x 6, and lim 0.125. c c cx 6 x 6 x 4x 12 (x 6)(x 2) x x 2 2 8 b b " " " " b c b c c # c c' ca b# x c' 70. (a) h(x) x 2x 3 / x 4x 3 c c c ba b a b# # x 2.9 2.99 2.999 2.9999 2.99999 2.999999 h(x) 2.052631 2.005025 2.000500 2.000050 2.000005 2.0000005 x 3.1 3.01 3.001 3.0001 3.00001 3.000001 h(x) 1.952380 1.995024 1.999500 1.999950 1.999995 1.999999 (b) (c) h(x) if x 3, and lim 2. x 2x 3 x 1 x 1 3 1 4 x 4x 3 (x 3)(x 1) x 1 x 1 3 1 (x 3)(x 1)# # c c b b b c b c c c c c # c b x $ 71. (a) f(x) x 1 / x 1 c ca b a bk k# x 1.1 1.01 1.001 1.0001 1.00001 1.000001 f(x) 2.1 2.01 2.001 2.0001 2.00001 2.000001 c c c c c c x .9 .99 .999 .9999 .99999 .999999 f(x) 1.9 1.99 1.999 1.9999 1.99999 1.999999 c c c c c c (b) (c) f(x) , and lim (1 x) 1 ( 1) 2. x 1, x 0 and x 1 1 x, x 0 and x 1 c c c b c c x x 1 (x 1)(x 1) x 1 (x 1)(x 1) (x 1) # c " c b c c b c c b k k J x 1 c Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 61. Section 2.2 Limit of a Function and Limit Laws 53 72. (a) F(x) x 3x 2 / 2 x b b ca b a bk k# x 2.1 2.01 2.001 2.0001 2.00001 2.000001 F(x) 1.1 1.01 1.001 1.0001 1.00001 1.000001 c c c c c c c c c c c c x 1.9 1.99 1.999 1.9999 1.99999 1.999999 F(x) .9 .99 .999 .9999 .99999 .999999 c c c c c c c c c c c c (b) (c) F(x) , and lim (x 1) 2 1 , x 0 x 1, x 0 and x 2 b c b b c x 3x 2 2 x (x 2)(x 1) x (x 2)(x ) 2 x # b b c b b # c b b " b k k J x c# c1. 73. (a) g( ) (sin )/) ) ) .1 .01 .001 .0001 .00001 .000001 g( ) .998334 .999983 .999999 .999999 .999999 .999999 ) ) .1 .01 .001 .0001 .00001 .000001 g( ) .998334 .999983 .999999 .999999 .999999 .999999 ) ) c c c c c c lim g( ) 1 ) ! ) (b) 74. (a) G(t) (1 cos t)/t c # t .1 .01 .001 .0001 .00001 .000001 G(t) .499583 .499995 .499999 .5 .5 .5 t .1 .01 .001 .0001 .00001 .000001 G(t) .499583 .499995 .499999 .5 .5 .5 c c c c c c lim G(t) 0.5 t ! (b) Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 62. 54 Chapter 2 Limits and Continuity 75. lim f(x) exists at those points c where lim x lim x . Thus, c c c 1 c 0 c 0, 1, or 1.x c x c x c % # % # # # c ca b Moreover, lim f(x) lim x 0 and lim f(x) lim f(x) 1. x x x 1 x 1 ! ! c # 76. Nothing can be concluded about the values of f, g, and h at x 2. Yes, f(2) could be 0. Since the conditions of the sandwich theorem are satisfied, lim f(x) 5 0. x # c 77. 1 lim lim f(x) 5 2(1) lim f(x) 2 5 7. c b x x x % % % f(x) 5 x 2 lim x lim 2 lim f(x) lim 5 lim f(x) 5c c c % c # c c x x x x x % % % % % 78. (a) 1 lim lim f(x) 4. x x c# c# f(x) x lim x lim f(x) lim f(x) # # x x x c# c# c# % (b) 1 lim lim lim lim lim 2. c x x x x x c# c# c# c# c# f(x) f(x) f(x) f(x) x x x x x# " " c# 79. (a) 0 3 0 lim lim (x 2) lim (x 2) lim [f(x) 5] lim f(x) 5 c c c c x x x x x # # # # # f(x) 5 f(x) 5 x x c c c # c # lim f(x) 5. x # (b) 0 4 0 lim lim (x 2) lim f(x) 5 as in part (a). c x x x # # # f(x) 5 x c c # 80. (a) 0 1 0 lim lim x lim lim x lim x lim f(x). That is, lim f(x x x x x x x x ! ! ! ! ! ! ! f(x) f(x) f(x) x x x# # # # # # ) 0. (b) 0 1 0 lim lim x lim x lim . That is, lim 0. x x x x x ! ! ! ! ! f(x) f(x) f(x) f(x) x x x x# # 81. (a) lim x sin 0 x ! " x (b) 1 sin 1 for x 0:c " x x 0 x x sin x lim x sin 0 by the sandwich theorem; c " " x xx ! x 0 x x sin x lim x sin 0 by the sandwich theorem. c " " x xx ! 82. (a) lim x cos 0 x ! # " x$ Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 63. Section 2.3 The Precise Definition of a Limit 55 (b) 1 cos 1 for x 0 x x cos x lim x cos 0 by the sandwichc c " " "# # # # x x x$ $ $ x ! theorem since lim x 0. x ! # 83-88. Example CAS commands: :Maple f := x -> (x^4 16)/(x 2);c c x0 := 2; plot( f(x), x x0-1..x0+1, color black, title "Section 2.2, #83(a)" ); limit( f(x), x x0 ); In Exercise 85, note that the standard cube root, x^(1/3), is not defined for x (surd(x+1, 3) 1)/x.c : (assigned function and values for x0 and h may vary)Mathematica Clear[f, x] f[x_]:=(x x 5x 3)/(x 1)3 2 2 c c c b x0= 1; h = 0.1;c Plot[f[x],{x, x0 h, x0 h}]c b Limit[f[x], x x0] 2.3 THE PRECISE DEFINITION OF A LIMIT 1. Step 1: x 5 x 5 5 x 5k kc c c c b b$ $ $ $ $ Step 2: 5 7 2, or 5 1 4.$ $ $ $b c b The value of which assures x 5 1 x 7 is the smaller value, 2.$ $ $k kc 2. Step 1: x 2 x 2 x 2k kc c c c b # b$ $ $ $ $ Step 2: 2 1 1, or 2 7 5.c b b $ $ $ $ The value of which assures x 2 1 x 7 is the smaller value, 1.$ $ $k kc 3. Step 1: x ( 3) x 3 x 3k kc c c b $ c c c$ $ $ $ $ Step 2: 3 , or .c c c c $ c $ $ $ $7 5 # # # # " " The value of which assures x ( 3) x is the smaller value, .$ $ $k kc c c c 7 # # # " " 4. Step 1: x x x c c c b c c c3 3 3 3 # # # #$ $ $ $ $ Step 2: , or 1.c c c # c c $ $ $ $3 7 3 # # # # " The value of which assures x x is the smaller value, .$ $ $ c c c c "3 7 # # # " 5. Step 1: x x x c c c c b b" " " " # # # #$ $ $ $ $ Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 64. 56 Chapter 2 Limits and Continuity Step 2: , or .c b b $ $ $ $" " " " # # 4 4 9 18 7 14 The value of which assures x x is the smaller value, .$ $ $ c " " # 4 4 9 7 18 6. Step 1: x 3 x 3 3 x 3k kc c c c b b$ $ $ $ $ Step 2: 2.7591 0.2409, or 3.2391 0.2391.c b $ b $ $ $ $ $ The value of which assures x 3 2.7591 x 3.2391 is the smaller value, 0.2391.$ $ $k kc 7. Step 1: x 5 x 5 5 x 5k kc c c c b b$ $ $ $ $ Step 2: From the graph, 5 4.9 0.1, or 5 5.1 0.1; thus 0.1 in either case.c b b $ $ $ $ $ 8. Step 1: x ( 3) x 3 3 x 3k kc c c b c c c$ $ $ $ $ Step 2: From the graph, 3 3.1 0.1, or 3 2.9 0.1; thus 0.1.c c c c c $ $ $ $ $ 9. Step 1: x 1 x 1 1 x 1k kc c c c b b$ $ $ $ $ Step 2: From the graph, 1 , or 1 ; thus .c b b $ $ $ $ $9 7 25 9 7 16 16 16 16 16 10. Step 1: x 3 x 3 3 x 3k kc c c c b b$ $ $ $ $ Step 2: From the graph, 3 2.61 0.39, or 3 3.41 0.41; thus 0.39.c b b $ $ $ $ $ 11. Step 1: x 2 x 2 2 x 2k kc c c c b b$ $ $ $ $ Step 2: From the graph, 2 3 2 3 0.2679, or 2 5 5 2 0.2361;c b c b c $ $ $ $ thus 5 2.$ c 12. Step 1: x ( 1) x 1 1 x 1k kc c c b c c c$ $ $ $ $ Step 2: From the graph, 1 0.1180, or 1 0.1340;c c c c c $ $ $ $ 5 5 2 3 2 3 # # # # c c thus .$ 5 2c # 13. Step 1: x ( 1) x 1 1 x 1k kc c c b c c c$ $ $ $ $ Step 2: From the graph, 1 0.77, or 1 0.36; thus 0.36.c c c c c $ $ $ $16 7 16 9 9 9 9 25 25 25 14. Step 1: x x x c c c c b b" " " " # # # #$ $ $ $ $ Step 2: From the graph, 0.00248, or 0.00251;c b c b c $ $ $ $" " " " # # # # 1 1 1 1 2.01 2 .01 1.99 1.99 thus 0.00248.$ 15. Step 1: (x 1) 5 0.01 x 4 0.01 0.01 x 4 0.01 3.99 x 4.01k k k kb c c c c Step 2: x 4 x 4 4 x 4 0.01.k kc c c c b b $ $ $ $ $ $ 16. Step 1: (2x 2) ( 6) 0.02 2x 4 0.02 0.02 2x 4 0.02 4.02 2x 3.98k k k kc c c b c b c c 2.01 x 1.99 c c Step 2: x ( 2) x 2 2 x 2 0.01.k kc c c b c c c $ $ $ $ $ $ 17. Step 1: x 1 0.1 0.1 x 1 0.1 0.9 x 1 1.1 0.81 x 1 1.21 b c " c b c " b b 0.19 x 0.21 c Step 2: x 0 x . Then, or ; thus, 0.19.k kc c c c!"* !"* !#" $ $ $ $ $ $ $ Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 65. Section 2.3 The Precise Definition of a Limit 57 18. Step 1: x 0.1 0.1 x 0.1 0.4 x 0.6 0.16 x 0.36 c c c " " # # Step 2: x x x . c c c c b b" " " " 4 4 4 4$ $ $ $ $ Then, 0.16 0.09 or 0.36 0.11; thus 0.09.c b b $ $ $ $ $" " 4 4 19. Step 1: 19 x 19 x 1 2 19 x 4 19 x 16 c c $ " c" c c $ c % c x 19 16 15 x 3 or 3 x 15 c% c c Step 2: x 10 x 10 10 x 10.k kc c c c b b$ $ $ $ $ Then 10 3 7, or 10 15 5; thus 5.c b b $ $ $ $ $ 20. Step 1: x 7 4 1 x 7 1 3 x 7 5 9 x 7 25 16 x 32 c c c" c c % c c Step 2: x 23 x 23 23 x 23.k kc c c c b b$ $ $ $ $ Then 23 16 7, or 23 32 9; thus 7.c b b $ $ $ $ $ 21. Step 1: 0.05 0.05 0.05 0.2 0.3 x or x 5. " " " " " #x 4 x 4 x 3 3 10 10 10 c c c Step 2: x 4 x 4 4 x 4.k kc c c c b b$ $ $ $ $ Then or , or 4 5 or 1; thus .c b % b $ $ $ $ $10 2 2 3 3 3 22. Step 1: x 3 .1 0.1 x 3 0.1 2.9 x 3.1 2.9 x 3.1k k # # # c ! c c Step 2: x 3 x 3 3 x 3. c c c c b b$ $ $ $ $ Then 3 2.9 3 2.9 0.0291, or 3 3.1 3.1 3 0.0286;c b c b c $ $ $ $ thus 0.0286.$ 23. Step 1: x 4 0.5 0.5 x 4 0.5 3.5 x 4.5 3.5 x 4.5 4.5 x 3.5,k k k k # # # c c c c c for x near 2.c Step 2: x ( 2) x 2 x 2.k kc c c b c c # c$ $ $ $ $ Then 4.5 4.5 0.1213, or 3.5 3.5 0.1292;c c # c c # c # c # c $ $ $ $ thus 4.5 2 0.12.$ c 24. Step 1: ( 1) 0.1 0.1 1 0.1 x or x . " " " x x 10 x 10 11 9 9 11 11 9 10 10 10 10 c c c b c c c c c c Step 2: x ( 1) x 1 x .k kc c c b c c " c "$ $ $ $ $ Then , or ; thus .c c " c c " c $ $ $ $ $10 10 9 9 11 11 11 " " " 25. Step 1: x 5 11 x 16 1 x 16 1 15 x 17 15 x 17.k k k ka b # # # # c c " c c" c Step 2: x 4 x 4 x .k kc c c c b % b %$ $ $ $ $ Then 15 15 0.1270, or 17 17 0.1231;c b % % c b % c % $ $ $ $ thus 17 4 0.12.$ c 26. Step 1: 5 1 4 6 30 x 20 or 20 x 30. 120 120 120 x x x x 4 120 6c " c" c & " " Step 2: x 24 x 24 24 x 24.k kc c c c b b$ $ $ $ $ Then 24 20 4, or 24 30 6; thus 4.c b b $ $ $ $ $ 27. Step 1: mx 2m 0.03 0.03 mx 2m 0.03 0.03 2m mx 0.03 2mk kc c c c b b 2 x 2 .c b0.03 0.03 m m Step 2: x 2 x 2 2 x 2.k kc c c c b b$ $ $ $ $ Then 2 2 , or 2 . In either case, .c b c b # b $ $ $ $ $0.03 0.03 0.03 0.03 0.03 m m m m m Copyright 2010 Pearson Education, Inc. Publishing as Addison-Wesley. 66. 58 Chapter 2 Limits and Continuity 28. Step 1: mx 3m c c mx 3m c c 3m mx c 3m 3 x 3k kc c c c b b c bc c m m Step 2: x 3 x 3 .k kc c c c b $ B b $$ $ $ $ $ Then , or . In either case, .c b $ $ c b $ $ b $ $ $ $ $c c c c c m m m m m 29. Step 1: (mx b) b c mx c c mx c x . b c b - c c c b b c bm m m m c c m m# # # # # # " " Step 2: x x x . c c c c b b" " " " # # # #$ $ $ $ $ Then , or . In either case, .c b c b b $ $ $ $ $" " " " # # # # c c c c c m m m m m 30. Step 1: (mx b) (m b) 0.05 0.05 mx m 0.05 0.05 m mx 0.05 mk kb c b c c c b b 1 x . c " b0.05 0.05 m m Step 2: x 1 x 1 x .k kc c c c b " b "$ $ $ $ $ Then , or . In either case, .c b " " c b " " b $ $ $ $ $0.05 0.05 0.05 0.05 0.05 m m m m m 31. lim (3 2x) 3 2(3) 3 x 3 c c c Step 1: 3 2x ( 3) 0.02 0.02 6 2x 0.02 6.02 2x 5.98 3.01 x 2.99 ork ka bc c c c c c c c 2.99 x 3.01. Step 2: 0 x 3