1. CHAPTER 0 0.1 Concepts Review 1. rational numbers Preliminaries 1 2 1 1 1 1 8. = 3 5 2 3 5 2 1 5 3 3 5 2 15 15 2. dense 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 3. If not Q then not P. 4. theorems 2 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. 6. 7. 5 1 65 7 58 = = 7 13 91 91 91 3 3 1 3 3 1 + = + 4 7 21 6 3 21 6 42 6 7 43 = + = 42 42 42 42 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 Instructors Resource Manual 2 2 14 2 14 2 14 6 = = 9. 21 5 1 21 14 21 14 3 3 2 = 14 3 2 9 6 = = 21 7 3 49 49 2 2 35 33 5 7 = 7 7 = 7 = 33 = 11 10. 6 2 1 7 1 6 1 7 7 7 7 7 11 12 11 4 7 21 = 7 7 = 7 = 7 11. 11 + 12 11 + 4 15 15 7 21 7 7 7 1 3 7 4 6 7 5 + + 5 12. 2 4 8 = 8 8 8 = 8 = 1 3 7 4 6 7 3 3 + + 2 4 8 8 8 8 8 13. 1 1 1 2 3 2 1 =1 =1 = = 1 3 3 3 3 3 1+ 2 2 14. 2 + 15. ( 3 5 1+ 2 5+ 3 3 3 = 2+ 2 5 7 2 2 2 6 14 6 20 = 2+ = + = 7 7 7 7 = 2+ )( ) ( 5) ( 3) 5 3 = 2 2 =53= 2 Section 0.1 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.

2. 16. ( 5 3 ) = ( 5) 2 2 2 ( 5 )( 3 ) + ( 3 ) 2 27. = 5 2 15 + 3 = 8 2 15 17. (3x 4)( x + 1) = 3 x 2 + 3 x 4 x 4 = 3x2 x 4 18. (2 x 3)2 = (2 x 3)(2 x 3) 12 4 2 + x + 2x x x + 2 12 4( x + 2) 2x = + + x( x + 2) x( x + 2) x( x + 2) 12 + 4 x + 8 + 2 x 6 x + 20 = = x( x + 2) x( x + 2) 2(3 x + 10) = x( x + 2) 2 + = 4 x2 6 x 6 x + 9 = 4 x 2 12 x + 9 19. 28. (3x 9)(2 x + 1) = 6 x 2 + 3 x 18 x 9 2 y + 2(3 y 1) (3 y + 1)(3 y 1) 2(3 y + 1) 2y = + 2(3 y + 1)(3 y 1) 2(3 y + 1)(3 y 1) = 2 = 6 x 15 x 9 20. (4 x 11)(3x 7) = 12 x 2 28 x 33 x + 77 = 21. (3t 2 t + 1) 2 = (3t 2 t + 1)(3t 2 t + 1) 3 2 3 2 2 = 9t 3t + 3t 3t + t t + 3t t + 1 6y + 2 + 2y 8y + 2 = 2(3 y + 1)(3 y 1) 2(3 y + 1)(3 y 1) = = 12 x 2 61x + 77 4 2 y + 6 y 2 9 y2 1 2(4 y + 1) 4y +1 = 2(3 y + 1)(3 y 1) (3 y + 1)(3 y 1) = 9t 4 6t 3 + 7t 2 2t + 1 00 = 0 b. 0 is undefined. 0 c. 0 =0 17 d. 3 is undefined. 0 e. 05 = 0 f. 170 = 1 29. a. 22. (2t + 3)3 = (2t + 3)(2t + 3)(2t + 3) = (4t 2 + 12t + 9)(2t + 3) = 8t 3 + 12t 2 + 24t 2 + 36t + 18t + 27 = 8t 3 + 36t 2 + 54t + 27 23. x 2 4 ( x 2)( x + 2) = = x+2, x 2 x2 x2 24. x 2 x 6 ( x 3)( x + 2) = = x+2, x 3 x3 ( x 3) 25. t 2 4t 21 (t + 3)(t 7) = = t 7 , t 3 t +3 t +3 26. 2 2x 2x 3 2 2 x 2x + x = 2 x(1 x) 2 x( x 2 x + 1) 2 x( x 1) = x( x 1)( x 1) 2 = x 1 Section 0.1 0 = a , then 0 = 0 a , but this is meaningless 0 because a could be any real number. No 0 single value satisfies = a . 0 30. If 31. .083 12 1.000 96 40 36 4 Instructors Resource Manual 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.

3. 32. .285714 7 2.000000 14 60 56 40 35 50 49 10 7 30 28 2 33. .142857 21 3.000000 21 90 84 60 42 180 168 120 105 150 147 3 34. .294117... 17 5.000000... 0.2941176470588235 34 160 153 70 68 20 17 30 17 130 119 11 Instructors Resource Manual 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 x = 123.123123... x = 0.123123... 999 x = 123 123 41 x= = 999 333 38. x = 0.217171717 1000 x = 217.171717... 10 x = 2.171717... 990 x = 215 215 43 x= = 990 198 39. x = 2.56565656... 100 x = 256.565656... x = 2.565656... 99 x = 254 254 x= 99 40. x = 3.929292 100 x = 392.929292... x = 3.929292... 99 x = 389 389 x= 99 Section 0.1 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.

4. 41. x = 0.199999... 100 x = 19.99999... 52. 10 x = 1.99999... 90 x = 18 18 1 x= = 90 5 54. 55. 10 x = 3.99999... 90 x = 36 36 2 x= = 90 5 56. 43. Those rational numbers that can be expressed by a terminating decimal followed by zeros. 1 p 1 = p , so we only need to look at . If q q q q = 2n 5m , then n m 1 1 1 = = (0.5)n (0.2)m . The product q 2 5 of any number of terminating decimals is also a n m terminating decimal, so (0.5) and (0.2) , and hence their product, decimal. Thus 1 , is a terminating q p has a terminating decimal q expansion. 45. Answers will vary. Possible answer: 0.000001, 1 0.0000010819... 12 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 , 2 and 2 51. ( 3 + 1)3 20.39230485 4 Section 0.1 2 3 ) 4 0.0102051443 53. 4 1.123 3 1.09 0.00028307388 42. x = 0.399999 100 x = 39.99999... 44. ( ( 3.1415 )1/ 2 0.5641979034 8.92 + 1 3 0.000691744752 4 (6 2 2) 3.661591807 57. Let a and b be real numbers with a < b . Let n be a natural number that satisfies 1 / n < b a . Let S = {k : k n > b} . Since a nonempty set of integers that is bounded below contains a least element, there is a k 0 S such that k 0 / n > b but (k 0 1) / n b . Then k0 1 k0 1 1 = >b > a n n n n k 0 1 k 0 1 Thus, a < n b . If n < b , then choose r= k 0 1 n . Otherwise, choose r = k0 2 n . 1

5. b. Every circle has area less than or equal to 9. The original statement is true. 62. V = (8.004) 2 (270) (8)2 (270) 54.3 ft.3 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. c. Some real number is less than or equal to its square. The negation is true. 71. a. True; If x is positive, then x 2 is positive. b. False; Take x = 2 . Then x 2 > 0 but x . x x e. b. b. The statement, converse, and contrapositive are all false. 69. a. True; x + ( x ) < x + 1 + ( x ) : 0 < 1 2 b. The statement, converse, and contrapositive are all true. 68. a. True; Let y be any positive number. Take y x = . Then 0 < x < y . 2 d. True; 1/ n can be made arbitrarily close to 0. b. If a < b then a < b. If a b then a b. 67. a. x 2 . If a triangle is a right triangle, then 2 1 2 . Then x = 2 d. True; Let x be any number. Take 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. False; Take x = Some natural number is larger than its square. The original statement is true. Prove the contrapositive. Suppose n is even. Then there is an integer k such that n = 2k . Then n 2 = (2k )2 = 4k 2 = 2(2k 2 ) . Thus n 2 is even. Parts (a) and (b) prove that n is odd if and 74. only if n 2 is odd. 75. a. b. 243 = 3 3 3 3 3 124 = 4 31 = 2 2 31 or 22 31 Some natural number is not rational. The original statement is true. Instructors Resource Manual Section 0.1 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.

6. 5100 = 2 2550 = 2 2 1275 c. 82. a. = 2 2 3 425 = 2 2 3 5 85 = 2 2 3 5 5 17 or 22 3 52 17 c. 76. For example, let A = b c 2 d 3 ; then A2 = b 2 c 4 d 6 , so the square of the number is the product of primes which occur an even number of times. 77. p p2 ;2 = ; 2q 2 = p 2 ; Since the prime 2 q q 2 must occur an even number of factors of p p times, 2q2 would not be valid and = 2 q must be irrational. 3= p p2 ; 3= ; 3q 2 = p 2 ; Since the prime q q2 factors of p 2 must occur an even number of times, 3q 2 would not be valid and e. f. 83. a. p = 3 q x = 2.4444...; 10 x = 24.4444... x = 2.4444... 9 x = 22 22 x= 9 2 3 n = 1: x = 0, n = 2: x = , n = 3: x = , 3 2 5 n = 4: x = 4 3 The upper bound is . 2 2 Answers will vary. Possible answer: An example is S = {x : x 2 < 5, x a rational number}. Here the least upper bound is 5, which is real but irrational. must be irrational. 79. Let a, b, p, and q be natural numbers, so b. 2 d. 1 2= 78. 2 a b p a p aq + bp are rational. + = This q b q bq sum is the quotient of natural numbers, so it is also rational. and b. True 0.2 Concepts Review 1. [1,5); (, 2] 2. b > 0; b < 0 p 80. Assume a is irrational, 0 is rational, and q p r qr is = is rational. Then a = q s ps rational, which is a contradiction. a 81. a. 9 = 3; rational b. 3 0.375 = ; rational 8 c. (1 + 3)2 = 1 + 2 3 + 3 = 4 + 2 3; irrational 4. 1 x 5 Problem Set 0.2 1. a. b. (3 2)(5 2) = 15 4 = 30; rational d. 3. (b) and (c) c. d. 6 Section 0.2 Instructors Resource Manual 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.

7. 3 < 1 6 x 4 9. 4 < 6 x 3 e. 2 1 1 2 > x ; , 3 2 2 3 f. 2. a. c. (2, 7) (, 2] b. d. [3, 4) [1, 3] 10. 3. x 7 < 2 x 5 2 < x;( 2, ) 4 < 5 3x < 7 1 < 3x < 2 1 2 2 1 > x > ; , 3 3 3 3 4. 3x 5 < 4 x 6 1 < x; (1, ) 11. x2 + 2x 12 < 0; x= 5. 7 x 2 9x + 3 5 2 x = 1 13 ( 7. 4 < 3 x + 2 < 5 6 < 3 x < 3 2 < x < 1; (2, 1) ) ( ) x 1 + 13 x 1 13 < 0; 5 5 x ; , 2 2 6. 5 x 3 > 6 x 4 1 > x;(,1) 2 (2)2 4(1)(12) 2 52 = 2(1) 2 ( 1 13, 1 + 13 ) 12. x 2 5 x 6 > 0 ( x + 1)( x 6) > 0; (, 1) (6, ) 13. 2x2 + 5x 3 > 0; (2x 1)(x + 3) > 0; 1 (, 3) , 2 8. 3 < 4 x 9 < 11 6 < 4 x < 20 3 3 < x < 5; ,5 2 2 14. 3 (4 x + 3)( x 2) < 0; , 2 4 15. Instructors Resource Manual 4 x2 5x 6 < 0 x+4 0; [4, 3) x3 Section 0.2 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.

8. 16. 3x 2 2 0; , (1, ) x 1 3 3 >2 x+5 20. 3 2 > 0 x+5 17. 2 5 < 0 x 2 5x < 0; x 2 ( , 0) , 5 18. 3 2( x + 5) >0 x+5 2 0; 5, 2 x+5 21. ( x + 2)( x 1)( x 3) > 0; (2,1) (3,8) 3 1 22. (2 x + 3)(3x 1)( x 2) < 0; , , 2 2 3 3 23. (2 x - 3)( x -1)2 ( x - 3) 0; , [3, ) 2 24. (2 x 3)( x 1) 2 ( x 3) > 0; 19. ( ,1) 1, 3 ( 3, ) 2 1 4 3x 2 1 4 0 3x 2 1 4(3 x 2) 0 3x 2 9 12 x 2 3 0; , , 3x 2 3 4 25. x3 5 x 2 6 x < 0 x( x 2 5 x 6) < 0 x( x + 1)( x 6) < 0; (, 1) (0, 6) 26. x3 x 2 x + 1 > 0 ( x 2 1)( x 1) > 0 ( x + 1)( x 1) 2 > 0; (1,1) (1, ) 27. a. c. 8 Section 0.2 False. False. b. True. Instructors Resource Manual 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.

9. 28. a. True. c. False. 29. a. b. True. 33. a. ( x + 1)( x 2 + 2 x 7) x 2 1 x3 + 3 x 2 5 x 7 x 2 1 x3 + 2 x 2 5 x 6 0 ( x + 3)( x + 1)( x 2) 0 [3, 1] [2, ) Let a < b , so ab < b 2 . Also, a 2 < ab . Thus, a 2 < ab < b 2 and a 2 < b 2 . Let a 2 < b 2 , so a b Then 0 < ( a b ) = a 2 2ab + b 2 2 x4 2 x2 8 b. < b 2 2ab + b 2 = 2b ( b a ) x4 2 x2 8 0 Since b > 0 , we can divide by 2b to get ba > 0. ( x 2 4)( x 2 + 2) 0 ( x 2 + 2)( x + 2)( x 2) 0 b. We can divide or multiply an inequality by any positive number. a 1 1 a < b 1 and 2x + 1 < 3 3x > 6 and 2x < 2 x > 2 and x < 1; (2, 1) ( x 2 4)( x 2 1) < 0 ( x + 2)( x + 1)( x 1)( x 2) < 0 (2, 1) (1, 2) 34. a. 32. a. 3x + 7 > 1 and 2x + 1 < 4 5 x > 2 and x < ; 2 1 1 , 2.01 1.99 2 x 7 > 1 or 2 x + 1 < 3 b. x > 4 or x < 1 (,1) (4, ) 2.99 < 1 < 3.01 x+2 2.99( x + 2) < 1 < 3.01( x + 2) 2.99 x + 5.98 < 1 and 1 < 3.01x + 6.02 4.98 and 5.02 x< x> 2 x 7 1 or 2 x + 1 < 3 2 x 8 or 2 x < 2 2.99 5.02 4.98 8 or 2 x < 2 b. 1.99 < 1.99 x < 1 < 2.01x 1.99 x < 1 and 1 < 2.01x 1 and x > 1 x< 1.99 2.01 b. 3x + 7 > 1 and 2x + 1 > 4 3x > 6 and 2x > 5 5 x > 2 and x > ; ( 2, ) 2 c. ( x 2 + 1)2 7( x 2 + 1) + 10 < 0 2 x 7 1 or 2 x + 1 > 3 3.01 2 x 8 or 2 x > 2 x 4 or x > 1 (, ) 35. x 2 5; x 2 5 or x 2 5 x 3 or x 7 (, 3] [7, ) Instructors Resource Manual Section 0.2 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.

10. 36. x + 2 < 1; 1 < x + 2 < 1 43. 3 < x < 1 (3, 1) 37. 4 x + 5 10; 10 4 x + 5 10 15 4 x 5 38. 15 5 15 5 x ; , 4 4 4 4 2 x 1 > 2; 2x 1 < 2 or 2x 1 > 2 2x < 1 or 2x > 3; 1 3 1 3 x < or x > , , , 2 2 2 2 39. 40. 2x 5 7 7 2x 2x 5 7 or 5 7 7 7 2x 2x 2 or 12 7 7 x 7 or x 42; (, 7] [42, ) x +1 < 1 4 x 1 < + 1 < 1 4 x 2 < < 0; 4 8 < x < 0; (8, 0) 41. 5 x 6 > 1; 5 x 6 < 1 or 5 x 6 > 1 5 x < 5 or 5 x > 7 7 7 x < 1 or x > ;(,1) , 5 5 42. 2 x 7 > 3; 2x 7 < 3 or 2x 7 > 3 2x < 4 or 2x > 10 x < 2 or x > 5; (, 2) (5, ) 44. 1 3 > 6; x 1 1 3 < 6 or 3 > 6 x x 1 1 + 3 < 0 or 9 > 0 x x 1 + 3x 1 9x < 0 or > 0; x x 1 1 , 0 0, 3 9 5 > 1; x 5 5 2 + < 1 or 2 + > 1 x x 5 5 3 + < 0 or 1 + > 0 x x 3x + 5 x+5 < 0 or > 0; x x 5 ( , 5) , 0 (0, ) 3 2+ 45. x 2 3x 4 0; x= 3 (3)2 4(1)(4) 3 5 = = 1, 4 2(1) 2 ( x + 1)( x 4) = 0; (, 1] [4, ) 4 (4)2 4(1)(4) =2 2(1) ( x 2)( x 2) 0; x = 2 46. x 2 4 x + 4 0; x = 47. 3x2 + 17x 6 > 0; x= 17 (17) 2 4(3)(6) 17 19 1 = = 6, 2(3) 6 3 1 (3x 1)(x + 6) > 0; ( , 6) , 3 48. 14 x 2 + 11x 15 0; 11 (11) 2 4(14)(15) 11 31 = 2(14) 28 3 5 x= , 2 7 3 5 3 5 x + x 0; , 2 7 2 7 x= 49. x 3 < 0.5 5 x 3 < 5(0.5) 5 x 15 < 2.5 50. x + 2 < 0.3 4 x + 2 < 4(0.3) 4 x + 18 < 1.2 10 Section 0.2 Instructors Resource Manual 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.

11. 51. x2 < 52. x+4 < 6 2 6 x 2 < 6 x 12 < 59. x 1 < 2 x 6 ( x 1) 2 < (2 x 6)2 2 x + 4 < 2x + 8 < x 2 2 x + 1 < 4 x 2 24 x + 36 3x 2 22 x + 35 > 0 53. 3x 15 < 3( x 5) < (3x 7)( x 5) > 0; 3 x5 < x5 < 54. 3 ; = 7 , (5, ) 3 3 4 x 8 < 4( x 2) < 60. 55. 4 ; = 4 x2 4 x + 1 x2 + 2 x + 1 4 3x2 6 x 0 3 x( x 2) 0 (, 0] [2, ) 6 x+6 < 6 ; = 2 6 x + 36 < 6( x + 6) < x+6 < 2x 1 x + 1 (2 x 1)2 ( x + 1) 4 x2 < x2 < x 1 < 2 x 3 6 61. 2 2 x 3 < x + 10 4 x 6 < x + 10 56. 5 x + 25 < 5( x + 5) < (4 x 6) 2 < ( x + 10)2 5 x+5 < 16 x 2 48 x + 36 < x 2 + 20 x + 100 x+5 < 5 ; = 15 x 2 68 x 64 < 0 5 (5 x + 4)(3 x 16) < 0; 57. C = d C 10 0.02 d 10 0.02 10 d 0.02 10 0.02 d 0.0064 We must measure the diameter to an accuracy of 0.0064 in. 58. C 50 1.5, 5 ( F 32 ) 50 1.5; 9 5 ( F 32 ) 90 1.5 9 F 122 2.7 4 16 , 5 3 3x 1 < 2 x + 6 62. 3x 1 < 2 x + 12 (3x 1) 2 < (2 x + 12)2 9 x 2 6 x + 1 < 4 x 2 + 48 x + 144 5 x 2 54 x 143 < 0 ( 5 x + 11)( x 13) < 0 11 ,13 5 We are allowed an error of 2.7 F. Instructors Resource Manual Section 0.2 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.

12. 63. x < y x x x y and x y < y y 2 x < y Order property: x < y xz < yz when z is positive. 2 Transitivity (x x2 < y 2 Conversely, 2 2 (x 2 x2 < y 2 x < y 2 = x 2 2 = x2 ) ) 2 2 x y 0 > 0, > 0. 2 x +2 x +3 x + (2) 4 + 4 + 7 = 15 1 and x 2 + 1 1 so 1. 2 x +1 1 = = x +9 x +2 2 + = 2 2 2 x + 9 x + 9 x + 9 x2 + 9 1 1 Since x 2 + 9 9, 2 x +9 9 x +2 x +2 9 x2 + 9 x +2 x2 9 x2 + 9 b a< b. of absolute values. c. x +9 2 a b a b a b Use Property 4 b. 2 x2 a b = a + (b) a + b = a + b 65. a. x2 69. 1 3 1 2 1 1 x + x + x+ 2 4 8 16 1 3 1 2 1 1 x4 + x + x + x + 2 4 8 16 1 1 1 1 1+ + + + since x 1. 2 4 8 16 1 1 1 1 1.9375 < 2. So x 4 + x3 + x 2 + x + 2 4 8 16 x4 + Instructors Resource Manual 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.

13. x < x2 70. a. 77. x x2 < 0 x(1 x) < 0 x < 0 or x > 1 1 11 R 60 2 x

14. Problem Set 0.3 5. d1 = (5 + 2) 2 + (3 4)2 = 49 + 1 = 50 d 2 = (5 10)2 + (3 8)2 = 25 + 25 = 50 1. d3 = (2 10)2 + (4 8)2 = 144 + 16 = 160 d1 = d 2 so the triangle is isosceles. 6. a = (2 4)2 + (4 0) 2 = 4 + 16 = 20 b = (4 8)2 + (0 + 2)2 = 16 + 4 = 20 c = (2 8)2 + (4 + 2) 2 = 36 + 4 = 40 d = (3 1)2 + (1 1)2 = 4 = 2 a 2 + b 2 = c 2 , so the triangle is a right triangle. 2. 7. (1, 1), (1, 3); (7, 1), (7, 3); (1, 1), (5, 1) 8. ( x 3) 2 + (0 1) 2 = ( x 6)2 + (0 4) 2 ; x 2 6 x + 10 = x 2 12 x + 52 6 x = 42 x = 7 ( 7, 0 ) d = (3 2)2 + (5 + 2)2 = 74 8.60 3. 2 + 4 2 + 3 1 9. , = 1, ; 2 2 2 2 25 1 d = (1 + 2)2 + 3 = 9 + 3.91 2 4 1+ 2 3 + 6 3 9 10. midpoint of AB = , = , 2 2 2 2 4 + 3 7 + 4 7 11 midpoint of CD = , = , 2 2 2 2 2 3 7 9 11 d = + 2 2 2 2 2 = 4 + 1 = 5 2.24 d = (4 5)2 + (5 + 8) 2 = 170 13.04 4. 11 (x 1)2 + (y 1)2 = 1 12. ( x + 2)2 + ( y 3)2 = 42 ( x + 2)2 + ( y 3)2 = 16 13. ( x 2) 2 + ( y + 1) 2 = r 2 (5 2)2 + (3 + 1) 2 = r 2 r 2 = 9 + 16 = 25 ( x 2) 2 + ( y + 1) 2 = 25 d = (1 6)2 + (5 3) 2 = 49 + 4 = 53 7.28 14 Section 0.3 Instructors Resource Manual 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.

15. 14. ( x 4) 2 + ( y 3) 2 = r 2 21. 4 x 2 + 16 x + 15 + 4 y 2 + 6 y = 0 (6 4) 2 + (2 3) 2 = r 2 3 9 9 4( x 2 + 4 x + 4) + 4 y 2 + y + = 15 + 16 + 2 16 4 2 r = 4 +1 = 5 2 3 13 4( x + 2)2 + 4 y + = 4 4 ( x 4)2 + ( y 3)2 = 5 1+ 3 3 + 7 15. center = , = (2, 5) 2 2 1 1 radius = (1 3)2 + (3 7)2 = 4 + 16 2 2 1 = 20 = 5 2 2 2 ( x 2) + ( y 5) = 5 2 3 13 ( x + 2)2 + y + = 4 16 3 center = 2, ; radius = 4 105 + 4 y2 + 3 y = 0 16 3 9 2 4( x + 4 x + 4) + 4 y 2 + y + 4 64 105 9 = + 16 + 16 16 22. 4 x 2 + 16 x + 16. Since the circle is tangent to the x-axis, r = 4. ( x 3)2 + ( y 4) 2 = 16 17. x 2 + 2 x + 10 + y 2 6 y 10 = 0 2 3 4( x + 2)2 + 4 y + = 10 8 x2 + 2 x + y 2 6 y = 0 2 ( x 2 + 2 x + 1) + ( y 2 6 y + 9) = 1 + 9 3 5 ( x + 2)2 + y + = 8 2 ( x + 1) 2 + ( y 3) 2 = 10 3 5 10 center = 2, ; radius = = 8 2 2 center = (1, 3); radius = 10 x 2 + y 2 6 y = 16 18. x 2 + ( y 2 6 y + 9) = 16 + 9 23. 2 1 =1 2 1 24. 75 =2 43 25. 6 3 9 = 5 2 7 26. 6 + 4 =1 02 27. 50 5 = 03 3 28. 60 =1 0+6 x 2 + ( y 3) 2 = 25 center = (0, 3); radius = 5 19. x 2 + y 2 12 x + 35 = 0 x 2 12 x + y 2 = 35 ( x 2 12 x + 36) + y 2 = 35 + 36 ( x 6) 2 + y 2 = 1 center = (6, 0); radius = 1 x 2 + y 2 10 x + 10 y = 0 20. 2 29. y 2 = 1( x 2) y 2 = x + 2 x+ y4 = 0 30. y 4 = 1( x 3) y 4 = x + 3 2 ( x 10 x + 25) + ( y + 10 y + 25) = 25 + 25 x+ y7 = 0 ( x 5) 2 + ( y + 5)2 = 50 center = ( 5, 5 ) ; radius = 50 = 5 2 13 4 31. y = 2x + 3 2x y + 3 = 0 32. Instructors Resource Manual y = 0x + 5 0x + y 5 = 0 Section 0.3 15 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. 33. m = 83 5 = ; 42 2 5 y 3 = ( x 2) 2 2 y 6 = 5 x 10 c. 3 y = 2 x + 6 y= x 4y + 0 = 0 2 x + 2; 3 2 m= ; 3 5x 2 y 4 = 0 2 1 1 34. m = = ; 84 4 1 y 1 = ( x 4) 4 4y 4 = x 4 2x + 3 y = 6 2 y + 3 = ( x 3) 3 2 y = x 1 3 d. 3 m= ; 2 3 ( x 3) 2 3 15 y= x 2 2 y+3= 2 1 2 35. 3y = 2x + 1; y = x + ; slope = ; 3 3 3 1 y -intercept = 3 1 2 3 = ; 3 +1 4 3 y + 3 = ( x 3) 4 3 3 y= x 4 4 36. 4 y = 5 x 6 5 3 y = x+ 4 2 5 3 slope = ; y -intercept = 4 2 e. m= 37. 6 2 y = 10 x 2 2 y = 10 x 8 y = 5 x + 4; slope = 5; y-intercept = 4 f. x=3 38. 4 x + 5 y = 20 5 y = 4 x 20 4 y = x4 5 4 slope = ; y -intercept = 4 5 39. a. b. m = 2; y + 3 = 2( x 3) y = 2x 9 1 m= ; 2 1 y + 3 = ( x 3) 2 1 3 y= x 2 2 16 Section 0.3 40. a. g. y = 3 3 x + cy = 5 3(3) + c(1) = 5 c = 4 b. c=0 c. 2 x + y = 1 y = 2 x 1 m = 2; 3x + cy = 5 cy = 3x + 5 3 5 y = x+ c c 3 2 = c 3 c= 2 d. c must be the same as the coefficient of x, so c = 3. Instructors Resource Manual 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.

17. e. y 2 = 3( x + 3); 1 perpendicular slope = ; 3 1 3 = 3 c c=9 3 ( x + 2) 2 3 y = x+2 2 y +1 = b. c. 2x + 3 y = 4 9 x 3 y = 15 = 11 11x x = 1 3(1) + y = 5 3 41. m = ; 2 42. a. 45. 2 x + 3 y = 4 3x + y = 5 m = 2; kx 3 y = 10 3 y = kx + 10 10 k y = x 3 3 k = 2; k = 6 3 1 m= ; 2 k 1 = 3 2 3 k= 2 2x + 3 y = 6 3 y = 2 x + 6 2 y = x + 2; 3 3 k 3 9 m= ; = ; k= 2 3 2 2 y=2 Point of intersection: (1, 2) 3 y = 2 x + 4 2 4 y = x+ 3 3 3 m= 2 3 y 2 = ( x + 1) 2 3 7 y = x+ 2 2 46. 4 x 5 y = 8 2 x + y = 10 4x 5 y = 8 4 x 2 y = 20 7 y = 28 y = 4 4 x 5(4) = 8 4 x = 12 x = 3 Point of intersection: ( 3, 4 ) ; 4x 5 y = 8 5 y = 4 x + 8 y= 43. y = 3(3) 1 = 8; (3, 9) is above the line. b0 b = 0a a b bx x y y = x + b; + y = b; + = 1 a a a b 44. (a, 0), (0, b); m = Instructors Resource Manual m= 4 8 x 5 5 5 4 5 y + 4 = ( x + 3) 4 5 31 y = x 4 4 Section 0.3 17 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. 47. 3x 4 y = 5 2x + 3y = 9 9 x 12 y = 15 8 x + 12 y = 36 17 x = 51 x=3 3(3) 4 y = 5 4 y = 4 y =1 Point of intersection: (3, 1); 3x 4y = 5; 4 y = 3x + 5 3 5 y= x 4 4 4 m= 3 4 y 1 = ( x 3) 3 4 y = x+5 3 48. 5 x 2 y = 5 2x + 3y = 6 15 x 6 y = 15 4 x + 6 y = 12 19 x = 27 27 x= 19 27 2 + 3y = 6 19 60 3y = 19 20 y= 19 27 20 Point of intersection: , ; 19 19 5x 2 y = 5 2 y = 5 x + 5 5 5 y= x 2 2 2 m= 5 20 2 27 y = x 19 5 19 2 54 20 y = x+ + 5 95 19 2 154 y = x+ 5 95 18 Section 0.3 2 + 6 1 + 3 , 49. center: = (4, 1) 2 2 2+ 6 3+3 midpoint = , = (4, 3) 2 2 inscribed circle: radius = (4 4)2 + (1 3)2 = 4=2 2 ( x 4) + ( y 1)2 = 4 circumscribed circle: radius = (4 2)2 + (1 3)2 = 8 ( x 4)2 + ( y 1)2 = 8 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. 1 1 L = 2 (4) + 2 (4) + 2 (1 + 9)2 + (2 10)2 2 2 = 8 + 2 100 + 144 56.37 51. Put the vertex of the right angle at the origin with the other vertices at (a, 0) and (0, b). The a b midpoint of the hypotenuse is , . The 2 2 distances from the vertices are 2 2 a b a +0 = 2 2 = 2 2 a b 0 + b = 2 2 = 2 2 a b 0 + 0 = 2 2 = a 2 b2 + 4 4 1 2 a + b2 , 2 a 2 b2 + 4 4 1 2 a + b 2 , and 2 a 2 b2 + 4 4 1 2 a + b2 , 2 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 ( x 4) 2 + ( y 3) 2 = 25. Instructors Resource Manual 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.

19. 53. x 2 + y 2 4 x 2 y 11 = 0 ( x 2 4 x + 4) + ( y 2 2 y + 1) = 11 + 4 + 1 ( x 2)2 + ( y 1)2 = 16 x 2 + y 2 + 20 x 12 y + 72 = 0 ( x 2 + 20 x + 100) + ( y 2 12 y + 36) = 72 + 100 + 36 2 2 ( x + 10) + ( y 6) = 64 center of first circle: (2, 1) center of second circle: (10, 6) d = (2 + 10)2 + (1 6) 2 = 144 + 25 = 169 = 13 However, the radii only sum to 4 + 8 = 12, so the circles must not intersect if the distance between their centers is 13. 54. x 2 + ax + y 2 + by + c = 0 2 a2 2 b2 x + ax + + y + by + 4 4 = c + a 2 b2 + 4 4 2 2 a b a 2 + b 2 4c x+ + y+ = 2 2 4 2 2 a + b 4c > 0 a 2 + b 2 > 4c 4 55. Label the points C, P, Q, and R as shown in the figure below. Let d = OP , h = OR , and a = PR . Triangles OPR and CQR are similar because each contains a right angle and they share angle QRC . For an angle of 30 , a 1 d 3 and = h = 2a . Using a = h 2 h 2 56. The equations of the two circles are ( x R)2 + ( y R)2 = R 2 ( x r )2 + ( y r )2 = r 2 Let ( a, a ) denote the point where the two circles touch. This point must satisfy (a R)2 + (a R)2 = R 2 R2 2 2 a = 1 R 2 (a R)2 = 2 Since a < R , a = 1 R. 2 At the same time, the point where the two circles touch must satisfy (a r )2 + (a r )2 = r 2 2 a = 1 r 2 2 Since a > r , a = 1 + r. 2 Equating the two expressions for a yields 2 2 1 R = 1 + r 2 2 2 2 1 2 R= r= 2 1+ 2 r= 1 2 + 2 1 2 R 2 2 1 + 1 2 2 1 2R 1 2 r = (3 2 2) R 0.1716 R 1 property of similar triangles, QC / RC = 3 / 2 , 2 3 4 = a = 2+ a2 2 3 By the Pythagorean Theorem, we have d = h 2 a 2 = 3a = 2 3 + 4 7.464 Instructors Resource Manual Section 0.3 19 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. 57. Refer to figure 15 in the text. Given ine l1 with slope m, draw ABC with vertical and horizontal sides m, 1. Line l2 is obtained from l1 by rotating it around the point A by 90 counter-clockwise. Triangle ABC is rotated into triangle AED . We read off 1 1 slope of l2 = = . m m 60. See the figure below. The angle at T is a right angle, so the Pythagorean Theorem gives ( PM + r )2 = ( PT )2 + r 2 ( PM )2 + 2rPM + r 2 = ( PT )2 + r 2 PM ( PM + 2r ) = ( PT )2 PM + 2r = PN so this gives ( PM )( PN ) = ( PT ) 2 58. 2 ( x 1)2 + ( y 1)2 = ( x 3) 2 + ( y 4)2 4( x 2 2 x + 1 + y 2 2 y + 1) = x 2 6 x + 9 + y 2 8 y + 16 3x 2 2 x + 3 y 2 = 9 + 16 4 4; 2 17 x + y2 = ; 3 3 1 17 1 2 2 2 x x+ + y = + 3 9 3 9 3x 2 2 x + 3 y 2 = 17; x 2 2 1 52 2 x + y = 3 9 B = (6)2 + (8)2 = 100 = 10 52 1 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 a 2 + b 2 = c 2 Thus, a 2 b 2 c 2 + = or 8 8 8 2 61. The lengths A, B, and C are the same as the corresponding distances between the centers of the circles: A = (2)2 + (8)2 = 68 8.2 2 1 a 1 b 1 c + = 2 2 2 2 2 2 C = (8)2 + (0)2 = 64 = 8 Each circle has radius 2, so the part of the belt around the wheels is 2(2 a ) + 2(2 b ) + 2(2 c ) = 2[3 - (a + b + c)] = 2(2 ) = 4 Since a + b + c = , the sum of the angles of a triangle. The length of the belt is 8.2 + 10 + 8 + 4 38.8 units. 2 2 1 x is the area of a semicircle with 2 2 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 a 2 + b 2 = c 2 , 3 2 3 2 3 2 a + b = c 4 4 4 3 2 x is the area of an equilateral triangle 4 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. 20 Section 0.3 62 As in Problems 50 and 61, the curved portions of the belt have total length 2 r. The lengths of the straight portions will be the same as the lengths of the sides. The belt will have length 2 r + d1 + d 2 + + d n . Instructors Resource Manual 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.

21. 63. A = 3, B = 4, C = 6 3(3) + 4(2) + (6) 7 d= = 5 (3) 2 + (4)2 64. A = 2, B = 2, C = 4 d= 2(4) 2(1) + 4) 2 (2) + (2) 2 = 14 8 = 7 2 2 65. A = 12, B = 5, C = 1 12(2) 5(1) + 1 18 d= = 13 (12) 2 + (5) 2 66. A = 2, B = 1, C = 5 d= 2(3) 1(1) 5 2 (2) + (1) 2 = 2 5 = 2 5 5 67. 2 x + 4(0) = 5 5 x= 2 d= 2 ( 5 ) + 4(0) 7 = 2 (2)2 + (4) 2 2 20 = 5 5 68. 7(0) 5 y = 1 1 y= 5 1 7(0) 5 6 7 7 74 5 d= = = 2 2 74 74 (7) + (5) 2 3 5 3 = ; m = ; passes through 1+ 2 3 5 2 + 1 3 2 1 1 , = , 2 2 2 2 1 3 1 y = x+ 2 5 2 3 4 y = x+ 5 5 69. m = Instructors Resource Manual 04 1 = 2; m = ; passes through 20 2 0+2 4+0 , = (1, 2) 2 2 1 y 2 = ( x 1) 2 1 3 y = x+ 2 2 60 1 m= = 3; m = ; passes through 42 3 2+4 0+6 , = (3, 3) 2 2 1 y 3 = ( x 3) 3 1 y = x+4 3 1 3 1 x+ = x+4 2 2 3 5 5 x= 6 2 x=3 1 3 y = (3) + = 3 2 2 center = (3, 3) 70. m = 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 ( x (4 r ))2 + ( y r )2 = r 2 . 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 x2 3 solution to ( x 4 + r ) 2 + x r = r 2 . 4 Solving for x in this equation gives 16 x = 16 r 24 r 2 + 7r 6 . There is 25 ( ) exactly one solution when r 2 + 7 r 6 = 0, that is, when r = 1 or r = 6 . The root r = 6 is extraneous. Thus, the largest circle that can be inscribed in this triangle has radius r = 1. Section 0.3 21 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. 72. The line tangent to the circle at ( a, b ) will be The slope of PS is 1 [ y1 + y4 ( y1 + y2 )] y y 2 2 = 4 . The slope of 1 x4 x2 x1 + x4 ( x1 + x2 ) ] [ 2 1 [ y3 + y4 ( y2 + y3 )] y y 2 . Thus QR is 2 = 4 1 x4 x2 [ x3 + x4 ( x2 + x3 )] 2 PS and QR are parallel. The slopes of SR and y y PQ are both 3 1 , so PQRS is a x3 x1 parallelogram. 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 0b b a r2 = ; ax + by = r 2 y = x + 0a a b b a so ax + by = r 2 has slope and is b perpendicular to the line through ( a, b ) and m= ( 0, 0 ) , so it is tangent to the circle at ( a, b ) . 73. 12a + 0b = 36 a=3 32 + b 2 = 36 b = 3 3 3x 3 3 y = 36 x 3 y = 12 3x + 3 3 y = 36 x + 3 y = 12 74. Use the formula given for problems 63-66, for ( x, y ) = ( 0, 0 ) . 77. x 2 + ( y 6) 2 = 25; passes through (3, 2) tangent line: 3x 4y = 1 The dirt hits the wall at y = 8. A = m, B = 1, C = B b;(0, 0) d= m(0) 1(0) + B b m2 + (1) 2 = Bb m2 + 1 75. The midpoint of the side from (0, 0) to (a, 0) is 0+a 0+0 a , = , 0 2 2 2 The midpoint of the side from (0, 0) to (b, c) is 0+b 0+c b c , = , 2 2 2 2 c0 c m1 = = ba ba c 0 c m2 = 2 = ; m1 = m2 ba ba 2 0.4 Concepts Review 1. y-axis 2. ( 4, 2 ) 3. 8; 2, 1, 4 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 2 76. See the figure below. The midpoints of the sides are x2 + x3 y2 + y3 x + x y + y2 P 1 2 , 1 , , , Q 2 2 2 2 x + x y + y4 R 3 4 , 3 , and 2 2 x + x y + y4 S 1 4 , 1 . 2 2 22 Section 0.4 Instructors Resource Manual 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.

23. 2. x = y 2 + 1; y -intercepts = 1,1; x-intercept = 1 . Symmetric with respect to the x-axis. 3. x = 4y2 1; x-intercept = 1 Symmetric with respect to the x-axis 4. y = 4 x 2 1; y -intercept = 1 1 1 y = (2 x + 1)(2 x 1); x-intercepts = , 2 2 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. y = x 2 2 x; y -intercept = 0 y = x(2 x); x-intercepts = 0, 2 7 7. 7x2 + 3y = 0; 3y = 7x2; y = x 2 3 x-intercept = 0, y-intercept = 0 Symmetric with respect to the y-axis 8. y = 3x 2 2 x + 2; y -intercept = 2 Instructors Resource Manual Section 0.4 23 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. 9. x2 + y2 = 4 x-intercepts = -2, 2; y-intercepts = -2, 2 Symmetric with respect to the x-axis, y-axis, and origin 10. 3x 2 + 4 y 2 = 12; y-intercepts = 3, 3 x-intercepts = 2, 2 Symmetric with respect to the x-axis, y-axis, and origin 11. y = x2 2x + 2: y-intercept = 2 2 4+8 22 3 x-intercepts = = = 1 3 2 2 24 Section 0.4 12. 4 x 2 + 3 y 2 = 12; y -intercepts = 2, 2 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. x 2 + ( y 1)2 = 9; y -intercepts = 2, 4 x-intercepts = 2 2, 2 2 Symmetric with respect to the y-axis Instructors Resource Manual 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.

25. 15. 4(x 1)2 + y2 = 36; y-intercepts = 32 = 4 2 x-intercepts = 2, 4 Symmetric with respect to the x-axis 18. x 4 + y 4 = 1; y -intercepts = 1,1 x-intercepts = 1,1 Symmetric with respect to the x-axis, y-axis, and origin 16. x 2 4 x + 3 y 2 = 2 19. x4 + y4 = 16; y-intercepts = 2, 2 x-intercepts = 2, 2 Symmetric with respect to the y-axis, x-axis and origin 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 Instructors Resource Manual 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 Section 0.4 25 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. 21. y = 1 2 ; y-intercept = 1 x +1 Symmetric with respect to the y-axis 2 24. 4 ( x 5 ) + 9( y + 2) 2 = 36; x-intercept = 5 25. y = (x 1)(x 2)(x 3); y-intercept = 6 x-intercepts = 1, 2, 3 22. y = x ; y -intercept = 0 x +1 x-intercept = 0 Symmetric with respect to the origin 2 26. y = x2(x 1)(x 2); y-intercept = 0 x-intercepts = 0, 1, 2 23. 2 x 2 4 x + 3 y 2 + 12 y = 2 2( x 2 2 x + 1) + 3( y 2 + 4 y + 4) = 2 + 2 + 12 2( x 1)2 + 3( y + 2)2 = 12 y-intercepts = 2 30 3 x-intercept = 1 27. y = x 2 ( x 1)2 ; y-intercept = 0 x-intercepts = 0, 1 26 Section 0.4 Instructors Resource Manual 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.

27. 28. y = x 4 ( x 1)4 ( x + 1)4 ; y -intercept = 0 x-intercepts = 1, 0,1 Symmetric with respect to the y-axis Intersection points: (0, 1) and (3, 4) 32. 2 x + 3 = ( x 1) 2 29. x + y = 1; y-intercepts = 1, 1; x-intercepts = 1, 1 Symmetric with respect to the x-axis, y-axis and origin 2 x + 3 = x2 + 2 x 1 x2 + 4 = 0 No points of intersection 33. 2 x + 3 = 2( x 4)2 30. x + y = 4; y-intercepts = 4, 4; x-intercepts = 4, 4 Symmetric with respect to the x-axis, y-axis and origin 2 x + 3 = 2 x 2 + 16 x 32 2 x 2 18 x + 35 = 0 x= 18 324 280 18 2 11 9 11 = = ; 4 4 2 9 11 Intersection points: 2 , 6 + 11 , 9 + 11 2 , 6 11 31. x + 1 = ( x + 1)2 x + 1 = x2 + 2 x + 1 x 2 + 3x = 0 x( x + 3) = 0 x = 0, 3 Instructors Resource Manual Section 0.4 27 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. 37. y = 3x + 1 34. 2 x + 3 = 3x 2 3 x + 12 x 2 + 2 x + (3 x + 1) 2 = 15 3x 2 x + 9 = 0 No points of intersection x 2 + 2 x + 9 x 2 + 6 x + 1 = 15 10 x 2 + 8 x 14 = 0 2(5 x 2 + 4 x 7) = 0 2 39 1.65, 0.85 5 Intersection points: 2 39 1 3 39 , and 5 5 2 + 39 1 + 3 39 , 5 5 [ or roughly (1.65, 3.95) and (0.85, 3.55) ] x= 35. x 2 + x 2 = 4 x2 = 2 x= 2 ( )( Intersection points: 2, 2 , 2, 2 ) 38. x 2 + (4 x + 3) 2 = 81 x 2 + 16 x 2 + 24 x + 9 = 81 17 x 2 + 24 x 72 = 0 12 38 2.88, 1.47 17 Intersection points: 12 38 3 24 38 , and 17 17 12 + 38 3 + 24 38 , 17 17 [ or roughly ( 2.88, 8.52 ) , (1.47,8.88 ) ] x= 36. 2 x 2 + 3( x 1)2 = 12 2 x 2 + 3 x 2 6 x + 3 = 12 5x2 6 x 9 = 0 6 36 + 180 6 6 6 3 3 6 = = 10 10 5 Intersection points: 3 3 6 2 3 6 3 + 3 6 2 + 3 6 , 5 , 5 , 5 5 x= 28 Section 0.4 Instructors Resource Manual 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.

29. 39. a. y = x 2 ; (2) ( b. ax3 + bx 2 + cx + d , with a > 0 : (1) c. ax3 + bx 2 + cx + d , with a < 0 : (3) d. y = ax3 , with a > 0 : (4) 40. x 2 + y 2 = 13;(2, 3), (2,3), (2, 3), (2,3) 2 2 2 2 2 d3 = (2 2) + (3 + 3) = 6 Three such distances. ( ) )( )( ) ( ) d1 = (2 2) 2 + 1 + 21 1 + 13 21 13 ) 21 + 13 ) f (2u ) = 3(2u ) 2 = 12u 2 ; f ( x + h) = 3( x + h)2 ) 2 = f (0) = 1 02 = 1 f (k ) = 1 k 2 f (5) = 1 (5) 2 = 24 f. 1 15 1 1 f =1 =1 = 16 16 4 4 g. f (1 + h ) = 1 (1 + h ) = 2h h 2 h. f (1 + h ) f (1) = 2h h 2 0 = 2h h 2 i. ( 2 21) 2 c. e. ) f (2) = 1 (2)2 = 3 f ( 2 + h ) f ( 2) = 1 ( 2 + h ) + 3 2 2 = 2 21 9.17 d 4 = (2 2)2 + 1 21 (1 + 13) ( f (1) = 1 12 = 0 d. ) 2 2 ( = 16 + 21 13 ) 2 2 2 2 = 4h h 2 = 50 + 2 273 9.11 ( ) d5 = (2 2)2 + 1 21 1 13 ( 2 b. d3 = (2 + 2)2 + 1 + 21 1 21 21 + 21 2. 1. a. = 50 + 2 273 9.11 = 16 + ( 2 13 ) 0.5 Concepts Review 2 ( ( = Problem Set 0.5 d 2 = (2 2)2 + 1 + 21 1 13 = 0+ 2 = 2 13 7.21 Four such distances ( d 2 = d 4 and d1 = d5 ). 2 = 50 2 273 4.12 ( ) 4. even; odd; y-axis; origin 21 , 2, 1 + 13 , 2, 1 13 = 16 + 13 + 13 3. asymptote 41. x2 + 2x + y2 2y = 20; 2, 1 + 21 , ( ( 1. domain; range d 2 = (2 + 2) + (3 3) = 52 = 2 13 = 16 + = 0+ 2 2 d1 = (2 + 2) + (3 + 3) = 4 ( 2, 1 ) d6 = (2 2)2 + 1 + 13 1 13 13 21 ) 2 2 2. a. b. F ( 2) = ( 2)3 + 3( 2) = 2 2 + 3 2 =5 2 = 50 2 273 4.12 3 c. Instructors Resource Manual F (1) = 13 + 3 1 = 4 1 1 1 1 3 49 F = + 3 = + = 4 4 4 64 4 64 Section 0.5 29 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. d. F (1 + h ) = (1 + h ) + 3 (1 + h ) 3 f. ( x 2 + x) = = 1 + 3h + 3h 2 + h3 + 3 + 3h = 4 + 6h + 3h 2 + h3 e. F (1 + h ) 1 = 3 + 6h + 3h + h f. = F ( 2 + h) F ( 2) 2 3 2 = 15h + 6h + h 3. a. b. G (0) = d. e. f. 4. a. b. 0.25 3 1 f ( x) = c. f (3 + 2) = 3 1 = 1 0 1 3 G( y ) = G ( x) = c. 1 1 = x 1 x +1 7. a. 2 1 x = 1 1 x2 1 2 0.841 3.293 (12.26) 2 + 9 12.26 3 1.199 1 t 1 2 1 = 2 d. (u + 1) = e. ( x2 ) = + (1) 2 = 2 1 2 = x2 = x2 + y2 = 1 c. x = 2 y +1 x2 = 2 y + 1 1.06 (u + 1) + (u + 1) 2 ( x2 ) + ( x2 )2 Section 0.5 t u +1 ; undefined b. xy + y + x = 1 y(x + 1) = 1 x 1 x 1 x y= ; f ( x) = x +1 x +1 t2 t 3 4 1 2 3 3 y = 1 x 2 ; not a function =2 t + ( t ) 2 ( 3)2 + 9 f ( 3) = y 2 = 1 x 2 1 x2 1 + 12 (t ) = is not y 2 1 1 G = x2 (1) = = 3+ 2 3 0.25 0.79 3 b. f(12.26) = 1 2.75 2.658 (0.79) 2 + 9 f(0.79) = 1 G (1.01) = = 100 1.01 1 2 1 = 1 =2 1 G (0.999) = = 1000 0.999 1 c. 30 1 f (0.25) = b. 6. a. c. x2 + x defined = ( 2 + h ) + 3 ( 2 + h ) 23 3 ( 2 ) = 8 + 12h + 6h 2 + h3 + 6 + 3h 14 x2 + x x 4 + 2 x3 + 2 x 2 + x 3 5. a. ( x 2 + x) + ( x 2 + x) 2 = y= u 2 + 3u + 2 x2 + x4 x u +1 d. x2 1 x2 1 ; f ( x) = 2 2 y y+1 xy + x = y x = y xy x = y(1 x) x x ; f ( x) = y= 1 x 1 x x= Instructors Resource Manual 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.

31. 8. The graphs on the left are not graphs of functions, the graphs on the right are graphs of functions. 9. f (a + h) f (a) [2(a + h) 2 1] (2a 2 1) = h h 4ah + 2h 2 = = 4a + 2h h x 2 9 0; x 2 9; x 3 Domain: {x d. F (a + h) F (a ) 4(a + h) 4a = h h 4a3 + 12a 2 h + 12ah 2 + 4h3 4a3 h 12a 2 h + 12ah 2 + 4h3 h 14. a. b. = 12a 2 + 12ah + 4h 2 f ( x) = 4 x2 = 4 x2 ( x 3)( x + 2) x2 x 6 Domain: {x : x 2, 3} G ( y ) = ( y + 1) 1 Domain: { y = x 4 x + hx 2h + 4 h 3h = 2 h( x 4 x + hx 2h + 4) 3 = 2 x 4 x + hx 2h + 4 : y > 1} c. (u ) = 2u + 3 (all real numbers) Domain: d. F (t ) = t 2 / 3 4 (all real numbers) Domain: 2 a+h a + h+ 4 : y 5} 1 0; y > 1 y +1 3 3 g ( x + h) g ( x) x + h 2 x 2 = h h 3x 6 3x 3h + 6 G ( a + h) G ( a ) = h H ( y ) = 625 y 4 Domain: { y = 12. : x 3} 3 = 11. ( x) = x 2 9 625 y 4 0; 625 y 4 ; y 5 3 10. c. 15. f(x) = 4; f(x) = 4; even function a a+4 h 2 a + 4a + ah + 4h a 2 ah 4a a 2 + 8a + ah + 4h + 16 h 4h = = = 13. a. h(a 2 + 8a + ah + 4h + 16) 4 a 2 + 8a + ah + 4h + 16 F ( z) = 2 z + 3 2z + 3 0; z Domain: z b. 16. f(x) = 3x; f(x) = 3x; odd function g (v ) = 3 2 3 :z 2 1 4v 1 4v 1 = 0; v = Domain: v 1 4 1 :v 4 Instructors Resource Manual Section 0.5 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.

32. 17. F(x) = 2x + 1; F(x) = 2x + 1; neither 20. g (u ) = 18. F ( x) = 3x 2; F ( x) = 3x 2; neither 21. g ( x) = 19. g ( x) = 3x 2 + 2 x 1; g ( x) = 3 x 2 2 x 1 ; neither 32 Section 0.5 22. ( z ) = u3 u3 ; g ( u ) = ; odd function 8 8 x 2 x 1 ; g ( x) = x 2 x 1 ; odd 2z +1 2 z + 1 ; ( z ) = ; neither z 1 z 1 Instructors Resource Manual 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.

33. 23. f ( w) = w 1; f ( w) = w 1; neither 2 2 24. h( x ) = x + 4; h( x) = x + 4; even function 26. F (t ) = t + 3 ; F ( t ) = t + 3 ; neither 27. g ( x) = x x ; g ( x ) = ; neither 2 2 28. G ( x) = 2 x 1 ; G ( x) = 2 x + 1 ; neither 25. f ( x) = 2 x ; f ( x) = 2 x = 2 x ; even function 1 if t 0 29. g (t ) = t + 1 if 0 < t < 2 2 t 1 if t 2 Instructors Resource Manual neither Section 0.5 33 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. x 2 + 4 if x 1 30. h( x ) = if x > 1 3x neither 35. Let y denote the length of the other leg. Then x2 + y 2 = h2 y 2 = h2 x 2 y = h2 x 2 L ( x ) = h2 x 2 36. The area is 1 1 A ( x ) = base height = x h 2 x 2 2 2 37. a. 31. T(x) = 5000 + 805x Domain: {x integers: 0 x 100} T ( x) 5000 u ( x) = = + 805 x x Domain: {x integers: 0 < x 100} P ( x) = 6 x (400 + 5 x( x 4)) 32. a. = 6 x 400 5 x( x 4) E(x) = 24 + 0.40x b. 120 = 24 + 0.40x 0.40x = 96; x = 240 mi 38. The volume of the cylinder is r 2 h, where h is the height of the cylinder. From the figure, 2 2 2 h 2 h 2 = 3r ; r + = (2r ) ; 2 4 h = 12r 2 = 2r 3. V (r ) = r 2 (2r 3) = 2r 3 3 P(200) 190 ; P (1000 ) 610 b. c. ABC breaks even when P(x) = 0; 6 x 400 5 x( x 4) = 0; x 390 33. E ( x) = x x 2 y 0.5 39. The area of the two semicircular ends is 0.5 1 x 0.5 1 exceeds its square by the maximum amount. 2 34. Each side has length p . The height of the 3 d 2 . 4 1 d . 2 2 d 2 d d 2 1 d d A(d ) = +d + = 4 4 2 2 The length of each parallel side is 2d d 2 4 Since the track is one mile long, d < 1, so 1 1 d < . Domain: d : 0 < d < = 3p . 6 1 p 3p 3 p2 A( p ) = = 2 3 6 36 triangle is 34 Section 0.5 Instructors Resource Manual 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.

35. 40. a. 1 3 A(1) = 1(1) + (1)(2 1) = 2 2 42. a. f(x + y) = 2(x + y) = 2x + 2y = f(x) + f(y) b. f ( x + y ) = ( x + y )2 = x 2 + 2 xy + y 2 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) b. 1 A(2) = 2(1) + (2)(3 1) = 4 2 c. A(0) = 0 d. 1 1 A(c) = c(1) + (c)(c + 1 1) = c 2 + c 2 2 e. 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, p = p 1 = 1 + 1 + ... + 1, so f(p) = f(1 + 1 + ... + 1) = f(1) + f(1) + ... + f(1) = pf(1) = pm. 1 1 1 1 1 = p = + + ... + , so p p p p 1 1 1 m = f (1) = f + + ... + p p p 1 1 1 1 = f + f + ... + f = pf , p p p p 1 m hence f = . Any rational number can p p be written as f. Domain: {c Range: { y 41. a. b. : c 0} : y 0} B (0) = 0 1 1 1 1 1 B = B (1) = = 2 6 12 2 2 p with p, q in N. q 1 1 1 p 1 = p = + + ... + , q q q q q p 1 1 1 so f = f + + ... + q q q q 1 1 1 = f + f + ... + f q q q 1 m p = pf = p = m q q q c. Instructors Resource Manual Section 0.5 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.

36. 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 46. a. x f(x) 4 6.1902 3 0.4118 2 13.7651 1 9.9579 0 0 1 7.3369 2 17.7388 if 18 < t 27 3 0.4521 if 27 < t 36 4 4.4378 b. 902 + (10t 90)2 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 902 + (270 10t ) 2 feet from home plate. The player is 10t 270 feet from third base when 27 t 36, thus he is 90 (10t 270) = 360 10t feet from home plate. a. b. 45. a. f(1.38) 76.8204 f(4.12) 6.7508 10t 902 + (10t 90)2 s= 902 + (270 10t ) 2 360 10t if 0 t 9 180 180 10t s = 902 + (10t 90) 2 2 2 90 + (270 10t ) if 0 t 9 if 9 < t 18 or 27 < t 36 47. if 9 < t 18 if 18 < t 27 f(1.38) 0.2994 f(4.12) 3.6852 x f(x) 4 4.05 3 3.1538 a. 2 2.375 b. f(x) = 0 when x 1.1, 1.7, 4.3 f(x) 0 on [1.1, 1.7] [4.3, 5] 1 1.8 0 1.25 1 0.2 2 1.125 3 2.3846 4 b. 3.55 48. a. 36 Section 0.5 Range: {y R: 22 y 13} f(x) = g(x) at x 0.6, 3.0, 4.6 Instructors Resource Manual 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.

37. b. f(x) g(x) on [-0.6, 3.0] [4.6, 5] c. f ( x) g ( x) = x3 5 x 2 + x + 8 2 x 2 + 8 x + 1 = x3 7 x 2 + 9 x + 9 b. On 6, 3) , g increases from 13 g ( 6 ) = 4.3333 to . On ( 2, 6 , g 3 26 2.8889 . On decreased from to 9 ( 3, 2 ) the maximum occurs around Largest value f (2) g (2) = 45 x = 0.1451 with value 0.6748 . Thus, the range is ( , 0.6748 2.8889, ) . 49. c. x 2 + x 6 = 0; (x + 3)(x 2) = 0 Vertical asymptotes at x = 3, x = 2 d. Horizontal asymptote at y = 3 0.6 Concepts Review 1. ( x 2 + 1)3 a. x-intercept: 3x 4 = 0; x = 4 3 3 0 4 2 = y-intercept: 2 0 +06 3 2. f(g(x)) 3. 2; left 4. a quotient of two polynomial functions b. c. x 2 + x 6 = 0; (x + 3)(x 2) = 0 Vertical asymptotes at x = 3, x = 2 Problem Set 0.6 1. a. ( f + g )(2) = (2 + 3) + 22 = 9 d. Horizontal asymptote at y = 0 b. ( f g )(1) = f (12 ) = 1 + 3 = 4 e. ( g f )(1) = g (1 + 3) = 4 2 = 16 f. ( g f )(8) = g (8 + 3) = (5) 2 = 25 2. a. x-intercepts: 3x 2 4 = 0; x = ( g f )(3) = d. a. ( f g )(0) = (0 + 3)(02 ) = 0 c. 50. 4 2 3 = 3 3 32 9 3 = = 3+3 6 2 ( f g )(2) = (22 + 2) 12 + 1 2 b. ( f g )(1) = c. 2 y-intercept: 3 1 2 1 g 2 (3) = = = 3 + 3 3 9 2 1+ 3 2 Instructors Resource Manual = 2 2 28 =6 = 2+3 5 5 2 4 =4 2 Section 0.6 37 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. 2 d. e. f. 3. a. (f 2 1 1 3 g )(1) = f = + = 1+ 3 2 2 4 5. = x2 + 2 x 3 2 2 = 2+3 5 2 ( g g )(3) = g = 3+3 2 2 3 = 10 = 1 +3 3 5 3 6. g 3 (x) = (x 2 +1) 3 = (x 4 + 2x 2 + 1)(x 2 + 1) 6 1 t b. c. ( )(r ) = (r + 1) = d. 3 ( z ) = ( z 3 + 1) 3 e. ( )(5t) = [(5t) 3 +1] 1 3 b. 2 = g[(x 2 + 1) 2 + 1] = g( x 4 + 2x 2 + 2) = ( x 4 + 2x 2 + 2) 2 + 1 = x 8 + 4x 6 + 8x 4 + 8x 2 + 5 7. g(3.141) 1.188 8. g(2.03) 0.000205 r 3 +1 1/ 3 9. g 2 ( ) g ( ) 4.789 1 5t 1/ 3 2 = (11 7 ) 11 7 10. [ g 3 () g ()]1/ 3 = [(6 11)3 (6 11)]1/ 3 7.807 1 5t 1 (( ) )(t ) = ( ) t 3 1 1 1 = + 1 1 = 3 + 1 t t t t 11. a. b. 12. a. 4. a. 4 = x + 3x + 3x + 1 ( g g g )( x) = ( g g )( x 2 + 1) 3 f. f ) ( x) = g x 2 4 = 1 + x 2 4 = 1 + x2 4 1 1 1 ( )(r ) = = + 1 = 3 + 1 r r r = 125t3 + 1 2 (g ( g f )(1) = g (12 + 1) = ( + )(t ) = t 3 + 1 + g ) ( x) = f ( 1 + x ) = 1 + x 4 (f g ( x) = x , f ( x) = x + 7 g (x) = x15 , f (x) = x 2 + x 2 f ( x) = x 2 x2 1 x Domain: ( , 1] [1, ) 3 , g ( x) = x 2 + x + 1 ( f g )( x) = b. 4 2 f 4 ( x) + g 4 ( x) = x 2 1 + x 16 = ( x 2 1)2 + x4 Domain: ( , 0 ) (0, ) 2 c. 2 2 g )( x) = f = 1 = x x Domain: [2, 0) (0, 2] d. ( g f )( x) = g x 2 1 = (f 4 f ( x) = 13. p = f 1 , g (x) = x 3 + 3 x x g h if f(x) =1/ x , g ( x) = x , h( x ) = x 2 + 1 p= f g h if f ( x) = 1/ x , g(x) = x + 1, h( x) = x 2 4 1 x2 14. p = f g h l if f ( x) = 1/ x , g ( x) = x , 2 h(x) = x + 1, l( x) = x 2 2 x 1 Domain: ( , 1) (1, ) 38 Section 0.6 Instructors Resource Manual 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.

39. 15. Translate the graph of g ( x) = x to the right 2 units and down 3 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. 16. Translate the graph of h( x) = x to the left 3 units and down 4 units. 19. ( f + g )( x) = x3 + x 2 20. ( f + g )( x) = x + x Instructors Resource Manual Section 0.6 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.

40. 21. F (t ) = t t 24. a. t 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. 22. G (t ) = t t F ( x ) F ( x) F ( x ) + F ( x) is odd and is 2 2 even. F ( x ) F ( x) F ( x) + F ( x) 2 F ( x) + = = F ( x) 2 2 2 25. Not every polynomial of even degree is an even function. For example f ( x) = x 2 + x is neither even nor odd. Not every polynomial of odd degree is an odd function. For example g ( x) = x 3 + x 2 is neither even nor odd. 26. a. Neither b. c. 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; ( f g )( x) = [ f ( x)][ g ( x)] = [ f ( x)][ g ( x)] = [ f ( x)][ g ( x)] = ( f g )( x) if f and g are both odd functions. e. 40 PF e. Even; (f + g)(x) = f(x) + g(x) = f(x) + g(x) = (f + g)(x) if f and g are both even functions. RF d. 23. a. PF RF f. Neither 27. a. P = 29 3(2 + t ) + (2 + t )2 = t + t + 27 b. When t = 15, P = 15 + 15 + 27 6.773 28. R(t) = (120 + 2t + 3t2 )(6000 + 700t ) = 2100 t3 + 19, 400t 2 + 96, 000t + 720, 000 400t 29. D(t ) = 2 2 (400t ) + [300(t 1)] if 0 < t < 1 if t 1 if 0 < t < 1 400t D(t ) = 2 250, 000t 180, 000t + 90, 000 if t 1 30. D(2.5) 1097 mi Odd; ( f g )( x) = [ f ( x)][ g ( x)] = [ f ( x)][ g ( x)] = [ f ( x)][ g ( x)] = ( f g )( x) if f is an even function and g is an odd function. Section 0.6 Instructors Resource Manual 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.

41. 31. (ax+ab ) + b cx (ax+ab ) a cx 36. ax + b a f ( f ( x)) = f = cx a c = a 2 x + ab + bcx ab = x(a 2 + bc) =x acx + bc acx + a 2 a 2 + bc 2 If a + bc = 0 , f(f(x)) is undefined, while if x = a , f(x) is undefined. c x 3 3 x 3 x +1 ( f ( f ( x))) = f f 32. f = f x 3 x +1 +1 x +1 x 3 3x 3 2 x 6 x 3 = f = f = f x 3 + x +1 2x 2 x 1 x3 3 x 3 3x + 3 4 x 1 = xx 3 = = =x x 3+ x 1 4 +1 x 1 If x = 1, f(x) is undefined, while if x = 1, f(f(x)) is undefined. 33. a. b. 1 f = x 34. a. b. 1 x 1 = 1 1 x 1 f1 ( f 2 ( x)) = ; x f1 ( f3 ( x)) = 1 x; 1 ; 1 x x 1 f1 ( f5 ( x)) = ; x x ; f1 ( f6 ( x)) = x 1 f1 ( f 4 ( x)) = f 2 ( f1 ( x)) = f 2 ( f 2 ( x)) = x x 1 x 1 x 1 = x; 1 ; 1 x 1 f 2 ( f 4 ( x)) = = 1 x; f 2 ( f 6 ( x)) = x =x x x +1 = 1 x x 1 x ; x 1 = 1 x 1 x x 1 ; x f3 ( f1 ( x)) = 1 x; 1 x 1 f f ( x) = f x = =1x 1/ x 1 / x 1 x 1 x x 1 1 x = x 1 x 1 x 1 x 1 ; = x x f3 ( f3 ( x)) = 1 (1 x) = x; f3 ( f 2 ( x)) = 1 1 x = ; 1 x x 1 x 1 1 = ; f3 ( f5 ( x)) = 1 x x x 1 f3 ( f 6 ( x)) = 1 = ; x 1 1 x f3 ( f 4 ( x)) = 1 1 = xx f ( f ( x)) = f ( x /( x 1)) = = 1 x f 2 ( f3 ( x)) = f 2 ( f 5 ( x )) = f (1 / x) = 1 ; x 1 1 1 x x f ( f ( x)) = f = x 1 = c. 1 x f1 ( f1 ( x)) = x; x /( x 1) x x( x 1) + 1 x 35. ( f1 ( f 2 f3 ))( x) = f1 (( f 2 f3 )( x)) = f1 ( f 2 ( f3 ( x))) (( f1 f 2 ) f3 )( x) = ( f1 f 2 )( f3 ( x)) = f1 ( f 2 ( f3 ( x))) = ( f1 ( f 2 f3 ))( x) x 1 x 1 1 ; 1 x 1 x ; f 4 ( f 2 ( x)) = = 1 1 x x 1 f 4 ( f1 ( x)) = f 4 ( f3 ( x)) = f 4 ( f 4 ( x)) = f 4 ( f5 ( x)) = f 4 ( f 6 ( x)) = Instructors Resource Manual 1 1 = ; 1 (1 x) x 1 1 1 1 x 1 1 x 1 x = 1 x x 1 = ; x 1 x 1 = x = x; x ( x 1) 1 x 1 = = 1 x; x 1 x 1 x 1 x Section 0.6 41 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. a. x 1 f5 ( f1 ( x)) = ; x 1 1 f5 ( f 2 ( x)) = x = 1 x; b. = f 5 ( f 5 ( x)) = x 1 1 x x 1 x f 5 ( f 6 ( x)) = x 1 x 1 x x 1 = 1 (1 x) = x; 1 f3 f4 f5 f6 f3 ) f4 ) f5 ) f6 ) = ( f 4 f 4 ) ( f5 f 6 ) = f5 f 2 = f3 c. x ( x 1) 1 = ; x x If F f 6 = f 1 , then F = f 6 . d. If G f 3 G = f5. e. 1 ; 1 x f 6 = f 1 , then G f 4 = f 1 so If f 2 f 5 H = f 5 , then f 6 H = f 5 so H = f3. 37. f 6 ( f3 ( x)) = x 1 1 x ; = 1 x 1 x f 6 ( f 4 ( x)) = 1 1 x 1 1 1 x = 1 1 = ; 1 (1 x) x f 6 ( f 5 ( x)) = x 1 x x 1 1 x = x 1 = 1 x; x 1 x f 6 ( f 6 ( x )) = x x 1 x 1 x 1 = x =x x ( x 1) 38. 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 42 f1 f 2 = (((( f 2 x ; x 1 = f3 ) f3 ) f3 ) f3 ) = ((((( f 1 f 2 ) f 3 ) f 4 ) f 5 ) f 6 ) x 1 x 1 = = ; x 1 1 x 1 f3 = f1 f 3 = f 3 1 1 1 x 1 1 x 1 x f3 = (( f3 f3 ) f3 ) f 5 ( f 4 ( x)) = f 6 ( f 2 ( x)) = f3 = ((( f1 f3 ) f3 ) f3 ) 1 x 1 x f5 ( f3 ( x)) = = ; 1 x x 1 1 x f3 = (((( f 3 1 x f 6 ( f1 ( x)) = f3 f6 f4 f5 f2 f3 f1 Section 0.6 39. Instructors Resource Manual 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.

43. Problem Set 0.7 40. 1. a. b. 60 = 3 180 d. 4 240 = 180 3 e. 37 370 = 18 180 f. b. 45 = 180 4 c. 41. a. 30 = 180 6 10 = 180 18 2. a. 7 180 = 210 6 180 = 135 b. c. 1 180 = 60 3 d. 4 180 = 240 3 e. f. c. 3 4 3 180 = 30 18 42. 3. a. 35 180 = 350 18 33.3 0.5812 180 b. c. 66.6 1.1624 180 d. 240.11 4.1907 180 e. 0.7 Concepts Review 46 0.8029 180 369 6.4403 180 f. 11 0.1920 180 1. ( , ); [1, 1] 2. 2 ; 2 ; 3. odd; even 4 x 4. r = (4) + 3 = 5; cos = = 5 r 2 2 Instructors Resource Manual Section 0.7 43 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. 4. a. 180 3.141 180 Thus b. 180 6. 28 359. 8 c. 180 286.5 5. 00 d. 180 0. 001 0 .057 e. 180 0.1 5.73 f. 180 36. 0 2062.6 5. a. 56. 4 tan34. 2 68.37 sin 34.1 b. cos = cos = and by the Pythagorean Identity, sin 3 = 3 . 2 sin (0.361) 0.3532 6. a. b. 7. a. 234.1sin(1.56) 248.3 cos(0.34 ) sin 2 (2.51) + cos(0.51) 1.2828 56. 3 tan34. 2 46.097 sin 56.1 Referring to Figure 2, it is clear that sin 3 b. sin 35 0. 0789 sin 26 + cos 26 Identity, cos 2 = 1 sin 2 9. a. =1 2 3 1 = 1 = . 6 4 2 1 = 1 cos() b. sec() = 1 3 sec = = 2 4 cos 3 4 d. Section 0.7 2 sin 6 = 3 tan = 3 6 cos 6 c. 6 = 0 . The rest of the values are 2 obtained using the same kind of reasoning in the second quadrant. and cos 8. Referring to Figure 2, it is clear that sin 0 = 0 and cos 0 = 1 . If the angle is / 6 , then the triangle in the figure below is 1 1 equilateral. Thus, PQ = OP = . This 2 2 1 implies that sin = . By the Pythagorean 6 2 44 tan (0.452) 0.4855 d. 6 3 . The results 2 = 2 were derived in the text. 4 4 2 If the angle is / 3 then the triangle in the 1 figure below is equilateral. Thus cos = 3 2 sin 5.34 tan 21.3 0.8845 sin 3.1+ cot 23.5 c. 1 csc = =1 2 sin ( ) (2) Instructors Resource Manual 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.

45. e. f. 10. a. b. ( ) ( ) b. cos 3t = cos(2t + t ) = cos 2t cos t sin 2t sin t cos 4 cot = =1 4 sin 4 = (2 cos 2 t 1) cos t 2sin 2 t cos t = 2 cos3 t cos t 2(1 cos 2 t ) cos t ( ) ( ) sin 4 tan = = 1 4 cos 4 ( ) ( ) = 2 cos3 t cos t 2 cos t + 2 cos3 t = 4 cos3 t 3cos t c. sin 3 tan = = 3 3 cos 3 = 2(2sin x cos x)(2 cos 2 x 1) = 2(4sin x cos3 x 2sin x cos x) = 8sin x cos3 x 4sin x cos x 1 sec = =2 3 cos 3 ( ) d. ( ) ( ) c. 3 cos 3 cot = = 3 3 sin 3 d. 1 csc = = 2 4 sin e. 3 sin 6 tan = = 6 cos 3 6 f. 1 cos = 3 2 13. a. b. (4) ( ) ( ) c. d. 11. a. (1 + sin z )(1 sin z ) = 1 sin 2 z 1 = cos 2 z = sin 4 x = sin[2(2 x)] = 2sin 2 x cos 2 x (1 + cos )(1 cos ) = 1 cos 2 = sin 2 sin u cos u + = sin 2 u + cos 2 u = 1 csc u sec u (1 cos 2 x)(1 + cot 2 x) = (sin 2 x)(csc2 x) 2 1 = sin x 2 = 1 sin x 1 sin t (csc t sin t ) = sin t sin t sin t 2 2 = 1 sin t = cos t 1 csc 2 t csc 2 t = cos 2 t = 2 sec z b. (sec t 1)(sec t + 1) = sec 2 t 1 = tan 2 t c. sec t sin t tan t = = d. 12. a. = 14. a. cot 2 t csc 2 t = cos 2 t sin 2 t 1 sin 2 t 1 sec 2 t y = sin 2x 1 sin 2 t cos t cos t 1 sin 2 t cos 2 t = = cos t cos t cos t sec2 t 1 sec 2 t sin 2 v + = tan 2 t sec 2 t 1 2 = sin 2 t cos 2 t 1 cos 2 t = sin 2 t = sin 2 v + cos 2 v = 1 sec v Instructors Resource Manual Section 0.7 45 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. b. y = 2 sin t b. y = 2 cos t c. y = cos x 4 c. y = cos 3t d. y = sec t d. y = cos t + 3 15. a. y = csc t 46 Section 0.7 x 2 Period = 4 , amplitude = 3 16. y = 3 cos Instructors Resource Manual 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.

47. 17. y = 2 sin 2x Period = , amplitude = 2 21. y = 21 + 7 sin( 2 x + 3) Period = , amplitude = 7, shift: 21 units up, 3 units left 2 22. y = 3cos x 1 2 18. y = tan x Period = Period = 2 , amplitude = 3, shifts: units 2 right and 1 unit down. 19. y = 2 + 1 cot(2 x) 6 Period = 2 , shift: 2 units up 23. y = tan 2 x 3 units right Period = , shift: 6 2 20. y = 3 + sec( x ) Period = 2 , shift: 3 units up, units right Instructors Resource Manual Section 0.7 47 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. 24. a. and g.: y = sin x + = cos x = cos( x) 2 b. and e.: y = cos x + = sin( x + ) 2 = sin( x ) c. and f.: y = cos x = sin x 2 = sin( x + ) d. and h.: y = sin x = cos( x + ) 2 = cos( x ) t sin (t) = t sin t; even 25. a. b. c. 1 csc( t ) = = csc t; odd sin( t ) ( ) = 32. a. sin(cos(t)) = sin(cos t); even f. x + sin(x) = x sin x = (x + sin x); odd cot(t) + sin(t) = cot t sin t = (cot t + sin t); odd 26. a. b. sin 3 (t ) = sin 3 t ; odd c. sec( t) = 1 = sec t; even cos(t ) sin 4 ( t ) = sin 4 t ; even d. e. cos(sin(t)) = cos(sin t) = cos(sin t); even f. ( x )2 + sin( x ) = x 2 sin x; neither 2 27. cos 2 1 1 = cos = = 3 3 4 2 2 28. sin 2 2 2 1 1 = sin = = 6 6 4 2 3 2 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( x y ) = 2 e. 3 2 2 4 = sin(t ) = sin t = sin t ; even d. 1 cos 4 1 2 1 cos 2 8 31. sin = = = 8 2 2 2 2 sin ( t ) = sin t ; even 2 () 1 + cos 6 1 + 2 1 + cos 2 12 = = = 30. cos 12 2 2 2 2+ 3 = 4 2 tan x + tan( y ) 1 tan x tan( y ) tan x tan y 1 + tan x tan y tan t + tan tan t + 0 = 1 tan t tan 1 (tan t )(0) = tan t 33. tan(t + ) = 34. cos( x ) = cos x cos( ) sin x sin( ) = cos x 0 sin x = cos x 35. s = rt = (2.5 ft)( 2 rad) = 5 ft, so the tire goes 5 feet per revolution, or 1 revolutions 5 per foot. ft 1 rev mi 1 hr 60 5280 5 ft hr 60 min mi 336 rev/min 36. s = rt = (2 ft)(150 rev)( 2 rad/rev) 1885 ft 37. r1t1 = r2 t2 ; 6(2)t1 = 8(2)(21) t1 = 28 rev/sec 38. y = sin and x = cos y sin m= = = tan x cos 3 1 1 3 29. sin 6 = sin 6 = 2 = 8 48 Section 0.7 Instructors Resource Manual 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.

49. 39. a. b. tan = 3 = 3 3x + 3 y = 6 3 y = 3x + 6 y= 3 3 x + 2; m = 3 3 3 3 tan = = 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 2 . the angle opposite each base has measure n Bisect this angle to divide the triangle into two right triangles (See figure). 5 6 40. m1 = tan 1 and m2 = tan 2 tan 2 + tan(1 ) tan = tan( 2 1 ) = 1 tan 2 tan(1 ) = tan 2 tan 1 m m1 = 2 1 + tan 2 tan 1 1 + m1m2 b h = so b = 2r sin and cos = so n 2r n n r h = r cos . n P = nb = 2rn sin n 1 A = n bh = nr 2 cos sin n n 2 sin 32 1 = 1 + 3(2 ) 7 0.1419 41. a. tan = b. tan = 1 1 2 1+ ( 1 ) (1) 2 = 3 1.8925 c. 2x 6y = 12 2x + y = 0 6y = 2x + 12y = 2x 1 y= x2 3 1 m1 = , m2 = 2 3 2 1 3 = 7; 1.7127 tan = 1 + 1 (2) 3 () 42. Recall that the area of the circle is r 2 . The measure of the vertex angle of the circle is 2 . Observe that the ratios of the vertex angles must equal the ratios of the areas. Thus, t A = , so 2 r 2 1 A = r 2t . 2 43. A = 45. The base of the triangle is the side opposite the t angle t. Then the base has length 2r sin 2 (similar to Problem 44). The radius of the t semicircle is r sin and the height of the 2 t triangle is r cos . 2 A= 1 t t t 2r sin r cos + r sin 2 2 2 2 2 2 t t r 2 t = r 2 sin cos + sin 2 2 2 2 2 1 (2)(5) 2 = 25cm 2 2 Instructors Resource Manual Section 0.7 49 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. x x x x 46. cos cos cos cos 2 4 8 16 1 3 1 1 3 1 = cos x + cos x cos x + cos x 2 2 4 4 16 16 1 3 1 3 1 = cos x + cos x cos x + cos x 4 4 4 16 16 1 3 3 3 1 = cos x cos x + cos x cos x 4 4 16 4 16 3 1 1 1 + cos x cos x + cos x cos x 16 4 16 4 1 1 15 9 1 13 11 = cos + cos x + cos x + cos x 4 2 16 16 2 16 16 1 7 1 1 5 3 + cos x + cos x + cos x + cos x 2 16 16 2 16 16 1 15 13 11 9 = cos x + cos x + cos x + cos x 8 16 16 16 16 7 5 3 1 + cos x + cos x + cos x + cos x 16 16 16 16 49. As t increases, the point on the rim of the wheel will move around the circle of radius 2. a. x(2) 1.902 y (2) 0.618 x(6) 1.176 y (6) 1.618 x(10) = 0 y (10) = 2 x(0) = 0 y (0) = 2 b. x(t ) = 2 sin t , y (t ) = 2 cos t 5 5 c. The point is at (2, 0) when is , when t = 5 t= 2 ; that 5 . 2 2 . When 10 you add functions that have the same frequency, the sum has the same frequency. 50. Both functions have frequency 47. The temperature function is 2 7 T (t ) = 80 + 25 sin 12 t 2 . The normal high temperature for November 15th is then T (10.5) = 67.5 F. a. y (t ) = 3sin( t / 5) 5cos( t / 5) +2sin(( t / 5) 3) 48. The water level function is 2 F (t ) = 8.5 + 3.5 sin (t 9) . 12 The water level at 5:30 P.M. is then F (17.5) 5.12 ft . b. 50 Section 0.7 y (t ) = 3cos( t / 5 2) + cos( t / 5) + cos(( t / 5) 3) Instructors Resource Manual 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.

51. a. C sin(t + ) = (C cos )sin t + (C sin ) cos t. Thus A = C cos and B = C sin . b. 51. A2 + B 2 = (C cos )2 + (C sin ) 2 = C 2 (cos 2 ) + C 2 (sin 2 ) = C 2 Also, c. B C sin = = tan A C cos A1 sin(t + 1 ) + A2 sin(t + 2 ) + A3 (sin t + 3 ) = A1 (sin t cos 1 + cos t sin 1 ) + A2 (sin t cos 2 + cos t sin 2 ) + A3 (sin t cos 3 + cos t sin 3 ) = ( A1 cos 1 + A2 cos 2 + A3 cos 3 ) sin t + ( A1 sin 1 + A2 sin 2 + A3 sin 3 ) cos t = C sin (t + ) where C and can be computed from A = A1 cos 1 + A2 cos 2 + A3 cos 3 B = A1 sin 1 + A2 sin 2 + A3 sin 3 as in part (b). d. Written response. Answers will vary. 52. ( a.), (b.), and (c.) all look similar to this: d. 53. a. b. c. 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. Instructors Resource Manual 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. Section 0.7 51 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. 54. a. h( x ) = ( f g ) ( x ) 3 cos(100 x) + 2 100 = 2 1 2 cos (100 x) + 1 100 j ( x) = ( g f )( x) = 56. 2 f ( x ) = ( x 2n ) , 0.0625, 1 1 x 2n , 2n + 4 4 otherwise where n is an integer. y 1 3x + 2 cos 100 100 x2 + 1 0.5 b. 0.25 2 c. 1 1 x 2 0.8 Chapter Review Concepts Test 1. False: 2. True: 1 4 x x + 1 : x n, n + 4 55. f ( x ) = 4 x x + 7 : x n + 1 , n + 1 3 3 4 where n is an integer. ( ) p and q must be integers. p1 p2 p1q2 p2 q1 = ; since q1 q2 q1q2 p1 , q1 , p2 , and q2 are integers, so are p1q2 p2 q1 and q1q2 . If the numbers are opposites ( and ) 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: ( am ) = ( an ) 7. False: (a * b) * c = abc ; a *(b * c) = ab 8. True: ( 3. False: Since x y z and x z , x = y = z ) y 2 1 1 1 x 9. True: 52 Section 0.8 n m = a mn c x would 2 be a positive number less than x . If x was not 0, then = Instructors Resource Manual 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.

53. 10. True: y x = ( x y ) so ( x y )( y x) = ( x y )(1)( x y ) 20. True: If r > 1, then 1 r < 0. Thus, since 1 + r 1 r , = (1)( x y ) . 2 1 1 . 1 r 1+ r ( x y ) 2 0 for all x and y, so ( x y ) 2 0. 11. True: 12. True: If r > 1, r = r , and 1 r = 1 r , so 1 1 1 . = 1 r 1 r 1+ r [ a, b] and [b, c ] If r < 1, r = r and 1 r = 1 + r , a 1 1 > 1; < b b a a < b < 0; a < b; so share point b in common. 13. True: 14. True: 21. True: If (a, b) and (c, d) share a point then c < b so they share the infinitely many points between b and c. 1 1 1 = . 1 r 1 r 1+ r 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 2 = x = x if x < 0. 16. False: 17. True: For example, if x = 3 , then x = ( 3) = 3 = 3 which does (x > 0, y < 0) or x y = x y = x + y not equal x. 15. False: (x < 0, y > 0). In either case x y = x+ y . For example, take x = 1 and y = 2 . 4 x < y x < y 4 If either x = 0 or y = 0, the inequality is easily seen to be true. 4 4 22. True: x = x and y = y , so x < y 18. True: 4 4 4 x + y = ( x + y ) 4 x2 = 23. True: = x + ( y ) = x + y 19. True: If r = 0, then 1 1 1 = = = 1. 1+ r 1 r 1 r For any r, 1 + r 1 r . Since r < 1, 1 r > 0 so 1 1 ; 1+ r 1 r If y is positive, then x = ( y) x3 = = y. (3 y ) 3 =y 24. True: For example x 2 0 has solution [0]. 25. True: x 2 + ax + y 2 + y = 0 x 2 + ax + If 1 < r < 0, then r = r and 2 a2 1 a2 1 + y2 + y + = + 4 4 4 4 2 a a2 + 1 1 x+ + y+ = 2 2 4 is a circle for all values of a. 1 r = 1 + r , so 1 1 1 = . 1+ r 1 r 1 r 1 r = 1 r , so y satisfies For every real number y, whether it is positive, zero, or negative, the cube root x = 3 y satisfies also, 1 < r < 1. If 0 < r < 1, then r = r and 2 26. False: If a = b = 0 and c < 0 , the equation does not represent a circle. 1 1 1 . = 1+ r 1 r 1 r Instructors Resource Manual Section 0.8 53 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. 27. True; 28. True: 3 ( x a) 4 3 3a y = x + b; 4 4 If x = a + 4: 3 3a y = (a + 4) +b 4 4 3a 3a = +3 +b = b+3 4 4 y b = 40. True: 41. True: The equation is (3 + 2m) x + (6m 2) y + 4 2m = 0 which is the equation of a straight line unless 3 + 2m and 6m 2 are both 0, and there is no real number m such that 3 + 2m = 0 and 6m 2 = 0. f ( x) = ( x 2 + 4 x + 3) = ( x + 3)( x + 1) 31. True: 32. True: If ab > 0, a and b have the same sign, so (a, b) is in either the first or third quadrant. The domain does not include n + where n is an integer. 2 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. 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: 30. True: 42. False: 49. True: 29. True: If the points are on the same line, they have equal slope. Then the reciprocals of the slopes are also equal. If f(x) and g(x) are odd functions, f ( g ( x)) = f(g(x)) = f(g(x)), so f(g(x)) is odd. Let x = / 2. If > 0 , then x > 0 and x < . 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. y1 = y2 , so ( x1 , y1 ) and ( x2 , y2 ) are on the same horizontal line. 33. True: ( x 2 + 4 x + 3) 0 on 3 x 1 . d = [(a + b) (a b)]2 + (a a) 2 = (2b) 2 = 2b 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 ( a, 0 ) , ( 0, b ) are the intercepts, the slope is 39. False: b which is rational. a f ( x) = 2( x)3 + ( x) ax + y = c y = ax + c ax y = c y = ax c (a )( a) 1. (unless a = 1 ) 54 52. True: Section 0.8 = ( x)2 + 1 = 2 x3 x x2 + 1 2 x3 + x x2 + 1 Instructors Resource Manual 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.

55. 53. True: f (t ) = = (sin(t )) 2 + cos( t ) tan( t ) csc( t ) ( sin t )2 + cos t (sin t )2 + cos t = tan t ( csc t ) tan t csc t 54. False: f(x) = c has domain ( , ) , yet the range has only one value, c. b. g (1.8) = 57. True: (f f )( x) = ( x 2 )3 = x 6 (f 2 = 4 25 2 (n 2 n + 1)2 ; (1)2 (1) + 1 = 1; 2 (2) 2 (2) + 1 = 9; g )( x) = ( x 3 ) 2 = x 6 (g 2 1 1 1 25 n + ; 1 + = 2; 2 + = ; 2 4 n 1 1 2 + 2 1.8 = 0.9 = 1 2 56. True: 1 n 1. a. f(x) = c has domain ( , ) and the only value of the range is c. 55. False: Sample Test Problems g )( x) = ( x 3 ) 2 = x 6 58. False: f ( x) g ( x) = x x = x 2 3 59. False: 60. True: 61. True: 5 2 (2)2 (2) + 1 = 49 c. 43 / n ; 43 /1 = 64; 43 / 2 = 8; 4 3 / 2 = d. n 1 1 ; 1 = 1; n 1 2 1 = 2 2 f The domain of excludes any g values where g = 0. f(a) = 0 Let F(x) = f(x + h), then F(a h) = f(a h + h) = f(a) = 0 cos x sin x cos( x) cot( x) = sin( x) cos x = = cot x sin x 2. a. 1 1 2 = = ; 2 2 2 1 1 1 1 1 + + 1 + m n m n 63. Fal