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
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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
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material is protected under all copyright laws as they 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
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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
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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
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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
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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
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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
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material is protected under all copyright laws as they currently
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or by any means, without permission in writing from the
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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
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they 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
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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
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reproduced, in any form or by any means, without permission in
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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
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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
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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
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material is protected under all copyright laws as they currently
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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,
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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
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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
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material is protected under all copyright laws as they currently
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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
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all copyright laws as they currently exist. No portion of this
material may be reproduced, in any form or by any means, without
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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
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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
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protected under all copyright laws as they 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
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under all copyright laws as they 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
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NJ. All rights reserved. This material is protected under all
copyright laws as they 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
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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
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under all copyright laws as they 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
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material is protected under all copyright laws as they currently
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or by any means, without permission in writing from the
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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
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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
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material is protected under all copyright laws as they 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
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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
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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
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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,
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protected under all copyright laws as they 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,
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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
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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
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material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form
or by any means, without permission in writing from the
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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
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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
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reserved. This material is protected under all copyright laws as
they 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
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material is protected under all copyright laws as they 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,
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protected under all copyright laws as they 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
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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
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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
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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
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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
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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
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reserved. This material is protected under all copyright laws as
they 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
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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
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material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form
or by any means, without permission in writing from the
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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
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under all copyright laws as they 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
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material is protected under all copyright laws as they currently
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or by any means, without permission in writing from the
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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
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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