Solucionario purcell, varbeg y rigdon   novena edicion
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Solucionario purcell, varbeg y rigdon novena edicion

Solucionario purcell, varbeg y rigdon novena edicion

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Solucionario purcell, varbeg y rigdon   novena edicion Solucionario purcell, varbeg y rigdon novena edicion Document Transcript

  • http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS.
  • Instructor’s Resource Manual Section 0.1 1 CHAPTER 0 Preliminaries 0.1 Concepts Review 1. rational numbers 2. dense 3. If not Q then not P. 4. theorems Problem Set 0.1 1. 4 2(8 11) 6 4 2( 3) 6 4 6 6 16 − − + = − − + = + + = 2. ( ) [ ] [ ] 3 2 4 7 12 3 2 4( 5) 3 2 20 3(22) 66 − − = − −⎡ ⎤⎣ ⎦ = + = = 3. –4[5(–3 12 – 4) 2(13 – 7)] –4[5(5) 2(6)] –4[25 12] –4(37) –148 + + = + = + = = 4. [ ] [ ] ( ) ( ) 5 1(7 12 16) 4 2 5 1(3) 4 2 5 3 4 2 5 1 2 5 2 7 − + − + + = − + + = − + + = + = + = 5. 5 1 65 7 58 – – 7 13 91 91 91 = = 6. 3 3 1 3 3 1 4 7 21 6 3 21 6 42 6 7 43 42 42 42 42 + − = + − − − = − + − = − 7. 1 1 1 1 1 1 1 3 – 4 1 – 3 2 4 3 6 3 2 12 6 1 1 1 1 – 3 2 12 6 1 1 4 – 3 24 24 1 3 1 3 24 24 ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ + = +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎡ ⎤⎛ ⎞ = +⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦ ⎡ ⎤ = +⎢ ⎥ ⎣ ⎦ ⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ 8. 1 2 1 1 1 1 3 5 2 3 5 2 1 5 3 3 5 2 15 15 1 2 1 2 1 2 1 3 5 2 15 3 5 15 1 6 1 1 5 1 3 15 15 3 15 9 ⎡ ⎤⎛ ⎞ − − − = −⎜ ⎟⎢ ⎥ ⎡ ⎤⎛ ⎞⎝ ⎠⎣ ⎦ − −⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦ ⎡ ⎤⎛ ⎞ ⎡ ⎤ = − − = − −⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦⎣ ⎦ ⎛ ⎞ ⎛ ⎞ = − − = − = −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 9. 2 2 2 1 14 3 3 2 14 2 14 2 14 6 21 21 21 145 14 3 2 9 6 21 7 3 49 49 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= = ⎜ ⎟⎜ ⎟ ⎜ ⎟− ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 10. 2 2 35 33 5 33 117 7 7 7 1 7 1 6 6 2 1 7 7 7 7 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠= = = − = − ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 11. 711 12 11 4– – 77 21 7 7 7 11 12 11 4 15 15 7 21 7 7 7 = = = + + 12. 1 3 7 4 6 7 5 52 4 8 8 8 8 8 1 3 7 4 6 7 3 3 2 4 8 8 8 8 8 − + − + = = = + − + − 13. 1 3 2 2 1 1 2 3 2 1 1– 1– 1– – 3 3 3 31 = = = = + 14. 3 3 3 2 2 2 5 2 5 7 1 2 2 2 2 6 14 6 20 2 7 7 7 7 + = + = + + − = + = + = 15. ( )( ) ( ) ( ) 2 2 5 3 5 – 3 5 – 3 5 – 3 2 + = = = © 2007 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.
  • 2 Section 0.1 Instructor’s Resource Manual 16. ( ) ( ) ( )( ) ( ) 2 2 2 5 3 5 2 5 3 3 5 2 15 3 8 2 15 − = − + = − + = − 17. 2 2 (3 4)( 1) 3 3 4 4 3 4 x x x x x x x − + = + − − = − − 18. 2 2 2 (2 3) (2 3)(2 3) 4 6 6 9 4 12 9 x x x x x x x x − = − − = − − + = − + 19. 2 2 (3 – 9)(2 1) 6 3 –18 – 9 6 –15 – 9 x x x x x x x + = + = 20. 2 2 (4 11)(3 7) 12 28 33 77 12 61 77 x x x x x x x − − = − − + = − + 21. 2 2 2 2 4 3 2 3 2 2 4 3 2 (3 1) (3 1)(3 1) 9 3 3 3 3 1 9 6 7 2 1 t t t t t t t t t t t t t t t t t t − + = − + − + = − + − + − + − + = − + − + 22. 3 2 3 2 2 3 2 (2 3) (2 3)(2 3)(2 3) (4 12 9)(2 3) 8 12 24 36 18 27 8 36 54 27 t t t t t t t t t t t t t t t + = + + + = + + + = + + + + + = + + + 23. 2 – 4 ( – 2)( 2) 2 – 2 – 2 x x x x x x + = = + , 2x ≠ 24. 2 6 ( 3)( 2) 2 3 ( 3) x x x x x x x − − − + = = + − − , 3x ≠ 25. 2 – 4 – 21 ( 3)( – 7) – 7 3 3 t t t t t t t + = = + + , 3t ≠ − 26. 2 3 2 2 2 2 2 (1 ) 2 ( 2 1) 2 ( 1) ( 1)( 1) 2 1 x x x x x x x x x x x x x x x x − − = − + − + − − = − − = − − 27. 2 12 4 2 22 x xx x + + ++ 12 4( 2) 2 ( 2) ( 2) ( 2) 12 4 8 2 6 20 ( 2) ( 2) 2(3 10) ( 2) x x x x x x x x x x x x x x x x x x + = + + + + + + + + + = = + + + = + 28. 2 2 6 2 9 1 y y y + − − 2 2(3 1) (3 1)(3 1) 2(3 1) 2 2(3 1)(3 1) 2(3 1)(3 1) y y y y y y y y y y = + − + − + = + + − + − 6 2 2 2(3 1)(3 1) y y y y + + = + − 8 2 2(3 1)(3 1) y y y + = + − 2(4 1) 4 1 2(3 1)(3 1) (3 1)(3 1) y y y y y y + + = = + − + − 29. a. 0 0 0⋅ = b. 0 0 is undefined. c. 0 0 17 = d. 3 0 is undefined. e. 5 0 0= f. 0 17 1= 30. If 0 0 a= , then 0 0 a= ⋅ , but this is meaningless because a could be any real number. No single value satisfies 0 0 a= . 31. .083 12 1.000 96 40 36 4 © 2007 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.
  • Instructor’s Resource Manual Section 0.1 3 32. .285714 7 2.000000 1 4 60 56 40 35 50 49 10 7 30 28 2 33. .142857 21 3.000000 2 1 90 84 60 42 180 168 120 105 150 147 3 34. .294117... 17 5.000000... 0.2941176470588235 3 4 160 153 70 68 20 17 30 17 130 119 11 → 35. 3.6 3 11.0 9 20 18 2 36. .846153 13 11.000000 10 4 60 52 80 78 20 13 70 65 50 39 11 37. x = 0.123123123... 1000 123.123123... 0.123123... 999 123 123 41 999 333 x x x x = = = = = 38. 0.217171717x = … 1000 217.171717... 10 2.171717... 990 215 215 43 990 198 x x x x = = = = = 39. x = 2.56565656... 100 256.565656... 2.565656... 99 254 254 99 x x x x = = = = 40. 3.929292x = … 100 392.929292... 3.929292... 99 389 389 99 x x x x = = = = © 2007 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.
  • 4 Section 0.1 Instructor’s Resource Manual 41. x = 0.199999... 100 19.99999... 10 1.99999... 90 18 18 1 90 5 x x x x = = = = = 42. 0.399999x = … 100 39.99999... 10 3.99999... 90 36 36 2 90 5 x x x x = = = = = 43. Those rational numbers that can be expressed by a terminating decimal followed by zeros. 44. 1 , p p q q ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ so we only need to look at 1 . q If 2 5 ,n m q = ⋅ then 1 1 1 (0.5) (0.2) . 2 5 n m n m q ⎛ ⎞ ⎛ ⎞ = ⋅ =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ The product of any number of terminating decimals is also a terminating decimal, so (0.5) and (0.2) ,n m and hence their product, 1 , q is a terminating decimal. Thus p q has a terminating decimal expansion. 45. Answers will vary. Possible answer: 0.000001, 12 1 0.0000010819... π ≈ 46. Smallest positive integer: 1; There is no smallest positive rational or irrational number. 47. Answers will vary. Possible answer: 3.14159101001... 48. There is no real number between 0.9999… (repeating 9's) and 1. 0.9999… and 1 represent the same real number. 49. Irrational 50. Answers will vary. Possible answers: and ,π π− 2 and 2− 51. 3 ( 3 1) 20.39230485+ ≈ 52. ( ) 4 2 3 0.0102051443− ≈ 53. 34 1.123 – 1.09 0.00028307388≈ 54. ( ) 1/ 2 3.1415 0.5641979034 − ≈ 55. 2 8.9 1 – 3 0.000691744752π + π ≈ 56. 24 (6 2) 3.661591807π π− ≈ 57. Let a and b be real numbers with ba < . Let n be a natural number that satisfies abn −</1 . Let }:{ bnkkS >= . Since a nonempty set of integers that is bounded below contains a least element, there is a Sk ∈0 such that bnk >/0 but bnk ≤− /)1( 0 . Then a n b nn k n k >−>−= − 111 00 Thus, ba n k ≤< −10 . If bn k <−10 , then choose n k r 10 − = . Otherwise, choose n k r 20 − = . Note that 1 a b r n < − < . Given a b< , choose r so that 1a r b< < . Then choose 2 3,r r so that 2 1 3a r r r b< < < < , and so on. 58. Answers will vary. Possible answer: 3 120 in≈ 59. ft 4000 mi 5280 21,120,000 ft mi r = × = equator 2 2 (21,120,000) 132,700,874 ft rπ π= = ≈ 60. Answers will vary. Possible answer: beats min hr day 70 60 24 365 20 yr min hr day year 735,840,000 beats × × × × = 61. 2 2 3 16 12 (270 12) 2 93,807,453.98 in. V r h ⎛ ⎞ = π = π ⋅ ⋅⎜ ⎟ ⎝ ⎠ ≈ volume of one board foot (in inches): 3 1 12 12 144 in.× × = number of board feet: 93,807,453.98 651,441 board ft 144 ≈ © 2007 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.
  • Instructor’s Resource Manual Section 0.1 5 62. 2 2 3 (8.004) (270) (8) (270) 54.3 ft.V π π= − ≈ 63. a. If I stay home from work today then it rains. If I do not stay home from work, then it does not rain. b. If the candidate will be hired then she meets all the qualifications. If the candidate will not be hired then she does not meet all the qualifications. 64. a. If I pass the course, then I got an A on the final exam. If I did not pass the course, thn I did not get an A on the final exam. b. If I take off next week, then I finished my research paper. If I do not take off next week, then I did not finish my research paper. 65. a. If a triangle is a right triangle, then 2 2 2 .a b c+ = If a triangle is not a right triangle, then 2 2 2 .a b c+ ≠ b. If the measure of angle ABC is greater than 0o and less than 90o , it is acute. If the measure of angle ABC is less than 0o or greater than 90o , then it is not acute. 66. a. If angle ABC is an acute angle, then its measure is 45o . If angle ABC is not an acute angle, then its measure is not 45o . b. If 2 2 a b< then .a b< If 2 2 a b≥ then .a b≥ 67. a. The statement, converse, and contrapositive are all true. b. The statement, converse, and contrapositive are all true. 68. a. The statement and contrapositive are true. The converse is false. b. The statement, converse, and contrapositive are all false. 69. a. Some isosceles triangles are not equilateral. The negation is true. b. All real numbers are integers. The original statement is true. c. Some natural number is larger than its square. The original statement is true. 70. a. Some natural number is not rational. The original statement is true. b. Every circle has area less than or equal to 9π. The original statement is true. c. Some real number is less than or equal to its square. The negation is true. 71. a. True; If x is positive, then 2 x is positive. b. False; Take 2x = − . Then 2 0x > but 0x < . c. False; Take 1 2 x = . Then xx <= 4 12 d. True; Let x be any number. Take 2 1y x= + . Then 2 y x> . e. True; Let y be any positive number. Take 2 y x = . Then 0 x y< < . 72. a. True; ( ) ( )1 : 0 1x x x x+ − < + + − < b. False; There are infinitely many prime numbers. c. True; Let x be any number. Take 1 1y x = + . Then 1 y x > . d. True; 1/ n can be made arbitrarily close to 0. e. True; 1/ 2n can be made arbitrarily close to 0. 73. a. If n is odd, then there is an integer k such that 2 1.n k= + Then 2 2 2 2 (2 1) 4 4 1 2(2 2 ) 1 n k k k k k = + = + + = + + b. Prove the contrapositive. Suppose n is even. Then there is an integer k such that 2 .n k= Then 2 2 2 2 (2 ) 4 2(2 )n k k k= = = . Thus 2 n is even. 74. Parts (a) and (b) prove that n is odd if and only if 2 n is odd. 75. a. 243 3 3 3 3 3= ⋅ ⋅ ⋅ ⋅ b. 2 124 4 31 2 2 31 or 2 31= ⋅ = ⋅ ⋅ ⋅ © 2007 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.
  • 6 Section 0.2 Instructor’s Resource Manual c. 2 2 5100 2 2550 2 2 1275 2 2 3 425 2 2 3 5 85 2 2 3 5 5 17 or 2 3 5 17 = ⋅ = ⋅ ⋅ = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 76. For example, let 2 3 ;A b c d= ⋅ ⋅ then 2 2 4 6 A b c d= ⋅ ⋅ , so the square of the number is the product of primes which occur an even number of times. 77. 2 2 2 2 2 ; 2 ; 2 ; p p q p q q = = = Since the prime factors of p2 must occur an even number of times, 2q2 would not be valid and 2 p q = must be irrational. 78. 2 2 2 2 3 ; 3 ; 3 ; p p q p q q = = = Since the prime factors of 2 p must occur an even number of times, 2 3q would not be valid and 3 p q = must be irrational. 79. Let a, b, p, and q be natural numbers, so a b and p q are rational. a p aq bp b q bq + + = This sum is the quotient of natural numbers, so it is also rational. 80. Assume a is irrational, 0 p q ≠ is rational, and p r a q s ⋅ = is rational. Then q r a p s ⋅ = ⋅ is rational, which is a contradiction. 81. a. – 9 –3;= rational b. 3 0.375 ; 8 = rational c. (3 2)(5 2) 15 4 30;= = rational d. 2 (1 3) 1 2 3 3 4 2 3;+ = + + = + irrational 82. a. –2 b. –2 c. x = 2.4444...; 10 24.4444... 2.4444... 9 22 22 9 x x x x = = = = d. 1 e. n = 1: x = 0, n = 2: 3 , 2 x = n = 3: 2 – , 3 x = n = 4: 5 4 x = The upper bound is 3 . 2 f. 2 83. a. Answers will vary. Possible answer: An example is 2 { : 5, a rational number}.S x x x= < Here the least upper bound is 5, which is real but irrational. b. True 0.2 Concepts Review 1. [ 1,5);( , 2]− −∞ − 2. b > 0; b < 0 3. (b) and (c) 4. 1 5x− ≤ ≤ Problem Set 0.2 1. a. b. c. d. © 2007 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.
  • Instructor’s Resource Manual Section 0.2 7 e. f. 2. a. (2,7) b. [ 3,4)− c. ( , 2]−∞ − d. [ 1,3]− 3. 7 2 5 2 ;( 2, ) x x x − < − − < − ∞ 4. ( ) 3 5 4 6 1 ; 1, x x x − < − < ∞ 5. 7 – 2 9 3 –5 2 5 5 – ; – , 2 2 x x x x ≤ + ≤ ⎡ ⎞ ≥ ∞⎟⎢ ⎣ ⎠ 6. 5 3 6 4 1 ;( ,1) x x x − > − > −∞ 7. 4 3 2 5 6 3 3 2 1;( 2, 1) x x x − < + < − < < − < < − − 8. 3 4 9 11 6 4 20 3 3 5; ,5 2 2 x x x − < − < < < ⎛ ⎞ < < ⎜ ⎟ ⎝ ⎠ 9. –3 1– 6 4 –4 –6 3 2 1 1 2 – ; – , 3 2 2 3 x x x < ≤ < ≤ ⎡ ⎞ > ≥ ⎟⎢ ⎣ ⎠ 10. 4 5 3 7 1 3 2 1 2 2 1 ; , 3 3 3 3 x x x < − < − < − < ⎛ ⎞ > > − −⎜ ⎟ ⎝ ⎠ 11. x2 + 2x – 12 < 0; 2 –2 (2) – 4(1)(–12) –2 52 2(1) 2 –1 13 x ± ± = = = ± ( ) ( )– –1 13 – –1– 13 0;x x⎡ ⎤ ⎡ ⎤+ < ⎣ ⎦ ⎣ ⎦ ( )–1– 13, –1 13+ 12. 2 5 6 0 ( 1)( 6) 0; x x x x − − > + − > ( , 1) (6, )−∞ − ∪ ∞ 13. 2x2 + 5x – 3 > 0; (2x – 1)(x + 3) > 0; 1 ( , 3) , 2 ⎛ ⎞ −∞ − ∪ ∞⎜ ⎟ ⎝ ⎠ 14. 2 4 5 6 0 3 (4 3)( 2) 0; ,2 4 x x x x − − < ⎛ ⎞ + − < −⎜ ⎟ ⎝ ⎠ 15. 4 0; – 3 x x + ≤ [–4, 3) © 2007 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.
  • 8 Section 0.2 Instructor’s Resource Manual 16. 3 2 2 0; , (1, ) 1 3 x x − ⎛ ⎤ ≥ −∞ ∪ ∞⎜ ⎥− ⎝ ⎦ 17. 2 5 2 5 0 2 5 0; x x x x < − < − < 2 (– , 0) , 5 ⎛ ⎞ ∞ ∪ ∞⎜ ⎟ ⎝ ⎠ 18. 7 7 4 7 7 0 4 7 28 0; 4 x x x x ≤ − ≤ − ≤ ( ) 1 ,0 , 4 ⎡ ⎞ −∞ ∪ ∞⎟⎢ ⎣ ⎠ 19. 1 4 3 2 1 4 0 3 2 1 4(3 2) 0 3 2 9 12 2 3 0; , , 3 2 3 4 x x x x x x ≤ − − ≤ − − − ≤ − − ⎛ ⎞ ⎡ ⎞ ≤ −∞ ∪ ∞⎜ ⎟ ⎟⎢− ⎝ ⎠ ⎣ ⎠ 20. 3 2 5 3 2 0 5 3 2( 5) 0 5 2 7 7 0; 5, 5 2 x x x x x x > + − > + − + > + − − ⎛ ⎞ > − −⎜ ⎟ + ⎝ ⎠ 21. ( 2)( 1)( 3) 0;( 2,1) (3,8)x x x+ − − > − ∪ 22. 3 1 (2 3)(3 1)( 2) 0; , ,2 2 3 x x x ⎛ ⎞ ⎛ ⎞ + − − < −∞ − ∪⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 23. 2 (2 -3)( -1) ( -3) 0;x x x ≥ [ ) 3 – , 3, 2 ⎛ ⎤ ∞ ∪ ∞⎜ ⎥ ⎝ ⎦ 24. ( ) ( ) 2 (2 3)( 1) ( 3) 0; 3 ,1 1, 3, 2 x x x− − − > ⎛ ⎞ −∞ ∪ ∪ ∞⎜ ⎟ ⎝ ⎠ 25. 3 2 – – 2 – 5 6 0 ( 5 – 6) 0 ( 1)( – 6) 0; x x x x x x x x x < < + < ( , 1) (0,6)−∞ − ∪ 26. 3 2 2 2 1 0 ( 1)( 1) 0 ( 1)( 1) 0; x x x x x x x − − + > − − > + − > ( 1,1) (1, )− ∪ ∞ 27. a. False. b. True. c. False. © 2007 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.
  • Instructor’s Resource Manual Section 0.2 9 28. a. True. b. True. c. False. 29. a. ⇒ Let a b< , so 2 ab b< . Also, 2 a ab< . Thus, 2 2 a ab b< < and 2 2 a b< . ⇐ Let 2 2 a b< , so a b≠ Then ( ) ( ) 2 2 2 2 2 0 2 2 2 a b a ab b b ab b b b a < − = − + < − + = − Since 0b > , we can divide by 2b to get 0b a− > . b. We can divide or multiply an inequality by any positive number. 1 1 1 a a b b b a < ⇔ < ⇔ < . 30. (b) and (c) are true. (a) is false: Take 1, 1a b= − = . (d) is false: if a b≤ , then a b− ≥ − . 31. a. 3x + 7 > 1 and 2x + 1 < 3 3x > –6 and 2x < 2 x > –2 and x < 1; (–2, 1) b. 3x + 7 > 1 and 2x + 1 > –4 3x > –6 and 2x > –5 x > –2 and 5 – ; 2 x > ( )2,− ∞ c. 3x + 7 > 1 and 2x + 1 < –4 x > –2 and 5 – ; 2 x < ∅ 32. a. 2 7 1 or 2 1 3 2 8 or 2 2 4 or 1 x x x x x x − > + < > < > < ( ,1) (4, )−∞ ∪ ∞ b. 2 7 1 or 2 1 3 2 8 or 2 2 4 or 1 ( ,4] x x x x x x − ≤ + < ≤ < ≤ < −∞ c. 2 7 1 or 2 1 3 2 8 or 2 2 4 or 1 ( , ) x x x x x x − ≤ + > ≤ > ≤ > −∞ ∞ 33. a. 2 2 – 3 2 2 – – 3 2 ( 1)( 2 – 7) 1 3 5 – 7 1 2 – 5 – 6 0 ( 3)( 1)( – 2) 0 x x x x x x x x x x x x x x + + ≥ + ≥ + ≥ + + ≥ [ 3, 1] [2, )− − ∪ ∞ b. 4 2 4 2 2 2 2 2 8 2 8 0 ( 4)( 2) 0 ( 2)( 2)( 2) 0 x x x x x x x x x − − − − ≥ ≥ + ≥ + + − ≥ ( , 2] [2, )−∞ − ∪ ∞ c. 2 2 2 2 2 2 2 ( 1) 7( 1) 10 0 [( 1) 5][( 1) 2] 0 ( 4)( 1) 0 ( 2)( 1)( 1)( 2) 0 x x x x x x x x x x − − − + + + < + − + − < < + + − − < ( 2, 1) (1,2)− − ∪ 34. a. 01.2 1 99.1 << x xx 01.2199.1 << 199.1 <x and x01.21< 99.1 1 <x and 01.2 1 >x 99.1 1 01.2 1 << x ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 99.1 1 , 01.2 1 b. 01.3 2 1 99.2 < + < x )2(01.31)2(99.2 +<<+ xx 198.599.2 <+x and 02.601.31 +< x 99.2 98.4− <x and 01.3 02.5− >x 99.2 98.4 01.3 02.5 −<<− x ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −− 99.2 98.4 , 01.3 02.5 35. 2 5; 2 5 or 2 5 3 or 7 x x x x x − ≥ − ≤ − − ≥ ≤ − ≥ ( , 3] [7, )−∞ − ∪ ∞ © 2007 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.
  • 10 Section 0.2 Instructor’s Resource Manual 36. 2 1;x + < –1 2 1 –3 –1 x x < + < < < (–3, –1) 37. 4 5 10; 10 4 5 10 15 4 5 15 5 15 5 ; , 4 4 4 4 x x x x + ≤ − ≤ + ≤ − ≤ ≤ ⎡ ⎤ − ≤ ≤ −⎢ ⎥ ⎣ ⎦ 38. 2 –1 2;x > 2x – 1 < –2 or 2x – 1 > 2 2x < –1 or 2x > 3; 1 3 1 3 – or , – , – , 2 2 2 2 x x ⎛ ⎞ ⎛ ⎞ < > ∞ ∪ ∞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 39. 2 5 7 7 2 2 5 7 or 5 7 7 7 2 2 2 or 12 7 7 7 or 42; x x x x x x x − ≥ − ≤ − − ≥ ≤ − ≥ ≤ − ≥ ( , 7] [42, )−∞ − ∪ ∞ 40. 1 1 4 x + < 1 1 1 4 2 0; 4 x x − < + < − < < –8 < x < 0; (–8, 0) 41. 5 6 1; 5 6 1 or 5 6 1 5 5 or 5 7 7 7 1 or ;( ,1) , 5 5 x x x x x x x − > − < − − > < > ⎛ ⎞ < > −∞ ∪ ∞⎜ ⎟ ⎝ ⎠ 42. 2 – 7 3;x > 2x – 7 < –3 or 2x – 7 > 3 2x < 4 or 2x > 10 x < 2 or x > 5; ( ,2) (5, )−∞ ∪ ∞ 43. 1 3 6; 1 1 3 6 or 3 6 1 1 3 0 or 9 0 x x x x x − > − < − − > + < − > 1 3 1 9 0 or 0; 1 1 ,0 0, 3 9 x x x x + − < > ⎛ ⎞ ⎛ ⎞ − ∪⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 44. 5 2 1; x + > 5 5 2 –1 or 2 1 5 5 3 0 or 1 0 3 5 5 0 or 0; x x x x x x x x + < + > + < + > + + < > 5 (– , – 5) – , 0 (0, ) 3 ⎛ ⎞ ∞ ∪ ∪ ∞⎜ ⎟ ⎝ ⎠ 45. 2 3 4 0;x x− − ≥ 2 3 (–3) – 4(1)(–4) 3 5 –1,4 2(1) 2 x ± ± = = = ( 1)( 4) 0;( , 1] [4, )x x+ − = −∞ − ∪ ∞ 46. 2 2 4 ( 4) 4(1)(4) 4 4 0; 2 2(1) ( 2)( 2) 0; 2 x x x x x x ± − − − + ≤ = = − − ≤ = 47. 3x2 + 17x – 6 > 0; 2 –17 (17) – 4(3)(–6) –17 19 1 –6, 2(3) 6 3 x ± ± = = = (3x – 1)(x + 6) > 0; 1 (– , – 6) , 3 ⎛ ⎞ ∞ ∪ ∞⎜ ⎟ ⎝ ⎠ 48. 2 14 11 15 0;x x+ − ≤ 2 11 (11) 4(14)( 15) 11 31 2(14) 28 3 5 , 2 7 x x − ± − − − ± = = = − 3 5 3 5 0; , 2 7 2 7 x x ⎛ ⎞⎛ ⎞ ⎡ ⎤ + − ≤ −⎜ ⎟⎜ ⎟ ⎢ ⎥ ⎝ ⎠⎝ ⎠ ⎣ ⎦ 49. 3 0.5 5 3 5(0.5) 5 15 2.5x x x− < ⇒ − < ⇒ − < 50. 2 0.3 4 2 4(0.3) 4 18 1.2x x x+ < ⇒ + < ⇒ + < © 2007 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.
  • Instructor’s Resource Manual Section 0.2 11 51. 2 6 2 6 12 6 x x x ε ε ε− < ⇒ − < ⇒ − < 52. 4 2 4 2 8 2 x x x ε ε ε+ < ⇒ + < ⇒ + < 53. 3 15 3( 5) 3 5 5 ; 3 3 x x x x ε ε ε ε ε δ − < ⇒ − < ⇒ − < ⇒ − < = 54. 4 8 4( 2) 4 2 2 ; 4 4 x x x x ε ε ε ε ε δ − < ⇒ − < ⇒ − < ⇒ − < = 55. 6 36 6( 6) 6 6 6 ; 6 6 x x x x ε ε ε ε ε δ + < ⇒ + < ⇒ + < ⇒ + < = 56. 5 25 5( 5) 5 5 5 ; 5 5 x x x x ε ε ε ε ε δ + < ⇒ + < ⇒ + < ⇒ + < = 57. C dπ= –10 0.02 –10 0.02 10 – 0.02 10 0.02 – 0.0064 C d d d ≤ π ≤ ⎛ ⎞ π ≤⎜ ⎟ π⎝ ⎠ ≤ ≈ π π We must measure the diameter to an accuracy of 0.0064 in. 58. ( ) 5 50 1.5, 32 50 1.5; 9 C F− ≤ − − ≤ ( ) 5 32 90 1.5 9 122 2.7 F F − − ≤ − ≤ We are allowed an error of 2.7 F. 59. 2 2 2 2 – – 2 – –1 2 – 3 –1 2 – 6 ( –1) (2 – 6) 2 1 4 24 36 3 22 35 0 (3 – 7)( – 5) 0; x x x x x x x x x x x x x x < < < + < + + > > 7 – , (5, ) 3 ⎛ ⎞ ∞ ∪ ∞⎜ ⎟ ⎝ ⎠ 60. ( )22 2 1 1 (2 1) 1 x x x x − ≥ + − ≥ + 2 2 2 4 4 1 2 1 3 6 0 3 ( 2) 0 x x x x x x x x − + ≥ + + − ≥ − ≥ ( ,0] [2, )−∞ ∪ ∞ 61. 2 2 2 2 2 2 2 3 10 4 6 10 (4 6) ( 10) 16 48 36 20 100 15 68 64 0 (5 4)(3 16) 0; x x x x x x x x x x x x x x − − − < + − < + − < + + < + + − < + − < 4 16 – , 5 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 62. ( )( ) 2 2 2 2 2 3 1 2 6 3 1 2 12 (3 1) (2 12) 9 6 1 4 48 144 5 54 143 0 5 11 13 0 x x x x x x x x x x x x x x − < + − < + − < + − + < + + − − < + − < 11 ,13 5 ⎛ ⎞ −⎜ ⎟ ⎝ ⎠ © 2007 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.
  • 12 Section 0.2 Instructor’s Resource Manual 63. ( ) 2 2 22 2 2 and Order property: when is positive. Transitivity x y x x x y x y y y x y xz yz z x y x y x x < ⇒ ≤ < < ⇔ < ⇒ < ⇒ < = Conversely, ( ) ( )( ) 2 2 22 2 2 2 2 2 – 0 Subtract from each side. – 0 Factor the difference of two squares. – 0 This is the only factor that can be negative. Add t x y x y x x x y y x y x y x y x y y < ⇒ < = ⇒ < ⇒ + < ⇒ < ⇒ < o each side. 64. ( ) ( ) 2 2 0 ,a b a a and b b< < ⇒ = = so ( ) ( ) 2 2 ,a b< and, by Problem 63, a b a b< ⇒ < . 65. a. – (– ) –a b a b a b a b= + ≤ + = + b. – – –a b a b a b≥ ≥ Use Property 4 of absolute values. c. ( )a b c a b c a b c a b c + + = + + ≤ + + ≤ + + 66. 2 2 2 2 2 1 1 1 1 2 23 3 1 1 23 1 1 23 1 1 23 x xx x xx xx xx ⎛ ⎞ − = + −⎜ ⎟⎜ ⎟+ ++ + ⎝ ⎠ ≤ + − ++ = + ++ = + ++ by the Triangular Inequality, and since 2 3 0,x + > 2 1 1 2 0 0, 0. 23 x xx + > ⇒ > > ++ 2 3 3x + ≥ and 2 2,x + ≥ so 2 1 1 33x ≤ + and 1 1 , 2 2x ≤ + thus, 2 1 1 1 1 2 3 23 xx + ≤ + ++ 67. 2 2 – 2 (–2) 9 9 x x x x + = + + 2 2 2 – 2 –2 9 9 9 x x x x x ≤ + + + + 2 2 2 2 2– 2 2 9 9 9 9 x xx x x x x + ≤ + = + + + + Since 2 2 1 1 9 9, 99 x x + ≥ ≤ + 2 2 2 99 x x x + + ≤ + 2 2– 2 99 xx x + ≤ + 68. 2 2 2 2 7 2 7 4 4 7 15 x x x x x≤ ⇒ + + ≤ + + ≤ + + = and 2 1 1x + ≥ so 2 1 1. 1x ≤ + Thus, 2 2 2 2 2 7 1 2 7 1 1 15 1 15 x x x x x x + + = + + + + ≤ ⋅ = 69. 4 3 21 1 1 1 2 4 8 16 x x x x+ + + + 4 3 2 4 3 2 1 1 1 1 2 4 8 16 1 1 1 1 1 since 1. 2 4 8 16 1 1 1 1 So 1.9375 2. 2 4 8 16 x x x x x x x x x ≤ + + + + ≤ + + + + ≤ + + + + ≤ < © 2007 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.
  • Instructor’s Resource Manual Section 0.3 13 70. a. 2 2 0 (1 ) 0 0 or 1 x x x x x x x x < − < − < < > b. 2 2 0 ( 1) 0 0 1 x x x x x x x < − < − < < < 71. 0a ≠ ⇒ 2 2 2 1 1 0 – – 2a a a a ⎛ ⎞ ≤ = +⎜ ⎟ ⎝ ⎠ so, 2 2 2 2 1 1 2 or 2a a a a ≤ + + ≥ . 72. and 2 2 2 a b a a a b a b b b a a b b a b a b < + < + + < + < + < + < < 73. 2 2 2 2 0 and a b a ab ab b a ab b a ab b < < < < < < < < 74. ( ) ( ) ( ) 2 2 2 2 2 2 2 1 1 2 2 4 1 1 1 1 0 2 4 2 4 4 1 0 ( ) which is always true. 4 ab a b ab a ab b a ab b a ab b a b ≤ + ⇔ ≤ + + ⇔ ≤ − + = − + ⇔ ≤ − 75. For a rectangle the area is ab, while for a square the area is 2 2 . 2 a b a +⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ From Problem 74, 2 1 ( ) 2 2 a b ab a b ab +⎛ ⎞ ≤ + ⇔ ≤ ⎜ ⎟ ⎝ ⎠ so the square has the largest area. 76. 2 3 99 1 0; ( , 1] x x x x+ + + + + ≤ −∞ − … 77. 1 1 1 1 10 20 30R ≤ + + 1 6 3 2 60 1 11 60 60 11 R R R + + ≤ ≤ ≥ 1 1 1 1 20 30 40 1 6 4 3 120 120 13 R R R ≥ + + + + ≥ ≤ Thus, 60 120 11 13 R≤ ≤ 78. 2 2 2 2 2 4 ; 4 (10) 400 4 400 0.01 4 100 0.01 0.01 100 4 A r A r r r π π π π π π π = = = − < − < − < 20.01 0.01 100 4 4 0.01 0.01 100 100 4 4 0.00004 in r r π π π π δ − < − < − < < + ≈ 0.3 Concepts Review 1. 2 2 ( 2) ( 3)x y+ + − 2. (x + 4)2 + (y – 2)2 = 25 3. 2 5 3 7 , (1.5,5) 2 2 − + +⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 4. d b c a − − © 2007 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.
  • 14 Section 0.3 Instructor’s Resource Manual Problem Set 0.3 1. 2 2 (3 –1) (1–1) 4 2d = + = = 2. 2 2 ( 3 2) (5 2) 74 8.60d = − − + + = ≈ 3. 2 2 (4 – 5) (5 8) 170 13.04d = + + = ≈ 4. 2 2 ( 1 6) (5 3) 49 4 53 7.28 d = − − + − = + = ≈ 5. 2 2 1 (5 2) (3 – 4) 49 1 50d = + + = + = 2 2(5 10) (3 8) 25 25 502 2 2( 2 10) (4 8)3 144 16 160 so the triangle is isosceles.1 2 d d d d = − + − = + = = − − + − = + = = 6. 2 2 (2 4) ( 4 0) 4 16 20a = − + − − = + = 2 2 (4 8) (0 2) 16 4 20b = − + + = + = 2 2 (2 8) ( 4 2) 36 4 40c = − + − + = + = 2 2 2 ,a b c+ = so the triangle is a right triangle. 7. (–1, –1), (–1, 3); (7, –1), (7, 3); (1, 1), (5, 1) 8. ( ) 2 2 2 2 2 2 ( 3) (0 1) ( 6) (0 4) ; 6 10 12 52 6 42 7 7,0 x x x x x x x x − + − = − + − − + = − + = = ⇒ 9. –2 4 –2 3 1 , 1, ; 2 2 2 + +⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 21 252(1 2) – 3 9 3.91 2 4 d ⎛ ⎞ = + + = + ≈⎜ ⎟ ⎝ ⎠ 10. 1 2 3 6 3 9 midpoint of , , 2 2 2 2 AB + +⎛ ⎞ ⎛ ⎞ = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 4 3 7 4 7 11 midpoint of , , 2 2 2 2 CD + +⎛ ⎞ ⎛ ⎞ = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 3 7 9 11 2 2 2 2 4 1 5 2.24 d ⎛ ⎞ ⎛ ⎞ = − + −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = + = ≈ 11 (x – 1)2 + (y – 1)2 = 1 12. 2 2 2 2 2 ( 2) ( 3) 4 ( 2) ( 3) 16 x y x y + + − = + + − = 13. 2 2 2 2 2 2 2 2 2 ( 2) ( 1) (5 2) (3 1) 9 16 25 ( 2) ( 1) 25 x y r r r x y − + + = − + + = = + = − + + = © 2007 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.
  • Instructor’s Resource Manual Section 0.3 15 14. 2 2 2 ( 4) ( 3)x y r− + − = 2 2 2 2 2 2 (6 4) (2 3) 4 1 5 ( 4) ( 3) 5 r r x y − + − = = + = − + − = 15. 1 3 3 7 center , (2, 5) 2 2 + +⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ 1 12 2radius (1– 3) (3 – 7) 4 16 2 2 1 20 5 2 = + = + = = 2 2 ( – 2) ( – 5) 5x y+ = 16. Since the circle is tangent to the x-axis, 4.r = 2 2 ( 3) ( 4) 16x y− + − = 17. 2 2 2 10 – 6 –10 0x x y y+ + + = 2 2 2 2 2 2 2 – 6 0 ( 2 1) ( – 6 9) 1 9 ( 1) ( – 3) 10 x x y y x x y y x y + + = + + + + = + + + = center (–1, 3); radius 10= = 18. 2 2 2 2 2 2 6 16 ( 6 9) 16 9 ( 3) 25 x y y x y y x y + − = + − + = + + − = center (0, 3); radius 5= = 19. 2 2 –12 35 0x y x+ + = 2 2 2 2 2 2 –12 –35 ( –12 36) –35 36 ( – 6) 1 x x y x x y x y + = + + = + + = center (6, 0); radius 1= = 20. 2 2 2 2 2 2 10 10 0 ( 10 25) ( 10 25) 25 25 ( 5) ( 5) 50 x y x y x x y y x y + − + = − + + + + = + − + + = ( )center 5, 5 ; radius 50 5 2= − = = 21. 2 2 4 16 15 4 6 0x x y y+ + + + = 2 2 3 9 9 4( 4 4) 4 15 16 2 16 4 x x y y ⎛ ⎞ + + + + + = − + +⎜ ⎟ ⎝ ⎠ 2 2 3 13 4( 2) 4 4 4 x y ⎛ ⎞ + + + =⎜ ⎟ ⎝ ⎠ 2 2 3 13 ( 2) 4 16 x y ⎛ ⎞ + + + =⎜ ⎟ ⎝ ⎠ center = 3 2, ; 4 ⎛ ⎞ − −⎜ ⎟ ⎝ ⎠ radius = 13 4 22. 2 2 2 2 105 4 16 4 3 0 16 3 9 4( 4 4) 4 4 64 105 9 16 16 16 x x y y x x y y + + + + = ⎛ ⎞ + + + + +⎜ ⎟ ⎝ ⎠ = − + + 2 2 2 2 3 4( 2) 4 10 8 3 5 ( 2) 8 2 x y x y ⎛ ⎞ + + + =⎜ ⎟ ⎝ ⎠ ⎛ ⎞ + + + =⎜ ⎟ ⎝ ⎠ 3 5 10 center 2, ; radius 8 2 2 ⎛ ⎞ = − − = =⎜ ⎟ ⎝ ⎠ 23. 2 –1 1 2 –1 = 24. 7 5 2 4 3 − = − 25. –6 – 3 9 –5 – 2 7 = 26. 6 4 1 0 2 − + = − 27. 5 – 0 5 – 0 – 3 3 = 28. 6 0 1 0 6 − = + 29. 2 1( 2) 2 2 4 0 y x y x x y − = − − − = − + + − = 30. 4 1( 3) 4 3 7 0 y x y x x y − = − − − = − + + − = 31. 2 3 2 – 3 0 y x x y = + + = 32. 0 5 0 5 0 y x x y = + + − = © 2007 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.
  • 16 Section 0.3 Instructor’s Resource Manual 33. 8 – 3 5 ; 4 – 2 2 m = = 5 – 3 ( – 2) 2 2 – 6 5 –10 5 – 2 – 4 0 y x y x x y = = = 34. 2 1 1 ; 8 4 4 m − = = − 1 1 ( 4) 4 4 4 4 4 0 0 y x y x x y − = − − = − − + = 35. 3y = –2x + 1; 2 1 – ; 3 3 y x= + 2 slope – ; 3 = 1 -intercept 3 y = 36. 4 5 6 5 3 4 2 y x y x − = − = − + 5 3 slope ; -intercept 4 2 y= − = 37. 6 – 2 10 – 2 –2 10 – 8 –5 4; y x y x y x = = = + slope = –5; y-intercept = 4 38. 4 5 20 5 4 20 4 4 5 x y y x y x + = − = − − = − − 4 slope ; -intercept = 4 5 y= − − 39. a. m = 2; 3 2( – 3) 2 – 9 y x y x + = = b. 1 – ; 2 m = 1 3 – ( – 3) 2 1 3 – – 2 2 y x y x + = = c. 2 3 6 3 –2 6 2 – 2; 3 x y y x y x + = = + = + 2 – ; 3 m = 2 3 – ( – 3) 3 2 – –1 3 y x y x + = = d. 3 ; 2 m = 3 3 ( – 3) 2 3 15 – 2 2 y x y x + = = e. –1– 2 3 – ; 3 1 4 m = = + 3 3 – ( – 3) 4 3 3 – – 4 4 y x y x + = = f. x = 3 g. y = –3 40. a. 3 5 3(3) (1) 5 4 x cy c c + = + = = − b. 0c = c. 2 1 2 1 x y y x + = − = − − 2;m = − 3 5 3 5 3 5 x cy cy x y x c c + = = − + = − + 3 2 3 2 c c − = − = d. c must be the same as the coefficient of x, so 3.c = © 2007 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.
  • Instructor’s Resource Manual Section 0.3 17 e. 2 3( 3);y x− = + 1 perpendicular slope ; 3 = − 1 3 3 9 c c − = − = 41. 3 ; 2 m = 3 1 ( 2) 2 3 2 2 y x y x + = + = + 42. a. 2;m = 3 10 3 10 10 3 3 kx y y kx k y x − = − = − + = − 2; 6 3 k k= = b. 1 ; 2 m = − 1 3 2 3 2 k k = − = − c. 2 3 6 3 2 6 2 2; 3 x y y x y x + = = − + = − + 3 3 9 ; ; 2 3 2 2 k m k= = = 43. y = 3(3) – 1 = 8; (3, 9) is above the line. 44. 0 ( ,0),(0, ); 0 ; ; 1 b b a b m a a b bx x y y x b y b a a a b − = = − − = − + + = + = 45. 2 3 4 –3 5 x y x y + = + = 2 3 4 9 – 3 –15 11 –11 –1 x y x y x x + = = = = –3(–1) 5 2 y y + = = Point of intersection: (–1, 2) 3 –2 4 2 4 – 3 3 y x y x = + = + 3 2 m = 3 2 ( 1) 2 3 7 2 2 y x y x − = + = + 46. 4 5 8 2 10 x y x y − = + = − 4 5 8 4 2 20 7 28 4 x y x y y y − = − − = − = = − 4 5( 4) 8 4 12 3 x x x − − = = − = − Point of intersection: ( )3, 4 ;− − 4 5 8 5 4 8 4 8 5 5 x y y x y x − = − = − + = − 5 4 m = − 5 4 ( 3) 4 5 31 4 4 y x y x + = − + = − − © 2007 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.
  • 18 Section 0.3 Instructor’s Resource Manual 47. 3 – 4 5 2 3 9 x y x y = + = 9 –12 15 8 12 36 17 51 3 x y x y x x = + = = = 3(3) – 4 5 –4 –4 1 y y y = = = Point of intersection: (3, 1); 3x – 4y = 5; –4 –3 5 3 5 – 4 4 y x y x = + = 4 – 3 m = 4 –1 – ( – 3) 3 4 – 5 3 y x y x = = + 48. 5 – 2 5 2 3 6 x y x y = + = 15 – 6 15 4 6 12 19 27 27 19 x y x y x x = + = = = 27 2 3 6 19 60 3 19 20 19 y y y ⎛ ⎞ + =⎜ ⎟ ⎝ ⎠ = = Point of intersection: 27 20 , ; 19 19 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 5 2 5 –2 –5 5 5 5 – 2 2 x y y x y x − = = + = 2 – 5 m = 20 2 27 – – 19 5 19 2 54 20 – 5 95 19 2 154 5 95 y x y x y x ⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ = + + = − + 49. 2 6 –1 3 center: , (4,1) 2 2 + +⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 2 6 3 3 midpoint , (4, 3) 2 2 + +⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ 2 2 inscribed circle: radius (4 – 4) (1– 3) 4 2 = + = = 2 2 ( – 4) ( –1) 4x y+ = circumscribed circle: 2 2 radius (4 – 2) (1– 3) 8= + = 2 2 ( – 4) ( –1) 8x y+ = 50. The radius of each circle is 16 4.= The centers are( ) ( )1, 2 and 9,10 .− − The length of the belt is the sum of half the circumference of the first circle, half the circumference of the second circle, and twice the distance between their centers. 2 21 1 2 (4) 2 (4) 2 (1 9) ( 2 10) 2 2 8 2 100 144 56.37 L π π π = ⋅ + ⋅ + + + − − = + + ≈ 51. Put the vertex of the right angle at the origin with the other vertices at (a, 0) and (0, b). The midpoint of the hypotenuse is , . 2 2 a b⎛ ⎞ ⎜ ⎟ ⎝ ⎠ The distances from the vertices are 2 2 2 2 2 2 – 0 – 2 2 4 4 1 , 2 a b a b a a b ⎛ ⎞ ⎛ ⎞ + = +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = + 2 2 2 2 2 2 0 – – 2 2 4 4 1 , and 2 a b a b b a b ⎛ ⎞ ⎛ ⎞ + = +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = + 2 2 2 2 2 2 0 – 0 – 2 2 4 4 1 , 2 a b a b a b ⎛ ⎞ ⎛ ⎞ + = +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = + which are all the same. 52. From Problem 51, the midpoint of the hypotenuse, ( )4,3, , is equidistant from the vertices. This is the center of the circle. The radius is 16 9 5.+ = The equation of the circle is 2 2 ( 4) ( 3) 25.x y− + − = © 2007 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.
  • Instructor’s Resource Manual Section 0.3 19 53. 2 2 – 4 – 2 –11 0x y x y+ = 2 2 2 2 2 2 2 2 2 2 ( – 4 4) ( – 2 1) 11 4 1 ( – 2) ( –1) 16 20 –12 72 0 ( 20 100) ( –12 36) –72 100 36 ( 10) ( – 6) 64 x x y y x y x y x y x x y y x y + + + = + + + = + + + = + + + + = + + + + = center of first circle: (2, 1) center of second circle: (–10, 6) 2 2 (2 10) (1– 6) 144 25 169 13 d = + + = + = = However, the radii only sum to 4 + 8 = 12, so the circles must not intersect if the distance between their centers is 13. 54. 2 2 2 2 2 2 2 2 0 4 4 4 4 x ax y by c a b x ax y by a b c + + + + = ⎛ ⎞ ⎛ ⎞ + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = − + + 2 2 2 2 2 2 2 2 4 2 2 4 4 0 4 4 a b a b c x y a b c a b c + −⎛ ⎞ ⎛ ⎞ + + + =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ + − > ⇒ + > 55. Label the points C, P, Q, and R as shown in the figure below. Let d OP= , h OR= , and a PR= . Triangles OPRΔ and CQRΔ are similar because each contains a right angle and they share angle QRC∠ . For an angle of 30 , 3 2 d h = and 1 2 2 a h a h = ⇒ = . Using a property of similar triangles, / 3 / 2QC RC = , 2 3 4 2 2 2 3 a a = → = + − By the Pythagorean Theorem, we have 2 2 3 2 3 4 7.464d h a a= − = = + ≈ 56. The equations of the two circles are 2 2 2 2 2 2 ( ) ( ) ( ) ( ) x R y R R x r y r r − + − = − + − = Let ( ),a a denote the point where the two circles touch. This point must satisfy 2 2 2 2 2 ( ) ( ) ( ) 2 2 1 2 a R a R R R a R a R − + − = − = ⎛ ⎞ = ±⎜ ⎟⎜ ⎟ ⎝ ⎠ Since a R< , 2 1 . 2 a R ⎛ ⎞ = −⎜ ⎟⎜ ⎟ ⎝ ⎠ At the same time, the point where the two circles touch must satisfy 2 2 2 ( ) ( ) 2 1 2 a r a r r a r − + − = ⎛ ⎞ = ±⎜ ⎟⎜ ⎟ ⎝ ⎠ Since ,a r> 2 1 . 2 a r ⎛ ⎞ = +⎜ ⎟⎜ ⎟ ⎝ ⎠ Equating the two expressions for a yields 2 2 2 1 1 2 2 22 11 22 2 2 2 1 1 1 2 2 2 R r r R R ⎛ ⎞ ⎛ ⎞ − = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ −⎜ ⎟− ⎜ ⎟ ⎝ ⎠= = ⎛ ⎞⎛ ⎞ + + −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 1 1 2 2 1 1 2 r R − + = − (3 2 2) 0.1716r R R= − ≈ © 2007 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.
  • 20 Section 0.3 Instructor’s Resource Manual 57. Refer to figure 15 in the text. Given ine 1l with slope m, draw ABC with vertical and horizontal sides m, 1. Line 2l is obtained from 1l by rotating it around the point A by 90° counter-clockwise. Triangle ABC is rotated into triangle AED . We read off 2 1 1 slope of l m m = = − − . 58. 2 2 2 2 2 ( 1) ( 1) ( 3) ( 4)x y x y− + − = − + − 2 2 2 2 4( 2 1 2 1) 6 9 8 16 x x y y x x y y − + + − + = − + + − + 2 2 2 2 2 2 2 2 2 2 3 2 3 9 16 4 4; 2 17 3 2 3 17; ; 3 3 2 1 17 1 3 9 3 9 1 52 3 9 x x y x x y x x y x x y x y − + = + − − − + = − + = ⎛ ⎞ − + + = +⎜ ⎟ ⎝ ⎠ ⎛ ⎞ − + =⎜ ⎟ ⎝ ⎠ 1 52 center: ,0 ; radius: 3 3 ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 59. Let a, b, and c be the lengths of the sides of the right triangle, with c the length of the hypotenuse. Then the Pythagorean Theorem says that 2 2 2 a b c+ = Thus, 2 2 2 8 8 8 a b cπ π π + = or 2 2 2 1 1 1 2 2 2 2 2 2 a b c⎛ ⎞ ⎛ ⎞ ⎛ ⎞ π + π = π⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 1 2 2 x⎛ ⎞ π⎜ ⎟ ⎝ ⎠ is the area of a semicircle with diameter x, so the circles on the legs of the triangle have total area equal to the area of the semicircle on the hypotenuse. From 2 2 2 ,a b c+ = 2 2 23 3 3 4 4 4 a b c+ = 23 4 x is the area of an equilateral triangle with sides of length x, so the equilateral triangles on the legs of the right triangle have total area equal to the area of the equilateral triangle on the hypotenuse of the right triangle. 60. See the figure below. The angle at T is a right angle, so the Pythagorean Theorem gives 2 2 2 2 2 2 2 2 ( ) ( ) ( ) 2 ( ) ( 2 ) ( ) PM r PT r PM rPM r PT r PM PM r PT + = + ⇔ + + = + ⇔ + = 2 2 so this gives ( )( ) ( )PM r PN PM PN PT+ = = 61. The lengths A, B, and C are the same as the corresponding distances between the centers of the circles: 2 2 2 2(–2) (8) 68 8.2 2 2(6) (8) 100 10 (8) (0) 64 8 A B C = + = ≈ = + = = = + = = Each circle has radius 2, so the part of the belt around the wheels is 2(2 a ) + 2(2 b ) + 2(2 c )π π π π π π− − − − − − 2[3 -( )] 2(2 ) 4a b cπ π π= + + = = Since a + b + c = π , the sum of the angles of a triangle. The length of the belt is 8.2 10 8 4 38.8 units. ≈ + + + π ≈ 62 As in Problems 50 and 61, the curved portions of the belt have total length 2 .rπ The lengths of the straight portions will be the same as the lengths of the sides. The belt will have length 1 22 .nr d d dπ + + + +… © 2007 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.
  • Instructor’s Resource Manual Section 0.3 21 63. A = 3, B = 4, C = –6 2 2 3(–3) 4(2) (–6) 7 5(3) (4) d + + = = + 64. 2, 2, 4A B C= = − = 2 2 2(4) 2( 1) 4) 14 7 2 28(2) (2) d − − + = = = + 65. A = 12, B = –5, C = 1 2 2 12(–2) – 5(–1) 1 18 13(12) (–5) d + = = + 66. 2 2 2, 1, 5 2(3) 1( 1) 5 2 2 5 55(2) ( 1) A B C d = = − = − − − − = = = + − 67. 2 4(0) 5 5 2 x x + = = ( )5 2 2 2 2 4(0) – 7 2 5 520(2) (4) d + = = = + 68. 7(0) 5 1 1 5 y y − = − = 2 2 1 7(0) 5 6 5 7 7 74 7474(7) ( 5) d ⎛ ⎞ − −⎜ ⎟ ⎝ ⎠ = = = + − 69. 2 3 5 ; 1 2 3 m − − = = − + 3 ; 5 m = passes through 2 1 3 2 1 1 , , 2 2 2 2 − + −⎛ ⎞ ⎛ ⎞ = −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 1 3 1 2 5 2 3 4 5 5 y x y x ⎛ ⎞ − = +⎜ ⎟ ⎝ ⎠ = + 70. 0 – 4 1 –2; ; 2 – 0 2 m m= = = passes through 0 2 4 0 , (1, 2) 2 2 + +⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 1 – 2 ( –1) 2 1 3 2 2 y x y x = = + 6 – 0 1 3; – ; 4 – 2 3 m m= = = passes through 2 4 0 6 , (3, 3) 2 2 + +⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 1 – 3 – ( – 3) 3 1 – 4 3 1 3 1 – 4 2 2 3 5 5 6 2 3 1 3 (3) 3 2 2 y x y x x x x x y = = + + = + = = = + = center = (3, 3) 71. Let the origin be at the vertex as shown in the figure below. The center of the circle is then ( )4 ,r r− , so it has equation 2 2 2 ( (4 )) ( ) .x r y r r− − + − = Along the side of length 5, the y-coordinate is always 3 4 times the x-coordinate. Thus, we need to find the value of r for which there is exactly one x- solution to 2 2 23 ( 4 ) . 4 x r x r r ⎛ ⎞ − + + − =⎜ ⎟ ⎝ ⎠ Solving for x in this equation gives ( )216 16 24 7 6 . 25 x r r r ⎛ ⎞ = − ± − + −⎜ ⎟ ⎝ ⎠ There is exactly one solution when 2 7 6 0,r r− + − = that is, when 1r = or 6r = . The root 6r = is extraneous. Thus, the largest circle that can be inscribed in this triangle has radius 1.r = © 2007 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.
  • 22 Section 0.4 Instructor’s Resource Manual 72. The line tangent to the circle at ( ),a b will be perpendicular to the line through ( ),a b and the center of the circle, which is ( )0,0 . The line through ( ),a b and ( )0,0 has slope 2 20 ; 0 b b a r m ax by r y x a a b b − = = + = ⇒ = − + − so 2 ax by r+ = has slope a b − and is perpendicular to the line through ( ),a b and ( )0,0 , so it is tangent to the circle at ( ), .a b 73. 12a + 0b = 36 a = 3 2 2 3 36b+ = 3 3b = ± 3 – 3 3 36 – 3 12 x y x y = = 3 3 3 36 3 12 x y x y + = + = 74. Use the formula given for problems 63-66, for ( ) ( ), 0,0x y = . 2 2 2 , 1, ;(0,0) (0) 1(0) ( 1) 1 A m B C B b m B b B b d m m = = − = − − + − − = = + − + 75. The midpoint of the side from (0, 0) to (a, 0) is 0 0 0 , , 0 2 2 2 a a+ +⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ The midpoint of the side from (0, 0) to (b, c) is 0 0 , , 2 2 2 2 b c b c+ +⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 1 2 2 1 2 2 2 – 0 – – – 0 ; –– c b a c c m b a b a c m m m b a = = = = = 76. See the figure below. The midpoints of the sides are 1 2 1 2, , 2 2 x x y y P + +⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 2 3 2 3 , , 2 2 x x y y Q + +⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 3 4 3 4 , , 2 2 x x y y R + +⎛ ⎞ ⎜ ⎟ ⎝ ⎠ and 1 4 1 4 , . 2 2 x x y y S + +⎛ ⎞ ⎜ ⎟ ⎝ ⎠ The slope of PS is [ ] [ ] 1 4 1 2 4 2 4 2 1 4 1 2 1 ( ) 2 . 1 ( ) 2 y y y y y y x xx x x x + − + − = −+ − + The slope of QR is [ ] [ ] 3 4 2 3 4 2 4 2 3 4 2 3 1 ( ) 2 . 1 ( ) 2 y y y y y y x xx x x x + − + − = −+ − + Thus PS and QR are parallel. The slopes of SR and PQ are both 3 1 3 1 , y y x x − − so PQRS is a parallelogram. 77. 2 2 ( – 6) 25;x y+ = passes through (3, 2) tangent line: 3x – 4y = 1 The dirt hits the wall at y = 8. 0.4 Concepts Review 1. y-axis 3. 8; –2, 1, 4 2. ( )4, 2− 4. line; parabola Problem Set 0.4 1. y = –x2 + 1; y-intercept = 1; y = (1 + x)(1 – x); x-intercepts = –1, 1 Symmetric with respect to the y-axis © 2007 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.
  • Instructor’s Resource Manual Section 0.4 23 2. 2 1;x y= − + -intercepts 1,1;y = − -intercept 1x = . Symmetric with respect to the x-axis. 3. x = –4y2 – 1; x-intercept = –1 Symmetric with respect to the x-axis 4. 2 4 1;y x= − -intercept 1y = − 1 1 (2 1)(2 1); -intercepts , 2 2 y x x x= + − = − Symmetric with respect to the y-axis. 5. x2 + y = 0; y = –x2 x-intercept = 0, y-intercept = 0 Symmetric with respect to the y-axis 6. 2 2 ; -intercept 0y x x y= − = (2 ); -intercepts 0,2y x x x= − = 7. 7x2 + 3y = 0; 3y = –7x2; 27 – 3 y x= x-intercept = 0, y-intercept = 0 Symmetric with respect to the y-axis 8. 2 3 2 2;y x x= − + -intercept 2y = © 2007 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.
  • 24 Section 0.4 Instructor’s Resource Manual 9. x2 + y2 = 4 x-intercepts = -2, 2; y-intercepts = -2, 2 Symmetric with respect to the x-axis, y-axis, and origin 10. 2 2 3 4 12;x y+ = y-intercepts 3, 3= − -intercepts 2,2x = − Symmetric with respect to the x-axis, y-axis, and origin 11. y = –x2 – 2x + 2: y-intercept = 2 2 4 8 2 2 3 -intercepts –1 3 –2 –2 x ± + ± = = = ± 12. 2 2 4 3 12;x y+ = -intercepts 2,2y = − x-intercepts 3, 3= − Symmetric with respect to the x-axis, y-axis, and origin 13. x2 – y2 = 4 x-intercept = -2, 2 Symmetric with respect to the x-axis, y-axis, and origin 14. 2 2 ( 1) 9; -intercepts 2,4x y y+ − = = − x-intercepts = 2 2,2 2− Symmetric with respect to the y-axis © 2007 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.
  • Instructor’s Resource Manual Section 0.4 25 15. 4(x – 1)2 + y2 = 36; y-intercepts 32 4 2= ± = ± x-intercepts = –2, 4 Symmetric with respect to the x-axis 16. 2 2 4 3 2x x y− + = − x-intercepts = 2 2± Symmetric with respect to the x-axis 17. x2 + 9(y + 2)2 = 36; y-intercepts = –4, 0 x-intercept = 0 Symmetric with respect to the y-axis 18. 4 4 1; -intercepts 1,1x y y+ = = − x-intercepts 1,1= − Symmetric with respect to the x-axis, y-axis, and origin 19. x4 + y4 = 16; y-intercepts 2,2= − x-intercepts 2,2= − Symmetric with respect to the y-axis, x-axis and origin 20. y = x3 – x; y-intercepts = 0; y = x(x2 – 1) = x(x + 1)(x – 1); x-intercepts = –1, 0, 1 Symmetric with respect to the origin © 2007 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.
  • 26 Section 0.4 Instructor’s Resource Manual 21. 2 1 ; 1 y x = + y-intercept = 1 Symmetric with respect to the y-axis 22. 2 ; -intercept 0 1 x y y x = = + x-intercept = 0 Symmetric with respect to the origin 23. 2 2 – 2 2 – 2 2 2 4 3 12 –2 2( 2 1) 3( 4 4) –2 2 12 2( –1) 3( 2) 12 x x y y x x y y x y + + = + + + + = + + + + = y-intercepts = 30 –2 3 ± x-intercept = 1 24. ( )2 2 4 5 9( 2) 36; -intercept 5x y x− + + = = 25. y = (x – 1)(x – 2)(x – 3); y-intercept = –6 x-intercepts = 1, 2, 3 26. y = x2(x – 1)(x – 2); y-intercept = 0 x-intercepts = 0, 1, 2 27. 2 2 ( 1) ;y x x= − y-intercept = 0 x-intercepts = 0, 1 © 2007 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.
  • Instructor’s Resource Manual Section 0.4 27 28. 4 4 4 ( 1) ( 1) ; -intercept 0y x x x y= − + = x-intercepts 1,0,1= − Symmetric with respect to the y-axis 29. 1;x y+ = y-intercepts = –1, 1; x-intercepts = –1, 1 Symmetric with respect to the x-axis, y-axis and origin 30. 4;x y+ = y-intercepts = –4, 4; x-intercepts = –4, 4 Symmetric with respect to the x-axis, y-axis and origin 31. 2 2 2 1 ( 1) 1 2 1 3 0 ( 3) 0 0, 3 x x x x x x x x x x − + = + − + = + + + = + = = − Intersection points: (0, 1) and (–3, 4) 32. 2 2 2 2 3 ( 1) 2 3 2 1 4 0 x x x x x x + = − − + = − + − + = No points of intersection 33. 2 2 2 2 3 2( 4) 2 3 2 16 32 2 18 35 0 x x x x x x x− − + = − − − + = − + − + = 18 324 – 280 18 2 11 9 11 ; 4 4 2 x ± ± ± = = = Intersection points: 9 – 11 , – 6 11 , 2 ⎛ ⎞ +⎜ ⎟⎜ ⎟ ⎝ ⎠ 9 11 , – 6 – 11 2 ⎛ ⎞+ ⎜ ⎟⎜ ⎟ ⎝ ⎠ © 2007 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.
  • 28 Section 0.4 Instructor’s Resource Manual 34. 2 2 2 3 3 3 12 3 9 0 x x x x x − + = − + − + = No points of intersection 35. 2 2 2 4 2 2 x x x x + = = = ± Intersection points: ( ) ( )– 2, – 2 , 2, 2 36. 2 2 2 2 2 2 3( 1) 12 2 3 6 3 12 5 6 9 0 6 36 180 6 6 6 3 3 6 10 10 5 x x x x x x x x + − = + − + = − − = ± + ± ± = = = Intersection points: 3 3 6 2 3 6 3 3 6 2 3 6 , , , 5 5 5 5 ⎛ ⎞ ⎛ ⎞− − − + − + ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 37. 2 2 2 2 2 2 3 1 2 (3 1) 15 2 9 6 1 15 10 8 14 0 2(5 4 7) 0 y x x x x x x x x x x x x = + + + + = + + + + = + − = + − = 2 39 1.65, 0.85 5 x − ± = ≈ − Intersection points: 2 39 1 3 39 , 5 5 ⎛ ⎞− − − − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ and 2 39 1 3 39 , 5 5 ⎛ ⎞− + − + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ [ or roughly (–1.65, –3.95) and (0.85, 3.55) ] 38. 2 2 2 2 2 (4 3) 81 16 24 9 81 17 24 72 0 x x x x x x x + + = + + + = + − = 12 38 2.88, 1.47 17 x − ± = ≈ − Intersection points: 12 38 3 24 38 , 17 17 ⎛ ⎞− − − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ and 12 38 3 24 38 , 17 17 ⎛ ⎞− + + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ [ or roughly ( ) ( )2.88, 8.52 , 1.47,8.88− − ] © 2007 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.
  • Instructor’s Resource Manual Section 0.5 29 39. a. 2 ; (2)y x= b. 3 2 , with 0 : (1)ax bx cx d a+ + + > c. 3 2 , with 0 : (3)ax bx cx d a+ + + < d. 3 , with 0 : (4)y ax a= > 40. 2 2 13;( 2, 3),( 2,3),(2, 3),(2,3)x y+ = − − − − 2 2 1 2 2 2 2 2 3 (2 2) ( 3 3) 4 (2 2) ( 3 3) 52 2 13 (2 2) (3 3) 6 d d d = + + − + = = + + − − = = = − + + = Three such distances. 41. x2 + 2x + y2 – 2y = 20; ( )–2,1 21 ,+ ( ) ( ) ( )–2,1– 21 , 2,1 13 , 2,1– 13+ ( ) ( ) 22 1 2 (–2 – 2) 1 21 – 1 13 16 21 – 13 50 – 2 273 4.12 d ⎡ ⎤= + + + ⎣ ⎦ = + = ≈ ( ) ( ) 22 2 2 (–2 – 2) 1 21 – 1– 13 16 21 13 50 2 273 9.11 d ⎡ ⎤= + + ⎣ ⎦ = + + = + ≈ ( ) ( ) ( ) 2 2 3 2 2 ( 2 2) 1 21 1 21 0 21 21 2 21 2 21 9.17 d ⎡ ⎤= − + + + − − ⎣ ⎦ = + + = = ≈ ( ) 22 4 2 ( 2 2) 1 21 (1 13) 16 21 13 50 2 273 9.11 d ⎡ ⎤= − − + − − +⎣ ⎦ = + − − = + ≈ ( ) ( ) 2 2 5 2 ( 2 2) 1 21 1 13 16 13 21 50 2 273 4.12 d ⎡ ⎤= − − + − − − ⎣ ⎦ = + − = − ≈ ( ) ( ) ( ) 2 2 6 2 2 (2 2) 1 13 1 13 0 13 13 2 13 2 13 7.21 d ⎡ ⎤= − + + − − ⎣ ⎦ = + + = = ≈ Four such distances ( 2 4 1 5andd d d d= = ). 0.5 Concepts Review 1. domain; range 2. 2 2 2 (2 ) 3(2 ) 12 ; ( ) 3( )f u u u f x h x h= = + = + 3. asymptote 4. even; odd; y-axis; origin Problem Set 0.5 1. a. 2 (1) 1–1 0f = = b. 2 (–2) 1– (–2) –3f = = c. 2 (0) 1– 0 1f = = d. 2 ( ) 1–f k k= e. 2 (–5) 1– (–5) –24f = = f. 2 1 1 1 15 1– 1– 4 4 16 16 f ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ g. ( ) ( ) 22 2111 hhhhf −−=+−=+ h. ( ) ( ) 2 2 1 1 2 0 2f h f h h h h+ − = − − − = − − i. ( ) ( ) ( )2 2 2 2 1 2 3 4 f h f h h h + − = − + + = − − 2. a. 3 (1) 1 3 1 4F = + ⋅ = b. 3 ( 2) ( 2) 3( 2) 2 2 3 2F = + = + 5 2= c. 3 1 1 1 1 3 49 3 4 4 4 64 4 64 F ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = + = + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ © 2007 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.
  • 30 Section 0.5 Instructor’s Resource Manual d. ( ) ( ) ( ) 3 2 3 2 3 1 1 3 1 1 3 3 3 3 4 6 3 F h h h h h h h h h h + = + + + = + + + + + = + + + e. ( ) 2 3 1 1 3 6 3F h h h h+ − = + + + f. ( ) ( ) ( ) ( ) ( )3 3 2 3 2 3 2 2 2 3 2 2 3 2 8 12 6 6 3 14 15 6 F h F h h h h h h h h h + − ⎡ ⎤= + + + − − ⎣ ⎦ = + + + + + − = + + 3. a. 1 (0) –1 0 –1 G = = b. 1 (0.999) –1000 0.999 –1 G = = c. 1 (1.01) 100 1.01–1 G = = d. 2 2 1 ( ) –1 G y y = e. 1 1 (– ) – – –1 1 G x x x = = + f. 2 2 1 2 2 1 1 –1 1– x x G x x ⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ 4. a. 2 1 1 (1) 2 1 + Φ = = b. 2 2 – (– ) – (– ) – – t t t t t t t + Φ = = c. ( ) 2 1 1 3 2 2 4 1 1 2 2 1 1.06 2 +⎛ ⎞ Φ = = ≈⎜ ⎟ ⎝ ⎠ d. 2 2 ( 1) ( 1) 3 2 ( 1) 1 1 u u u u u u u + + + + + Φ + = = + + e. 2 2 2 2 4 2 2 ( ) ( ) ( ) x x x x x xx + + Φ = = f. 2 2 2 2 2 ( ) ( ) ( ) x x x x x x x x + + + Φ + = + 4 3 2 2 2 2x x x x x x + + + = + 5. a. 75.2 1 325.0 1 )25.0( − = − =f is not defined b. 1 ( ) 2.658 3 f x π = ≈ − c. 0.25 1 1 (3 2) 3 2 3 2 2 0.841 f − + = = + − = ≈ 6. a. f(0.79) 3–79.0 9)79.0( 2 + = 3.293≈ − b. f(12.26) 2 (12.26) 9 12.26 – 3 + = 1.199≈ c. 2 ( 3) 9 ( 3) ; 3 – 3 f + = undefined 7. a. x2 + y2 = 1 y2 = 1– x2 2 1– ;y x= ± not a function b. xy + y + x = 1 y(x + 1) = 1 – x 1– 1– ; ( ) 1 1 x x y f x x x = = + + c. 2 1x y= + x2 = 2y + 1 2 2 –1 –1 ; ( ) 2 2 x x y f x= = d. x = y y +1 xy + x = y x = y – xy x = y(1 – x) ; ( ) 1– 1– x x y f x x x = = © 2007 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.
  • Instructor’s Resource Manual Section 0.5 31 8. The graphs on the left are not graphs of functions, the graphs on the right are graphs of functions. 9. 2 2 ( ) – ( ) [2( ) –1] – (2 –1)f a h f a a h a h h + + = 2 4 2 4 2 ah h a h h + = = + 10. 3 3 ( ) – ( ) 4( ) – 4F a h F a a h a h h + + = 3 2 2 3 3 4 12 12 4 – 4a a h ah h a h + + + = 2 2 3 12 12 4a h ah h h + + = 2 2 12 12 4a ah h= + + 11. 3 3 –2 –2 –( ) – ( ) x h xg x h g x h h ++ = 2 2 3 6 3 3 6 4 2 4 3 ( 4 2 4) x x h x x hx h h h h x x hx h − − − + − + − += − = − + − + 2 3 – – 4 – 2 4x x hx h = + + 12. 4 4 –( ) – ( ) a h a a h aG a h G a h h + + + ++ = 2 2 2 2 4 4 4 8 4 16 4 ( 8 4 16) a a ah h a ah a a a ah h h h h a a ah h + + + − − − + + + += = + + + + 2 4 8 4 16a a ah h = + + + + 13. a. ( ) 2 3F z z= + 2z + 3 ≥ 0; z ≥ – 3 2 Domain: 3 : 2 z z ⎧ ⎫ ∈ ≥ −⎨ ⎬ ⎩ ⎭ b. 1 ( ) 4 –1 g v v = 4v – 1 = 0; v = 1 4 Domain: 1 : 4 v v ⎧ ⎫ ∈ ≠⎨ ⎬ ⎩ ⎭ c. 2 ( ) – 9x xψ = x2 – 9 ≥ 0; x2 ≥ 9; x ≥3 Domain: { : 3}x x∈ ≥ d. 4 ( ) – 625 –H y y= 4 4 625 – 0; 625 ; 5y y y≥ ≥ ≤ Domain: { : 5}y y∈ ≤ 14. a. 2 2 2 4 – 4 – ( ) ( – 3)( 2)– – 6 x x f x x xx x = = + Domain: { : 2, 3}x x∈ ≠ − b. –1 ( ) ( 1)G y y= + 1 0; –1 1 y y ≥ > + Domain: { : 1}y y∈ > − c. ( ) 2 3u uφ = + Domain: (all real numbers) d. 2/3 ( ) – 4F t t= Domain: (all real numbers) 15. f(x) = –4; f(–x) = –4; even function 16. f(x) = 3x; f(–x) = –3x; odd function © 2007 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.
  • 32 Section 0.5 Instructor’s Resource Manual 17. F(x) = 2x + 1; F(–x) = –2x + 1; neither 18. ( ) 3 – 2; (– ) –3 – 2;F x x F x x= = neither 19. 2 2 ( ) 3 2 –1; (– ) 3 – 2 –1g x x x g x x x= + = ; neither 20. 3 3 ( ) ; (– ) – ; 8 8 u u g u g u= = odd function 21. 2 2 – ( ) ; (– ) ; –1 –1 x x g x g x x x = = odd 22. 2 1 –2 1 ( ) ; (– ) ; –1 – –1 z z z z z z φ φ + + = = neither © 2007 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.
  • Instructor’s Resource Manual Section 0.5 33 23. ( ) –1; (– ) – –1;f w w f w w= = neither 24. 2 2 ( ) 4; (– ) 4;h x x h x x= + = + even function 25. ( ) 2 ; (– ) –2 2 ;f x x f x x x= = = even function 26. ( ) – 3 ; (– ) – – 3 ;F t t F t t= + = + neither 27. ( ) ; ( ) ; 2 2 x x g x g x= − = − neither 28. ( ) 2 1 ; ( ) 2 1 ;G x x G x x= − − = − + neither 29. 2 1 if 0 ( ) 1 if 0 2 –1 if 2 t g t t t t t ⎧ ≤ ⎪ = + < <⎨ ⎪ ≥⎩ neither © 2007 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.
  • 34 Section 0.5 Instructor’s Resource Manual 30. 2 – 4 if 1 ( ) 3 if 1 x x h x x x ⎧ + ≤⎪ = ⎨ >⎪⎩ neither 31. T(x) = 5000 + 805x Domain: { integers: 0 100}x x∈ ≤ ≤ ( ) 5000 ( ) 805 T x u x x x = = + Domain: { integers: 0 100}x x∈ < ≤ 32. a. ( ) 6 – (400 5 ( – 4))P x x x x= + 6 – 400 – 5 ( – 4)x x x= b. P(200) 190≈ − ; ( )1000 610P ≈ c. ABC breaks even when P(x) = 0; 6 – 400 – 5 ( – 4) 0; 390x x x x= ≈ 33. 2 ( ) –E x x x= 1 x y −0.5 0.5 0.5 1 2 exceeds its square by the maximum amount. 34. Each side has length . 3 p The height of the triangle is 3 . 6 p 2 1 3 3 ( ) 2 3 6 36 p p p A p ⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ 35. Let y denote the length of the other leg. Then ( ) 2 2 2 2 2 2 2 2 2 2 x y h y h x y h x L x h x + = = − = − = − 36. The area is ( ) 2 21 1 base height 2 2 A x x h x= × = − 37. a. E(x) = 24 + 0.40x b. 120 = 24 + 0.40x 0.40x = 96; x = 240 mi 38. The volume of the cylinder is 2 ,r hπ where h is the height of the cylinder. From the figure, r 2 + h 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 = (2r) 2 ; h2 4 = 3r 2 ; 2 12 2 3.h r r= = 2 3 ( ) (2 3) 2 3V r r r r= π = π 39. The area of the two semicircular ends is 2 . 4 dπ The length of each parallel side is 1– . 2 dπ 2 2 2 1– – ( ) 4 2 4 2 d d d d d A d d π π π π⎛ ⎞ = + = +⎜ ⎟ ⎝ ⎠ 2 2 – 4 d dπ = Since the track is one mile long, dπ < 1, so 1 .d < π Domain: 1 :0d d ⎧ ⎫ ∈ < <⎨ ⎬ π⎩ ⎭ © 2007 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.
  • Instructor’s Resource Manual Section 0.5 35 40. a. 1 3 (1) 1(1) (1)(2 1) 2 2 A = + − = b. 1 (2) 2(1) (2)(3 1) 4 2 A = + − = c. 0)0( =A d. 21 1 ( ) (1) ( )( 1 1) 2 2 A c c c c c c= + + − = + e. f. Domain: { }: 0c c∈ ≥ Range: { }: 0y y∈ ≥ 41. a. (0) 0B = b. 1 1 1 1 1 (1) 2 2 2 6 12 B B ⎛ ⎞ = = ⋅ =⎜ ⎟ ⎝ ⎠ c. 42. a. f(x + y) = 2(x + y) = 2x + 2y = f(x) + f(y) b. 2 2 2 ( ) ( ) 2f x y x y x xy y+ = + = + + ( ) ( )f x f y≠ + c. f(x + y) = 2(x + y) + 1 = 2x + 2y + 1 ≠ f(x) + f(y) d. f(x + y) = –3(x + y) = –3x – 3y = f(x) + f(y) 43. For any x, x + 0 = x, so f(x) = f(x + 0) = f(x) + f(0), hence f(0) = 0. Let m be the value of f(1). For p in N, 1 1 1 ... 1,p p= ⋅ = + + + so f(p) = f(1 + 1 + ... + 1) = f(1) + f(1) + ... + f(1) = pf(1) = pm. 1 1 1 1 1 ... ,p p p p p ⎛ ⎞ = = + + +⎜ ⎟ ⎝ ⎠ so 1 1 1 (1) ...m f f p p p ⎛ ⎞ = = + + +⎜ ⎟ ⎝ ⎠ 1 1 1 1 ... ,f f f pf p p p p ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = + + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ hence 1 . m f p p ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ Any rational number can be written as p q with p, q in N. 1 1 1 1 ... , p p q q q q q ⎛ ⎞ = = + + +⎜ ⎟ ⎝ ⎠ so 1 1 1 ... p f f q q q q ⎛ ⎞ ⎛ ⎞ = + + +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 1 1 1 ...f f f q q q ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1 m p pf p m q q q ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ © 2007 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.
  • 36 Section 0.5 Instructor’s Resource Manual 44. The player has run 10t feet after t seconds. He reaches first base when t = 9, second base when t = 18, third base when t = 27, and home plate when t = 36. The player is 10t – 90 feet from first base when 9 ≤ t ≤ 18, hence 2 2 90 (10 90)t+ − feet from home plate. The player is 10t – 180 feet from second base when 18 ≤t ≤ 27, thus he is 90 – (10t – 180) = 270 – 10t feet from third base and 2 2 90 (270 10 )t+ − feet from home plate. The player is 10t – 270 feet from third base when 27 36,t≤ ≤ thus he is 90 – (10t – 270) = 360 – 10t feet from home plate. a. 2 2 2 2 10 if 0 9 90 (10 90) if 9 18 90 (270 10 ) if 18 27 360 –10 if 27 36 t t t t s t t t t ≤ ≤⎧ ⎪ ⎪ + − < ≤⎪ = ⎨ ⎪ + − < ≤ ⎪ < ≤⎪⎩ b. 2 2 2 2 180 180 10 if 0 9 or 27 36 90 (10 90) if 9 18 90 (270 10 ) if 18 27 t t t s t t t t ⎧ − − ≤ ≤ ⎪ < ≤⎪ ⎪⎪ = + − < ≤⎨ ⎪ + − < ≤⎪ ⎪ ⎪⎩ 45. a. f(1.38) ≈ 0.2994 f(4.12) ≈ 3.6852 b. x f(x) –4 –4.05 –3 –3.1538 –2 –2.375 –1 –1.8 0 –1.25 1 –0.2 2 1.125 3 2.3846 4 3.55 46. a. f(1.38) ≈ –76.8204 f(4.12) ≈ 6.7508 b. x f(x) –4 –6.1902 –3 0.4118 –2 13.7651 –1 9.9579 0 0 1 –7.3369 2 –17.7388 3 –0.4521 4 4.4378 47. a. Range: {y ∈ R: –22 ≤ y ≤ 13} b. f(x) = 0 when x ≈ –1.1, 1.7, 4.3 f(x) ≥ 0 on [–1.1, 1.7] ∪ [4.3, 5] 48. a. f(x) = g(x) at x ≈ –0.6, 3.0, 4.6 © 2007 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.
  • Instructor’s Resource Manual Section 0.6 37 b. f(x) ≥ g(x) on [-0.6, 3.0] [4.6, 5]∪ c. ( ) – ( )f x g x 3 2 2 – 5 8 – 2 8 1x x x x x= + + + + 3 2 – 7 9 9x x x= + + Largest value (–2) – (–2) 45f g = 49. a. x-intercept: 3x – 4 = 0; x = 4 3 y-intercept: 2 3 0 – 4 2 30 0 – 6 ⋅ = + b. c. 2 – 6 0;x x+ = (x + 3)(x – 2) = 0 Vertical asymptotes at x = –3, x = 2 d. Horizontal asymptote at y = 0 50. a. x-intercepts: 2 4 2 3 3 – 4 0; 3 3 x x= = ± = ± y-intercept: 2 3 b. On )6, 3− −⎡⎣ , g increases from ( ) 13 6 4.3333 3 g − = ≈ to ∞ . On (2,6⎤⎦ , g decreased from ∞ to 26 2.8889 9 ≈ . On ( )3,2− the maximum occurs around 0.1451x = with value 0.6748 . Thus, the range is ( ),0.6748 2.8889,−∞ ∪ ∞⎤ ⎡⎦ ⎣ . c. 2 – 6 0;x x+ = (x + 3)(x – 2) = 0 Vertical asymptotes at x = –3, x = 2 d. Horizontal asymptote at y = 3 0.6 Concepts Review 1. 2 3 ( 1)x + 2. f(g(x)) 3. 2; left 4. a quotient of two polynomial functions Problem Set 0.6 1. a. 2 ( )(2) (2 3) 2 9f g+ = + + = b. 2 ( )(0) (0 3)(0 ) 0f g⋅ = + = c. 2 3 9 3 ( )(3) 3 3 6 2 g f = = = + d. 431)1()1)(( 2 =+== fgf e. 164)31()1)(( 2 ==+= gfg f. 25)5(–)38(–)8)(–( 2 ==+= gfg 2. a. 2 2 2 28 ( – )(2) (2 2) – 6 – 2 3 5 5 f g = + = = + b. 4 211 )1)(( 4 2 31 2 2 == + = + gf c. 9 1 3 1 33 2 )3( 22 2 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + =g © 2007 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.
  • 38 Section 0.6 Instructor’s Resource Manual d. 4 3 2 1 2 1 31 2 )1)(( 2 =+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = fgf e. 5 2 32 2 )11()1)(( 2 = + =+= gfg f. 5 32 3 2 33 2 )3)(( 3 10 3 1 == + =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ggg 3. a. 3 1 ( )( ) 1t t t Φ + Ψ = + + b. 1 1 1 11 ))(( 3 3 +=+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Φ=ΨΦ rrr r c. 1 1 )1())(( 3 3 + =+Ψ=ΦΨ r rr d. Φ3 (z) = (z3 +1)3 e. (Φ – Ψ)(5t) = [(5t)3 +1] – 1 5t = 125t3 +1– 1 5t f. ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ΨΦ=ΨΨΦ t t 1 )–())()–(( = 1 t ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 + 1– 1 1 t = 1 t3 +1 – t 4. a. 2 2 –1 ( )( ) x f g x x ⋅ = Domain: (–∞, –1]∪[1, ∞) b. 44 4 4 2 2 ( ) ( ) –1f x g x x x ⎛ ⎞⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 4 16 ( –1)x x = + Domain: (–∞, 0) ∪(0, ∞) c. 1– 4 1– 22 ))(( 2 2 xxx fxgf =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = Domain: [–2, 0) ∪ (0, 2] d. 1– 2 1–))(( 2 2 x xgxfg =⎟ ⎠ ⎞ ⎜ ⎝ ⎛= Domain: (– ∞ , –1) ∪ (1, ∞ ) 5. ( ) ( ) 2 ( ) 1 1 4f g x f x x= + = + − 2 2 – 3x x= + ( ) 2 2 ( ) 4 1 4g f x g x x⎛ ⎞= − = + −⎜ ⎟ ⎝ ⎠ 2 1 – 4x= + 6. g3 (x) = (x2 +1)3 = (x4 +2x 2 + 1)(x2 +1) = x 6 + 3x 4 +3x 2 + 1 )1)(())(( 2 += xggxggg = g[(x 2 + 1)2 + 1] = g( x4 + 2x2 + 2) = ( x4 + 2x2 + 2)2 + 1 = x8 + 4x6 + 8x 4 +8x2 + 5 7. g(3.141) ≈ 1.188 8. g(2.03) ≈ 0.000205 9. ( ) 1/31/3 22 ( ) ( ) 11 7 11 7 4.789 g gπ π π π⎡ ⎤⎡ ⎤− = − − −⎢ ⎥⎣ ⎦ ⎣ ⎦ ≈ 10. 3 1/ 3 3 1/ 3 [ ( ) – ( )] [(6 –11) – (6 –11)]g gπ π = π π ≈ 7.807 11. a. ( ) , ( ) 7g x x f x x= = + b. g(x) = x15 , f (x) = x2 + x 12. a. 3 2 ( )f x x = , 2 ( ) 1g x x x= + + b. 1 ( )f x x = , g(x) = x3 + 3x 13. hgfp = if f(x) =1/ x , ( ) ,g x x= 2 ( ) 1h x x= + hgfp = if ( ) 1/f x x= , g(x) = x + 1, 2 ( )h x x= 14. lhgfp = if ( ) 1/f x x= , ( ) ,g x x= h(x) = x + 1, l( x) = x 2 © 2007 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.
  • Instructor’s Resource Manual Section 0.6 39 15. Translate the graph of xxg =)( to the right 2 units and down 3 units. 16. Translate the graph of ( )h x x= to the left 3 units and down 4 units. 17. Translate the graph of y = x 2 to the right 2 units and down 4 units. 18. Translate the graph of y = x 3 to the left 1 unit and down 3 units. 19. – 3 ( )( ) 2 x f g x x+ = + 20. ( )( )f g x x x+ = + © 2007 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.
  • 40 Section 0.6 Instructor’s Resource Manual 21. – ( ) t t F t t = 22. ( )G t t t= − 23. a. Even; (f + g)(–x) = f(–x) + g(–x) = f(x) + g(x) = (f + g)(x) if f and g are both even functions. b. Odd; (f + g)(–x) = f(–x) + g(–x) = –f(x) – g(x) = –(f + g)(x) if f and g are both odd functions. c. Even; ( )( ) [ ( )][ ( )] [ ( )][ ( )] ( )( ) f g x f x g x f x g x f g x ⋅ − = − − = = ⋅ if f and g are both even functions. d. Even; )]()][([))(( xgxfxgf −−=−⋅ [ ( )][ ( )] [ ( )][ ( )] ( )( ) f x g x f x g x f g x = − − = = ⋅ if f and g are both odd functions. e. Odd; )]()][([))(( xgxfxgf −−=−⋅ [ ( )][ ( )] [ ( )][ ( )] ( )( ) f x g x f x g x f g x = − = − = − ⋅ if f is an even function and g is an odd function. 24. a. F(x) – F(–x) is odd because F(–x) – F(x) = –[F(x) – F(–x)] b. F(x) + F(–x) is even because F(–x) + F(–(–x)) = F(–x) + F(x) = F(x) + F(–x) c. ( ) – (– ) 2 F x F x is odd and ( ) (– ) 2 F x F x+ is even. ( ) ( ) ( ) ( ) 2 ( ) ( ) 2 2 2 F x F x F x F x F x F x − − + − + = = 25. Not every polynomial of even degree is an even function. For example xxxf += 2 )( is neither even nor odd. Not every polynomial of odd degree is an odd function. For example 23 )( xxxg += is neither even nor odd. 26. a. Neither b. PF c. RF d. PF e. RF f. Neither 27. a. 2 29 – 3(2 ) (2 )P t t= + + + 27t t= + + b. When t = 15, 15 15 27 6.773P = + + ≈ 28. R(t) = (120 + 2t + 3t 2 )(6000 + 700t) = 2100t3 + 19, 400t2 + 96, 000t + 720,000 29. 2 2 400 if 0 1 ( ) (400 ) [300( 1)] if 1 t t D t t t t < <⎧⎪ = ⎨ + − ≥⎪⎩ 2 400 if 0 1 ( ) 250,000 180,000 90,000 if 1 t t D t t t t < <⎧⎪ = ⎨ − + ≥⎪⎩ 30. D(2.5) ≈ 1097 mi © 2007 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.
  • Instructor’s Resource Manual Section 0.6 41 31. ( ) ( ) ac ba acx bax fxff acx bax acx bax –– ))(( – – + + + =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = 2 2 2 2 – ( ) – a x ab bcx ab x a bc x acx bc acx a a bc + + + = = = + + + If a2 + bc = 0 , f(f(x)) is undefined, while if x = a c , f(x) is undefined. 32. –3 1 –3 1 – 3– 3 ( ( ( ))) 1 1 x x x x x f f f x f f f x + + ⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟= =⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠ ⎝ ⎠ – 3 – 3 – 3 –2 – 6 – – 3 – 3 1 2 – 2 –1 x x x x f f f x x x x ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ x x xx xx x x x x == + + = + = 4– 4– 1–3–– 33–3–– 1 3– 1– 3–– 1– 3–– If x = –1, f(x) is undefined, while if x = 1, f(f(x)) is undefined. 33. a. xx f x x –1 1 1– 1 1 1 ==⎟ ⎠ ⎞ ⎜ ⎝ ⎛ b. 1–1– ))(( 1– 1– x x x x x x fxff =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = – 1 x x x x = = + c. xx x x x f xf f x x x x –1– 1– 1– 1– )( 1 1– 1– ==⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 1 – x 34. a. xxx x xf − = − = 1 1/1 /1 )/1( b. /( 1) ( ( )) ( /( 1)) 1 1 ( 1) 1 x x f f x f x x x x x x x x − = − = − − = − + − 35. 1 2 3 1 2 3 1 2 3 ( ( ))( ) (( )( )) ( ( ( ))) f f f x f f f x f f f x = = 1 2 3 1 2 3 1 2 3 1 2 3 (( ) )( ) ( )( ( )) ( ( ( ))) ( ( ))( ) f f f x f f f x f f f x f f f x = = = 36. 1 1 1 2 ( ( )) ; 1 ( ( )) ; f f x x f f x x = = 1 3 1 4 ( ( )) 1 ; 1 ( ( )) ; 1 f f x x f f x x = − = − 1 5 1 6 1 ( ( )) ; ( ( )) ; 1 x f f x x x f f x x − = = − 2 1 2 2 1 1 ( ( )) ; 1 ( ( )) ; x f f x x f f x x = = = 2 3 2 4 1 1 1 ( ( )) ; 1 1 ( ( )) 1 ; x f f x x f f x x − = − = = − ; 1– 1 ))(( 152 x x xff x x == − ; 1–1 ))(( 1 62 x x xff x x == − 3 1 3 2 ( ( )) 1 ; 1 1 ( ( )) 1 ; f f x x x f f x x x = − − = − = 3 3( ( )) 1– (1– ) ;f f x x x= = 3 4 1 ( ( )) 1– ; 1– –1 x f f x x x = = 3 5 –1 1 ( ( )) 1– ; x f f x x x = = 3 6 1 ( ( )) 1– ; –1 1– x f f x x x = = 4 1 4 2 1 1 ( ( )) ; 1 1 ( ( )) ; 11 x f f x x x f f x x = − = = −− 4 3 1 1 ( ( )) ; 1– (1– ) f f x x x = = 4 4 1 1– 1 1 –1 ( ( )) ; 1 11– x x x f f x x x − = = = − − 4 5 –1 1 ( ( )) ; ( 1)1– x x x f f x x x x = = = − − ;–1 1 1 –1 1 ))(( 1– 64 x xx x xff x x = −− − == © 2007 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.
  • 42 Section 0.6 Instructor’s Resource Manual 5 1 1 5 2 1 1 ( ( )) ; 1 ( ( )) 1 ;x x x f f x x f f x x − = − = = − 5 3 1– –1 ( ( )) ; 1– –1 x x f f x x x = = ; 1 )1(11– ))(( –1 1 –1 1 45 x x xff x x = −− == ; –1 1 1 11– ))(( 1– 1– 55 xx xx xff x x x x = − −− == ; 1)1(1– ))(( 1– 1– 65 xx xx xff x x x x = −− == 6 1 1 6 2 1 ( ( )) ; –1 1 ( ( )) ; 1––1 x x x f f x x f f x x = = = 6 3 1– –1 ( ( )) ; 1– –1 x x f f x x x = = ; 1 )1(1 1 1– ))(( –1 1 –1 1 46 xx xff x x = −− == ;–1 1 1 1– ))(( 1– 1– 56 x xx x xff x x x x = −− − == x xx x xff x x x x = −− == )1(1– ))(( 1– 1– 66 f1 f2 f3 f4 f5 f6 f1 f1 f2 f3 f4 f5 f6 f2 f2 f1 f4 f3 f6 f5 f3 f3 f5 f1 f6 f2 f4 f4 f4 f6 f2 f5 f1 f3 f5 f5 f3 f6 f1 f4 f2 f6 f6 f4 f5 f2 f3 f1 a. 33333 fffff ))))(((( 33333 fffff= 1 3 3 3 3 3 3 ((( ) ) ) (( ) ) f f f f f f f = = 331 fff == b. 654321 ffffff )))))((((( 654321 ffffff= ))))(((( 65432 fffff= 4 4 5 6 5 2 3 ( ) ( )f f f f f f f = = = c. If 616 then, fFffF == . d. If ,163 fffG = then 14 ffG = so G = f5. e. If ,552 fHff = then 56 fHf = so H = f3. 37. 38. 39. © 2007 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.
  • Instructor’s Resource Manual Section 0.7 43 40. 41. a. b. c. 42. 0.7 Concepts Review 1. (– ∞ , ∞ ); [–1, 1] 2. 2π ; 2π ; π 3. odd; even 4. 2 2 4 (–4) 3 5; cos – 5 x r r θ= + = = = Problem Set 0.7 1. a. 30 180 6 π π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ b. 45 180 4 π π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ c. –60 – 180 3 π π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ d. 4 240 180 3 π π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ e. 37 –370 – 180 18 π π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ f. 10 180 18 π π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 2. a. 7 6 π 180 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 210° b. 3 4 π 180 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 135° c. – 1 3 π 180 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = –60° d. 4 3 π 180 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 240° e. – 35 18 π 180 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = –350° f. 3 180 30 18 π π ⎛ ⎞ = °⎜ ⎟ ⎝ ⎠ 3. a. 33.3 0.5812 180 π⎛ ⎞ ≈⎜ ⎟ ⎝ ⎠ b. 46 0.8029 180 π⎛ ⎞ ≈⎜ ⎟ ⎝ ⎠ c. –66.6 –1.1624 180 π⎛ ⎞ ≈⎜ ⎟ ⎝ ⎠ d. 240.11 4.1907 180 π⎛ ⎞ ≈⎜ ⎟ ⎝ ⎠ e. –369 –6.4403 180 π⎛ ⎞ ≈⎜ ⎟ ⎝ ⎠ f. 11 0.1920 180 π⎛ ⎞ ≈⎜ ⎟ ⎝ ⎠ © 2007 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.
  • 44 Section 0.7 Instructor’s Resource Manual 4. a. 3.141 180 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≈ 180° b. 6. 28 180 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≈ 359. 8° c. 5. 00 180 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≈ 286.5° d. 0. 001 180 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≈ 0.057° e. –0.1 180 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≈ –5.73° f. 36.0 180 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≈ 2062.6° 5. a. 56. 4 tan34.2° sin34.1° ≈ 68.37 b. 5.34 tan 21.3° sin3.1°+ cot 23.5° ≈ 0.8845 c. tan (0.452) ≈ 0.4855 d. sin (–0.361) ≈ –0.3532 6. a. 234.1sin(1.56) cos(0.34) ≈ 248.3 b. 2 sin (2.51) cos(0.51) 1.2828+ ≈ 7. a. 56. 3tan34.2° sin56.1° ≈ 46.097 b. sin35° sin26° + cos 26° ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 ≈ 0. 0789 8. Referring to Figure 2, it is clear that 00sin = and 10cos = . If the angle is 6/π , then the triangle in the figure below is equilateral. Thus, 1 1 2 2 PQ OP= = . This implies that 1 sin 6 2 π = . By the Pythagorean Identity, 2 2 2 1 3 cos 1 sin 1 6 6 2 4 π π ⎛ ⎞ = − = − =⎜ ⎟ ⎝ ⎠ . Thus 3 cos 6 2 π = . The results 2 sin cos 4 4 2 π π = = were derived in the text. If the angle is /3π then the triangle in the figure below is equilateral. Thus 1 cos 3 2 π = and by the Pythagorean Identity, 3 sin 3 2 π = . Referring to Figure 2, it is clear that sin 1 2 π = and cos 0 2 π = . The rest of the values are obtained using the same kind of reasoning in the second quadrant. 9. a. sin 36 tan 6 3 cos 6 π π ⎛ ⎞ ⎜ ⎟π⎛ ⎞ ⎝ ⎠= =⎜ ⎟ ⎛ ⎞⎝ ⎠ ⎜ ⎟ ⎝ ⎠ b. 1 sec( ) –1 cos( ) π = = π c. ( )3 4 3 1 sec – 2 4 cos π π⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ d. ( )2 1 csc 1 2 sin π π⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ © 2007 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.
  • Instructor’s Resource Manual Section 0.7 45 e. ( ) ( ) 4 4 cos cot 1 4 sin π π π⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ f. ( ) ( ) 4 4 sin – tan – –1 4 cos – π π π⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ 10. a. ( ) ( ) 3 3 sin tan 3 3 cos π π π⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ b. ( )3 1 sec 2 3 cos π π⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ c. ( ) ( ) 3 3 cos 3 cot 3 3sin π π π⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ d. ( )4 1 csc 2 4 sin π π⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ e. ( ) ( ) 6 6 sin – 3 tan – – 6 3cos – π π π⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ f. 1 cos – 3 2 π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 11. a. 2 (1 sin )(1– sin ) 1– sinz z z+ = 2 2 1 cos sec z z = = b. 2 2 (sec –1)(sec 1) sec –1 tant t t t+ = = c. 2 1 sin sec – sin tan – cos cos t t t t t t = 2 2 1– sin cos cos cos cos t t t t t = = = d. 2sin 2 2 2 2cos 2 2 1 2cos sec –1 tan sin sec sec t t t t t t t t = = = 12. a. 2 2 2 2 1 sin sin cos 1 sec v v v v + = + = b. cos3 cos(2 ) cos2 cos – sin 2 sint t t t t t t= + = 2 2 (2cos –1)cos – 2sin cost t t t= 3 2 2cos – cos – 2(1– cos )cost t t t= 3 3 2cos – cos – 2cos 2cost t t t= + 3 4cos – 3cost t= c. sin 4 sin[2(2 )] 2sin 2 cos2x x x x= = 2 2(2sin cos )(2cos –1)x x x= 3 2(4sin cos – 2sin cos )x x x x= 3 8sin cos – 4sin cosx x x x= d. 2 2 (1 cos )(1 cos ) 1 cos sinθ θ θ θ+ − = − = 13. a. 2 2sin cos sin cos 1 csc sec u u u u u u + = + = b. 2 2 2 2 (1 cos )(1 cot ) (sin )(csc )x x x x− + = = sin 2 x 1 sin2 x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1 c. 1 sin (csc – sin ) sin – sin sin t t t t t t ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = 1– sin2 t = cos2 t d. 2cos 2 2 2sin 2 2 1 2sin 1– csc cot – – csc csc t t t t t t t = = 2 2 1 – cos – sec t t = = 14. a. y = sin 2x © 2007 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.
  • 46 Section 0.7 Instructor’s Resource Manual b. ty sin2= c. cos 4 y x π⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ d. secy t= 15. a. y = csc t b. y = 2 cos t c. cos3y t= d. cos 3 y t π⎛ ⎞ = +⎜ ⎟ ⎝ ⎠ 16. y = 3cos x 2 Period = 4π , amplitude = 3 © 2007 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.
  • Instructor’s Resource Manual Section 0.7 47 17. y = 2 sin 2x Period = π , amplitude = 2 18. y = tan x Period = π 19. )2cot( 6 1 2 xy += Period = 2 π , shift: 2 units up 20. 3 sec( )y x π= + − Period = 2π , shift: 3 units up, π units right 21. )32sin(721 ++= xy Period = π , amplitude = 7, shift: 21 units up, 2 3 units left 22. 3cos – –1 2 y x π⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ Period = 2π , amplitude = 3, shifts: π 2 units right and 1 unit down. 23. y = tan 2x – π 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Period = π 2 , shift: π 6 units right © 2007 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.
  • 48 Section 0.7 Instructor’s Resource Manual 24. a. and g.: sin cos – cos( – ) 2 y x x x π⎛ ⎞ = + = = π⎜ ⎟ ⎝ ⎠ b. and e.: cos sin( ) 2 sin( ) y x x x π⎛ ⎞ = + = + π⎜ ⎟ ⎝ ⎠ = − π − c. and f.: cos sin 2 sin( ) y x x x π⎛ ⎞ = − =⎜ ⎟ ⎝ ⎠ = − + π d. and h.: sin cos( ) 2 cos( ) y x x x π⎛ ⎞ = − = + π⎜ ⎟ ⎝ ⎠ = − π 25. a. –t sin (–t) = t sin t; even b. 2 2 sin (– ) sin ;t t= even c. csc(–t) = 1 sin(–t) = –csc t; odd d. sin( ) –sin sin ;t t t− = = even e. sin(cos(–t)) = sin(cos t); even f. –x + sin(–x) = –x – sin x = –(x + sin x); odd 26. a. cot(–t) + sin(–t) = –cot t – sin t = –(cot t + sin t); odd b. 3 3 sin (– ) –sin ;t t= odd c. sec(–t) = 1 cos(–t) = sect; even d. 4 4 sin (– ) sin ;t t= even e. cos(sin(–t)) = cos(–sin t) = cos(sin t); even f. (–x )2 + sin(–x ) = x2 – sinx; neither 27. 2 2 2 1 1 cos cos 3 3 2 4 ππ ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 28. 2 2 2 1 1 sin sin 6 6 2 4 ππ ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 29. 3 3 3 1 1 sin sin 6 6 2 8 π π⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 30. ( ) 3 122 6 2 1 cos2 1 cos 1 cos 12 2 2 2 π π+ + +π = = = 2 3 4 + = 31. ( ) 2 82 4 2 1– cos2 1– cos 1– sin 8 2 2 2 π π π = = = 2 – 2 4 = 32. a. sin(x – y) = sin x cos(–y) + cos x sin(–y) = sin x cos y – cos x sin y b. cos(x – y) = cos x cos(–y) – sin x sin (–y) = cos x cos y + sin x sin y c. tan tan(– ) tan( – ) 1– tan tan(– ) x y x y x y + = tan – tan 1 tan tan x y x y = + 33. tan tan tan 0 tan( ) 1– tan tan 1– (tan )(0) t t t t t + π + + π = = π tant= 34. cos( ) cos cos( ) sin sin( )x x xπ π π− = − − − = –cos x – 0 · sin x = –cos x 35. s = rt = (2.5 ft)( 2π rad) = 5π ft, so the tire goes 5π feet per revolution, or 1 5π revolutions per foot. 1 rev mi 1 hr ft 60 5280 5 ft hr 60 min miπ ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ ≈ 336 rev/min 36. s = rt = (2 ft)(150 rev)( 2π rad/rev) ≈ 1885 ft 37. 1 1 2 2 1; 6(2 ) 8(2 )(21)r t r t t= π = π t1 = 28 rev/sec 38. Δy = sinα and Δx = cos α m = Δy Δx = sinα cos α = tanα © 2007 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.
  • Instructor’s Resource Manual Section 0.7 49 39. a. tan 3α = α = π 3 b. 3 3 6x y+ = 3 – 3 6y x= + 3 3 – 2; – 3 3 y x m= + = 3 tan – 3 α = 5 6 α π = 40. m1 = tan θ1 and m2 = tanθ2 2 1 2 1 2 1 2 1 2 1 2 1 1 2 tan tan( ) tan tan( ) 1 tan tan( ) tan tan 1 tan tan 1 m m m m θ θ θ θ θ θ θ θ θ θ θ + − = − = − − − − = = + + 41. a. tan θ = 3 – 2 1+ 3(2) = 1 7 θ ≈ 0.1419 b. ( ) 1 2 1 2 –1– tan –3 1 (–1) θ = = + θ ≈1.8925 c. 2x – 6y = 12 2x + y = 0 –6y = –2x + 12y = –2x y = 1 3 x – 2 m1 = 1 3 , m2 = –2 ( ) 1 3 1 3 –2 – tan –7; 1.7127 1 (–2) θ θ= = ≈ + 42. Recall that the area of the circle is 2 rπ . The measure of the vertex angle of the circle is 2π . Observe that the ratios of the vertex angles must equal the ratios of the areas. Thus, 2 , 2 t A rπ π = so 21 2 A r t= . 43. A = 1 2 (2)(5)2 = 25cm2 44. Divide the polygon into n isosceles triangles by drawing lines from the center of the circle to the corners of the polygon. If the base of each triangle is on the perimeter of the polygon, then the angle opposite each base has measure 2 . n π Bisect this angle to divide the triangle into two right triangles (See figure). sin so 2 sin and cos 2 b h b r n r n n r π π π = = = so h = r cos π n . 2 sinP nb rn n π = = 21 cos sin 2 A n bh nr n n π π⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ 45. The base of the triangle is the side opposite the angle t. Then the base has length 2 sin 2 t r (similar to Problem 44). The radius of the semicircle is sin 2 t r and the height of the triangle is cos . 2 t r 2 1 2 sin cos sin 2 2 2 2 2 t t t A r r r π⎛ ⎞⎛ ⎞ ⎛ ⎞ = +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ 2 2 2 sin cos sin 2 2 2 2 t t r t r π = + © 2007 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.
  • 50 Section 0.7 Instructor’s Resource Manual 46. cos cos cos cos 2 4 8 16 x x x x = 1 2 cos 3 4 x + cos 1 4 x ⎡ ⎣⎢ ⎤ ⎦⎥ 1 2 cos 3 16 x + cos 1 16 x ⎡ ⎣⎢ ⎤ ⎦⎥ 1 3 1 3 1 cos cos cos cos 4 4 4 16 16 x x x x ⎡ ⎤ ⎡ ⎤ = + +⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 1 3 3 3 1 cos cos cos cos 4 4 16 4 16 x x x x ⎡ = +⎢ ⎣ + cos 1 4 x cos 3 16 x + cos 1 4 x cos 1 16 x ⎤ ⎦⎥ 1 1 15 9 1 13 11 cos cos cos cos 4 2 16 16 2 16 16 x x x ⎡ ⎛ ⎞ ⎛ ⎞ = + + +⎜ ⎟ ⎜ ⎟⎢ ⎝ ⎠ ⎝ ⎠⎣ 1 7 1 1 5 3 cos cos cos cos 2 16 16 2 16 16 x x x x ⎤⎛ ⎞ ⎛ ⎞ + + + +⎜ ⎟ ⎜ ⎟⎥ ⎝ ⎠ ⎝ ⎠⎦ = 1 8 cos 15 16 x + cos 13 16 x + cos 11 16 x + cos 9 16 x ⎡ ⎣⎢ 7 5 3 1 cos cos cos cos 16 16 16 16 x x x x ⎤ + + + + ⎥ ⎦ 47. The temperature function is ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+= 2 7 12 2 sin2580)( ttT π . The normal high temperature for November 15th is then 5.67)5.10( =T °F. 48. The water level function is ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+= )9( 12 2 sin5.35.8)( ttF π . The water level at 5:30 P.M. is then (17.5) 5.12 ftF ≈ . 49. As t increases, the point on the rim of the wheel will move around the circle of radius 2. a. 902.1)2( ≈x 618.0)2( ≈y 176.1)6( −≈x 618.1)6( −≈y 0)10( =x 2)10( =y 0)0( =x 2)0( =y b. ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= ttyttx 5 cos2)(, 5 sin2)( ππ c. The point is at (2, 0) when 25 ππ =t ; that is , when 2 5 =t . 50. Both functions have frequency 10 2π . When you add functions that have the same frequency, the sum has the same frequency. a. ( ) 3sin( /5) 5cos( /5) 2sin(( /5) 3) y t t t t π π π = − + − b. ( ) 3cos( /5 2) cos( /5) cos(( /5) 3) y t t t t π π π = − + + − © 2007 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.
  • Instructor’s Resource Manual Section 0.7 51 51. a. sin( ) ( cos )sin ( sin )cos .C t C t C tω φ φ ω φ ω+ = + Thus cosA C φ= ⋅ and sinB C φ= ⋅ . b. 2 2 2 2 2 2 2 2 2 ( cos ) ( sin ) (cos ) (sin )A B C C C C Cφ φ φ φ+ = + = + = Also, sin tan cos B C A C φ φ φ ⋅ = = ⋅ c. tAAA tAAA ttA ttA ttA tAtAtA ωφφφ ωφφφ φωφω φωφω φωφω φωφωφω cos)sinsinsin( sin)coscoscos( )sincoscos(sin )sincoscos(sin )sincoscos(sin )(sin)sin()sin( 332211 332211 333 222 111 332211 +++ ++= ++ ++ += +++++ ( )sinC tω φ= + where C and φ can be computed from 1 1 2 2 3 3 1 1 2 2 3 3 cos cos cos sin sin sin A A A A B A A A φ φ φ φ φ φ = + + = + + as in part (b). d. Written response. Answers will vary. 52. ( a.), (b.), and (c.) all look similar to this: d. e. The windows in (a)-(c) are not helpful because the function oscillates too much over the domain plotted. Plots in (d) or (e) show the behavior of the function. 53. a. b. c. The plot in (a) shows the long term behavior of the function, but not the short term behavior, whereas the plot in (c) shows the short term behavior, but not the long term behavior. The plot in (b) shows a little of each. © 2007 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.
  • 52 Section 0.8 Instructor’s Resource Manual 54. a. ( ) 2 2 ( ) ( ) 3 cos(100 ) 2 100 1 cos (100 ) 1 100 h x f g x x x = + = ⎛ ⎞ +⎜ ⎟ ⎝ ⎠ 2 1 3 2 ( ) ( )( ) cos 100 100 1 x j x g f x x +⎛ ⎞ = = ⎜ ⎟ +⎝ ⎠ b. c. 55. ( ) ( ) ( ) 1 4 1 : , 4 4 7 1 : , 1 3 3 4 x x x n n f x x x x n n ⎧ ⎡ ⎞ − + ∈ +⎪ ⎟⎢ ⎪ ⎣ ⎠ = ⎨ ⎡ ⎞⎪− − + ∈ + + ⎟⎢⎪ ⎣ ⎠⎩ where n is an integer. 2 1 x y −1 1 56. ( ) ( )2 1 1 2 , 2 ,2 4 4 0.0625, otherwise f x x n x n n ⎧ ⎡ ⎤ = − ∈ − +⎪ ⎢ ⎥ ⎨ ⎣ ⎦ ⎪ ⎩ where n is an integer. −2 0.5 1 x y −1 2 0.25 0.8 Chapter Review Concepts Test 1. False: p and q must be integers. 2. True: 1 2 1 2 2 1 1 2 1 2 ; p p p q p q q q q q − − = since 1 1 2 2, , , andp q p q are integers, so are 1 2 2 1 1 2and .p q p q q q− 3. False: If the numbers are opposites (–π and π ) then the sum is 0, which is rational. 4. True: Between any two distinct real numbers there are both a rational and an irrational number. 5. False: 0.999... is equal to 1. 6. True: ( ) ( ) n mm n mn a a a= = 7. False: ( * )* ; *( * ) cbc b a b c a a b c a= = 8. True: Since and ,x y z x z x y z≤ ≤ ≥ = = 9. True: If x was not 0, then 2 x ε = would be a positive number less than x . © 2007 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.
  • Instructor’s Resource Manual Section 0.8 53 10. True: ( )y x x y− = − − so 2 ( )( ) ( )( 1)( ) ( 1)( ) . x y y x x y x y x y − − = − − − = − − 2 ( ) 0x y− ≥ for all x and y, so 2 ( ) 0.x y− − ≤ 11. True: a < b < 0; a < b; 1 1 1; a b b a > < 12. True: [ ],a b and [ ],b c share point b in common. 13. True: If (a, b) and (c, d) share a point then c < b so they share the infinitely many points between b and c. 14. True: 2 x x x= = − if 0.x < 15. False: For example, if 3x = − , then ( )3 3 3x− = − − = = which does not equal x. 16. False: For example, take 1x = and 2y = − . 17. True: 4 4 x y x y< ⇔ < 4 44 4 4 4 and , sox x y y x y= = < 18. True: ( ) ( ) x y x y x y x y + = − + = − + − = + 19. True: If r = 0, then 1 1 1 1. 1 1– 1–r r r = = = + For any r, 1 1– .r r+ ≥ Since 1,1– 0r r< > so 1 1 ; 1 1–r r ≤ + also, –1 < r < 1. If –1 < r < 0, then –r r= and 1– 1 ,r r= + so 1 1 1 . 1 1– 1–r r r = ≤ + If 0 < r < 1, then r r= and 1– 1– ,r r= so 1 1 1 . 1 1– 1–r r r ≤ = + 20. True: If 1,r > then 1 0.r− < Thus, since 1 1 ,r r+ ≥ − 1 1 . 1 1r r ≤ − + If 1,r > ,r r= and 1 1 ,r r− = − so 1 1 1 . 1 1 1r r r = ≤ − − + If 1,r r r< − = − and 1 1 ,r r− = + so 1 1 1 . 1 1 1r r r ≤ = − − + 21. True: If x and y are the same sign, then – – .x y x y= –x y x y≤ + when x and y are the same sign, so – .x y x y≤ + If x and y have opposite signs then either – – (– )x y x y x y= = + (x > 0, y < 0) or – – –x y x y x y= = + (x < 0, y > 0). In either case – .x y x y= + If either x = 0 or y = 0, the inequality is easily seen to be true. 22. True: If y is positive, then x y= satisfies ( ) 22 .x y y= = 23. True: For every real number y, whether it is positive, zero, or negative, the cube root 3x y= satisfies ( ) 33 3x y y= = 24. True: For example 2 0x ≤ has solution [0]. 25. True: 2 2 2 2 2 2 2 2 2 0 1 1 4 4 4 4 1 1 2 2 4 x ax y y a a x ax y y a a x y + + + = + + + + + = + +⎛ ⎞ ⎛ ⎞ + + + =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ is a circle for all values of a. 26. False: If 0a b= = and 0c < , the equation does not represent a circle. © 2007 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.
  • 54 Section 0.8 Instructor’s Resource Manual 27. True; 3 ( ) 4 3 3 ; 4 4 y b x a a y x b − = − = − + If x = a + 4: 3 3 ( 4) – 4 4 3 3 3 – 3 4 4 a y a b a a b b = + + = + + = + 28. True: If the points are on the same line, they have equal slope. Then the reciprocals of the slopes are also equal. 29. True: If ab > 0, a and b have the same sign, so (a, b) is in either the first or third quadrant. 30. True: Let / 2.x ε= If 0ε > , then 0x > and .x ε< 31. True: If ab = 0, a or b is 0, so (a, b) lies on the x-axis or the y-axis. If a = b = 0, (a, b) is the origin. 32. True: 1 2,y y= so ( ) ( )1 1 2 2, and ,x y x y are on the same horizontal line. 33. True: 2 2 2 [( ) – ( – )] ( – ) (2 ) 2 d a b a b a a b b = + + = = 34. False: The equation of a vertical line cannot be written in point-slope form. 35. True: This is the general linear equation. 36. True: Two non-vertical lines are parallel if and only if they have the same slope. 37. False: The slopes of perpendicular lines are negative reciprocals. 38. True: If a and b are rational and ( ) ( ),0 , 0,a b are the intercepts, the slope is b a − which is rational. 39. False: ( )( ) 1. ax y c y ax c ax y c y ax c a a + = ⇒ = − + − = ⇒ = − − ≠ − (unless 1a = ± ) 40. True: The equation is (3 2 ) (6 2) 4 2 0m x m y m+ + − + − = which is the equation of a straight line unless 3 2 and 6 2m m+ − are both 0, and there is no real number m such that 3 2 0 and 6 2 0.m m+ = − = 41. True: 2 ( ) –( 4 3)f x x x= + + –( 3)( 1)x x= + + 2 ( 4 3) 0x x− + + ≥ on 3 1x− ≤ ≤ − . 42. False: The domain does not include nπ + π 2 where n is an integer. 43. True: The domain is ( , )− ∞ ∞ and the range is[ 6, )− ∞ . 44. False: The range is ( , )− ∞ ∞ . 45. False: The range ( , )− ∞ ∞ . 46. True: If f(x) and g(x) are even functions, f(x) + g(x) is even. f(–x) + g(–x) = f(x) + g(x) 47. True: If f(x) and g(x) are odd functions, f(–x) + g(–x) = –f(x) – g(x) = –[f(x) + g(x)], so f(x) + g(x) is odd 48. False: If f(x) and g(x) are odd functions, f(–x)g(–x) = –f(x)[–g(x)] = f(x)g(x), so f(x)g(x) is even. 49. True: If f(x) is even and g(x) is odd, f(–x)g(–x) = f(x)[–g(x)] = –f(x)g(x), so f(x)g(x) is odd. 50. False: If f(x) is even and g(x) is odd, f(g(–x)) = f(–g(x)) = f(g(x)); while if f(x) is odd and g(x) is even, f(g(–x)) = f(g(x)); so f(g(x)) is even. 51. False: If f(x) and g(x) are odd functions, ( ( ))f g x− = f(–g(x)) = –f(g(x)), so f(g(x)) is odd. 52. True: 3 3 2 2 3 2 2(– ) (– ) –2 – (– ) (– ) 1 1 2 1 x x x x f x x x x x x + = = + + + = − + © 2007 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.
  • Instructor’s Resource Manual Section 0.8 55 53. True: 2 2 2 (sin(– )) cos(– ) (– ) tan(– )csc(– ) ( sin ) cos (sin ) cos – tan (– csc ) tan csc t t f t t t t t t t t t t t + = − + + = = 54. False: f(x) = c has domain ( , )− ∞ ∞ and the only value of the range is c. 55. False: f(x) = c has domain ( , )− ∞ ∞ , yet the range has only one value, c. 56. True: 1.8 ( 1.8) 0.9 1 2 g − − = = − = − 57. True: 623 )())(( xxxgf == 632 )())(( xxxfg == 58. False: 623 )())(( xxxgf == 532 )()( xxxxgxf ==⋅ 59. False: The domain of f g excludes any values where g = 0. 60. True: f(a) = 0 Let F(x) = f(x + h), then F(a – h) = f(a – h + h) = f(a) = 0 61. True: cos cot sin cos( ) cot( ) sin( ) cos cot sin x x x x x x x x x = − − = − = = − − 62. False: The domain of the tangent function excludes all nπ + π 2 where n is an integer. 63. False: The cosine function is periodic, so cos s = cos t does not necessarily imply s = t; e.g., cos0 cos 2 1π= = , but 0 2 .π≠ Sample Test Problems 1. a. 1 2 1 1 1 25 ; 1 2; 2 ; 1 2 4 n n n ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + + = + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ –2 1 4 –2 –2 25 ⎛ ⎞ + =⎜ ⎟ ⎝ ⎠ b. 22 2 2 ( – 1) ; (1) – (1) 1 1;n n ⎡ ⎤+ + = ⎣ ⎦ 22 (2) – (2) 1 9;⎡ ⎤+ = ⎣ ⎦ 22 (–2) – (–2) 1 49⎡ ⎤+ = ⎣ ⎦ c. 3/ 3/1 3/ 2 –3/ 2 1 4 ; 4 64; 4 8; 4 8 n = = = d. 1 1 1 1 1 2 ; 1; ; 1 2 22 n n = = = 2 1 2 2 − = − 2. a. 1 1 1 1 1 1 1 1 1 1 1 1 1 m n m n m n m n mn n m mn n m − + + ⎛ ⎞⎛ ⎞ + + − + =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ − + + + = − + b. 2 2( 2) 3 ( 2) 2( 1) 4 8 22 1 ( 2)( 1)1 2 3 2 3 2 1 2 1 2 x x x x x x xx x x xx x x x x x x − − = − − + − = − −− + − ++ − − = − − + − + − c. 3 2 2( 1) ( 1)( 1) 1 1 1 t t t t t t t t − − + + = = + + − − 3. Let a, b, c, and d be integers. 2 2 2 2 a c b d a c ad bc b d bd + + = + = which is rational. © 2007 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.
  • 56 Section 0.8 Instructor’s Resource Manual 4. 4.1282828x = … 1000 4128.282828 10 41.282828 990 4087 4087 990 x x x x = = = = … … 5. Answers will vary. Possible answer: 13 0.50990... 50 ≈ 6. 2 3 4 8.15 10 1.32 545.39 3.24 ⎛ ⎞× −⎜ ⎟ ⎝ ⎠ ≈ 7. ( ) 2.5 3 – 2.0 – 2.0 2.66π ≈ 8. ( ) ( )2 2 sin 2.45 cos 2.40 1.00 0.0495+ − ≈ − 9. 1– 3 0 3 1 1 3 x x x > < < 1 – , 3 ⎛ ⎞ ∞⎜ ⎟ ⎝ ⎠ 10. ( ) 6 3 2 5 4 8 2; 2, x x x x + > − > − > − − ∞ 11. 3 2 4 1 2 7 3 2 4 1 and 4 1 2 7 6 2 and 2 6 x x x x x x x x x − ≤ + ≤ + − ≤ + + ≤ + ≥ ≥ 1 and 3; 3 x x≥ ≤ 1 , 3 3 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ 12. 2 2 5 3 0;(2 1)( 3) 0; 1 1 3 ; 3, 2 2 x x x x x + − < − + < ⎛ ⎞ − < < −⎜ ⎟ ⎝ ⎠ 13. 2 2 21 – 44 12 –3; 21 – 44 15 0;t t t t+ ≤ + ≤ 2 44 44 – 4(21)(15) 44 26 3 5 , 2(21) 42 7 3 t ± ± = = = 3 5 3 5 – – 0; , 7 3 7 3 t t ⎛ ⎞⎛ ⎞ ⎡ ⎤ ≤⎜ ⎟⎜ ⎟ ⎢ ⎥ ⎝ ⎠⎝ ⎠ ⎣ ⎦ 14. ( ) 2 1 1 0; , 2, 2 2 x x − ⎛ ⎞ > −∞ ∪ ∞⎜ ⎟ − ⎝ ⎠ 15. 2 ( 4)(2 1) ( 3) 0;[ 4,3]x x x+ − − ≤ − 16. 3 4 6; 6 3 4 6; 2 3 10; 2 10 2 10 ; , 3 3 3 3 x x x x − < − < − < − < < ⎛ ⎞ − < < −⎜ ⎟ ⎝ ⎠ 17. 3 2 1– 3 – 2 0 1– 3 – 2(1– ) 0 1– 2 1 0; 1– x x x x x x ≤ ≤ ≤ + ≤ ( ) 1 – , – 1, 2 ⎛ ⎤ ∞ ∪ ∞⎜ ⎥ ⎝ ⎦ 18. 2 2 2 2 12 3 (12 3 ) 144 72 9 x x x x x x x − ≥ − ≥ − + ≥ 2 8 72 144 0 8( 3)( 6) 0 ( ,3] [6, ) x x x x − + ≥ − − ≥ −∞ ∪ ∞ 19. For example, if x = –2, ( 2) 2 2− − = ≠ − x x− ≠ for any x < 0 20. If ,x x− = then .x x= 0x ≥ © 2007 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.
  • Instructor’s Resource Manual Section 0.8 57 21. |t – 5| = |–(5 – t)| = |5 – t| If |5 – t| = 5 – t, then 5 0.t− ≥ 5t ≤ 22. ( )t a a t a t− = − − = − If ,a t a t− = − then 0.a t− ≥ t a≤ 23. If 2,x ≤ then 2 2 0 2 3 2 2 3 2 8 6 2 16x x x x≤ + + ≤ + + ≤ + + = also 2 2 1 1 2 2 so . 22 x x + ≥ ≤ + Thus 2 2 2 2 2 3 2 1 1 2 3 2 16 22 2 8 x x x x x x + + ⎛ ⎞ = + + ≤ ⎜ ⎟ ⎝ ⎠+ + = 24. a. The distance between x and 5 is 3. b. The distance between x and –1 is less than or equal to 2. c. The distance between x and a is greater than b. 25. 2 2 2 2 2 2 ( , ) (1 2) (2 6) 9 16 5 ( , ) (5 1) (5 2) 16 9 5 ( , ) (5 2) (5 6) 49 1 50 5 2 d A B d B C d A C = + + − = + = = − + − = + = = + + − = + = = 2 2 2 ( ) ( ) ( ) ,AB BC AC+ = so ABCΔ is a right triangle. 26. midpoint: ( ) 1 7 2 8 , 4,5 2 2 + +⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 2 2 (4 3) (5 6) 1 121 122d = − + + = + = 27. 2 10 0 4 center , (6, 2) 2 2 + +⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ 2 21 1 radius (10 – 2) (4 – 0) 64 16 2 2 2 5 = + = + = 2 2 circle: ( – 6) ( – 2) 20x y+ = 28. 2 2 2 2 2 2 8 6 0 8 16 6 9 16 9 ( 4) ( 3) 25; x y x y x x y y x y + − + = − + + + + = + − + + = center = ( )4, 3− , radius = 5 29. 2 2 2 2 2 2 2 2 2 2 1 2 1 2 1 1 ( 1) ( 1) 4 x x y y x x y y x y − − + + = + + + + = + + − + + = center = (1, –1) 2 2 – 2 2 – 2 2 6 4 –7 6 9 4 4 –7 9 4 ( 3) ( – 2) 6 x x y y x x y y x y + + = + + + + = + + + + = center = (–3, 2) 2 2 (–3 –1) (2 1) 16 9 5d = + + = + = 30. a. 3 2 6 2 3 6 3 3 2 x y y x y x + = = − + = − + 3 2 3 2 ( 3) 2 3 13 2 2 m y x y x = − − = − − = − + © 2007 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.
  • 58 Section 0.8 Instructor’s Resource Manual b. 2 ; 3 2 1 ( 1) 3 2 5 3 3 m y x y x = + = − = − c. 9y = d. x = –3 31. a. 3 –1 2 ; 7 2 9 m = = + 2 –1 ( 2) 9 2 13 9 9 y x y x = + = + b. 3 – 2 5 –2 –3 5 3 5 – ; 2 2 x y y x y x = = + = 3 2 m = 3 –1 ( 2) 2 3 4 2 y x y x = + = + c. 3x + 4y = 9 4y = –3x + 9; 3 9 – ; 4 4 y x= + 4 3 m = 4 –1 ( 2) 3 4 11 3 3 y x y x = + = + d. x = –2 e. contains (–2, 1) and (0, 3); 3 –1 ; 0 2 m = + y = x + 3 32. 1 2 3 3 1 4 11 3 8 4 ; ; 5 2 3 11 5 6 3 11 1 12 4 11 2 9 3 m m m + − = = = = = − − + = = = − 1 2 3,m m m= = so the points lie on the same line. 33. The figure is a cubic with respect to y. The equation is (b) 3 x y= . 34. The figure is a quadratic, opening downward, with a negative y-intercept. The equation is (c) 2 .y ax bx c= + + with a < 0, b > 0, and c < 0. 35. © 2007 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.
  • Instructor’s Resource Manual Section 0.8 59 36. 2 2 2 2 2 2 2 3 2 1 4 ( 1) 4 x x y x x y x y − + = − + + = − + = 37. 38. 39. y = x2 – 2x + 4 and y – x = 4; 2 2 4 2 4 3 0 ( – 3) 0 x x x x x x x + = − + − = = points of intersection: (0, 4) and (3, 7) 40. 4 2 4 2; x y y x − = = − 1 4 m = − contains ( ) ( ),0 , 0, ;a b 2 8 2 16 16 0 1 ; 0 4 4 16 4 64 8 16 1 2; 2 8 4 ab ab b a b b a a a b a a a a b y x = = = − = − = − − = ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = = = = = − + 41. a. 1 1 1 (1) – – 1 1 1 2 f = = + b. 1 1 2 2 1 1 1 – – 4 2 – 1 – f ⎛ ⎞ = =⎜ ⎟ +⎝ ⎠ c. f(–1) does not exist. d. 1 1 1 1 ( –1) – – –1 1 –1 –1 f t t t t t = = + e. 1 1 1 1 1 – – 11t t t f t t t ⎛ ⎞ = =⎜ ⎟ ++⎝ ⎠ 42. a. 2 1 3 (2) 2 2 g + = = b. 3 1 2 1 2 1 2 1 = + =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ g c. hh ghg h h 2 12 2 12 –)2(–)2( + + ++ = + )2(2 1–– )2(2)2(2 6–3–62 + === ++ + hhh h h h hh 43. a. { : –1,1}x x∈ ≠ b. { }: 2x x∈ ≤ © 2007 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.
  • 60 Section 0.8 Instructor’s Resource Manual 44. a. 2 2 3(– ) 3 (– ) – ; (– ) 1 1 x x f x x x = = + + odd b. (– ) sin(– ) cos(– )g x x x= + sin cos sin cos ;x x x x= − + = + even c. 3 3 (– ) (– ) sin(– ) – – sinh x x x x x= + = ; odd d. 2 2 4 4 (– ) 1 1 (– ) ; – (– ) x x k x x x x x + + = = + + even 45. a. f (x) = x2 – 1 b. g(x) = x x2 +1 c. h(x) = x2 if 0 ≤ x ≤ 2 6 – x if x > 2 ⎧ ⎨ ⎩ 46. 47. V(x) = x(32 – 2x)(24 – 2x) Domain [0, 12] 48. a. 21 13 ( )(2) 2 – (2 1) 2 2 f g ⎛ ⎞ + = + + =⎜ ⎟ ⎝ ⎠ b. 3 15 ( )(2) (5) 2 2 f g ⎛ ⎞ ⋅ = =⎜ ⎟ ⎝ ⎠ c. 5 24 5 1 –5)5()2)(( === fgf d. 4 13 1 2 3 2 3 )2)(( 2 =+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = gfg e. 3 3 1 (–1) –1 0 1 f ⎛ ⎞ = + =⎜ ⎟ ⎝ ⎠ f. 2 2 2 23 (2) (2) (5) 2 9 109 25 4 4 f g ⎛ ⎞ + = +⎜ ⎟ ⎝ ⎠ = + = 49. a. y = 1 4 x2 © 2007 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.
  • Instructor’s Resource Manual Review and Preview 61 b. y = 1 4 ( x + 2)2 c. 21 –1 ( 2) 4 y x= + + 50. a. ( ,16]−∞ b. ;–16 4 xgf = domain [–2, 2] c. ;)–16()–16( 24 xxfg == domain ( ,16]−∞ (note: the simplification 4 2 ( 16 – ) (16 – )x x= is only true given the restricted domain) 51. 2 ( ) , ( ) 1 , ( ) ,f x x g x x h x x= = + = k(x) = sin x, khgfxxF =+= 2 sin1)( 52. a. 1 sin(570 ) sin(210 ) – 2 ° = ° = b. 9 cos cos 0 2 2 π π⎛ ⎞ ⎛ ⎞ = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ c. 13 3 cos – cos 6 6 2 π π⎛ ⎞ ⎛ ⎞ = − =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 53. a. sin (–t) = –sin t = –0.8 b. 2 2 sin cos 1t t+ = 2 2 cos 1– (0.8) 0.36t = = cos t = –0.6 c. sin 2t = 2 sin t cos t = 2(0.8)(–0.6) = –0.96 d. sin 0.8 4 tan – –1.333 cos –0.6 3 t t t = = = ≈ e. cos – sin 0.8 2 t t π⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ f. sin( ) sin 0.8t tπ + = − = − 54. sin3 sin(2 ) sin 2 cos cos2 sint t t t t t t= + = + 2 2 2sin cos (1– 2sin )sint t t t= + 2 3 2sin (1– sin ) sin – 2sint t t t= + 3 3 2sin – 2sin sin – 2sint t t t= + 3 3sin – 4sint t= 55. s = rt rev rad 1 min 9 20 2 (1 sec) 6 min rev 60 sec ⎛ ⎞⎛ ⎞⎛ ⎞ = π = π⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ ≈ 18.85 in. Review and Preview Problems 1. a) 0 2 4; 0 2x x< < < < b) 6 16x− < < 2. a) 13 2 14; 6.5 7x x< < < < b) 4 / 2 7; 14 8x x− < − < − < < 3. 7 3 or 7 3 10 or 4 x x x x − = − = − = = 4. 3 2 or 3 2 1 or 5 x x x x + = + = − = − = − 5. 7 3 or 7 3 10 or 4 x x x x − = − = − = = 6. 7 or 7 7 or 7 x d x d x d x d − = − = − = + = − 7. a) 7 3 and 7 3 10 and 4 4 10 x x x x x − < − > − < > < < © 2007 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.
  • 62 Review and Preview Instructor’s Resource Manual b) 7 3 and 7 3 10 and 4 4 10 x x x x x − ≤ − ≥ − ≤ ≥ ≤ ≤ c) 7 1 and 7 1 8 and 6 6 8 x x x x x − ≤ − ≥ − ≤ ≥ ≤ ≤ d) 7 0.1 and 7 0.1 7.1 and 6.9 6.9 7.1 x x x x x − < − > − < > < < 8. a) 2 1 and 2 1 3 and 1 1 3 x x x x x − < − > − < > < < b) 2 1 or 2 1 3 or 1 x x x x − ≥ − ≤ − ≥ ≤ c) 2 0.1 and 2 0.1 2.1 and 1.9 1.9 2.1 x x x x x − < − > − < > < < d) 2 0.01 and 2 0.01 2.01 and 1.99 1.99 2.01 x x x x x − < − > − < > < < 9. a) 1 0; 1x x− ≠ ≠ b) 2 2 1 0; 1, 0.5x x x− − ≠ ≠ − 10. a) 0x ≠ b) 0x ≠ 11. a) ( ) ( ) 0 1 0 1 0 1 0.81 1 0.9 1.9 0.9 1 f f − = = − − = = − ( ) ( ) ( ) ( ) ( ) ( ) 0.9801 1 0.99 1.99 0.99 1 0.998001 1 0.999 1.999 .999 1 1.002001 1 1.001 2.001 1.001 1 1.0201 1 1.01 2.01 1.01 1 1.21 1 1.1 2.1 1.1 1 4 1 2 3 2 1 f f f f f f − = = − − = = − − = = − − = = − − = = − − = = − b) ( )0 1g = − ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0.9 0.0357143 0.99 0.0033557 0.999 0.000333556 1.001 0.000333111 1.01 0.00331126 1.1 0.03125 1 2 5 g g g g g g g = − = − = − = = = = 12. a) ( ) 1 1 1 1 F − = = − − ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0.1 0.1 1 0.1 0.01 0.01 1 0.01 0.001 0.001 1 0.001 0.001 0.001 1 0.001 0.01 0.01 1 0.01 0.01 0.1 1 0.01 1 1 1 1 F F F F F F F − = = − − − = = − − − = = − − = = = = = = = = b) ( )1 0.841471G − = ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0.1 0.998334 0.01 0.999983 0.001 0.99999983 0.001 0.99999983 0.01 0.999983 0.1 0.998334 1 0.841471 G G G G G G G − = − = − = = = = = 13. 5 0.1 and 5 0.1 5.1 and 4.9 4.9 5.1 x x x x x − < − > − < > < < 14. 5 and 5 5 and 5 5 5 x x x x x ε ε ε ε ε ε − < − > − < + > − − < < + 15. a. True. b. False: Choose a = 0. c. True. d. True 16. ( )sin sin cos cos sinc h c h c h+ = + © 2007 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.
  • Instructor’s Resource Manual Section 1.1 63 CHAPTER 1 Limits 1.1 Concepts Review 1. L; c 2. 6 3. L; right 4. lim ( ) x c f x M → = Problem Set 1.1 1. 3 lim( – 5) –2 x x → = 2. –1 lim (1– 2 ) 3 t t → = 3. 2 2 2 lim ( 2 1) ( 2) 2( 2) 1 1 x x x →− + − = − + − − = − 4. 2 2 2 lim ( 2 1) ( 2) 2 1 3 2 x x t t t →− + − = − + − = + 5. ( ) ( )( )22 1 lim 1 1 1 0 t t →− − = − − = 6. ( ) ( )( )22 2 2 2 1 lim 1 1 t t x x x →− − = − − = − 7. 2 2 2 2 – 4 ( – 2)( 2) lim lim – 2 – 2 lim( 2) x x x x x x x x x → → → + = = + = 2 + 2 = 4 8. 2 –7 –7 4 – 21 lim 7 ( 7)( – 3) lim 7 t t t t t t t t → → + + + = + –7 lim ( – 3) –7 – 3 –10 t t → = = = 9. 3 2 –1 – 4 6 lim 1x x x x x→ + + + 2 –1 ( 1)( – 5 6) lim 1x x x x x→ + + = + 2 –1 2 lim ( – 5 6) (–1) – 5(–1) 6 12 x x x → = + = + = 10. 4 3 2 20 2 0 2 – lim lim( 2 –1) –1 x x x x x x x x → → + = + = 11. 2 2 – – – – ( )( – ) lim lim lim ( – ) x t x t x t x t x t x t x t x t x t → → → + = + + = = –t – t = –2t 12. 2 3 3 3 – 9 lim – 3 ( – 3)( 3) lim – 3 lim( 3) x x x x x x x x x → → → + = = + = 3 + 3 = 6 13. 4 22 2 22 ( 4)( 2) lim (3 6) ( 2) 4 lim 9( 2) t t t t t t t t → → + − − − + = − 2 4 lim 9 2 4 6 9 9 t t → + = + = = 14. 3 7 7 7 ( 7) lim 7 ( 7) 7 lim 7 lim 7 t t t t t t t t t + + + → → → − − − − = − = − 7 7 0= − = © 2007 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.
  • 64 Section 1.1 Instructor’s Resource Manual 15. 4 2 2 2 2 23 3 –18 81 ( – 9) lim lim ( – 3) ( – 3)x x x x x x x→ → + = 2 2 2 2 23 3 ( – 3) ( 3) lim lim( 3) (3 3) ( – 3)x x x x x x→ → + = = + = + 36= 16. 3 3 2 21 1 (3 4)(2 – 2) 8(3 4)( –1) lim lim ( –1) ( –1)u u u u u u u u→ → + + = 1 lim8(3 4)( –1) 8[3(1) 4](1–1) 0 u u u → = + = + = 17. 2 2 0 0 2 0 0 (2 ) 4 4 4 4 lim lim 4 lim lim( 4) 4 h h h h h h h h h h h h h → → → → + − + + − = + = = + = 18. 2 2 2 2 2 0 0 2 0 0 ( ) 2 lim lim 2 lim lim( 2 ) 2 h h h h x h x x xh h x h h h xh h x x h → → → → + − + + − = + = = + = 19. x sin 2 x x 1. 0.420735 0.1 0.499167 0.01 0.499992 0.001 0.49999992 –1. 0.420735 –0.1 0.499167 –0.01 0.499992 –0.001 0.49999992 0 sin lim 0.5 2x x x→ = 20. t 1 cos 2 t t − 1. 0.229849 0.1 0.0249792 0.01 0.00249998 0.001 0.00024999998 –1. –0.229849 –0.1 –0.0249792 –0.01 –0.00249998 –0.001 –0.00024999998 0 1 cos lim 0 2t t t→ − = 21. x 2 2 ( sin ) /x x x− 1. 0.0251314 0.1 6 2.775 10− × 0.01 10 2.77775 10− × 0.001 14 2.77778 10− × –1. 0.0251314 –0.1 6 2.775 10− × –0.01 10 2.77775 10− × –0.001 14 2.77778 10− × 2 20 ( – sin ) lim 0 x x x x→ = 22. x 2 2 (1 cos ) /x x− 1. 0.211322 0.1 0.00249584 0.01 0.0000249996 0.001 7 2.5 10− × –1. 0.211322 –0.1 0.00249584 –0.01 0.0000249996 –0.001 7 2.5 10− × 2 20 (1– cos ) lim 0 x x x→ = 23. t 2 ( 1) /(sin( 1))t t− − 2. 3.56519 1.1 2.1035 1.01 2.01003 1.001 2.001 0 1.1884 0.9 1.90317 0.99 1.99003 0.999 1.999 2 1 1 lim 2 sin( 1)t t t→ − = − © 2007 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.
  • Instructor’s Resource Manual Section 1.1 65 24. x sin( 3) 3 3 x x x − − − − 4. 0.158529 3.1 0.00166583 3.01 0.0000166666 3.001 7 1.66667 10− × 2. 0.158529 2.9 0.00166583 2.99 0.0000166666 2.999 7 1.66667 10− × 3 – sin( – 3) – 3 lim 0 – 3x x x x→ = 25. x (1 sin( 3 / 2)) /( )x xπ π+ − − 1. + π 0.4597 0.1 + π 0.0500 0.01 + π 0.0050 0.001 + π 0.0005 –1. + π –0.4597 –0.1 + π –0.0500 –0.01 + π –0.0050 –0.001 + π –0.0005 ( )3 2 1 sin lim 0 x x x π →π + − = − π 26. t (1 cot ) /(1/ )t t− 1. 0.357907 0.1 –0.896664 0.01 –0.989967 0.001 –0.999 –1. –1.64209 –0.1 –1.09666 –0.01 –1.00997 –0.001 –1.001 10 1– cot lim –1 t t t → = 27. x 2 2 ( / 4) /(tan 1)x xπ− − 4 1. π + 0.0320244 4 0.1 π + 0.201002 4 0.01 π + 0.245009 4 0.001 π + 0.2495 4 1. π − + 0.674117 4 0.1 π − + 0.300668 4 0.01 π − + 0.255008 4 0.001 π − + 0.2505 ( ) 4 2 4 2 lim 0.25 (tan 1)x x xπ π → − = − 28. u (2 2sin ) /3u u− 2 1. π + 0.11921 2 0.1 π + 0.00199339 2 0.01 π + 0.0000210862 2 0.001 π + 7 2.12072 10− × 2 1. π − + 0.536908 2 0.1 π − + 0.00226446 2 0.01 π − + 0.0000213564 2 0.001 π − + 7 2.12342 10− × 2 2 2sin lim 0 3u u uπ→ − = 29. a. –3 lim ( ) 2 x f x → = b. f(–3) = 1 c. f(–1) does not exist. d. –1 5 lim ( ) 2x f x → = e. f(1) = 2 f. 1 lim x→ f(x) does not exist. g. – 1 lim ( ) 2 x f x → = h. 1 lim ( ) 1 x f x + → = i. ( ) 1 5 lim 2x f x + →− = © 2007 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.
  • 66 Section 1.1 Instructor’s Resource Manual 30. a. –3 lim ( ) x f x → does not exist. b. f(–3) = 1 c. f(–1) = 1 d. –1 lim ( ) 2 x f x → = e. f(1) = 1 f. 1 lim ( ) x f x → does not exist. g. – 1 lim ( ) 1 x f x → = h. 1 lim ( ) x f x + → does not exist. i. ( ) 1 lim 2 x f x + →− = 31. a. f(–3) = 2 b. f(3) is undefined. c. –3 lim ( ) 2 x f x − → = d. –3 lim ( ) 4 x f x + → = e. –3 lim ( ) x f x → does not exist. f. 3 lim ( ) x f x + → does not exist. 32. a. –1 lim ( ) 2 x f x − → = − b. –1 lim ( ) 2 x f x + → = − c. –1 lim ( ) 2 x f x → = − d. f (–1) = –2 e. 1 lim ( ) 0 x f x → = f. f (1) = 0 33. a. 0 lim ( ) 0 x f x → = b. 1 lim ( ) x f x → does not exist. c. f(1) = 2 d. 1 lim ( ) 2 x f x + → = 34. a. 1 lim ( ) 0 x g x → = b. g(1) does not exist. c. 2 lim ( ) 1 x g x → = d. 2 lim ( ) 1 x g x + → = 35. [ ]( ) –f x x x⎡ ⎤= ⎣ ⎦ a. f(0) = 0 b. 0 lim ( ) x f x → does not exist. c. – 0 lim ( ) 1 x f x → = d. 1 2 1 lim ( ) 2x f x → = © 2007 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.
  • Instructor’s Resource Manual Section 1.2 67 36. ( ) x f x x = a. (0)f does not exist. b. 0 lim ( ) x f x → does not exist. c. – 0 lim ( ) –1 x f x → = d. 1 2 lim ( ) 1 x f x → = 37. 2 1 1 lim 1x x x→ − − does not exist. 2 1 1 lim 2 1x x x− → − = − − and 2 1 1 lim 2 1x x x+ → − = − 38. 0 2 2 lim x x x→ + − 0 ( 2 2)( 2 2) lim ( 2 2)x x x x x→ + − + + = + + 0 0 2 2 lim lim ( 2 2) ( 2 2)x x x x x x x x→ → + − = = + + + + 0 1 1 1 2 lim 42 2 0 2 2 2 2x x→ = = = = + + + + 39. a. 1 lim ( ) x f x → does not exist. b. 0 lim ( ) 0 x f x → = 40. 41. lim ( ) x a f x → exists for a = –1, 0, 1. 42. The changed values will not change lim ( ) x a f x → at any a. As x approaches a, the limit is still 2 .a 43. a. 1 1 lim 1x x x→ − − does not exist. 1 1 lim 1 1x x x− → − = − − and 1 1 lim 1 1x x x+ → − = − b. 1 1 lim 1 1x x x− → − = − − c. 2 1 1 1 lim 3 1x x x x− → − − − = − − d. 1 1 1 lim 1 1x x x− → ⎡ ⎤ −⎢ ⎥ − −⎢ ⎥⎣ ⎦ does not exist. 44. a. 1 lim 0 x x x +→ − = b. 0 1 lim x x+ → does not exist. c. 0 1/ lim ( 1) 0 x x x + → − = d. 0 1/ lim ( 1) 0 x x x + → − = 45. a) 1 b) 0 c) 1− d) 1− 46. a) Does not exist b) 0 c) 1 d) 0.556 47. 0 lim x x → does not exist since x is not defined for x < 0. 48. 0 lim 1x x x + → = 49. 0 lim 0 x x → = 50. 0 lim 1 x x x → = 51. 0 sin 2 1 lim 4 2x x x→ = © 2007 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.
  • 68 Section 1.2 Instructor’s Resource Manual 52. 0 sin5 5 lim 3 3x x x→ = 53. 0 1 lim cos x x→ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ does not exist. 54. 0 1 lim cos 0 x x x→ ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 55. 3 1 1 lim 6 2 2 2x x x→ − = + − 56. 20 sin 2 lim 2 sin( )x x x x→ = 57. – 2 2 – – 2 lim –3 – 2x x x x→ = 58. 1/( 1) 1 2 lim 0 1 2 x x + − → = + 59. 0 lim ; x x → The computer gives a value of 0, but 0 lim x x − → does not exist. 1.2 Concepts Review 1. L – ε; L + ε 2. 0 < x – a < δ; f( x) – L < ε 3. ε 3 4. ma + b Problem Set 1.2 1. εδ <⇒<< Mtfat –)(–0 2. 0 – ( ) –u b g u Lδ ε< < ⇒ < 3. εδ <⇒<< Pzhdz –)(–0 4. εφδ <⇒<< Byey –)(–0 5. εδ <⇒<< Lxfxc –)(–0 6. εδ <⇒<< Dtgat –)(–0 7. If x is within 0.001 of 2, then 2x is within 0.002 of 4. 8. If x is within 0.0005 of 2, then x2 is within 0.002 of 4. 9. If x is within 0.0019 of 2, then x8 is within 0.002 of 4. 10. If x is within 0.001 of 2, then x 8 is within 0.002 of 4. 11. 0 – 0 (2 –1) – (–1)x xδ ε< < ⇒ < 2 –1 1 2x xε ε+ < ⇔ < 2 x ε⇔ < 2 x ε ⇔ < ; 0 – 0 2 x ε δ δ= < < (2 –1) – (–1) 2 2 2x x x δ ε= = < = © 2007 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.
  • Instructor’s Resource Manual Section 1.2 69 12. 0 21 (3 –1) – (–64)x xδ ε< + < ⇒ < 3 –1 64 3 63x xε ε+ < ⇔ + < 3( 21)x ε⇔ + < 3 21x ε⇔ + < 21 3 x ε ⇔ + < ; 0 21 3 x ε δ δ= < + < (3 –1) – (–64) 3 63 3 21 3x x x δ ε= + = + < = 13. 2 – 25 0 – 5 –10 – 5 x x x δ ε< < ⇒ < 2 – 25 ( – 5)( 5) –10 –10 – 5 – 5 x x x x x ε ε + < ⇔ < 5 –10x ε⇔ + < – 5x ε⇔ < ; 0 – 5xδ ε δ= < < 2 – 25 ( – 5)( 5) –10 –10 5 –10 – 5 – 5 x x x x x x + = = + – 5x δ ε= < = 14. 2 2 – 0 – 0 ( 1) x x x x δ ε< < ⇒ − − < 2 2 – (2 –1) 1 1 x x x x x x ε ε+ < ⇔ + < 2 –1 1x ε⇔ + < 2x ε⇔ < 2 x ε⇔ < 2 x ε ⇔ < ; 0 – 0 2 x ε δ δ= < < 2 2 – (2 –1) ( 1) 1 2 –1 1 x x x x x x x − − = + = + 2 2 2x x δ ε= = < = 15. 2 2 –11 5 0 – 5 – 9 – 5 x x x x δ ε + < < ⇒ < 2 2 –11 5 (2 –1)( – 5) – 9 – 9 – 5 – 5 x x x x x x ε ε + < ⇔ < 2 –1– 9x ε⇔ < 2( – 5)x ε⇔ < – 5 2 x ε ⇔ < ; 0 – 5 2 x ε δ δ= < < 2 2 –11 5 (2 –1)( – 5) – 9 – 9 – 5 – 5 x x x x x x + = 2 –1– 9 2( – 5) 2 – 5 2x x x δ ε= = = < = 16. 0 –1 2 – 2x xδ ε< < ⇒ < ε ε < + + ⇔ < 22 )22)(2–2( 2–2 x xx x 2 – 2 2 2 x x ε⇔ < + –1 2 2 2 x x ε⇔ < + 2 ; 0 –1 2 x ε δ δ= < < ( 2 – 2)( 2 2) 2 2 2 2 x x x x + − = + 2 – 2 2 2 x x = + 2 –1 2 –1 2 2 2 2 2 x x x δ ε≤ < = + 17. 2 –1 0 – 4 – 7 – 3 x x x δ ε< < ⇒ < 2 –1 – 7( – 3)2 –1 – 7 – 3 – 3 x xx x x ε ε< ⇔ < ( 2 –1 – 7( – 3))( 2 –1 7( – 3)) – 3( 2 –1 7( – 3)) x x x x x x x ε + ⇔ < + 2 –1– (7 – 21) – 3( 2 –1 7( – 3)) x x x x x ε⇔ < + –5( – 4) – 3( 2 –1 7( – 3)) x x x x ε⇔ < + © 2007 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.
  • 70 Section 1.2 Instructor’s Resource Manual 5 4 3( 2 1 7( 3)) x x x x ε⇔ − ⋅ < − − + − To bound 5 – 3( 2 –1 7( – 3))x x x+ , agree that 1 . 2 δ ≤ If 1 , 2 δ ≤ then 7 2 < x < 9 2 , so 5 0.65 1.65 – 3( 2 –1 7( – 3))x x x < < + and hence 5 4 3( 2 1 7( 3)) x x x x ε− ⋅ < − − + − – 4 1.65 x ε ⇔ < For whatever ε is chosen, let δ be the smaller of 1 2 and ε 1.65 . δ = min 1 2 , ε 1. 65 ⎧ ⎨ ⎩ ⎫ ⎬ ⎭ , 0 < x – 4 < δ 2 1 5 7 4 3 3( 2 1 7( 3)) x x x x x x − − = − ⋅ − − − + − < x – 4 (1.65) < 1. 65δ ≤ ε since δ = 1 2 only when 1 2 ≤ ε 1. 65 so 1.65δ ≤ ε. 18. 2 14 – 20 6 0 –1 – 8 –1 x x x x δ ε + < < ⇒ < 2 14 – 20 6 2(7 – 3)( –1) – 8 – 8 –1 –1 x x x x x x ε ε + < ⇔ < 2(7 – 3) – 8x ε⇔ < 14( –1)x ε⇔ < 14 –1x ε⇔ < –1 14 x ε ⇔ < ; 0 –1 14 x ε δ δ= < < 2 14 – 20 6 2(7 – 3)( –1) – 8 – 8 –1 –1 x x x x x x + = 2(7 – 3) – 8x= 14( –1) 14 –1 14x x δ ε= = < = 19. 3 2 2 10 – 26 22 – 6 0 –1 – 4 ( –1) x x x x x δ ε + < < ⇒ < 3 2 2 10 – 26 22 – 6 – 4 ( –1) x x x x ε + < 2 2 (10 – 6)( –1) – 4 ( –1) x x x ε⇔ < 10 – 6 – 4x ε⇔ < 10( –1)x ε⇔ < 10 –1x ε⇔ < –1 10 x ε ⇔ < ; 0 –1 10 x ε δ δ= < < 3 2 2 2 2 10 – 26 22 – 6 (10 – 6)( –1) – 4 – 4 ( –1) ( –1) x x x x x x x + = 10 6 4 10( 1) 10 1 10 x x x δ ε = − − = − = − < = 20. 2 0 –1 (2 1) – 3x xδ ε< < ⇒ + < 2 2 2 1– 3 2 – 2 2 1 –1x x x x+ = = + To bound 2 2x + , agree that 1δ ≤ . –1x δ< implies 2 2 2 – 2 4x x+ = + 2 – 2 4x≤ + < 2 + 4 = 6 δ ε δ ε δ << ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ =≤ 1–0; 6 ,1min; 6 x 2 2 (2 1) – 3 2 – 2x x+ = 2 2 1 6 6 x x ε ε ⎛ ⎞ = + − < ⋅ =⎜ ⎟ ⎝ ⎠ © 2007 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.
  • Instructor’s Resource Manual Section 1.2 71 21. 2 0 1 ( – 2 –1) – 2x x xδ ε< + < ⇒ < 2 2 – 2 –1– 2 – 2 – 3 1 – 3x x x x x x= = + To bound – 3x , agree that 1δ ≤ . 1x δ+ < implies – 3 1– 4x x= + 1 –4x≤ + + < 1 + 4 = 5 ; min 1, ; 0 1 5 5 x ε ε δ δ δ ⎧ ⎫ ≤ = < + <⎨ ⎬ ⎩ ⎭ 2 2 ( – 2 –1) – 2 – 2 – 3x x x x= 1 – 3 5 5 x x ε ε= + < ⋅ = 22. 4 4 0 – 0x x xδ ε< < ⇒ = < 4 3 .x x x= To bound 3 x , agree that δ ≤ 1. x < δ ≤ 1 implies 33 1x x= ≤ so δ ≤ ε. 4 3 min{1, }; 0 1x x x xδ ε δ ε= < < ⇒ = < ⋅ = ε 23. Choose ε > 0. Then since lim ( ) , x c f x L → = there is some δ1 > 0 such that 10 – ( ) –x c f x Lδ ε< < ⇒ < . Since lim x→c f (x) = M, there is some δ2 > 0 such that 20 ( )x c f x Mδ ε< − < ⇒ − < . Let δ = min{δ1,δ2} and choose x0 such that 0 < x0 – c < δ . Thus, 0 0( ) – ( )f x L f x Lε ε ε< ⇒ − < − < 0 0( ) ( )f x L f xε ε⇒ − − < − < − + 0 0( ) ( ) .f x L f xε ε⇒ − < < + Similarly, 0 0( ) ( )f x M f xε ε− < < + . Thus, 2 2 .L Mε ε− < − < As ε ⇒ 0, L − M → 0, so L = M. 24. Since lim x→c G(x) = 0, then given any ε > 0, we can find δ > 0 such that whenever – , ( ) .x c G xδ ε< < Take any ε > 0 and the corresponding δ that works for G(x), then –x c δ< implies ( ) – 0 ( ) ( )F x F x G x ε= ≤ < since lim x→c G(x) = 0. Thus, lim x→c F( x) = 0. 25. For all 0x ≠ , 2 1 0 sin 1 x ⎛ ⎞ ≤ ≤⎜ ⎟ ⎝ ⎠ so 4 2 41 sinx x x ⎛ ⎞ ≤⎜ ⎟ ⎝ ⎠ for all 0x ≠ . By Problem 18, lim x→0 x 4 = 0, so, by Problem 20, lim x→0 x 4 sin 2 1 x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0. 26. εδ <==⇒<< xxxx 0–0 For x > 0, .)( 2 xx = 22 )( εε <=⇔< xxx εεδδεδ ==<⇒<<= 22 0; xx 27. εδ <⇒<<+ → 0–0:lim 0 xxx x For 0x ≥ , x x= . εδδεδ =<==⇒<<= xxxx 0–0; Thus, lim x→0 + x = 0. –0 lim :0 0 – – 0 x x x xδ ε → < < ⇒ < For x < 0, x = –x; note also that xx = since x ≥ 0. ;0 x x x xδ ε δ δ ε= < − < ⇒ = = − < = Thus, lim x→0 – x = 0, since – 00 0 lim lim 0, lim 0. xx x x x x + →→ → = = = 28. Choose ε > 0. Since lim x→a g( x) = 0 there is some δ1 > 0 such that 0 < x – a < δ1 ⇒ g(x ) − 0 < ε B . Let δ = min{1,δ1}, then ( )f x B< for x a δ− < or ( ) .x a f x Bδ− < ⇒ < Thus, ( ) ( ) 0 ( ) ( )x a f x g x f x g xδ− < ⇒ − = ( ) ( )f x g x B B ε ε= < ⋅ = so lim x→a f( x)g(x) = 0. © 2007 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.
  • 72 Section 1.3 Instructor’s Resource Manual 29. Choose ε > 0. Since lim x→a f( x) = L, there is a δ > 0 such that for 0 < x – a < δ, ( ) – .f x L ε< That is, for ora x a a x aδ δ− < < < < + , ( )L f x Lε ε− < < + . Let f(a) = A, { }max , , ,M L L Aε ε= − + c = a – δ, d = a + δ. Then for x in (c, d), ( ) ,f x M≤ since either x = a, in which case ( ) ( )f x f a A M= = ≤ or 0 –x a δ< < so ( )L f x Lε ε− < < + and ( ) .f x M< 30. Suppose that L > M. Then L – M = α > 0. Now take ε < α 2 and δ = min{δ1, δ2} where 10 – ( ) –x a f x Lδ ε< < ⇒ < and 20 – ( ) – .x a g x Mδ ε< < ⇒ < Thus, for 0 – ,x a δ< < L – ε < f(x) < L + ε and M – ε < g(x) < M + ε. Combine the inequalities and use the fact that ( ) ( )f x g x≤ to get L – ε < f(x) ≤ g(x) < M + ε which leads to L – ε < M + ε or L – M < 2ε. However, L – M = α > 2ε which is a contradiction. Thus L M≤ . 31. (b) and (c) are equivalent to the definition of limit. 32. For every ε > 0 and δ > 0 there is some x with 0 –x c δ< < such that f (x ) – L > ε. 33. a. g(x) = x3 – x2 – 2x – 4 x4 – 4x3 + x2 + x + 6 b. No, because x + 6 x4 – 4x3 + x2 + x + 6 + 1 has an asymptote at x ≈ 3.49. c. If δ ≤ 1 4 , then 2.75 < x < 3 or 3 < x < 3.25 and by graphing 3 2 4 3 2 2 4 ( ) 4 6 x x x y g x x x x x − − − = = − + + + on the interval [2.75, 3.25], we see that 3 2 4 3 2 – – 2 – 4 0 3 – 4 6 x x x x x x x < < + + + so m must be at least three. 1.3 Concepts Review 1. 48 2. 4 3. – 8; – 4 + 5c 4. 0 Problem Set 1.3 1. lim x→1 (2x + 1) 4 = lim x→1 2x + lim x→1 1 3 = 2 lim x →1 x + lim x→1 1 2,1 = 2(1) + 1 = 3 2. lim x→–1 (3x 2 – 1) 5 = lim x→–1 3x2 – lim x→–1 1 3 = 3 lim x→–1 x2 – lim x→–1 1 8 = 3 lim x→–1 x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 – lim x →–1 1 2, 1 = 3(–1) 2 –1 = 2 3. lim x→0 [(2x +1)( x – 3)] 6 = lim x→0 (2x +1) ⋅ lim x→0 (x – 3) 4, 5 = lim x→0 2x + lim x→0 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ lim x→0 x – lim x→0 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 = 2 lim x →0 x + lim x→0 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ lim x→0 x – lim x→0 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2, 1 = [2(0) +1](0 – 3) = –3 4. 2 2 2 lim [(2 1)(7 13)] x x x → + + 6 2 2 2 2 lim (2 1) lim (7 13) x x x x → → = + ⋅ + 4, 3 2 2 2 2 2 2 2 lim lim 1 7 lim lim 13 x x x x x x → → → → ⎛ ⎞ ⎛ ⎞ = + ⋅ +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 8,1 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = →→ 13lim71lim2 2 2 2 2 xx xx 2 ]13)2(7][1)2(2[ 22 ++= = 135 © 2007 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.
  • Instructor’s Resource Manual Section 1.3 73 5. 2 2 1 lim 5 – 3x x x→ + 7 )3–5(lim )12(lim 2 2 x x x x → → + = 4, 5 x x xx xx 3lim–5lim 1lim2lim 22 22 →→ →→ + = 3, 1 x x x x 2 2 lim3–5 1lim2 → → + = 2 = 2(2) +1 5 – 3(2) = –5 6. 3 2–3 4 1 lim 7 – 2x x x→ + 7 )2–7(lim )14(lim 2 3– 3 3– x x x x → → + = 4, 5 2 3–3– 3– 3 3– 2lim–7lim 1lim4lim x x xx xx →→ →→ + = 3, 1 2 3– 3 3– lim2–7 1lim4 x x x x → → + = 8 2 3– 3 3– lim2–7 1lim4 ⎟ ⎠ ⎞⎜ ⎝ ⎛ +⎟ ⎠ ⎞⎜ ⎝ ⎛ = → → x x x x 2 = 4(–3)3 + 1 7 – 2(–3)2 = 107 11 7. 3 lim 3 – 5 x x → 9 3 lim (3 – 5) x x → = 5, 3 3 3 3 lim – lim 5 x x x → → = 2, 1 3(3) – 5= = 2 8. 2 –3 lim 5 2 x x x → + 9 2 –3 lim (5 2 ) x x x → = + 4, 3 2 –3 –3 5 lim 2 lim x x x x → → = + 8 2 –3 –3 5 lim 2 lim x x x x → → ⎛ ⎞= +⎜ ⎟ ⎝ ⎠ 2 2 5(–3) 2(–3)= + 39= 9. lim t→–2 (2t3 +15)13 8 = lim t→–2 (2t3 + 15) ⎡ ⎣⎢ ⎤ ⎦⎥ 13 4, 3 = 2 lim t→–2 t3 + lim t→–2 15 ⎡ ⎣⎢ ⎤ ⎦⎥ 13 8 133 –2 –2 2 lim lim 15 t t t → → ⎡ ⎤⎛ ⎞= +⎢ ⎥⎜ ⎟ ⎝ ⎠⎢ ⎥⎣ ⎦ 2, 1 = [2(–2)3 +15]13 = –1 10. 3 2 –2 lim –3 7 w w w → + 9 3 2 –2 lim (–3 7 ) w w w → = + 4, 3 3 2 –2 –2 –3 lim 7 lim w w w w → → = + 8 3 2 –2 –2 –3 lim 7 lim w w w w → → ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 3 2 –3(–2) 7(–2)= + 2 13= 11. 1/33 2 4 8 lim 4y y y y→ ⎛ ⎞+ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ 9 1/33 2 4 8 lim 4y y y y→ ⎛ ⎞+ = ⎜ ⎟ ⎜ ⎟+⎝ ⎠ 7 1 33 2 2 lim (4 8 ) lim ( 4) y y y y y → → ⎡ ⎤+ ⎢ ⎥ = ⎢ ⎥+⎢ ⎥ ⎣ ⎦ 4, 3 1 33 2 2 2 2 4 lim 8 lim lim lim 4 y y y y y y y → → → → ⎛ ⎞+ ⎜ ⎟ = ⎜ ⎟ +⎜ ⎟ ⎝ ⎠ 8, 1 1/33 2 2 2 4 lim 8 lim lim 4 y y y y y y → → → ⎡ ⎤⎛ ⎞ ⎢ ⎥+⎜ ⎟ ⎢ ⎥⎝ ⎠= ⎢ ⎥+ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 2 1/33 4(2) 8(2) 2 4 ⎡ ⎤+ = ⎢ ⎥ +⎢ ⎥⎣ ⎦ = 2 © 2007 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.
  • 74 Section 1.3 Instructor’s Resource Manual 12. lim w→5 (2w4 – 9w3 +19)–1/2 4 35 1 lim 2 9 19w w w→ = − + 7 5 4 3 5 lim 1 lim 2 – 9 19 w w w w → → = + 1, 9 4 3 5 1 lim (2 – 9 19) w w w → = + 4,5 4 3 5 5 5 1 lim 2 lim 9 lim 19 w w w w w → → → = − + 1,3 4 3 5 5 1 2 lim 9 lim 19 w w w w → → = − + 8 4 3 5 5 1 2 lim 9 lim 19 w w w w → → = ⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 4 3 1 2(5) 9(5) 19 = − + 1 1 12144 = = 13. 2 22 4 lim 4x x x→ − + ( ) ( ) 2 2 2 2 lim 4 lim 4 x x x x → → − = + 4 4 0 4 4 − = = + 14. 2 2 5 6 lim 2x x x x→ − + − ( )( ) ( )2 3 2 lim 2x x x x→ − − = − ( )2 lim 3 1 x x → = − = − 15. 2 1 2 3 lim 1x x x x→− − − + ( )( ) ( )1 3 1 lim 1x x x x→− − + = + ( )1 lim 3 4 x x →− = − = − 16. 2 21 lim 1x x x x→− + + ( ) ( ) 2 1 2 1 lim lim 1 x x x x x →− →− + = + 0 0 2 = = 17. 1 1 ( 1)( 2)( 3) 3 lim lim ( 1)( 2)( 7) 7x x x x x x x x x x→− →− − − − − = − − + + 1 3 2 1 7 3 − − = = − − + 18. 2 2 2 7 10 ( 2)( 5) lim lim 2 2x x x x x x x x→ → + + + + = + + 2 lim( 5) 7 x x → = + = 19. 2 21 1 2 ( 2)( 1) lim lim ( 1)( 1)1x x x x x x x xx→ → + − + − = + −− 1 2 1 2 3 lim 1 1 1 2x x x→ + + = = = + + 20. lim x→–3 x 2 –14x – 51 x 2 – 4x – 21 = lim x→–3 (x + 3)(x – 17) (x + 3)(x – 7) = lim x→–3 x –17 x – 7 = –3 –17 –3 – 7 = 2 21. lim u→–2 u2 – ux + 2u – 2x u2 – u – 6 = lim u→–2 (u + 2)(u – x) (u + 2)(u – 3) = lim u→–2 u – x u – 3 = x + 2 5 22. lim x→1 x2 + ux – x – u x 2 +2x – 3 = lim x →1 (x – 1)(x + u) ( x –1)(x + 3) = lim x→1 x + u x + 3 = 1+ u 1+ 3 = u +1 4 23. lim x→π 2 x2 – 6xπ+ 4π2 x 2 – π 2 = lim x→π 2(x – π)( x – 2π) (x – π)( x + π) = lim x→π 2(x – 2π) x + π = 2(π – 2π) π + π = –1 24. lim w→–2 (w + 2)(w2 – w – 6) w2 + 4w + 4 = lim w→–2 (w + 2)2(w – 3) (w + 2)2 = lim w→–2 (w – 3) = –2 – 3 = –5 25. 2 2 lim ( ) ( ) x a f x g x → + 2 2 lim ( ) lim ( ) x a x a f x g x → → = + 2 2 lim ( ) lim ( ) x a x a f x g x → → ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 (3) (–1) 10= + = 26. )]()([lim )](3–)(2[lim )()( )(3–)(2 lim xgxf xgxf xgxf xgxf ax ax ax + = + → → → 2 9 )1(–3 )1(–3–)3(2 )(lim)(lim )(lim3–)(lim2 = + = + = →→ →→ xgxf xgxf axax axax © 2007 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.
  • Instructor’s Resource Manual Section 1.3 75 27. ]3)([lim)(lim]3)([)(lim 33 +⋅=+ →→→ xfxgxfxg axaxax )33(1–3lim)(lim)(lim 3 3 +⋅= ⎥⎦ ⎤ ⎢⎣ ⎡ +⋅= →→→ axaxax xfxg = –6 28. 4 4 lim[ ( ) – 3] lim ( ( ) – 3) x a x a f x f x → → ⎡ ⎤= ⎢ ⎥⎣ ⎦ 4 4 lim ( ) – lim 3 (3 – 3) 0 x a x a f x → → ⎡ ⎤= = =⎢ ⎥⎣ ⎦ 29. lim ( ) 3 ( ) lim ( ) 3 lim ( ) t a t a t a f t g t f t g t → → → ⎡ + ⎤ = +⎣ ⎦ lim ( ) 3 lim ( ) t a t a f t g t → → = + 3 3 –1 6= + = 30. lim u→a [ f (u) + 3g(u)]3 = lim u→a [ f(u) + 3g(u)] ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 = lim u→a f (u) + 3 lim u→a g(u) ⎡ ⎣⎢ ⎤ ⎦⎥ 3 = [3+ 3( –1)]3 = 0 31. lim x→2 3x 2 –12 x – 2 = lim x→2 3( x – 2)(x + 2) x – 2 = 3 lim x→2 (x + 2) = 3(2 + 2) = 12 32. 2 2 2 2 (3 2 1) –17 3 2 –16 lim lim – 2 – 2x x x x x x x x→ → + + + = 2 2 (3 8)( – 2) lim lim (3 8) – 2x x x x x x→ → + = = + 2 3 lim 8 3(2) 8 14 x x → = + = + = 33. 2– – lim 2– lim 2– – lim 2 2– 2 2 –2 2 2 11 2 xxx x x x x x x x x →→→ == 2 2 1 –1 –1 1 lim – – 2 2 lim 2(2) 4x x x x→ → = = = = 34. 2– lim 2– lim 2– – lim 22 2 2 4 )2–)(2(3– 2 4 )–4(3 2 4 33 2 xxx x xx x x x x x x + →→→ == 2 2 2 22 2 –3 lim 2 –3( 2) –3(2 2) lim 4 4(2) 4 lim x x x x x x x → → → ⎛ ⎞+⎜ ⎟+ +⎝ ⎠= = = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = – 3 4 35. Suppose lim x→c f (x) = L and lim x→c g(x) = M. MxgLLxfxgLMxgxf –)(–)()(–)()( +≤ as shown in the text. Choose ε1 = 1. Since lim ( ) , x c g x M → = there is some δ1 > 0 such that if 10 –x c δ< < , 1( ) – 1g x M ε< = or M – 1 < g(x) < M + 1 –1 1M M≤ + and M +1 ≤ M +1 so for 10 – , ( ) 1.x c g x Mδ< < < + Choose ε > 0. Since lim x→c f (x) = L and lim x→c g(x) = M, there exist δ2 and δ3 such that 20 –x c δ< < ⇒ ( ) – 1 f x L L M ε < + + and 30 –x c δ< < ⇒ ( ) – . 1 g x M L M ε < + + Let 1 2 3min{ , , },δ δ δ δ= then 0 –x c δ< < ⇒ MxgLLxfxgLMxgxf –)(–)()(–)()( +≤ ( )1 1 1 M L L M L M ε ε ε< + + = + + + + Hence, lim ( ) ( ) lim ( ) lim ( ) x c x c x c f x g x LM f x g x → → → ⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 36. Say ( )lim x c g x M → = , 0M ≠ , and choose 1 1 2 Mε = . There is some δ1 > 0 such that 1 1 1 0 ( ) 2 x c g x M Mδ ε< − < ⇒ − < = or ( ) 1 1 2 2 M M g x M M− < < + . 1 1 2 2 M M M− ≥ and 1 1 2 2 M M M+ ≥ so 1 ( ) 2 g x M> and 1 2 ( )g x M < Choose ε > 0. Since lim x→c g(x) = M there is δ2 > 0 such that 2 2 1 0 ( ) 2 x c g x M Mδ< − < ⇒ − < . Let δ = min{δ1, δ2}, then 1 1 – ( ) 0 – – ( ) ( ) M g x x c g x M g x M δ< < ⇒ = ( ) 2 2 2 1 2 2 1 ( ) ( ) 2 g x M g x M M M g x M M ε= − < − = ⋅ = ε © 2007 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.
  • 76 Section 1.3 Instructor’s Resource Manual Thus, lim x→c 1 g(x) = 1 M = 1 lim x→c g(x) . Using statement 6 and the above result, ( ) 1 lim lim ( ) lim ( ) ( )x c x c x c f x f x g x g x→ → → = ⋅ lim ( ) 1 lim ( ) . lim ( ) lim ( ) x c x c x c x c f x f x g x g x → → → → = ⋅ = 37. lim x→c f (x) = L ⇔ lim x→c f( x) = lim x→c L ⇔ lim x→c f (x) – lim x →c L = 0 ⇔ lim x→c [ f (x) – L] = 0 38. lim x→c f (x) = 0 ⇔ lim x→c f (x) ⎡ ⎣⎢ ⎤ ⎦⎥ 2 = 0 ⇔ lim x→c f 2 ( x) = 0 2 lim ( ) 0 x c f x → ⇔ = 2 lim ( ) 0 x c f x → ⇔ = ⇔ lim x→c f( x) = 0 39. 2 2 2 lim lim lim lim x c x c x c x c x x x x → → → → ⎛ ⎞= = =⎜ ⎟ ⎝ ⎠ 2 2 lim x c x c c → ⎛ ⎞= = =⎜ ⎟ ⎝ ⎠ 40. a. If 1 – 5 ( ) , ( ) – 2 – 2 x x f x g x x x + = = and c = 2, then lim x→c [ f (x) + g(x)] exists, but neither lim x→c f (x) nor lim x→c g(x) exists. b. If 2 ( ) , ( ) ,f x g x x x = = and c = 0, then lim x→c [ f (x) ⋅ g( x)] exists, but lim x→c f (x) does not exist. 41. –3 3 3 – 3 lim 0 –3x x x+→ + = = 42. 3 33 3 – (– ) lim 0 –x x x+→ π π + ππ + = = π 43. 2 223 3 – 3 ( – 3) – 9 lim lim – 9– 9x x x x x xx+ +→ → = 2 2 3 3 ( – 3) – 9 – 9 lim lim ( – 3)( 3) 3x x x x x x x x+ +→ → = = + + 2 3 – 9 0 3 3 = = + 44. –1 1 1 1 2 lim 4 4 4 4(1) 8x x x→ + + = = + + 45. 2 2 2 2 2 2 ( 1) (2 1) 2 5 2 2 lim 5(3 1) (3 2 1) 5x x x x+→ + + ⋅ = = = − ⋅ − 46. 3 3 3 lim ( ) lim lim 3 2 1 x x x x x x x − − −→ → → − = − = − = 47. –0 lim –1 x x x→ = 48. 2 2 3 lim 2 3 2 3 15 x x x +→ + = + ⋅ = 49. 1 ( ) ( ) 1; ( ) ( ) f x g x g x f x = = 1 lim ( ) 0 lim 0 ( )x a x a g x f x→ → = ⇔ = 1 0 lim ( ) x a f x → ⇔ = No value satisfies this equation, so lim x→a f( x) must not exist. 50. R has the vertices ± x 2 , ± 1 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Each side of Q has length 2 1x + so the perimeter of Q is 2 4 1.x + R has two sides of length 1 and two sides of length 2 x so the perimeter of R is 2 2 2 .x+ 2 20 0 perimeter of 2 2 lim lim perimeter of 4 1x x R x Q x+ +→ → + = + 2 2 2 0 2 2 1 4 24 0 1 + = = = + © 2007 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.
  • Instructor’s Resource Manual Section 1.4 77 51. a. 2 2 (0 – 0) (1– 0) 1NO = + = 2 2 2 2 ( – 0) ( – 0)OP x y x y= + = + 2 x x= + 2 2 2 2 ( – 0) ( –1) – 2 1NP x y x y y= + = + + 2 2 1x x x= + − + 2 2 (1– 0) (0 – 0) 1MO = + = 2 2 2 2 ( –1) ( – 0) – 2 1MP x y y x x= + = + + 2 1x x= − + 0 perimeter of lim perimeter ofx NOP MOP+→ Δ Δ 2 2 2 20 1 – 2 1 lim 1 – 1x x x x x x x x x x+→ + + + + + = + + + + 1 1 1 1 1 + = = + b. Area of ΔNOP = 1 2 (1)( x) = x 2 Area of 1 (1)( ) 2 2 x MOP yΔ = = 2 0 0 0 2 area of lim lim lim area of x xx x x NOP x MOP x+ + +→ → → Δ = = Δ 0 lim 0 x x +→ = = 1.4 Concepts Review 1. 0 2. 1 3. the denominator is 0 when 0=t . 4. 1 Problem Set 1.4 1. 0 cos 1 lim 1 1 1x x x→ = = + 2. / 2 lim cos 0 0 2θ π π θ θ → = ⋅ = 3. 2 2 0 cos cos 0 1 lim 1 1 sin 1 sin 0 1 0t t t→ = = = + + + 4. 0 0 0 3 tan 3 (sin / cos ) 3 lim lim lim sin sin cos 0 0 1 x x x x x x x x x x x x→ → → = = = = 5. 2 1 1 2 1sin lim 2 1 2 sin lim 00 =⋅== →→ x x x x xx 6. 0 0 0 sin3 3 sin3 3 sin3 lim lim lim 2 2 3 2 3θ θ θ θ θ θ θ θ θ→ → → = ⋅ = = 3 2 ⋅1 = 3 2 7. θ θθθ θ θ θ θ θθθ sin 3sincos lim 3sin lim tan 3sin lim 0 cos sin00 →→→ == ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⋅⋅⋅= → θ θθ θ θ θ sin0 1 3 3sin 3coslim sin0 sin3 1 3 lim cos 3 1 1 1 3 3 θθ θ θ θ θ→ ⎡ ⎤ = ⋅ ⋅ = ⋅ ⋅ ⋅ =⎢ ⎥ ⎢ ⎥⎣ ⎦ 8. θθ θ θθ θ θ θ θ θθ 2sin5cos 5sin lim 2sin lim 2sin 5tan lim 0 5cos 5sin 00 →→→ == 0 1 sin5 1 2 lim 5 cos5 5 2 sin 2θ θ θ θ θ θ→ ⎡ ⎤ = ⋅ ⋅ ⋅ ⋅⎢ ⎥ ⎣ ⎦ 0 5 1 sin5 2 lim 2 cos5 5 sin 2θ θ θ θ θ θ→ ⎡ ⎤ = ⋅ ⋅⎢ ⎥ ⎣ ⎦ = 5 2 ⋅1⋅1⋅1 = 5 2 9. θ θ θ θθ θ θ θθ cos 2 sin cos 00 sin lim sec2 sincot lim π π →→ = π 0 cos sin cos lim 2sinθ θ θ θ θ→ π = π 0 cos cos sin 1 lim 2 sinθ θ θ θ θ θ θ→ π π⎡ ⎤ = ⋅ ⋅ ⋅⎢ ⎥π π⎣ ⎦ 0 1 sin lim cos cos 2 sinθ θ θ θ θ θ θ→ π⎡ ⎤ = π ⋅ ⋅⎢ ⎥π π⎣ ⎦ = 1 2π ⋅1⋅1⋅1⋅1 = 1 2π 10. 2 0 0 sin 3 9 sin3 sin3 lim lim 0 1 1 0 2 2 3 3t t t t t t t t t→ → = ⋅ ⋅ = ⋅ ⋅ = 11. 2 2 20 0 20 tan 3 sin 3 lim lim 2 (2 )(cos 3 ) 3(sin3 ) sin3 lim 0 1 0 32cos 3 t t t t t t t t t t tt → → → = = ⋅ = ⋅ = 12. 0 1 0 12sin 2tan lim 0 = − = −→ t t t © 2007 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.
  • 78 Section 1.4 Instructor’s Resource Manual 13. 0 0 0 0 0 0 sin(3 ) 4 sin3 4 lim lim sec sec sec sin3 4 lim lim sec sec sin3 lim3cos lim4cos 3 3 1 4 7 t t t t t t t t t t t t t t t t t t t t t t t t t t → → → → → → + ⎛ ⎞ = +⎜ ⎟ ⎝ ⎠ = + = ⋅ + = ⋅ + = 14. 2sin sin sin lim lim 20 0 sin sin lim lim 1 1 1 0 0 θ θ θ θ θθ θθ θ θ θ θθ θ = → → = × = × = → → 15. ( )0 lim sin 1/ 0 x x x → = 16. ( )2 0 lim sin 1/ 0 x x x → = 17. ( )2 0 lim 1 cos / 0 x x x → − = 18. 2 0 lim cos 1 x x → = 19. 0 sin lim1 2 x x x→ + = 20. The result that 0 lim cos 1 t t → = was established in the proof of the theorem. Then 0 0 0 0 0 lim cos lim cos( ) lim (cos cos sin sin ) lim cos lim cos sin lim sin cos t c h h h h h t c h c h c h c h c h c → → → → → → = + = − = − = 21. c c c t t t t t ct ct ctct tan cos sin coslim sinlim cos sin limtanlim ==== → → →→ c c c t t t t t ct ct ctct cot sin cos sinlim coslim sin cos limcotlim ==== → → →→ 22. c ct t ctct sec cos 1 cos 1 limseclim === →→ c ct t ctct csc sin 1 sin 1 limcsclim === →→ 23. sin , cosBP t OB t= = area( ) area (sector )OBP OAPΔ ≤ area ( ) area( )OBP ABPQ≤ Δ + 21 1 1 (1) (1– ) 2 2 2 OB BP t OB BP OB BP⋅ ≤ ≤ ⋅ + 1 1 1 sin cos sin cos (1– cos )sin 2 2 2 t t t t t t t≤ ≤ + cos 2 – cos sin t t t t ≤ ≤ 1 sin 1 2 – cos cos t t t t ≤ ≤ for 2 2 t π π − < < . 0 0 0 1 sin 1 lim lim lim 2 – cos cost t t t t t t→ → → ≤ ≤ 0 sin 1 lim 1 t t t→ ≤ ≤ Thus, 0 sin lim 1. t t t→ = © 2007 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.
  • Instructor’s Resource Manual Section 1.5 79 24. a. Written response b. 1 1 (1 cos )sin 2 2 sin (1 cos ) 2 D AB BP t t t t = ⋅ = − − = 21 1 sin cos (1) – – 2 2 2 2 t t t E t OB BP= ⋅ = sin (1– cos ) – sin cos D t t E t t t = c. 0 lim 0.75 t D E+→ ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 1.5 Concepts Review 1. x increases without bound; f(x) gets close to L as x increases without bound 2. f(x) increases without bound as x approaches c from the right; f(x) decreases without bound as x approaches c from the left 3. y = 6; horizontal 4. x = 6; vertical Problem Set 1.5 1. 5 1 lim lim 1 – 5 1–x x x x x→∞ →∞ = = 2. 12 3 5 3 lim lim 0 –15 – x x x x x x→∞ →∞ = = 3. 2 2 7– – 2 1 lim lim 1 17t t t t t→ ∞ → ∞ = = − −− 4. 5– – 1 lim lim 1 – 5 1–t t t t t→ ∞ → ∞ = = 5. 2 2 2 lim lim ( – 5)(3 – ) 8 15x x x x x x x x→∞ →∞ = − + − 1 lim –1 8 151 2 x x x = = → ∞ − + − 6. 2 1 lim lim 1 2 8 151–– 8 15 2 x x xx x x x = = → ∞ → ∞ ++ 7. 2 1 –2 1 lim 100–2 lim 10023 3 == ∞→∞→ x xx xx x 8. 5 5 4 5– – lim lim 1–– 5θ θ θ θ θ θ→ ∞ → ∞ π π = = π 9. 13 2 3 –3 – 3 lim lim 3 2 5–– 5 x x x x xx x x = = π→ ∞ → ∞ ππ 10. 2 2 2 sin lim ; 0 sin 1 – 5θ θ θ θ→∞ ≤ ≤ for all θ and 1 2 2 5 2 1 lim lim 0 1–– 5 θ θ θ θ θ→∞ →∞ = = so 2 2 sin lim 0 – 5θ θ θ→∞ = 11. 3 3/ 2 3/ 23 3 3 3 3 lim lim 22x x x x x x xx→∞ →∞ + + = 2 3 2 3 lim 3 = + = ∞→ x x 12. 3 3 3 3 3 3 3 3 lim lim 2 7 2 7x x x x x x x x x x→∞ →∞ π + π + = + + 33 7 3 22 lim 2 2 π = + +π = ∞→ x x x 13. 2 2 3 3 2 2 1 8 1 8 lim lim 4 4x x x x x x→∞ →∞ + + = + + 1 2 3 3 4 2 8 lim 8 2 1 x x x →∞ + = = = + 14. 2 2 2 3 3 lim lim ( –1)( 1) –1x x x x x x x x x→∞ →∞ + + + + = + 31 2 1 2 1 lim 1 1 1– x x x x →∞ + + = = = 15. 1 1 lim lim 12 1 22 n n n n n →∞ →∞ = = + + © 2007 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.
  • 80 Section 1.5 Instructor’s Resource Manual 16. 2 2 2 1 1 lim lim 1 1 1 01 1 n n n n n →∞ →∞ = = = ++ + 17. 2 lim lim lim 1 11 1 01 lim 1 n n n n nn n n n n →∞ →∞ →∞ →∞ ∞ = = = = ∞ + +⎛ ⎞+ +⎜ ⎟ ⎝ ⎠ 18. 2 2 1 0 lim lim 0 1 1 01 1 n n n n n n →∞ →∞ = = = ++ + 19. For x > 0, 2 .x x= 2 2 2 3 1 3 1 2 1 2 lim 2 lim 3 12 lim x x x x x x xx x x + + = + = + + ∞→+∞→∞→ 2 2 1 = = 20. 2 1 2 1 22 4 4 2 1 lim lim lim 0 4 1 1 x x xx x x x x x x x + →∞ →∞ →∞ + + = = = + + + 21. 2 2 lim 2 3 – 2 – 5 x x x →∞ ⎛ ⎞+⎜ ⎟ ⎝ ⎠ 2 2 2 2 2 2 2 3 – 2 – 5 2 3 2 – 5 lim 2 3 2 – 5x x x x x x x→∞ ⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠= + + 2 2 2 2 2 3 – (2 – 5) lim 2 3 2 – 5x x x x x→∞ + = + + 8 2 22 2 2 3 2 –5 2 8 lim lim 2 3 2 5 x x x x x x x x→∞ →∞ + + = = + + − 2 2 8 3 5 lim 2 2 – x x x x →∞ = + + = 0 22. 2 lim 2 x x x x →∞ ⎛ ⎞+ −⎜ ⎟ ⎝ ⎠ 2 2 2 2 – 2 lim 2x x x x x x x x x x→∞ ⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠= + + 2 2 2 2 2 – 2 lim lim 2 2x x x x x x x x x x x x→∞ →∞ + = = + + + + 2 2 2 lim 1 21 1x x →∞ = = = + + 23. 1 3 2 2 2 2– – 2 9 9 1 lim lim – 1–– 2 2 y y y y y y y y y→ ∞ → ∞ + + = = ∞ ++ 24. –1 0 1 –1 –1 0 1 –1 lim n n n n n nx n n a x a x a x a b x b x b x b→∞ + +…+ + + +…+ + –11 0 –1 0 –11 00 –1 lim a aa n n x n nx x b bb n nx x n nx x a a bb→∞ + +…+ + = = + +…+ + 25. 2 2 1 1 lim lim 1 1 1 01 1 n n n n n →∞ →∞ = = = ++ + 26. 2 2 3/ 2 3 2 3 lim lim 12 12 1 1 n n n n n n n n n →∞ →∞ ∞ = = = ∞ + + + + 27. As 4 , 4x x+ → → while x – 4 → 0 + . 4 lim – 4x x x+→ = ∞ 28. 2 –3 –3 – 9 ( 3)( – 3) lim lim 3 3t t t t t t t+ +→ → + = + + –3 lim ( – 3) –6 t t +→ = = 29. As – 2 3 , 9t t→ → while 2 9 – 0 .t + → 2 2–3 lim 9 –t t t→ = ∞ 30. As 2 2/33 5 , 5x x + → → while 3 – 5 – 0 .x → 3 2 3 5 lim – 5 –x x x+ → = ∞ 31. As – 2 – 5 , 25, – 5 0 ,x x x→ → → and 3 – x → –2. 2 –5 lim ( – 5)(3 – )x x x x→ = ∞ 32. As 2 2 ,θ θ+ → π → π while sinθ → 0− . 2 lim sinθ θ θ+→π = −∞ © 2007 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.
  • Instructor’s Resource Manual Section 1.5 81 33. As 3 3 , 27,x x− → → while x − 3 → 0− . 3 3 lim 3x x x−→ = −∞ − 34. As 2 , 2 2 θ θ + π π → π → while cos θ → 0– . 2 lim – cos θ θ θ+π→ π = ∞ 35. 2 – –3 3 – – 6 ( 2)( – 3) lim lim – 3 – 3x x x x x x x x→ → + = –3 lim ( 2) 5 x x → = + = 36. 2 2 2 2 2 – 8 ( 4)( – 2) lim lim ( 2)( – 2)– 4x x x x x x x xx+ +→ → + + = + 2 4 6 3 lim 2 4 2x x x+→ + = = = + 37. For 0 1x≤ < , 0x = , so for 0 < x < 1, 0 x x = thus 0 lim 0 x x x+→ = 38. For 1 0x− ≤ < , 1x = − , so for –1 < x < 0, 1x x x = − thus 0 lim . x x x−→ = ∞ (Since x < 0, – 1 x > 0. ) 39. For x < 0, – ,x x= thus – –0 0 – lim lim –1 x x x x x x→ → = = 40. For x > 0, ,x x= thus 0 0 lim lim 1 x x x x x x+ +→ → = = 41. As – 0 ,1 cos 2x x→ + → while – sin 0 .x → –0 1 cos lim – sinx x x→ + = ∞ 42. –1 sin 1x≤ ≤ for all x, and 1 sin lim 0, so lim 0. x x x x x→∞ →∞ = = 43. – 3 3 lim 0, lim 0; 1 1x xx x→∞ → ∞ = = + + Horizontal asymptote y = 0. ––1 –1 3 3 lim , lim – ; 1 1x xx x+→ → = ∞ = ∞ + + Vertical asymptote x = –1 44. 2 2– 3 3 lim 0, lim 0; ( 1) ( 1)x xx x→∞ → ∞ = = + + Horizontal asymptote y = 0. 2 2––1 –1 3 3 lim , lim ; ( 1) ( 1)x xx x+→ → = ∞ = ∞ + + Vertical asymptote x = –1 45. 3 2 2 lim lim 2, – 3 1–x x x x x→∞ →∞ = = 3 2 2 lim lim 2, – 3 1–x x x x x→−∞ →−∞ = = Horizontal asymptote y = 2 –3 3 2 2 lim , lim – ; – 3 – 3x x x x x x+→ → = ∞ = ∞ Vertical asymptote x = 3 © 2007 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.
  • 82 Section 1.5 Instructor’s Resource Manual 46. 2 2– 3 3 lim 0, lim 0; 9 – 9 –x xx x→∞ → ∞ = = Horizontal asymptote y = 0 2 2–3 3 3 3 lim – , lim , 9 – 9 –x xx x+→ → = ∞ = ∞ 2 2––3 –3 3 3 lim , lim – ; 9 – 9 –x xx x+→ → = ∞ = ∞ Vertical asymptotes x = –3, x = 3 47. 2 2– 14 14 lim 0, lim 0; 2 7 2 7x xx x→∞ → ∞ = = + + Horizontal asymptote y = 0 Since 2x 2 + 7 > 0 for all x, g(x) has no vertical asymptotes. 48. 2 5 2 2 2 2 lim lim 2, 115x x x x x→∞ →∞ = = = ++ 2 5– – 2 2 2 2 lim lim –2 – 1– 15x x x x x→ ∞ → ∞ = = = ++ Since 2 5 0x + > for all x, g(x) has no vertical asymptotes. 49. f (x) = 2x + 3 – 1 x3 – 1 , thus 3 1 lim [ ( ) – (2 3)] lim – 0 –1x x f x x x→∞ →∞ ⎡ ⎤ + = =⎢ ⎥ ⎣ ⎦ The oblique asymptote is y = 2x + 3. 50. 2 4 3 ( ) 3 4 – , 1 x f x x x + = + + thus 2 4 3 lim [ ( ) – (3 4)] lim – 1x x x f x x x→∞ →∞ +⎡ ⎤ + = ⎢ ⎥ +⎣ ⎦ 34 2 1 2 lim – 0 1 x x x x →∞ ⎡ ⎤+ ⎢ ⎥= = ⎢ ⎥+ ⎢ ⎥⎣ ⎦ . The oblique asymptote is y = 3x + 4. 51. a. We say that lim ( ) – x c f x +→ = ∞ if to each negative number M there corresponds a δ > 0 such that 0 < x – c < δ ⇒ f(x) < M. b. We say that – lim ( ) x c f x → = ∞ if to each positive number M there corresponds a δ > 0 such that 0 < c – x < δ ⇒ f(x) > M. 52. a. We say that lim ( ) x f x →∞ = ∞ if to each positive number M there corresponds an N > 0 such that N < x ⇒ f(x) > M. b. We say that lim x→–∞ f (x ) = ∞ if to each positive number M there corresponds an N < 0 such that x < N ⇒ f(x) > M. 53. Let ε > 0 be given. Since lim x→∞ f (x ) = A, there is a corresponding number M1 such that 1 ( ) – . 2 x M f x A ε > ⇒ < Similarly, there is a number M2 such that 2 ( ) – . 2 x M g x B ε > ⇒ < Let 1 2max{ , }M M M= , then ( ) ( ) – ( )x M f x g x A B> ⇒ + + ( ) – ( ) – ( ) – ( ) –f x A g x B f x A g x B= + ≤ + 2 2 ε ε ε< + = Thus, lim [ ( ) ( )] x f x g x A B →∞ + = + 54. Written response © 2007 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.
  • Instructor’s Resource Manual Section 1.5 83 55. a. lim x→∞ sin x does not exist as sin x oscillates between –1 and 1 as x increases. b. Let u = 1 x , then as , 0 .x u +→ ∞ → 0 1 lim sin lim sin 0 x u u x +→∞ → = = c. Let 1 ,u x = then as , 0 .x u +→ ∞ → 0 0 1 1 sin lim sin lim sin lim 1 x u u u x u x u u+ +→∞ → → = = = d. Let u = 1 x , then u ux x ux sin 1 lim 1 sinlim 2/3 0 2/3 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = + →∞→ ∞= ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = + → u u uu sin1 lim 0 e. As x → ∞, sin x oscillates between –1 and 1, while –1/ 2 1 0.x x = → –1/ 2 lim sin 0 x x x →∞ = f. Let u = 1 x , then ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + π =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + π + →∞→ u x ux 6 sinlim 1 6 sinlim 0 1 sin 6 2 π = = g. As 1 , ,x x x → ∞ + → ∞ so 1 lim sin x x x→∞ ⎛ ⎞ +⎜ ⎟ ⎝ ⎠ does not exist. (See part a.) h. 1 1 1 sin sin cos cos sinx x x x x x ⎛ ⎞ + = +⎜ ⎟ ⎝ ⎠ 1 lim sin – sin x x x x→∞ ⎡ ⎤⎛ ⎞ +⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦ 1 1 lim sin cos –1 cos sin x x x x x→∞ ⎡ ⎤⎛ ⎞ = +⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦ As 1 , cos 1x x → ∞ → so 1 cos –1 0. x → From part b., lim x→∞ sin 1 x = 0. As x → ∞ both sin x and cos x oscillate between –1 and 1. 1 lim sin – sin 0. x x x x→∞ ⎡ ⎤⎛ ⎞ + =⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦ 56. ∞= − = −− →→ 22 0 /1 lim)(lim cv m vm cvcv 57. lim x→∞ 3x2 + x +1 2x2 –1 = 3 2 58. 2 2– 2 – 3 2 lim 55 1x x x x→ ∞ = + 59. 2 2 – 3 lim 2 3 – 2 – 5 – 2 2x x x x → ∞ ⎛ ⎞+ =⎜ ⎟ ⎝ ⎠ 60. 2 2 1 2 lim 33 1x x x→∞ + = + 61. 10 1 lim 1 1 x x→∞ ⎛ ⎞ + =⎜ ⎟ ⎝ ⎠ 62. 1 lim 1 2.718 x x e x→∞ ⎛ ⎞ + = ≈⎜ ⎟ ⎝ ⎠ 63. 2 1 lim 1 x x x→∞ ⎛ ⎞ + = ∞⎜ ⎟ ⎝ ⎠ 64. sin 1 lim 1 1 x x x→∞ ⎛ ⎞ + =⎜ ⎟ ⎝ ⎠ 65. –3 sin – 3 lim –1 – 3x x x→ = 66. –3 sin – 3 lim –1 tan( – 3)x x x→ = 67. –3 cos( – 3) lim – – 3x x x→ = ∞ 68. 2 2 cos lim –1 –x x x π+π→ = 69. 1 0 lim (1 ) 2.718x x x e +→ + = ≈ 70. 1/ 0 lim (1 ) x x x +→ + = ∞ 71. 0 lim (1 ) 1x x x +→ + = © 2007 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.
  • 84 Section 1.6 Instructor’s Resource Manual 1.6 Concepts Review 1. lim ( ) x c f x → 2. every integer 3. – lim ( ) ( ); lim ( ) ( ) x a x b f x f a f x f b +→ → = = 4. a; b; f(c) = W Problem Set 1.6 1. 3 lim[( – 3)( – 4)] 0 (3); x x x f → = = continuous 2. 2 3 lim ( – 9) 0 (3); x x g → = = continuous 3. 3 3 lim – 3x x→ and h(3) do not exist, so h(x) is not continuous at 3. 4. 3 lim – 4 t t → and g(3) do not exist, so g(t) is not continuous at 3. 5. 3 – 3 lim – 3t t t→ and h(3) do not exist, so h(t) is not continuous at 3. 6. h(3) does not exist, so h(t) is not continuous at 3. 7. 3 lim 3 (3); t t f → = = continuous 8. 3 lim – 2 1 (3); t t g → = = continuous 9. h(3) does not exist, so h(t) is not continuous at 3. 10. f(3) does not exist, so f(x) is not continuous at 3. 11. 3 2 3 3 – 27 ( – 3)( 3 9) lim lim – 3 – 3t t t t t t t t→ → + + = 2 2 3 lim( 3 9) (3) 3(3) 9 27 (3) t t t r → = + + = + + = = continuous 12. From Problem 11, 3 lim ( ) 27, t r t → = so r(t) is not continuous at 3 because 3 lim ( ) (3). t r t r → ≠ 13. 3 3 lim ( ) lim (3 – ) 0 t t f t t + +→ → = = – –3 3 lim ( ) lim ( – 3) 0 t t f t t → → = = 3 lim ( ) (3); t f t f → = continuous 14. 2 3 3 lim ( ) lim (3 – ) 0 t t f t t + +→ → = = 2 – –3 3 lim ( ) lim ( – 9) 0 t t f t t → → = = 3 lim ( ) (3); t f t f → = continuous 15. 3 lim ( ) 2 (3); t f x f → = − = continuous 16. g is discontinuous at x = –3, 4, 6, 8; g is left continuous at x = 4, 8; g is right continuous at x = –3, 6 17. h is continuous on the intervals [ ] [ ]( , 5), 5,4 , (4,6), 6,8 , (8, )−∞ − − ∞ 18. 2 7 7 7 – 49 ( – 7)( 7) lim lim lim ( 7) – 7 – 7x x x x x x x x x→ → → + = = + = 7 + 7 = 14 Define f(7) = 14. 19. 2 3 3 2 –18 2( 3)( – 3) lim lim 3 – 3 –x x x x x x x→ → + = 3 lim[–2( 3)] –2(3 3) –12 x x → = + = + = Define f(3) = –12. 20. 0 sin( ) lim 1 θ θ θ→ = Define g(0) = 1 21. 1 1 –1 ( –1)( 1) lim lim –1 ( –1)( 1)t t t t t t t t→ → + = + 1 1 –1 1 1 lim lim 2( –1)( 1) 1t t t t t t→ → = = = + + Define H(1) = 1 2 . 22. 4 2 2 2 –1 –1 2 – 3 ( –1)( 3) lim lim 1 1x x x x x x x x→ → + + = + + 2 –1 2 –1 ( 1)( –1)( 3) lim 1 lim [( –1)( 3)] x x x x x x x x → → + + = + = + 2 (–1–1)[(–1) 3] –8= + = Define φ(–1) = –8. © 2007 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.
  • Instructor’s Resource Manual Section 1.6 85 23. 2 –1 –1 –1 ( –1)( 1) lim sin lim sin 1 1x x x x x x x→ → ⎛ ⎞ +⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠ –1 lim sin( –1) sin(–1–1) sin( 2) –sin 2 x x → = = = − = Define F(–1) = –sin 2. 24. Discontinuous at ,30x π= 25. 2 33 – ( ) ( – )( – 3) x f x x x = π Discontinuous at 3,x π= 26. Continuous at all points 27. Discontinuous at all θ = nπ + π 2 where n is any integer. 28. Discontinuous at all 5u ≤ − 29. Discontinuous at u = –1 30. Continuous at all points 31. 1 ( ) (2 – )(2 ) G x x x = + Discontinuous on ( , 2] [2, )−∞ − ∪ ∞ 32. Continuous at all points since 0 lim ( ) 0 (0) x f x f → = = and 1 lim ( ) 1 (1). x f x f → = = 33. 0 lim ( ) 0 (0) x g x g → = = –1 1 lim ( ) 1, lim ( ) –1 x x g x g x +→ → = = lim x→1 g(x ) does not exist, so g(x) is discontinuous at x = 1. 34. Discontinuous at every integer 35. Discontinuous at t = 1 2 n + where n is any integer 36. 37. 38. 39. 40. Discontinuous at all points except x = 0, because lim ( ) ( ) x c f x f c → ≠ for 0c ≠ . lim ( ) x c f x → exists only at c = 0 and 0 lim ( ) 0 (0) x f x f → = = . 41. Continuous. 42. Discontinuous: removable, define (10) 20f = 43. Discontinuous: removable, define (0) 1f = 44. Discontinuous: nonremovable. 45. Discontinuous, removable, redefine (0) 1g = 46. Discontinuous: removable, define (0) 0F = © 2007 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.
  • 86 Section 1.6 Instructor’s Resource Manual 47. Discontinuous: nonremovable. 48. Discontinuous: removable, define (4) 4f = 49. The function is continuous on the intervals ( ]0,1 ,(1,2], (2,3], … Length of call in minutes Cost $ 0.24 0.36 0.12 0.48 0.60 0.72 1 2 3 4 5 6 50. The function is continuous on the intervals [0,200], (200,300], (300,400], … Cost $ 20 40 60 80 100 200 300 400 500 Miles Driven 51. The function is continuous on the intervals (0,0.25], (0.25,0.375], (0.375,0.5], … Cost $ 1 2 3 4 0.25 0.5 0.75 1 Miles Driven 52. Let 3 ( ) 3 2.f x x x= + − f is continuous on [0, 1]. f(0) = –2 < 0 and f(1) = 2 > 0. Thus, there is at least one number c between 0 and 1 such that x3 + 3x −2 = 0. 53. Because the function is continuous on[ ]0,2π and 3 5 (cos0)0 6sin 0 – 3 –3 0,+ = < 3 5 3 (cos2 )(2 ) 6sin (2 ) – 3 8 – 3 0,π π + π = π > there is at least one number c between 0 and 2π such that 3 5 (cos ) 6sin – 3 0.t t t+ = 54. Let ( ) 8147 23 −+−= xxxxf . f(x) is continuous at all values of x. f(0) = –8, f(5) = 12 Because 0 is between –8 and 12, there is at least one number c between 0 and 5 such that ( ) 08147 23 =−+−= xxxxf . This equation has three solutions (x = 1,2,4) 55. Let ( ) cos .f x x x= − . f(x) is continuous at all values of x ≥ 0. f(0) = –1, f(π/2) = / 2π Because 0 is between –1 and / 2π , there is at least one number c between 0 and π/2 such that ( ) cos 0.f x x x= − = The interval [0.6,0.7] contains the solution. 56. Let 5 3 ( ) 4 – 7 14f x x x x= + + f(x) is continuous at all values of x. f(–2) = –36, f(0) = 14 Because 0 is between –36 and 14, there is at least one number c between –2 and 0 such that 5 3 ( ) 4 – 7 14 0.f x x x x= + + = © 2007 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.
  • Instructor’s Resource Manual Section 1.6 87 57. Suppose that f is continuous at c, so lim ( ) ( ). x c f x f c → = Let x = c + t, so t = x – c, then as x → c , t → 0 and the statement lim ( ) ( ) x c f x f c → = becomes 0 lim ( ) ( ). t f t c f c → + = Suppose that lim t→0 f (t + c) = f (c) and let x = t + c, so t = x – c. Since c is fixed, t → 0 means that x → c and the statement 0 lim ( ) ( ) t f t c f c → + = becomes lim ( ) ( ) x c f x f c → = , so f is continuous at c. 58. Since f(x) is continuous at c, lim ( ) ( ) 0. x c f x f c → = > Choose ( )f cε = , then there exists a 0δ > such that ( ) ( )0 x c f x f cδ ε< − < ⇒ − < . Thus, ( ) ( ) ( )f x f c f cε− > − = − , or ( ) 0f x > . Since also ( ) 0f c > , ( ) 0f x > for all x in ( , ).c cδ δ− + 59. Let g(x) = x – f(x). Then, g(0) = 0 – f(0) = –f(0) ≤ 0 and g(1) = 1 – f(1) ≥ 0 since 0 ≤ f(x) ≤ 1 on [0, 1]. If g(0) = 0, then f(0) = 0 and c = 0 is a fixed point of f. If g(1) = 0, then f(1) = 1 and c = 1 is a fixed point of f. If neither g(0) = 0 nor g(1) = 0, then g(0) < 0 and g(1) > 0 so there is some c in [0, 1] such that g(c) = 0. If g(c) = 0 then c – f(c) = 0 or f(c) = c and c is a fixed point of f. 60. For f(x) to be continuous everywhere, f(1) = a(1) + b = 2 and f(2) = 6 = a(2) + b a + b = 2 2a + b = 6 – a = –4 a = 4, b = –2 61. For x in [0, 1], let f(x) indicate where the string originally at x ends up. Thus f(0) = a, f(1) = b. f(x) is continuous since the string is unbroken. Since 0 , 1a b≤ ≤ , f(x) satisfies the conditions of Problem 59, so there is some c in [0, 1] with f(c) = c, i.e., the point of string originally at c ends up at c. 62. The Intermediate Value Theorem does not imply the existence of a number c between –2 and 2 such that ( ) 0.f c = The reason is that the function ( )f x is not continuous on [ ]2,2 .− 63. Let f(x) be the difference in times on the hiker’s watch where x is a point on the path, and suppose 0x = at the bottom and x = 1 at the top of the mountain. So f(x) = (time on watch on the way up) – (time on watch on the way down). f(0) = 4 – 11 = –7, f(1) = 12 – 5 = 7. Since time is continuous, f(x) is continuous, hence there is some c between 0 and 1 where f(c) = 0. This c is the point where the hiker’s watch showed the same time on both days. 64. Let f be the function on ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 ,0 π such that f(θ) is the length of the side of the rectangle which makes angle θ with the x-axis minus the length of the sides perpendicular to it. f is continuous on ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 ,0 π . If f(0) = 0 then the region is circumscribed by a square. If f(0) ≠ 0, then observe that (0) . 2 f f π⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ Thus, by the Intermediate Value Theorem, there is an angle θ0 between 0 and 2 π such that 0( ) 0.f θ = Hence, D can be circumscribed by a square. 65. Yes, g is continuous at R . ( ) ( )2 lim lim r R r R GMm R g r g r= − +→ → = 66. No. By the Intermediate Value Theorem, if f were to change signs on [a,b], then f must be 0 at some c in [a,b]. Therefore, f cannot change sign. 67. a. f(x) = f(x + 0) = f(x) + f(0), so f(0) = 0. We want to prove that lim x→c f (x) = f (c), or, equivalently, lim x→c [ f (x) – f (c)] = 0. But f(x) – f(c) = f(x – c), so lim[ ( ) – ( )] lim ( – ). x c x c f x f c f x c → → = Let h = x – c then as x → c, h → 0 and 0 lim ( – ) lim ( ) (0) 0. x c h f x c f h f → → = = = Hence lim x→c f (x) = f (c) and f is continuous at c. Thus, f is continuous everywhere, since c was arbitrary. b. By Problem 43 of Section 0.5, f(t) = mt for all t in Q. Since g(t) = mt is a polynomial function, it is continuous for all real numbers. f(t) = g(t) for all t in Q, thus f(t) = g(t) for all t in R, i.e. ( ) .f t mt= © 2007 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.
  • 88 Section 1.7 Instructor’s Resource Manual 68. If f(x) is continuous on an interval then lim ( ) ( ) x c f x f c → = for all points in the interval: lim ( ) ( ) x c f x f c → = ⇒ lim ( ) x c f x → 2 2 lim ( ) lim ( ) x c x c f x f x → → ⎛ ⎞= = ⎜ ⎟ ⎝ ⎠ 2 ( ( )) ( )f c f c= = 69. Suppose 1 if 0 ( ) . 1 if 0 x f x x ≥⎧ = ⎨ − <⎩ f(x) is discontinuous at x = 0, but g(x) = ( )f x = 1 is continuous everywhere. 70. a. b. If r is any rational number, then any deleted interval about r contains an irrational number. Thus, if f (r) = 1 q , any deleted interval about r contains at least one point c such that 1 1 ( ) – ( ) – 0 .f r f c q q = = Hence, lim x→r f (x) does not exist. If c is any irrational number in (0, 1), then as p x c q = → (where p q is the reduced form of the rational number) ,q → ∞ so ( ) 0f x → as .x c→ Thus, lim ( ) 0 ( ) x c f x f c → = = for any irrational number c. 71. a. Suppose the block rotates to the left. Using geometry, 3 ( ) – . 4 f x = Suppose the block rotates to the right. Using geometry, 3 ( ) . 4 f x = If x = 0, the block does not rotate, so f(x) = 0. Domain: 3 3 – , ; 4 4 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ Range: 3 3 – , 0, 4 4 ⎧ ⎫ ⎨ ⎬ ⎩ ⎭ b. At x = 0 c. If 0, ( ) 0x f x= = , if 3 3 – , ( ) – 4 4 x f x= = and if 3 3 , ( ) , 4 4 x f x= = so 3 3 ,0, 4 4 x = − are fixed points of f. 1.7 Chapter Review Concepts Test 1. False. Consider ( )f x x= at 2.x = 2. False: c may not be in the domain of f(x), or it may be defined separately. 3. False: c may not be in the domain of f(x), or it may be defined separately. 4. True. By definition, where 0, 0.c L= = 5. False: If f(c) is not defined, lim x→c f (x) might exist; e.g., 2 – 4 ( ) . 2 x f x x = + f(–2) does not exist, but 2 2 – 4 lim 4. 2x x x→− = − + 6. True: 2 5 5 5 25 ( 5)( 5) lim lim 5 5 lim ( 5) 5 5 10 x x x x x x x x x → → → − − + = − − = + = + = 7. True: Substitution Theorem 8. False: 1 sin lim 0 = → x x x 9. False: The tangent function is not defined for all values of c. 10. True: If x is in the domain of sin tan cos x x x = , then cos 0x ≠ , and Theorem A.7 applies.. © 2007 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.
  • Instructor’s Resource Manual Section 1.7 89 11. True: Since both xsin and xcos are continuous for all real numbers, by Theorem C we can conclude that xxxf cossin2)( 2 −= is also continuous for all real numbers. 12. True. By definition, ( ) ( )lim x c f x f c → = . 13. True. [ ]2 1,3∈ 14. False: − →0 lim x may not exist 15. False: Consider .sin)( xxf = 16. True. By the definition of continuity on an interval. 17. False: Since 1sin1 ≤≤− x for all x and 0 1 lim = ∞→ xx , we get 0 sin lim = ∞→ x x x . 18. False. It could be the case where ( )lim 2 x f x →−∞ = 19. False: The graph has many vertical asymptotes; e.g., x = ± π/2, ± 3π/2, ± 5π/2, … 20. True: x = 2 ; x = –2 21. True: As + →1x both the numerator and denominator are positive. Since the numerator approaches a constant and the denominator approaches zero, the limit goes to + ∞ . 22. False: lim ( ) x c f x → must equal f(c) for f to be continuous at x = c. 23. True: lim ( ) lim ( ), x c x c f x f x f c → → ⎛ ⎞= =⎜ ⎟ ⎝ ⎠ so f is continuous at x = c. 24. True: ( ) 2.3 lim 1 2.3 2x x f → = = 25. True: Choose ε = 0. 001f (2) then since lim x→2 f( x) = f (2), there is some δ such that 0 < x − 2 < δ ⇒ f (x ) − f (2) < 0.001f (2), or −0. 001f (2) < f (x) − f(2) < 0.001f(2) Thus, 0.999f(2) < f(x) < 1.001f(2) and f(x) < 1.001f(2) for 0 < x − 2 < δ. Since f(2) < 1.001f(2), as f(2) > 0, f(x) < 1.001f(2) on (2 −δ , 2 + δ ). 26. False: That lim x→c [ f (x) + g(x)] exists does not imply that lim x→c f (x) and lim x→c g(x) exist; e.g., – 3 ( ) 2 x f x x = + and 7 ( ) 2 x g x x + = + for 2c = − . 27. True: Squeeze Theorem 28. True: A function has only one limit at a point, so if lim x→a f( x) = L and lim x→a f( x) = M, L M= 29. False: That f(x) ≠ g(x) for all x does not imply that lim ( ) lim ( ). x c x c f x g x → → ≠ For example, if 2 – 6 ( ) – 2 x x f x x + = and 5 ( ) , 2 g x x= then f(x) ≠ g(x) for all x, but lim x→2 f( x) = lim x→2 g(x) = 5. 30. False: If f(x) < 10, lim x→2 f( x) could equal 10 if there is a discontinuity point (2, 10). For example, 3 2 – 6 2 12 ( ) 10 – 2 x x x f x x + − − = < for all x, but 2 lim ( ) 10. x f x → = 31. True: 2 2 2 lim ( ) lim ( ) lim ( ) ( ) x a x a x a f x f x f x b b → → → = ⎡ ⎤= = =⎢ ⎥⎣ ⎦ 32. True: If f is continuous and positive on [a, b], the reciprocal is also continuous, so it will assume all values between 1 f (a) and 1 f (b) . © 2007 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.
  • 90 Section 1.7 Instructor’s Resource Manual Sample Test Problems 1. 2 2 2 2 0 lim 0 2 2 2 4x x x→ − − = = = + + 2. 2 2 1 –1 1 1 lim 0 1 1 1u u u→ − = = + + 3. 2 1 1 1 –1 ( –1)( 1) lim lim lim( 1) –1 –1u u u u u u u u u→ → → + = = + = 1 + 1 = 2 4. 21 1 1 1 1 1 lim lim lim ; ( 1)( –1) –1–1u u u u u u u uu→ → → + + = = + does not exist 5. –22 22 2 2 1– 1 lim lim lim ( – 2)( 2) ( 2)– 4 x x x x x xx x x xx→ → → = = + + = 1 2(2 + 2) = 1 8 6. 2 22 2 – 4 ( 2)( – 2) lim lim ( 3)( – 2)– 6z z z z z z zz z→ → + = ++ = lim z→2 z + 2 z + 3 = 2 + 2 2 +3 = 4 5 7. xxxx x x x x xx 20 cos sin 00 cos2 1 lim cossin2 lim 2sin tan lim →→→ == = 1 2 cos2 0 = 1 2 8. 3 2 21 1 –1 ( –1)( 1) lim lim ( –1)( 1)–1y y y y y y y yy→ → + + = + 2 2 1 1 1 1 1 3 lim 1 1 1 2y y y y→ + + + + = = = + + 9. 4 4 – 4 ( – 2)( 2) lim lim – 2 – 2x x x x x x x→ → + = 4 lim ( 2) 4 2 4 x x → = + = + = 10. 0 cos lim x x x→ does not exist. 11. – – –0 0 0 – lim lim lim (–1) –1 x x x x x x x→ → → = = = 12. (1/ 2) lim 4 2 x x +→ = 13. ( )– – –2 2 2 lim lim lim 1 2 1 t t t t t t t → → → − = − = − = − 14. 1 1 1 1 lim lim 1 1 1x x x x x x− −→ → − − = = − − − since 1 0x − < as 1x − → 15. 0 0 0 sin5 5 sin5 lim lim 3 3 5 5 sin5 5 5 lim 1 3 5 3 3 x x x x x x x x x → → → = = = × = 16. 0 0 0 1 cos2 2 1 cos2 lim lim 3 3 2 2 1 cos2 2 lim 0 0 3 2 3 x x x x x x x x x → → → − − = − = = × = 17. 1 1 1 1 0 lim lim 1 22 1 01 x x x x x x →∞ →∞ − − + = = = + ++ 18. Since 1 sin 1t− ≤ ≤ for all t and 1 lim 0 t t→∞ = , we get sin lim 0 t t t→∞ = . 19. ( )22 2 lim 2t t t→ + = ∞ − because as 0,t → 2 4t + → while the denominator goes to 0 from the right. 20. 0 cos lim x x x+→ = ∞ , because as 0x + → , cos 1x → while the denominator goes to 0 from the right. 21. / 4 lim tan 2 x x π −→ = ∞ because as ( )/ 4 ,x π − → ( )2 / 2x π − → , so tan 2 .x → ∞ 22. 0 1 sin lim x x x+→ + = ∞ , because as 0x + → , 1 sin 1x+ → while the denominator goes to 0 from the right. 23. Preliminary analysis: Let ε > 0. We need to find a δ > 0 such that ( )0 | 3| | 2 1 7 | . | 2 6 | 2 | 3| | 3| . Choose . 2 2 x x x x x δ ε ε ε ε ε δ < − < ⇒ + − < − < ⇔ − < ⇔ − < = Let ε > 0. Choose / 2.δ ε= Thus, ( ) ( )2 1 7 2 6 2 3 2 / 2 .x x x ε ε+ − = − = − < = © 2007 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.
  • Instructor’s Resource Manual Section 1.7 91 24. a. f(1) = 0 b. 1 1 lim ( ) lim (1– ) 0 x x f x x + +→ → = = c. – –1 1 lim ( ) lim 1 x x f x x → → = = d. –1 lim ( ) –1 x f x → = because 3 – ––1 –1 lim ( ) lim –1 x x f x x → → = = and –1 –1 lim ( ) lim –1 x x f x x + +→ → = = 25. a. f is discontinuous at x = 1 because f(1) = 0, but lim x→1 f (x ) does not exist. f is discontinuous at x = –1 because f(–1) does not exist. b. Define f(–1) = –1 26. a. 0 – ( ) –u a g u Mδ ε< < ⇒ < b. 0 – ( ) –a x f x Lδ ε< < ⇒ < 27. a. 3 lim[2 ( ) – 4 ( )] x f x g x → 3 3 2 lim ( ) – 4 lim ( ) x x f x g x → → = = 2(3) – 4(–2) = 14 b. 2 3 3 – 9 lim ( ) lim ( )( 3) – 3x x x g x g x x x→ → = + 3 3 lim ( ) lim ( 3) –2 (3 3) –12 x x g x x → → = ⋅ + = ⋅ + = c. g(3) = –2 d. 3 3 lim ( ( )) lim ( ) (3) –2 x x g f x g f x g → → ⎛ ⎞= = =⎜ ⎟ ⎝ ⎠ e. 2 3 lim ( ) – 8 ( ) x f x g x → 2 3 3 lim ( ) – 8 lim ( ) x x f x g x → → ⎡ ⎤= ⎢ ⎥⎣ ⎦ 2 (3) – 8(–2) 5= = f. 3 ( ) – (3) –2 – (3) 2 ( 2) lim ( ) 3 3x g x g g f x→ − − − = = = 0 28. 29. a(0) + b = –1 and a(1) + b = 1 b = –1; a + b = 1 a – 1 = 1 a = 2 30. Let 5 3 ( ) – 4 – 3 1f x x x x= + f(2) = –5, f(3) = 127 Because f(x) is continuous on [2, 3] and f(2) < 0 < f(3), there exists some number c between 2 and 3 such that f(c) = 0. 31. Vertical: None, denominator is never 0. Horizontal: 2 2 lim lim 0 1 1x x x x x x→∞ →−∞ = = + + , so 0y = is a horizontal asymptote. 32. Vertical: None, denominator is never 0. Horizontal: 2 2 2 2 lim lim 1 1 1x x x x x x→∞ →−∞ = = + + , so 1y = is a horizontal asymptote. 33. Vertical: 1, 1x x= = − because 2 2 1 lim 1x x x+→ = ∞ − and 2 2 1 lim 1x x x−→− = ∞ − Horizontal: 2 2 2 2 lim lim 1 1 1x x x x x x→∞ →−∞ = = − − , so 1y = is a horizontal asymptote. © 2007 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.
  • 92 Review and Preview Instructor’s Resource Manual 34. Vertical: 2, 2x x= = − because 3 2 2 lim 4x x x+→ = ∞ − and 3 2 2 lim 4x x x−→− = ∞ − Horizontal: 3 2 lim 4x x x→∞ = ∞ − and 3 2 lim 4x x x→−∞ = −∞ − , so there are no horizontal asymptotes. 35. Vertical: / 4, 3 / 4, 5 / 4,x π π π= ± ± ± … because / 4 lim tan 2 x x π −→ = ∞ and similarly for other odd multiples of / 4.π Horizontal: None, because lim tan 2 x x →∞ and lim tan 2 x x →−∞ do not exist. 36. Vertical: 0,x = because 2 0 0 sin 1 sin lim lim x x x x x xx+ +→ → = = ∞ . Horizontal: 0,y = because 2 2 sin sin lim lim 0. x x x x x x→∞ →−∞ = = Review and Preview Problems 1. a. ( ) 2 2 2 4f = = b. ( ) 2 2.1 2.1 4.41f = = c. ( ) ( )2.1 2 4.41 4 0.41f f− = − = d. ( ) ( )2.1 2 0.41 4.1 2.1 2 0.1 f f− = = − e. ( ) ( )2 2 2 2f a h a h a ah h+ = + = + + f. ( ) ( ) 2 2 2 2 2 2 f a h f a a ah h a ah h + − = + + − = + g. ( ) ( ) ( ) 2 2 2 f a h f a ah h a h a h a h + − + = = + + − h. ( ) ( ) ( ) ( )0 0 lim lim 2 2 h h f a h f a a h a a h a→ → + − = + = + − 2. a. ( )2 1/ 2g = b. ( )2.1 1/ 2.1 0.476g = ≈ c. ( ) ( )2.1 2 0.476 0.5 0.024g g− = − = − d. ( ) ( )2.1 2 0.024 0.24 2.1 2 0.1 g g− − = = − − e. ( ) ( )1/g a h a h+ = + f. ( ) ( ) ( ) ( ) 1/ 1/ h g a h g a a h a a a h − + − = + − = + g. ( ) ( ) ( ) ( ) ( ) 1 h g a h g a a a h a h a h a a h − + − + − = = + − + h. ( ) ( ) ( ) 20 1 lim h g a h g a a h a a→ + − − = + − 3. a. ( )2 2 1.414F = ≈ b. ( )2.1 2.1 1.449F = ≈ c. ( ) ( )2.1 2 1.449 1.414 0.035F F− = − = d. ( ) ( )2.1 2 0.035 0.35 2.1 2 0.1 F F− = = − e. ( )F a h a h+ = + f. ( ) ( )F a h F a a h a+ − = + − g. ( ) ( ) ( ) F a h F a a h a a h a h + − + − = + − h. ( ) ( ) ( )0 0 lim lim h h F a h F a a h a a h a h→ → + − + − = + − ( )( ) ( ) ( ) ( ) 0 0 0 0 lim lim lim 1 1 lim 22 h h h h a h a a h a h a h a a h a h a h a h h a h a a aa h a a → → → → + − + + = + + + − = + + = + + = = = + + © 2007 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.
  • Instructor’s Resource Manual Review and Preview 93 4. a. ( ) ( )3 2 2 1 8 1 9G = + = + = b. ( ) ( )3 2.1 2.1 1 9.261 1 10.261G = + = + = c. ( ) ( )2.1 2 10.261 9 1.261G G− = − = d. ( ) ( )2.1 2 1.261 12.61 2.1 2 0.1 G G− = = − e. ( ) ( )3 3 2 2 3 1 3 3 1 G a h a h a a h ah h + = + + = + + + + f. ( ) ( ) ( ) ( ) ( ) 3 3 3 2 2 3 3 2 2 3 1 1 3 3 1 1 3 3 G a h G a a h a a a h ah h a a h ah h ⎡ ⎤ ⎡ ⎤+ − = + + − +⎣ ⎦⎣ ⎦ = + + + + − + = + + g. ( ) ( ) ( ) 2 2 3 2 2 3 3 3 3 G a h G a a h ah h a h a h a ah h + − + + = + − = + + h. ( ) ( ) ( ) 2 2 0 0 2 lim lim 3 3 3 h h G a h G a a ah h a h a a → → + − = + + + − = 5. a. ( )3 3 2 3a b a a b+ = + + b. ( )4 4 3 4a b a a b+ = + + c. ( )5 5 4 5a b a a b+ = + + 6. ( ) 1n n n a b a na b− + = + + 7. ( )sin sin cos cos sinx h x h x h+ = + 8. ( )cos cos cos sin sinx h x h x h+ = − 9. a. The point will be at position ( )10,0 in all three cases ( 1,2,3t = ) because it will have made 4, 8, and 12 revolutions respectively. b. Since the point is rotating at a rate of 4 revolutions per second, it will complete 1 revolution after 1 4 second. Therefore, the point will first return to its starting position at time 1 4 t = . 10. ( ) ( ) 3 3 0 3 3 1 3 3 1 0 3 3 4 32 2 cm 3 3 4 62.5 125 2.5 = cm 3 3 6 125 32 cm cm 6 3 61 cm 31.940cm 6 V V V V V π π π π π π π π = = = = Δ = − = − = ≈ 11. a. North plane has traveled 600miles. East plane has traveled 400 miles. b. 2 2 600 400 721 miles d = + = c. 2 2 675 500 840 miles d = + = © 2007 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.
  • 94 Section 2.1 Instructor’s Resource Manual CHAPTER 2 The Derivative 2.1 Concepts Review 1. tangent line 2. secant line 3. ( ) ( )f c h f c h + − 4. average velocity Problem Set 2.1 1. Slope 3 2 5 – 3 4 2 – = = 2. 6 – 4 Slope –2 4 – 6 = = 3. Slope 2≈ − 4. Slope 1.5≈ 5. 5 Slope 2 ≈ 6. 3 Slope – 2 ≈ © 2007 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.
  • Instructor’s Resource Manual Section 2.1 95 7. y = x 2 + 1 a., b. c. mtan = 2 d. 2 sec (1.01) 1.0 2 1.01 1 0.0201 .01 2.01 m + − = − = = e. tan 0 (1 ) – (1) lim h f h f m h→ + = 2 2 0 [(1 ) 1] – (1 1) lim h h h→ + + + = 2 0 0 2 2 2 lim (2 ) lim h h h h h h h h → → + + − = + = 0 lim (2 ) 2 h h → = + = 8. y = x 3 –1 a., b. c. mtan = 12 d. 3 sec [(2.01) 1.0] 7 2.01 2 0.120601 0.01 m − − = − = = 12.0601 e. tan 0 (2 ) – (2) lim h f h f m h→ + = 3 3 0 [(2 ) –1] – (2 1) lim h h h→ + − = 2 3 0 2 0 12 6 lim (12 6 ) lim h h h h h h h h h h → → + + = + + = = 12 9. f (x) = x2 – 1 tan 0 ( ) – ( ) lim h f c h f c m h→ + = 2 2 0 [( ) –1] – ( –1) lim h c h c h→ + = 2 2 2 0 2 –1– 1 lim h c ch h c h→ + + + = 0 (2 ) lim 2 h h c h c h→ + = = At x = –2, mtan = –4 x = –1, mtan = –2 x = 1, mtan = 2 x = 2, mtan = 4 10. f (x) = x3 – 3x tan 0 ( ) – ( ) lim h f c h f c m h→ + = 3 3 0 [( ) – 3( )] – ( – 3 ) lim h c h c h c c h→ + + = 3 2 2 3 3 0 3 3 – 3 – 3 – 3 lim h c c h ch h c h c c h→ + + + + = 2 2 2 0 (3 3 3) lim 3 – 3 h h c ch h c h→ + + − = = At x = –2, mtan = 9 x = –1, mtan = 0 x = 0, mtan = –3 x = 1, mtan = 0 x = 2, mtan = 9 © 2007 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.
  • 96 Section 2.1 Instructor’s Resource Manual 11. 1 ( ) 1 f x x = + tan 0 (1 ) – (1) lim h f h f m h→ + = 1 1 2 2 0 2(2 ) 0 0 lim lim 1 lim 2(2 ) h h h h h h h h h + → + → → − = − = = − + 1 – 4 = 1 1 – – ( –1) 2 4 y x= 12. f (x) = 1 x –1 tan 0 1 1 0 1 0 0 (0 ) (0) lim 1 lim lim 1 lim 1 1 h h h h h h h f h f m h h h h → − → − → → + − = + = = = − = − y + 1 = –1(x – 0); y = –x – 1 13. a. 2 2 16(1 ) –16(0 ) 16 ft= b. 2 2 16(2 ) –16(1 ) 48 ft= c. ave 144 – 64 V 80 3 – 2 = = ft/sec d. 2 2 ave 16(3.01) 16(3) V 3.01 3 0.9616 0.01 − = − = = 96.16 ft/s e. 2 ( ) 16 ; 32f t t v c= = v = 32(3) = 96 ft/s 14. a. 2 2 ave (3 1) – (2 1) V 5 3 – 2 + + = = m/sec b. 2 2 ave [(2.003) 1] (2 1) V 2.003 2 0.012009 0.003 + − + = − = = 4.003 m/sec c. 2 2 ave 2 [(2 ) 1] – (2 1) V 2 – 2 4 h h h h h + + + = + + = = 4 +h d. f (t) = t2 + 1 0 (2 ) – (2) lim h f h f v h→ + = 2 2 0 [(2 ) 1] – (2 1) lim h h h→ + + + = 2 0 0 4 lim lim (4 ) 4 h h h h h h → → + = = + = © 2007 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.
  • Instructor’s Resource Manual Section 2.1 97 15. a. 0 ( ) – ( ) lim h f h f v h α α → + = 0 2( ) 1 – 2 1 lim h h h α α → + + + = 0 2 2 1 – 2 1 lim h h h α α → + + + = 0 ( 2 2 1 – 2 1)( 2 2 1 2 1) lim ( 2 2 1 2 1)h h h h h α α α α α α→ + + + + + + + = + + + + 0 2 lim ( 2 2 1 2 1)h h h hα α→ = + + + + 2 1 2 1 2 1 2 1α α α = = + + + + ft/s b. 1 1 22 1α = + 2 1 2α + = 2α +1= 4; α = 3 2 The object reaches a velocity of 1 2 ft/s when t = 3 2 . 16. f (t) = –t2 + 4t 2 2 0 [–( ) 4( )] – (– 4 ) lim h c h c h c c v h→ + + + + = 2 2 2 0 – – 2 – 4 4 – 4 lim h c ch h c h c c h→ + + + = 0 (–2 – 4) lim –2 4 h h c h c h→ + = = + –2c + 4 = 0 when c = 2 The particle comes to a momentary stop at t = 2. 17. a. 2 21 1 (2.01) 1 – (2) 1 0.02005 2 2 ⎡ ⎤ ⎡ ⎤ + + =⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ g b. ave 0.02005 2.005 2.01– 2 r = = g/hr c. 21 ( ) 1 2 f t t= + 2 21 1 2 2 0 (2 ) 1 – 2 1 lim h h r h→ ⎡ ⎤ ⎡ ⎤+ + + ⎣ ⎦ ⎣ ⎦= 21 2 0 2 2 1 2 1 lim h h h h→ + + + − − = ( )1 2 0 2 lim 2 h h h h→ + = = At t = 2, r = 2 18. a. 2 2 1000(3) –1000(2) 5000= b. 2 2 1000(2.5) –1000(2) 2250 4500 2.5 – 2 0.5 = = c. f (t) = 1000t2 2 2 0 1000(2 ) 1000(2) lim h h r h→ + − = 2 0 4000 4000 1000 – 4000 lim h h h h→ + + = 0 (4000 1000 ) lim 4000 h h h h→ + = = 19. a. 3 3 ave 5 – 3 98 49 5 – 3 2 d = = = g/cm b. f (x) = x3 3 3 0 (3 ) – 3 lim h h d h→ + = 2 3 0 27 27 9 – 27 lim h h h h h→ + + + = 2 0 (27 9 ) lim 27 h h h h h→ + + = = g/cm © 2007 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.
  • 98 Section 2.1 Instructor’s Resource Manual 20. 0 ( ) – ( ) lim h R c h R c MR h→ + = 2 2 0 [0.4( ) – 0.001( ) ] – (0.4 – 0.001 ) lim h c h c h c c h→ + + = 2 2 2 0 0.4 0.4 – 0.001 – 0.002 – 0.001 – 0.4 0.001 lim h c h c ch h c c h→ + + = 0 (0.4 – 0.002 – 0.001 ) lim 0.4 – 0.002 h h c h c h→ = = When n = 10, MR = 0.38; when n = 100, MR = 0.2 21. 2 2 0 2(1 ) – 2(1) lim h h a h→ + = 2 0 0 2 4 2 – 2 lim (4 2 ) lim 4 h h h h h h h h → → + + = + = = 22. 0 ( ) – ( ) lim h p c h p c r h→ + = 2 3 2 3 0 [120( ) – 2( ) ] – (120 – 2 ) lim h c h c h c c h→ + + = 2 2 0 (240 – 6 120 – 6 – 2 ) lim h h c c h ch h h→ + = 2 240 – 6c c= When t = 10, 2 240(10) – 6(10) 1800r = = t = 20, 2 240(20) – 6(20) 2400r = = t = 40, 2 240(40) – 6(40) 0r = = 23. ave 100 – 800 175 – –29.167 24 – 0 6 r = = ≈ 29,167 gal/hr At 8 o’clock, 700 – 400 75 6 10 r ≈ ≈ − − 75,000 gal/hr 24. a. The elevator reached the seventh floor at time 80=t . The average velocity is 05.180/)084( =−=avgv feet per second b. The slope of the line is approximately 2.1 1555 1260 = − − . The velocity is approximately 1.2 feet per second. c. The building averages 84/7=12 feet from floor to floor. Since the velocity is zero for two intervals between time 0 and time 85, the elevator stopped twice. The heights are approximately 12 and 60. Thus, the elevator stopped at floors 1 and 5. 25. a. A tangent line at 91=t has slope approximately 5.0)6191/()4863( =−− . The normal high temperature increases at the rate of 0.5 degree F per day. b. A tangent line at 191=t has approximate slope 067.030/)8890( ≈− . The normal high temperature increases at the rate of 0.067 degree per day. c. There is a time in January, about January 15, when the rate of change is zero. There is also a time in July, about July 15, when the rate of change is zero. d. The greatest rate of increase occurs around day 61, that is, some time in March. The greatest rate of decrease occurs between day 301 and 331, that is, sometime in November. 26. The slope of the tangent line at 1930=t is approximately (8 6)/(1945 1930) 0.13− − ≈ . The rate of growth in 1930 is approximately 0.13 million, or 130,000, persons per year. In 1990, the tangent line has approximate slope (24 16) /(20000 1980) 0.4− − ≈ . Thus, the rate of growth in 1990 is 0.4 million, or 400,000, persons per year. The approximate percentage growth in 1930 is 0.107 / 6 0.018≈ and in 1990 it is approximately 02.020/4.0 ≈ . 27. In both (a) and (b), the tangent line is always positive. In (a) the tangent line becomes steeper and steeper as t increases; thus, the velocity is increasing. In (b) the tangent line becomes flatter and flatter as t increases; thus, the velocity is decreasing. © 2007 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.
  • Instructor’s Resource Manual Section 2.2 99 28. 31 ( ) 3 f t t t= + 0 ( ) – ( ) current lim h f c h f c h→ + = ( )3 31 1 3 3 0 ( ) ( ) – lim h c h c h c c h→ ⎡ ⎤+ + + + ⎣ ⎦= ( )2 21 3 2 0 1 lim 1 h h c ch h c h→ + + + = = + When t = 3, the current =10 2 1 20c + = c 2 = 19 19 4.4c = ≈ A 20-amp fuse will blow at t = 4.4 s. 29. 2 ,A r= π r = 2t A = 4πt2 2 2 0 4 (3 ) – 4 (3) rate lim h h h→ π + π = 0 (24 4 ) lim 24 h h h h→ π + π = = π km2/day 30. 3 3 3 3 0 3 4 1 , 3 4 1 48 1 (3 ) 3 27 rate lim 48 48 9 inch /sec 16 h V r r t V t h h π π π π π → = = = + − = = = 31. 3 2 ( ) – 2 1y f x x x= = + a. mtan = 7 b. mtan = 0 c. mtan = –1 d. mtan = 17. 92 32. 2 ( ) sin sin 2y f x x x= = a. mtan = –1.125 b. mtan ≈ –1.0315 c. mtan = 0 d. mtan ≈1.1891 33. 2 ( ) coss f t t t t= = + At t = 3, v ≈ 2.818 34. 3 ( 1) ( ) 2 t s f t t + = = + At t = 1.6, v ≈ 4.277 2.2 Concepts Review 1. ( ) – ( ) ( ) – ( ) ; – f c h f c f t f c h t c + 2. ( )cf ′ 3. continuous; ( )f x x= 4. ( )' ; dy f x dx Problem Set 2.2 1. 0 (1 ) – (1) (1) lim h f h f f h→ + ′ = 2 2 2 0 0 (1 ) –1 2 lim lim h h h h h h h→ → + + = = = lim h→0 (2 + h) = 2 2. 0 (2 ) – (2) (2) lim h f h f f h→ + ′ = 2 2 0 [2(2 )] –[2(2)] lim h h h→ + = 2 0 0 16 4 lim lim (16 4 ) 16 h h h h h h→ → + = = + = 3. 0 (3 ) – (3) (3) lim h f h f f h→ + ′ = 2 2 0 [(3 ) – (3 )] – (3 – 3) lim h h h h→ + + = 2 0 0 5 lim lim (5 ) 5 h h h h h h→ → + = = + = 4. 0 (4 ) – (4) (4) lim h f h f f h→ + ′ = 3–(3 )1 1 3(3 )3 4–1 0 0 0 – –1 lim lim lim 3(3 ) h hh h h hh h h + ++ → → → = = = + = – 1 9 5. 0 ( ) – ( ) ( ) lim h s x h s x s x h→ + ′ = 0 [2( ) 1] – (2 1) lim h x h x h→ + + + = 0 2 lim 2 h h h→ = = © 2007 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.
  • 100 Section 2.2 Instructor’s Resource Manual 6. 0 ( ) – ( ) ( ) lim h f x h f x f x h→ + ′ = 0 [ ( ) ] – ( ) lim h x h x h α β α β → + + + = 0 lim h h h α α → = = 7. 0 ( ) – ( ) ( ) lim h r x h r x r x h→ + ′ = 2 2 0 [3( ) 4] – (3 4) lim h x h x h→ + + + = 2 0 0 6 3 lim lim (6 3 ) 6 h h xh h x h x h→ → + = = + = 8. 0 ( ) – ( ) ( ) lim h f x h f x f x h→ + ′ = 2 2 0 [( ) ( ) 1] – ( 1) lim h x h x h x x h→ + + + + + + = 2 0 0 2 lim lim (2 1) 2 1 h h xh h h x h x h→ → + + = = + + = + 9. 0 ( ) – ( ) ( ) lim h f x h f x f x h→ + ′ = 2 2 0 [ ( ) ( ) ] – ( ) lim h a x h b x h c ax bx c h→ + + + + + + = 2 0 0 2 lim lim (2 ) h h axh ah bh ax ah b h→ → + + = = + + = 2ax + b 10. 0 ( ) – ( ) ( ) lim h f x h f x f x h→ + ′ = 4 4 0 ( ) – lim h x h x h→ + = 3 2 2 3 4 0 4 6 4 lim h hx h x h x h h→ + + + = 3 2 2 3 3 0 lim (4 6 4 ) 4 h x hx h x h x → = + + + = 11. 0 ( ) – ( ) ( ) lim h f x h f x f x h→ + ′ = 3 2 3 2 0 [( ) 2( ) 1] – ( 2 1) lim h x h x h x x h→ + + + + + + = 2 2 3 2 0 3 3 4 2 lim h hx h x h hx h h→ + + + + = 2 2 2 0 lim (3 3 4 2 ) 3 4 h x hx h x h x x → = + + + + = + 12. 0 ( ) – ( ) ( ) lim h g x h g x g x h→ + ′ = 4 2 4 2 0 [( ) ( ) ] – ( ) lim h x h x h x x h→ + + + + = 3 2 2 3 4 2 0 4 6 4 2 lim h hx h x h x h hx h h→ + + + + + = 3 2 2 3 0 lim (4 6 4 2 ) h x hx h x h x h → = + + + + + = 4x3 +2x 13. 0 ( ) – ( ) ( ) lim h h x h h x h x h→ + ′ = 0 2 2 1 lim – h x h x h→ ⎡ ⎤⎛ ⎞ = ⋅⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦ 0 0 –2 1 –2 lim lim ( ) ( )h h h x x h h x x h→ → ⎡ ⎤ = ⋅ =⎢ ⎥+ +⎣ ⎦ 2 2 – x = 14. 0 ( ) – ( ) ( ) lim h S x h S x S x h→ + ′ = 0 1 1 1 lim – 1 1h x h x h→ ⎡ ⎤⎛ ⎞ = ⋅⎜ ⎟⎢ ⎥+ + +⎝ ⎠⎣ ⎦ 0 – 1 lim ( 1)( 1)h h x x h h→ ⎡ ⎤ = ⋅⎢ ⎥ + + +⎣ ⎦ 20 –1 1 lim ( 1)( 1) ( 1)h x x h x→ = = − + + + + 15. 0 ( ) – ( ) ( ) lim h F x h F x F x h→ + ′ = 2 20 6 6 1 lim – ( ) 1 1h hx h x→ ⎡ ⎤⎛ ⎞ = ⋅⎢ ⎥⎜ ⎟⎜ ⎟+ + +⎢ ⎥⎝ ⎠⎣ ⎦ 2 2 2 2 2 20 6( 1) – 6( 2 1) 1 lim ( 1)( 2 1)h x x hx h hx x hx h→ ⎡ ⎤+ + + + = ⋅⎢ ⎥ + + + +⎢ ⎥⎣ ⎦ 2 2 2 20 –12 – 6 1 lim ( 1)( 2 1)h hx h hx x hx h→ ⎡ ⎤ = ⋅⎢ ⎥ + + + +⎢ ⎥⎣ ⎦ 2 2 2 2 20 –12 – 6 12 lim ( 1)( 2 1) ( 1)h x h x x x hx h x→ = = − + + + + + 16. 0 ( ) – ( ) ( ) lim h F x h F x F x h→ + ′ = 0 –1 –1 1 lim – 1 1h x h x x h x h→ ⎡ + ⎤⎛ ⎞ = ⋅⎜ ⎟⎢ ⎥+ + +⎝ ⎠⎣ ⎦ 2 2 0 –1– ( – –1) 1 lim ( 1)( 1)h x hx h x hx h x h x h→ ⎡ ⎤+ + + = ⋅⎢ ⎥ + + +⎢ ⎥⎣ ⎦ 20 2 1 2 lim ( 1)( 1) ( 1)h h x h x h x→ ⎡ ⎤ = ⋅ =⎢ ⎥+ + + +⎣ ⎦ © 2007 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.
  • Instructor’s Resource Manual Section 2.2 101 17. 0 ( ) – ( ) ( ) lim h G x h G x G x h→ + ′ = 0 2( ) –1 2 –1 1 lim – – 4 – 4h x h x x h x h→ ⎡ + ⎤⎛ ⎞ = ⋅⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦ 2 2 2 2 9 8 4 (2 2 9 4) 1 lim ( 4)( 4)0 x hx x h x hx x h hx h xh + − − + − + − − + = ⋅ + − −→ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦ 0 –7 1 lim ( – 4)( – 4)h h x h x h→ ⎡ ⎤ = ⋅⎢ ⎥+⎣ ⎦ 20 –7 7 lim – ( – 4)( – 4) ( – 4)h x h x x→ = = + 18. 0 ( ) – ( ) ( ) lim h G x h G x G x h→ + ′ = 2 20 2( ) 2 1 lim – ( ) – ( ) –h x h x hx h x h x x→ ⎡ ⎤⎛ ⎞+ = ⋅⎢ ⎥⎜ ⎟⎜ ⎟+ +⎢ ⎥⎝ ⎠⎣ ⎦ 2 2 2 2 2 20 (2 2 )( – ) – 2 ( 2 – – ) 1 lim ( 2 – – )( – )h x h x x x x xh h x h hx hx h x h x x→ ⎡ ⎤+ + + = ⋅⎢ ⎥ + +⎢ ⎥⎣ ⎦ 2 2 2 2 20 –2 – 2 1 lim ( 2 – – )( – )h h x hx hx hx h x h x x→ ⎡ ⎤ = ⋅⎢ ⎥ + +⎢ ⎥⎣ ⎦ 2 2 2 20 –2 – 2 lim ( 2 – – )( – )h hx x x hx h x h x x→ = + + 2 2 2 2 –2 2 – ( – ) ( –1) x x x x = = 19. 0 ( ) – ( ) ( ) lim h g x h g x g x h→ + ′ = 0 3( ) – 3 lim h x h x h→ + = 0 ( 3 3 – 3 )( 3 3 3 ) lim ( 3 3 3 )h x h x x h x h x h x→ + + + = + + 0 0 3 3 lim lim ( 3 3 3 ) 3 3 3h h h h x h x x h x→ → = = + + + + 3 2 3x = 20. 0 ( ) – ( ) ( ) lim h g x h g x g x h→ + ′ = 0 1 1 1 lim – 3( ) 3h hx h x→ ⎡ ⎤⎛ ⎞ = ⋅⎢ ⎥⎜ ⎟⎜ ⎟+⎢ ⎥⎝ ⎠⎣ ⎦ 0 3 – 3 3 1 lim 9 ( )h x x h hx x h→ ⎡ ⎤+ = ⋅⎢ ⎥ +⎢ ⎥⎣ ⎦ 0 ( 3 – 3 3 )( 3 3 3 ) 1 lim 9 ( )( 3 3 3 )h x x h x x h hx x h x x h→ ⎡ ⎤+ + + = ⋅⎢ ⎥ + + +⎢ ⎥⎣ ⎦ 0 –3 –3 lim 9 ( )( 3 3 3 ) 3 2 3h h h x x h x x h x x→ = = + + + ⋅ 1 – 2 3x x = © 2007 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.
  • 102 Section 2.2 Instructor’s Resource Manual 21. 0 ( ) – ( ) ( ) lim h H x h H x H x h→ + ′ = 0 3 3 1 lim – – 2 – 2h hx h x→ ⎡ ⎤⎛ ⎞ = ⋅⎢ ⎥⎜ ⎟ +⎝ ⎠⎣ ⎦ 0 3 – 2 – 3 – 2 1 lim ( – 2)( – 2)h x x h hx h x→ ⎡ ⎤+ = ⋅⎢ ⎥ +⎢ ⎥⎣ ⎦ 0 3( – 2 – – 2)( – 2 – 2) lim ( – 2)( – 2)( – 2 – 2)h x x h x x h h x h x x x h→ + + + = + + + 0 3 lim [( – 2) – 2 ( – 2) – 2]h h h x x h x h x→ − = + + + 0 –3 lim ( – 2) – 2 ( – 2) – 2h x x h x h x→ = + + + 3 2 3 3 – 2( – 2) – 2 2( 2)x x x = = − − 22. 0 ( ) – ( ) ( ) lim h H x h H x H x h→ + ′ = 2 2 0 ( ) 4 – 4 lim h x h x h→ + + + = 2 2 2 2 2 2 0 2 2 2 2 4 – 4 2 4 4 lim 2 4 4h x hx h x x hx h x h x hx h x→ ⎛ ⎞⎛ ⎞+ + + + + + + + +⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠= ⎛ ⎞+ + + + +⎜ ⎟ ⎝ ⎠ 2 0 2 2 2 2 lim 2 4 4h hx h h x hx h x→ + = ⎛ ⎞+ + + + +⎜ ⎟ ⎝ ⎠ 2 2 20 2 lim 2 4 4h x h x hx h x→ + = + + + + + 2 2 2 2 4 4 x x x x = = + + 23. ( ) – ( ) ( ) lim –t x f t f x f x t x→ ′ = 2 2 ( 3 ) – ( – 3 ) lim –t x t t x x t x→ − = 2 2 – – (3 – 3 ) lim –t x t x t x t x→ = ( – )( ) – 3( – ) lim –t x t x t x t x t x→ + = ( – )( – 3) lim lim( – 3) –t x t x t x t x t x t x→ → + = = + = 2 x – 3 24. ( ) – ( ) ( ) lim –t x f t f x f x t x→ ′ = 3 3 ( 5 ) – ( 5 ) lim –t x t t x x t x→ + + = 3 3 – 5 – 5 lim –t x t x t x t x→ + = 2 2 ( – )( ) 5( – ) lim –t x t x t tx x t x t x→ + + + = 2 2 ( – )( 5) lim –t x t x t tx x t x→ + + + = 2 2 2 lim( 5) 3 5 t x t tx x x → = + + + = + © 2007 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.
  • Instructor’s Resource Manual Section 2.2 103 25. ( ) – ( ) ( ) lim –t x f t f x f x t x→ ′ = 1 lim – – 5 – 5 –t x t x t x t x→ ⎡ ⎤⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟⎢ ⎥ ⎝ ⎠⎝ ⎠⎣ ⎦ – 5 – 5 lim ( – 5)( – 5)( – )t x tx t tx x t x t x→ + = –5( – ) –5 lim lim ( – 5)( – 5)( – ) ( – 5)( – 5)t x t x t x t x t x t x→ → = = ( ) 2 5 5x = − − 26. ( ) – ( ) ( ) lim –t x f t f x f x t x→ ′ = 3 3 1 lim – –t x t x t x t x→ ⎡ + + ⎤⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟⎢ ⎥ ⎝ ⎠⎝ ⎠⎣ ⎦ 2 3 – 3 –3 3 lim lim – ( – )t x t x x t xt t x xt x→ → = = = 27. f (x) = 2x3 at x = 5 28. f (x) = x2 + 2x at x = 3 29. f (x) = x2 at x = 2 30. f (x) = x3 + x at x = 3 31. f (x) = x2 at x 32. f (x) = x3 at x 33. f (t) = 2 t at t 34. f(y) = sin y at y 35. f(x) = cos x at x 36. f(t) = tan t at t 37. The slope of the tangent line is always 2. 38. The slope of the tangent line is always 1− . 39. The derivative is positive until 0x = , then becomes negative. 40. The derivative is negative until 1x = , then becomes positive. 41. The derivative is 1− until 1x = . To the right of 1x = , the derivative is 1. The derivative is undefined at 1x = . 42. The derivative is 2− to the left of 1x = − ; from 1− to 1, the derivative is 2, etc. The derivative is not defined at 1, 1, 3x = − . © 2007 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.
  • 104 Section 2.2 Instructor’s Resource Manual 43. The derivative is 0 on ( )3, 2− − , 2 on ( )2, 1− − , 0 on ( )1,0− , 2− on ( )0,1 , 0 on ( )1,2 , 2 on ( )2,3 and 0 on ( )3,4 . The derivative is undefined at 2, 1, 0, 1, 2, 3x = − − . 44. The derivative is 1 except at 2, 0, 2x = − where it is undefined. 45. Δy = [3(1.5) + 2] – [3(1) + 2] = 1.5 46. 2 2 [3(0.1) 2(0.1) 1] –[3(0.0) 2(0.0) 1]yΔ = + + + + = 0.23 47. Δy = 1/1.2 – 1/1 = – 0.1667 48. Δy = 2/(0.1+1) – 2/(0+1) = – 0.1818 49. 3 3 – 0.0081 2.31 1 2.34 1 yΔ = ≈ + + 50. cos[2(0.573)] – cos[2(0.571)] –0.0036yΔ = ≈ 51. 2 2 2 ( ) – 2 ( ) 2 y x x x x x x x x x x x Δ + Δ Δ + Δ = = = + Δ Δ Δ Δ 0 lim (2 ) 2 x dy x x x dx Δ → = + Δ = 52. 3 2 3 2 [( ) – 3( ) ] – ( – 3 )y x x x x x x x x Δ + Δ + Δ = Δ Δ 2 2 2 3 3 3 ( ) – 6 – 3( )x x x x x x x x x Δ + Δ Δ Δ + Δ = Δ 2 2 3 3 – 6 – 3 ( )x x x x x x= + Δ Δ + Δ 2 2 0 lim (3 3 – 6 – 3 ( ) ) x dy x x x x x x dx Δ → = + Δ Δ + Δ 2 3 – 6x x= 53. 1 1 1 1 –x x xy x x +Δ + +Δ = Δ Δ 1– ( 1) 1 ( 1)( 1) x x x x x x x ⎛ ⎞+ + Δ + ⎛ ⎞ = ⎜ ⎟⎜ ⎟ + Δ + + Δ⎝ ⎠⎝ ⎠ – ( 1)( 1) x x x x x Δ = + Δ + + Δ 1 – ( 1)( 1)x x x = + Δ + + 20 1 1 lim ( 1)( 1) ( 1)x dy dx x x x xΔ → ⎡ ⎤ = − = −⎢ ⎥+ Δ + + +⎣ ⎦ 54. ( ) ( ) ( ) 20 1 1 1 1 1 1 1 1 1 lim x y x x x x x x x x xx x x x x x x x dy dx x x x xΔ → ⎛ ⎞ + − +⎜ ⎟Δ + Δ ⎝ ⎠= Δ Δ −Δ − + Δ+ Δ= = = − Δ Δ + Δ = − = − + Δ 55. ( )( ) ( )( ) ( )( ) ( ) 2 2 2 2 2 2 2 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 2 1 1 2 2 2 lim 1 2 1 1x x x x y x x x x x x x x x x x x x x x x x x x x x x x x x x x xx x x x x x x xx x x x x x x x x x x x dy dx x x x x x x x x xΔ → + Δ − − − Δ + Δ + += Δ Δ + + Δ − − − + Δ + = × + Δ + + Δ ⎡ ⎤+ Δ − + + Δ − − + Δ − + − Δ − ⎣ ⎦= × Δ+ Δ + + + Δ + Δ = × = Δ+ Δ + + + Δ + + Δ + + + Δ + = = = + Δ + + + Δ + + + + 56. ( )2 21 1x x x y x x x x x + Δ − − − Δ + Δ= Δ Δ ( ) ( )( ) ( ) ( )( ) ( ) ( ) 2 2 2 2 3 2 2 22 2 2 2 2 2 2 20 1 1 2 1 1 1 1 1 lim x x x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x xx x x x x x dy x x x x dx x x x xΔ → ⎡ ⎤+ Δ − − + Δ − ⎢ ⎥= × ⎢ ⎥+ Δ Δ ⎣ ⎦ ⎡ ⎤+ Δ + Δ − − + Δ − − Δ ⎢ ⎥= × ⎢ ⎥ Δ+ Δ ⎢ ⎥⎣ ⎦ Δ + Δ + Δ + Δ + = × = Δ+ Δ + Δ + Δ + + = = + Δ © 2007 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.
  • Instructor’s Resource Manual Section 2.2 105 57. 1 (0) – ; (2) 1 2 f f′ ′≈ ≈ 2 (5) ; (7) –3 3 f f′ ′≈ ≈ 58. (–1) 2; (1) 0g g′ ′≈ ≈ 1 (4) –2; (6) – 3 g g′ ′≈ ≈ 59. 60. 61. a. 5 3 (2) ; (2) 2 2 f f ′≈ ≈ (0.5) 1.8; (0.5) –0.6f f ′≈ ≈ b. 2.9 1.9 0.5 2.5 0.5 − = − c. x = 5 d. x = 3, 5 e. x = 1, 3, 5 f. x = 0 g. 3 0.7, 2 x ≈ − and 5 < x < 7 62. The derivative fails to exist at the corners of the graph; that is, at 80,60,55,15,10=t . The derivative exists at all other points on the interval )85,0( . 63. The derivative is 0 at approximately 15=t and 201=t . The greatest rate of increase occurs at about 61=t and it is about 0.5 degree F per day. The greatest rate of decrease occurs at about 320=t and it is about 0.5 degree F per day. The derivative is positive on (15,201) and negative on (0,15) and (201,365). 64. The slope of a tangent line for the dashed function is zero when x is approximately 0.3 or 1.9. The solid function is zero at both of these points. The graph indicates that the solid function is negative when the dashed function has a tangent line with a negative slope and positive when the dashed function has a tangent line with a positive slope. Thus, the solid function is the derivative of the dashed function. 65. The short-dash function has a tangent line with zero slope at about 1.2=x , where the solid function is zero. The solid function has a tangent line with zero slope at about 2.1,4.0=x and 3.5. The long-dash function is zero at these points. The graph shows that the solid function is positive (negative) when the slope of the tangent line of the short-dash function is positive (negative). Also, the long-dash function is positive (negative) when the slope of the tangent line of the solid function is positive (negative). Thus, the short-dash function is f, the solid function is gf =' , and the dash function is 'g . 66. Note that since x = 0 + x, f(x) = f(0 + x) = f(0)f(x), hence f(0) = 1. 0 ( ) – ( ) ( ) lim h f a h f a f a h→ + ′ = 0 ( ) ( ) – ( ) lim h f a f h f a h→ = 0 0 ( ) –1 ( ) – (0) ( ) lim ( ) lim h h f h f h f f a f a h h→ → = = ( ) (0)f a f ′= ′f (a) exists since ′f (0) exists. © 2007 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.
  • 106 Section 2.2 Instructor’s Resource Manual 67. If f is differentiable everywhere, then it is continuous everywhere, so lim x→2 − f( x) = lim x→2 − (mx + b) = 2m + b = f (2) = 4 and b = 4 – 2m. For f to be differentiable everywhere, 2 ( ) (2) (2) lim 2x f x f f x→ − ′ = − must exist. 2 2 2 2 ( ) (2) 4 lim lim lim ( 2) 4 2 2x x x f x f x x x x+ + +→ → → − − = = + = − − 2 2 ( ) (2) 4 lim lim 2 2x x f x f mx b x x− −→ → − + − = − − 2 2 4 2 4 ( 2) lim lim 2 2x x mx m m x m x x− −→ → + − − − = = = − − Thus m = 4 and b = 4 – 2(4) = –4 68. 0 ( ) – ( ) ( ) – ( – ) ( ) lim 2 s h f x h f x f x f x h f x h→ + + = 0 ( ) – ( ) ( – ) – ( ) lim 2 –2h f x h f x f x h f x h h→ +⎡ ⎤ = +⎢ ⎥ ⎣ ⎦ 0 – 0 1 ( ) – ( ) 1 [ (– )] – ( ) lim lim 2 2 –h h f x h f x f x h f x h h→ → + + = + 1 1 ( ) ( ) ( ). 2 2 f x f x f x′ ′ ′= + = For the converse, let f (x) = x . Then 0 0 – – – (0) lim lim 0 2 2 s h h h h h h f h h→ → = = = but ′f (0) does not exist. 69. 0 0 00 ( ) ( ) ( ) lim , t x f t f x f x t x→ − ′ = − so 0 0 00 ( ) ( ) ( ) lim ( )t x f t f x f x t x→− − − ′ − = − − 0 00 ( ) ( ) lim t x f t f x t x→− − − = + a. If f is an odd function, 0 0 0 0 ( ) [ ( )] ( ) lim t x f t f x f x t x→− − − − ′ − = + 0 0 0 ( ) ( ) lim t x f t f x t x→− + − = + . Let u = –t. As t → −x0 , u → x0 and so 0 0 0 0 ( ) ( ) ( ) lim u x f u f x f x u x→ − + ′ − = − + 0 0 0 00 0 ( ) ( ) [ ( ) ( )] lim lim ( ) ( )u x u x f u f x f u f x u x u x→ → − + − − = = − − − − 0 0 0 0 ( ) ( ) lim ( ) . u x f u f x f x m u x→ − ′= = = − b. If f is an even function, 0 0 0 0 ( ) ( ) (– ) lim t x f t f x f x t x→− − ′ = + . Let u = –t, as above, then 0 0 0 0 ( ) ( ) ( ) lim u x f u f x f x u x→ − − ′ − = − + 0 0 0 00 0 ( ) ( ) ( ) ( ) lim lim ( )u x u x f u f x f u f x u x u x→ → − − = = − − − − = − ′f (x0 ) = −m. 70. Say f(–x) = –f(x). Then 0 (– ) – (– ) (– ) lim h f x h f x f x h→ + ′ = 0 0 – ( – ) ( ) ( – ) – ( ) lim – lim h h f x h f x f x h f x h h→ → + = = – 0 [ (– )] ( ) lim ( ) –h f x h f x f x h→ + − ′= = so ′f (x) is an even function if f(x) is an odd function. Say f(–x) = f(x). Then 0 (– ) – (– ) (– ) lim h f x h f x f x h→ + ′ = 0 ( – ) – ( ) lim h f x h f x h→ = – 0 [ (– )] – ( ) – lim – ( ) –h f x h f x f x h→ + ′= = so ′f (x) is an odd function if f(x) is an even function. 71. a. 8 0 3 x< < ; 8 0, 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ b. 8 0 3 x≤ ≤ ; 8 0, 3 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ c. A function f(x) decreases as x increases when ′f (x) < 0. 72. a. 6.8xπ < < b. 6.8xπ < < c. A function f(x) increases as x increases when ′f (x) > 0. © 2007 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.
  • Instructor’s Resource Manual Section 2.3 107 2.3 Concepts Review 1. the derivative of the second; second; f (x) ′g (x ) + g(x) ′f ( x) 2. denominator; denominator; square of the denominator; 2 ( ) ( ) – ( ) ( ) ( ) g x f x f x g x g x ′ ′ 3. nx n–1 h; nxn–1 4. kL(f); L(f) + L(g); Dx Problem Set 2.3 1. 2 2 (2 ) 2 ( ) 2 2 4x xD x D x x x= = ⋅ = 2. 3 3 2 2 (3 ) 3 ( ) 3 3 9x xD x D x x x= = ⋅ = 3. ( ) ( ) 1x xD x D xπ = π = π⋅ = π 4. 3 3 2 2 ( ) ( ) 3 3x xD x D x x xπ = π = π⋅ = π 5. –2 –2 –3 –3 (2 ) 2 ( ) 2(–2 ) –4x xD x D x x x= = = 6. –4 –4 –5 –5 (–3 ) –3 ( ) –3(–4 ) 12x xD x D x x x= = = 7. –1 –2 –2 ( ) (–1 ) –x xD D x x x x π⎛ ⎞ = π = π = π⎜ ⎟ ⎝ ⎠ = – π x2 8. –3 –4 –4 3 ( ) (–3 ) –3x xD D x x x x α α α α ⎛ ⎞ = = =⎜ ⎟ ⎝ ⎠ 4 3 – x α = 9. –5 –6 5 100 100 ( ) 100(–5 )x xD D x x x ⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ –6 6 500 –500 –x x = = 10. –5 –6 5 3 3 3 ( ) (–5 ) 4 44 x xD D x x x α α α⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ –6 6 15 15 – – 4 4 x x α α = = 11. 2 2 ( 2 ) ( ) 2 ( ) 2 2x x xD x x D x D x x+ = + = + 12. 4 3 4 3 (3 ) 3 ( ) ( )x x xD x x D x D x+ = + 3 2 3 2 3(4 ) 3 12 3x x x x= + = + 13. 4 3 2 ( 1)xD x x x x+ + + + 4 3 2 ( ) ( ) ( ) ( ) (1)x x x x xD x D x D x D x D= + + + + 3 2 4 3 2 1x x x= + + + 14. 4 3 2 2 (3 – 2 – 5 )xD x x x x+ π + π 4 3 2 2 3 ( ) – 2 ( ) – 5 ( ) ( ) ( ) x x x x x D x D x D x D x D = + π + π 3 2 3(4 ) – 2(3 ) – 5(2 ) (1) 0x x x= + π + 3 2 12 – 6 –10x x x= + π 15. 7 5 –2 ( – 2 – 5 )xD x x xπ 7 5 –2 ( ) – 2 ( ) – 5 ( )x x xD x D x D x= π 6 4 –3 (7 ) – 2(5 ) – 5(–2 )x x x= π 6 4 –3 7 –10 10x x x= π + 16. 12 2 10 ( 5 )xD x x x− − + − π 12 2 10 ( ) 5 ( ) ( )x x xD x D x D x− − = + − π 11 3 11 12 5( 2 ) ( 10 )x x x− − = + − − π − 11 3 11 12 10 10x x x− − = − + π 17. –4 –3 –4 3 3 3 ( ) ( )x x xD x D x D x x ⎛ ⎞ + = +⎜ ⎟ ⎝ ⎠ –4 –5 –5 4 9 3(–3 ) (–4 ) – – 4x x x x = + = 18. –6 –1 –6 –1 (2 ) 2 ( ) ( )x x xD x x D x D x+ = + –7 –2 –7 –2 2(–6 ) (–1 ) –12 –x x x x= + = 19. –1 –2 2 2 1 – 2 ( ) – ( )x x xD D x D x x x ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ –2 –3 2 3 2 2 2(–1 ) – (–2 ) –x x x x = = + 20. –3 –4 3 4 3 1 – 3 ( ) – ( )x x xD D x D x x x ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ –4 –5 4 5 9 4 3(–3 ) – (–4 ) –x x x x = = + 21. –11 1 2 ( ) 2 ( ) 2 2 x x xD x D x D x x ⎛ ⎞ + = +⎜ ⎟ ⎝ ⎠ –2 2 1 1 (–1 ) 2(1) – 2 2 2 x x = + = + © 2007 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.
  • 108 Section 2.3 Instructor’s Resource Manual 22. –12 2 2 2 – ( ) – 3 3 3 3 x x xD D x D x ⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ –2 2 2 2 (–1 ) – 0 – 3 3 x x = = 23. 2 2 2 [ ( 1)] ( 1) ( 1) ( )x x xD x x x D x x D x+ = + + + 2 2 (2 ) ( 1)(1) 3 1x x x x= + + = + 24. 3 3 3 [3 ( –1)] 3 ( –1) ( –1) (3 )x x xD x x x D x x D x= + 2 3 3 3 (3 ) ( –1)(3) 12 – 3x x x x= + = 25. 2 [(2 1) ]xD x + (2 1) (2 1) (2 1) (2 1)x xx D x x D x= + + + + + (2 1)(2) (2 1)(2) 8 4x x x= + + + = + 26. 2 [(–3 2) ]xD x + (–3 2) (–3 2) (–3 2) (–3 2)x xx D x x D x= + + + + + = (–3x + 2)(–3) + (–3x + 2)(–3) = 18x – 12 27. 2 3 [( 2)( 1)]xD x x+ + 2 3 3 2 ( 2) ( 1) ( 1) ( 2)x xx D x x D x= + + + + + 2 2 3 ( 2)(3 ) ( 1)(2 )x x x x= + + + 4 2 4 3 6 2 2x x x x= + + + 4 2 5 6 2x x x= + + 28. 4 2 [( –1)( 1)]xD x x + 4 2 2 4 ( –1) ( 1) ( 1) ( –1)x xx D x x D x= + + + 4 2 3 ( –1)(2 ) ( 1)(4 )x x x x= + + 5 5 3 5 3 2 – 2 4 4 6 4 – 2x x x x x x x= + + = + 29. 2 3 [( 17)( – 3 1)]xD x x x+ + 2 3 3 2 ( 17) ( – 3 1) ( – 3 1) ( 17)x xx D x x x x D x= + + + + + 2 2 3 ( 17)(3 – 3) ( – 3 1)(2 )x x x x x= + + + 4 2 4 2 3 48 – 51 2 – 6 2x x x x x= + + + 4 2 5 42 2 – 51x x x= + + 30. 4 3 2 [( 2 )( 2 1)]xD x x x x+ + + 4 3 2 3 2 4 ( 2 ) ( 2 1) ( 2 1) ( 2 )x xx x D x x x x D x x= + + + + + + + 4 2 3 2 3 ( 2 )(3 4 ) ( 2 1)(4 2)x x x x x x x= + + + + + + 6 5 3 2 7 12 12 12 2x x x x= + + + + 31. 2 2 [(5 – 7)(3 – 2 1)]xD x x x + 2 2 2 2 (5 – 7) (3 – 2 1) (3 – 2 1) (5 – 7)x xx D x x x x D x= + + + 2 2 (5 – 7)(6 – 2) (3 – 2 1)(10 )x x x x x= + + 3 2 60 – 30 – 32 14x x x= + 32. 2 4 [(3 2 )( – 3 1)]xD x x x x+ + 2 4 4 2 (3 2 ) ( – 3 1) ( – 3 1) (3 2 )x xx x D x x x x D x x= + + + + + 2 3 4 (3 2 )(4 – 3) ( – 3 1)(6 2)x x x x x x= + + + + 5 4 2 18 10 – 27 – 6 2x x x x= + + 33. 2 2 2 2 2 (3 1) (1) – (1) (3 1)1 3 1 (3 1) x x x x D D x D x x + +⎛ ⎞ =⎜ ⎟ + +⎝ ⎠ 2 2 2 2 2 (3 1)(0) – (6 ) 6 – (3 1) (3 1) x x x x x + = = + + 34. 2 2 2 2 2 (5 –1) (2) – (2) (5 –1)2 5 –1 (5 –1) x x x x D D x D x x ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 2 2 2 2 2 (5 –1)(0) – 2(10 ) 20 – (5 –1) (5 –1) x x x x x = = © 2007 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.
  • Instructor’s Resource Manual Section 2.3 109 35. 2 1 4 – 3 9 xD x x ⎛ ⎞ ⎜ ⎟ +⎝ ⎠ 2 2 2 2 (4 – 3 9) (1) – (1) (4 – 3 9) (4 – 3 9) x xx x D D x x x x + + = + 2 2 2 2 2 (4 – 3 9)(0) – (8 – 3) 8 3 – (4 – 3 9) (4 – 3 9) x x x x x x x x + − = = + + 2 2 8 3 (4 – 3 9) x x x − + = + 36. 3 4 2 – 3 xD x x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 3 3 3 2 (2 – 3 ) (4) – (4) (2 – 3 ) (2 – 3 ) x xx x D D x x x x = 3 2 2 3 2 3 2 (2 – 3 )(0) – 4(6 – 3) –24 12 (2 – 3 ) (2 – 3 ) x x x x x x x x + = = 37. 2 ( 1) ( –1) – ( –1) ( 1)–1 1 ( 1) x x x x D x x D xx D x x + +⎛ ⎞ =⎜ ⎟ +⎝ ⎠ + 2 2 ( 1)(1) – ( –1)(1) 2 ( 1) ( 1) x x x x + = = + + 38. 2 –1 –1 x x D x ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 2 ( –1) (2 –1) – (2 –1) ( –1) ( –1) x xx D x x D x x = 2 2 ( –1)(2) – (2 –1)(1) 1 – ( –1) ( –1) x x x x = = 39. 2 2 –1 3 5 x x D x ⎛ ⎞ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ 2 2 2 (3 5) (2 –1) – (2 –1) (3 5) (3 5) x xx D x x D x x + + = + 2 2 (3 5)(4 ) – (2 –1)(3) (3 5) x x x x + = + 2 2 6 20 3 (3 5) x x x + + = + 40. 2 5 – 4 3 1 x x D x ⎛ ⎞ ⎜ ⎟ +⎝ ⎠ 2 2 2 2 (3 1) (5 – 4) – (5 – 4) (3 1) (3 1) x xx D x x D x x + + = + 2 2 2 (3 1)(5) – (5 – 4)(6 ) (3 1) x x x x + = + 2 2 2 15 24 5 (3 1) x x x − + + = + 41. 2 2 – 3 1 2 1 x x x D x ⎛ ⎞+ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ 2 2 2 (2 1) (2 – 3 1) – (2 – 3 1) (2 1) (2 1) x xx D x x x x D x x + + + + = + 2 2 (2 1)(4 – 3) – (2 – 3 1)(2) (2 1) x x x x x + + = + 2 2 4 4 – 5 (2 1) x x x + = + © 2007 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.
  • 110 Section 2.3 Instructor’s Resource Manual 42. 2 5 2 – 6 3 –1 x x x D x ⎛ ⎞+ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 2 2 2 (3 –1) (5 2 – 6) – (5 2 – 6) (3 –1) (3 –1) x xx D x x x x D x x + + = 2 2 (3 –1)(10 2) – (5 2 – 6)(3) (3 –1) x x x x x + + = 2 2 15 –10 16 (3 –1) x x x + = 43. 2 2 – 1 1 x x x D x ⎛ ⎞+ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ 2 2 2 2 2 2 ( 1) ( – 1) – ( – 1) ( 1) ( 1) x xx D x x x x D x x + + + + = + 2 2 2 2 ( 1)(2 –1) – ( – 1)(2 ) ( 1) x x x x x x + + = + 2 2 2 –1 ( 1) x x = + 44. 2 2 – 2 5 2 – 3 x x x D x x ⎛ ⎞+ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ 2 2 2 2 2 2 ( 2 – 3) ( – 2 5) – ( – 2 5) ( 2 – 3) ( 2 – 3) x xx x D x x x x D x x x x + + + + = + 2 2 2 2 ( 2 – 3)(2 – 2) – ( – 2 5)(2 2) ( 2 – 3) x x x x x x x x + + + = + 2 2 2 4 –16 – 4 ( 2 – 3) x x x x = + 45. a. ( ) (0) (0) (0) (0) (0)f g f g g f′ ′ ′⋅ = + = 4(5) + (–3)(–1) = 23 b. ( ) (0) (0) (0) –1 5 4f g f g′ ′ ′+ = + = + = c. 2 (0) (0) – (0) (0) ( ) (0) (0) g f f g f g g ′ ′ ′ = 2 –3(–1) – 4(5) 17 – 9(–3) = = 46. a. ( – ) (3) (3) – (3) 2 – (–10) 12f g f g′ ′ ′= = = b. ( ) (3) (3) (3) (3) (3)f g f g g f′ ′ ′⋅ = + = 7(–10) + 6(2) = –58 c. 2 (3) (3) – (3) (3) ( ) (3) (3) f g g f g f f ′ ′ ′ = 2 7(–10) – 6(2) 82 – 49(7) = = 47. 2 [ ( )] [ ( ) ( )]x xD f x D f x f x= ( ) [ ( )] ( ) [ ( )]x xf x D f x f x D f x= + 2 ( ) ( )xf x D f x= ⋅ ⋅ 48. [ ( ) ( ) ( )] ( ) [ ( ) ( )] ( ) ( ) ( )x x xD f x g x h x f x D g x h x g x h x D f x= + ( )[ ( ) ( ) ( ) ( )] ( ) ( ) ( )x x xf x g x D h x h x D g x g x h x D f x= + + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )x x xf x g x D h x f x h x D g x g x h x D f x= + + © 2007 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.
  • Instructor’s Resource Manual Section 2.4 111 49. 2 ( – 2 2) 2 – 2xD x x x+ = At x = 1: mtan = 2(1) – 2 = 0 Tangent line: y = 1 50. 2 2 2 2 2 ( 4) (1) – (1) ( 4)1 4 ( 4) x x x x D D x D x x + +⎛ ⎞ =⎜ ⎟ + +⎝ ⎠ 2 2 2 2 2 ( 4)(0) – (2 ) 2 – ( 4) ( 4) x x x x x + = = + + At x = 1: tan 2 2 2(1) 2 – 25(1 4) m = − = + Tangent line: 1 2 – – ( –1) 5 25 y x= 2 7 – 25 25 y x= + 51. 3 2 2 ( – ) 3 – 2xD x x x x= The tangent line is horizontal when mtan = 0: 2 tan 3 – 2 0m x x= = (3 2) 0x x − = x = 0 and x = 2 3 (0, 0) and 2 4 , – 3 27 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 52. 3 2 21 – 2 –1 3 xD x x x x x ⎛ ⎞ + = +⎜ ⎟ ⎝ ⎠ 2 tan 2 –1 1m x x= + = 2 2 – 2 0x x+ = –2 4 – 4(1)(–2) –2 12 2 2 x ± ± = = –1– 3, –1 3= + –1 3x = ± 5 5 1 3, 3 , 1 3, 3 3 3 ⎛ ⎞ ⎛ ⎞ − + − − − +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 53. 5 5 6 100/ 100 ' 500 y x x y x − − = = = − Set 'y equal to 1− , the negative reciprocal of the slope of the line xy = . Solving for x gives 1/ 6 5/ 6 500 2.817 100(500) 0.563 x y − = ± ≈ ± = ± ≈ ± The points are (2.817,0.563) and )563.0,817.2( −− . 54. Proof #1: [ ] [ ] [ ] [ ] ( ) ( ) ( ) ( 1) ( ) ( ) ( 1) ( ) ( ) ( ) x x x x x x D f x g x D f x g x D f x D g x D f x D g x − = + − = + − = − Proof #2: Let ( ) ( ) ( )F x f x g x= − . Then [ ] [ ] 0 0 ( ) ( ) ( ) ( ) '( ) lim ( ) ( ) ( ) ( ) lim '( ) '( ) h h f x h g x h f x g x F x h f x h f x g x h g x h h f x g x → → + − + − − = + − + −⎡ ⎤ = −⎢ ⎥ ⎣ ⎦ = − 55. a. 2 (–16 40 100) –32 40tD t t t+ + = + v = –32(2) + 40 = –24 ft/s b. v = –32t + 40 = 0 t = 5 4 s 56. 2 (4.5 2 ) 9 2tD t t t+ = + 9t + 2 = 30 t = 28 9 s 57. 2 tan (4 – ) 4 – 2xm D x x x= = The line through (2,5) and (x0, y0) has slope 0 0 5 . 2 y x − − 2 0 0 0 0 4 – – 5 4 – 2 – 2 x x x x = 2 2 0 0 0 0–2 8 – 8 – 4 – 5x x x x+ = + 2 0 0– 4 3 0x x + = 0 0( – 3)( –1) 0x x = x0 = 1, x0 = 3 At x0 = 1: 2 0 4(1) – (1) 3y = = tan 4 – 2(1) 2m = = Tangent line: y – 3 = 2(x – 1); y = 2x + 1 At 2 0 03: 4(3) – (3) 3x y= = = tan 4 – 2(3) –2m = = Tangent line: y – 3 = –2(x – 3); y = –2x + 9 © 2007 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.
  • 112 Section 2.4 Instructor’s Resource Manual 58. 2 ( ) 2xD x x= The line through (4, 15) and 0 0( , )x y has slope 0 0 15 . 4 y x − − If (x0, y0) is on the curve y = x 2 , then 2 0 tan 0 0 –15 2 – 4 x m x x = = . 2 2 0 0 02 – 8 –15x x x= 2 0 0– 8 15 0x x + = 0 0( – 3)( – 5) 0x x = At 2 0 03: (3) 9x y= = = She should shut off the engines at (3, 9). (At x0 = 5 she would not go to (4, 15) since she is moving left to right.) 59. 2 (7 – ) –2xD x x= The line through (4, 0) and 0 0( , )x y has slope 0 0 0 . 4 y x − − If the fly is at 0 0( , )x y when the spider sees it, then 2 0 tan 0 0 7 – – 0 –2 – 4 x m x x = = . 2 2 0 0 0–2 8 7 –x x x+ = x0 2 – 8x0 + 7 = 0 0 0( – 7)( –1) 0x x = At 0 01: 6x y= = 2 2 (4 –1) (0 – 6) 9 36 45 3 5d = + = + = = ≈ 6. 7 They are 6.7 units apart when they see each other. 60. P(a, b) is 1 , .a a ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 2 1 –xD y x = so the slope of the tangent line at P is 2 1 – . a The tangent line is 2 1 1 – – ( – )y x a a a = or 2 1 – ( – 2 )y x a a = which has x-intercept (2a, 0). 2 2 2 2 1 1 ( , ) , ( , ) ( – 2 )d O P a d P A a a a a = + = + 2 2 1 ( , )a d O P a = + = so AOP is an isosceles triangle. The height of AOP is a while the base, OA has length 2a, so the area is 1 2 (2a)(a) = a2 . 61. The watermelon has volume 4 3 πr3 ; the volume of the rind is 3 3 34 4 271 – – . 3 3 10 750 r V r r r ⎛ ⎞ = π π = π⎜ ⎟ ⎝ ⎠ At the end of the fifth week r = 10, so 2 2271 271 542 (10) 340 250 250 5 rD V r π = π = π = ≈ cm3 per cm of radius growth. Since the radius is growing 2 cm per week, the volume of the rind is growing at the rate of 3542 (2) 681 cm 5 π ≈ per week. 2.4 Concepts Review 1. sin( ) – sin( )x h x h + 2. 0; 1 3. cos x; –sin x 4. 1 3 1 cos ; – – 3 2 2 2 3 y x π π⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ Problem Set 2.4 1. Dx(2 sin x + 3 cos x) = 2 Dx(sin x) + 3 Dx(cos x) = 2 cos x – 3 sin x 2. 2 (sin ) sin (sin ) sin (sin )x x xD x x D x x D x= + = sin x cos x + sin x cos x = 2 sin x cos x = sin 2x 3. 2 2 (sin cos ) (1) 0x xD x x D+ = = 4. 2 2 (1– cos ) (sin )x xD x D x= sin (sin ) sin (sin )x xx D x x D x= + = sin x cos x + sin x cos x = 2 sin x cos x = sin 2x 5. 2 1 (sec ) cos cos (1) – (1) (cos ) cos x x x x D x D x x D D x x ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = 2 sin 1 sin sec tan cos coscos x x x x x xx = = ⋅ = © 2007 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.
  • Instructor’s Resource Manual Section 2.4 113 6. 2 1 (csc ) sin sin (1) (1) (sin ) sin x x x x D x D x x D D x x ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − = 2 – cos –1 cos – csc cot sin sinsin x x x x x xx = = ⋅ = 7. 2 sin (tan ) cos cos (sin ) sin (cos ) cos x x x x x D x D x x D x x D x x ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − = 2 2 2 2 2 cos sin 1 sec cos cos x x x x x + = = = 8. 2 cos (cot ) sin sin (cos ) cos (sin ) sin x x x x x D x D x x D x x D x x ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − = 2 2 2 2 2 2 sin – cos –(sin cos ) sin sin x x x x x x − + = = 2 2 1 – – csc sin x x = = 9. 2 sin cos cos cos (sin cos ) (sin cos ) (cos ) cos x x x x x D x x D x x x x D x x +⎛ ⎞ ⎜ ⎟ ⎝ ⎠ + − + = 2 2 cos (cos – sin ) – (–sin – sin cos ) cos x x x x x x x = 2 2 2 2 2 cos sin 1 sec cos cos x x x x x + = = = 10. 2 sin cos tan tan (sin cos ) (sin cos ) (tan ) tan x x x x x D x x D x x x x D x x +⎛ ⎞ ⎜ ⎟ ⎝ ⎠ + − + = 2 2 tan (cos – sin ) – sec (sin cos ) tan x x x x x x x + = 2 2 2 2 2 2 2 2 sin sin 1 sin sin – – – cos coscos cos sin sin 1 cos sin cos coscos sin x x x x x xx x x x x x x xx x ⎛ ⎞ ⎛ ⎞ = ÷⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞⎛ ⎞ = − − −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 2 2 cos 1 cos cos sin sin sin x x x x x x = − − − 11. ( ) [ ] [ ] ( ) ( ) 2 2 sin cos sin cos cos sin sin sin cos cos cos sin x x xD x x xD x xD x x x x x x x = + = − + = − 12. ( ) [ ] [ ] ( ) ( ) ( ) 2 2 sin tan sin tan tan sin sin sec tan cos 1 sin sin cos coscos tan sec sin x x xD x x xD x xD x x x x x x x x xx x x x = + = + ⎛ ⎞ = +⎜ ⎟ ⎝ ⎠ = + 13. ( ) ( ) 2 2 sin sinsin cos sin x x x xD x xD xx D x x x x x x −⎛ ⎞ =⎜ ⎟ ⎝ ⎠ − = 14. ( ) ( ) ( ) 2 2 1 cos 1 cos1 cos sin cos 1 x x x xD x x D xx D x x x x x x − − −−⎛ ⎞ =⎜ ⎟ ⎝ ⎠ + − = 15. 2 2 2 2 ( cos ) (cos ) cos ( ) sin 2 cos x x xD x x x D x x D x x x x x = + = − + 16. 2 2 2 2 2 cos sin 1 ( 1) ( cos sin ) ( cos sin ) ( 1) ( 1) x x x x x x D x x D x x x x x x D x x +⎛ ⎞ ⎜ ⎟ +⎝ ⎠ + + − + + = + 2 2 2 ( 1)(– sin cos cos ) – 2 ( cos sin ) ( 1) x x x x x x x x x x + + + + = + 3 2 2 – sin – 3 sin 2cos ( 1) x x x x x x + = + © 2007 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.
  • 114 Section 2.4 Instructor’s Resource Manual 17. 2 2 2 2 tan (tan )(tan ) (tan )(sec ) (tan )(sec ) 2tan sec x y x x x D y x x x x x x = = = + = 18. 3 2 2 2 3 3 3 2 sec (sec )(sec ) (sec )sec tan (sec ) (sec ) sec tan sec (sec sec tan sec sec tan ) sec tan 2sec tan 3sec tan x x y x x x D y x x x x D x x x x x x x x x x x x x x x x = = = + = + ⋅ + ⋅ = + = 19. Dx(cos x) = –sin x At x = 1: tan –sin1 –0.8415m = ≈ y = cos 1 ≈ 0.5403 Tangent line: y – 0.5403 = –0.8415(x – 1) 20. 2 (cot ) – cscxD x x= tanAt : –2; 4 x m π = = y = 1 Tangent line: –1 –2 – 4 y x π⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 21. 2 2 sin 2 (2sin cos ) 2 sin cos cos sin 2sin 2cos x x x x D x D x x x D x x D x x x = = +⎡ ⎤⎣ ⎦ = − + 22. 2 2 cos2 (2cos 1) 2 cos 1 2sin cos x x x xD x D x D x D x x = − = − = − 23. ( )2 2 (30sin 2 ) 30 (2sin cos ) 30 2sin 2cos 60cos2 t tD t D t t t t t = = − + = 30sin 2 15 1 sin 2 2 2 6 12 t t t t π π = = = → = At ; 60cos 2 30 3 ft/sec 12 12 t ππ ⎛ ⎞ = ⋅ =⎜ ⎟ ⎝ ⎠ The seat is moving to the left at the rate of 30 3 ft/s. 24. The coordinates of the seat at time t are (20 cos t, 20 sin t). a. 20cos , 20sin (10 3,10) 6 6 π π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ ≈ (17.32, 10) b. Dt(20 sin t) = 20 cos t At :rate 20cos 10 3 6 6 t π π = = = ≈ 17.32 ft/s c. The fastest rate 20 cos t can obtain is 20 ft/s. 25. 2 tan ' sec y x y x = = When 0y = , tan 0 0y = = and 2 ' sec 0 1y = = . The tangent line at 0x = is y x= . 26. 2 2 2 2 tan (tan )(tan ) ' (tan )(sec ) (tan )(sec ) 2tan sec y x x x y x x x x x x = = = + = Now, 2 sec x is never 0, but tan 0x = at x kπ= where k is an integer. 27. [ ] [ ] 2 2 9sin cos ' 9 sin ( sin ) cos (cos ) 9 sin cos 9 cos2 y x x y x x x x x x x = = − + ⎡ ⎤= − ⎣ ⎦ = − The tangent line is horizontal when ' 0y = or, in this case, where cos2 0x = . This occurs when 4 2 x k π π = + where k is an integer. 28. ( ) sin '( ) 1 cos f x x x f x x = − = − '( ) 0f x = when cos 1x = ; i.e. when 2x kπ= where k is an integer. '( ) 2f x = when (2 1)x k π= + where k is an integer. 29. The curves intersect when 2 sin 2 cos ,x x= sin x = cos x at x = π 4 for 0 < x < π 2 . ( 2 sin ) 2 cosxD x x= ; 2 cos 1 4 π = ( 2 cos ) – 2 sinxD x x= ; 2 sin 1 4 π − = − 1(–1) = –1 so the curves intersect at right angles. 30. v = (3sin 2 ) 6cos2tD t t= At t = 0: v = 6 cm/s t = π 2 : v = 6− cm/s t π= : v = 6 cm/s © 2007 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.
  • Instructor’s Resource Manual Section 2.4 115 31. 2 2 2 0 2 2 2 0 sin( ) – sin (sin ) lim sin( 2 ) – sin lim x h h x h x D x h x xh h x h → → + = + + = 2 2 2 2 2 0 sin cos(2 ) cos sin(2 ) – sin lim h x xh h x xh h x h→ + + + = 2 2 2 2 0 sin [cos(2 ) –1] cos sin(2 ) lim h x xh h x xh h h→ + + + = 2 2 2 2 2 20 cos(2 ) –1 sin(2 ) lim(2 ) sin cos 2 2h xh h xh h x h x x xh h xh h→ ⎡ ⎤+ + = + +⎢ ⎥ + +⎢ ⎥⎣ ⎦ 2 2 2 2 (sin 0 cos 1) 2 cosx x x x x= ⋅ + ⋅ = 32. 0 sin(5( )) – sin5 (sin5 ) limx h x h x D x h→ + = 0 sin(5 5 ) – sin5 lim h x h x h→ + = 0 sin5 cos5 cos5 sin5 – sin5 lim h x h x h x h→ + = 0 cos5 –1 sin5 lim sin5 cos5 h h h x x h h→ ⎡ ⎤ = +⎢ ⎥ ⎣ ⎦ 0 cos5 –1 sin5 lim 5sin5 5cos5 5 5h h h x x h h→ ⎡ ⎤ = +⎢ ⎥ ⎣ ⎦ 0 5cos5 1 5cos5x x= + ⋅ = 33. f(x) = x sin x a. b. f(x) = 0 has 6 solutions on [ ,6 ]π π ′f (x) = 0 has 5 solutions on [ ,6 ]π π c. f(x) = x sin x is a counterexample. Consider the interval [ ]0,π . ( ) ( ) 0f fπ π− = = and ( ) 0f x = has exactly two solutions in the interval (at 0 and π ). However, ( )' 0f x = has two solutions in the interval, not 1 as the conjecture indicates it should have. d. The maximum value of ( ) – ( )f x f x′ on [ ,6 ]π π is about 24.93. 34. ( ) 3 2 cos 1.25cos 0.225f x x x= − + 0 1.95x ≈ ′f (x0) ≈ –1.24 2.5 Concepts Review 1. ; ( ( )) ( )tD u f g t g t′ ′ 2. ; ( ( )) ( )vD w G H s H s′ ′ 3. 2 2 ( ( )) ;( ( ))f x f x 4. 2 2 2 cos( );6(2 1)x x x + Problem Set 2.5 1. y = u15 and u = 1 + x x u xD y D y D u= ⋅ 14 (15 )(1)u= 14 15(1 )x= + 2. y = u5 and u = 7 + x x u xD y D y D u= ⋅ = (5u4 )(1) 4 5(7 )x= + 3. y = u5 and u = 3 – 2x x u xD y D y D u= ⋅ 4 4 (5 )(–2) –10(3 – 2 )u x= = © 2007 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.
  • 116 Section 2.5 Instructor’s Resource Manual 4. y = u7 and 2 4 2u x= + x u xD y D y D u= ⋅ 6 2 6 (7 )(4 ) 28 (4 2 )u x x x= = + 5. y = u11 and 3 2 – 2 3 1u x x x= + + x u xD y D y D u= ⋅ 10 2 (11 )(3 – 4 3)u x x= + 2 3 2 10 11(3 – 4 3)( – 2 3 1)x x x x x= + + + 6. –7 2 and – 1y u u x x= = + x u xD y D y D u= ⋅ –8 (–7 )(2 –1)u x= 2 –8 –7(2 –1)( – 1)x x x= + 7. –5 and 3y u u x= = + x u xD y D y D u= ⋅ –6 –6 6 5 (–5 )(1) –5( 3) – ( 3) u x x = = + = + 8. –9 2 and 3 – 3y u u x x= = + x u xD y D y D u= ⋅ –10 (–9 )(6 1)u x= + 2 –10 –9(6 1)(3 – 3)x x x= + + 2 10 9(6 1) – (3 – 3) x x x + = + 9. y = sin u and 2 u x x= + x u xD y D y D u= ⋅ = (cos u)(2x + 1) 2 (2 1)cos( )x x x= + + 10. y = cos u and 2 3 – 2u x x= x u xD y D y D u= ⋅ = (–sin u)(6x – 2) 2 –(6 – 2)sin(3 – 2 )x x x= 11. 3 y u= and u = cos x x u xD y D y D u= ⋅ 2 (3 )(–sin )u x= 2 –3sin cosx x= 12. 4 y u= , u = sin v, and 2 3v x= x u v xD y D y D u D v= ⋅ ⋅ 3 (4 )(cos )(6 )u v x= 3 2 2 24 sin (3 )cos(3 )x x x= 13. 3 1 and –1 x y u u x + = = x u xD y D y D u= ⋅ 2 2 ( –1) ( 1) – ( 1) ( –1) (3 ) ( –1) x xx D x x D x u x + + = 2 2 2 4 1 –2 6( 1) 3 –1 ( –1) ( –1) x x x x x ⎛ ⎞+ +⎛ ⎞ = = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 14. 3 y u− = and 2x u x − = − π x u xD y D y D u= ⋅ 4 2 ( ) ( 2) ( 2) ( ) 3 ) ( ) x xx D x x D x u x − − π − − − − π = (− ⋅ − π 4 2 2 4 2 (2 ) ( ) 3 3 (2 ) ( ) ( 2) x x x x x − − − π − π⎛ ⎞ = − = − − π⎜ ⎟ − π⎝ ⎠ − π − 15. y = cos u and 2 3 2 x u x = + x u xD y D y D u= ⋅ 2 2 2 ( 2) (3 ) – (3 ) ( 2) (–sin ) ( 2) x xx D x x D x u x + + = + 2 2 2 3 ( 2)(6 ) – (3 )(1) –sin 2 ( 2) x x x x x x ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ 2 2 2 3 12 3 – sin 2( 2) x x x xx ⎛ ⎞+ = ⎜ ⎟ ⎜ ⎟++ ⎝ ⎠ 16. 2 3 , cos , and 1– x y u u v v x = = = x u v xD y D y D u D v= ⋅ ⋅ 2 2 2 2 (1– ) ( ) – ( ) (1 ) (3 )( sin ) (1– ) x xx D x x D x u v x − = − 2 2 2 2 2 (1– )(2 ) – ( )(–1) –3cos sin 1– 1– (1– ) x x x x x x x x ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 2 2 2 –3(2 – ) cos sin 1– 1–(1– ) x x x x x xx ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ © 2007 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.
  • Instructor’s Resource Manual Section 2.5 117 17. 2 2 2 2 2 2 2 2 2 [(3 – 2) (3 – ) ] (3 – 2) (3 – ) (3 – ) (3 – 2)x x xD x x x D x x D x= + 2 2 2 2 (3 – 2) (2)(3 – )(–2 ) (3 – ) (2)(3 – 2)(3)x x x x x= + 2 2 2(3 2)(3 )[(3 2)( 2 ) (3 )(3)]x x x x x= − − − − + − 2 2 2(3 2)(3 )(9 4 9 )x x x x= − − + − 18. 2 4 7 3 2 4 7 3 7 3 2 4 [(2 – 3 ) ( 3) ] (2 – 3 ) ( 3) ( 3) (2 – 3 )x x xD x x x D x x D x+ = + + + 2 4 7 2 6 7 3 2 3 (2 – 3 ) (3)( 3) (7 ) ( 3) (4)(2 – 3 ) (–6 )x x x x x x= + + + 2 3 7 2 7 5 3 (3 – 2) ( 3) (29 –14 24)x x x x x= + + 19. 2 22 2 (3 – 4) ( 1) – ( 1) (3 – 4)( 1) 3 – 4 (3 – 4) x x x x D x x D xx D x x ⎡ ⎤ + ++ =⎢ ⎥ ⎢ ⎥⎣ ⎦ 2 2 2 2 (3 – 4)(2)( 1)(1) – ( 1) (3) 3 – 8 –11 (3 – 4) (3 – 4) x x x x x x x + + = = 2 ( 1)(3 11) (3 4) x x x + − = − 20. 2 2 2 2 2 2 2 4 ( 4) (2 – 3) – (2 – 3) ( 4)2 – 3 ( 4) ( 4) x x x x D x x D xx D x x ⎡ ⎤ + + =⎢ ⎥ + +⎢ ⎥⎣ ⎦ 2 2 2 2 4 ( 4) (2) – (2 – 3)(2)( 4)(2 ) ( 4) x x x x x + + = + 2 2 3 6 12 8 ( 4) x x x − + + = + 21. ( )( ) ( )( ) ( )44242442 2222 +=+= ′ ++=′ xxxxxxy 22. ( )( ) ( )( )xxxxxxxy cos1sin2sinsin2 ++=′++=′ 23. 3 2 2 ( 5) (3 – 2) – (3 – 2) ( 5)3 – 2 3 – 2 ` 3 5 5 ( 5) t t t t D t t D tt t D t t t + +⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟ + +⎝ ⎠ ⎝ ⎠ + 2 2 3 – 2 ( 5)(3) – (3 – 2)(1) 3 5 ( 5) t t t t t +⎛ ⎞ = ⎜ ⎟ +⎝ ⎠ + 2 4 51(3 – 2) ( 5) t t = + 24. 2 22 2 ( 4) ( – 9) – ( – 9) ( 4)– 9 4 ( 4) s s s s D s s D ss D s s ⎛ ⎞ + + =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ 2 2 2 2 ( 4)(2 ) – ( – 9)(1) 8 9 ( 4) ( 4) s s s s s s s + + + = = + + 25. 3 3 3 2 ( 5) (3 2) (3 2) ( 5) (3 2) 5 ( 5) d d t t t t d t dt dt dt t t + − − − +⎛ ⎞− =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ 2 3 2 ( 5)(3)(3 – 2) (3) – (3 – 2) (1) ( 5) t t t t + = + 2 2 (6 47)(3 – 2) ( 5) t t t + = + 26. 3 2 (sin ) 3sin cos d d θ θ θ θ = 27. 3 2 2 2 2 2 3 2 4 2 (cos2 ) (sin ) (sin ) (cos2 ) sin sin sin sin 3 3 cos2 cos2 cos2 cos2 cos 2 sin cos cos2 2sin sin 2 3sin cos cos2 6sin sin 2 3 cos2 cos 2 cos 2 3(sin ) d d x x x x dy d x x d x x dx dx dx dx x x dx x x x x x x x x x x x x x x x x x − ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = = ⋅ = ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ + +⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ = 4 (cos cos2 2sin sin 2 ) cos 2 x x x x x + © 2007 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.
  • 118 Section 2.5 Instructor’s Resource Manual 28. 2 2 2 2 2 2 2 2 2 [sin tan( 1)] sin [tan( 1)] tan( 1) (sin ) (sin )[sec ( 1)](2 ) tan( 1)cos 2 sin sec ( 1) cos tan( 1) dy d d d t t t t t t dt dt dt dt t t t t t t t t t t = + = ⋅ + + + ⋅ = + + + = + + + 29. 2 2 22 2 ( 2) ( 1) – ( 1) ( 2)1 ( ) 3 2 ( 2) x xx D x x D xx f x x x ⎛ ⎞ + + + ++ ′ = ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ 22 2 2 2 1 2 4 – –1 3 2 ( 2) x x x x x x ⎛ ⎞+ + = ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ 2 2 2 4 3( 1) ( 4 –1) ( 2) x x x x + + = + (3) 9.6f ′ = 30. 2 3 2 4 2 4 2 3 ( ) ( 9) ( – 2) ( – 2) ( 9)t tG t t D t t D t′ = + + + 2 3 2 3 2 4 2 2 ( 9) (4)( – 2) (2 ) ( – 2) (3)( 9) (2 )t t t t t t= + + + 2 2 2 2 3 2 (7 30)( 9) ( – 2)t t t t= + + (1) –7400G′ = 31. 2 ( ) [cos( 3 1)](2 3)F t t t t′ = + + + 2 (2 3)cos( 3 1)t t t= + + + ; (1) 5cos5 1.4183F′ = ≈ 32. 2 2 ( ) (cos ) (sin ) (sin ) (cos )s sg s s D s s D s′ = π π + π π 2 (cos )(2sin )(cos )( ) (sin )(–sin )( )s s s s s= π π π π + π π π 2 2 sin [2cos – sin ]s s s= π π π π 1 – 2 g ⎛ ⎞′ = π⎜ ⎟ ⎝ ⎠ 33. 4 2 3 2 2 [sin ( 3 )] 4sin ( 3 ) sin( 3 )x xD x x x x D x x+ = + + 3 2 2 2 4sin ( 3 )cos( 3 ) ( 3 )xx x x x D x x= + + + 3 2 2 4sin ( 3 )cos( 3 )(2 3)x x x x x= + + + 3 2 2 4(2 3)sin ( 3 )cos( 3 )x x x x x= + + + 34. 5 4 [cos (4 –19)] 5cos (4 –19) cos(4 –19)t tD t t D t= 4 5cos (4 –19)[–sin(4 –19)] (4 –19)tt t D t= 4 –5cos (4 –19)sin(4 –19)(4)t t= 4 –20cos (4 –19)sin(4 –19)t t= 35. 3 2 [sin (cos )] 3sin (cos ) sin(cos )t tD t t D t= 2 3sin (cos )cos(cos ) (cos )tt t D t= 2 3sin (cos )cos(cos )(–sin )t t t= 2 –3sin sin (cos )cos(cos )t t t= 36 . 4 31 1 1 cos 4cos cos –1 –1 –1 u u u u u D D u u u ⎡ + ⎤ + +⎛ ⎞ ⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ 3 1 1 1 4cos –sin –1 –1 –1 u u u u D u u u + ⎡ + ⎤ +⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ 3 2 ( –1) ( 1) – ( 1) ( –1)1 1 –4cos sin –1 –1 ( –1) u uu D u u D uu u u u u + ++ +⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 3 2 8 1 1 cos sin –1 –1( –1) u u u uu + +⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 37. 4 2 3 2 2 [cos (sin )] 4cos (sin ) cos(sin )D Dθ θθ θ θ= 3 2 2 2 4cos (sin )[–sin(sin )] (sin )Dθθ θ θ= 3 2 2 2 2 –4cos (sin )sin(sin )(cos ) ( )Dθθ θ θ θ= 3 2 2 2 –8 cos (sin )sin(sin )(cos )θ θ θ θ= 38. 2 2 2 [ sin (2 )] sin (2 ) sin (2 )x x xD x x x D x x D x= + 2 [2sin(2 ) sin(2 )] sin (2 )(1)xx x D x x= + 2 [2sin(2 )cos(2 ) (2 )] sin (2 )xx x x D x x= + 2 [4sin(2 )cos(2 )] sin (2 )x x x x= + 2 2 sin(4 ) sin (2 )x x x= + 39. {sin[cos(sin 2 )]} cos[cos(sin 2 )] cos(sin 2 )x xD x x D x= cos[cos(sin 2 )][–sin(sin 2 )] (sin 2 )xx x D x= – cos[cos(sin 2 )]sin(sin 2 )(cos2 ) (2 )xx x x D x= –2cos[cos(sin 2 )]sin(sin 2 )(cos2 )x x x= 40. 2 {cos [cos(cos )]} 2cos[cos(cos )] cos[cos(cos )]t tD t t D t= 2cos[cos(cos )]{–sin[cos(cos )]} cos(cos )tt t D t= –2cos[cos(cos )]sin[cos(cos )][–sin(cos )] (cos )tt t t D t= 2cos[cos(cos )]sin[cos(cos )]sin(cos )(–sin )t t t t= –2sin cos[cos(cos )]sin[cos(cos )]sin(cos )t t t t= © 2007 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.
  • Instructor’s Resource Manual Section 2.5 119 41. ( ) (4) (4) (4)f g f g′ ′ ′+ = + 1 3 2 2 2 ≈ + ≈ 42. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 1 2 0 1 f g f g f g ′ ′ ′− = − ′′= − = − = 43. ( ) ( ) ( )( ) ( ) ( ) 1110222 =+=′+′=′ fggffg 44. 2 (2) (2) – (2) (2) ( ) (2) (2) g f f g f g g ′ ′ ′ = 2 (1)(1) – (3)(0) 1 (1) ≈ = 45. ( ) (6) ( (6)) (6)f g f g g′ ′ ′= (2) (6) (1)( 1) –1f g′ ′= ≈ − = 46. ( ) (3) ( (3)) (3)g f g f f′ ′ ′= 3 3 (4) (3) (1) 2 2 g f ⎛ ⎞′ ′= ≈ =⎜ ⎟ ⎝ ⎠ 47. ( ) ( ) ( ) ( )xFxDxFxFD xx 22222 ′=′= 48. ( ) ( ) ( ) ( )12 111 2 222 +′= ++′=+ xFx xDxFxFD xx 49. ( )( )[ ] ( )( ) ( )tFtFtFDt ′−= −− 32 2 50. ( )( ) ( )( ) ( )zFzF zFdz d ′−= ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ −3 2 2 1 51. ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) 2 1 2 2 1 2 1 2 2 1 2 2 2 4 1 2 2 d d F z F z F z dz dz F z F z F z F z ⎡ ⎤+ = + + ⎢ ⎥⎣ ⎦ ′ ′= + = + 52. ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )( ) 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 1 d d y y F y dy dyF y yF yd y F y y y dy F y F y y F y −⎡ ⎤ ⎡ ⎤⎢ ⎥+ = + ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ′ ′= − = − ⎛ ⎞ ′⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 53. ( ) ( ) ( ) ( ) cos cos cos sin cos d d F x F x x dx dx xF x ′= ′= − 54. ( )( ) ( )( ) ( ) ( ) ( )( ) cos sin sin d d F x F x F x dx dx F x F x = − ′= − 55. ( )( ) ( )( ) ( ) ( )( ) ( ) [ ] ( ) ( )( ) 2 2 2 tan 2 sec 2 2 sec 2 2 2 2 2 sec 2 x x x D F x F x D F x F x F x D x F x F x ⎡ ⎤ = ⎡ ⎤⎣ ⎦⎣ ⎦ ′= × × ′= 56. ( ) ( ) ( )( ) ( ) 2 2 tan 2 ' tan 2 tan 2 ' tan 2 sec 2 2 2 ' tan 2 sec 2 d d g x g x x dx dx g x x g x x = ⋅⎡ ⎤⎣ ⎦ = ⋅ = 57. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 sin sin sin x x x D F x F x F x D F x F x D F x ⎡ ⎤ ⎣ ⎦ ⎡ ⎤= × + ×⎣ ⎦ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2sin sin sin 2sin cos sin 2 sin cos sin x x F x F x D F x F x F x F x F x F x D F x F x F x F x F x F x F x F x F x = × × ⎡ ⎤⎣ ⎦ ′+ = × × × ⎡ ⎤⎣ ⎦ ′+ ′= ′+ 58. ( ) ( ) ( ) ( ) ( ) ( ) [ ] ( ) ( ) ( ) 3 2 2 3 sec 3sec sec 3sec sec tan 3 sec tan x x x D F x F x D F x F x F x F x D x F x F x F x ⎡ ⎤ = ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦ = ⎡ ⎤⎣ ⎦ ′= 59. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ' sin sin 0 0 sin 0 2sin1 1.683 xg x f x D f x f x f x g f f ′= − = − ′ ′= − = − ≈ − 60. ( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )( ) 2 2 1 sec 2 1 sec 2 1 sec 2 1 sec 2 2 2 sec 2 tan 2 1 sec 2 d d F x x x F x dx dxG x F x F x xF x F x F x F x + − + ′ = + ′+ − = + ( ) ( ) ( )( ) ( ) ( )( ) ( ) 2 2 1 sec 0 0 1 sec 0 0 1 sec 0 1 sec 0 1 1 0.713 1 sec 0 1 sec2 F F G F F F + − + ′ = = + + = = ≈ − + + © 2007 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.
  • 120 Section 2.5 Instructor’s Resource Manual 61. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) sin cos 1 1 1 sin 1 1 cos 1 2 1 sin 0 1cos0 1 F x f x g x g x f x g x F f g g f g ′ ′ ′= − + ′ ′ ′= − + = − + − = − 62. ( ) 1 sin3 ; 3 cos3 sin3 /3 3 cos3 sin 0 3 3 3 1 /3 /3 1 y x x y x x x y y x y x π π π π π π π π π π ′= + = + ′ = + = − + = − − = − − = − − + The line crosses the x-axis at 3 3 x π− = . 63. 2 sin ; 2sin cos sin 2 1 / 4 , 0, 1, 2,... y x y x x x x k kπ π ′= = = = = + = ± ± 64. ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) 3 2 22 4 3 2 4 3 2 23 4 2 4 2 3 2 2 1 2 1 3 1 1 2 1 1 3 1 1 1 2 2 2 3 1 2 2 32 48 80 32 80 1, 80 31 y x x x x x x x x x x x x y y x y x ′ = + + + + + = + + + + + ′ = + = + = − = − = + 65. ( ) ( ) ( ) ( ) ( )( ) 3 32 2 3 2 1 2 4 1 1 4 1 1 1 1/ 2 1 1 1 1 3 , 4 2 2 2 4 y x x x x y y x y x − − − ′ = − + = − + ′ = − + = − − = − + = − + 66. ( ) ( ) ( ) ( ) ( ) 2 2 2 3 2 1 2 6 2 1 0 6 1 6 1 6 0, 6 1 y x x y y x y x ′ = + = + ′ = = − = − = + The line crosses the x-axis at 1/ 6x = − . 67. ( ) ( ) ( ) ( ) ( ) 3 32 2 3 2 1 2 4 1 1 4 2 1/ 2 1 1 1 1 3 , 4 2 2 2 4 y x x x x y y x y x − − − ′ = − + = − + ′ = − = − − = − + = − + Set 0y = and solve for x. The line crosses the x-axis at 3/ 2x = . 68. a. 2 2 2 2 2 2 4cos2 7sin 2 4 7 4 7 cos 2 sin 2 1 x y t t t t ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = + = b. 2 2 2 2 ( – 0) ( – 0)L x y x y= + = + 2 2 (4cos2 ) (7sin 2 )t t= + 2 2 16cos 2 49sin 2t t= + c. 2 2 2 2 1 (16cos 2 49sin 2 ) 2 16cos 2 49sin 2 tt D t t t t D L = + + 2 2 32cos2 (cos2 ) 98sin 2 (sin 2 ) 2 16cos 2 49sin 2 t tt D t t D t t t + = + 2 2 –64cos2 sin 2 196sin 2 cos2 2 16cos 2 49sin 2 t t t t t t + = + 2 2 16sin 4 49sin 4 16cos 2 49sin 2 t t t t − + = + 2 2 33sin 4 16cos 2 49sin 2 t t t = + At t = π 8 : rate = 1 1 2 2 33 16 49⋅ + ⋅ ≈ 5.8 ft/sec. 69. a. (10cos8 ,10sin8 )t tπ π b. (10sin8 ) 10cos(8 ) (8 )t tD t t D tπ = π π 80 cos(8 )t= π π At t = 1: rate = 80π ≈ 251 cm/s P is rising at the rate of 251 cm/s. 70. a. (cos 2t, sin 2t) b. 2 2 2 (0 – cos2 ) ( – sin 2 ) 5 ,t y t+ = so 2 sin 2 25 – cos 2y t t= + c. 2 2 sin 2 25 cos 2 1 2cos2 4cos2 sin 2 2 25 cos 2 tD t t t t t t ⎛ ⎞+ −⎜ ⎟ ⎝ ⎠ = + ⋅ − 2 sin 2 2cos2 1 25 – cos 2 t t t ⎛ ⎞ ⎜ ⎟= + ⎜ ⎟ ⎝ ⎠ 71. 60 revolutions per minute is 120π radians per minute or 2π radians per second. a. (cos2 ,sin 2 )t tπ π b. 2 2 2 (0 – cos2 ) ( – sin 2 ) 5 ,t y tπ + π = so 2 sin 2 25 – cos 2y t t= π + π c. 2 2 sin 2 25 cos 2 2 cos2 1 4 cos2 sin 2 2 25 cos 2 tD t t t t t t ⎛ ⎞π + − π⎜ ⎟ ⎝ ⎠ = π π + ⋅ π π π − π 2 sin 2 2 cos2 1 25 – cos 2 t t t ⎛ ⎞π ⎜ ⎟= π π + ⎜ ⎟ π⎝ ⎠ © 2007 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.
  • Instructor’s Resource Manual Section 2.5 121 72. The minute hand makes 1 revolution every hour, so at t minutes after the hour, it makes an angle of 30 tπ radians with the vertical. By the Law of Cosines, the length of the elastic string is 2 2 10 10 – 2(10)(10)cos 30 t s π = + 10 2 – 2cos 30 tπ = 1 10 sin 15 30 2 2 – 2cos 30 ds t dt t π π = ⋅ ⋅ π 30 30 sin 3 2 – 2cos t t π π π = At 12:15, the string is stretching at the rate of 2 2 sin 0.74 3 23 2 – 2cos π π π π = ≈ cm/min 73. The minute hand makes 1 revolution every hour, so at t minutes after noon it makes an angle of 30 tπ radians with the vertical. Similarly, at t minutes after noon the hour hand makes an angle of 360 tπ with the vertical. Thus, by the Law of Cosines, the distance between the tips of the hands is 2 2 6 8 – 2 6 8cos – 30 360 t t s π π⎛ ⎞ = + ⋅ ⋅ ⎜ ⎟ ⎝ ⎠ 11 100 – 96cos 360 tπ = 11 360 1 44 11 sin 15 3602 100 – 96cos t ds t dt π π π = ⋅ 11 360 11 360 22 sin 15 100 – 96cos t t π π π = At 12:20, 11 18 11 18 22 sin 0.38 15 100 – 96cos ds dt π π π = ≈ in./min 74. From Problem 73, 11 360 11 360 22 sin . 15 100 – 96cos t t ds dt π π π = Using a computer algebra system or graphing utility to view ds dt for 0 60t≤ ≤ , ds dt is largest when t ≈ 7.5. Thus, the distance between the tips of the hands is increasing most rapidly at about 12:08. 75. 0 0sin sin 2x x= 0 0 0sin 2sin cosx x x= 0 0 1 cos [if sin 0] 2 x x= ≠ x0 = π 3 Dx(sin x) = cos x, Dx(sin 2x) = 2cos 2x, so at x0, the tangent lines to y = sin x and y = sin 2x have slopes of m1 = 1 2 and 2 1 2 – –1, 2 m ⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ respectively. From Problem 40 of Section 0.7, 2 1 1 2 – tan 1 m m m m θ = + where θ is the angle between the tangent lines. ( ) 31 2 2 11 22 –1– – tan –3, 1 (–1) θ = = = + so θ ≈ –1.25. The curves intersect at an angle of 1.25 radians. 76. 1 sin 2 2 t AB OA= 21 cos cos sin 2 2 2 2 t t t D OA AB OA= ⋅ = E = D + area (semi-circle) 2 2 1 1 cos sin 2 2 2 2 t t OA AB ⎛ ⎞ = + π⎜ ⎟ ⎝ ⎠ 2 2 21 cos sin sin 2 2 2 2 t t t OA OA= + π 2 1 sin cos sin 2 2 2 2 t t t OA ⎛ ⎞ = + π⎜ ⎟ ⎝ ⎠ 2 1 2 2 cos cos sin 2 t t D tE = + π 0 1 lim 1 1 0t D E+→ = = + cos( / 2) lim lim cos( / 2) sin( / 2) 2 0 0 0 2 t t D t E t tπ π π π − −→ → = + = = + © 2007 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.
  • 122 Section 2.5 Instructor’s Resource Manual 77. 2 andy u u x= = Dx y = Duy ⋅ Dxu 2 1 2 2 2 2 xx x x x xu x = ⋅ = = = 78. 2 2 2 2 –1 –1 ( –1) –1 x x x D x D x x = 2 2 2 2 –1 2 –1 (2 ) –1 –1 x x x x x x = = 79. sin sin (sin ) sin x x x D x D x x = sin cos cot sin sin x x x x x = = 80. a. ( ) ( ) ( )2 2 2 2 1 2 ' 2x xD L x L x D x x xx = = ⋅ = b. 4 4 4 (cos ) sec (cos )x xD L x x D x= 4 3 sec (4cos ) (cos )xx x D x= ( ) 4 3 3 4 4sec cos ( sin ) 1 4 cos sin cos x x x x x x = − = ⋅ ⋅ ⋅ − –4sec sin 4tanx x x= = − 81. [ ( ( ( (0))))] ( ( ( (0)))) ( ( (0))) ( (0)) (0) f f f f f f f f f f f f f f ′ ′ ′ ′ ′= ⋅ ⋅ ⋅ = 2 ⋅2 ⋅2 ⋅ 2 = 16 82. a. [2] [1] [1] '( ( )) '( ) '( ) ( ) d f f f x f x dx d f f f x dx = ⋅ = ⋅ b. [3] [2] [1] [1] [2] [2] '( ( ( ))) '( ( )) '( ) '( ( )) '( ( )) ( ) '( ( )) ( ) d f f f f x f f x f x dx d f f x f f x f x dx d f f x f x dx = ⋅ ⋅ = ⋅ ⋅ = ⋅ c. Conjecture: [ ] [ 1] [ 1] ( ) '( ( )) ( )n n nd d f x f f x f x dx dx − − = ⋅ 83. ( ) ( )1 1 1 2 1 2 1 2 2 ( ) 1 ( ) ( ) ( ( )) ( ) ( ( )) ( ( )) ( ) ( ) ( ) ( ) ( 1)( ( )) ( ) ( ( )) ( ) ( )( ( )) ( ) ( ( )) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) x x x x x x x x x x x x f x D D f x D f x g x f x D g x g x D f x g x g x f x g x D g x g x D f x f x g x D g x g x D f x f x D g x D f x f x D g x g xg x g x − − − − − − − ⎛ ⎞ ⎛ ⎞ = ⋅ = ⋅ = +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = ⋅ − + = − + − − = + = 2 2 2 ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) x x x x x D f x f x D g x g x D f xg x g x g x g x g x g x D f x f x D g x g x − + ⋅ = + − = 84. ( ) ( )( )( )( ) ( )( )( ) ( )( ) ( )g x f f f f x f f f x f f x f x′ ′ ′ ′ ′= ( ) ( )( )( )( ) ( )( )( ) ( )( ) ( ) ( )( )( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )( )( ) ( )( )( ) ( )( ) ( ) ( )( )( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 2 2 2 2 2 1 1 1 2 2 2 1 2 2 2 1 2 1 g x f f f f x f f f x f f x f x f f f x f f x f x f x f f x f x f x f x f x f x g x f f f f x f f f x f f x f x f f f x f f x f x f x f f x f x f x f x f x f x g x ′ ′ ′ ′ ′= ′ ′ ′ ′ ′ ′ ′ ′= = ′ ′= ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ′ ′ ′ ′ ′= ′ ′ ′ ′ ′ ′ ′ ′= = ′ ′ ′= =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ © 2007 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.
  • Instructor’s Resource Manual Section 2.6 123 2.6 Concepts Review 1. 3 3 3 ( ), ,x d y f x D y dx ′′′ , '''y 2. 2 2 ; ; ds ds d s dt dt dt 3. ( ) 0f t′ > 4. 0; < 0 Problem Set 2.6 1. 2 3 6 6 dy x x dx = + + 2 2 6 6 d y x dx = + 3 3 6 d y dx = 2. 4 3 5 4 dy x x dx = + d2y dx 2 = 20x3 +12x 2 3 2 3 60 24 d y x x dx = + 3. 2 2 3(3 5) (3) 9(3 5) dy x x dx = + = + 2 2 18(3 5)(3) 162 270 d y x x dx = + = + 3 3 162 d y dx = 4. 4 4 5(3 – 5 ) (–5) –25(3 – 5 ) dy x x dx = = 2 3 3 2 –100(3 – 5 ) (–5) 500(3 – 5 ) d y x x dx = = 3 2 2 3 1500(3 – 5 ) (–5) –7500(3 – 5 ) d y x x dx = = 5. 7cos(7 ) dy x dx = 2 2 2 –7 sin(7 ) d y x dx = 3 3 3 –7 cos(7 ) –343cos(7 ) d y x x dx = = 6. 2 3 3 cos( ) dy x x dx = 2 2 2 3 3 2 3 [–3 sin( )] 6 cos( ) d y x x x x x dx = + 4 3 3 –9 sin( ) 6 cos( )x x x x= + 3 4 3 2 3 3 3 2 3 3 –9 cos( )(3 ) sin( )(–36 ) 6 [–sin( )(3 )] 6cos( ) d y x x x x x x x x x dx = + + + 6 3 3 3 3 3 3 –27 cos( ) – 36 sin( ) –18 sin( ) 6cos( )x x x x x x x= + 6 3 3 3 (6 – 27 )cos( ) – 54 sin( )x x x x= 7. 2 2 ( –1)(0) – (1)(1) 1 – ( –1) ( –1) dy x dx x x = = 2 2 2 4 3 ( –1) (0) – 2( –1) 2 ( –1) ( –1) d y x x dx x x = − = 3 3 2 3 6 4 ( 1) (0) 2[3( 1) ] ( 1) 6 ( 1) d y x x dx x x − − − = − = − − 8. 2 2 (1– )(3) – (3 )(–1) 3 (1– ) ( –1) dy x x dx x x = = 2 2 2 4 3 ( –1) (0) – 3[2( –1)] 6 – ( –1) ( –1) d y x x dx x x = = 3 3 2 3 6 4 ( 1) (0) 6(3)( 1) ( 1) 18 ( 1) d y x x dx x x − − − = − − = − © 2007 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.
  • 124 Section 2.6 Instructor’s Resource Manual 9. ( ) 2 ; ( ) 2; (2) 2f x x f x f′ ′′ ′′= = = 10. 2 ( ) 15 4 1f x x x′ = + + ( ) 30 4f x x′′ = + (2) 64f ′′ = 11. 2 2 ( ) –f t t ′ = 3 4 ( )f t t ′′ = 4 1 (2) 8 2 f ′′ = = 12. 2 2 2 2 (5 – )(4 ) – (2 )(–1) 20 – 2 ( ) (5 – ) (5 – ) u u u u u f u u u ′ = = 2 2 4 (5 – ) (20 – 4 ) – (20 – 2 )2(5 – )(–1) ( ) (5 – ) u u u u u f u u ′′ = 3 100 (5 – )u = 3 100 100 (2) 273 f ′′ = = 13. –3 ( ) –2(cos ) (–sin )f θ θ θ′ = π π π = 2π(cosθπ)–3 (sinθπ) –3 –4 ( ) 2 [(cos ) ( )(cos ) (sin )(–3)(cos ) (–sin )( )]f θ θ θ θ θ θ′′ = π π π π + π π π π 2 2 2 4 2 [(cos ) 3sin (cos ) ]θ θ θ− − = π π + π π 2 2 (2) 2 [1 3(0)(1)] 2f ′′ = π + = π 14. 2 ( ) cos – sinf t t t tt π π π⎛ ⎞⎛ ⎞ ⎛ ⎞′ = +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ – cos sin t t t π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 2 2 ( ) – –sin – cos – cosf t t t t tt t t ⎡ ⎤π π π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ′′ = + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ 2 3 – sin tt π π⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 2 2 (2) – sin – –1.23 8 2 8 f π π π⎛ ⎞′′ = = ≈⎜ ⎟ ⎝ ⎠ 15. 2 2 2 3 ( ) (3)(1– ) (–2 ) (1– )f s s s s s′ = + 2 2 2 2 3 –6 (1– ) (1– )s s s= + 6 4 2 –7 15 – 9 1s s s= + + 5 3 ( ) –42 60 –18f s s s s′′ = + (2) –900f ′′ = 16. 2 2 ( –1)2( 1) – ( 1) ( ) ( –1) x x x f x x + + ′ = 2 2 – 2 – 3 ( –1) x x x = 2 2 4 ( –1) (2 – 2) – ( – 2 – 3)2( –1) ( ) ( –1) x x x x x f x x ′′ = 2 3 3 ( –1)(2 – 2) – ( – 2 – 3)(2) 8 ( –1) ( –1) x x x x x x = = 3 8 (2) 8 1 f ′′ = = 17. –1 ( )n n xD x nx= 2 –2 ( ) ( –1)n n xD x n n x= 3 –3 ( ) ( –1)( – 2)n n xD x n n n x= 4 –4 ( ) ( –1)( – 2)( – 3)n n xD x n n n n x= 1 ( ) ( –1)( – 2)( – 3)...(2)n n xD x n n n n x− = 0 ( ) ( –1)( – 2)( – 3)...2(1)n n xD x n n n n x= = n! 18. Let k < n. ( ) [ ( )] ( !) 0n k n k k k x x x xD x D D x D k− = = = so –1 1 0[ ] 0n n x nD a x a x a+…+ + = 19. a. 4 3 (3 2 –19) 0xD x x+ = b. 12 11 10 (100 79 ) 0xD x x− = c. 11 2 5 ( – 3) 0xD x = © 2007 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.
  • Instructor’s Resource Manual Section 2.6 125 20. 2 1 1 –xD x x ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 2 –2 –3 3 1 2 (– ) 2x xD D x x x x ⎛ ⎞ = = =⎜ ⎟ ⎝ ⎠ 3 –3 4 1 3(2) (2 ) –x xD D x x x ⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ 4 5 1 4(3)(2) xD x x ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 1 1 ( 1) !n n x n n D x x + −⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 21. 2 ( ) 3 6 – 45 3( 5)( 3)f x x x x x′ = + = + − 3(x + 5)(x – 3) = 0 x = –5, x = 3 ( ) 6 6f x x′′ = + (–5) –24f ′′ = (3) 24f ′′ = 22. ( ) 2g t at b′ = + ( ) 2g t a′′ = (1) 2 4 2 g a a ′′ = = − = − (1) 2 3g a b′ = + = 2(–2) + b = 3 b = 7 ( ) ( ) ( ) 1 5 2 7 5 0 g a b c c c = + + = − + + = = 23. a. ( ) 12 – 4 ds v t t dt = = 2 2 ( ) –4 d s a t dt = = b. 12 – 4t > 0 4t < 12 t < 3; ( ),3−∞ c. 12 – 4t < 0 t > 3; (3, )∞ d. a(t) = –4 < 0 for all t e. 24. a. 2 ( ) 3 –12 ds v t t t dt = = 2 2 ( ) 6 –12 d s a t t dt = = b. 2 3 –12 0t t > 3t(t – 4) > 0; ( ,0) (4, )−∞ ∪ ∞ c. 2 3 –12 0t t < (0, 4) d. 6t – 12 < 0 6t < 12 t < 2; ( ,2)−∞ e. 25. a. 2 ( ) 3 –18 24 ds v t t t dt = = + 2 2 ( ) 6 –18 d s a t t dt = = b. 2 3 –18 24 0t t + > 3(t – 2)(t – 4) > 0 ( ,2) (4, )−∞ ∪ ∞ c. 3t2 –18t +24 < 0 (2, 4) d. 6t – 18 < 0 6t < 18 t < 3; ( ,3)−∞ e. 26. a. 2 ( ) 6 – 6 ds v t t dt = = 2 2 ( ) 12 d s a t t dt = = b. 2 6 – 6 0t > 6(t + 1)(t – 1) > 0 ( , 1) (1, )−∞ − ∪ ∞ c. 2 6 – 6 0t < (–1, 1) d. 12t < 0 t < 0 The acceleration is negative for negative t. © 2007 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.
  • 126 Section 2.6 Instructor’s Resource Manual e. 27. a. 2 16 ( ) 2 – ds v t t dt t = = 2 2 3 32 ( ) 2 d s a t dt t = = + b. 2 16 2 – 0t t > 3 2 2 –16 0; t t > (2, )∞ c. 2t – 16 t2 < 0; (0, 2) d. 2 + 32 t3 < 0 2t3 + 32 t3 < 0; The acceleration is not negative for any positive t. e. 28. a. 2 4 ( ) 1– ds v t dt t = = 2 2 3 8 ( ) d s a t dt t = = b. 2 4 1– 0 t > 2 2 – 4 0; t t > (2, )∞ c. 2 4 1– 0; t < (0, 2) d. 3 8 0; t < The acceleration is not negative for any positive t. e. 29. 3 2 ( ) 2 –15 24 ds v t t t t dt = = + 2 2 2 ( ) 6 – 30 24 d s a t t t dt = = + 2 6 – 30 24 0t t + = 6(t – 4)(t – 1) = 0 t = 4, 1 v(4) = –16, v(1) = 11 30. 3 21 ( ) (4 – 42 120 ) 10 ds v t t t t dt = = + 2 2 2 1 ( ) (12 – 84 120) 10 d s a t t t dt = = + 21 (12 – 84 120) 0 10 t t + = 12 ( 2)( 5) 0 10 t t− − = t = 2, t = 5 v(2) = 10.4, v(5) = 5 31. 1 1( ) 4 – 6 ds v t t dt = = 2 2 ( ) 2 – 2 ds v t t dt = = a. 4 – 6t = 2t – 2 8t = 6 3 4 t = sec b. 4 – 6 2 – 2 ;t t= 4 – 6t = –2t + 2 1 3 sec and 2 4 t t= = sec c. 2 2 4 – 3 – 2t t t t= 2 4 – 6 0t t = 2t(2t – 3) = 0 3 0 sec and sec 2 t t= = 32. 21 1( ) 9 – 24 18 ds v t t t dt = = + 22 2 ( ) –3 18 –12 ds v t t t dt = = + 2 2 9 – 24 18 –3 18 –12t t t t+ = + 2 12 – 42 30 0t t + = 2 2 – 7 5 0t t + = (2t – 5)(t – 1) = 0 t =1, 5 2 © 2007 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.
  • Instructor’s Resource Manual Section 2.6 127 33. a. v(t) = –32t + 48 initial velocity = v0 = 48 ft/sec b. –32t + 48 = 0 t = 3 2 sec c. 2 –16(1.5) 48(1.5) 256 292s = + + = ft d. 2 –16 48 256 0t t+ + = 2 –48 48 – 4(–16)(256) –2.77, 5.77 –32 t ± = ≈ The object hits the ground at t = 5.77 sec. e. v(5.77) ≈ –137 ft/sec; speed = 137 137ft/sec.− = 34. v(t) = 48 –32t a. 48 – 32t = 0 t = 1.5 2 48(1.5) –16(1.5) 36s = = ft b. v(1) = 16 ft/sec upward c. 2 48 –16 0t t = –16t(–3 + t) = 0 t = 3 sec 35. 0( ) – 32v t v t= 0 – 32 0v t = 0 32 v t = 2 0 0 0 –16 5280 32 32 v v v ⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 0 0 – 5280 32 64 v v = 2 0 5280 64 v = 0 337,920 581v = ≈ ft/sec 36. 0( ) 32v t v t= + 0 32 140v t+ = 0 32(3) 140v + = 0 44v = 2 44(3) 16(3) 276s = + = ft 37. 2 ( ) 3 – 6 – 24v t t t= 2 2 2 3 – 6 – 24 3 – 6 – 24 (6 – 6) 3 – 6 – 24 t td t t t dt t t = ( – 4)( 2) (6 – 6) ( – 4)( 2) t t t t t + = + ( – 4)( 2) (6 – 6) 0 ( – 4)( 2) t t t t t + < + t < –2, 1 < t < 4; ( , 2)−∞ − ∪ (1, 4) 38. Point slowing down when ( ) 0 d v t dt < ( ) ( ) ( ) ( ) v t a td v t dt v t = ( ) ( ) 0 ( ) v t a t v t < when a(t) and v(t) have opposite signs. 39. ( )xD uv uv u v′ ′= + 2 ( ) 2 xD uv uv u v u v u v uv u v u v ′′ ′ ′ ′ ′ ′′= + + + ′′ ′ ′ ′′= + + 3 ( ) 2( )xD uv uv u v u v u v u v u v′′′ ′ ′′ ′ ′′ ′′ ′ ′′ ′ ′′′= + + + + + 3 3uv u v u v u v′′′ ′ ′′ ′′ ′ ′′′= + + + 0 ( ) ( ) ( ) n n n k k x x x k n D uv D u D v k − = ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ∑ where n k ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ is the binomial coefficient ! . ( – )! ! n n k k 40. 4 4 4 4 04 ( sin ) ( ) (sin ) 0 x x xD x x D x D x ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 3 4 1 2 4 24 4 ( ) (sin ) ( ) (sin ) 1 2 x x x xD x D x D x D x ⎛ ⎞ ⎛ ⎞ + +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 1 4 3 0 4 44 4 ( ) (sin ) ( ) (sin ) 3 4 x x x xD x D x D x D x ⎛ ⎞ ⎛ ⎞ + +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 3 4 24sin 96 cos 72 sin 16 cos sin x x x x x x x x x = + − − + © 2007 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.
  • 128 Section 2.7 Instructor’s Resource Manual 41. a. b. ′′′f (2.13) ≈ –1.2826 42. a. b. (2.13) 0.0271f ′′′ ≈ 2.7 Concepts Review 1. 3 9 – 3x 2. 2 3 dy y dx 3. 2 2 2 (2 ) 3 – 3 dy dy dy x y y y x dx dx dx + + = 4. –1 2 2/35 ; ( – 5 ) (2 – 5) 3 p qp x x x x q Problem Set 2.7 1. 2 – 2 0xy D y x = 2 2 x x x D y y y = = 2. 18 8 0xx y D y+ = –18 9 – 8 4 x x x D y y y = = 3. 0xx D y y+ = –x y D y x = 4. 2 2 2 0xx yD yα+ = 2 2 2 – – 2 x x x D y y yα α = = 5. 2 (2 ) 1xx y D y y+ = 2 1– 2 x y D y xy = 6. 2 2 2 4 3 3 0x xx x D y xy x D y y+ + + + = 2 (2 3 ) –2 – 4 – 3xD y x x x xy y+ = 2 –2 – 4 – 3 2 3 x x xy y D y x x = + 7. 2 2 2 12 7 (2 ) 7 6x xx x y D y y y D y+ + = 2 2 2 12 7 6 –14x xx y y D y xyD y+ = 2 2 2 12 7 6 –14 x x y D y y xy + = 8. 2 2 2 (2 )x xx D y xy y x y D y+ = + 2 2 – 2 – 2x xx D y xyD y y xy= 2 2 – 2 – 2 x y xy D y x xy = 9. 2 3 1 (5 5 ) 2 2 5 2 (3 ) x x x x x D y y D y xy y D y x y D y y ⋅ + + = + + 2 3 5 2 – 2 – 3 2 5 5 – 2 5 x x x x x D y D y y D y xy D y xy y y xy + = 53 2 5 25 2 5 – 2 – 2 – 3 y xy x x xy y D y y xy = + 10. 1 1 2 1 x xx D y y x D y y y + + = + + – – 1 2 1 x x x D y x D y y y y = + + 2 1 – 1 – x x y y y D y x + + = © 2007 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.
  • Instructor’s Resource Manual Section 2.7 129 11. cos( )( ) 0x xx D y y xy x D y y+ + + = cos( ) – – cos( )x xx D y x xy D y y y xy+ = – – cos( ) – cos( ) x y y xy y D y x x xy x = = + 12. 2 2 –sin( )(2 ) 2 1x xxy xy D y y yD y+ = + 2 2 2 –2 sin( ) – 2 1 sin( )x xxy xy D y y D y y xy= + 2 2 2 1 sin( ) –2 sin( ) – 2 x y xy D y xy xy y + = 13. 3 2 3 2 3 3 0x y x y y xy y′ ′+ + + = 3 2 2 3 ( 3 ) –3 –y x xy x y y′ + = 2 3 3 2 –3 – 3 x y y y x xy ′ = + At (1, 3), 36 9 – – 28 7 y′ = = Tangent line: 9 – 3 – ( –1) 7 y x= 14. 2 2 (2 ) 2 4 4 12x y y xy xy y y′ ′ ′+ + + = 2 2 (2 4 –12) –2 – 4y x y x xy y′ + = 2 2 2 2 –2 – 4 – – 2 2 4 –12 2 – 6 xy y xy y y x y x x y x ′ = = + + At (2, 1), –2y′ = Tangent line: –1 –2( – 2)y x= 15. cos( )( )xy xy y y′ ′+ = [ cos( ) –1] – cos( )y x xy y xy′ = – cos( ) cos( ) cos( ) –1 1– cos( ) y xy y xy y x xy x xy ′ = = At ,1 , 0 2 y π⎛ ⎞ ′ =⎜ ⎟ ⎝ ⎠ Tangent line: –1 0 – 2 y x π⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ y = 1 16. 2 2 [–sin( )][2 ] 6 0y xy xyy y x′ ′+ + + = 2 2 2 [1– 2 sin( )] sin( ) – 6y xy xy y xy x′ = 2 2 2 sin( ) – 6 1– 2 sin( ) y xy x y xy xy ′ = At (1, 0), 6 – –6 1 y′ = = Tangent line: y – 0 = –6(x – 1) 17. –1/3 –1/32 2 – – 2 0 3 3 x y y y′ ′ = –1/3 –1/32 2 2 3 3 x y y ⎛ ⎞′= +⎜ ⎟ ⎝ ⎠ –1/32 3 –1/ 32 3 2 x y y ′ = + At (1, –1), 2 3 4 3 1 2 y′ = = Tangent line: 1 1 ( –1) 2 y x+ = 18. 21 2 0 2 y xyy y y ′ ′+ + = 21 2 – 2 y xy y y ⎛ ⎞ ′ + =⎜ ⎟⎜ ⎟ ⎝ ⎠ 2 1 2 – 2 y y y xy ′ = + At (4, 1), 17 2 –1 2 – 17 y′ = = Tangent line: 2 –1 – ( – 4) 17 y x= 19. 2/3 1 5 2 dy x dx x = + 20. –2/3 5/ 2 5/ 2 3 2 1 1 – 7 – 7 3 3 dy x x x dx x = = 21. –2/3 –4/3 3 32 4 1 1 1 1 – – 3 3 3 3 dy x x dx x x = = 22. –3/ 4 34 1 1 (2 1) (2) 4 2 (2 1) dy x dx x = + = + 23. 2 –3/ 41 (3 – 4 ) (6 – 4) 4 dy x x x dx = 2 3 2 34 4 6 – 4 3 – 2 4 (3 – 4 ) 2 (3 – 4 ) x x x x x x = = 24. 3 –2/3 21 ( – 2 ) (3 – 2) 3 dy x x x dx = 25. 3 2/3 [( 2 ) ] dy d x x dx dx − = + 2 3 –5/3 2 3 53 2 6 4 – ( 2 ) (3 2) 3 3 ( 2 ) x x x x x x + = + + = − + © 2007 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.
  • 130 Section 2.7 Instructor’s Resource Manual 26. –8/3 –8/35 – (3 – 9) (3) –5(3 – 9) 3 dy x x dx = = 27. 2 1 (2 cos ) 2 sin dy x x dx x x = + + 2 2 cos 2 sin x x x x + = + 28. 2 2 1 [ (–sin ) 2 cos ] 2 cos dy x x x x dx x x = + 2 2 2 cos – sin 2 cos x x x x x x = 29. 2 –1/3 [( sin ) ] dy d x x dx dx = 2 –4/3 21 – ( sin ) ( cos 2 sin ) 3 x x x x x x= + 2 2 43 cos 2 sin – 3 ( sin ) x x x x x x + = 30. –3/ 41 (1 sin5 ) (cos5 )(5) 4 dy x x dx = + 34 5cos5 4 (1 sin5 ) x x = + 31. 2 –3/ 4 2 [1 cos( 2 )] [–sin( 2 )(2 2)] 4 dy x x x x x dx + + + + = 2 2 34 ( 1)sin( 2 ) 2 [1 cos( 2 )] x x x x x + + = − + + 32. 2 2 –1/ 2 2 (tan sin ) (2 tan sec 2sin cos ) 2 x x x x x xdy dx + + = 2 2 2 tan sec sin cos tan sin x x x x x x + = + 33. 2 2 2 3 0 ds s st t dt + + = 2 2 2 2 – – 3 3 2 2 ds s t s t dt st st + = = − 2 2 2 3 0 dt dt s st t ds ds + + = 2 2 ( 3 ) –2 dt s t st ds + = 2 2 2 3 dt st ds s t = − + 34. 2 2 1 cos( )(2 ) 6 dx dx x x x dy dy = + 2 2 1 2 cos( ) 6 dx dy x x x = + 35. 5 −5 5−5 y x (x + 2)2 + y2 = 1 2 4 2 0 dy x y dx + + = 2 4 2 2 dy x x dx y y + + = − = − The tangent line at 0 0( , )x y has equation 0 0 0 0 2 – ( – ) x y y x x y + = − which simplifies to 2 2 0 0 0 0 02 – – 2 – 0.x yy x xx y x+ + = Since 0 0( , )x y is on the circle, 2 2 0 0 0–3 – 4 ,x y x+ = so the equation of the tangent line is 0 0 0– – 2 – 2 – 3.yy x x xx = If (0, 0) is on the tangent line, then 0 3 – . 2 x = Solve for 0y in the equation of the circle to get 0 3 . 2 y = ± Put these values into the equation of the tangent line to get that the tangent lines are 3 0y x+ = and 3 – 0.y x = 36. 2 2 16( )(2 2 ) 100(2 – 2 )x y x yy x yy′ ′+ + = 3 2 2 3 32 32 32 32 200 – 200x x yy xy y y x yy′ ′ ′+ + + = 2 3 3 2 (4 4 25 ) 25 – 4 – 4y x y y y x x xy′ + + = 3 2 2 3 25 – 4 – 4 4 4 25 x x xy y x y y y ′ = + + The slope of the normal line 1 – y = ′ 2 3 3 2 4 4 25 4 4 – 25 x y y y x xy x + + = + At (3, 1), 65 13 slope 45 9 = = Normal line: 13 –1 ( – 3) 9 y x= © 2007 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.
  • Instructor’s Resource Manual Section 2.7 131 37. a. 2 3 0xy y y y′ ′+ + = 2 ( 3 ) –y x y y′ + = 2 – 3 y y x y ′ = + b. 2 2 2 2 2 – – 3 3 3 – 6 0 3 y y xy y y x y x y y y x y ⎛ ⎞ ⎛ ⎞ ′′ ′′+ + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎛ ⎞ + =⎜ ⎟⎜ ⎟+⎝ ⎠ 3 2 2 2 2 2 6 3 – 0 3 ( 3 ) y y xy y y x y x y ′′ ′′+ + = + + 3 2 2 2 2 2 6 ( 3 ) – 3 ( 3 ) y y y x y x y x y ′′ + = + + 2 2 2 2 ( 3 ) ( 3 ) xy y x y x y ′′ + = + 2 3 2 ( 3 ) xy y x y ′′ = + 38. 2 3 – 8 0x yy′ = 2 3 8 x y y ′ = 2 6 – 8( ( ) ) 0x yy y′′ ′+ = 22 3 6 – 8 – 8 0 8 x x yy y ⎛ ⎞ ′′ =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 4 2 9 6 – 8 – 0 8 x x yy y ′′ = 2 4 2 48 9 8 8 xy x yy y − ′′= 2 4 3 48 – 9 64 xy x y y ′′ = 39. 2 2 2( 2 ) –12 0x y xy y y′ ′+ = 2 2 2 –12 –4x y y y xy′ ′ = 2 2 2 6 – xy y y x ′ = 2 2 2 2( 2 2 2 ) –12[ 2 ( ) ] 0x y xy xy y y y y y′′ ′ ′ ′′ ′+ + + + = 2 2 2 2 12 8 4 24 ( )x y y y xy y y y′′ ′′ ′ ′− = − − + 2 2 3 2 2 2 2 2 2 2 16 96 (2 –12 ) – 4 6 – (6 ) x y x y y x y y y x y x ′′ = − + − 4 2 3 5 2 2 2 2 2 12 48 144 (2 –12 ) (6 – ) x y x y y y x y y x + − ′′ = 5 4 2 3 2 2 2 2 2 72 6 24 (6 – ) (6 – ) y x y x y y y x y x − − ′′ = 5 4 2 3 2 2 3 72 6 24 (6 – ) y x y x y y y x − − ′′ = At (2, 1), 120 15 8 y − ′′ = = − 40. 2 2 0x yy′+ = 2 – – 2 x x y y y ′ = = 2 2 2[ ( ) ] 0yy y′′ ′+ + = 2 2 2 2 – 0 x yy y ⎛ ⎞ ′′+ + =⎜ ⎟ ⎝ ⎠ 2 2 2 2 2 x yy y ′′ = − − 2 2 2 3 3 1 x y x y y y y + ′′ = − − = − At (3, 4), 25 64 y′′ = − 41. 2 2 3 3 3( )x y y xy y′ ′+ = + 2 2 (3 – 3 ) 3 – 3y y x y x′ = 2 2 – – y x y y x ′ = At 3 3 , , 2 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ –1y′ = Slope of the normal line is 1. Normal line: 3 3 – 1 – ; 2 2 y x y x ⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ This line includes the point (0, 0). 42. 0xy y′+ = – y y x ′ = 2 2 0x yy′− = x y y ′ = The slopes of the tangents are negative reciprocals, so the hyperbolas intersect at right angles. © 2007 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.
  • 132 Section 2.7 Instructor’s Resource Manual 43. Implicitly differentiate the first equation. 4 2 0x yy′+ = 2 – x y y ′ = Implicitly differentiate the second equation. 2 4yy′ = 2 y y ′ = Solve for the points of intersection. 2 2 4 6x x+ = 2 2( 2 – 3) 0x x+ = (x + 3)(x – 1) = 0 x = –3, x = 1 x = –3 is extraneous, and y = –2, 2 when x = 1. The graphs intersect at (1, –2) and (1, 2). At (1, –2): 1 21, –1m m= = At (1, 2): 1 2–1, 1m m= = 44. Find the intersection points: 2 2 2 2 1 1x y y x+ = → = − ( ) ( ) ( ) 2 2 2 2 2 2 1 1 1 1 1 1 2 1 1 1 2 x y x x x x x x − + = − + − = − + + − = ⇒ = Points of intersection: 1 3 1 3 , and , – 2 2 2 2 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Implicitly differentiate the first equation. 2 2 0x yy′+ = – x y y ′ = Implicitly differentiate the second equation. 2( –1) 2 0x yy′+ = 1– x y y ′ = At 1 2 1 3 1 1 , : – , 2 2 3 3 m m ⎛ ⎞ = =⎜ ⎟⎜ ⎟ ⎝ ⎠ ( )( ) 1 1 2 3 3 3 21 1 3 3 3 tan 3 31 θ θ + π = = = → = + − At 1 2 1 3 1 1 , – : , – 2 2 3 3 m m ⎛ ⎞ = =⎜ ⎟⎜ ⎟ ⎝ ⎠ ( )( ) 1 1 2 3 3 3 21 1 3 3 3 tan 3 1 – θ − − − = = = − + 2 3 θ π = 45. 2 2 – (2 ) 2(2 ) 28x x x x+ = 2 7 28x = 2 4x = x = –2, 2 Intersection point in first quadrant: (2, 4) 1 2y′ = 2 22 – – 4 0x xy y yy′ ′+ = 2 (4 – ) – 2y y x y x′ = 2 – 2 4 – y x y y x ′ = At (2, 4): 1 22, 0m m= = –10 – 2 tan –2; tan (–2) 2.034 1 (0)(2) θ θ= = = π + ≈ + 46. The equation is 2 2 2 2 0 0– – .mv mv kx kx= Differentiate implicitly with respect to t to get 2 –2 . dv dx mv kx dt dt = Since dx v dt = this simplifies to 2 –2 dv mv kxv dt = or – . dv m kx dt = 47. 2 2 – 16x xy y+ = , when y = 0, 2 16x = x = –4, 4 The ellipse intersects the x-axis at (–4, 0) and (4, 0). 2 – – 2 0x xy y yy′ ′+ = (2 – ) – 2y y x y x′ = – 2 2 – y x y y x ′ = At (–4, 0), 2y′ = At (4, 0), 2y′ = Tangent lines: y = 2(x + 4) and y = 2(x – 4) © 2007 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.
  • Instructor’s Resource Manual Section 2.8 133 48. 2 2 2 – 2 – 0 dx dx x xy xy y dy dy + = 2 2 (2 – ) 2 – ; dx xy y xy x dy = 2 2 2 – 2 – dx xy x dy xy y = 2 2 2 – 0 2 – xy x xy y = if x(2y – x) = 0, which occurs when x = 0 or . 2 x y = There are no points on 2 2 – 2x y xy = where x = 0. If , 2 x y = then 2 3 3 3 2 2 – – 2 2 2 4 4 x x x x x x x ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ so x = 2, 2 1. 2 y = = The tangent line is vertical at (2, 1). 49. 2 2 0; – dy dy x x y dx dx y + = = The tangent line at 0 0( , )x y has slope 0 0 – , x y hence the equation of the tangent line is 0 0 0 0 – – ( – ) x y y x x y = which simplifies to 2 2 0 0 0 0– ( ) 0yy xx x y+ + = or 0 0 1yy xx+ = since 0 0( , )x y is on 2 2 1x y+ = . If (1.25, 0) is on the tangent line through 0 0( , )x y , 0 0.8.x = Put this into 2 2 1x y+ = to get 0 0.6,y = since 0 0.y > The line is 6y + 8x = 10. When x = –2, 13 , 3 y = so the light bulb must be 13 3 units high. 2.8 Concepts Review 1. ; 2 du t dt = 2. 400 mi/hr 3. negative 4. negative; positive Problem Set 2.8 1. 3 ; 3 dx V x dt = = 2 3 dV dx x dt dt = When x = 12, 2 3(12) (3) 1296 dV dt = = in.3/s. 2. 34 ; 3 3 dV V r dt = π = 2 4 dV dr r dt dt = π When r = 3, 2 3 4 (3) dr dt = π . 1 0.027 12 dr dt = ≈ π in./s 3. 2 2 2 1 ; 400 dx y x dt = + = 2 2 dy dx y x dt dt = dy x dx dt y dt = mi/hr When 5 5, 26, (400) 26 dy x y dt = = = ≈ 392 mi/h. 4. 21 3 3 ; ; 3 10 10 r h V r h r h = π = = 2 3 1 3 3 ; 3, 5 3 10 100 h h dV V h h dt π⎛ ⎞ = π = = =⎜ ⎟ ⎝ ⎠ 2 9 100 dV h dh dt dt π = When h = 5, 2 9 (5) 3 100 dh dt π = 4 0.42 3 dh dt = ≈ π cm/s 5. 2 2 2 ( 300) ; 300, 400, dx dy s x y dt dt = + + = = 2 2( 300) 2 ds dx dy s x y dt dt dt = + + ( 300) ds dx dy s x y dt dt dt = + + When x = 300, y = 400, 200 13s = , so 200 13 (300 300)(300) 400(400) ds dt = + + 471 ds dt ≈ mi/h © 2007 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.
  • 134 Section 2.8 Instructor’s Resource Manual 6. 2 2 2 (10) ; 2 dy y x dt = + = 2 2 dy dx y x dt dt = When y = 25, x ≈ 22.9, so 25 (2) 2.18 22.9 dx y dy dt x dt = ≈ ≈ ft/s 7. 2 2 2 20 ; 1 dx x y dt = + = 0 2 2 dx dy x y dt dt = + When x = 5, 375 5 15y = = , so 5 – – (1) –0.258 5 15 dy x dx dt y dt = = ≈ ft/s The top of the ladder is moving down at 0.258 ft/s. 8. –4 dV dt = ft3/h; 2 ; –0.0005 dh V hr dt = π = ft/h 2 –1 , V A r Vh h = π = = so –1 2 – dA dV V dh h dt dt dth = . When h = 0.001 ft, 2 (0.001)(250) 62.5V = π = π and 1000(–4) –1,000,000(62.5 )(–0.0005) dA dt = π = –4000 + 31,250π ≈ 94,175 ft2/h. (The height is decreasing due to the spreading of the oil rather than the bacteria.) 9. 21 ; , 2 3 4 2 d r V r h h r h= π = = = 2 31 4 (2 ) ; 16 3 3 dV V h h h dt = π = π = 2 4 dV dh h dt dt = π When h = 4, 2 16 4 (4) dh dt = π 1 0.0796 4 dh dt = ≈ π ft/s 10. 2 2 2 (90) ; 5 dx y x dt = + = 2 2 dy dx y x dt dt = When y = 150, x = 120, so 120 (5) 4 150 dy x dx dt y dt = = = ft/s 11. 40 (20); , 8 2 5 hx x V x h h = = = 2 10 (8 ) 80 ; 40 dV V h h h dt = = = 160 dV dh h dt dt = When h = 3, 40 160(3) dh dt = 1 12 dh dt = ft/min 12. 2 – 4; 5 dx y x dt = = 2 2 1 (2 ) 2 – 4 – 4 dy dx x dx x dt dt dtx x = = When x = 3, 2 3 15 (5) 6.7 53 – 4 dy dt = = ≈ units/s 13. 2 ; 0.02 dr A r dt = π = 2 dA dr r dt dt = π When r = 8.1, 2 (0.02)(8.1) 0.324 dA dt = π = π ≈ 1.018 in.2/s 14. 2 2 2 ( 48) ; 30, 24 dx dy s x y dt dt = + + = = 2 2 2( 48) ds dx dy s x y dt dt dt = + + ( 48) ds dx dy s x y dt dt dt = + + At 2:00 p.m., x = 3(30) = 90, y = 3(24) = 72, so s = 150. (150) 90(30) (72 48)(24) ds dt = + + 5580 37.2 150 ds dt = = knots/h © 2007 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.
  • Instructor’s Resource Manual Section 2.8 135 15. Let x be the distance from the beam to the point opposite the lighthouse and θ be the angle between the beam and the line from the lighthouse to the point opposite. tan ; 2(2 ) 4 1 x d dt θ θ = = π = π rad/min, 2 sec d dx dt dt θ θ = At –1 21 1 5 , tan and sec 2 2 4 x θ θ= = = . 5 (4 ) 15.71 4 dx dt = π ≈ km/min 16. 4000 tan x θ = 2 2 4000 sec d dx dt dtx θ θ = − When 1 2 1 1 4000 , and 7322. 2 10 tan d x dt θ θ = = = ≈ 2 2 1 1 (7322) sec 2 10 4000 dx dt ⎡ ⎤⎛ ⎞ ≈ −⎢ ⎥⎜ ⎟ ⎝ ⎠⎢ ⎥⎣ ⎦ ≈ –1740 ft/s or –1186 mi/h The plane’s ground speed is 1186 mi/h. 17. a. Let x be the distance along the ground from the light pole to Chris, and let s be the distance from Chris to the tip of his shadow. By similar triangles, 6 30 , s x s = + so 4 x s = and 1 . 4 ds dx dt dt = 2 dx dt = ft/s, hence 1 2 ds dt = ft/s no matter how far from the light pole Chris is. b. Let l = x + s, then 1 5 2 2 2 dl dx ds dt dt dt = + = + = ft/s. c. The angular rate at which Chris must lift his head to follow his shadow is the same as the rate at which the angle that the light makes with the ground is decreasing. Let θ be the angle that the light makes with the ground at the tip of Chris' shadow. 6 tan s θ = so 2 2 6 sec – d ds dt dts θ θ = and 2 2 6cos – . d ds dt dts θ θ = 1 2 ds dt = ft/s When s = 6, , 4 θ π = so ( ) 2 1 2 2 6 1 1 – – . 2 246 d dt θ ⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ Chris must lift his head at the rate of 1 24 rad/s. 18. Let θ be the measure of the vertex angle, a be the measure of the equal sides, and b be the measure of the base. Observe that 2 sin 2 b a θ = and the height of the triangle is cos . 2 a θ 21 1 2 sin cos sin 2 2 2 2 A a a a θ θ θ ⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 21 1 (100) sin 5000sin ; 2 10 d A dt θ θ θ= = = 5000cos dA d dt dt θ θ= When 1 , 5000 cos 250 3 6 6 10 dA dt θ π π⎛ ⎞⎛ ⎞ = = =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 2 433 cm min≈ . 19. Let p be the point on the bridge directly above the railroad tracks. If a is the distance between p and the automobile, then 66 da dt = ft/s. If l is the distance between the train and the point directly below p, then 88 dl dt = ft/s. The distance from the train to p is 2 2 100 ,l+ while the distance from p to the automobile is a. The distance between the train and automobile is 2 2 2 2 2 2 2 100 100 .D a l a l⎛ ⎞= + + = + +⎜ ⎟ ⎝ ⎠ 2 2 2 1 2 2 2 100 dD da dl a l dt dt dta l ⎛ ⎞ = ⋅ +⎜ ⎟ ⎝ ⎠+ + 2 2 2 . 100 da dl dt dt a l a l + = + + After 10 seconds, a = 660 and l = 880, so 2 2 2 660(66) 880(88) 110 660 880 100 dD dt + = ≈ + + ft/s. © 2007 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.
  • 136 Section 2.8 Instructor’s Resource Manual 20. 2 21 ( ); 20, 20, 3 4 h V h a ab b a b= π ⋅ + + = = + 2 1 400 5 400 10 400 3 16 h V h h h ⎛ ⎞ = π + + + + +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 3 21 1200 15 3 16 h h h ⎛ ⎞ = π + +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 2 1 3 1200 30 3 16 dV h dh h dt dt ⎛ ⎞ = π + +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ When h = 30 and 2000, dV dt = 1 675 3025 2000 1200 900 3 4 4 dh dh dt dt π⎛ ⎞ = π + + =⎜ ⎟ ⎝ ⎠ 320 0.84 121 dh dt = ≈ π cm/min. 21. 2 – ; –2, 8 3 h dV V h r r dt ⎡ ⎤ = π = =⎢ ⎥ ⎣ ⎦ 3 3 2 2 – 8 – 3 3 h h V rh h π π = π = π 2 16 – dV dh dh h h dt dt dt = π π When h = 3, 2 –2 [16 (3) – (3) ] dh dt = π π –2 –0.016 39 dh dt = ≈ π ft/hr 22. 2 2 2 2 cos ;s a b ab θ= + − a = 5, b = 4, 11 2 – 6 6 d dt θ π π = π = rad/h 2 41– 40coss θ= 2 40sin ds d s dt dt θ θ= At 3:00, and 41 2 sθ π = = , so 11 220 2 41 40sin 2 6 3 ds dt π π π⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 18 ds dt ≈ in./hr 23. Let P be the point on the ground where the ball hits. Then the distance from P to the bottom of the light pole is 10 ft. Let s be the distance between P and the shadow of the ball. The height of the ball t seconds after it is dropped is 2 64 –16 .t By similar triangles, 2 48 10 64 –16 s st + = (for t > 1), so 2 2 10 – 40 . 1– t s t = 2 2 2 2 20 (1– ) – (10 – 40)(–2 ) (1– ) ds t t t t dt t = 2 2 60 – (1– ) t t = The ball hits the ground when t = 2, 120 – . 9 ds dt = The shadow is moving 120 13.33 9 ≈ ft/s. 24. 2 – ; 20 3 h V h r r ⎛ ⎞ = π =⎜ ⎟ ⎝ ⎠ 2 2 3 20 – 20 3 3 h V h h h π⎛ ⎞ = π = π −⎜ ⎟ ⎝ ⎠ 2 (40 ) dV dh h h dt dt = π − π At 7:00 a.m., h = 15, 3, dh dt ≈ − so 2 (40 (15) (15) )( 3) 1125 3534. dV dt = π − π − ≈ − π ≈ − Webster City residents used water at the rate of 2400 + 3534 = 5934 ft3/h. 25. Assuming that the tank is now in the shape of an upper hemisphere with radius r, we again let t be the number of hours past midnight and h be the height of the water at time t. The volume, V, of water in the tank at that time is given by ( )3 22 ( ) 2 3 3 V r r h r h π π= − − + and so ( )216000 (20 ) 40 3 3 V h h π π= − − + from which ( )2 2 (20 ) (20 ) 40 3 3 dV dh dh h h h dt dt dt π π = − − + − + At 7t = , 525 1649 dV dt π≈ − ≈ − Thus Webster City residents were using water at the rate of 2400 1649 4049+ = cubic feet per hour at 7:00 A.M. 26. The amount of water used by Webster City can be found by: usage beginning amount added amount remaining amount = + − Thus the usage is 2 2 3 (20) (9) 2400(12) (20) (10.5) 26,915 ftπ π≈ + − ≈ over the 12 hour period. © 2007 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.
  • Instructor's Resource Manual Section 2.8 137 27. a. Let x be the distance from the bottom of the wall to the end of the ladder on the ground, so 2 dx dt = ft/s. Let y be the height of the opposite end of the ladder. By similar triangles, 2 18 , 12 144 y x = + so 2 216 . 144 y x = + 2 3/ 2 216 – 2 2(144 ) dy dx x dt dtx = + 2 3/ 2 216 – (144 ) x dx dtx = + When the ladder makes an angle of 60° with the ground, 4 3x = and 3/ 2 216(4 3) – 2 –1.125 (144 48) dy dt = ⋅ = + ft/s. b. 2 2 2 3/ 2 216 – (144 ) d y d x dx dt dtdt x ⎛ ⎞ = ⎜ ⎟⎜ ⎟+⎝ ⎠ 2 2 3/ 2 2 3/ 2 2 216 216 – – (144 ) (144 ) d x dx x d x dt dtx x dt ⎛ ⎞ = ⋅⎜ ⎟⎜ ⎟+ +⎝ ⎠ Since 2 2 2, 0, dx d x dt dt = = thus ( )2 3/ 2 232 2 2 2 3 –216(144 ) 216 144 (2 ) (144 ) dx dx dt dt x x x xd y dx dtdt x ⎡ ⎤+ + + ⎢ ⎥= ⎢ ⎥+⎢ ⎥⎣ ⎦ 22 2 2 5/ 2 –216(144 ) 648 (144 ) x x dx dtx + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠+ 22 2 5/ 2 432 – 31,104 (144 ) x dx dtx ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠+ When the ladder makes an angle of 60° with the ground, 2 2 2 5/ 2 432 48 – 31,104 (2) –0.08 (144 48) d y dt ⋅ = ≈ + ft/s2 28. a. If the ball has radius 6 in., the volume of the water in the tank is 33 2 4 1 8 – – 3 3 2 h V h π ⎛ ⎞ = π π⎜ ⎟ ⎝ ⎠ 3 2 8 – – 3 6 h h π π = π 2 16 – dV dh dh h h dt dt dt = π π This is the same as in Problem 21, so dh dt is again –0.016 ft/hr. b. If the ball has radius 2 ft, and the height of the water in the tank is h feet with 2 3h≤ ≤ , the part of the ball in the water has volume 2 3 24 4 – (6 – ) (2) – (4 – ) 2 – . 3 3 3 h h h h π⎡ ⎤ π π =⎢ ⎥ ⎣ ⎦ The volume of water in the tank is 3 2 2 2(6 – ) 8 – – 6 3 3 h h h V h h π π = π = π 12 dV dh h dt dt = π 1 12 dh dV dt h dt = π When h = 3, 1 (–2) –0.018 36 dh dt = ≈ π ft/hr. 29. 2 (4 ) dV k r dt = π a. 34 3 V r= π 2 4 dV dr r dt dt = π 2 2 (4 ) 4 dr k r r dt π = π dr k dt = b. If the original volume was 0V , the volume after 1 hour is 0 8 . 27 V The original radius was 3 0 0 3 4 r V= π while the radius after 1 hour is 3 1 0 0 8 3 2 . 27 4 3 r V r= ⋅ = π Since dr dt is constant, 0 1 – 3 dr r dt = unit/hr. The snowball will take 3 hours to melt completely. © 2007 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.
  • 138 Section 2.9 Instructor’s Resource Manual 30. PV = k 0 dV dP P V dt dt + = At t = 6.5, P ≈ 67, –30, dP dt ≈ V = 300 300 – – (–30) 134 67 dV V dP dt P dt = = ≈ in.3/min 31. Let l be the distance along the ground from the brother to the tip of the shadow. The shadow is controlled by both siblings when 3 5 4l l = + or l = 6. Again using similar triangles, this occurs when 6 , 20 3 y = so y = 40. Thus, the girl controls the tip of the shadow when y ≥ 40 and the boy controls it when y < 40. Let x be the distance along the ground from the light pole to the girl. –4 dx dt = When y ≥ 40, 20 5 –y y x = or 4 . 3 y x= When y < 40, 20 3 – ( 4)y y x = + or 20 ( 4). 17 y x= + x = 30 when y = 40. Thus, 4 if 30 3 20 ( 4) if 30 17 x x y x x ⎧ ≥⎪⎪ = ⎨ ⎪ + < ⎪⎩ and 4 if 30 3 20 if 30 17 dx x dy dt dxdt x dt ⎧ ≥⎪⎪ = ⎨ ⎪ < ⎪⎩ Hence, the tip of the shadow is moving at the rate of 4 16 (4) 3 3 = ft/s when the girl is at least 30 feet from the light pole, and it is moving 20 80 (4) 17 17 = ft/s when the girl is less than 30 ft from the light pole. 2.9 Concepts Review 1. ( )f x dx′ 2. ;y dyΔ 3. Δx is small. 4. larger ; smaller Problem Set 2.9 1. dy = (2x + 1)dx 2. 2 (21 6 )dy x x dx= + 3. –5 –5 –4(2 3) (2) –8(2 3)dy x dx x dx= + = + 4. 2 –3 –2(3 1) (6 1)dy x x x dx= + + + 2 –3 –2(6 1)(3 1)x x x dx= + + + 5. 2 3(sin cos ) (cos – sin )dy x x x x dx= + 6. 2 2 3(tan 1) (sec )dy x x dx= + 2 2 3sec (tan 1)x x dx= + 7. 2 –5/ 23 – (7 3 –1) (14 3) 2 dy x x x dx= + + 2 5 23 (14 3)(7 3 1) 2 x x x dx− = − + + − 8. 10 9 1 2( sin 2 )[10 (cos 2 )(2)] 2 sin 2 x x x x x dy dx+ + ⋅= 9 10cos2 2 10 ( sin 2 ) sin 2 x x x x dx x ⎛ ⎞ = + +⎜ ⎟ ⎝ ⎠ 9. 2 1/ 2 23 ( – cot 2) (2 csc ) 2 ds t t t t dt= + + 2 23 (2 csc ) – cot 2 2 t t t t dt= + + 10. a. 2 2 3 3(0.5) (1) 0.75dy x dx= = = b. 2 2 3 3(–1) (0.75) 2.25dy x dx= = = 11. 12. a. 2 2 0.5 – – –0.5 (1) dx dy x = = = b. 2 2 0.75 – – –0.1875 (–2) dx dy x = = = © 2007 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.
  • Instructor’s Resource Manual Section 2.9 139 13. 14. a. 3 3 (1.5) – (0.5) 3.25yΔ = = b. 3 3 (–0.25) – (–1) 0.984375yΔ = = 15. a. 1 1 1 – – 1.5 1 3 yΔ = = b. 1 1 –0.3 –1.25 2 yΔ = + = 16. a. 2 2 [(2.5) – 3] –[(2) – 3] 2.25yΔ = = dy = 2xdx = 2(2)(0.5) = 2 b. 2 2 [(2.88) – 3] –[(3) – 3] –0.7056yΔ = = dy = 2xdx = 2(3)(–0.12) = –0.72 17. a. 4 4 [(3) 2(3)] –[(2) 2(2)] 67yΔ = + + = 3 3 (4 2) [4(2) 2](1) 34dy x dx= + = + = b. 4 4 [(2.005) 2(2.005)] –[(2) 2(2)]yΔ = + + ≈ 0.1706 3 3 (4 2) [4(2) 2](0.005) 0.17dy x dx= + = + = 18. 1 ; ; 400, 2 2 y x dy dx x dx x = = = = 1 (2) 0.05 2 400 dy = = 402 400 20 0.05 20.05dy≈ + = + = 19. 1 ; ; 36, –0.1 2 y x dy dx x dx x = = = = 1 (–0.1) –0.0083 2 36 dy = ≈ 35.9 36 6 – 0.0083 5.9917dy≈ + = = 20. –2/33 3 2 1 1 ; ; 3 3 y x dy x dx dx x = = = x = 27, dx = –0.09 23 1 (–0.09) –0.0033 3 (27) dy = ≈ 3 3 26.91 27 3 – 0.0033 2.9967dy≈ + = = 21. 34 ; 5, 0.125 3 V r r dr= π = = 2 2 4 4 (5) (0.125) 39.27dV r dr= π = π ≈ cm3 22. 3 3 ; 40, 0.5V x x dx= = = 2 23 3 3( 40) (0.5) 17.54dV x dx= = ≈ in.3 23. 34 ; 6ft 72in., –0.3 3 V r r dr= π = = = 2 2 4 4 (72) (–0.3) –19,543dV r dr= π = π ≈ 3 3 3 4 (72) –19,543 3 1,543,915 in 893 ft V ≈ π ≈ ≈ 24. 2 ; 6ft 72in., 0.05,V r h r dr= π = = = − 8ft 96in.h = = 3 2 2 (72)(96)( 0.05) 2171in.dV rhdr= π = π − ≈ − About 9.4 gal of paint are needed. 25. 2C rπ= ; r = 4000 mi = 21,120,000 ft, dr = 2 2 2 (2) 4 12.6dC dr ftπ π π= = = ≈ 26. 2 ; 4, –0.03 32 L T L dL= π = = 32 2 1 32 322 L dT dL dL L π π = ⋅ ⋅ = (–0.03) –0.0083 32(4) dT π = ≈ The time change in 24 hours is (0.0083)(60)(60)(24) ≈ 717 sec 27. 3 34 4 (10) 4189 3 3 V r= π = π ≈ 2 2 4 4 (10) (0.05) 62.8dV r dr= π = π ≈ The volume is 4189 ± 62.8 cm3. The absolute error is ≈ 62.8 while the relative error is 62.8/ 4189 0.015 or 1.5%≈ . © 2007 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.
  • 140 Section 2.9 Instructor’s Resource Manual 28. 2 2 (3) (12) 339V r h= π = π ≈ 24 24 (3)(0.0025) 0.565dV rdr= π = π ≈ The volume is 339 ± 0.565 in.3 The absolute error is ≈ 0.565 while the relative error is 0.565/339 0.0017 or 0.17%≈ . 29. 2 2 – 2 coss a b ab θ= + 2 2 151 151 – 2(151)(151)cos0.53= + ≈ 79.097 45,602 – 45,602coss θ= 1 45,602sin 2 45,602 – 45,602cos ds dθ θ θ = ⋅ 22,801sin 45,602 – 45,602cos d θ θ θ = 22,801sin 0.53 (0.005) 45,602 – 45,602cos0.53 = ≈ 0.729 s ≈ 79.097 ± 0.729 cm The absolute error is ≈ 0.729 while the relative error is 0.729/ 79.097 0.0092 or 0.92%≈ . 30. 1 1 sin (151)(151)sin 0.53 5763.33 2 2 A ab θ= = ≈ 22,801 sin ; 0.53, 0.005 2 A dθ θ θ= = = 22,801 (cos ) 2 dA dθ θ= 22,801 (cos0.53)(0.005) 49.18 2 = ≈ A ≈ 5763.33 ± 49.18 cm2 The absolute error is ≈ 49.18 while the relative error is 49.18/5763.33 0.0085 or 0.85%≈ . 31. 2 3 – 2 11; 2, 0.001y x x x dx= + = = dy = (6x – 2)dx = [6(2) – 2](0.001) = 0.01 2 2 6, d y dx = so with Δx = 0.001, 21 – (6)(0.001) 0.000003 2 y dyΔ ≤ = 32. Using the approximation ( ) ( ) '( )f x x f x f x x+ Δ ≈ + Δ we let 1.02x = and 0.02xΔ = − . We can rewrite the above form as ( ) ( ) '( )f x f x x f x x≈ + Δ − Δ which gives (1.02) (1) '(1.02)( 0.02) 10 12(0.02) 10.24 f f f≈ − − = + = 33. Using the approximation ( ) ( ) '( )f x x f x f x x+ Δ ≈ + Δ we let 3.05x = and 0.05xΔ = − . We can rewrite the above form as ( ) ( ) '( )f x f x x f x x≈ + Δ − Δ which gives (3.05) (3) '(3.05)( 0.05) 1 8 (0.05) 8.0125 4 f f f≈ − − = + = 34. From similar triangles, the radius at height h is 2 . 5 h Thus, 2 31 4 , 3 75 V r h h= π = π so 24 . 25 dV h dh= π h = 10, dh = –1: 34 (100)( 1) 50cm 25 dV = π − ≈ − The ice cube has volume 3 3 3 27cm ,= so there is room for the ice cube without the cup overflowing. 35. 2 34 3 V r h r= π + π 2 34 100 ; 10, 0.1 3 V r r r dr= π + π = = 2 (200 4 )dV r r dr= π + π (2000 400 )(0.1) 240 754= π + π = π ≈ cm3 36. The percent increase in mass is . dm m –3/ 22 0 2 2 2 – 1– – 2 m v v dm dv c c ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ –3/ 22 0 2 2 1– m v v dv c c ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ –12 2 2 2 2 2 2 1– dm v v v c dv dv m c c c c v ⎛ ⎞ ⎛ ⎞ = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ 2 2 v dv c v = − v = 0.9c, dv = 0.02c 2 2 0.9 0.018 (0.02 ) 0.095 0.190.81 dm c c m c c = = ≈ − The percent increase in mass is about 9.5. © 2007 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.
  • Instructor’s Resource Manual Section 2.9 141 37. 2 ( ) ; '( ) 2 ; 2f x x f x x a= = = The linear approximation is then ( ) (2) '(2)( 2) 4 4( 2) 4 4 L x f f x x x = + − = + − = − 38. 2 2 ( ) cos ; '( ) sin 2 cos / 2 g x x x g x x x x x a π = = − + = The linear approximation is then 2 2 3 ( ) 0 2 2 4 8 L x x x π π π π ⎛ ⎞ ⎛ ⎞ = + − −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = − + 2 2 3 ( ) 0 4 2 4 8 L x x x π π π π ⎛ ⎞ = + − −⎜ ⎟ ⎝ ⎠ = − + 39. ( ) sin ; '( ) cos ; 0h x x h x x a= = = The linear approximation is then ( ) 0 1( 0)L x x x= + − = 40. ( ) 3 4; '( ) 3; 3F x x F x a= + = = The linear approximation is then ( ) 13 3( 3) 13 3 9 3 4 L x x x x = + − = + − = + © 2007 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.
  • 142 Section 2.9 Instructor’s Resource Manual 41. ( ) ( ) ( ) 2 1/ 2 2 2 1 ; 1 1 ( 2 ) 2 , 0 1 f x x f x x x x a x − = − ′ = − − − = = − The linear approximation is then ( ) ( )1 0 0 1L x x= + − = 42. ( ) 2 1 x g x x = − ; ( ) ( ) ( ) ( ) ( ) 2 2 2 22 2 1 2 1 1 ' , 21 1 x x x x g x a x x − − − + = = = − − The linear approximation is then ( ) 9 4 9 20 2 1 9 20 3 2 −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+= xxxL 43. ( ) ( ) 0,tansecsec;sec =+=′= axxxxxhxxxh The linear approximation is then ( ) ( ) xxxL =−+= 010 44. ( ) ( ) 2/,2cos21;2sin π=+=′+= axxGxxxG The linear approximation is then ( ) ( ) π ππ +−=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −−+= xxxL 2 1 2 45. ( ) ( ) mxfbmxxf =′+= ; The linear approximation is then ( ) ( ) ( ) ( )xLxfbmx mamxbamaxmbmaxL =+= −++=−++= 46. ( ) ( ) ( ) ( ) 2 1 2 2 22 2 0 2 L x f x a x a x a x a x a x a x a a x a a − = + − − − + = − + = − = ≥ 47. The linear approximation to ( )f x at a is 2 2 ( ) ( ) '( )( ) 2 ( ) 2 L x f a f a x a a a x a ax a = + − = + − = − Thus, ( )2 2 2 2 2 ( ) ( ) 2 2 ( ) 0 f x L x x ax a x ax a x a − = − − = − + = − ≥ 48. ( ) ( ) ( ) ( ) 0,1,1 1 =+=′+= − axxfxxf αα α The linear approximation is then ( ) ( ) 11 +=+= xxxL αα 5 −5 −5 5 x y 2α = − 5 −5 −5 5 x y 1α = − © 2007 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.
  • Instructor’s Resource Manual Section 2.10 143 5 −5 −5 5 x y 0.5α = − 5 −5 −5 5 x y 0α = 5 −5 −5 5 x y 0.5α = 5 −5 −5 5 x y 1α = 5 −5 −5 5 x y 2α = 49. a. ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 0 0 lim lim 0 0 h h h f x h f x f x h f x f x f x ε → → ′= + − − ′= − − = b. ( ) ( ) ( ) ( ) ( ) ( ) 0 lim lim 0 h h f x h f x f x h h f x f x ε → + −⎡ ⎤ ′= −⎢ ⎥ ⎣ ⎦ ′ ′= − = 2.10 Chapter Review Concepts Test 1. False: If 3 2 ( ) , '( ) 3f x x f x x= = and the tangent line 0 at 0y x= = crosses the curve at the point of tangency. 2. False: The tangent line can touch the curve at infinitely many points. 3. True: 3 tan 4 ,m x= which is unique for each value of x. 4. False: tan –sin ,m x= which is periodic. 5. True: If the velocity is negative and increasing, the speed is decreasing. 6. True: If the velocity is negative and decreasing, the speed is increasing. 7. True: If the tangent line is horizontal, the slope must be 0. 8. False: 2 2 ( ) , ( ) ,f x ax b g x ax c= + = + b c≠ . Then ( ) 2 ( ),f x ax g x′ ′= = but f(x) ≠ g(x). 9. True: ( ( )) ( ( )) ( );xD f g x f g x g x′ ′= since g(x) = x, ( ) 1,g x′ = so ( ( )) ( ( )).xD f g x f g x′= 10. False: 0xD y = because π is a constant, not a variable. 11. True: Theorem 3.2.A 12. True: The derivative does not exist when the tangent line is vertical. 13. False: ( ) ( ) ( ) ( ) ( ) ( )f g x f x g x g x f x′ ′ ′⋅ = + 14. True: Negative acceleration indicates decreasing velocity. © 2007 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.
  • 144 Section 2.10 Instructor’s Resource Manual 15. True: If 3 ( ) ( ),f x x g x= then 3 2 ( ) ( ) 3 ( )xD f x x g x x g x′= + 2 [ ( ) 3 ( )].x xg x g x′= + 16. False: 2 3 ;xD y x= At (1, 1): 2 tan 3(1) 3m = = Tangent line: y – 1 = 3(x – 1) 17. False: ( ) ( ) ( ) ( )xD y f x g x g x f x′ ′= + 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) xD y f x g x g x f x g x f x f x g x ′′ ′ ′= + ′′ ′ ′+ + ( ) ( ) 2 ( ) ( ) ( ) ( )f x g x f x g x f x g x′′ ′ ′ ′′= + + 18. True: The degree of 3 8 ( )y x x= + is 24, so 25 0.xD y = 19. True: –1 ( ) ; ( )n n f x ax f x anx′= = 20. True: 2 ( ) ( ) ( ) – ( ) ( ) ( ) ( ) x f x g x f x f x g x D g x g x ′ ′ = 21. True: ( ) ( ) ( ) ( ) ( )h x f x g x g x f x′ ′ ′= + ( ) ( ) ( ) ( ) ( )h c f c g c g c f c′ ′ ′= + = f(c)(0) + g(c)(0) = 0 22. True: ( )2 22 sin – sin lim 2 –x x f x π ππ→ π⎛ ⎞ ′ =⎜ ⎟ ⎝ ⎠ 22 sin –1 lim –x x x ππ→ = 23. True: 2 2 ( )D kf kD f= and 2 2 2 ( )D f g D f D g+ = + 24. True: ( ) ( ( )) ( )h x f g x g x′ ′ ′= ⋅ ( ) ( ( )) ( ) 0h c f g c g c′ ′ ′= ⋅ = 25. True: ( ) (2) ( (2)) (2)f g f g g′ ′ ′= ⋅ (2) (2) 2 2 4f g′ ′= ⋅ = ⋅ = 26. False: Consider ( ) .f x x= The curve always lies below the tangent. 27. False: The rate of volume change depends on the radius of the sphere. 28. True: 2c rπ= ; 4 dr dt = 2 2 (4) 8 dc dr dt dt = π = π = π 29. True: (sin ) cos ;xD x x= 2 (sin ) –sin ;xD x x= 3 (sin ) – cos ;xD x x= 4 (sin ) sin ;xD x x= 5 (sin ) cosxD x x= 30. False: (cos ) –sin ;xD x x= 2 (cos ) – cos ;xD x x= 3 (cos ) sin ;xD x x= 4 3 (cos ) [ (cos )] (sin )x x x xD x D D x D x= = Since 1 3 1 (cos ) (sin ),x xD x D x+ = 3 (cos ) (sin ).n n x xD x D x+ = 31. True: 0 0 tan 1 sin lim lim 3 3 cosx x x x x x x→ → = 1 1 1 3 3 = ⋅ = 32. True: 2 15 6 ds v t dt = = + which is greater than 0 for all t. 33. True: 34 3 V r= π 2 4 dV dr r dt dt = π If 3, dV dt = then 2 3 4 dr dt r = π so 0. dr dt > 2 2 3 3 – 2 d r dr dtdt r = π so 2 2 0 d r dt < 34. True: When h > r, then 2 2 0 d h dt > 35. True: 34 , 3 V r= π 2 4S r= π 2 4dV r dr S dr= π = ⋅ If Δr = dr, then dV S r= ⋅Δ 36. False: 4 5 ,dy x dx= so dy > 0 when dx > 0, but dy < 0 when dx < 0. 37. False: The slope of the linear approximation is equal to '( ) '(0) sin(0) 0f a f= = − = . © 2007 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.
  • Instructor’s Resource Manual Section 2.10 145 Sample Test Problems 1. a. 3 3 0 3( ) – 3 ( ) lim h x h x f x h→ + ′ = 2 2 3 0 9 9 3 lim h x h xh h h→ + + = 2 2 2 0 lim (9 9 3 ) 9 h x xh h x → = + + = b. 5 5 0 [2( ) 3( )] – (2 3 ) ( ) lim h x h x h x x f x h→ + + + + ′ = 4 3 2 2 3 4 5 0 10 20 20 10 2 3 lim h x h x h x h xh h h h→ + + + + + = 4 3 2 2 3 4 0 lim (10 20 20 10 2 3) h x x h x h xh h → = + + + + + 4 10 3x= + c. 1 1 3( ) 3 0 0 – 1 ( ) lim lim – 3( ) x h x h h h f x h x h x h + → → ⎡ ⎤ ′ = = ⎢ ⎥ +⎣ ⎦ 20 1 1 lim – – 3 ( ) 3h x x h x→ ⎛ ⎞ = =⎜ ⎟ +⎝ ⎠ d. 2 20 1 1 1 ( ) lim – 3( ) 2 3 2h f x hx h x→ ⎡ ⎤⎛ ⎞ ′ = ⎢ ⎥⎜ ⎟⎜ ⎟+ + +⎢ ⎥⎝ ⎠⎣ ⎦ 2 2 2 20 3 2 – 3( ) – 2 1 lim (3( ) 2)(3 2)h x x h hx h x→ ⎡ ⎤+ + = ⋅⎢ ⎥ + + +⎢ ⎥⎣ ⎦ 2 2 20 –6 – 3 1 lim (3( ) 2)(3 2)h xh h hx h x→ ⎡ ⎤ = ⋅⎢ ⎥ + + +⎢ ⎥⎣ ⎦ 2 2 2 20 –6 – 3 6 lim – (3( ) 2)(3 2) (3 2)h x h x x h x x→ = = + + + + e. 0 3( ) – 3 ( ) lim h x h x f x h→ + ′ = 0 ( 3 3 – 3 )( 3 3 3 ) lim ( 3 3 3 )h x h x x h x h x h x→ + + + = + + 0 0 3 3 lim lim ( 3 3 3 ) 3 3 3h h h h x h x x h x→ → = = + + + + 3 2 3x = f. 0 sin[3( )] – sin3 ( ) lim h x h x f x h→ + ′ = 0 sin(3 3 ) – sin3 lim h x h x h→ + = 0 sin3 cos3 sin3 cos3 – sin3 lim h x h h x x h→ + = 0 0 sin3 (cos3 –1) sin3 cos3 lim lim h h x h h x h h→ → = + 0 0 cos3 –1 sin3 3sin3 lim cos3 lim 3h h h h x x h h→ → = + 0 sin3 (3sin3 )(0) (cos3 )3 lim 3h h x x h→ = + (cos3 )(3)(1) 3cos3x x= = g. 2 2 0 ( ) 5 – 5 ( ) lim h x h x f x h→ + + + ′ = 2 2 2 2 0 2 2 ( ) 5 – 5 ( ) 5 5 lim ( ) 5 5h x h x x h x h x h x→ ⎛ ⎞⎛ ⎞+ + + + + + +⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠= ⎛ ⎞+ + + +⎜ ⎟ ⎝ ⎠ 2 0 2 2 2 lim ( ) 5 5h xh h h x h x→ + = ⎛ ⎞+ + + +⎜ ⎟ ⎝ ⎠ 2 2 2 20 2 2 lim ( ) 5 5 2 5 5h x h x x x h x x x→ + = = = + + + + + + h. 0 cos[ ( )] – cos ( ) lim h x h x f x h→ π + π ′ = 0 cos( ) – cos lim h x h x h→ π + π π = 0 cos cos – sin sin – cos lim h x h x h x h→ π π π π π = 0 0 1– cos sin lim – cos lim sin h h h h x x h h→ → π π⎛ ⎞ ⎛ ⎞ = π π − π π⎜ ⎟ ⎜ ⎟ π π⎝ ⎠ ⎝ ⎠ (– cos )(0) – ( sin ) – sinx x xπ= π π π = π π 2. a. 2 2 2 – 2 2( – )( ) ( ) lim lim – –t x t x t x t x t x g x t x t x→ → + ′ = = 2 lim( ) 2(2 ) 4 t x t x x x → = + = = © 2007 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.
  • 146 Section 2.10 Instructor’s Resource Manual b. 3 3 ( ) – ( ) ( ) lim –t x t t x x g x t x→ + + ′ = 2 2 ( – )( ) ( – ) lim –t x t x t tx x t x t x→ + + + = 2 2 lim( 1) t x t tx x → = + + + 2 3 1x= + c. 1 1– – ( ) lim lim – ( – ) t x t x t x x t g x t x tx t x→ → ′ = = 2 –1 1 lim – t x tx x→ = = d. 2 2 1 1 1 ( ) lim – –1 1t x g x t xt x→ ⎡ ⎤⎛ ⎞⎛ ⎞ ′ = ⎢ ⎥⎜ ⎟⎜ ⎟ ⎝ ⎠+ +⎝ ⎠⎣ ⎦ 2 2 2 2 – lim ( 1)( 1)( – )t x x t t x t x→ = + + 2 2 –( )( – ) lim ( 1)( 1)( – )t x x t t x t x t x→ + = + + 2 2 2 2 –( ) 2 lim – ( 1)( 1) ( 1)t x x t x t x x→ + = = + + + e. – ( ) lim –t x t x g x t x→ ′ = ( – )( ) lim ( – )( )t x t x t x t x t x→ + = + – 1 lim lim ( – )( )t x t x t x t x t x t x→ → = = + + 1 2 x = f. sin – sin ( ) lim –t x t x g x t x→ π π ′ = Let v = t – x, then t = v + x and as , 0.t x v→ → 0 sin – sin sin ( ) – sin lim lim –t x v t x v x x t x v→ → π π π + π = 0 sin cos sin cos – sin lim v v x x v x v→ π π + π π π = 0 sin cos –1 lim cos sin v v v x x v v→ π π⎡ ⎤ = π π + π π⎢ ⎥π π⎣ ⎦ cos 1 sin 0 cosx x x= π π ⋅ + π π ⋅ = π π Other method: Use the subtraction formula ( ) ( ) sin – sin 2cos sin 2 2 t x t x t x π + π − π π = g. 3 3 – ( ) lim –t x t C x C g x t x→ + + ′ = 3 3 3 3 3 3 – lim ( – )t x t C x C t C x C t x t C x C→ ⎛ ⎞⎛ ⎞+ + + + +⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠= ⎛ ⎞+ + +⎜ ⎟ ⎝ ⎠ 3 3 3 3 – lim ( – )t x t x t x t C x C→ = ⎛ ⎞+ + +⎜ ⎟ ⎝ ⎠ 2 2 2 3 3 3 3 lim 2t x t tx x x t C x C x C→ + + = = + + + + h. cos2 – cos2 ( ) lim –t x t x g x t x→ ′ = Let v = t – x, then t = v + x and as , 0.t x v→ → 0 cos2 – cos2 cos2( ) – cos2 lim lim –t x v t x v x x t x v→ → + = 0 cos2 cos2 – sin 2 sin 2 – cos2 lim v v x v x x v→ = 0 cos2 –1 sin 2 lim 2cos2 – 2sin 2 2 2v v v x x v v→ ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ 2cos2 0 – 2sin 2 1 –2sin 2x x x= ⋅ ⋅ = Other method: Use the subtraction formula cos2 cos2 2sin( )sin( ).t x t x t x− = − + − 3. a. f(x) = 3x at x = 1 b. 3 ( ) 4f x x= at x = 2 c. 3 ( )f x x= at x = 1 d. f(x) = sin x at x π= e. 4 ( )f x x = at x f. f(x) = –sin 3x at x g. f(x) = tan x at 4 x π = h. 1 ( )f x x = at x = 5 4. a. 3 (2) – 4 f ′ ≈ b. 3 (6) 2 f ′ ≈ © 2007 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.
  • Instructor’s Resource Manual Section 2.10 147 c. 3 2 avg 6 – 9 7 – 3 8 V = = d. 2 2 ( ) ( )(2 ) d f t f t t dt ′= At t = 2, 2 8 4 (4) 4 3 3 f ⎛ ⎞′ ≈ =⎜ ⎟ ⎝ ⎠ e. 2 [ ( )] 2 ( ) ( ) d f t f t f t dt ′= At t = 2, 2 (2) (2)f f ′ 3 2(2) – –3 4 ⎛ ⎞ ≈ =⎜ ⎟ ⎝ ⎠ f. ( ( ( ))) ( ( )) ( ) d f f t f f t f t dt ′ ′= At t = 2, ( (2)) (2) (2) (2)f f f f f′ ′ ′ ′= 3 3 9 – – 4 4 16 ⎛ ⎞⎛ ⎞ ≈ =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 5. 5 4 (3 ) 15xD x x= 6. 3 2 –2 2 –3 ( – 3 ) 3 – 6 (–2)xD x x x x x x+ = + 2 –3 3 – 6 – 2x x x= 7. 3 2 2 ( 4 2 ) 3 8 2zD z z z z z+ + = + + 8. 2 2 2 2 3 – 5 ( 1)(3) – (3 – 5)(2 ) 1 ( 1) x x x x x D x x +⎛ ⎞ =⎜ ⎟ + +⎝ ⎠ 2 2 2 3 10 3 ( 1) x x x − + + = + 9. 2 2 2 2 4 5 (6 2 )(4) – (4 – 5)(12 2) 6 2 (6 2 ) t t t t t t D t t t t − + +⎛ ⎞ =⎜ ⎟ + +⎝ ⎠ 2 2 2 24 60 10 (6 2 ) t t t t − + + = + 10. 2/3 –1/32 (3 2) (3 2) (3) 3 xD x x+ = + –1/3 2(3 2)x= + 2 2/3 –4/32 (3 2) – (3 2) (3) 3 xD x x+ = + –4/3 –2(3 2)x= + 11. 2 3 2 2 3 3 2 4 – 2 ( )(8 ) – (4 – 2)(3 1) ( ) d x x x x x x dx x x x x ⎛ ⎞ + + =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ 4 2 3 2 4 10 2 ( ) x x x x − + + = + 12. 1 ( 2 6) (2) 2 6 2 2 6 tD t t t t t + = + + + 2 6 2 6 t t t = + + + 13. 2 –1/ 2 2 2 –3/ 2 1 ( 4) 4 1 – ( 4) (2 ) 2 d d x dx dxx x x ⎛ ⎞ ⎜ ⎟ = + ⎜ ⎟ +⎝ ⎠ = + 2 3 – ( 4) x x = + 14. 2 1 2 3 3 2 –1 1 1 – 2 d x d d x dx dx dxxx x x − = = = − 15. 3 2 (sin cos ) cos 3cos (–sin )Dθ θ θ θ θ θ+ = + 2 cos – 3sin cosθ θ θ= 2 3 3 (sin cos ) –sin – 3[sin (2)(cos )(–sin ) cos ] Dθ θ θ θ θ θ θ θ + = + 2 3 –sin 6sin cos – 3cosθ θ θ θ= + 16. 2 2 2 [sin( ) – sin ( )] cos( )(2 ) – (2sin )(cos ) d t t t t t t dt = 2 2 cos( ) – sin(2 )t t t= 17. 2 2 2 [sin( )] cos( )(2 ) 2 cos( )Dθ θ θ θ θ θ= = 18. 3 2 2 (cos 5 ) (3cos 5 )(–sin5 )(5) –15cos 5 sin5 d x x x dx x x = = © 2007 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.
  • 148 Section 2.10 Instructor’s Resource Manual 19. 2 [sin (sin( ))] 2sin(sin( ))cos(sin( ))(cos( ))( ) d d θ θ θ θ θ π = π π π π 2 sin(sin( ))cos(sin( ))cos( )θ θ θ= π π π π 20. ( )2 [sin (cos4 )] 2sin(cos4 ) cos(cos4 ) (–sin 4 )(4) d t t t t dt = –8sin(cos4 )cos(cos4 )sin 4t t t= 21. 2 2 tan3 (sec 3 )(3) 3sec 3Dθ θ θ θ= = 22. 2 2 2 2 2 sin3 (cos5 )(cos3 )(3) – (sin3 )(–sin5 )(10 ) cos5 cos 5 d x x x x x x dx x x ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 2 2 2 2 3cos5 cos3 10 sin3 sin5 cos 5 x x x x x x + = 23. 2 2 2 3 2 ( ) ( –1) (9 – 4) (3 – 4 )(2)( –1)(2 )f x x x x x x x′ = + 2 2 2 2 3 ( –1) (9 – 4) 4 ( –1)(3 – 4 )x x x x x x= + (2) 672f ′ = 24. ( ) 3cos3 2(sin3 )(cos3 )(3)g x x x x′ = + 3cos3 3sin 6x x= + ( ) –9sin3 18cos6g x x x′′ = + (0) 18g′′ = 25. 2 2 2 2 2 2 2 cot (sec )(– csc ) – (cot )(sec )(tan )(2 ) sec sec d x x x x x x x dx x x ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 2 2 2 – csc – 2 cot tan sec x x x x x = 26. 2 4 sin (cos – sin )(4 cos 4sin ) – (4 sin )(–sin – cos ) cos – sin (cos – sin ) t t t t t t t t t t t t D t t t t +⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 2 2 2 2 4 cos 2sin 2 – 4sin 4 sin (cos – sin ) t t t t t t t t + + = 2 2 4 2sin 2 – 4sin (cos – sin ) t t t t t + = 27. 3 2 2 ( ) ( –1) 2(sin – )( cos –1) (sin – ) 3( –1)f x x x x x x x x′ = π π π + π 3 2 2 2( –1) (sin – )( cos –1) 3(sin – ) ( –1)x x x x x x x= π π π + π (2) 16 4 3.43f ′ = − π ≈ 28. 4 ( ) 5(sin(2 ) cos(3 )) (2cos(2 ) – 3sin(3 ))h t t t t t′ = + 4 3 2 ( ) 5(sin(2 ) cos(3 )) ( 4sin(2 ) – 9cos(3 )) 20(sin(2 ) cos(3 )) (2cos(2 ) – 3sin(3 ))h t t t t t t t t t′′ = + − + + 4 3 2 (0) 5 1 ( 9) 20 1 2 35h′′ = ⋅ ⋅ − + ⋅ ⋅ = 29. 2 ( ) 3(cos 5 )(–sin5 )(5)g r r r′ = 2 –15cos 5 sin5r r= 2 ( ) –15[(cos 5 )(cos5 )(5) (sin5 )2(cos5 )(–sin5 )(5)]g r r r r r r′′ = + 3 2 –15[5cos 5 –10(sin 5 )(cos5 )]r r r= 2 2 ( ) –15[5(3)(cos 5 )(–sin5 )(5) (10sin 5 )( sin5 )(5) (cos5 )(20sin5 )(cos5 )(5)]g r r r r r r r r′′′ = − − − 2 3 –15[ 175(cos 5 )(sin5 ) 50sin 5 ]r r r= − + (1) 458.8g′′′ ≈ 30. ( ) ( ( )) ( ) 2 ( ) ( )f t h g t g t g t g t′ ′ ′ ′= + 31. ( ) ( ( ) ( ))( ( ) ( )) ( )G x F r x s x r x s x s x′ ′ ′ ′ ′= + + + ( ) ( ( ) ( ))( ( ) ( )) ( ( ) ( )) ( ( ) ( ))( ( ) ( )) ( )G x F r x s x r x s x r x s x F r x s x r x s x s x′′ ′ ′′ ′′ ′ ′ ′′ ′ ′ ′′= + + + + + + + 2 ( ( ) ( ))( ( ) ( )) ( ( ) ( )) ( ( ) ( )) ( )F r x s x r x s x r x s x F r x s x s x′ ′′ ′′ ′ ′ ′′ ′′= + + + + + + © 2007 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.
  • Instructor’s Resource Manual Section 2.10 149 32. 2 ( ) ( ( )) ( ) 3[ ( )] (–sin )F x Q R x R x R x x′ ′ ′= = 2 –3cos sinx x= 33. 2 ( ) ( ( )) ( ) [3cos(3 ( ))](9 )F z r s z s z s z z′ ′ ′= = 2 3 27 cos(9 )z z= 34. 2( – 2) dy x dx = 2x – y + 2 = 0; y = 2x + 2; m = 2 1 2( – 2) – 2 x = 7 4 x = 2 7 1 7 1 – 2 ; , 4 16 4 16 y ⎛ ⎞ ⎛ ⎞ = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 35. 34 3 V r= π 2 4 dV r dr = π When r = 5, 2 4 (5) 100 314 dV dr = π = π ≈ m3 per meter of increase in the radius. 36. 34 ; 10 3 dV V r dt = π = 2 4 dV dr r dt dt = π When r = 5, 2 10 4 (5) dr dt = π 1 0.0318 10 dr dt = ≈ π m/h 37. 1 6 3 (12); ; 2 4 2 b h V bh b h = = = 23 6 9 ; 9 2 h dV V h h dt ⎛ ⎞ = = =⎜ ⎟ ⎝ ⎠ 18 dV dh h dt dt = When h = 3, 9 18(3) dh dt = 1 0.167 6 dh dt = ≈ ft/min 38. a. v = 128 – 32t v = 0, when 4t s= 2 128(4) –16(4) 256s = = ft b. 2 128 –16 0t t = –16t(t – 8) = 0 The object hits the ground when t = 8s v = 128 – 32(8) = –128 ft/s 39. 3 2 – 6 9s t t t= + 2 ( ) 3 –12 9 ds v t t t dt = = + 2 2 ( ) 6 –12 d s a t t dt = = a. 2 3 –12 9 0t t + < 3(t – 3)(t – 1) < 0 1 < t < 3; (1,3) b. 2 3 –12 9 0t t + = 3(t – 3)(t – 1) = 0 t = 1, 3 a(1) = –6, a(3) = 6 c. 6t – 12 > 0 t > 2; (2, )∞ 40. a. 20 19 12 5 ( 100) 0xD x x x+ + + = b. 20 20 19 18 ( ) 20!xD x x x+ + = c. 20 21 20 (7 3 ) (7 21!) (3 20!)xD x x x+ = ⋅ + ⋅ d. 20 4 (sin cos ) (sin cos )x xD x x D x x+ = + = sin x + cos x e. 20 20 (sin 2 ) 2 sin 2xD x x= = 1,048,576 sin 2x f. 20 20 21 21 1 (–1) (20!) 20! xD x x x ⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ 41. a. 2( –1) 2 0 dy x y dx + = –( –1) 1–dy x x dx y y = = b. 2 2 (2 ) (2 ) 0 dy dy x y y y x x dx dx + + + = 2 2 (2 ) –( 2 ) dy xy x y xy dx + = + 2 2 2 2 dy y xy dx x xy + = − + © 2007 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.
  • 150 Section 2.10 Instructor’s Resource Manual c. 2 2 3 2 2 3 3 3 (3 ) 3 dy dy x y x y x y dx dx + = + 2 3 2 2 3 2 (3 – 3 ) 3 – 3 dy y x y x y x dx = 2 3 2 2 3 2 2 3 2 2 3 2 3 – 3 – 3 – 3 – dy x y x x y x dx y x y y x y = = d. cos( ) sin( ) 2 dy x xy x y xy x dx ⎡ ⎤ + + =⎢ ⎥ ⎣ ⎦ 2 cos( ) 2 – sin( ) – cos( ) dy x xy x xy xy xy dx = 2 2 – sin( ) – cos( ) cos( ) dy x xy xy xy dx x xy = e. 2 sec ( ) tan( ) 0 dy x xy x y xy dx ⎛ ⎞ + + =⎜ ⎟ ⎝ ⎠ 2 2 2 sec ( ) –[tan( ) sec ( )] dy x xy xy xy xy dx = + 2 2 2 tan( ) sec ( ) – sec ( ) dy xy xy xy dx x xy + = 42. 2 12 12yy x′ = 2 1 6x y y ′ = At (1, 2): 1 3y′ = 24 6 0x yy′+ = 2 2 – 3 x y y ′ = At (1, 2): 2 1 – 3 y′ = Since 1 2( )( ) –1y y′ ′ = at (1, 2), the tangents are perpendicular. 43. [ cos( ) 2 ]dy x x dxπ π= + ; x = 2, dx = 0.01 [ cos(2 ) 2(2)](0.01) (4 )(0.01)dy π π π= + = + ≈ 0.0714 44. 2 2 (2 ) 2 [2( 2)] ( 2) (2) 0 dy dy x y y y x x dx dx + + + + + = 2 2 [2 2( 2) ] –[ 2 (2 4)] dy xy x y y x dx + + = + + 2 2 –( 4 8 ) 2 2( 2) dy y xy y dx xy x + + = + + 2 2 4 8 – 2 2( 2) y xy y dy dx xy x + + = + + When x = –2, y = ±1 a. 2 2 (1) 4(–2)(1) 8(1) – (–0.01) 2(–2)(1) 2(–2 2) dy + + = + + = –0.0025 b. 2 2 (–1) 4(–2)(–1) 8(–1) – (–0.01) 2(–2)(–1) 2(–2 2) dy + + = + + = 0.0025 45. a. 2 3 [ ( ) ( )] d f x g x dx + 2 2 ( ) ( ) 3 ( ) ( )f x f x g x g x′ ′= + 2 2 (2) (2) 3 (2) (2)f f g g′ ′+ 2 2(3)(4) 3(2) (5) 84= + = b. [ ( ) ( )] ( ) ( ) ( ) ( ) d f x g x f x g x g x f x dx ′ ′= + (2) (2) (2) (2) (3)(5) (2)(4) 23f g g f′ ′+ = + = c. [ ( ( ))] ( ( )) ( ) d f g x f g x g x dx ′ ′= ( (2)) (2) (2) (2) (4)(5) 20f g g f g′ ′ ′ ′= = = d. 2 [ ( )] 2 ( ) ( )xD f x f x f x′= 2 2 [ ( )] 2[ ( ) ( ) ( ) ( )]xD f x f x f x f x f x′′ ′ ′= + 2 2 (2) (2) 2[ (2)]f f f′′ ′= + 2 2(3)(–1) 2(4) 26= + = 46. 2 2 2 (13) ; 2 dx x y dt = + = 0 2 2 dx dy x y dt dt = + – dy x dx dt y dt = When y= 5, x = 12, so 12 24 – (2) – –4.8 5 5 dy dt = = = ft/s 47. sin15 , 400 y dx x dt ° = = sin15y x= ° sin15 dy dx dt dt = ° 400sin15 dy dt = ° ≈ 104 mi/hr 48. a. 2 2 2 2( ) 2 ( ) 2 2x x x x D x x x x x x = ⋅ = = = b. 2 2 2 0 x x x x x x x x x D x D x x x ⎛ ⎞ −⎜ ⎟⎛ ⎞ −⎝ ⎠= = = =⎜ ⎟ ⎝ ⎠ © 2007 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.
  • Instructor’s Resource Manual Review and Preview 151 c. 3 2 ( ) (0) 0x x x xD x D D x D= = = d. 22 ( ) (2 ) 2x xD x D x= = 49. a. sin sin cos cot sin sin Dθ θ θ θ θ θ θ = = b. cos cos ( sin ) tan cos cos Dθ θ θ θ θ θ θ = − = − 50. a. ( ) 1/ 2 1/ 2 1 ( ) 1; '( ) 1 ; 3 2 ( ) (3) '(3)( 3) 1 4 (4) ( 3) 2 1 3 1 11 2 4 4 4 4 f x x f x x a L x f f x x x x − − = + = − + = = + − = + − − = − + = − + b. ( ) cos ; '( ) sin cos ; 1 ( ) (1) '(1)( 1) cos1 ( sin1 cos1)( 1) cos1 (sin1) sin1 (cos1) cos1 (cos1 sin1) sin1 0.3012 0.8415 f x x x f x x x x a L x f f x x x x x x = = − + = = + − = + − + − = − + + − = − + ≈ − + Review and Preview Problems 1. ( )( )2 3 0x x− − < ( )( )2 3 0 2 or 3 x x x x − − = = = The split points are 2 and 3. The expression on the left can only change signs at the split points. Check a point in the intervals ( ),2−∞ , ( )2,3 , and ( )3,∞ . The solution set is { }| 2 3x x< < or ( )2,3 . −2 76 853−1 10 2 4 2. ( )( ) 2 6 0 3 2 0 x x x x − − > − + > ( )( )3 2 0 3 or 2 x x x x − + = = = − The split points are 3 and 2− . The expression on the left can only change signs at the split points. Check a point in the intervals ( ), 2−∞ − , ( )2,3− , and ( )3,∞ . The solution set is { }| 2 or 3x x x< − > , or ( ) ( ), 2 3,−∞ − ∪ ∞ . −5 −3−4 −2 53−1 10 2 4 3. ( )( )1 2 0x x x− − ≤ ( )( )1 2 0x x x− − = 0, 1 or 2x x x= = = The split points are 0, 1, and 2. The expression on the left can only change signs at the split points. Check a point in the intervals ( ),0−∞ , ( )0,1 , ( )1,2 , and ( )2,∞ . The solution set is { }| 0 or 1 2x x x≤ ≤ ≤ , or ( ] [ ],0 1,2−∞ ∪ . −5 −3−4 −2 53−1 10 2 4 4. ( ) ( )( ) 3 2 2 3 2 0 3 2 0 1 2 0 x x x x x x x x x + + ≥ + + ≥ + + ≥ ( )( )1 2 0 0, 1, 2 x x x x x x + + = = = − = − The split points are 0, 1− , and 2− . The expression on the left can only change signs at the split points. Check a point in the intervals ( ), 2−∞ − , ( )2, 1− − , ( )1,0− , and ( )0,∞ . The solution set is { }| 2 1 or 0x x x− ≤ ≤ − ≥ , or [ ] [ )2, 1 0,− − ∪ ∞ . −5 −3−4 −2 53−1 10 2 4 5. ( ) ( ) ( )( ) 2 2 0 4 2 0 2 2 x x x x x x x − ≥ − − ≥ − + The expression on the left is equal to 0 or undefined at 0x = , 2x = , and 2x = − . These are the split points. The expression on the left can only change signs at the split points. Check a point in the intervals: ( ), 2−∞ − , ( )2,0− , ( )0,2 , and ( )2,∞ . The solution set is { }| 2 or 0 2 or 2x x x x< − ≤ < > , or ( ) [ ) ( ), 2 0,2 2,−∞ − ∪ ∪ ∞ . −5 −3−4 −2 53−1 10 2 4 © 2007 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.
  • 152 Review and Preview Instructor’s Resource Manual 6. ( )( ) 2 2 2 9 0 2 3 3 0 2 x x x x x − > + − + > + The expression on the left is equal to 0 at 3x = , and 3x = − . These are the split points. The expression on the left can only change signs at the split points. Check a point in the intervals: ( ), 3−∞ − , ( )3,3− , and ( )3,∞ . The solution set is { }| 3 or 3x x x< − > , or ( ) ( ), 3 3,−∞ − ∪ ∞ . −5 −3−4 −2 53−1 10 2 4 7. ( ) ( ) ( ) ( )3 3 ' 4 2 1 2 8 2 1f x x x= + = + 8. ( ) ( ) ( )' cos cosf x x xπ π π π= ⋅ = 9. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 ' 1 sin 2 2 cos 2 2 2 1 sin 2 2 cos 2 f x x x x x x x x x = − ⋅− ⋅ + ⋅ = − − + 10. ( ) ( ) 2 2 sec tan sec 1 ' sec tan 1 x x x x f x x x x x x ⋅ − ⋅ = − = 11. ( ) ( ) ( )( ) 2 2 ' 2 tan3 sec 3 3 6 sec 3 tan3 f x x x x x = ⋅ ⋅ = 12. ( ) ( ) ( )( ) 1/ 22 2 1 ' 1 sin 2sin cos 2 sin cos 1 sin f x x x x x x x − = + = + 13. ( ) ( ) 1/ 21 cos ' cos 2 2 x f x x x x − = ⋅ = (note: you cannot cancel the x here because it is not a factor of both the numerator and denominator. It is the argument for the cosine in the numerator.) 14. ( ) ( ) 1/ 21 cos2 ' sin 2 cos2 2 2 sin 2 x f x x x x − = ⋅ ⋅ = 15. The tangent line is horizontal when the derivative is 0. 2 ' 2tan secy x x= ⋅ 2 2tan sec 0 2sin 0 cos x x x x = = The tangent line is horizontal whenever sin 0x = . That is, for x kπ= where k is an integer. 16. The tangent line is horizontal when the derivative is 0. ' 1 cosy x= + The tangent line is horizontal whenever cos 1x = − . That is, for ( )2 1x k π= + where k is an integer. 17. The line 2y x= + has slope 1, so any line parallel to this line will also have a slope of 1. For the tangent line to siny x x= + to be parallel to the given line, we need its derivative to equal 1. ' 1 cos 1y x= + = cos 0x = The tangent line will be parallel to 2y x= + whenever ( )2 1 2 x k π = + . 18. Length: 24 2x− Width: 9 2x− Height: x Volume: ( )( ) ( )( ) 24 2 9 2 9 2 24 2 l w h x x x x x x ⋅ ⋅ = − − = − − 19. Consider the diagram: 1 4 x− x His distance swimming will be 2 2 2 1 1x x+ = + kilometers. His distance running will be 4 x− kilometers. Using the distance traveled formula, d r t= ⋅ , we solve for t to get d t r = . Andy can swim at 4 kilometers per hour and run 10 kilometers per hour. Therefore, the time to get from A to D will be 2 1 4 4 10 x x+ − + hours. © 2007 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.
  • Instructor’s Resource Manual Review and Preview 153 20. a. ( ) ( ) ( ) ( ) ( ) 0 0 cos 0 0 1 1 cos 1 1 f f π π π π π = − = − = − = − = − − = + Since cosx x− is continuous, ( )0 0f < , and ( ) 0f π > , there is at least one point c .in the interval ( )0,π where ( ) 0f c = . (Intermediate Value Theorem) b. cos 2 2 2 2 f π π π π⎛ ⎞ ⎛ ⎞ = − =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ( )' 1 sinf x x= + ' 1 sin 1 1 2 2 2 f π π⎛ ⎞ ⎛ ⎞ = + = + =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ The slope of the tangent line is 2m = at the point , 2 2 π π⎛ ⎞ ⎜ ⎟ ⎝ ⎠ . Therefore, 2 2 2 y x π π⎛ ⎞ − = −⎜ ⎟ ⎝ ⎠ or 2 2 y x π = − . c. 2 0 2 x π − = . 2 2 4 x x π π = = The tangent line will intersect the x-axis at 4 x π = . 21. a. The derivative of 2 x is 2x and the derivative of a constant is 0. Therefore, one possible function is ( ) 2 3f x x= + . b. The derivative of cos x− is sin x and the derivative of a constant is 0. Therefore, one possible function is ( ) ( )cos 8f x x= − + . c. The derivative of 3 x is 2 3x , so the derivative of 31 3 x is 2 x . The derivative of 2 x is 2x , so the derivative of 21 2 x is x . The derivative of x is 1, and the derivative of a constant is 0. Therefore, one possible function is 3 21 1 2 3 2 x x x+ + + . 22. Yes. Adding 1 only changes the constant term in the function and the derivative of a constant is 0. Therefore, we would get the same derivative regardless of the value of the constant. © 2007 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.
  • 154 Section 3.1 Instructor’s Resource Manual CHAPTER 3 Applications of the Derivative 3.1 Concepts Review 1. continuous; closed and bounded 2. extreme 3. endpoints; stationary points; singular points 4. ( ) 0; ( )f c f c′ ′= does not exist Problem Set 3.1 1. Endpoints: 2− , 4 Singular points: none Stationary points: 0, 2 Critical points: 2,0,2,4− 2. Endpoints: 2− , 4 Singular points: 2 Stationary points: 0 Critical points: 2,0,2,4− 3. Endpoints: 2− , 4 Singular points: none Stationary points: 1,0,1,2,3− Critical points: 2, 1,0,1,2,3,4− − 4. Endpoints: 2− , 4 Singular points: none Stationary points: none Critical points: 2,4− 5. ( ) 2 4; 2 4 0f x x x′ = + + = when x = –2. Critical points: –4, –2, 0 f(–4) = 4, f(–2) = 0, f(0) = 4 Maximum value = 4, minimum value = 0 6. ( ) 2 1; 2 1 0h x x x′ = + + = when 1 – . 2 x = Critical points: –2, 1 – , 2 2 h(–2) = 2, 1 1 – – , 2 4 h ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ h(2) = 6 Maximum value = 6, minimum value = 1 – 4 7. ( ) 2 3;x x′Ψ = + 2x + 3 = 0 when 3 – . 2 x = Critical points: –2, 3 – , 2 1 3 9 (–2) –2, – – , (1) 4 2 4 ⎛ ⎞ Ψ = Ψ = Ψ =⎜ ⎟ ⎝ ⎠ Maximum value = 4, minimum value = 9 – 4 8. 2 21 6 ( ) (6 6 –12) ( – 2); 5 5 G x x x x x′ = + = + 2 – 2 0x x+ = when x = –2, 1 Critical points: –3, –2, 1, 3 9 7 (–3) , (–2) 4, (1) – , (3) 9 5 5 G G G G= = = = Maximum value = 9, minimum value 7 – 5 = 9. 2 2 ( ) 3 – 3;3 – 3 0f x x x′ = = when x = –1, 1. Critical points: –1, 1 f(–1) = 3, f(1) = –1 No maximum value, minimum value = –1 (See graph.) 10. 2 2 ( ) 3 – 3;3 – 3 0f x x x′ = = when x = –1, 1. Critical points: 3 – , –1, 1, 3 2 3 17 – , (–1) 3, (1) –1, (3) 19 2 8 f f f f ⎛ ⎞ = = = =⎜ ⎟ ⎝ ⎠ Maximum value = 19, minimum value = –1 © 2007 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.
  • Instructor’s Resource Manual Section 3.1 155 11. 2 1 ( ) ; ( )h r h r r ′ ′= − is never 0; ( )h r′ is not defined when r = 0, but r = 0 is not in the domain on [–1, 3] since h(0) is not defined. Critical points: –1, 3 Note that 0 0 lim ( ) and lim ( ) . x x h r h x − +→ → = −∞ = ∞ No maximum value, no minimum value. 12. 2 2 2 ( ) ; (1 ) x g x x ′ = − + 2 2 2 0 (1 ) x x − = + when x = 0 Critical points: –3, 0, 1 g(–3) = 1 10 , g(0) = 1, g(1) = 1 2 Maximum value = 1, minimum value = 1 10 13. ( ) ( ) ( )( ) 3 2 ' 4 4 4 1 4 1 1 f x x x x x x x x = − = − = − + ( )( )4 1 1 0x x x− + = when 0,1, 1x = − . Critical points: 2, 1,0,1,2− − ( )2 10f − = ; ( )1 1f − = ; ( )0 2f = ; ( )1 1f = ; ( )2 10f = Maximum value: 10 Minimum value: 1 14. ( ) ( ) ( )( ) ( )( )( )( ) 4 2 4 2 2 2 ' 5 25 20 5 5 4 5 4 1 5 2 2 1 1 f x x x x x x x x x x x = − + = − + = − − = − + − + ( )( )( )( )5 2 2 1 1 0x x x x− + − + = when 2, 1,1,2x = − − Critical points: 3, 2, 1,1,2− − − ( )3 79f − = − ; ( ) 19 2 3 f − = − ; ( ) 41 1 3 f − = − ; ( ) 35 1 3 f = ; ( ) 13 2 3 f = Maximum value: 35 3 Minimum value: 79− 15. 2 2 2 ( ) ; (1 ) x g x x ′ = − + 2 2 2 0 (1 ) x x − = + when x = 0. Critical point: 0 g(0) = 1 As , ( ) 0 ;x g x + → ∞ → as , ( ) 0 .x g x + → −∞ → Maximum value = 1, no minimum value (See graph.) 16. 2 2 2 1 ( ) ; (1 ) x f x x − ′ = + 2 2 2 1 0 (1 ) x x − = + when x = –1, 1 Critical points: –1, 1, 4 1 1 4 ( 1) , (1) , (4) 2 2 17 f f f− = − = = Maximum value = 1 2 , minimum value 1 – 2 = 17. ( ) cos ;r θ θ′ = cos 0θ = when 2 kθ π = + π Critical points: – , 4 6 π π 1 1 , 4 6 22 r r π π⎛ ⎞ ⎛ ⎞ − = − =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Maximum value 1 , 2 = minimum value 1 – 2 = 18. ( ) cos sin ;s t t t′ = + cos t + sin t = 0 when tan t = –1 or – . 4 t k π = + π Critical points: 3 0, , 4 π π s(0) = –1, 3 2, 4 s π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ ( ) 1s π = . Maximum value 2,= minimum value = –1 © 2007 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.
  • 156 Section 3.1 Instructor’s Resource Manual 19. –1 ( ) ; ( ) –1 x a x a x x ′ ′= does not exist when x = 1. Critical points: 0, 1, 3 a(0) = 1, a(1) = 0, a(3) = 2 Maximum value = 2, minimum value = 0 20. 3(3 – 2) ( ) ; ( ) 3 – 2 s f s f s s ′ ′= does not exist when 2. 3 s = Critical points: 21, , 4 3 − f(–1) = 5, ( )2 0, 3 f = f(4) = 10 Maximum value = 10, minimum value = 0 21. 2/3 1 ( ) ; ( ) 3 g x f x x ′ ′= does not exist when x = 0. Critical points: –1, 0, 27 g(–1) = –1, g(0) = 0, g(27) = 3 Maximum value = 3, minimum value = –1 22. 3/5 2 ( ) ; ( ) 5 s t s t t ′ ′= does not exist when t = 0. Critical points: –1, 0, 32 s(–1) = 1, s(0) = 0, s(32) = 4 Maximum value = 4, minimum value = 0 23. ( )' sinH t t= − sin 0t− = when 0, ,2 ,3 ,4 ,5 ,6 ,7 ,8t π π π π π π π π= Critical points: 0, ,2 ,3 ,4 ,5 ,6 ,7 ,8π π π π π π π π ( )0 1H = ; ( ) 1H π = − ; ( )2 1H π = ; ( )3 1H π = − ; ( )4 1H π = ; ( )5 1H π = − ; ( )6 1H π = ; ( )7 1H π = − ; ( )8 1H π = Maximum value: 1 Minimum value: 1− 24. ( )' 1 2cosg x x= − 1 1 2cos 0 cos 2 x x− = → = when 5 , , , 3 3 3 3 x π π π π5 = − − Critical points: 5 2 , , , , ,2 3 3 3 3 π π π π π π 5 − − − ( )2 2g π π− = − ; 5 5 3 3 3 g π π−⎛ ⎞ − = −⎜ ⎟ ⎝ ⎠ ; 3 3 3 g π π⎛ ⎞ − = − +⎜ ⎟ ⎝ ⎠ ; 3 3 3 g π π⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ ; 5 5 3 3 3 g π π⎛ ⎞ = +⎜ ⎟ ⎝ ⎠ ; ( )2 2g π π= Maximum value: 5 3 3 π + Minimum value: 5 3 3 π − − 25. ( ) ( ) ( ) 2 ' sec tan 2 sec sec tan 2 g θ θ θ θ θ θ θ θ θ θ = + = + ( )sec tan 2 0θ θ θ θ + = when 0θ = . Consider the graph: 1 x y −1 4 π 4 π − Critical points: ,0, 4 4 π π − 2 2 4 16 g π π⎛ ⎞ − =⎜ ⎟ ⎝ ⎠ ; ( )0 0g = ; 2 2 4 16 g π π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ Maximum value: 2 2 16 π ; Minimum value: 0 26. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2/3 5/3 2 2/3 2/3 2 2 2/3 2 5 2 1 3 ' 2 5 10 2 2 3 3 3 2 2 2 5 3 2 t t t h t t t t t t t t t t t t ⎛ ⎞ + −⎜ ⎟ ⎝ ⎠= + ⎛ ⎞ ⎛ ⎞ + − +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠= = + + + = + ( )'h t is undefined when 2t = − and ( )' 0h t = when 0t = or 5t = − . Since 5− is not in the interval of interest, it is not a critical point. Critical points: 1,0,8− ( )1 1h − = − ; ( )0 0h = ; ( ) 16 5 8h = Maximum value: 16 5 ; Minimum value: 1− 27. a. 2 2 ( ) 3 –12 1;3 –12 1 0f x x x x x′ = + + = when 33 2 – 3 x = and 33 2 . 3 x = + Critical points: 33 33 –1, 2 – , 2 , 5 3 3 + f(–1) = –6, 33 2 – 2.04, 3 f ⎛ ⎞ ≈⎜ ⎟⎜ ⎟ ⎝ ⎠ 33 2 –26.04, 3 f ⎛ ⎞ + ≈⎜ ⎟⎜ ⎟ ⎝ ⎠ f(5) = –18 Maximum value ≈ 2.04; minimum value 26.04≈ − © 2007 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.
  • Instructor’s Resource Manual Section 3.1 157 b. 3 2 2 3 2 ( – 6 2)(3 –12 1) ( ) ; – 6 2 x x x x x g x x x x + + + ′ = + + '( ) 0g x = when 33 2 – 3 x = and 33 2 . 3 x = + ( )g x′ does not exist when f(x) = 0; on [–1, 5], f(x) = 0 when x ≈ –0.4836 and x ≈ 0.7172 Critical points: –1, –0.4836, 33 2 – , 3 0.7172, 33 2 , 3 + 5 g(–1) = 6, g(–0.4836) = 0, 33 2 – 2.04, 3 g ⎛ ⎞ ≈⎜ ⎟⎜ ⎟ ⎝ ⎠ g(0.7172) = 0, 33 2 26.04, 3 g ⎛ ⎞ + ≈⎜ ⎟⎜ ⎟ ⎝ ⎠ g(5) = 18 Maximum value ≈ 26.04, minimum value = 0 28. a. ( ) cos ;f x x x′ = on [–1, 5], x cos x = 0 when x = 0, 3 , 2 2 x x π π = = Critical points: 3 –1, 0, , , 5 2 2 π π f(–1) ≈3.38, f(0) = 3, 3.57, 2 f π⎛ ⎞ ≈⎜ ⎟ ⎝ ⎠ 3 –2.71, 2 f π⎛ ⎞ ≈⎜ ⎟ ⎝ ⎠ f(5) 2.51≈ − Maximum value ≈ 3.57, minimum value ≈–2.71 b. (cos sin 2)( cos ) ( ) ; cos sin 2 x x x x x g x x x x + + ′ = + + 3 ( ) 0 when 0, , 2 2 g x x x x π π ′ = = = = ( )g x′ does not exist when f(x) = 0; on [–1, 5], f(x) = 0 when x ≈ 3.45 Critical points: 3 –1, 0, , 3.45, , 5 2 2 π π g(–1) ≈ 3.38, g(0) = 3, 3.57, 2 g π⎛ ⎞ ≈⎜ ⎟ ⎝ ⎠ g(3.45) = 0, 3 2.71, 2 g π⎛ ⎞ ≈⎜ ⎟ ⎝ ⎠ g(5) ≈ 2.51 Maximum value ≈ 3.57; minimum value = 0 29. Answers will vary. One possibility: −5 5 5−5 x y 30. Answers will vary. One possibility: −5 5 5 x y 31. Answers will vary. One possibility: −5 5 5 x y 32. Answers will vary. One possibility: −5 5 5 x y © 2007 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.
  • 158 Section 3.2 Instructor’s Resource Manual 33. Answers will vary. One possibility: −5 5 5 x y 34. Answers will vary. One possibility: −5 5 5 x y 35. Answers will vary. One possibility: −5 5 5 x y 36. Answers will vary. One possibility: −5 5 5 x y 3.2 Concepts Review 1. Increasing; concave up 2. ( ) 0; ( ) 0f x f x′ ′′> < 3. An inflection point 4. ( ) 0; ( )f c f c′′ ′′= does not exist. Problem Set 3.2 1. ( ) 3;f x′ = 3 > 0 for all x. f(x) is increasing for all x. 2. ( ) 2 –1;g x x′ = 2x – 1 > 0 when 1 . 2 x > g(x) is increasing on 1 , 2 ⎡ ⎞ ∞⎟⎢ ⎣ ⎠ and decreasing on 1 – , . 2 ⎛ ⎤ ∞⎜ ⎥ ⎝ ⎦ 3. ( ) 2 2;h t t′ = + 2t + 2 > 0 when t > –1. h(t) is increasing on [–1, ∞ ) and decreasing on ( −∞ , –1]. 4. 2 2 ( ) 3 ;3 0f x x x′ = > for 0x ≠ . f(x) is increasing for all x. 5. 2 ( ) 6 –18 12 6( – 2)( –1)G x x x x x′ = + = Split the x-axis into the intervals (– ∞ , 1), (1, 2), (2, ∞ ). Test points: 3 0, , 3; (0) 12, 2 x G′= = 3 3 – , 2 2 G ⎛ ⎞′ =⎜ ⎟ ⎝ ⎠ (3) 12G′ = G(x) is increasing on (– ∞ , 1] ∪ [2, ∞ ) and decreasing on [1, 2]. 6. 2 ( ) 3 6 3 ( 2)f t t t t t′ = + = + Split the x-axis into the intervals (– ∞ , –2), (–2, 0), (0, ∞ ). Test points: t = –3, –1, 1; (–3) 9,f ′ = (–1) –3,f ′ = (1) 9f ′ = f(t) is increasing on (– ∞ , –2] ∪ [0, ∞ ) and decreasing on [–2, 0]. 7. 3 2 2 ( ) – 2 ( – 2)h z z z z z′ = = Split the x-axis into the intervals (– ∞ , 0), (0, 2), (2, ∞ ). Test points: z = –1, 1, 3; (–1) –3,h′ = (1) –1,h′ = (3) 9h′ = h(z) is increasing on [2, ∞ ) and decreasing on (– ∞ , 2]. © 2007 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.
  • Instructor’s Resource Manual Section 3.2 159 8. 3 2 – ( ) x f x x ′ = Split the x-axis into the intervals (– ∞ , 0), (0, 2), (2, ∞ ). Test points: –1, 1, 3; (–1) –3,f ′ = (1) 1,f ′ = 1 (3) – 27 f ′ = f(x) is increasing on (0, 2] and decreasing on (– ∞ , 0) ∪ [2, ∞ ). 9. ( ) cos ; ( ) 0H t t H t′ ′= > when 0 2 t π ≤ < and 3 2 . 2 t π < ≤ π H(t) is increasing on 3 0, , 2 2 2 π π⎡ ⎤ ⎡ ⎤ ∪ π⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ and decreasing on 3 , . 2 2 π π⎡ ⎤ ⎢ ⎥ ⎣ ⎦ 10. ( ) –2cos sin ; ( ) 0R Rθ θ θ θ′ ′= > when 2 θ π < < π and 3 2 . 2 θ π < < π R(θ ) is increasing on 3 , ,2 2 2 π π⎡ ⎤ ⎡ ⎤ π ∪ π⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ and decreasing on 3 0, , . 2 2 π π⎡ ⎤ ⎡ ⎤ ∪ π⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 11. ( ) 2;f x′′ = 2 > 0 for all x. f(x) is concave up for all x; no inflection points. 12. ( ) 2;G w′′ = 2 > 0 for all w. G(w) is concave up for all w; no inflection points. 13. ( ) 18 ;T t t′′ = 18t > 0 when t > 0. T(t) is concave up on (0, ∞ ) and concave down on (– ∞ , 0); (0, 0) is the only inflection point. 14. 4 4 4 6 2 ( ) 2 – ( – 3);f z z z z ′′ = = 4 – 3 0z > for 4 – 3z < and 4 3.z > f(z) is concave up on 4 4 (– , – 3) ( 3, )∞ ∪ ∞ and concave down on 4 4 (– 3, 0) (0, 3);∪ inflection points are 4 1 – 3, 3 – 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ and 4 1 3, 3 – . 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 15. 2 ( ) 12 – 36 – 48; ( ) 0q x x x q x′′ ′′= > when x < –1 and x > 4. q(x) is concave up on (– ∞ , –1) ∪ (4, ∞ ) and concave down on (–1, 4); inflection points are (–1, –19) and (4, –499). 16. 2 ( ) 12 48 12 ( 4); ( ) 0f x x x x x f x′′ ′′= + = + > when x < –4 and x > 0. f(x) is concave up on (– ∞ , –4) ∪ (0, ∞ ) and concave down on (–4, 0); inflection points are (–4, –258) and (0, –2). 17. 2 2 2 ( ) 2sin – 2cos 4 6 – 4cos ;F x x x x′′ = + = 2 6 – 4cos 0x > for all x since 2 0 cos 1.x≤ ≤ F(x) is concave up for all x; no inflection points. 18. 2 2 ( ) 48 24cos – 24sinG x x x′′ = + 2 24 48cos ;x= + 2 24 48cos 0x+ > for all x. G(x) is concave up for all x; no inflection points. 19. 2 2 ( ) 3 –12;3 –12 0f x x x′ = > when x < –2 or x > 2. f(x) is increasing on (– ∞ , –2] ∪ [2, ∞ ) and decreasing on [–2, 2]. ( ) 6 ;f x x′′ = 6x > 0 when x > 0. f(x) is concave up on (0, ∞ ) and concave down on (– ∞ , 0). 20. 2 ( ) 12 – 6 – 6 6(2 1)( –1);g x x x x x′ = = + ( ) 0g x′ > when 1 – 2 x < or x > 1. g(x) is increasing on 1 – , – [1, ) 2 ⎛ ⎤ ∞ ∪ ∞⎜ ⎥ ⎝ ⎦ and decreasing on 1 – ,1 . 2 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ ( ) 24 – 6 6(4 –1);g x x x′′ = = ( ) 0g x′′ > when 1 . 4 x > g(x) is concave up on 1 , 4 ⎛ ⎞ ∞⎜ ⎟ ⎝ ⎠ and concave down on 1 – , . 4 ⎛ ⎞ ∞⎜ ⎟ ⎝ ⎠ © 2007 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.
  • 160 Section 3.2 Instructor’s Resource Manual 21. 3 2 2 ( ) 12 –12 12 ( –1);g x x x x x′ = = ( ) 0g x′ > when x > 1. g(x) is increasing on [1, ∞ ) and decreasing on ( ,1].−∞ 2 ( ) 36 – 24 12 (3 – 2);g x x x x x′′ = = ( ) 0g x′′ > when x < 0 or 2 . 3 x > g(x) is concave up on 2 (– , 0) , 3 ⎛ ⎞ ∞ ∪ ∞⎜ ⎟ ⎝ ⎠ and concave down on 2 0, . 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 22. 5 3 3 2 ( ) 6 –12 6 ( – 2)F x x x x x′ = = Split the x-axis into the intervals (– ∞ , 2)− , ( 2, 0), (0, 2), ( 2, )− ∞ . Test points: x = –2, –1, 1, 2; (–2) –96,F′ = (–1) 6, (1) –6, (2) 96F F F′ ′ ′= = = F(x) is increasing on [– 2, 0] [ 2, )∪ ∞ and decreasing on (– , – 2] [0, 2]∞ ∪ 4 2 2 2 ( ) 30 – 36 6 (5 – 6);F x x x x x′′ = = 2 5 – 6 0x > when 6 – 5 x < or 6 . 5 x > F(x) is concave up on 6 6 – , – , 5 5 ⎛ ⎞ ⎛ ⎞ ∞ ∪ ∞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ and concave down on 6 6 – , . 5 5 ⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠ 23. 4 2 2 2 ( ) 15 –15 15 ( –1);G x x x x x′ = = ( ) 0G x′ > when x < –1 or x > 1. G(x) is increasing on (– ∞ , –1] ∪ [1, ∞ ) and decreasing on [–1, 1]. 3 2 ( ) 60 – 30 30 (2 –1);G x x x x x′′ = = Split the x-axis into the intervals 1 , , 2 ⎛ ⎞ −∞ −⎜ ⎟ ⎝ ⎠ 1 1 1 , 0 , 0, , , . 2 2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − ∞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Test points: 1 1 –1, – , ,1; 2 2 x = (–1) –30,G′′ = 1 15 1 15 – , – , (1) 30. 2 2 2 2 G G G ⎛ ⎞ ⎛ ⎞′′ ′′ ′′= = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ G(x) is concave up on 1 1 – , 0 , 2 2 ⎛ ⎞ ⎛ ⎞ ∪ ∞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ and concave down on 1 1 – , – 0, . 2 2 ⎛ ⎞ ⎛ ⎞ ∞ ∪⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 24. 2 2 2 ( ) ; ( 1) x H x x ′ = + ( ) 0H x′ > when x > 0. H(x) is increasing on [0, ∞ ) and decreasing on (– ∞ , 0]. 2 2 3 2(1– 3 ) ( ) ; ( ) 0 ( 1) x H x H x x ′′ ′′= > + when 1 1 – . 3 3 x< < H(x) is concave up on 1 1 – , 3 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ and concave down on 1 1 – , – , . 3 3 ⎛ ⎞ ⎛ ⎞ ∞ ∪ ∞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ © 2007 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.
  • Instructor’s Resource Manual Section 3.2 161 25. cos ( ) ; ( ) 0 2 sin x f x f x x ′ ′= > when 0 . 2 x π < < f(x) is increasing on 0, 2 π⎡ ⎤ ⎢ ⎥ ⎣ ⎦ and decreasing on , . 2 π⎡ ⎤ π⎢ ⎥ ⎣ ⎦ 2 2 3/ 2 – cos – 2sin ( ) ; ( ) 0 4sin x x f x f x x ′′ ′′= < for all x in (0, ∞ ). f(x) is concave down on (0, π ). 26. 3 – 4 ( ) ; 2 – 2 x g x x ′ = 3x – 4 > 0 when 4 . 3 x > g(x) is increasing on [2, ∞ ). 3/ 2 3 – 8 ( ) ; 4( – 2) x g x x ′′ = 3x – 8 > 0 when 8 . 3 x > g(x) is concave up on 8 , 3 ⎛ ⎞ ∞⎜ ⎟ ⎝ ⎠ and concave down on 8 2, . 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 27. 1/3 2 – 5 ( ) ; 3 x f x x ′ = 2 – 5x > 0 when 2 , 5 x < ( )f x′ does not exist at x = 0. Split the x-axis into the intervals ( , 0),− ∞ 2 2 0, , , . 5 5 ⎛ ⎞ ⎛ ⎞ ∞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Test points: 1 7 –1, ,1; ( 1) – , 5 3 f ′ − = 3 1 5 , (1) –1. 5 3 f f ⎛ ⎞′ ′= =⎜ ⎟ ⎝ ⎠ f(x) is increasing on 2 0, 5 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ and decreasing on 2 (– , 0] , . 5 ⎡ ⎞ ∞ ∪ ∞⎟⎢ ⎣ ⎠ 4/3 –2(5 1) ( ) ; 9 x f x x + ′′ = –2(5x + 1) > 0 when 1 – , ( ) 5 x f x′′< does not exist at x = 0. Test points: 1 8 –1, – ,1; (–1) , 10 9 f ′′ = 4/3 1 10 4 – – , (1) – . 10 9 3 f f ⎛ ⎞′′ = =⎜ ⎟ ⎝ ⎠ f(x) is concave up on 1 – , – 5 ⎛ ⎞ ∞⎜ ⎟ ⎝ ⎠ and concave down on 1 – , 0 (0, ). 5 ⎛ ⎞ ∪ ∞⎜ ⎟ ⎝ ⎠ 28. 2/3 4( 2) ( ) ; 3 x g x x + ′ = x + 2 > 0 when x > –2, ( )g x′ does not exist at x = 0. Split the x-axis into the intervals ( ), 2−∞ − , (–2, 0), (0, ∞ ). Test points: –3, –1, 1; 5/3 4 (–3) – , 3 g′ = 4 (–1) , (1) 4. 3 g g′ ′= = g(x) is increasing on [–2, ∞ ) and decreasing on (– ∞ , –2]. 5/3 4( – 4) ( ) ; 9 x g x x ′′ = x – 4 > 0 when x > 4, ( )g x′′ does not exist at x = 0. Test points: –1, 1, 5; 20 (–1) , 9 g′′ = 5/3 4 4 (1) – , (5) . 3 9(5) g g′′ ′′= = g(x) is concave up on (– ∞ , 0) ∪ (4, ∞ ) and concave down on (0, 4). © 2007 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.
  • 162 Section 3.2 Instructor’s Resource Manual 29. 30. 31. 32. 33. 34. 35. 2 ( ) ; ( ) 2 ;f x ax bx c f x ax b′= + + = + ( ) 2f x a′′ = An inflection point would occur where ( ) 0f x′′ = , or 2a = 0. This would only occur when a = 0, but if a = 0, the equation is not quadratic. Thus, quadratic functions have no points of inflection. 36. 3 2 ( ) ;f x ax bx cx d= + + + 2 ( ) 3 2 ;f x ax bx c′ = + + ( ) 6 2f x ax b′′ = + An inflection point occurs where ( ) 0f x′′ = , or 6ax + 2b = 0. The function will have an inflection point at – , 0. 3 b x a a = ≠ 37. Suppose that there are points 1x and 2x in I where 1( ) 0f x′ > and 2( ) 0.f x′ < Since f ′ is continuous on I, the Intermediate Value Theorem says that there is some number c between 1x and 2x such that ( ) 0,f c′ = which is a contradiction. Thus, either ( ) 0f x′ > for all x in I and f is increasing throughout I or ( ) 0f x′ < for all x in I and f is decreasing throughout I. 38. Since 2 1 0x + = has no real solutions, ( )f x′ exists and is continuous everywhere. 2 – 1 0x x + = has no real solutions. 2 – 1 0x x + > and 2 1 0x + > for all x, so ( ) 0f x′ > for all x. Thus f is increasing everywhere. 39. a. Let 2 ( )f x x= and let [ ]0, ,I a a y= > . ( ) 2 0f x x′ = > on I. Therefore, f(x) is increasing on I, so f(x) < f(y) for x < y. b. Let ( )f x x= and let [ ]0, ,I a a y= > . 1 ( ) 2 f x x ′ = > 0 on I. Therefore, f(x) is increasing on I, so f(x) < f(y) for x < y. c. Let 1 ( )f x x = and let I = [0, a], a > y. 2 1 ( ) 0f x x ′ = − < on I. Therefore f(x) is decreasing on I, so f(x) > f(y) for x < y. 40. 2 ( ) 3 2f x ax bx c′ = + + In order for f(x) to always be increasing, a, b, and c must meet the condition 2 3 2 0ax bx c+ + > for all x. More specifically, a > 0 and 2 3 0.b ac− < © 2007 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.
  • Instructor’s Resource Manual Section 3.2 163 41. 5/ 2 3 – ( ) . 4 b ax f x x ′′ = If (4, 13) is an inflection point then 13 2 2 b a= + and 3 – 4 0. 4 32 b a = ⋅ Solving these equations simultaneously, 39 8 a = and 13 . 2 b = 42. 1 2 3( ) ( )( )( )f x a x r x r x r= − − − 1 2 3 2 3( ) [( )(2 ) ( )( )]f x a x r x r r x r x r′ = − − − + − − 2 1 2 3 1 2 2 3 1 3( ) [3 2 ( ) ]f x a x x r r r r r r r r r′ = − + + + + + 1 2 3( ) [6 2( )]f x a x r r r′′ = − + + 1 2 3[6 2( )] 0a x r r r− + + = 1 2 3 1 2 36 2( ); 3 r r r x r r r x + + = + + = 43. a. [ ( ) ( )] ( ) ( ).f x g x f x g x′ ′ ′+ = + Since ( ) 0f x′ > and ( ) 0g x′ > for all x, ( ) ( ) 0f x g x′ ′+ > for all x. No additional conditions are needed. b. [ ( ) ( )] ( ) ( ) ( ) ( ).f x g x f x g x f x g x′ ′ ′⋅ = + ( ) ( ) ( ) ( ) 0f x g x f x g x′ ′+ > if ( ) ( ) ( ) ( ) f x f x g x g x ′ > − ′ for all x. c. [ ( ( ))] ( ( )) ( ).f g x f g x g x′ ′ ′= Since ( ) 0f x′ > and ( ) 0g x′ > for all x, ( ( )) ( ) 0f g x g x′ ′ > for all x. No additional conditions are needed. 44. a. [ ( ) ( )] ( ) ( ).f x g x f x g x′′ ′′ ′′+ = + Since ( ) 0f x′′ > and 0g′′ > for all x, ( ) ( ) 0f x g x′′ ′′+ > for all x. No additional conditions are needed. b. [ ( ) ( )] [ ( ) ( ) ( ) ( )]f x g x f x g x f x g x′′ ′ ′ ′⋅ = + ( ) ( ) ( ) ( ) 2 ( ) ( ).f x g x f x g x f x g x′′ ′′ ′ ′= + + The additional condition is that ( ) ( ) ( ) ( ) 2 ( ) ( ) 0f x g x f x g x f x g x′′ ′′ ′ ′+ + > for all x is needed. c. [ ( ( ))] [ ( ( )) ( )]f g x f g x g x′′ ′ ′ ′= 2 ( ( )) ( ) ( ( ))[ ( )] .f g x g x f g x g x′ ′′ ′′ ′= + The additional condition is that 2 ( ( ))[ ( )] ( ( )) ( ) f g x g x f g x g x ′′ ′ ′ > − ′′ for all x. 45. a. b. ( ) 0 :(1.3, 5.0)f x′ < c. ( ) 0 :( 0.25, 3.1) (6.5, 7]f x′′ < − ∪ d. 1 ( ) cos – sin 2 2 x f x x′ = e. 1 ( ) sin cos 4 2 x f x x′′ = − − 46. a. b. ( ) 0 :(2.0, 4.7) (9.9, 10]f x′ < ∪ c. ( ) 0 :[0, 3.4) (7.6,10]f x′′ < ∪ d. 22 ( ) cos sin cos 3 3 3 3 x x x f x x ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ′ = − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ 2 2 cos sin 3 3 3 x x x⎛ ⎞ ⎛ ⎞ = −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ © 2007 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.
  • 164 Section 3.2 Instructor’s Resource Manual e. 2 2 2 2 ( ) cos sin 9 3 3 3 x x x f x ⎛ ⎞ ⎛ ⎞′′ = − −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 47. ( ) 0f x′ > on (–0.598, 0.680) f is increasing on [–0.598, 0.680]. 48. ( ) 0 when 1.63f x x′′ < > in [–2, 3] f is concave down on (1.63, 3). 49. Let s be the distance traveled. Then ds dt is the speed of the car. a. , ds ks dt = k a constant t s Concave up. b. 2 2 0 d s dt > t s Concave up. c. 3 2 3 2 0, 0 d s d s dt dt < > t s Concave up. d. 2 2 10 d s dt = mph/min t s Concave up. e. 2 2 and ds d s dt dt are approaching zero. t s Concave down. © 2007 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.
  • Instructor’s Resource Manual Section 3.2 165 f. ds dt is constant. t s Neither concave up nor down. 50. a. dV k dt = <0, V is the volume of water in the tank, k is a constant. Neither concave up nor down. t v t( ) b. 1 1 3 – 2 2 2 dV dt = = gal/min Neither concave up nor down. t v t( ) c. 2 2 , 0, 0 dV dh d h k dt dt dt = > < Concave down. t h t( ) d. I(t) = k now, but 2 2 , 0 dI d I dt dt > in the future where I is inflation. t I t( ) e. 2 2 0, but 0 dp d p dt dt < > and at t = 2: 0 dp dt > . where p is the price of oil. Concave up. t P t( ) 2 © 2007 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.
  • 166 Section 3.2 Instructor’s Resource Manual f. 2 2 0, 0 dT d T dt dt > < , where T is David’s temperature. Concave down. t T t( ) 51. a. 2 2 0, 0 dC d C dt dt > > , where C is the car’s cost. Concave up. t C t( ) b. f(t) is oil consumption at time t. 2 2 0, 0 df d f dt dt < > Concave up. t f t( ) c. 2 2 0, 0 dP d P dt dt > < , where P is world population. Concave down. t P t( ) d. 2 2 0, 0 d d dt dt θ θ > > , where θ is the angle that the tower makes with the vertical. Concave up. t θ( )t e. P = f(t) is profit at time t. 2 2 0, 0 dP d P dt dt > < Concave down. t P t( ) © 2007 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.
  • Instructor’s Resource Manual Section 3.2 167 f. R is revenue at time t. P < 0, 0 dP dt > Could be either concave up or down. t P t P 52. a. R(t) ≈ 0.28, t < 1981 b. On [1981, 1983], 2 2 0, 0 dR d R dt dt > > , R(1983) ≈ 0.36 53. 3 2 in /sec dV dt = The cup is a portion of a cone with the bottom cut off. If we let x represent the height of the missing cone, we can use similar triangles to show that 5 3 3.5 3.5 3 15 0.5 15 30 x x x x x x + = = + = = Similar triangles can be used again to show that, at any given time, the radius of the cone at water level is 30 20 h r + = Therefore, the volume of water can be expressed as 3 ( 30) 45 1200 2 h V π π+ = − . We also know that 2V t= from above. Setting the two volume equations equal to each other and solving for h gives 3 2400 27000 30h t π = + − . 54. The height is always increasing so '( ) 0h t > . The rate of change of the height decreases for the first 50 minutes and then increases over the next 50 minutes. Thus ''( ) 0h t < for 0 50t≤ ≤ and ''( ) 0h t > for 50 100t< ≤ . 55. 3 , 0 8V t t= ≤ ≤ . The height is always increasing, so '( ) 0.h t > The rate of change of the height decreases from time 0t = until time 1t when the water reaches the middle of the rounded bottom part. The rate of change then increases until time 2t when the water reaches the middle of the neck. Then the rate of change decreases until 8t = and the vase is full. Thus, 1 2''( ) 0 forh t t t t> < < and 2''( ) 0 for 8h t t t< < < . h t( ) t1t 2t 8 24 © 2007 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.
  • 168 Section 3.3 Instructor’s Resource Manual 56. 20 .1V t= − , 0 200t≤ ≤ . The height of the water is always decreasing so '( ) 0h t < . The rate of change in the height increases (the rate is negative, and its absolute value decreases) for the first 100 days and then decreases for the remaining time. Therefore we have ''( ) 0h t > for 0 100t< < , and ''( ) 0h t < for 100 200t< < . 57. a. The cross-sectional area of the vase is approximately equal to ΔV and the corresponding radius is r = /V πΔ . The table below gives the approximate values for r. The vase becomes slightly narrower as you move above the base, and then gets wider as you near the top. Depth V A ≈ ΔV /r V π= Δ 1 4 4 1.13 2 8 4 1.13 3 11 3 0.98 4 14 3 0.98 5 20 6 1.38 6 28 8 1.60 b. Near the base, this vase is like the one in part (a), but just above the base it becomes larger. Near the middle of the vase it becomes very narrow. The top of the vase is similar to the one in part (a). Depth V A ≈ ΔV /r V π= Δ 1 4 4 1.13 2 9 5 1.26 3 12 3 0.98 4 14 2 0.80 5 20 6 1.38 6 28 8 1.60 3.3 Concepts Review 1. maximum 2. maximum; minimum 3. maximum 4. local maximum, local minimum, 0 Problem Set 3.3 1. 2 ( ) 3 –12 3 ( – 4)f x x x x x′ = = Critical points: 0, 4 ( )f x′ > 0 on (– ∞ , 0), ( ) 0f x′ < on (0, 4), ( ) 0f x′ > on (4, ∞ ) ( ) 6 –12; (0) –12, (4) 12.f x x f f′′ ′′ ′′= = = Local minimum at x = 4; local maximum at x = 0 2. 2 2 ( ) 3 –12 3( – 4)f x x x′ = = Critical points: –2, 2 ( ) 0f x′ > on (– ∞ , –2), ( ) 0f x′ < on (–2, 2), ( ) 0f x′ > on (2, ∞ ) ( ) 6 ; (–2) –12, (2) 12f x x f f′′ ′′ ′′= = = Local minimum at x = 2; local maximum at x = –2 3. ( ) 2cos2 ; 2cos2 0f θ θ θ′ = ≠ on 0, 4 π⎛ ⎞ ⎜ ⎟ ⎝ ⎠ No critical points; no local maxima or minima on 0, . 4 π⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 4. 1 1 ( ) cos ; cos 0 2 2 f x x x′ = + + = when 1 cos – . 2 x = Critical points: 2 4 , 3 3 π π 2 ( ) 0 on 0, , 3 f x π⎛ ⎞′ > ⎜ ⎟ ⎝ ⎠ 2 4 ( ) 0 on , , 3 3 f x π π⎛ ⎞′ < ⎜ ⎟ ⎝ ⎠ 4 ( ) 0 on , 2 3 f x π⎛ ⎞′ > π⎜ ⎟ ⎝ ⎠ 2 3 4 3 ( ) –sin ; – , 3 2 3 2 f x x f f π π⎛ ⎞ ⎛ ⎞′′ ′′ ′′= = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Local minimum at 4 ; 3 x π = local maximum at 2 . 3 x π = © 2007 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.
  • Instructor’s Resource Manual Section 3.3 169 5. ( ) 2sin cosθ θ θ′Ψ = 2 2 π π θ− < < Critical point: 0 ( ) 0 on , 0 , 2 θ π⎛ ⎞′Ψ < −⎜ ⎟ ⎝ ⎠ ( ) 0 on 0, , 2 θ π⎛ ⎞′Ψ > ⎜ ⎟ ⎝ ⎠ 2 2 ( ) 2cos – 2sin ;θ θ θ′′Ψ = (0) 2′′Ψ = Local minimum at x = 0 6. 3 ( ) 4r z z′ = Critical point: 0 ( ) 0 on ( , 0);r z′ < −∞ ( ) 0 on (0, )r z′ > ∞ 2 ( ) 12 ; (0) 0;r x x r′′ ′′= = the Second Derivative Test fails. Local minimum at z = 0; no local maxima 7. ( ) ( ) ( ) ( ) ( ) 2 2 2 22 2 4 1 2 4 ' 4 4 x x x x f x x x + ⋅ − − = = + + Critical points: 2,2− ( )' 0f x < on ( ), 2−∞ − and ( )2,∞ ; ( )' 0f x > on ( )2,2− ( ) ( ) ( ) 2 32 2 12 '' 4 x x f x x − = + ( ) 1 '' 2 16 f − = ; ( ) 1 '' 2 16 f = − Local minima at 2x = − ; Local maxima at 2x = 8. ( ) ( )( ) ( ) ( ) ( ) 2 2 2 22 2 1 2 2 2 ' 1 1 z z z z z g z z z + − = = + + Critical point: 0z = ( )' 0g z < on ( ),0−∞ ( )' 0g z > on ( )0,∞ ( ) ( ) ( ) 2 32 2 3 1 '' 1 z g z z − − = + ( )'' 0 2g = Local minima at 0z = . 9. ( ) 2 1 ' 2h y y y = + Critical point: 3 4 2 − ( )' 0h y < on 3 4 , 2 ⎛ ⎞ −∞ −⎜ ⎟ ⎝ ⎠ ( )' 0h y > on 3 4 ,0 2 ⎛ ⎞ −⎜ ⎟ ⎝ ⎠ and ( )0,∞ ( ) 3 2 '' 2h y y = − ( )3 3 3 4 2 4 2 16 2 2 6 2 4 h ⎛ ⎞ − = − = + =⎜ ⎟ ⎝ ⎠ − Local minima at 3 4 2 − 10. ( ) ( )( ) ( )( ) ( ) ( ) 2 2 2 22 2 1 3 3 1 2 3 2 3 ' 1 1 x x x x x f x x x + − + − − = = + + The only critical points are stationary points. Find these by setting the numerator equal to 0 and solving. 2 3 2 3 0x x− − = 3, 2, 3a b c= − = − = ( ) ( )( ) ( ) 2 2 2 4 3 3 2 40 1 10 2 3 6 3 x ± − − − ± − ± = = = − − Critical points: 1 10 3 − − and 1 10 3 − + ( )' 0f x < on 1 10 , 3 ⎛ ⎞− − −∞⎜ ⎟⎜ ⎟ ⎝ ⎠ and 1 10 , 3 ⎛ ⎞− + ∞⎜ ⎟⎜ ⎟ ⎝ ⎠ . ( )' 0 0f > on 1 10 1 10 , 3 3 ⎛ ⎞− − − + ⎜ ⎟⎜ ⎟ ⎝ ⎠ ( ) ( ) ( ) 3 2 32 2 3 3 9 1 '' 1 x x x f x x + − − = + 1 10 '' 0.739 3 f ⎛ ⎞− − ≈⎜ ⎟⎜ ⎟ ⎝ ⎠ 1 10 '' 2.739 3 f ⎛ ⎞− + ≈ −⎜ ⎟⎜ ⎟ ⎝ ⎠ Local minima at 1 10 3 x − − = ; Local maxima at 1 10 3 x − + = © 2007 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.
  • 170 Section 3.3 Instructor’s Resource Manual 11. 2 2 ( ) 3 – 3 3( –1)f x x x′ = = Critical points: –1, 1 ( ) 6 ; (–1) –6, (1) 6f x x f f′′ ′′ ′′= = = Local minimum value f(1) = –2; local maximum value f(–1) = 2 12. 3 2 ( ) 4 2 2 (2 1)g x x x x x′ = + = + Critical point: 0 2 ( ) 12 2; (0) 2g x x g′′ ′′= + = Local minimum value g(0) = 3; no local maximum values 13. 3 2 2 ( ) 4 – 6 2 (2 – 3)H x x x x x′ = = Critical points: 3 0, 2 2 ( ) 12 –12 12 ( –1);H x x x x x′′ = = (0) 0,H′′ = 3 9 2 H ⎛ ⎞′′ =⎜ ⎟ ⎝ ⎠ 3 ( ) 0 on ( , 0), ( ) 0 on 0, 2 H x H x ⎛ ⎞′ ′< −∞ < ⎜ ⎟ ⎝ ⎠ Local minimum value 3 27 – ; 2 16 H ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ no local maximum values (x = 0 is neither a local minimum nor maximum) 14. 4 ( ) 5( – 2)f x x′ = Critical point: 2 3 ( ) 20( – 2) ; (2) 0f x x f′′ ′′= = ( ) 0 on ( , 2), ( ) 0 on (2, )f x f x′ ′> −∞ > ∞ No local minimum or maximum values 15. 1/3 2 ( ) – ; ( ) 3( – 2) g t g t t ′ ′= does not exist at t = 2. Critical point: 2 2 2 (1) , (3) – 3 3 g g′ ′= = No local minimum values; local maximum value g(2) = π . 16. 3/5 3/5 3/5 2 15 2 ( ) 3 ; ( ) 0 5 5 s r s r s s s + ′ ′= + = = when 5/3 2 – , ( ) 15 s r s ⎛ ⎞ ′= ⎜ ⎟ ⎝ ⎠ does not exist at s = 0. Critical points: 5/3 2 – , 0 15 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 5/3 8/3 8/5 6 2 6 15 ( ) – ; – – 15 25 225 r s r s ⎛ ⎞⎛ ⎞ ⎛ ⎞ ′′ ′′⎜ ⎟= =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ 5 3 2 ( ) 0 on , 0 , ( ) 0 on (0, ) 15 r s r s ⎛ ⎞⎛ ⎞ ′ ′⎜ ⎟< − > ∞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ Local minimum value r(0) = 0; local maximum value 5/3 5/3 2/3 2/3 2 2 2 3 2 – –3 15 15 15 5 15 r ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ = + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ 17. 2 1 ( ) 1f t t ′ = + No critical points No local minimum or maximum values 18. 2 2 3/ 2 ( 8) ( ) ( 4) x x f x x + ′ = + Critical point: 0 ( ) 0 on ( , 0), ( ) 0 on (0, )f x f x′ ′< −∞ > ∞ Local minimum value f(0) = 0, no local maximum values 19. 1 ( ) – ; ( ) 1 sin θ θ θ ′ ′Λ = Λ + does not exist at 3 , 2 θ π = but ( )θΛ does not exist at that point either. No critical points No local minimum or maximum values 20. sin cos ( ) ; ( ) 0 sin g g θ θ θ θ θ ′ ′= = when 3 , 2 2 θ π π = ; ( )g θ′ does not exist at x = π . Split the x -axis into the intervals 0, , 2 π⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 3 3 , , , , , 2 . 2 2 2 π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ π π π⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Test points: 3 5 7 1 , , , ; , 4 4 4 4 4 2 g π π π π π⎛ ⎞′ =⎜ ⎟ ⎝ ⎠ 3 1 5 1 7 1 – , , – 4 4 42 2 2 g g g π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞′ ′ ′= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Local minimum value g(π ) = 0; local maximum values 1 2 g π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ and 3 1 2 g π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ © 2007 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.
  • Instructor’s Resource Manual Section 3.3 171 21. ( ) ( )( )' 4 sin 2 cos2f x x x= ( )( )4 sin 2 cos2 0x x = when ( )2 1 4 k x π− = or 2 k x π = where k is an integer. Critical points: 0, 4 π , 2 π , 2 ( )0 0f = ; 1 4 f π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ ; 0 2 f π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ ; ( )2 0.5728f ≈ Minimum value: ( )0 0 2 f f π⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ Maximum value: 1 4 f π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 22. ( ) ( ) ( ) 2 22 2 4 ' 4 x f x x − − = + ( )' 0f x = when 2x = or 2x = − . (there are no singular points) Critical points: 0, 2 (note: 2− is not in the given domain) ( )0 0f = ; ( ) 1 2 2 f = ; ( ) 0f x → as x → ∞ . Minimum value: ( )0 0f = Maximum value: ( ) 1 2 2 f = 23. ( ) ( ) ( ) 3 23 64 ' 32 x x g x x − − = + ( )' 0g x = when 0x = or 4x = . Critical points: 0, 4 ( )0 0g = ; ( ) 1 4 6 g = As x approaches ∞ , the value of g approaches 0 but never actually gets there. Maximum value: ( ) 1 4 6 g = Minimum value: ( )0 0g = 24. ( ) ( ) 22 2 ' 4 x h x x − = + ( )' 0h x = when 0x = . (there are no singular points) Critical points: 0 Since ( )' 0h x < for 0x > , the function is always decreasing. Thus, there is no minimum value. Maximum value: ( ) 1 0 4 h = 25. 3 3 ( ) – 4; – 4 0F x x x ′ = = when 9 16 x = Critical points: 9 0, , 4 16 F(0) = 0, 9 9 , 16 4 F ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ F(4) = –4 Minimum value F(4) = –4; maximum value 9 9 16 4 F ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 26. From Problem 25, the critical points are 0 and 9 . 16 9 9 ( ) 0 on 0, , ( ) 0 on , 16 16 F x F x ⎛ ⎞ ⎛ ⎞′ ′> < ∞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ F decreases without bound on 9 , . 16 ⎛ ⎞ ∞⎜ ⎟ ⎝ ⎠ No minimum values; maximum value 9 9 16 4 F ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 27. 2 2 ( ) 64( 1)(sin ) cos 27( 1)(cos ) ( sin ) f x x x x x − − ′ = − + − − 2 2 2 2 2 2 (3sin 4cos )(9sin 12cos sin 16cos ) sin cos 64cos 27sin sin cos x x x x x x x x x x x x − + + = − + = On 0, , ( ) 0 2 f x π⎛ ⎞ ′ =⎜ ⎟ ⎝ ⎠ only where 3sin x = 4cos x; 4 tan ; 3 x = 1 4 tan 0.9273 3 x − = ≈ Critical point: 0.9273 For 0 < x < 0.9273, ( ) 0,f x′ < while for 0.9273 < x < 2 π , '( ) 0f x > Minimum value 1 4 3 5 5 4 64 27 tan 125; 3 f −⎛ ⎞ = + =⎜ ⎟ ⎝ ⎠ no maximum value © 2007 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.
  • 172 Section 3.3 Instructor’s Resource Manual 28. 2 2 4 (8 ) (32 ) (16 )2(8 )( 1) ( ) 2 (8 ) x x x x g x x x − − − − ′ = + − 3 3 3 256 2 [(8 ) 128] 2 (8 ) (8 ) x x x x x x − + = + = − − For x > 8, ( ) 0g x′ = when 3 (8 ) 128 0;x− + = 3 3 (8 ) 128;8 128x x− = − − = − ; 3 8 4 2 13.04x = + ≈ 3 ( ) 0 on (8,8 4 2),g x′ < + 3 ( ) 0 on (8 4 2, )g x′ > + ∞ g(13.04) ≈ 277 is the minimum value 29. ( ) ( )2 2 2 1 ' 1 x x H x x − = − ( )' 0H x = when 0x = . ( )'H x is undefined when 1x = − or 1x = Critical points: 2− , 1− , 0, 1, 2 ( )2 3H − = ; ( )1 0H − = ; ( )0 1H = ; ( )1 0H = ; ( )2 3H = Minimum value: ( ) ( )1 1 0H H− = = Maximum value: ( ) ( )2 2 3H H− = = 30. ( ) 2 ' 2 cosh t t t= ( )' 0h t = when 0t = , 2 2 t π = , 6 2 t π = , and 10 2 t π = (Consider 2 2 t π = , 2 3 2 t π = , and 2 5 2 t π = ) Critical points: 0, 2 2 π , 6 2 π , 10 2 π , π ( )0 0h = ; 2 1 2 h π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ ; 6 1 2 h π⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ ; 10 1 2 h π⎛ ⎞ =⎜ ⎟ ⎝ ⎠ ; ( ) 0.4303h π ≈ − Minimum value: 6 1 2 h π⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ Maximum value: 2 10 1 2 2 h h π π⎛ ⎞ ⎛ ⎞ = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 31. '( ) 0f x = when 0x = and 1x = . On the interval ( ,0)−∞ we get '( ) 0f x < . On (0, )∞ , we get '( ) 0f x > . Thus there is a local min at 0x = but no local max. 32. '( ) 0f x = at 1,2,3,4x = ; '( )f x is negative on ( ,1) (2,3) (4, )−∞ ∪ ∪ ∞ and positive on (1,2) (3,4).∪ Thus, the function has a local minimum at 1,3x = and a local maximum at 2,4x = . 33. '( ) 0f x = at 1,2,3,4x = ; '( )f x is negative on (3,4) and positive on ( ,1) (1,2) (2,3) (4, )−∞ ∪ ∪ ∪ ∞ Thus, the function has a local minimum at 4x = and a local maximum at 3x = . 34. Since ( )' 0f x ≥ for all x, the function is always increasing. Therefore, there are no local extrema. 35. Since ( )' 0f x ≥ for all x, the function is always increasing. Therefore, there are no local extrema. 36. ( )' 0f x = at 0, , andx A B= . ( )'f x is negative on ( ),0−∞ and ( ),A B ( )'f x is positive on ( )0, A and ( ),B ∞ Therefore, the function has a local minimum at 0x = and x B= , and a local maximum at x A= . 37. Answers will vary. One possibility: −5 5 x y 3 6 38. Answers will vary. One possibility: −5 5 x y 3 6 © 2007 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.
  • Instructor’s Resource Manual Section 3.3 173 39. Answers will vary. One possibility: −5 5 x y 3 6 40. Answers will vary. One possibility: −5 5 x y 3 6 41. Answers will vary. One possibility: −5 5 x y 3 6 42. Answers will vary. One possibility: −5 5 x y 3 6 43. The graph of f is a parabola which opens up. ( ) ( ) ' 2 0 2 '' 2 B f x Ax B x A f x A = + = → = − = Since 0A > , the graph of f is always concave up. There is exactly one critical point which yields the minimum of the graph. 2 2 2 2 2 2 2 2 2 2 4 2 2 4 4 4 4 4 4 B B B f A B C A A A B B C A A B B AC A AC B B AC A A ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − = − + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = − + − + = − − = = − If ( ) 0f x ≥ with 0A > , then ( )2 4 0B AC− − ≥ , or 2 4 0B AC− ≤ . If 2 4 0B AC− ≤ , then we get 0 2 B f A ⎛ ⎞ − ≥⎜ ⎟ ⎝ ⎠ Since ( )0 2 B f f x A ⎛ ⎞ ≤ − ≤⎜ ⎟ ⎝ ⎠ for all x, we get ( ) 0f x ≥ for all x. 44. A third degree polynomial will have at most two extrema. ( ) ( ) 2 ' 3 2 '' 6 2 f x Ax Bx C f x Ax B = + + = + Critical points are obtained by solving ( )' 0f x = . 2 3 2 0Ax Bx C+ + = 2 2 2 2 4 12 6 2 2 3 6 3 3 B B AC x A B B AC A B B AC A − ± − = − ± − = − ± − = To have a relative maximum and a relative minimum, we must have two solutions to the above quadratic equation. That is, we must have 2 3 0B AC− > . The two solutions would be 2 3 3 B B AC A − − − and 2 3 3 B B AC A − + − . Evaluating the second derivative at each of these values gives: © 2007 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.
  • 174 Section 3.4 Instructor’s Resource Manual 2 2 2 2 3 '' 3 3 6 2 3 2 2 3 2 2 3 B B AC f A B B AC A B A B B AC B B AC ⎛ ⎞− − − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞− − − ⎜ ⎟= + ⎜ ⎟ ⎝ ⎠ = − − − + = − − and 2 2 2 2 3 '' 3 3 6 2 3 2 2 3 2 2 3 B B AC f A B B AC A B A B B AC B B AC ⎛ ⎞− + − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞− + − ⎜ ⎟= + ⎜ ⎟ ⎝ ⎠ = − + − + = − If 2 3 0B AC− > , then 2 2 3B AC− − exists and is negative, and 2 2 3B AC− exists and is positive. Thus, from the Second Derivative Test, 2 3 3 B B AC A − − − would yield a local maximum and 2 3 3 B B AC A − + − would yield a local minimum. 45. ( ) 0f c′′′ > implies that f ′′ is increasing at c, so f is concave up to the right of c (since ( ) 0f x′′ > to the right of c) and concave down to the left of c (since ( ) 0f x′′ < to the left of c). Therefore f has a point of inflection at c. 3.4 Concepts Review 1. 0 < x < ∞ 2. 200 2x x + 3. ( )2 1 n i i i S y bx = = −∑ 4. marginal revenue; marginal cost Problem Set 3.4 1. Let x be one number, y be the other, and Q be the sum of the squares. xy = –16 16 –y x = The possible values for x are in (– ∞ , 0) or (0, )∞ . 2 2 2 2 256 Q x y x x = + = + 3 512 2 – dQ x dx x = 3 512 2 – 0x x = 4 256x = x = ±4 The critical points are –4, 4. 0 dQ dx < on (– ∞ , –4) and (0, 4). 0 dQ dx > on (–4, 0) and (4, ∞ ). When x = –4, y = 4 and when x = 4, y = –4. The two numbers are –4 and 4. 2. Let x be the number. – 8Q x x= x will be in the interval (0, ∞ ). –1/ 21 – 8 2 dQ x dx = –1/ 21 – 8 0 2 x = –1/ 2 16x = 1 256 x = 0 dQ dx > on 1 0, 256 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ and 0 dQ dx < on 1 , . 256 ⎛ ⎞ ∞⎜ ⎟ ⎝ ⎠ Q attains its maximum value at 1 . 256 x = 3. Let x be the number. 4 – 2Q x x= x will be in the interval (0, ∞ ). –3/ 41 – 2 4 dQ x dx = –3/ 41 – 2 0 4 x = –3/ 4 8x = 1 16 x = 0 dQ dx > on 1 0, 16 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ and 0 dQ dx < on 1 , 16 ⎛ ⎞ ∞⎜ ⎟ ⎝ ⎠ Q attains its maximum value at 1 . 16 x = © 2007 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.
  • Instructor’s Resource Manual Section 3.4 175 4. Let x be one number, y be the other, and Q be the sum of the squares. xy = –12 12 –y x = The possible values for x are in (– ∞ , 0) or (0, )∞ . 2 2 2 2 144 Q x y x x = + = + 3 288 2 – dQ x dx x = 3 288 2 – 0x x = 4 144x = 2 3x = ± The critical points are –2 3, 2 3 0 dQ dx < on (– , – 2 3)∞ and (0, 2 3). 0 dQ dx > on (–2 3, 0) and (2 3, ).∞ When –2 3, 2 3x y= = and when 2 3, –2 3.x y= = The two numbers are –2 3 and 2 3. 5. Let Q be the square of the distance between (x, y) and (0, 5). 2 2 2 2 2 ( – 0) ( – 5) ( – 5)Q x y x x= + = + 4 2 – 9 25x x= + 3 4 –18 dQ x x dx = 3 4 –18 0x x = 2 2 (2 – 9) 0x x = 3 0, 2 x = ± 0 dQ dx < on 3 – , – 2 ⎛ ⎞ ∞⎜ ⎟ ⎝ ⎠ and 3 0, . 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 0 dQ dx > on 3 3 – , 0 and , . 2 2 ⎛ ⎞ ⎛ ⎞ ∞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ When 3 9 3 – , and when , 22 2 x y x= = = 9 . 2 y = The points are 3 9 3 9 – , and , . 2 22 2 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 6. Let Q be the square of the distance between (x, y) and (10, 0). 2 2 2 2 2 ( –10) ( – 0) (2 –10)Q x y y y= + = + 4 2 4 – 39 100y y= + 3 16 – 78 dQ y y dy = 3 16 – 78 0y y = 2 2 (8 – 39) 0y y = 39 0, 2 2 y = ± 39 39 0 on – , – and 0, . 2 2 2 2 dQ dy ⎛ ⎞ ⎛ ⎞ < ∞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 39 39 0 on – , 0 and , . 2 2 2 2 dQ dy ⎛ ⎞ ⎛ ⎞ > ∞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ When 39 39 – , 42 2 y x= = and when 39 39 , . 42 2 y x= = The points are 39 39 39 39 , – and , . 4 42 2 2 2 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 7. 2 x x≥ if 0 1x≤ ≤ 2 ( ) ; ( ) 1 2 ;f x x x f x x′= − = − 1 ( ) 0 when 2 f x x′ = = Critical points: 1 0, ,1 2 f(0) = 0, f(1) = 0, 1 1 ; 2 4 f ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ therefore, 1 2 exceeds its square by the maximum amount. 8. For a rectangle with perimeter K and width x, the length is 2 K x− . Then the area is 2 2 2 K Kx A x x x ⎛ ⎞ = − = −⎜ ⎟ ⎝ ⎠ . 2 ; 0 when 2 4 dA K dA K x x dx dx = − = = Critical points: 0, , 4 2 K K At x = 0 or 2 K , A = 0; at x = , 4 K 2 16 K A = . The area is maximized when the width is one fourth of the perimeter, so the rectangle is a square. © 2007 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.
  • 176 Section 3.4 Instructor’s Resource Manual 9. Let x be the width of the square to be cut out and V the volume of the resulting open box. 2 3 2 (24 2 ) 4 96 576V x x x x x= − = − + 2 12 192 576 12( 12)( 4); dV x x x x dx = − + = − − 12(x – 12)(x – 4) = 0; x = 12 or x = 4. Critical points: 0, 4, 12 At x = 0 or 12, V = 0; at x = 4, V = 1024. The volume of the largest box is 3 1024 in. 10. Let A be the area of the pen. 2 (80 2 ) 80 2 ; 80 4 ; dA A x x x x x dx = − = − = − 80 4 0; 20x x− = = Critical points: 0, 20, 40. At x = 0 or 40, A = 0; at x = 20, A = 800. The dimensions are 20 ft by 80 – 2(20) = 40 ft, with the length along the barn being 40 ft. 11. Let x be the width of each pen, then the length along the barn is 80 – 4x. 2 (80 4 ) 80 4 ; 80 8 ; dA A x x x x x dx = − = − = − 0 when 10. dA x dx = = Critical points: 0, 10, 20 At x = 0 or 20, A = 0; at x = 10, A = 400. The area is largest with width 10 ft and length 40 ft. 12. Let A be the area of the pen. The perimeter is 100 + 180 = 280 ft. y + y – 100 + 2x = 180; y = 140 – x 2 (140 ) 140 ; 140 2 ; dA A x x x x x dx = − = − = − 140 2 0; 70x x− = = Since 0 40x≤ ≤ , the critical points are 0 and 40. When x = 0, A = 0. When x = 40, A = 4000. The dimensions are 40 ft by 100 ft. 13. 900 900;xy y x = = The possible values for x are in (0, ∞ ). 900 2700 4 3 4 3 4Q x y x x x x ⎛ ⎞ = + = + = +⎜ ⎟ ⎝ ⎠ 2 2700 4 dQ dx x = − 2 2700 4 – 0 x = 2 675x = 15 3x = ± 15 3x = is the only critical point in (0, ∞ ). 0 dQ dx < on (0,15 3) and 0 on (15 3, ). dQ dx > ∞ When 900 15 3, 20 3. 15 3 x y= = = Q has a minimum when 15 3 25.98x = ≈ ft and 20 3 34.64y = ≈ ft. 14. xy = 300; 300 y x = The possible values for x are in (0, ∞ ). 1200 6 4 6Q x y x x = + = + 2 1200 6 – dQ dx x = 2 1200 6 – 0 x = 2 200x = 10 2x = ± 10 2x = is the only critical point in (0, ∞ ). 0 on (0, 10 2) dQ dx < and 0 on (10 2, ) dQ dx > ∞ When 300 10 2, 15 2. 10 2 x y= = = Q has a minimum when 10 2 14.14x = ≈ ft and 15 2 21.21y = ≈ ft. 15. xy = 300; 300 y x = The possible values for x are in (0, ∞). Q = 3(6x + 2y) + 2(2y) = 18x + 10y 3000 18x x = + 2 3000 18 – dQ dx x = 2 3000 18 – 0 x = 2 500 3 x = 10 5 3 x = ± 10 5 3 x = is the only critical point in (0, ∞). 10 5 0 on 0, 3 dQ dx ⎛ ⎞ < ⎜ ⎟⎜ ⎟ ⎝ ⎠ and 10 5 0 on , . 3 dQ dx ⎛ ⎞ > ∞⎜ ⎟⎜ ⎟ ⎝ ⎠ © 2007 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.
  • Instructor’s Resource Manual Section 3.4 177 When 10 5 3 10 5 300 , 6 15 3 x y= = = Q has a minimum when 10 5 12.91 3 x = ≈ ft and 6 15 23.24y = ≈ ft. 16. xy = 900; 900 y x = The possible values for x are in (0, ∞ ). 3600 6 4 6Q x y x x = + = + 2 3600 6 – dQ dx x = 2 3600 6 – 0 x = 2 600x = 10 6x = ± 10 6x = is the only critical point in (0, ∞ ). 0 on (0,10 6) dQ dx < and 0 on (10 6, ). dQ dx > ∞ When 900 10 6, 15 6 10 6 x y= = = Q has a minimum when 10 6 24.49x = ≈ ft and 15 6 36.74.y = ≈ It appears that 2 . 3 x y = Suppose that each pen has area A. xy = A; A y x = The possible values for x are in (0, ∞ ). 4 6 4 6 A Q x y x x = + = + 2 4 6 – dQ A dx x = 2 4 6 – 0 A x = 2 2 3 A x = 2 3 A x = ± 2 3 A x = is the only critical point on (0, ∞ ). 2A 0 on 0, 3 dQ dx ⎛ ⎞ < ⎜ ⎟⎜ ⎟ ⎝ ⎠ and 2 0 on , . 3 dQ A dx ⎛ ⎞ > ∞⎜ ⎟⎜ ⎟ ⎝ ⎠ When 2 3 2 3 , 3 2A A A A x y= = = 2 3 3 2 2 3 A A x y = = 17. Let D be the square of the distance. 22 2 2 2 ( 0) ( 4) 4 4 x D x y x ⎛ ⎞ = − + − = + −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 4 2 16 16 x x= − + 3 3 2 2 ; 2 0; ( 8) 0 4 4 dD x x x x x x dx = − − = − = 0, 2 2x x= = ± Critical points: 0, 2 2, 2 3 Since D is continuous and we are considering a closed interval for x, there is a maximum and minimum value of D on the interval. These extrema must occur at one of the critical points. At x = 0, y = 0, and D = 16. At 2 2, 2,x y= = and D = 12. At 2 3x = , y = 3, and D = 13. Therefore, the point on 2 4 x y = closest to ( )0,4 is ( )2 2,2P and the point farthest from ( )0,4 is ( )0,0Q . 18. Let 1r and 1h be the radius and altitude of the outer cone; 2r and 2h the radius and altitude of the inner cone. 2 1 1 1 1 3 V r h= π is fixed. 1 1 1 3V r h = π By similar triangles 1 2 2 1 1 –h h r h r = (see figure). 2 1 2 2 1 1 1 1 3 1– 1– h V h r r h h h ⎛ ⎞ ⎛ ⎞ = =⎜ ⎟ ⎜ ⎟ π⎝ ⎠ ⎝ ⎠ © 2007 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.
  • 178 Section 3.4 Instructor’s Resource Manual 2 2 1 2 2 2 2 2 1 1 31 1 1– 3 3 V h V r h h h h ⎡ ⎤⎛ ⎞ = π = π⎢ ⎥⎜ ⎟ π⎢ ⎥⎝ ⎠⎣ ⎦ 2 2 1 2 2 2 2 1 1 1 1 1 3 1– 1– 3 V h h h h V h h h h ⎛ ⎞ ⎛ ⎞π = ⋅ =⎜ ⎟ ⎜ ⎟ π ⎝ ⎠ ⎝ ⎠ Let 2 1 , h k h = the ratio of the altitudes of the cones, then 2 2 1 (1– ) .V V k k= 22 1 1 1(1– ) – 2 (1– ) (1– )(1– 3 ) dV V k kV k V k k dk = = 0 < k < 1 so 2 0 dV dk = when 1 . 3 k = 2 2 2 2 12 2 (6 4); 0 d V d V V k dk dk = − < when 1 3 k = The altitude of the inner cone must be 1 3 the altitude of the outer cone. 19. Let x be the distance from P to where the woman lands the boat. She must row a distance of 2 4x + miles and walk 10 – x miles. This will take her 2 4 10 – ( ) 3 4 x x T x + = + hours; 0 ≤ x ≤ 10. 2 1 ( ) – ; ( ) 0 43 4 x T x T x x ′ ′= = + when 6 . 7 x = 19 (0) 6 T = hr = 3 hr 10 min 3.17 hr≈ , 6 15 7 2.94 67 T +⎛ ⎞ = ≈⎜ ⎟ ⎝ ⎠ hr, 104 (10) 3.40 3 T = ≈ hr She should land the boat 6 2.27 7 ≈ mi down the shore from P. 20. 2 4 10 – ( ) , 0 10. 3 50 x x T x x + = + ≤ ≤ 2 1 ( ) – ; ( ) 0 503 – 4 x T x T x x ′ ′= = when 6 2491 x = 13 (0) 0.867 hr; 15 T = ≈ 6 0.865 hr; 2491 T ⎛ ⎞ ≈⎜ ⎟ ⎝ ⎠ (10) 3.399 hrT ≈ She should land the boat 6 2491  ≈ 0.12 mi down the shore from P. 21. 2 4 10 – ( ) , 0 10. 20 4 x x T x x + = + ≤ ≤ 2 1 ( ) – ; ( ) 0 420 4 x T x T x x ′ ′= = + has no solution. 2 10 13 (0) 20 4 5 T = + = hr = 2 hr, 36 min 104 (10) 0.5 20 T = ≈ hr She should take the boat all the way to town. 22. Let x be the length of cable on land, 0 ≤ x ≤ L. Let C be the cost. 2 2 ( )C a L x w bx= − + + 2 2 ( ) ( ) dC a L x b dx L x w − = − + − + 2 2 ( ) 0 ( ) a L x b L x w − − + = − + when 2 2 2 2 2 [( ) ] ( )b L x w a L x− + = − 2 2 2 2 2 ( )( )a b L x b w− − = 2 2 – bw x L a b = − ft on land; 2 2 – aw a b ft under water 2 2 2 2 2 3 2 0 [( ) ] d C aw dx L x w = > − + for all x, so this minimizes the cost. © 2007 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.
  • Instructor’s Resource Manual Section 3.4 179 23. Let the coordinates of the first ship at 7:00 a.m. be (0, 0). Thus, the coordinates of the second ship at 7:00 a.m. are (–60, 0). Let t be the time in hours since 7:00 a.m. The coordinates of the first and second ships at t are (–20t, 0) and ( )60 15 2 , 15 2t t− + − respectively. Let D be the square of the distances at t. ( ) ( ) 2 2 20 60 15 2 0 15 2D t t t= − + − + + ( ) ( )2 1300 600 2 2400 1800 2 3600t t= + − + + ( ) ( )2 1300 600 2 2400 1800 2 dD t dt = + − + ( ) ( )2 1300 600 2 2400 1800 2 0t+ − + = when 12 9 2 1.15 13 6 2 t + = ≈ + hrs or 1 hr, 9 min D is the minimum at 12 9 2 13 6 2 t + = + since 2 2 0 d D dt > for all t. The ships are closest at 8:09 A.M. 24. Write y in terms of x: 2 2b y a x a = − (positive square root since the point is in the first quadrant). Compute the slope of the tangent line: 2 2 bx y a a x ′ = − − . Find the y-intercept, 0y , of the tangent line through the point (x, y): 0 2 20 y y bx x a a x − = − − − 2 2 2 2 0 2 2 2 2 bx bx b y y a x aa a x a a x = + = + − − − 2 2 ab a x = − Find the x-intercept, 0,x of the tangent line through the point (x, y): 2 20 – 0 – – – y bx x x a a x = 2 2 2 2 2 0 – –ay a x a x a x x x bx x x = + = + = Compute the Area A of the resulting triangle and maximize: 3 3 1 2 2 0 0 2 2 1 2 22 a b a b A x y x a x x a x − ⎛ ⎞= = = −⎜ ⎟ ⎝ ⎠− 3 22 2 2 2 2 2 22 dA a b x x a x a x dx a x − ⎛ ⎞⎛ ⎞ ⎜ ⎟= − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ −⎝ ⎠ 3 2 2 2 2 2 3/ 2 (2 – ) 2 ( ) a b x a x a x = − 3 2 2 2 2 2 3/ 2 (2 ) 0 2 ( ) a b x a x a x − = − when 2 2 ; 2 2 2 a b a b x y a a ⎛ ⎞ = = − =⎜ ⎟ ⎝ ⎠ 2 2 2 2 a b b y a a a a ⎛ ⎞ ⎜ ⎟ ⎝ ⎠′ = − = − ⎛ ⎞ −⎜ ⎟ ⎝ ⎠ Note that 0 on 0, 2 dA a dx ⎛ ⎞ < ⎜ ⎟ ⎝ ⎠ and 0 on , , 2 dA a a dx ⎛ ⎞ > ⎜ ⎟ ⎝ ⎠ so A is a minimum at . 2 a x = Then the equation of the tangent line is or 2 0 2 2 b a b y x bx ay ab a ⎛ ⎞ = − − + + − =⎜ ⎟ ⎝ ⎠ . 25. Let x be the radius of the base of the cylinder and h the height. 2 2 2 2 2 2 2 ; ; 2 4 h h V x h r x x r ⎛ ⎞ = π = + = −⎜ ⎟ ⎝ ⎠ 2 3 2 2 4 4 h h V r h hr ⎛ ⎞ π = π − = π −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 2 2 3 2 3 ; 0 when 4 3 dV h r r V h dh π = π − ′ = = ± Since 2 2 3 , 2 d V h dh π = − the volume is maximized when 2 3 3 r h = . ( ) 3 2 3 322 3 3 4 r V r r π⎛ ⎞ = π −⎜ ⎟⎜ ⎟ ⎝ ⎠ 3 3 32 3 2 3 4 3 3 9 9 r r r π π π = − = 26. Let r be the radius of the circle, x the length of the rectangle, and y the width of the rectangle. P = 2x + 2y; 2 2 2 2 2 2 ; ; 2 2 4 4 x y x y r r ⎛ ⎞ ⎛ ⎞ = + = +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 4 ;y r x= − 2 2 2 2 4P x r x= + − 2 2 2 2 ; 4 dP x dx r x = − − © 2007 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.
  • 180 Section 3.4 Instructor’s Resource Manual 2 2 2 2 2 2 0; 2 4 2 ; 4 x r x x r x − = − = − 2 2 2 16 4 4 ; 2r x x x r− = = ± 2 2 2 2 2 3 2 8 0 (4 ) d P r dx r x = − < − when 2 ;x r= 2 2 4 2 2y r r r= − = The rectangle with maximum perimeter is a square with side length 2r . 27. Let x be the radius of the cylinder, r the radius of the sphere, and h the height of the cylinder. A = 2 xhπ ; 2 2 2 2 2 ; 4 4 h h r x x r= + = − 2 4 2 2 2 2 2 4 4 h h A r h h r= π − = π − ( )2 3 42 2 4 2 ; 0 when 0, 2 h r h hdA A h r dh h r π − = ′ = = ± − 0 on (0, 2 ) and 0 on ( 2 , 2 ), dA dA r r r dh dh > < so A is a maximum when 2 .h r= The dimensions are 2 , 2 r h r x= = . 28. Let x be the distance from I1. 1 2 2 2 ( ) kI kI Q x s x = + − 1 2 3 3 2 2 ( ) kI kIdQ dx x s x − = + − 3 1 2 1 3 3 3 2 2 2 0; ; ( ) ( ) kI kI Ix Ix s x s x − + = = − − 3 1 3 3 1 2 s I x I I = + 2 1 2 2 4 4 6 6 0, ( ) kI kId Q dx x s x = + > − so this point minimizes the sum. 29. Let x be the length of a side of the square, so 100 4 3 x− is the side of the triangle, 0 ≤ x ≤ 25 2 1 100 4 3 100 4 2 3 2 3 x x A x − −⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 3 10,000 800 16 4 9 x x x ⎛ ⎞− + = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 200 3 8 3 2 9 9 dA x x dx = − + ( )' 0A x = when 300 3 400 10.874 11 11 x = − ≈ . Critical points: x = 0, 10.874, 25 At x = 0, A ≈ 481; at x = 10.874, A ≈ 272; at x = 25, A = 625. a. For minimum area, the cut should be approximately 4(10.874) ≈ 43.50 cm from one end and the shorter length should be bent to form the square. b. For maximum area, the wire should not be cut; it should be bent to form a square. 30. Let x be the length of the sides of the base, y be the height of the box, and k be the cost per square inch of the material in the sides of the box. 2 ;V x y= The cost is 2 2 1.2 1.5 4C kx kx kxy= + + 2 2 2 4 2.7 4 2.7 V kV kx kx kx xx ⎛ ⎞ = + = +⎜ ⎟ ⎝ ⎠ 3 2 4 5.4 ; 0 when 0.905 dC kV dC kx x V dx dxx = − = ≈ 3 23 1.22 (0.905 ) V y V V ≈ ≈ 31. Let r be the radius of the cylinder and h the height of the cylinder. 32 2 3 3 2 2 2 2 ; 3 3 V r V V r h r h r r r − π = π + π = = − π π Let k be the cost per square foot of the cylindrical wall. The cost is 2 (2 ) 2 (2 )C k rh k r= π + π 2 2 2 2 2 8 2 4 3 3 V V r k r r r k rr ⎛ ⎞⎛ ⎞ π⎛ ⎞ = π − + π = +⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟π⎝ ⎠⎝ ⎠ ⎝ ⎠ 2 2 2 16 2 16 ; 0 3 3 dC V r V r k k dr r r π π⎛ ⎞ ⎛ ⎞ = − + − + =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ when 1/3 3 3 1 3 , 8 2 V V r r ⎛ ⎞ = = ⎜ ⎟ π π⎝ ⎠ ( ) 1/ 3 1 3 2/33 4 1 3 3 3V V V V h π ⎛ ⎞ ⎛ ⎞ = − =⎜ ⎟ ⎜ ⎟ π π⎝ ⎠ ⎝ ⎠π For a given volume V, the height of the cylinder is 1/3 3V⎛ ⎞ ⎜ ⎟ π⎝ ⎠ and the radius is 1/3 1 3 2 V⎛ ⎞ ⎜ ⎟ π⎝ ⎠ . © 2007 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.
  • Instructor’s Resource Manual Section 3.4 181 32. 2cos2 2 3sin 2 ; dx t t dt = − 1 0 when tan2t= ; 3 dx dt = 2 6 t n π = + π for any integer n 12 2 t n π π = + When 12 2 t n π π = + , sin 3 cos 6 6 x n n π π⎛ ⎞ ⎛ ⎞ = + π + + π⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ sin cos cos sin 6 6 n n π π = π + π 3(cos cos sin sin ) 6 6 n n π π + π − π 1 3 ( 1) ( 1) 2. 2 2 n n = − + − = The farthest the weight gets from the origin is 2 units. 33. 2 ; 2 r A θ = 2 2A r θ = The perimeter is 2 2 2 2 2 2 Ar A Q r r r r rr θ= + = + = + 2 2 2 ; 0 when dQ A Q r A dr r = − ′ = = 2 2 2 ( ) A A θ = = 2 2 3 4 0, d Q A dr r = > so this minimizes the perimeter. 34. The distance from the fence to the base of the ladder is tan h θ . The length of the ladder is x. tan cos ; cos ; tan h w h x w x θθ θ θ + = = + sin cos h w x θ θ = + 3 3 2 2 2 2 cos sin sin cos ; 0 sin cos sin cos dx h w w h d θ θ θ θ θ θ θ θ θ − = − + = when 3 tan h w θ = 1 3tan h w θ − = 3 3 3 2/ 3 2/ 3 tan ;sin , h h w h w θ θ= = + 3 2/3 2/3 cos w h w θ = + 2/3 2/3 2/ 3 2/ 3 3 3 h w h w x h w h w ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟= + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2/3 2/3 3/ 2 ( )h w= + 35. x is limited by 0 12x≤ ≤ . 2 3 2 2 (12 ) 24 2 ; 24 6 ; dA A x x x x x dx = − = − = − 2 24 6 0; 2, 2x x− = = − Critical points: 0, 2, 12. When x = 0 or 12, A = 0. When x = 2, 2 12 (2) 8.y = − = The dimensions are 2x = 2(2) = 4 by 8. 36. Let the x-axis lie on the diameter of the semicircle and the y-axis pass through the middle. Then the equation 2 2 y r x= − describes the semicircle. Let (x, y) be the upper-right corner of the rectangle. x is limited by 0 x r≤ ≤ . 2 2 2 2A xy x r x= = − 2 2 2 2 2 2 2 2 2 2 2 2 ( 2 ) dA x r x r x dx r x r x = − − = − − − 2 2 2 2 2 ( 2 ) 0; 2 r r x x r x − = = − Critical points: 0, , 2 r r When x = 0 or r, A = 0. When x = 2 , . 2 r A r= 2 2 2 2 r r y r ⎛ ⎞ = − =⎜ ⎟ ⎝ ⎠ The dimensions are 2 by 2 2 r r . © 2007 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.
  • 182 Section 3.4 Instructor’s Resource Manual 37. If the end of the cylinder has radius r and h is the height of the cylinder, the surface area is 2 2 2A r rh= π + π so – . 2 A h r r = π The volume is 2 2 3 – – 2 2 A Ar V r h r r r r ⎛ ⎞ = π = π = π⎜ ⎟ π⎝ ⎠ . 2 ( ) – 3 ; ( ) 0 2 A V r r V r′ ′= π = when , 6 A r = π ( ) 6 ,V r r′′ = − π so the volume is maximum when . 6 A r = π – 2 2 2 6 A A h r r r = = = π π 38. The ellipse has equation 2 2 2 2 2 2 – – b x b y b a x aa = ± = ± Let 2 2 ( , ) , – b x y x a x a ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ be the upper right- hand corner of the rectangle (use a and b positive). Then the dimensions of the rectangle are 2x by 2 22 – b a x a and the area is 2 24 ( ) – . bx A x a x a = 2 2 2 2 2 2 2 2 2 4 4 4 ( – 2 ) ( ) – – ; – – b bx b a x A x a x a a a x a a x ′ = = ( ) 0A x′ = when , 2 a x = so the corner is at , . 2 2 a b⎛ ⎞ ⎜ ⎟ ⎝ ⎠ The corners of the rectangle are at , , – , , – , – 2 2 2 2 2 2 a b a b a b⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ , , 2 2 a b⎛ ⎞ −⎜ ⎟ ⎝ ⎠ . The dimensions are 2a and 2b . 39. If the rectangle has length l and width w, the diagonal is 2 2 ,d l w= + so 2 2 – .l d w= The area is 2 2 – .A lw w d w= = 2 2 2 2 2 2 2 2 2 – 2 ( ) – – ; – – w d w A w d w d w d w ′ = = ( ) 0 when 2 d A w w′ = = and so 2 2 – . 2 2 d d l d= = ( ) 0 on 0, 2 d A w ⎛ ⎞ ′ > ⎜ ⎟ ⎝ ⎠ and ( ) 0 on , 2 d A w d ⎛ ⎞ ′ < ⎜ ⎟ ⎝ ⎠ . Maximum area is for a square. 40. Note that cos , h t r = so h = r cos t, 2 21 sin –t r h r = , and 2 2 – sinr h r t= Area of submerged region 2 2 2 – –tr h r h= 2 2 – ( cos )( sin ) ( – cos sin )tr r t r t r t t t= = A = area of exposed wetted region 2 2 2 – – ( – cos sin )r h r t t t= π π 2 2 ( – cos – cos sin )r t t t t= π π + 2 2 2 (2 cos sin –1 cos – sin ) dA r t t t t dt = π + 2 2 (2 cos sin – 2sin )r t t t= π 2 2 sin ( cos – sin )r t t t= π Since 0 < t < π , 0 dA dt = only when cos sin or tant t tπ π= = . In terms of r and h, this is 2 21 –r h r r h = π or 2 1 r h = + π . 41. The carrying capacity of the gutter is maximized when the area of the vertical end of the gutter is maximized. The height of the gutter is 3sinθ . The area is 1 3(3sin ) 2 (3cos )(3sin ) 2 A θ θ θ ⎛ ⎞ = + ⎜ ⎟ ⎝ ⎠ 9sin 9cos sinθ θ θ= + . 9cos 9( sin )sin 9cos cos dA d θ θ θ θ θ θ = + − + 2 2 9(cos sin cos )θ θ θ= − + 2 9(2cos cos 1)θ θ= + − 2 2cos cos 1 0;θ θ+ − = 1 cos 1, ; , 2 3 θ θ π = − = π Since 0 , 2 θ π ≤ ≤ the critical points are 0, , and . 3 2 π π When 0θ = , A = 0. When 27 3 , 3 4 Aθ π = = ≈ 11.7. When , 2 θ π = A = 9. The carrying capacity is maximized when . 3 θ π = © 2007 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.
  • Instructor’s Resource Manual Section 3.4 183 42. The circumference of the top of the tank is the circumference of the circular sheet minus the arc length of the sector, 20 10θπ − meters. The radius of the top of the tank is 20 10 5 (2 ) 2 r θ θ π − = = π − π π meters. The slant height of the tank is 10 meters, so the height of the tank is 2 2 25 5 10 10 4h θ θ θ ⎛ ⎞ = − − = π −⎜ ⎟ π π⎝ ⎠ meters. 2 2 21 1 5 5 (2 ) 4 3 3 V r h θ θ θ ⎡ ⎤ ⎡ ⎤ = π = π π − π −⎢ ⎥ ⎢ ⎥π π⎣ ⎦ ⎣ ⎦ 2 2 2 125 (2 ) 4 3 θ θ θ= π − π − π ( ) 2 2 2 1 2 2 125 2(2 )( 1) 4 3 (2 ) (4 2 ) 4 dV d θ θ θ θ θ θ θ θ ⎛= π − − π −⎜ ⎝π ⎞π − π − ⎟+ ⎟π − ⎠ 2 2 2 2 125(2 ) (3 12 4 ) 3 4 θ θ θ θ θ π − = − π + π π π − ; 2 2 2 2 125(2 ) (3 12 4 ) 0 3 4 θ θ θ θ θ π − − π + π = π π − 2 0π θ− = or 2 2 3 12 4 0θ θ− π + π = 2 6 2 6 2 , 2 , 2 3 3 θ θ θ= π = π − π = π + π Since 0 2 ,θ< < π the only critical point is 2 6 2 3 π − π . A graph shows that this maximizes the volume. 43. Let V be the volume. y = 4 – x and z = 5 – 2x. x is limited by 0 2.5x≤ ≤ . 2 3 (4 )(5 2 ) 20 13 2V x x x x x x= − − = − + 2 2 20 26 6 ; 2(3 13 10) 0; dV x x x x dx = − + − + = 2(3 10)( 1) 0;x x− − = 10 1, 3 x = Critical points: 0, 1, 2.5 At x = 0 or 2.5, V = 0. At x = 1, V = 9. Maximum volume when x = 1, y = 4 – 1 = 3, and z = 5 – 2(1) = 3. 44. Let x be the length of the edges of the cube. The surface area of the cube is 2 6x so 1 0 . 6 x≤ ≤ The surface area of the sphere is 2 4 ,rπ so 2 2 2 1– 6 6 4 1, 4 x x r r+ π = = π 3 3 3 2 3/ 24 1 (1– 6 ) 3 6 V x r x x= + π = + π 2 2 23 1– 6 3 – 1– 6 3 – dV x x x x x x dx ⎛ ⎞ ⎜ ⎟= = ⎜ ⎟ππ ⎝ ⎠ 1 0 when 0, 6 dV x dx = = + π 31 (0) 0.094 m . 6 V = ≈ π 3/ 2 –3/ 21 1 6 (6 ) 1– 66 6 V ⎛ ⎞ ⎛ ⎞ = + π +⎜ ⎟ ⎜ ⎟ + π+ π π ⎝ ⎠⎝ ⎠ –3/ 2 31 1 (6 ) 0.055 m 6 6 6 π⎛ ⎞ = + + π = ≈⎜ ⎟ + π⎝ ⎠ For maximum volume: no cube, a sphere of radius 1 0.282 meters. 2 ≈ π For minimum volume: cube with sides of length 1 0.331 meters, 6 ≈ + π sphere of radius 1 0.165 meters 2 6 ≈ + π 45. Consider the figure below. a. 2 2 2 ( ) 2y x a x ax a= − − = − Area of A 1 ( ) 2 A a x y= = − 21 ( ) 2 2 a x ax a= − − ( )1 1 2 22 2 ( ) (2 )1 2 2 2 a x adA ax a dx ax a − = − − + − 2 3 2 2 2 a ax ax a − = − © 2007 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.
  • 184 Section 3.4 Instructor’s Resource Manual 2 3 2 2 2 0 when . 32 a ax a x ax a − = = − 2 2 0 on , and 0 on , , 2 3 3 dA a a dA a a dx dx ⎛ ⎞ ⎛ ⎞ > <⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ so 2 3 a x = maximizes the area of triangle A. b. Triangle A is similar to triangle C, so 2 2 ax ax w y ax a = = − 2 2 1 Area of 2 2 2 ax B B xw ax a = = = − 2 2 22 2 2 2 2 2 a ax a x ax a x dB a dx ax a − ⎛ ⎞− −⎜ ⎟ ⎜ ⎟= ⎜ ⎟− ⎜ ⎟ ⎝ ⎠ 2 2 2 2 2 3/ 2 2 3/ 2 2 (2 ) 3 2 2 2(2 ) (2 ) a x ax a ax a ax xa ax a ax a ⎛ ⎞ ⎛ ⎞− − − = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ 2 2 2 3/ 2 3 2 2 0 when 0, 2 3(2 ) a x xa a x ax a ⎛ ⎞− = =⎜ ⎟ ⎜ ⎟−⎝ ⎠ Since x = 0 is not possible, 2 3 a x = . 2 2 0 on , and 0 on , , 2 3 3 dB a a dB a a dx dx ⎛ ⎞ ⎛ ⎞ < >⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ so 2 3 a x = minimizes the area of triangle B. c. 2 2 2 2 2 2 2 – a x z x w x ax a = + = + 3 2 2 2 – ax ax a = 2 2 2 3 3 2 2 1 2 6 (2 ) 2 (2 ) 2 2 (2 ) dz ax a ax ax a ax a dx ax ax a ⎛ ⎞− − − = ⎜ ⎟ ⎜ ⎟−⎝ ⎠ 2 3 3 2 3 2 3 4 3 2 (2 ) a x a x ax ax a − = − 3 0 when 0, 4 dz a x dx = = → 3 4 a x = 3 3 0 on , and 0 on , , 2 4 4 dz a a dz a a dx dx ⎛ ⎞ ⎛ ⎞ < >⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ so 3 4 a x = minimizes length z. 46. Let 2x be the length of a bar and 2y be the width of a bar. 1 1 cos cos sin cos sin 4 2 2 2 2 22 2 2 a x a a θ θ θ θ θπ ⎛ ⎞⎛ ⎞ ⎛ ⎞ = − = + = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ 1 1 sin cos sin cos sin 4 2 2 2 2 22 2 2 a y a a θ θ θ θ θπ ⎛ ⎞⎛ ⎞ ⎛ ⎞ = − = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ Compute the area A of the cross and maximize. 2 2(2 )(2 ) (2 )A x y y= − 2 8 cos sin cos sin 4 cos sin 2 2 2 2 2 22 2 2 a a aθ θ θ θ θ θ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = + − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 2 2 2 2 2 2 4 cos sin 2 1 2cos sin 4 cos 2 (1 sin ) 2 2 2 2 a a a a θ θ θ θ θ θ ⎛ ⎞ ⎛ ⎞ = − − − = − −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 4 sin 2 cos dA a a d θ θ θ = − + ; 2 2 1 4 sin 2 cos 0 when tan 2 a aθ θ θ− + = = ; 1 2 sin , cos 5 5 θ θ= = 2 2 0 d A dθ < when 1 tan 2 θ = , so this maximizes the area. 2 2 2 2 22 1 10 4 – 2 1– – 2 2 ( 5 –1) 5 5 5 a A a a a a ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ © 2007 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.
  • Instructor’s Resource Manual Section 3.4 185 47. a. 1/ 2 1/ 2 ( ) 15(9 25 30cos ) sin 15(34 30cos ) sinL θ θ θ θ θ− −′ = + − = − 3/ 2 1/ 215 ( ) (34 30cos ) (30sin )sin 15(34 30cos ) cos 2 L θ θ θ θ θ θ− −′′ = − − + − 3/ 2 2 1/ 2 225(34 30cos ) sin 15(34 30cos ) cosθ θ θ θ− − = − − + − 3/ 2 2 15(34 30cos ) [ 15sin (34 30cos )cos ]θ θ θ θ− = − − + − 3/ 2 2 2 15(34 30cos ) [ 15sin 34cos 30cos ]θ θ θ θ− = − − + − 3/ 2 2 15(34 30cos ) [ 15 34cos 15cos ]θ θ θ− = − − + − 3/ 2 2 15(34 30cos ) [15cos 34cos 15]θ θ θ− = − − − + 2 34 (34) 4(15)(15) 5 3 0 when cos , 2(15) 3 5 L θ ± − ′′ = = = 1 3 cos 5 θ − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 1/ 2 1 3 3 4 cos 15 9 25 30 3 5 5 5 L − −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞′ = + − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ 1/ 2 1 3 3 cos 9 25 30 4 5 5 L −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ = + − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ 90φ = ° since the resulting triangle is a 3-4-5 right triangle. b. 1/ 2 1/ 2 ( ) 65(25 169 130cos ) sin 65(194 130cos ) sinL θ θ θ θ θ− −′ = + − = − 3/ 2 1/ 265 ( ) (194 130cos ) (130sin )sin 65(194 130cos ) cos 2 L θ θ θ θ θ θ− −′′ = − − + − 3/ 2 2 1/ 2 4225(194 130cos ) sin 65(194 130cos ) cosθ θ θ θ− − = − − + − 3/ 2 2 65(194 130cos ) [ 65sin (194 130cos )cos ]θ θ θ θ− = − − + − 3/ 2 2 2 65(194 130cos ) [ 65sin 194cos 130cos ]θ θ θ θ− = − − + − 3/ 2 2 65(194 130cos ) [ 65cos 194cos 65]θ θ θ− = − − + − 3/ 2 2 65(194 130cos ) [65cos 194cos 65]θ θ θ− = − − − + 0L′′ = when 2 194 (194) 4(65)(65) 13 5 cos , 2(65) 5 13 θ ± − = = 1 5 cos 13 θ − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 1/ 2 1 5 5 12 cos 65 25 169 130 5 13 13 13 L −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞′ = + − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ 1/ 2 1 5 5 cos 25 169 130 12 13 13 L −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ = + − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ 90φ = ° since the resulting triangle is a 5-12-13 right triangle. c. When the tips are separating most rapidly, 2 2 90 , ,L m h L hφ ′= ° = − = © 2007 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.
  • 186 Section 3.4 Instructor’s Resource Manual d. 2 2 1/ 2 ( ) ( 2 cos ) sinL hm h m hmθ θ θ−′ = + − 2 2 2 2 3/ 2 2 2 2 1/ 2 ( ) ( 2 cos ) sin ( 2 cos ) cosL h m h m hm hm h m hmθ θ θ θ θ− −′′ = − + − + + − 2 2 3/ 2 2 2 2 2 ( 2 cos ) [ sin ( )cos 2 cos ]hm h m hm hm h m hmθ θ θ θ− = + − − + + − 2 2 3/ 2 2 2 2 ( 2 cos ) [ cos ( )cos ]hm h m hm hm h m hmθ θ θ− = + − − + + − 2 2 3/ 2 2 2 2 ( 2 cos ) [ cos ( )cos ]hm h m hm hm h m hmθ θ θ− = − + − − + + 2 2 2 0 when cos ( )cos 0L hm h m hmθ θ′′ = − + + = ( cos )( cos ) 0h m m hθ θ− − = cos , m h h m θ = Since h < m, 1 cos so cos h h m m θ θ − ⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ . 1/ 2 2 2 1 2 2 cos 2 h h m h L hm h m hm m m m − −⎛ ⎞ ⎛ ⎞ −⎛ ⎞ ⎛ ⎞ ′ = + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ 2 2 2 2 1/ 2 ( ) m h hm m h h m − − = − = 1/ 2 1 2 2 2 2 cos 2 h h L h m hm m h m m −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ = + − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ Since 2 2 2 , 90 .h L m φ+ = = ° 48. We are interested in finding the global extrema for the distance of the object from the observer. We will obtain this result by considering the squared distance instead. The squared distance can be expressed as 2 2 21 ( ) ( 2) 100 10 D x x x x ⎛ ⎞ = − + + −⎜ ⎟ ⎝ ⎠ The first and second derivatives are given by ( ) 3 2 2 1 3 '( ) 36 196 and 25 5 3 ''( ) 10 300 25 D x x x x D x x x = − − + = − − Using a computer package, we can solve the equation '( ) 0D x = to find the critical points. The critical points are 5.1538,36.148x ≈ . Using the second derivative we see that ''(5.1538) 38.9972 (max)D ≈ − and ''(36.148) 77.4237 (min)D ≈ Therefore, the position of the object closest to the observer is ( )36.148,5.48≈ while the position of the object farthest from the person is (5.1538,102.5)≈ . (Remember to go back to the original equation for the path of the object once you find the critical points.) 49. Here we are interested in minimizing the distance between the earth and the asteroid. Using the coordinates P and Q for the two bodies, we can use the distance formula to obtain a suitable equation. However, for simplicity, we will minimize the squared distance to find the critical points. The squared distance between the objects is given by 2 2 ( ) (93cos(2 ) 60cos[2 (1.51 1)]) (93sin(2 ) 120sin[2 (1.51 1)]) D t t t t t π π π π = − − + − − The first derivative is [ ][ ] [ ] '( ) 34359 cos(2 ) sin(9.48761 ) cos(9.48761 ) [(204932sin(9.48761 ) 141643sin(2 ))] D t t t t t t π π ≈ − + − Plotting the function and its derivative reveal a periodic relationship due to the orbiting of the objects. Careful examination of the graphs reveals that there is indeed a minimum squared distance (and hence a minimum distance) that occurs only once. The critical value for this occurrence is 13.82790355t ≈ . This value gives a squared distance between the objects of 0.0022743≈ million miles. The actual distance is 0.047851≈ million miles 47,851≈ miles. © 2007 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.
  • Instructor’s Resource Manual Section 3.4 187 50. Let x be the width and y the height of the flyer. 2 inches 1inch 1inch 2 inches We wish to minimize the area of the flyer, A xy= . As it stands, A is expressed in terms of two variables so we need to write one in terms of the other. The printed area of the flyer has an area of 50 square inches. The equation for this area is ( )( )2 4 50x y− − = We can solve this equation for y to obtain 50 4 2 y x = + − Substituting this expression for y in our equation for A, we get A in terms of a single variable, x. 50 50 4 4 2 2 A xy x x x x x = ⎛ ⎞ = + = +⎜ ⎟ − −⎝ ⎠ The allowable values for x are 2 x< < ∞ ; we want to minimize A on the open interval ( )2,∞ . ( ) ( ) ( ) ( ) ( )( ) ( ) 2 2 2 2 2 2 50 50 100 4 4 2 2 4 7 34 16 84 2 2 x xdA dx x x x xx x x x − − − = + = + − − − +− − = = − − The only critical points are obtained by solving 0 dA dx = ; this yields 7x = and 3x = − . We reject 3x = − because it is not in the feasible domain ( )2,∞ . Since 0 dA dx < for x in ( )2,7 and 0 dA dx > for x in ( )7,∞ , we conclude that A attains its minimum value at 7x = . This value of x makes 14y = . So, the dimensions for the flyer that will use the least amount of paper are 7 inches by 14 inches. 51. Consider the following sketch. By similar triangles, 2 2 27 64 64 x t t t = − + + . 2 27 64 t x t t = − + 22 27 2 64 2 2 3/ 2 27 64 1728 1 1 64 ( 64) t t t dx dt t t + + − = − = − + + 2 3/ 2 1728 1 0 when 4 5 ( 64) t t − = = + 2 2 2 2 5/ 2 2 4 5 5184 ; 0 ( 64) t d x t d x dt t dt = − = < + Therefore ( ) ( ) 2 27 4 5 4 5 5 5 11.18 4 5 64 x = − = ≈ + ft is the maximum horizontal overhang. 52. a. b. There are only a few data points, but they do seem fairly linear. c. The data values can be entered into most scientific calculators to utilize the Least Squares Regression feature. Alternately one could use the formulas for the slope and intercept provided in the text. The resulting line should be 0.56852 2.6074y x= + © 2007 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.
  • 188 Section 3.4 Instructor’s Resource Manual d. Using the result from c., the predicted number of surface imperfections on a sheet with area 2.0 square feet is 0.56852 2.6074(2.0) 5.7833 6 since we can't have partial imperfections y = + = ≈ 53. a. [ ] [ ] 2 1 2 1 (5 ) (5 ) n i i i n i i i dS d y bx db db d y bx db = = = − + = − + ∑ ∑ ( ) 1 2 1 2 1 1 1 2( 5 )( ) 2 5 2 10 2 n i i i i n i i i i i n n n i i i i i i i y bx x x y x bx x y x b x = = = = = = − − − ⎡ ⎤ = − + +⎢ ⎥ ⎢ ⎥⎣ ⎦ = − + + ∑ ∑ ∑ ∑ ∑ Setting 0 dS db = gives 2 1 1 1 2 1 1 1 0 2 10 2 0 5 n n n i i i i i i i n n n i i i i i i i x y x b x x y x b x = = = = = = = − + + = − + + ∑ ∑ ∑ ∑ ∑ ∑ 2 1 1 1 1 1 2 1 5 5 n n n i i i i i i i n n i i i i i n i i b x x y x x y x b x = = = = = = = − − = ∑ ∑ ∑ ∑ ∑ ∑ You should check that this is indeed the value of b that minimizes the sum. Taking the second derivative yields 2 2 2 1 2 n i i d S x db = = ∑ which is always positive (unless all the x values are zero). Therefore, the value for b above does minimize the sum as required. b. Using the formula from a., we get that (2037) 5(52) 3.0119 590 b − = ≈ c. The Least Squares Regression line is 5 3.0119y x= + Using this line, the predicted total number of labor hours to produce a lot of 15 brass bookcases is 5 3.0119(15) 50.179 hoursy = + ≈ 54. C(x) = 7000 + 100x 55. 250 – ( ) 100 10 5 p n n = + so ( ) 300 – 2 n p n = 2 ( ) ( ) 300 – 2 n R n np n n= = 56. 2 ( ) ( ) – ( ) 300 – – (7000 100 ) 2 P n R n C n n n n = = + 2 7000 200 – 2 n n= − + 57. Estimate 200n ≈ ( ) 200 – ; 200 – 0P n n n′ = = when n = 200. ( ) –1,P n′′ = so profit is maximum at n = 200. 58. ( ) 100 3.002 – 0.0001 C x x x x = + When x = 1600, ( ) 2.9045 C x x = or $2.90 per unit. 3.002 0.0002 dC x dx = − (1600) 2.682C′ = or $2.68 59. ( ) 1000 1200 C n n n n = + When n = 800, ( ) 1.9167 C n n ≈ or $1.92 per unit. 600 dC n dn = (800) 1.333C′ ≈ or $1.33 60. a. 2 33 18 3 dC x x dx = − + 2 2 2 2 18 6 ; 0 when 3 d C d C x x dx dx = − + = = 2 2 0 on (0, 3), d C dx < 2 2 0 on (3, ) d C dx > ∞ Thus, the marginal cost is a minimum when x = 3 or 300 units. b. 2 33 18(3) 3(3) 6− + = © 2007 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.
  • Instructor’s Resource Manual Section 3.4 189 61. a. 3 2 ( ) ( ) 20 4 3 x R x xp x x x= = + − ( )( )2 20 8 10 2 dR x x x x dx = + − = − + b. Increasing when 0 dR dx > 2 20 8 0 on [0,10)x x+ − > Total revenue is increasing if 0 ≤ x ≤ 10. c. 2 2 2 2 8 – 2 ; 0 when 4 d R d R x x dx dx = = = 3 3 2; d R dR dxdx = − is maximum at x = 4. 62. 1/ 2 ( ) 182 36 x R x x ⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ 1/ 2 1/ 2 1 1 182 182 2 36 36 36 dR x x x dx − ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = − − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1 2 182 182 36 24 x x − ⎛ ⎞ ⎛ ⎞ = − −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 0 when 4368 dR x dx = = 1 4368; (4368) 34,021.83x R= ≈ At 1, 0 dR x dx = . 63. 800 ( ) 3 3 x R x x x = − + 2 2 ( 3)(800) 800 2400 3 3; ( 3) ( 3) dR x x dx x x + − = − = − + + 0 when 20 2 3 25 dR x dx = = − ≈ 1 25; (25) 639.29x R= ≈ At 1, 0 dR x dx = . 64. ( 400) ( ) 12 (0.20) 20 0.02 10 x p x x − = − = − 2 ( ) 20 0.02R x x x= − 20 0.04 ; 0 when 500 dR dR x x dx dx = − = = Total revenue is maximized at 1 500x = . 65. The revenue function would be ( ) ( ) 2 200 0.15R x x p x x x= ⋅ = − . This, together with the cost function yields the following profit function: ( ) 2 2 5000 194 0.148 if 0 500 9000 194 0.148 if 500 750 x x x P x x x x ⎧− + − ≤ ≤⎪ = ⎨ − + − < ≤⎪⎩ a. The only difference in the two pieces of the profit function is the constant. Since the derivative of a constant is 0, we can say that on the interval 0 750x< < , 194 0.296 dP x dx = − There are no singular points in the given interval. To find stationary points, we solve 0 194 0.296 0 0.296 194 655 dP dx x x x = − = − = − ≈ Thus, the critical points are 0, 500, 655, and 750. ( )0 5000P = − ; ( )500 55,000P = ; ( )655 54,574.30P = ; ( )750 53,250P = The profit is maximized if the company produces 500 chairs. The current machine can handle this work, so they should not buy the new machine. b. Without the new machine, a production level of 500 chairs would yield a maximum profit of $55,000. 66. The revenue function would be ( ) ( ) 2 200 0.15R x x p x x x= ⋅ = − . This, together with the cost function yields the following profit function: ( ) 2 2 5000 194 0.148 if 0 500 8000 194 0.148 if 500 750 x x x P x x x x ⎧− + − ≤ ≤⎪ = ⎨ − + − < ≤⎪⎩ a. The only difference in the two pieces of the profit function is the constant. Since the derivative of a constant is 0, we can say that on the interval 0 750x< < , 194 0.296 dP x dx = − There are no singular points in the given interval. To find stationary points, we solve 0 194 0.296 0 0.296 194 655 dP dx x x x = − = − = − ≈ © 2007 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.
  • 190 Section 3.4 Instructor’s Resource Manual Thus, the critical points are 0, 500, 655, and 750. ( )0 5000P = − ; ( )500 55,000P = ; ( )655 55,574.30P = ; ( )750 54,250P = The profit is maximized if the company produces 655 chairs. The current machine cannot handle this work, so they should buy the new machine. b. With the new machine, a production level of 655 chairs would yield a maximum profit of $55,574.30. 67. 2 ( ) 10 0.001 ; 0 300R x x x x= − ≤ ≤ 2 2 ( ) (10 – 0.001 ) – (200 4 – 0.01 )P x x x x x= + 2 –200 6 0.009x x= + + 6 0.018 ; 0 when 333 dP dP x x dx dx = + = ≈ − Critical numbers: x = 0, 300; P(0) = –200; P(300) = 2410; Maximum profit is $2410 at x = 300. 68. 2 2 200 4 0.01 if 0 300 ( ) 800 3 0.01 if 300 450 x x x C x x x x ⎧ + − ≤ ≤⎪ = ⎨ + − < ≤⎪⎩ 2 2 200 6 0.009 if 0 300 ( ) 800 7 0.009 if 300 450 x x x P x x x x ⎧− + + ≤ ≤⎪ = ⎨ − + + < ≤⎪⎩ There are no stationary points on the interval [0, 300]. On [300, 450]: 7 0.018 ; 0 when 389 dP dP x x dx dx = + = ≈ − The critical numbers are 0, 300, 450. P(0) = –200, P(300) = 2410, P(450) = 4172.5 Monthly profit is maximized at x = 450, P(450) = 4172.50 69. a. 2 2 2 2 2 4 a b a ab b ab + + +⎛ ⎞ ≤ =⎜ ⎟ ⎝ ⎠ 2 2 1 4 2 4 a b ab= + + This is true if 2 22 2 1 – 0 – – 4 2 4 2 2 2 a b a b a b ab ⎛ ⎞ ⎛ ⎞ ≤ + = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Since a square can never be negative, this is always true. b. 2 2 2 ( ) 4 a ab b F b b + + = As 2 2 2 0 , 2b a ab b a+ → + + → while 4 0 ,b + → thus 0 lim ( ) b F b +→ = ∞ which is not close to a. 2 2 2 22 lim lim 4 4 a b b b a ba ab b b→∞ →∞ + ++ + = = ∞ , so when b is very large, F(b) is not close to a. 2 2 2( )(4 ) – 4( ) ( ) 16 a b b a b F b b + + ′ = 2 2 2 2 2 2 4 – 4 – ; 16 4 b a b a b b = = 2 2 ( ) 0 whenF b b a′ = = or b = a since a and b are both positive. 2 2 ( ) 4 ( ) 4 4 a a a F a a a a + = = = Thus 2 ( ) 4 a b a b + ≤ for all b > 0 or 2 ( ) 4 a b ab + ≤ which leads to . 2 a b ab + ≤ c. Let 3 3 1 ( ) ( ) 3 27 a b c a b c F b b b + + + +⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ 2 3 2 2 3( ) (27 ) – 27( ) ( ) 27 a b c b a b c F b b + + + + ′ = 2 2 ( ) [3 – ( )] 27 a b c b a b c b + + + + = 2 2 ( ) (2 – – ) ; 27 a b c b a c b + + = ( ) 0F b′ = when . 2 a c b + = 3 2 2 3 6 a c a c a c F a c + + +⎛ ⎞ ⎛ ⎞ = ⋅ +⎜ ⎟ ⎜ ⎟ +⎝ ⎠ ⎝ ⎠ 3 3 2 2 3( ) 2 6 2 2 a c a c a c a c a c + + +⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Thus 2 3 1 2 3 a c a b c b + + +⎛ ⎞ ⎛ ⎞ ≤⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ for all b > 0. From (b), 2 2 a c ac +⎛ ⎞ ≤ ⎜ ⎟ ⎝ ⎠ , thus 3 1 3 a b c ac b + +⎛ ⎞ ≤ ⎜ ⎟ ⎝ ⎠ or 3 3 a b c abc + +⎛ ⎞ ≤ ⎜ ⎟ ⎝ ⎠ which gives the desired result 1/3 ( ) 3 a b c abc + + ≤ . © 2007 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.
  • Instructor’s Resource Manual Section 3.5 191 70. Let a = lw, b = lh, and c = hw, then S = 2(a + b + c) while 2 .V abc= By problem 69(c), 1/3 ( ) 3 a b c abc + + ≤ so 2 1/3 2( ) ( ) . 2 3 6 a b c S V + + ≤ = ⋅ In problem 1c, the minimum occurs, hence equality holds, when . 2 a c b + = In the result used from Problem 69(b), equality holds when c = a, thus , 2 a a b a + = = so a = b = c. For the boxes, this means l = w = h, so the box is a cube. 3.5 Concepts Review 1. f(x); –f(x) 2. decreasing; concave up 3. x = –1, x = 2, x = 3; y = 1 4. polynomial; rational. Problem Set 3.5 1. Domain: ( , )−∞ ∞ ; range: ( , )−∞ ∞ Neither an even nor an odd function. y-intercept: 5; x-intercept: ≈ –2.3 2 2 ( ) 3 – 3;3 – 3 0f x x x′ = = when x = –1, 1 Critical points: –1, 1 ( ) 0f x′ > when x < –1 or x > 1 f(x) is increasing on (– ∞ , –1] ∪ [1, ∞ ) and decreasing on [–1, 1]. Local minimum f(1) = 3; local maximum f(–1) = 7 ( ) 6 ; ( ) 0f x x f x′′ ′′= > when x > 0. f(x) is concave up on (0, ∞ ) and concave down on (– ∞ , 0); inflection point (0, 5). 2. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ ) Neither an even nor an odd function. y-intercept: –10; x-intercept: 2 2 2 2 ( ) 6 – 3 3(2 –1); 2 –1 0f x x x x′ = = = when 1 1 – , 2 2 x = Critical points: 1 1 – , 2 2 ( ) 0f x′ > when 1 – 2 x < or 1 2 x > f(x) is increasing on 1 1 – , – , 2 2 ⎛ ⎤ ⎡ ⎞ ∞ ∪ ∞⎜ ⎟⎥ ⎢ ⎝ ⎦ ⎣ ⎠ and decreasing on 1 1 – , 2 2 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ . Local minimum 1 – 2 –10 –11.4 2 f ⎛ ⎞ = ≈⎜ ⎟ ⎝ ⎠ Local maximum 1 – 2 –10 –8.6 2 f ⎛ ⎞ = ≈⎜ ⎟ ⎝ ⎠ ( ) 12 ; ( ) 0f x x f x′′ ′′= > when x > 0. f(x) is concave up on (0, ∞ ) and concave down on (– ∞ , 0); inflection point (0, –10). 3. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ ) Neither an even nor an odd function. y-intercept: 3; x-intercepts: ≈ –2.0, 0.2, 3.2 2 ( ) 6 – 6 –12 6( – 2)( 1);f x x x x x′ = = + ( ) 0f x′ = when x = –1, 2 Critical points: –1, 2 ( ) 0f x′ > when x < –1 or x > 2 f(x) is increasing on (– ∞ , –1] ∪ [2, ∞ ) and decreasing on [–1, 2]. Local minimum f(2) = –17; local maximum f(–1) = 10 ( ) 12 6 6(2 1);f x x x′′ = − = − ( ) 0f x′′ > when 1 . 2 x > f(x) is concave up on 1 , 2 ⎛ ⎞ ∞⎜ ⎟ ⎝ ⎠ and concave down on 1 – , ; 2 ⎛ ⎞ ∞⎜ ⎟ ⎝ ⎠ inflection point: 1 7 , – 2 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ © 2007 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.
  • 192 Section 3.5 Instructor’s Resource Manual 4. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ ) Neither an even nor an odd function y-intercept: –1; x-intercept: 1 2 ( ) 3( –1) ; ( ) 0f x x f x′ ′= = when x = 1 Critical point: 1 ( ) 0f x′ > for all x ≠ 1 f(x) is increasing on (– ∞ , ∞ ) No local minima or maxima ( ) 6( –1); ( ) 0f x x f x′′ ′′= > when x > 1. f(x) is concave up on (1, ∞ ) and concave down on (– ∞ , 1); inflection point (1, 0) 5. Domain: (– ∞ , ∞ ); range: [0, ∞ ) Neither an even nor an odd function. y-intercept: 1; x-intercept: 1 3 ( ) 4( –1) ; ( ) 0G x x G x′ ′= = when x = 1 Critical point: 1 ( ) 0G x′ > for x > 1 G(x) is increasing on [1, ∞ ) and decreasing on (– ∞ , 1]. Global minimum f(1) = 0; no local maxima 2 ( ) 12( –1) ; ( ) 0G x x G x′′ ′′= > for all x ≠ 1 G(x) is concave up on (– ∞ , 1) ∪ (1, ∞ ); no inflection points 6. Domain: (– ∞ , ∞ ); range: 1 – , 4 ⎡ ⎞ ∞⎟⎢ ⎣ ⎠ 2 2 2 2 (– ) (– ) [(– ) –1] ( –1) ( );H t t t t t H t= = = even function; symmetric with respect to the y-axis. y-intercept: 0; t-intercepts: –1, 0, 1 3 2 ( ) 4 – 2 2 (2 –1); ( ) 0H t t t t t H t′ ′= = = when 1 1 – , 0, 2 2 t = Critical points: 1 1 – , 0, 2 2 ( ) 0H t′ > for 1 – 0 2 t< < or 1 . 2 t< H(t) is increasing on 1 1 – , 0 , 2 2 ⎡ ⎤ ⎡ ⎞ ∪ ∞⎟⎢ ⎥ ⎢ ⎣ ⎦ ⎣ ⎠ and decreasing on 1 1 – , – 0, 2 2 ⎛ ⎤ ⎡ ⎤ ∞ ∪⎜ ⎥ ⎢ ⎥ ⎝ ⎦ ⎣ ⎦ Global minima 1 1 1 1 – – , – ; 4 42 2 f f ⎛ ⎞ ⎛ ⎞ = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Local maximum f(0) = 0 2 2 ( ) 12 – 2 2(6 –1); 0H t t t H′′ ′′= = > when 1 1 – or 6 6 t t< > H(t) is concave up on 1 1 – , – , 6 6 ⎛ ⎞ ⎛ ⎞ ∞ ∪ ∞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ and concave down on 1 1 – , ; 6 6 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ inflection points 1 5 1 5 – , and , 36 366 6 H H ⎛ ⎞ ⎛ ⎞ −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 7. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ ) Neither an even nor an odd function. y-intercept: 10; x-intercept: 1/3 1–11 –1.2≈ 2 2 ( ) 3 – 6 3 3( –1) ; ( ) 0f x x x x f x′ ′= + = = when x = 1. Critical point: 1 ( ) 0f x′ > for all x ≠ 1. f(x) is increasing on (– ∞ , ∞ ) and decreasing nowhere. No local maxima or minima ( ) 6 – 6 6( –1); ( ) 0f x x x f x′′ ′′= = > when x > 1. f(x) is concave up on (1, ∞ ) and concave down on (– ∞ , 1); inflection point (1, 11) © 2007 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.
  • Instructor’s Resource Manual Section 3.5 193 8. Domain: (– ∞ , ∞ ); range: 16 – , 3 ⎡ ⎞ ∞⎟⎢ ⎣ ⎠ 4 2 4 2 4(– ) – 8(– ) –12 4 – 8 –12 (– ) 3 3 s s s s F s = = ( );F s= even function; symmetric with respect to the y-axis y-intercept: –4; s-intercepts: – 3, 3 3 216 16 16 ( ) – ( –1); ( ) 0 3 3 3 F s s s s s F s′ ′= = = when s = –1, 0, 1. Critical points: –1, 0, 1 ( ) 0F s′ > when –1 < x < 0 or x > 1. F(s) is increasing on [–1, 0] ∪ [1, ∞ ) and decreasing on (– ∞ , –1] ∪ [0, 1] Global minima 16 16 ( 1) , (1) ; 3 3 F F− = − = − local maximum F(0) = –4 2 216 1 ( ) 16 16 ; ( ) 0 3 3 F s s s F s ⎛ ⎞′′ ′′= − = − >⎜ ⎟ ⎝ ⎠ when 1 1 – or 3 3 s s< > F(s) is concave up on 1 1 , , 3 3 ⎛ ⎞ ⎛ ⎞ −∞ − ∪ ∞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ and concave down on 1 1 – , ; 3 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ inflection points 1 128 1 128 – , , , 27 273 3 F F ⎛ ⎞ ⎛ ⎞ − −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 9. Domain: (– ∞ , –1) ∪ (–1, ∞ ); range: (– ∞ , 1) ∪ (1, ∞ ) Neither an even nor an odd function y-intercept: 0; x-intercept: 0 2 1 ( ) ; ( ) ( 1) g x g x x ′ ′= + is never 0. No critical points ( ) 0g x′ > for all x ≠ –1. g(x) is increasing on (– ∞ , –1) ∪ (–1, ∞ ). No local minima or maxima 3 2 ( ) – ; ( ) 0 ( 1) g x g x x ′′ ′′= > + when x < –1. g(x) is concave up on (– ∞ , –1) and concave down on (–1, ∞ ); no inflection points (–1 is not in the domain of g). 1 1 lim lim 1; 1 1x x x x x→∞ →∞ = = + + – lim 1x x x→ ∞ + 1– 1 lim 1; 1x x → ∞ = = + horizontal asymptote: y = 1 As – – –1 , 1 0x x→ + → so ––1 lim ; 1x x x→ = ∞ + as –1 , 1 0x x+ + → + → so –1 lim – ; 1x x x+→ = ∞ + vertical asymptote: x = –1 10. Domain: (– ∞ , 0) ∪ (0, ∞ ); range: (– ∞ , –4π ] ∪ [0, ∞ ) Neither an even nor an odd function No y-intercept; s-intercept: π 2 2 2 – ( ) ; ( ) 0 s g s g s s π ′ ′= = when s = –π , π Critical points: ,π π− ( ) 0g s′ > when s < –π or s > π g(s) is increasing on ( , ]π−∞ − ∪ [ , )π ∞ and decreasing on [–π , 0) ∪ (0, π ]. Local minimum g(π ) = 0; local maximum g(–π ) = –4π 2 3 2 ( ) ; ( ) 0g s g s s π ′′ ′′= > when s > 0 g(s) is concave up on (0, ∞ ) and concave down on (– ∞ , 0); no inflection points (0 is not in the domain of g(s)). 2 ( ) – 2 ; – 2g s s y s s π = π + = π is an oblique asymptote. As – 2 2 0 , ( – ) ,s s→ π → π so –0 lim ( ) – ; s g s → = ∞ © 2007 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.
  • 194 Section 3.5 Instructor’s Resource Manual as 2 2 0 , ( – ) ,s s+ → π → π so 0 lim ( ) ; s g s +→ = ∞ s = 0 is a vertical asymptote. 11. Domain: (– ∞ , ∞ ); range: 1 1 – , 4 4 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ 2 2 – (– ) – – ( ); (– ) 4 4 x x f x f x x x = = = + + odd function; symmetric with respect to the origin. y-intercept: 0; x-intercept: 0 2 2 2 4 – ( ) ; ( ) 0 ( 4) x f x f x x ′ ′= = + when x = –2, 2 Critical points: –2, 2 ( ) 0f x′ > for –2 < x < 2 f(x) is increasing on [–2, 2] and decreasing on (– ∞ , –2] ∪ [2, ∞ ). Global minimum 1 (–2) – ; 4 f = global maximum 1 (2) 4 f = 2 2 3 2 ( –12) ( ) ; ( ) 0 ( 4) x x f x f x x ′′ ′′= > + when –2 3 0x< < or 2 3x > f(x) is concave up on (–2 3, 0) (2 3, )∪ ∞ and concave down on (– , – 2 3) (0, 2 3);∞ ∪ inflection points 3 2 3, , (0, 0) 8 ⎛ ⎞ − −⎜ ⎟⎜ ⎟ ⎝ ⎠ , 3 2 3, 8 ⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠ 1 2 4 2 lim lim 0; 14 x x x x x x→∞ →∞ = = ++ 1 2 4– – 2 lim lim 0; 14 x x x x x x→ ∞ → ∞ = = ++ y = 0 is a horizontal asymptote. No vertical asymptotes 12. Domain: (– ∞ , ∞ ); range: [0, 1) 2 2 2 2 (– ) (– ) ( ); (– ) 1 1 θ θ θ θ θ θ Λ = = = Λ + + even function; symmetric with respect to the y-axis. y-intercept: 0; θ -intercept: 0 2 2 2 ( ) ; ( ) 0 ( 1) θ θ θ θ ′ ′Λ = Λ = + when θ = 0 Critical point: 0 ( ) 0θ′Λ > when θ > 0 Λ(θ) is increasing on [0, ∞ ) and decreasing on (– ∞ , 0]. Global minimum Λ(0) = 0; no local maxima 2 2 3 2(1– 3 ) ( ) ; ( ) 0 ( 1) θ θ θ θ ′′ ′′Λ = Λ > + when 1 1 – 3 3 θ< < Λ(θ) is concave up on 1 1 – , 3 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ and concave down on 1 1 – , – , ; 3 3 ⎛ ⎞ ⎛ ⎞ ∞ ∪ ∞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ inflection points 1 1 1 1 – , , , 4 43 3 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 1 2 1 lim lim 1; 11θ θ θ θ θ→∞ →∞ = = ++ 2 2 1– – 2 1 lim lim 1; 11θ θ θ θ θ→ ∞ → ∞ = = ++ y = 1 is a horizontal asymptote. No vertical asymptotes © 2007 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.
  • Instructor’s Resource Manual Section 3.5 195 13. Domain: (– ∞ , 1) ∪ (1, ∞ ); range (– ∞ , 1) ∪ (1, ∞ ) Neither an even nor an odd function y-intercept: 0; x-intercept: 0 2 1 ( ) ; ( ) ( 1) h x h x x ′= − − is never 0. No critical points ( ) 0h x′ < for all x ≠ 1. h(x) is increasing nowhere and decreasing on (– ∞ , 1) ∪ (1, ∞ ). No local maxima or minima 3 2 ( ) ; ( ) 0 ( –1) h x h x x ′′ ′′= > when x > 1 ( )h x is concave up on (1, ∞ ) and concave down on (– ∞ , 1); no inflection points (1 is not in the domain of ( )h x ) 1 1 lim lim 1; –1 1–x x x x x→∞ →∞ = = 1 1 lim lim 1; 1 1x x x x x→−∞ →−∞ = = − − y = 1 is a horizontal asymptote. As – – 1 , –1 0x x→ → so –1 lim – ; –1x x x→ = ∞ as 1 , –1 0x x+ + → → so 1 lim ; –1x x x+→ = ∞ x = 1 is a vertical asymptote. 14. Domain: ( ),−∞ ∞ Range: ( ]0,1 Even function since 2 2 1 1 ( ) ( ) ( ) 1 1 P x P x x x − = = = − + + so the function is symmetric with respect to the y-axis. y-intercept: 1y = x-intercept: none 2 2 2 '( ) ( 1) x P x x − = + ; '( )P x is 0 when 0x = . critical point: 0x = '( ) 0P x > when 0x < so ( )P x is increasing on ( ,0]−∞ and decreasing on [0, )∞ . Global maximum (0) 1P = ; no local minima. 2 2 3 6 2 ''( ) ( 1) x P x x − = + ''( ) 0P x > on ( , 1/ 3) (1/ 3, )−∞ − ∪ ∞ (concave up) and ''( ) 0 on ( 1/ 3,1/ 3)P x < − (concave down). Inflection points: 1 3 , 43 ⎛ ⎞ ±⎜ ⎟ ⎝ ⎠ No vertical asymptotes. lim ( ) 0; lim ( ) 0 x x P x P x →∞ →−∞ = = 0y = is a horizontal asymptote. 15. Domain: (– ∞ , –1) ∪ (–1, 2) ∪ (2, ∞ ); range: (– ∞ , ∞ ) Neither an even nor an odd function y-intercept: 3 – ; 2 x-intercepts: 1, 3 2 2 2 3 –10 11 ( ) ; ( ) ( 1) ( – 2) x x f x f x x x + ′ ′= + is never 0. No critical points ( ) 0f x′ > for all x ≠ –1, 2 f(x) is increasing on (– ∞ , –1) ∪ (–1, 2) ∪ ( )2,∞ . No local minima or maxima 3 2 3 3 –6 30 – 66 42 ( ) ; ( ) 0 ( 1) ( – 2) x x x f x f x x x + + ′′ ′′= > + when x < –1 or 1 < x < 2 f(x) is concave up on (– ∞ , –1) ∪ (1, 2) and concave down on (–1, 1) ∪ (2, ∞ ); inflection point f(1) = 0 2 2 ( –1)( – 3) – 4 3 lim lim ( 1)( – 2) – – 2x x x x x x x x x x→∞ →∞ + = + 34 2 1 2 2 1– lim 1; 1– – x x x x x →∞ + = = 34 2 1 2– – 2 1– ( –1)( – 3) lim lim 1; ( 1)( – 2) 1– – x x x x x x x x x x→ ∞ → ∞ + = = + y =1 is a horizontal asymptote. As – –1 , –1 –2, – 3 –4,x x x→ → → – – 2 –3, and 1 0x x→ + → so ––1 lim ( ) ; x f x → = ∞ © 2007 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.
  • 196 Section 3.5 Instructor’s Resource Manual as –1 , –1 –2, – 3 –4,x x x+ → → → – 2 –3,x → and 1 0 ,x + + → so –1 lim ( ) – x f x +→ = ∞ As – 2 , –1 1, – 3 –1, 1 3,x x x x→ → → + → and – – 2 0 ,x → so –2 lim ( ) ; x f x → = ∞ as 2 , –1 1, – 3 –1, 1 3,x x x x+ → → → + → and 2 – 2 0 , so lim ( ) – x x f x+ +→ → = ∞ x = –1 and x = 2 are vertical asymptotes. 16. Domain: ( ) ( ),0 0,−∞ ∪ ∞ Range: ( , 2] [2, )−∞ − ∪ ∞ Odd function since 2 2 ( ) 1 1 ( ) ( ) z z w z w z z z − + + − = = − = − − ; symmetric with respect to the origin. y-intercept: none x-intercept: none 2 1 '( ) 1w z z = − ; '( ) 0w z = when 1z = ± . critical points: 1z = ± . '( ) 0w z > on ( , 1) (1, )−∞ − ∪ ∞ so the function is increasing on ( , 1] [1, )−∞ − ∪ ∞ . The function is decreasing on [ 1,0) (0,1)− ∪ . local minimum (1) 2w = and local maximum ( 1) 2w − = − . No global extrema. 3 2 ''( ) 0w z z = > when 0z > . Concave up on (0, )∞ and concave down on ( ),0−∞ . No horizontal asymptote; 0x = is a vertical asymptote; the line y z= is an oblique (or slant) asymptote. No inflection points. 17. Domain: ( ) ( ),1 1,−∞ ∪ ∞ Range: ( ),−∞ ∞ Neither even nor odd function. y-intercept: 6y = ; x-intercept: 3,2x = − 2 2 2 5 '( ) ( 1) x x g x x − + = − ; '( )g x is never zero. No critical points. '( ) 0g x > over the entire domain so the function is always increasing. No local extrema. 3 8 ''( ) ( 1) f x x − = − ; ''( ) 0f x > when 1x < (concave up) and ''( ) 0f x < when 1x > (concave down); no inflection points. No horizontal asymptote; 1x = is a vertical asymptote; the line 2y x= + is an oblique (or slant) asymptote. 18. Domain: (– ∞ , ∞ ); range [0, ∞ ) 3 3 (– ) – ( );f x x x f x= = = even function; symmetric with respect to the y-axis. y-intercept: 0; x-intercept: 0 2 ( ) 3 3 ; ( ) 0 x f x x x x f x x ⎛ ⎞ ′ ′= = =⎜ ⎟⎜ ⎟ ⎝ ⎠ when x = 0 Critical point: 0 ( ) 0f x′ > when x > 0 f(x) is increasing on [0, ∞ ) and decreasing on (– ∞ , 0]. Global minimum f(0) = 0; no local maxima 2 3 ( ) 3 6 x f x x x x ′′ = + = as 22 ;x x= ( ) 0f x′′ > when x ≠ 0 f(x) is concave up on (– ∞ , 0) ∪ (0, ∞ ); no inflection points © 2007 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.
  • Instructor’s Resource Manual Section 3.5 197 19. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ ) (– ) – – – – ( );R z z z z z R z= = = odd function; symmetric with respect to the origin. y-intercept: 0; z-intercept: 0 2 ( ) 2 z R z z z z ′ = + = since 22 z z= for all z; ( ) 0R z′ = when z = 0 Critical point: 0 ( ) 0R z′ > when z ≠ 0 R(z) is increasing on (– ∞ , ∞ ) and decreasing nowhere. No local minima or maxima 2 ( ) ; ( ) 0 z R z R z z ′′ ′′= > when z > 0. R(z) is concave up on (0, ∞ ) and concave down on (– ∞ , 0); inflection point (0, 0). 20. Domain: (– ∞ , ∞ ); range: [0, ∞ ) 2 2 (– ) (– ) – ( );H q q q q q H q= = = even function; symmetric with respect to the y-axis. y-intercept: 0; q-intercept: 0 3 3 3 ( ) 2 3 q q H q q q q q q q ′ = + = = as 2 2 q q= for all q; ( ) 0H q′ = when q = 0 Critical point: 0 ( ) 0H q′ > when q > 0 H(q) is increasing on [0, ∞ ) and decreasing on (– ∞ , 0]. Global minimum H(0) = 0; no local maxima 2 3 ( ) 3 6 ; ( ) 0 q H q q q H q q ′′ ′′= + = > when q ≠ 0. H(q) is concave up on (– ∞ , 0) ∪ (0, ∞ ); no inflection points. 21. Domain: (– ∞ , ∞ ); range: [0, ∞ ) Neither an even nor an odd function. Note that for x ≤ 0, –x x= so 0,x x+ = while for x > 0, x x= so . 2 x x x + = 2 0 if 0 ( ) 3 2 if 0 x g x x x x ≤⎧⎪ = ⎨ + >⎪⎩ y-intercept: 0; x-intercepts: ( , 0]− ∞ 0 if 0 ( ) 6 2 if 0 x g x x x ≤⎧ ′ = ⎨ + >⎩ No critical points for x > 0. g(x) is increasing on [0, ∞ ) and decreasing nowhere. 0 if 0 ( ) 6 if 0 x g x x ≤⎧ ′′ = ⎨ >⎩ g(x) is concave up on (0, ∞ ); no inflection points 22. Domain: (– ∞ , ∞ ); range: [0, ∞ ) Neither an even nor an odd function. Note that for x < 0, –x x= so – – 2 x x x= , while for x ≥ 0, x x= so 0. 2 x x− = 3 2 6 if 0 ( ) 0 if 0 x x x x h x x ⎧− + − <⎪ = ⎨ ≥⎪⎩ y-intercept: 0; x-intercepts: [0, ∞ ) 2 3 2 6 if 0 ( ) 0 if 0 x x x h x x ⎧− + − <⎪ ′ = ⎨ ≥⎪⎩ No critical points for x < 0 h(x) is increasing nowhere and decreasing on (– ∞ , 0]. 6 2 if 0 ( ) 0 if 0 x x h x x − + <⎧ ′′ = ⎨ ≥⎩ h(x) is concave up on (– ∞ , 0); no inflection © 2007 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.
  • 198 Section 3.5 Instructor’s Resource Manual points 23. Domain: (– ∞ , ∞ ); range: [0, 1] ( ) sin( ) sin sin ( );f x x x x f x− = − = − = = even function; symmetric with respect to the y-axis. y-intercept: 0; x-intercepts: kπ where k is any integer. sin ( ) cos ; ( ) 0 sin x f x x f x x ′ ′= = when 2 x k π = + π and ( )f x′ does not exist when x = kπ , where k is any integer. Critical points: 2 kπ and 2 k π π + , where k is any integer; ( ) 0f x′ > when sin x and cos x are either both positive or both negative. f(x) is increasing on , 2 k k π⎡ ⎤ π π +⎢ ⎥ ⎣ ⎦ and decreasing on , ( 1) 2 k k π⎡ ⎤ π + + π⎢ ⎥ ⎣ ⎦ where k is any integer. Global minima f(kπ ) = 0; global maxima 1, 2 f k π⎛ ⎞ π + =⎜ ⎟ ⎝ ⎠ where k is any integer. 2 2 2 cos sin ( ) sin sin 1 sin sin cos (cos ) sinsin x x f x x x x x x x xx ′′ = − ⎛ ⎞⎛ ⎞ ⎜ ⎟+ − ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ 2 2 2 2 cos sin cos sin sin sin sin sin sin x x x x x x x x x = − − = − = − ( ) 0f x′′ < when x ≠ kπ , k any integer f(x) is never concave up and concave down on (kπ , (k + 1)π ) where k is any integer. No inflection points 24. Domain: [2kπ , (2k + 1)π ] where k is any integer; range: [0, 1] Neither an even nor an odd function y-intercept: 0; x-intercepts: kπ , where k is any integer. cos ( ) ; ( ) 0 2 sin x f x f x x ′ ′= = when 2 2 x k π = π + while ( )f x′ does not exist when x = kπ , k any integer. Critical points: , 2 2 k k π π π + where k is any integer ( ) 0f x′ > when 2 2 2 k x k π π < < π + f(x) is increasing on 2 , 2 2 k k π⎡ ⎤ π π +⎢ ⎥ ⎣ ⎦ and decreasing on 2 , (2 1) , 2 k k π⎡ ⎤ π + + π⎢ ⎥ ⎣ ⎦ k any integer. Global minima f(kπ ) = 0; global maxima 2 1, 2 f k π⎛ ⎞ π + =⎜ ⎟ ⎝ ⎠ k any integer 2 2 2 3/ 2 3/ 2 –cos – 2sin –1– sin ( ) 4sin 4sin x x x f x x x ′′ = = 2 3/ 2 1 sin – ; 4sin x x + = ( ) 0f x′′ < for all x. f(x) is concave down on (2kπ , (2k + 1)π ); no inflection points 25. Domain: ( , )−∞ ∞ Range: [0,1] Even function since 2 2 ( ) cos ( ) cos ( )h t t t h t− = − = = so the function is symmetric with respect to the y-axis. y-intercept: 1y = ; t-intercepts: 2 x k π π= + where k is any integer. '( ) 2cos sinh t t t= − ; '( ) 0h t = at 2 k t π = . Critical points: 2 k t π = © 2007 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.
  • Instructor’s Resource Manual Section 3.5 199 '( ) 0 when ( 1) 2 h t k t k π π π> + < < + . The function is increasing on the intervals [ ]( / 2),( 1)k kπ π π+ + and decreasing on the intervals [ ], ( / 2)k kπ π π+ . Global maxima ( ) 1h kπ = Global minima 0 2 h k π π ⎛ ⎞ + =⎜ ⎟ ⎝ ⎠ 2 2 ''( ) 2sin 2cos 2(cos2 )h t t t t= − = − ''( ) 0h t < on , 4 4 k k π π π π ⎛ ⎞ − +⎜ ⎟ ⎝ ⎠ so h is concave down, and ''( ) 0h t > on 3 , 4 4 k k π π π π ⎛ ⎞ + +⎜ ⎟ ⎝ ⎠ so h is concave up. Inflection points: 1 , 2 4 2 kπ π⎛ ⎞ +⎜ ⎟ ⎝ ⎠ No vertical asymptotes; no horizontal asymptotes. 26. Domain: all reals except 2 t k π π= + Range: [0, )∞ y-intercepts: 0y = ; t-intercepts: t kπ= where k is any integer. Even function since 2 2 2 ( ) tan ( ) ( tan ) tang t t t t− = − = − = so the function is symmetric with respect to the y-axis. 2 3 2sin '( ) 2sec tan cos t g t t t t = = ; '( ) 0g t = when t kπ= . Critical points: kπ ( )g t is increasing on , 2 k k π π π ⎡ ⎞ + ⎟⎢ ⎣ ⎠ and decreasing on , 2 k k π π π ⎛ ⎤ −⎜ ⎥ ⎝ ⎦ . Global minima ( ) 0g kπ = ; no local maxima 4 2 6 2 2 4 2 4 cos sin (3)cos sin '( ) 2 cos cos 3sin 2 cos 1 2sin 2 0 cos t t t t g t t t t t t t + = + = + = > over the entire domain. Thus the function is concave up on , 2 2 k k π π π π ⎛ ⎞ − +⎜ ⎟ ⎝ ⎠ ; no inflection points. No horizontal asymptotes; 2 t k π π= + are vertical asymptotes. 27. Domain: ≈ (– ∞ , 0.44) ∪ (0.44, ∞ ); range: (– ∞ , ∞ ) Neither an even nor an odd function y-intercept: 0; x-intercepts: 0, ≈ 0.24 3 2 2 74.6092 – 58.2013 7.82109 ( ) ; (7.126 – 3.141) x x x f x x + ′ = ( ) 0f x′ = when x = 0, ≈ 0.17, ≈ 0.61 Critical points: 0, ≈ 0.17, ≈ 0.61 ( ) 0f x′ > when 0 < x < 0.17 or 0.61 < x f(x) is increasing on ≈ [0, 0.17] ∪ [0.61, ∞ ) and decreasing on (– ∞ , 0] ∪ [0.17, 0.44) ∪ (0.44, 0.61] Local minima f(0) = 0, f(0.61) ≈ 0.60; local maximum f(0.17) ≈ 0.01 3 2 3 531.665 – 703.043 309.887 – 24.566 (7.126 – 3.141) ( ) ; x x x x f x + ′′ = ( ) 0f x′′ > when x < 0.10 or x > 0.44 © 2007 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.
  • 200 Section 3.5 Instructor’s Resource Manual f(x) is concave up on (– ∞ , 0.10) ∪ (0.44, ∞ ) and concave down on (0.10, 0.44); inflection point ≈ (0.10, 0.003) 3 2 2 3.141 5.235 1.245 5.235 1.245 lim lim 7.126 3.141 7.126x x x x x x x x→∞ →∞ − − = = ∞ − − so f(x) does not have a horizontal asymptote. As – 3 2 0.44 , 5.235 –1.245 0.20x x x→ → while – 7.126 – 3.141 0 ,x → so –0.44 lim ( ) – ; x f x → = ∞ as 3 2 0.44 , 5.235 –1.245 0.20x x x+ → → while 7.126 – 3.141 0 ,x + → so 0.44 lim ( ) ; x f x +→ = ∞ x ≈ 0.44 is a vertical asymptote of f(x). 28. 29. 30. 31. 32. 33. 34. © 2007 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.
  • Instructor’s Resource Manual Section 3.5 201 35. 36. 4 3 5( –1) ; 20( –1) ; ( ) 0y x y x y x′ ′′ ′′= = > when x > 1; inflection point (1, 3) At x = 1, 0,y′ = so the linear approximation is a horizontal line. 3 37. 38. 39. 40. Let 2 ( ) ,f x ax bx c= + + then ( ) 2f x ax b′ = + and ( ) 2 .f x a′′ = An inflection point occurs where ( )f x′′ changes from positive to negative, but 2a is either always positive or always negative, so f(x) does not have any inflection points. ( ( )f x′′ = 0 only when a = 0, but then f(x) is not a quadratic curve.) 41. Let 3 2 ( ) ,f x ax bx cx d= + + + then 2 ( ) 3 2f x ax bx c′ = + + and ( ) 6 2 .f x ax b′′ = + As long as 0a ≠ , ( )f x′′ will be positive on one side of 3 b x a = and negative on the other side. 3 b x a = is the only inflection point. 42. Let 4 3 2 ( ) ,f x ax bx cx dx c= + + + + then 3 2 ( ) 4 3 2f x ax bx cx d′ = + + + and 2 2 ( ) 12 6 2 2(6 3 )f x ax bx c ax bx c′′ = + + = + + Inflection points can only occur when ( )f x′′ changes sign from positive to negative and ( ) 0.f x′′ = ( )f x′′ has at most 2 zeros, thus f(x) has at most 2 inflection points. 43. Since the c term is squared, the only difference occurs when c = 0. When c = 0, 32 2 y x x x= = which has domain (– ∞ , ∞ ) and range [0, ∞ ). When c ≠ 0, 2 2 2 –y x x c= has domain (– ∞ , –|c|] ∪ [|c|, ∞ ) and range [0, ∞ ). © 2007 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.
  • 202 Section 3.5 Instructor’s Resource Manual The only extremum points are c± . For 0c = , there is one minimum, for 0c ≠ there are two. No maxima, independent of c. No inflection points, independent of c. 44. 2 2 2 ( ) 4 ( ) 4 cx cx f x cx c x = = + + 2 2 2 2 2 (4 – ) ( ) ; ( ) 0 (4 ) c c x f x f x c x ′ ′= = + when 2 x c = ± unless c = 0, in which case f(x) = 0 and ( ) 0.f x′ = If c > 0, f(x) is increasing on 2 2 – , c c ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ and decreasing on 2 2 – , – , , c c ⎛ ⎤ ⎡ ⎞ ∞ ∪ ∞⎜ ⎟⎥ ⎢ ⎝ ⎦ ⎣ ⎠ thus, f(x) has a global minimum at 2 1 – – 4 f c ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ and a global maximum of 2 1 . 4 f c ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ If c < 0, f(x) is increasing on 2 2 – , – , c c ⎛ ⎤ ⎡ ⎞ ∞ ∪ ∞⎜ ⎟⎥ ⎢ ⎝ ⎦ ⎣ ⎠ and decreasing on 2 2 , – . c c ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ Thus, f(x) has a global minimum at 2 1 – – 4 f c ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ and a global maximum at 2 1 . 4 f c ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 3 2 2 2 2 3 2 ( –12) ( ) , (4 ) c x c x f x c x ′′ = + so f(x) has inflection points at x = 0, 2 3 , c ± c ≠ 0 45. 2 2 2 1 ( ) , ( – 4) f x cx cx = + then 2 2 2 2 2 2 (7 2 ) ( ) ; [( – 4) ] cx cx f x cx cx − ′ = + If c > 0, ( ) 0f x′ = when x = 0, 7 . 2c ± If c < 0, ( ) 0f x′ = when x = 0. Note that f(x) = 1 16 (a horizontal line) if c = 0. If c > 0, ( ) 0f x′ > when 7 2 x c < − and 7 0 , 2 x c < < so f(x) is increasing on 7 7 , 0, 2 2c c ⎛ ⎤ ⎡ ⎤ −∞ − ∪⎜ ⎥ ⎢ ⎥⎜ ⎝ ⎦ ⎣ ⎦ and decreasing on 7 7 , 0 , 2 2c c ⎡ ⎤ ⎡ ⎞ − ∪ ∞⎟⎢ ⎥ ⎢ ⎟ ⎣ ⎦ ⎣ ⎠ . Thus, f(x) has local maxima 7 4 , 2 15 f c ⎛ ⎞ − =⎜ ⎟⎜ ⎟ ⎝ ⎠ 7 4 2 15 f c ⎛ ⎞ =⎜ ⎟⎜ ⎟ ⎝ ⎠ and local minimum 1 (0) 16 f = . If c < 0, ( ) 0f x′ > when x < 0, so f(x) is increasing on (– ∞ , 0] and decreasing on [0, ∞ ). Thus, f(x) has a local maximum 1 (0) 16 f = . Note that f(x) > 0 and has horizontal asymptote y = 0. © 2007 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.
  • Instructor’s Resource Manual Section 3.5 203 46. 2 1 ( ) . 4 f x x x c = + + By the quadratic formula, 2 4 0x x c+ + = when –2 4 – .x c= ± Thus f(x) has vertical asymptote(s) at –2 4 –x c= ± when c ≤ 4. 2 2 –2 – 4 ( ) ; ( ) 0 ( 4 ) x f x f x x x c ′ ′= = + + when x = –2, unless c = 4 since then x = –2 is a vertical asymptote. For c ≠ 4, ( ) 0f x′ > when x < –2, so f(x) is increasing on (– ∞ , –2] and decreasing on [–2, ∞ ) (with the asymptotes excluded). Thus f(x) has a local maximum at 1 (–2) . – 4 f c = For c = 4, 3 2 ( ) – ( 2) f x x ′ = + so f(x) is increasing on (– ∞ , –2) and decreasing on (–2, ∞ ). 47. ( ) sinf x c cx= + . Since c is constant for all x and sin cx is continuous everywhere, the function ( )f x is continuous everywhere. ( )' cosf x c cx= ⋅ ( )' 0f x = when ( )1 2 cx k π= + or ( )1 2 cx k π= + where k is an integer. ( ) 2 '' sinf x c cx= − ⋅ ( )( ) ( )( ) ( )2 21 1 2 2 '' sin 1 k c cf k c c k cπ π+ = − ⋅ ⋅ + = − ⋅ − In general, the graph of f will resemble the graph of siny x= . The period will decrease as c increases and the graph will shift up or down depending on whether c is positiv