- 1. CHAPTER00.1 Concepts Review 1. rational numbersPreliminaries
1 2 1 1 1 1 8. = 3 5 2 3 5 2 1 5 3 3 5 2 15 15 2. dense1 2 1 2 1 2
1 = = 3 5 2 15 3 5 15 1 6 1 1 5 1 = = = 3 15 15 3 15 93. If not Q
then not P. 4. theorems2Problem 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) = 1484.5 [ 1(7 + 12 16) + 4] + 2 = 5 [ 1(3) + 4] + 2 = 5 (
3 + 4 ) + 2 = 5 (1) + 2 = 5 + 2 = 75.6.7.5 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 24Instructors Resource Manual22 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
492 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 8 813. 1 1 1 2 3 2 1 =1 =1 = = 1 3 3 3 3 3 1+ 2 214. 2 +15.(3 5
1+ 25+ 33 3 = 2+ 2 5 7 2 2 2 6 14 6 20 = 2+ = + = 7 7 7 7= 2+)() (
5) ( 3)5 3 =22=53= 2Section 0.11 2007 Pearson Education, Inc.,
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means, without permission in writing from the publisher.
2. 16.(5 3) = ( 5) 222( 5 )( 3 ) + ( 3 )227.= 5 2 15 + 3 = 8 2
1517. (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)124 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 +
919.28.(3x 9)(2 x + 1) = 6 x 2 + 3 x 18 x 92 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 920. (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) 32322= 9t 3t + 3t 3t + t t + 3t t +
16y + 2 + 2y 8y + 2 = 2(3 y + 1)(3 y 1) 2(3 y + 1)(3 y 1)== 12 x 2
61x + 7742 y + 6 y 2 9 y2 12(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 + 100 = 0b.0 is undefined. 0c.0
=0 17d.3 is undefined. 0e.05 = 0f. 170 = 129. 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 + 2723.x 2 4 ( x 2)(
x + 2) = = x+2, x 2 x2 x224.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 +326.22x 2x
322x 2x + x=2 x(1 x) 2x( x 2 x + 1) 2 x( x 1) = x( x 1)( x 1) 2 = x
1Section 0.10 = a , then 0 = 0 a , but this is meaningless 0
because a could be any real number. No 0 single value satisfies = a
. 030. If31..083 12 1.000 96 40 36 4Instructors 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 28 233..142857 21 3.000000 21 90 84 60 42 180 168 120
105 150 147 334..294117... 17 5.000000... 0.2941176470588235 34 160
153 70 68 20 17 30 17 130 119 11Instructors Resource Manual35.3.6 3
11.0 9 20 18 236..846153 13 11.000000 10 4 60 52 80 78 20 13 70 65
50 39 1137. 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= 99Section 0.13 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= = 90 554. 55.10 x = 3.99999... 90 x = 36 36 2 x= = 90
556.43. 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 nm1 1 1 = = (0.5)n (0.2)m .
The product q 2 5 of any number of terminating decimals is also a
nmterminating decimal, so (0.5) and (0.2) , and hence their
product, decimal. Thus1 , is a terminating qp has a terminating
decimal qexpansion. 45. Answers will vary. Possible answer:
0.000001, 1 0.0000010819... 1246. 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
20.392304854Section 0.12 3)4 0.010205144353. 4 1.123 3 1.09
0.0002830738842. x = 0.399999 100 x = 39.99999...44.(( 3.1415 )1/ 2
0.5641979034 8.92 + 1 3 0.000691744752 4 (6 2 2) 3.66159180757. 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
. Thenk0 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 0 but x . x xe.b.b. The statement, converse, and contrapositive
are all false. 69. a.True; x + ( x ) < x + 1 + ( x ) : 0 <
12b. The statement, converse, and contrapositive are all true. 68.
a.True; Let y be any positive number. Take y x = . Then 0 < x
< y . 2d. 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 21 2 . Then x = 2d. True; Let x be any
number. Takeb. 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 and74.only if n 2 is
odd. 75. a. b.243 = 3 3 3 3 3 124 = 4 31 = 2 2 31 or 22 31Some
natural number is not rational. The original statement is
true.Instructors Resource ManualSection 0.15 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.
6. 5100 = 2 2550 = 2 2 1275c.82. a.= 2 2 3 425 = 2 2 3 5 85 = 2 2 3
5 5 17 or 22 3 52 17c.76. For example, let A = b c 2 d 3 ; thenA2 =
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 q2factors of p 2 must occur
an even number of times, 3q 2 would not be valid ande.f. 83. a.p =
3 qx = 2.4444...; 10 x = 24.4444... x = 2.4444... 9 x = 22 22 x= 92
3 n = 1: x = 0, n = 2: x = , n = 3: x = , 3 2 5 n = 4: x = 4 3 The
upper bound is . 2 2Answers 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, sob. 2d. 12=78.2a bp a p aq
+ bp are rational. + = This q b q bq sum is the quotient of natural
numbers, so it is also rational.andb. True0.2 Concepts Review 1.
[1,5); (, 2] 2. b > 0; b < 0p 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. a81. a. 9 = 3; rationalb.3
0.375 = ; rational 8c.(1 + 3)2 = 1 + 2 3 + 3 = 4 + 2 3;
irrational4. 1 x 5Problem Set 0.2 1. a.b.(3 2)(5 2) = 15 4 = 30;
rationald.3. (b) and (c)c.d.6Section 0.2Instructors 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 3e.2 1
1 2 > x ; , 3 2 2 3f.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 34. 3x 5 < 4 x 61 < 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 Manual4 x2 5x 6 < 0x+4 0; [4, 3) x3Section 0.27 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. 8. 16.3x 2 2 0; , (1, ) x 1 3 3 >2
x+520.3 2 > 0 x+517.2 5 < 0 x 2 5x < 0; x 2 ( , 0) , 5
18.3 2( x + 5) >0 x+52 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, ) 21 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.
8Section 0.2False. 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 1x3 + 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
2x4 2 x2 8b.< b 2 2ab + b 2 = 2b ( b a )x4 2 x2 8 0Since 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) 0b. 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 < ; 21 1 , 2.01 1.99 2
x 7 > 1 or 2 x + 1 < 3b.x > 4 or x < 1 (,1) (4, )2.99 2
x 7 1 or 2 x + 1 < 3 2 x 8 or 2 x < 22.99 5.02 4.98 8 or 2 x
< 2b.1.99 1 x< 1.99 2.01b. 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 < 02 x 7 1 or 2 x + 1 > 33.012 x 8 or 2 x
> 2x 4 or x > 1 (, )35.x 2 5; x 2 5 or x 2 5 x 3 or x 7 (, 3]
[7, )Instructors Resource ManualSection 0.29 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.
10. 36.x + 2 < 1; 1 < x + 2 < 143.3 < 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 42
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 = 246. 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 31 (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 7x=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 10Section
0.2Instructors 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 053. 3x
15 < 3( x 5) < (3x 7)( x 5) > 0; 3 x5 < x5 0, > 0. 2
x +2 x +3x + (2) 4 + 4 + 7 = 15 1 and x 2 + 1 1 so 1. 2 x +11==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 + 9b a< b.of absolute
values. c.x +9 2a b a b a b Use Property 4b.2x2a b = a + (b) a + b
= a + b65. a.x269.1 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 < x270. a.77.x x2 < 0
x(1 x) < 0 x < 0 or x > 11 11 R 602x 0 a 2 + b 2 > 4c
455. 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 aresimilar
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 256. The
equations of the two circles are ( x R)2 + ( y R)2 = R 2 ( x r )2 +
( y r )2 = r 2Let ( 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 22 1 2 R= r= 2 1+ 2
r=1 2 + 2 1 2 R 2 2 1 + 1 2 2 1 2R1 2 r = (3 2 2) R 0.1716 R
1property 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
7.464Instructors Resource ManualSection 0.319 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.
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 m60. 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 )
258. 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 + 163x 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 21 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 Theoremsays that a 2 + b 2 = c 2 Thus,a 2 b 2
c 2 + = or 8 8 8 261. 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 8.221 a 1 b 1 c + = 2 2 2 2 2 2C = (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.221 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.20Section 0.362 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 2 265. 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 5 567. 2 x + 4(0) = 5 5
x= 2 d=2( 5 ) + 4(0) 7 = 2 (2)2 + (4) 22 20=5 568. 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+ 5 569. m =Instructors Resource Manual04 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 oflength 5, the y-coordinate is always3
4timesthe x-coordinate. Thus, we need to find the value of r for
which there is exactly one x23 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 0.321 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 beThe 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 =
1274. 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 + 175. 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 20.4 Concepts Review 1.
y-axis 2.( 4, 2 )3. 8; 2, 1, 4 4. line; parabolaProblem 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-axis276. 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 22Section 0.4Instructors 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-axis4.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-axis6. y = x 2 2 x;
y -intercept = 0 y = x(2 x); x-intercepts = 0, 27 7. 7x2 + 3y = 0;
3y = 7x2; y = x 2 3 x-intercept = 0, y-intercept = 0 Symmetric with
respect to the y-axis8. y = 3x 2 2 x + 2; y -intercept =
2Instructors Resource ManualSection 0.423 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 origin10. 3x 2 + 4 y 2 = 12;
y-intercepts = 3, 3 x-intercepts = 2, 2 Symmetric with respect to
the x-axis, y-axis, and origin11. y = x2 2x + 2: y-intercept = 2 2
4+8 22 3 x-intercepts = = = 1 3 2 224Section 0.412. 4 x 2 + 3 y 2 =
12; y -intercepts = 2, 2x-intercepts = 3, 3 Symmetric with respect
to the x-axis, y-axis, and origin13. x2 y2 = 4 x-intercept = -2, 2
Symmetric with respect to the x-axis, y-axis, and origin14. x 2 + (
y 1)2 = 9; y -intercepts = 2, 4x-intercepts = 2 2, 2 2 Symmetric
with respect to the y-axisInstructors 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-axis18. x 4 + y 4 = 1; y
-intercepts = 1,1 x-intercepts = 1,1 Symmetric with respect to the
x-axis, y-axis, and origin16. x 2 4 x + 3 y 2 = 219. x4 + y4 = 16;
y-intercepts = 2, 2 x-intercepts = 2, 2 Symmetric with respect to
the y-axis, x-axis and originx-intercepts = 2 2 Symmetric with
respect to the x-axis17. x2 + 9(y + 2)2 = 36; y-intercepts = 4, 0
x-intercept = 0 Symmetric with respect to the y-axisInstructors
Resource Manual20. 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
originSection 0.425 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 = 1x +1 Symmetric with respect to the y-axis224. 4 ( x
5 ) + 9( y + 2) 2 = 36; x-intercept = 525. 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 226. y =
x2(x 1)(x 2); y-intercept = 0 x-intercepts = 0, 1, 223. 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 = 12y-intercepts = 2 30 3x-intercept = 1 27.
y = x 2 ( x 1)2 ; y-intercept = 0 x-intercepts = 0, 126Section
0.4Instructors 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-axisIntersection 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 origin2 x + 3
= x2 + 2 x 1 x2 + 4 = 0 No points of intersection33. 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 origin2 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,
3Instructors Resource ManualSection 0.427 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 + 134. 2 x + 3 = 3x 2 3 x + 12x 2 + 2 x + (3 x + 1) 2 = 153x 2
x + 9 = 0 No points of intersectionx 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=28Section 0.4Instructors 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) 22222d3 = (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 = 24f.1 15 1 1 f =1 =1 = 16 16 4 4g.f
(1 + h ) = 1 (1 + h ) = 2h h 2h.f (1 + h ) f (1) = 2h h 2 0 = 2h h
2i.( 2 21)2c.e.)f (2) = 1 (2)2 = 3f ( 2 + h ) f ( 2) = 1 ( 2 + h )
+ 322= 2 21 9.17 d 4 = (2 2)2 + 1 21 (1 + 13) (f (1) = 1 12 =
0d.)22(= 16 + 21 13)2222= 4h h 2= 50 + 2 273 9.11()d5 = (2 2)2 + 1
21 1 13 (2b.d3 = (2 + 2)2 + 1 + 21 1 21 21 + 212.1. a.= 50 + 2 273
9.11= 16 +( 2 13 )0.5 Concepts Review2((=Problem Set 0.5d 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; origin21 , 2,
1 + 13 , 2, 1 13= 16 +13 + 133. asymptote41. x2 + 2x + y2 2y = 20;
2, 1 + 21 ,((1. domain; ranged 2 = (2 + 2) + (3 3) = 52 = 2 13= 16
+= 0+22d1 = (2 + 2) + (3 + 3) = 4( 2, 1 )d6 = (2 2)2 + 1 + 13 1 13
13 21)222. a. b.F ( 2) = ( 2)3 + 3( 2) = 2 2 + 3 2 =5 2= 50 2 273
4.123c.Instructors Resource ManualF (1) = 13 + 3 1 = 41 1 1 1 3 49
F = + 3 = + = 4 4 4 64 4 64Section 0.529 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 ) 3f. ( x 2 + x) == 1 + 3h + 3h 2 + h3
+ 3 + 3h = 4 + 6h + 3h 2 + h3e.F (1 + h ) 1 = 3 + 6h + 3h + hf.=F (
2 + h) F ( 2)232= 15h + 6h + h3. a.b.G (0) =d.e.f.4. a.b.0.25 3 1f
( x) =c.f (3 + 2) =31 = 1 0 1 3G( y ) =G ( x) =c.1 1 = x 1 x +17.
a.21 x = 1 1 x21 2 0.841 3.293(12.26) 2 + 9 12.26 3 1.1991t 1 21 =
2d. (u + 1) =e.( x2 ) =+(1) 2=21 2=x2=x2 + y2 = 1c.x = 2 y +1x2 = 2
y + 1 1.06(u + 1) + (u + 1) 2( x2 ) + ( x2 )2Section 0.5tu +1;
undefinedb. xy + y + x = 1 y(x + 1) = 1 x 1 x 1 x y= ; f ( x) = x
+1 x +1t2 t3 4 1 23 3y = 1 x 2 ; not a function=2t + ( t ) 2( 3)2 +
9f ( 3) =y 2 = 1 x 21 x21 + 12 (t ) =is noty 2 1 1 G = x2 (1) ==3+
2 3 0.250.79 3b. f(12.26) =1 2.75 2.658(0.79) 2 + 9f(0.79) =1 G
(1.01) = = 100 1.01 1 21=1=21 G (0.999) = = 1000 0.999 1c.301f
(0.25) =b.6. a. c.x2 + xdefined= ( 2 + h ) + 3 ( 2 + h ) 23 3 ( 2 )
= 8 + 12h + 6h 2 + h3 + 6 + 3h 14x2 + xx 4 + 2 x3 + 2 x 2 + x35.
a.( x 2 + x) + ( x 2 + x) 2=y=u 2 + 3u + 2x2 + x4 xu +1d.x2 1 x2 1
; f ( x) = 2 2y 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 hx 2 9 0; x 2 9; x 3Domain: {x
d.F (a + h) F (a ) 4(a + h) 4a = h h4a3 + 12a 2 h + 12ah 2 + 4h3
4a3 h 12a 2 h + 12ah 2 + 4h3 h14. a.b.= 12a 2 + 12ah + 4h 2f ( x)
=4 x2=4 x2 ( x 3)( x + 2)x2 x 6 Domain: {x : x 2, 3} G ( y ) = ( y
+ 1) 1Domain: { 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:2a+h
a + h+ 4: y 5}1 0; y > 1 y +13 3 g ( x + h) g ( x) x + h 2 x 2 =
h h 3x 6 3x 3h + 6G ( a + h) G ( a ) = hH ( y ) = 625 y 4Domain: {
y =12.: x 3}3=11. ( x) = x 2 9625 y 4 0; 625 y 4 ; y 5 310.c.15.
f(x) = 4; f(x) = 4; even functiona a+4h2a + 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 + 16F ( z) = 2 z + 32z + 3 0; z Domain: z b.16.
f(x) = 3x; f(x) = 3x; odd functiong (v ) =3 23 :z 21 4v 14v 1 = 0;
v = Domain: v 1 4 1 :v 4Instructors Resource ManualSection 0.531
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;
neither20. 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 ;
neither32Section 0.522. ( z ) =u3 u3 ; g ( u ) = ; odd function 8
8x 2x 1; g ( x) =x 2x 1; odd2z +1 2 z + 1 ; ( z ) = ; neither z 1 z
1Instructors 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; neither2224. h( x ) = x + 4; h( x) = x + 4; even
function26. F (t ) = t + 3 ; F ( t ) = t + 3 ; neither27. g ( x) =x
x ; g ( x ) = ; neither 2 228. G ( x) = 2 x 1 ; G ( x) = 2 x + 1 ;
neither25.f ( x) = 2 x ; f ( x) = 2 x = 2 x ; evenfunction1 if t 0
29. g (t ) = t + 1 if 0 < t < 2 2 t 1 if t 2Instructors
Resource ManualneitherSection 0.533 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 neither35. 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 236. 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 +
0.40xb. 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 3P(200) 190 ; P (1000 ) 610b.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 0.539. The area of the two semicircular ends is 0.51x0.51
exceeds its square by the maximum amount. 234. Each side has
lengthp . The height of the 3d 2 . 41 d . 2 2 d 2 d d 2 1 d d A(d )
= +d + = 4 4 2 2 The length of each parallel side is2d 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
is34Section 0.5Instructors 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) = 2 242. 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 2c.A(0) = 0d.1 1 A(c) = c(1) + (c)(c + 1 1) = c 2 + c 2 2e.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 pbe written asf.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 2p with p, q in N. q1 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 qc.Instructors Resource ManualSection 0.535
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, hence46.
a.xf(x)46.190230.4118213.765119.95790017.3369217.7388if 18 < t
2730.4521if 27 < t 3644.4378b.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 baseand 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) 6.750810t 902 +
(10t 90)2 s= 902 + (270 10t ) 2 360 10t if 0 t 9180 180 10t s = 902
+ (10t 90) 2 2 2 90 + (270 10t ) if 0 t 9if 9 < t 18or 27 < t
3647.if 9 < t 18 if 18 < t 27f(1.38) 0.2994 f(4.12) 3.6852
xf(x)44.0533.1538a.22.375b. f(x) = 0 when x 1.1, 1.7, 4.3 f(x) 0 on
[1.1, 1.7] [4.3, 5]11.801.2510.221.12532.38464b.3.5548.a. 36Section
0.5Range: {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 + 9b. 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 aroundLargest value f (2) g (2) = 45x = 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 = 2d.
Horizontal asymptote at y = 30.6 Concepts Review 1. ( x 2 + 1)3
a.x-intercept: 3x 4 = 0; x =4 33 0 42 = y-intercept: 2 0 +06 32.
f(g(x)) 3. 2; left 4. a quotient of two polynomial functionsb. c.x
2 + x 6 = 0; (x + 3)(x 2) = 0 Vertical asymptotes at x = 3, x =
2Problem Set 0.6 1. a.( f + g )(2) = (2 + 3) + 22 = 9d. Horizontal
asymptote at y = 0 b.( f g )(1) = f (12 ) = 1 + 3 = 4e.( g f )(1) =
g (1 + 3) = 4 2 = 16f.( g f )(8) = g (8 + 3) = (5) 2 = 252.
a.x-intercepts: 3x 2 4 = 0; x = ( g f )(3) =d.a.( f g )(0) = (0 +
3)(02 ) = 0c.50.4 2 3 = 3 332 9 3 = = 3+3 6 2( f g )(2) = (22 + 2)
12 + 12b.( f g )(1) =c.2 y-intercept: 31 2 1 g 2 (3) = = = 3 + 3 3
9 2 1+ 3 2Instructors Resource Manual=2 2 28 =6 = 2+3 5 52
4=42Section 0.637 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. 2d.e.f.3. a.(f 2 1 1 3 g )(1) =
f = + = 1+ 3 2 2 45.= x2 + 2 x 32 2 = 2+3 5 2 ( g g )(3) = g = 3+32
2 3 = 10 = 1 +3 3 5 36. g 3 (x) = (x 2 +1) 3 = (x 4 + 2x 2 + 1)(x 2
+ 1) 61 tb.c.( )(r ) = (r + 1) =d. 3 ( z ) = ( z 3 + 1) 3e.( )(5t)
= [(5t) 3 +1] 13b.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 + 57. g(3.141) 1.188
8. g(2.03) 0.000205r 3 +11/ 39. g 2 ( ) g ( ) 4.7891 5t1/ 32 = (11
7 ) 11 7 10. [ g 3 () g ()]1/ 3 = [(6 11)3 (6 11)]1/ 3 7.8071 5t1
(( ) )(t ) = ( ) t 3 1 1 1 = + 1 1 = 3 + 1 t t t t11. a. b. 12.
a.4. a.4= x + 3x + 3x + 1 ( g g g )( x) = ( g g )( x 2 + 1)3f. f )
( x) = g x 2 4 = 1 + x 2 4 = 1 + x2 41 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(fg ( x) = x , f ( x) = x + 7g (x)
= x15 , f (x) = x 2 + x 2f ( x) =x2 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, ) 2c.2 2 g )( x) = f = 1
= x x Domain: [2, 0) (0, 2]d.( g f )( x) = g x 2 1 = (f4f ( x) =13.
p = f1 , g (x) = x 3 + 3 x x g h if f(x) =1/ x , g ( x) = x ,h( x )
= x 2 + 1 p= fg h if f ( x) = 1/ x , g(x) = x + 1,h( x) = x 24 1
x214. p = f g h l if f ( x) = 1/ x , g ( x) = x , 2 h(x) = x + 1,
l( x) = x2 2x 1 Domain: ( , 1) (1, )38Section 0.6Instructors
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 3units and
down 4 units. 19. ( f + g )( x) =x3 + x 220. ( f + g )( x) = x +
xInstructors Resource ManualSection 0.639 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 t24. a.tF(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 tF ( 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
2 225. 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.Neitherb. 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.40PFe.Even; (f + g)(x) = f(x) + g(x) = f(x) + g(x) = (f + g)(x)
if f and g are both even functions.RFd.23. a.PFRFf.Neither27. a.P =
29 3(2 + t ) + (2 + t )2 = t + t + 27b. 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, 000400t 29. D(t ) = 2 2 (400t ) + [300(t
1)] if 0 < t < 1 if t 1if 0 < t < 1 400t D(t ) = 2 250,
000t 180, 000t + 90, 000 if t 1 30. D(2.5) 1097 miOdd; ( 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
0.6Instructors 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 cx36. 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 1If x = 1, f(x) is undefined, while if x =
1, f(f(x)) is undefined. 33. a.b.1 f = x34. a. b.1 x1=1 1 x1 f1 ( f
2 ( x)) = ; x f1 ( f3 ( x)) = 1 x; 1 ; 1 x x 1 f1 ( f5 ( x)) = ; x
x ; f1 ( f6 ( x)) = x 1f1 ( 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 1x ; x 1=1 x 1 xx 1 ; xf3 ( f1 ( x)) = 1 x;
1 x 1 f f ( x) = f x = =1x 1/ x 1 / x 1x 1 x x 1 1 x=x 1 x 1 x1 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=xxf ( f ( x)) = f ( x /( x 1)) ==1 xf 2 (
f3 ( x)) =f 2 ( f 5 ( x )) =f (1 / x) =1 ; x 11 1 x x f ( f ( x)) =
f = x 1=c.1 xf1 ( f1 ( x)) = x;x /( x 1)x x( x 1) + 1 x35. ( 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 11 ; 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 Manual1 1 = ; 1 (1 x) x 1 1 1 1 x1
1x 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 0.641 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 xf 5 ( f 6 (
x)) =x 1 x 1 x x 1=1 (1 x) = x; 1f3f4f5f6f3 )f4 )f5 )f6 )= ( f 4 f
4 ) ( f5 f 6 ) = f5 f 2 = f3 c.x ( x 1) 1 = ; x xIf Ff 6 = f 1 ,
then F = f 6 .d. If G f 3 G = f5. e.1 ; 1 xf 6 = f 1 , then G f 4 =
f 1 soIf 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 xf 6 ( f 4 ( x)) =1 1 x 1 1 1 x=1 1 = ; 1 (1
x) xf 6 ( f 5 ( x)) =x 1 x x 1 1 x=x 1 = 1 x; x 1 xf 6 ( f 6 ( x ))
=x x 1 x 1 x 1=x =x x ( x
1)38.f1f2f3f4f5f6f1f1f2f3f4f5f6f2f2f1f4f3f6f5f3f3f5f1f6f2f4f4f4f6f2f5f1f3f5f5f3f6f1f4f2f642f1
f 2 = (((( f 2x ; 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 x1f3= f1 f 3 = f 31 1 1 x 1 1 x1
xf3= (( 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 11 xf3= (((( f 31 xf 6
( f1 ( x)) =f3f6f4f5f2f3f1Section 0.639.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 0.740.1. a.b. 60 = 3
180 d. 4 240 = 180 3e.37 370 = 18 180 f.b. 45 = 180 4c.41. a. 30 =
180 6 10 = 180 182. a.7 180 = 210 6 180 = 135 b. c.1 180 = 60 3 d.4
180 = 240 3 e.f.c.3 43 180 = 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; even4 x 4. r = (4) + 3 = 5; cos = = 5 r
22Instructors Resource ManualSection 0.743 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 Thusb. 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 34.1b.cos= cos=and by the Pythagorean Identity, sin3=3 .
2sin (0.361) 0.35326. 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
56.1Referring to Figure 2, it is clear that sin3b.sin 35 0. 0789
sin 26 + cos 26 Identity, cos 2= 1 sin 29. a.=123 1 = 1 = . 6 4 21
= 1 cos()b.sec() =1 3 sec = = 2 4 cos 3 4d. Section 0.72 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 cos8.
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 244tan (0.452) 0.4855d.63 . 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 sin5.34 tan 21.3
0.8845 sin 3.1+ cot 23.5c.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 x1 sec = =2 3 cos 3( )d.( ) ( )c. 3 cos 3 cot = = 3 3
sin 3d.1 csc = = 2 4 sin e. 3 sin 6 tan = = 6 cos 3 6f. 1 cos = 3
213. 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 = 2sec zb.(sec t 1)(sec t + 1) =
sec 2 t 1 = tan 2 tc.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 t1 sec 2 ty = sin 2x1 sin 2 t cos t
cos t1 sin 2 t cos 2 t = = cos t cos t cos tsec2 t 1 sec 2 t sin 2
v +=tan 2 t sec 2 t1 2=sin 2 t cos 2 t 1 cos 2 t= sin 2 t= sin 2 v
+ cos 2 v = 1sec vInstructors Resource ManualSection 0.745 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 tb. y = 2 cos tc. y =
cos x 4 c.y = cos 3td.y = sec td. y = cos t + 315. a.y = csc
t46Section 0.7x 2 Period = 4 , amplitude = 316. y = 3
cosInstructors 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 = 221. 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
2right and 1 unit down.19. y = 2 +1 cot(2 x) 6Period = 2, shift: 2
units up 23. y = tan 2 x 3 units right Period = , shift: 6 220. y =
3 + sec( x ) Period = 2 , shift: 3 units up, units rightInstructors
Resource ManualSection 0.747 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; even25.
a. b. c.1 csc( t ) = = csc t; odd sin( t )( )=32. a.sin(cos(t)) =
sin(cos t); evenf.x + sin(x) = x sin x = (x + sin x); odd cot(t) +
sin(t) = cot t sin t = (cot t + sin t); odd26. a.b.sin 3 (t ) = sin
3 t ; oddc.sec( t) =1 = sec t; even cos(t )sin 4 ( t ) = sin 4 t ;
evend. e.cos(sin(t)) = cos(sin t) = cos(sin t); evenf.( x )2 + sin(
x ) = x 2 sin x; neither 227. cos 2 1 1 = cos = = 3 3 4 2 228. sin
222 1 1 = sin = = 6 6 4 2 32sin(x y) = sin x cos(y) + cos x sin(y)
= sin x cos y cos x sin yb.cos(x y) = cos x cos(y) sin x sin (y) =
cos x cos y + sin x sin yc.tan( x y ) =2e.32 2 4=sin(t ) = sin t =
sin t ; evend.1 cos 4 1 2 1 cos 2 8 31. sin = = = 8 2 2 2 2sin ( t
) = sin t ; even 2()1 + cos 6 1 + 2 1 + cos 2 12 = = = 30. cos 12 2
2 2 2+ 3 = 4 2tan x + tan( y ) 1 tan x tan( y ) tan x tan y 1 + tan
x tan ytan t + tan tan t + 0 = 1 tan t tan 1 (tan t )(0) = tan t33.
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 31 1 3 29.
sin 6 = sin 6 = 2 = 8 48Section 0.7Instructors 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 3tan = =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).5 640. 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 sin32 1 = 1 + 3(2 ) 7 0.141941. 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 22t t r 2 t = r 2 sin cos + sin 2 2 2 2 21
(2)(5) 2 = 25cm 2 2Instructors Resource ManualSection 0.749 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) = 2b. x(t ) = 2 sin t , y (t ) = 2 cos t 5 5 c.The
point is at (2, 0) when is , when t = 5t= 2; that5 . 22 . 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.50Section 0.7y
(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 2Also, 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 ManualThe 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
0.751 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
otherwisewhere n is an integer. y1 3x + 2 cos 100 100 x2 + 1
0.5b.0.252c.11x20.8 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) = ab8. True:(3. False:Since x y z and x z
, x = y = z)y2111x9. True:52Section 0.8nm= a mn cx 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 ) . 21 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 aa < b
< 0; a < b;so share point b incommon. 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+ rIf
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 + yx 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 + ynot equal x.15.
False:(x < 0, y > 0). In either case x y = x+ y .For example,
take x = 1 and y = 2 . 4x < y x < y 4If either x = 0 or y =
0, the inequality is easily seen to be true.4422. True:x = x and y
= y , so x < y18. True:444x + y = ( x + y )4x2 =23. True:= x + (
y ) = x + y19. 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 so1 1 ; 1+ r 1 rIf y
is positive, then x =( y)x3 == y.(3 y )3=y24. 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 and2a2 1 a2 1 + y2 + y + = + 4 4 4 4 2a
a2 + 1 1 x+ + y+ = 2 2 4 is a circle for all values of a.1 r = 1 +
r , so1 1 1 = . 1+ r 1 r 1 r 1 r = 1 r , soy satisfiesFor every
real number y, whether it is positive, zero, or negative, the cube
root x = 3 y satisfiesalso, 1 < r < 1.If 0 < r < 1,
then r = r and226. False:If a = b = 0 and c < 0 , the equation
does not represent a circle.1 1 1 . = 1+ r 1 r 1 rInstructors
Resource ManualSection 0.853 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.
243. 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 odd48. 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 = 2b34. 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. af ( x) =2( x)3 + ( x)ax + y = c
y = ax + c ax y = c y = ax c (a )( a) 1. (unless a = 1 )5452.
True:Section 0.8=( x)2 + 1=2 x3 x x2 + 12 x3 + x x2 + 1Instructors
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 t54. False:f(x) = c has domain ( , ) ,
yet the range has only one value, c.b.g (1.8) =57. True:(ff )( x) =
( x 2 )3 = x 6(f2=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(g21 1 1 25 n + ; 1 + = 2; 2
+ = ; 2 4 n 1 1 2 + 2 1.8 = 0.9 = 1 256. True:1n1. a.f(x) = c has
domain ( , ) and the only value of the range is c.55. False:Sample
Test Problemsg )( x) = ( x 3 ) 2 = x 658. False:f ( x) g ( x) = x x
= x 2 359. False:60. True:61. True:52 (2)2 (2) + 1 = 49 c.43 / n ;
43 /1 = 64; 43 / 2 = 8; 4 3 / 2 =d.n1 1 ; 1 = 1; n 121 = 2 2f 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 x2. a.1 1 2 = = ; 2 2 21 1 1 1 1 +
+ 1 + m n m n 63. False:1cot x =b.The domain of the tangent
function excludes all n + where n is an 2 integer. The cosine
function is periodic, so cos s = cos t does not necessarily imply s
= t; e.g., cos 0 = cos 2 = 1 , but 0 2 .Instructors Resource
Manual=c.1 1 + m n = 1 1 1 + m n mn + n + m = mn n + m 1+2 x 2 x 2
x + 1 x x 2 x + 1 ( x 2)( x + 1) = 3 2 3 2 x +1 x 2 x +1 x 2 =62.
False:1 82( x 2) x 3 ( x 2) 2( x + 1) x4 x 8(t 3 1) (t 1)(t 2 + t +
1) 2 = = t + t +1 t 1 t 13. Let a, b, c, and d be integers. a+ c a
c ad + bc b d which is rational. = + = 2 2b 2d 2bdSection 0.855
2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights
reserved. This material is protected under all copyright laws as
they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in
writing from the publisher. 56. 4. x = 4.1282828 1000 x =
4128.282828 10 x = 41.28282813. 21t 2 44t + 12 3; 21t 2 44t + 15 0;
t=990 x = 4087 4087 x= 99044 442 4(21)(15) 44 26 3 5 = = , 2(21) 42
7 3 3 5 3 5 t t 0; , 7 3 7 35. Answers will vary. Possible answer:
13 0.50990... 502x 1 1 > 0; , ( 2, ) x2 2 14.2 3 8.15 104 1.32
545.39 6. 3.247.( 2.0)2.515. ( x + 4)(2 x 1) 2 ( x 3) 0;[4,3] 3 2.0
2.6616. 3x 4 < 6; 6 < 3 x 4 < 6; 2 < 3x < 10; 8.
sin2( 2.45 ) + cos ( 2.40 ) 1.00 0.0495 29. 1 3 x > 0 3x < 1
1 x< 3 1 , 3 10. 6 x + 3 > 2 x 5 4 x > 8 x > 2; ( 2,
)11. 3 2 x 4 x + 1 2 x + 7 3 2 x 4 x + 1 and 4 x + 1 2 x + 7 6 x 2
and 2 x 6 1 1 x and x 3; , 3 3 3 12. 2 x 2 + 5 x 3 < 0;(2 x 1)(
x + 3) < 0; 1 1 3 < x < ; 3, 2 217.2 10 2 10 < x < ;
, 3 3 3 33 2 1 x 3 20 1 x 3 2(1 x) 0 1 x 2x +1 0; 1 x 1 , (1, ) 2
18. 12 3 x x (12 3 x)2 x 2 144 72 x + 9 x 2 x 2 8 x 2 72 x + 144 0
8( x 3)( x 6) 0 (,3] [6, )19. For example, if x = 2, (2) = 2 2 x x
for any x < 020. If x = x, then x = x. x056Section
0.8Instructors 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. 57. 2 + 10 0 + 4
, 27. center = = (6, 2) 2 2 1 1 radius = (10 2) 2 + (4 0) 2 = 64 +
16 2 2 =2 521. |t 5| = |(5 t)| = |5 t| If |5 t| = 5 t, then 5 t 0.
t 5 22. t a = (a t ) = a tIf a t = a t , then a t 0. tacircle: ( x
6)2 + ( y 2) 2 = 2023. If x 2, then28. x 2 + y 2 8 x + 6 y = 0 x 2
8 x + 16 + y 2 + 6 y + 9 = 16 + 90 2 x 2 + 3 x + 2 2 x 2 + 3 x + 2
8 + 6 + 2 = 16( x 4) 2 + ( y + 3) 2 = 25;11 . Thus also x + 2 2 so
2 x +2 2 22 x2 + 3x + 2 x2 + 2= 2 x2 + 3x + 21 16 2 2 x +2 1=824.
a.The distance between x and 5 is 3.b. The distance between x and 1
is less than or equal to 2. c.The distance between x and a is
greater than b.center = ( 4, 3) , radius = 5 x2 2 x + y 2 + 2 y =
229.x2 2 x + 1 + y2 + 2 y + 1 = 2 + 1 + 1 ( x 1) 2 + ( y + 1) 2 = 4
center = (1, 1) x 2 + 6 x + y 2 4 y = 7 x 2 + 6 x + 9 + y 2 4 y + 4
= 7 + 9 + 4 ( x + 3)2 + ( y 2)2 = 6 center = (3, 2) d = (3 1) 2 +
(2 + 1)2 = 16 + 9 = 525.30. a.d ( A, B ) = (1 + 2) 2 + (2 6)23x + 2
y = 6 2 y = 3 x + 6 3 y = x+3 2 3 m= 2 3 y 2 = ( x 3) 2 3 13 y = x+
2 2= 9 + 16 = 5 d ( B, C ) = (5 1)2 + (5 2)2 = 16 + 9 = 5 d ( A, C
) = (5 + 2)2 + (5 6) 2 = 49 + 1 = 50 = 5 2 ( AB) + ( BC )2 = ( AC )
2 , so ABC is a right triangle. 21+ 7 2 + 8 , 26. midpoint: = ( 4,5
) 2 2 d = (4 3)2 + (5 + 6)2 = 1 + 121 = 122Instructors Resource
ManualSection 0.857 2007 Pearson Education, Inc., Upper Saddle
River, NJ. All rights reserved. This material is protected under
all copyright laws as they currently exist. No portion of this
material may be reproduced, in any form or by any means, without
permission in writing from the publisher. 58. b.2 m= ; 33x 2 y = 5
2 y = 3 x + 5 3 5 y = x ; 2 2 3 m= 2 3 y 1 = ( x + 2) 2 3 y = x+4
2c.b.3 x + 4y = 9 4y = 3x + 9; 4 3 9 y = x+ ; m = 3 4 4 4 y 1 = ( x
+ 2) 3 4 11 y = x+ 3 32 ( x 1) 3 2 5 y = x 3 3 y +1 =c.y=9d. x = 2
e.contains (2, 1) and (0, 3); m =3 1 ; 0+2y=x+3 3 +1 4 11 3 8 4 = ;
m2 = = = ; 52 3 11 5 6 3 11 + 1 12 4 m3 = = = 11 2 9 3 m1 = m2 = m3
, so the points lie on the same line.32. m1 = d. x = 333. The
figure is a cubic with respect to y.The equation is (b) x = y 3 .
34. The figure is a quadratic, opening downward, with a negative
y-intercept. The equation is (c) y = ax 2 + bx + c. with a < 0,
b > 0, and c < 0.35. 31. a.583 1 2 m= = ; 7+2 9 2 y 1 = ( x +
2) 9 2 13 y = x+ 9 9Section 0.8Instructors 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. 59. 36.x2 2 x + y 2 = 3 x2 2 x + 1 + y
2 = 4 ( x 1)2 + y 2 = 440. 4 x y = 2 y = 4 x 2; 1 4 contains ( a, 0
) , ( 0, b ) ; m=ab =8 2 ab = 16 16 b= a 1 b0 b = = ; 0a 4 a a = 4b
16 a = 4 a37.a 2 = 64 a =8 b=41. a.b.38.c.16 1 = 2; y = x + 2 8 4 f
(1) =1 1 1 = 1+1 1 21 1 1 f = =4 2 1 +1 1 2 2f(1) does not exist. 1
1 1 1 = t 1+1 t 1 t t 1d.e.39. y = x2 2x + 4 and y x = 4; x + 4 =
x2 2 x + 4 x 2 3x = 0 x( x 3) = 0 points of intersection: (0, 4)
and (3, 7)f (t 1) =1 1 t 1 f = = t 1 +1 1 1+ t t t t42. a.b.c.g (2)
=2 +1 3 = 2 21 g = 21 2+1 1 2=32 + h +1 + 22 1 g ( 2 + h ) g ( 2) =
2+ h h h h 2 h + 6 3h 6 2( h + 2) 1 2( h + 2) = = = h h 2(h + 2)43.
a.: x 1, 1}b.Instructors Resource Manual{x {x : x 2}Section 0.859
2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights
reserved. This material is protected under all copyright laws as
they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in
writing from the publisher. 60. 44. a.b.f ( x) =3( x) ( x) + 1 2=3x
x +1 2; odd46.g ( x) = sin( x) + cos( x)= sin x + cos x = sin x +
cos x; evenc.h( x) = ( x)3 + sin( x) = x3 sin x ; oddd.k ( x) =45.
a.( x )2 + 1 x + ( x) 4=x2 + 1 x + x4; even47. V(x) = x(32 2x)(24
2x) Domain [0, 12]2f (x) = x 148. a.1 13 ( f + g )(2) = 2 + (22 +
1) = 2 2 b.15 3 ( f g )(2) = (5) = 2 2c.(fg )(2) = f (5) = 5 d.(g13
3 3 f )(2) = g = + 1 = 2 2 4 e.1 f 3 (1) = 1 + = 0 1 1 24 = 5 5
2b.x g(x) = 2 x +132f.49. a. c.60 x2 h(x) = 6 xSection 0.83 f 2 (2)
+ g 2 (2) = + (5) 2 2 9 109 = + 25 = 4 4 y=1 2 x 4if 0 x 2 if x
> 2Instructors 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. 61. b.y=1
( x + 2)2 453. a. b.sin (t) = sin t = 0.8 sin 2 t + cos2 t = 1 cos
2 t = 1 (0.8)2 = 0.36 cos t = 0. 6c. d.c.tan t =e. cos t = sin t =
0.8 2 f.1 y = 1 + ( x + 2) 2 4sin 2t = 2 sin t cos t = 2(0.8)(0.6)
= 0.96sin( + t ) = sin t = 0.8sin t 0.8 4 = = 1.333 cos t 0.6 354.
sin 3t = sin(2t + t ) = sin 2t cos t + cos 2t sin t = 2sin t cos 2
t + (1 2sin 2 t ) sin t = 2sin t (1 sin 2 t ) + sin t 2sin 3 t=
2sin t 2sin 3 t + sin t 2sin 3 t = 3sin t 4sin 3 t 55. s = rt rev
rad 1 min = 9 20 2 (1 sec) = 6 min rev 60 sec 18.85 in. 50. a. b.
c.(,16] fReview and Preview Problemsg = 16 x 4 ; domain [2, 2]g f =
( 16 x ) 4 = (16 x) 2 ; domain (,16] (note: the simplification ( 16
x ) 4 = (16 x) 2 is only true given the restricted domain)51.f ( x)
= x , g ( x) = 1 + x, h( x) = x 2 , k(x) = sin x, F ( x) = 1 + sin
2 x = f g h k52. a.sin(570) = sin(210) = 1 2b. 9 cos = cos = 0 2
2c.3 13 cos = cos = 6 6 2Instructors Resource Manual1. a) b) 2. a)
b)0 < 2 x < 4; 0 < x < 2 6 < x < 16 13 < 2 x
< 14; 6.5 < x < 74 < x / 2 < 7; 14 < x < 83. x
7 = 3 or x 7 = 3 x = 10 or x=4 4. x + 3 = 2 or x = 1 orx + 3 = 2 x
= 55. x 7 = 3 or x 7 = 3 x = 10 or x=4 6. x 7 = d or x 7 = d x = 7
+ d or x = 7 d 7. a)x 7 < 3 and x 7 > 3 x < 10 and x>4
4 < x < 10Review and Preview61 2007 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is
protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any
means, without permission in writing from the publisher. 62.
b)c)d)8. a)x 7 3 and x 7 3 x 10 and x4 4 x 10 x 7 1 and x 8 and 6
x8b)g ( 0.9 ) = 0.0357143 g ( 0.99 ) = 0.0033557 g ( 0.999 ) =
0.000333556x 7 1 x6g (1.001) = 0.000333111 g (1.01) = 0.00331126 g
(1.1) = 0.03125x 7 < 0.1 and x 7 > 0.1 x < 7.1 and x >
6.9 6.9 < x < 7.1 x 2 < 1 and x 2 > 1 x < 3 and x
>1 1< x < 3b)c)12. a)x 2 < 0.1 and x 2 > 0.1 x <
2.1 and x > 1.9 1.9 < x < 2.1d)g ( 2) =x 2 1 or x 2 1 x 3
or x 1x 2 < 0.01 and x 2 > 0.01 x < 2.01 and x > 1.99
1.99 < x < 2.019. a) b) 10. a) 11. a)x 1 0; x 1 2 x 2 x 1 0;
x 1, 0.5 x0b)g ( 0 ) = 1b)1 = 1 1 0.1 = 1 F ( 0.1) = 0.1 0.01 F (
0.01) = = 1 0.01 0.001 F ( 0.001) = = 1 0.001 0.001 F ( 0.001) = =1
0.001 0.01 F ( 0.01) = =1 0.01 0.01 F ( 0.1) = =1 0.01 1 F (1) = =
1 1 F ( 1) =G ( 1) = 0.841471 G ( 0.1) = 0.998334 G ( 0.01) =
0.999983x00 1 f ( 0) = =1 0 1 0.81 1 f ( 0.9 ) = = 1.9 0.9 1 0.9801
1 = 1.99 f ( 0.99 ) = 0.99 1 0.998001 1 = 1.999 f ( 0.999 ) = .999
1 1.002001 1 = 2.001 f (1.001) = 1.001 1 1.0201 1 = 2.01 f (1.01) =
1.01 1 1.21 1 = 2.1 f (1.1) = 1.1 1 4 1 =3 f ( 2) = 2 11 5G (
0.001) = 0.99999983 G ( 0.001) = 0.99999983 G ( 0.01) = 0.999983 G
( 0.1) = 0.998334 G (1) = 0.84147113. x 5 < 0.1 and x 5 > 0.1
x < 5.1 and x > 4.9 4.9 < x < 5.1 14. x 5 < and x 5
> x < 5 + and x > 5 5 < x < 5+ 15. a.True.b. False:
Choose a = 0.c.True.d. True16. sin ( c + h ) = sin c cos h + cos c
sin h 62Review and PreviewInstructors 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.
63. 1CHAPTERLimits 9.1.1 Concepts Reviewx3 4 x 2 + x + 6 x 1 x +1
lim( x + 1)( x 2 5 x + 6) x 1 x +11. L; c= lim2. 6= lim ( x 2 5 x +
6) x 1 23. L; right= (1) 5(1) + 64. lim f ( x) = M= 12x cProblem
Set 1.1x2 = lim( x 2 + 2 x 1) = 1 x 01. lim( x 5) = 2x 0x 32. lim
(1 2t ) = 311.t 13. 4.lim ( x 2 + 2 x 1) = (2) 2 + 2(2) 1 = 1= t t
= 2tlim ( x 2 + 2t 1) = (2) 2 + 2t 1 = 3 + 2tx 2(2) ( ( 1)6. lim t
2 x 2 = t 112.) ( ( 1) 1) = 05. lim t 2 1 = t 12)x2 4 ( x 2)( x +
2) = lim x2 x 2 x2 x2 = lim( x + 2)x2 9 x 3 x 3 ( x 3)( x + 3) =
lim x 3 x3 = lim( x + 3) limx 3 x2 = 1 x27. lim=3+3=6 13.x2lim(t +
4)(t 2) 4 (3t 6) 2t 2= lim=2+2=4 8.x2 t 2 ( x + t )( x t ) = lim xt
x + t x t x+t = lim ( x t ) limx tx 2(x 4 + 2 x3 x 210. lim(t 2) 2
t + 4 9(t 2) 2t 2t 2 + 4t 21 t 7 t+7 (t + 7)(t 3) = lim t 7 t+7 =
lim (t 3)t+4 9= limlimt 2=2+4 6 = 9 9t 7= 7 3 = 1014.(t 7)3 t 7limt
7+= lim t 7+= limt 7+(t 7) t 7 t 7 t 7= 77 = 0Instructors Resource
ManualSection 1.163 2007 Pearson Education, Inc., Upper Saddle
River, NJ. All rights reserved. This material is protected under
all copyright laws as they currently exist. No portion of this
material may be reproduced, in any form or by any means, without
permission in writing from the publisher. 64. 15. limx 4 18 x 2 +
81x 3( x 3)2= limx 3( x 3) 2 ( x + 3) 2= limx 3( x 3)( x 2 9) 2 ( x
3)= lim( x + 3)2 = (3 + 3) 22limt 02( x sin x ) 2 / x 2x21.x
31.0.02513140.12.775 1068(3u + 4)(u 1)30.012.77775 1010(u 1)
20.0012.77778 10141. 0.10.0251314 2.775 1060.012.77775
10100.0012.77778 1014= 36 (3u + 4)(2u 2)316. lim1 cos t =0 2t(u 1)
2u 1= limu 1= lim 8(3u + 4)(u 1) = 8[3(1) + 4](1 1) = 0 u 117.(2 +
h) 2 4 4 + 4h + h 2 4 = lim h0 h0 h h limh 2 + 4h = lim(h + 4) = 4
h 0 h 0 h= limlim( x sin x) 2 x2x 018.( x + h) 2 x 2 x 2 + 2 xh + h
2 x 2 = lim h0 h 0 h h limh 2 + 2 xh = lim(h + 2 x) = 2 x h 0 h 0
h2(1 cos x ) / xx22.=0 21.x19.0.002495840.01 0.0010.0000249996 2.5
1071.0.2113220.1sin x 2x0.2113220.1= lim0.00249584 0.0000249996 2.5
1071.0.4207350.10.4991670.010.4999920.010.0010.499999920.0011.0.4207350.10.4991670.010.4999920.0010.499999920.01
0.0012(t 1) /(sin(t
1))t23.=01.12.10351.012.010031.0012.0010.22984901.18840.02497920.91.903170.002499980.991.990030.9991.9991
cos t 2tt0.1x2x 03.56519sin x = 0.5 x 0 2 x1.(1 cos x)
22.lim20.lim0.00024999998t 1 =2 1) 2lim1.
0.10.02497920.010.002499980.001640.229849t 1
sin(t0.00024999998Section 1.1Instructors 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. 65. x sin( x 3) 3 x 3x24.4.1. + 4 0.1
+0.1585293.12.limx 4(1 + sin( x 3 / 2)) /( x )x 1. + 0.6741170.1 +
0.01 + 0.00500.001 + (x ) 2 4(tan x 1)20.119210.001 +0.1 +1 + sin (
x 32 ) xx =00.0000210862 2.12072 107 20.536908 20.01 +0.0500
0.0050lim0.001993390.01 + 20.45970.0005= 0.250.1 + 21. + 20.001 +
0.2505(2 2sin u ) / 3u0.00050.01 + 0.2550081. + 20.05000.1 +
0.300668u28.0.45971. + 0.249540.001 + 4x sin( x 3) 3 =0 lim x 3 x3
25.0.245009 0.01 + 40.0000166666 1.66667 1072.99940.1 +
40.001665832.990.201002 1. + 40.1585292.940.001 +0.0000166666
1.66667 1073.0010.0320244 0.01 +0.001665833.01( x / 4) 2 /(tan x 1)
2x27.0.00226446 20.001 + 20.0000213564 2.12342 1072 2sin u lim =0 u
3u 229. a.t(1 cot t ) /(1 / t
)1.0.3579070.10.8966640.010.9899670.0010.999d.1.1.64209e.0.11.09666f.0.011.009970.0011.001lim
f ( x) = 2x 326.limt 01 cot t 1 tb. f(3) = 1 c.g.= 1
h.i.Instructors Resource Manualf(1) does not exist. lim f ( x) =x
15 2f(1) = 2 lim f(x) does not exist. x1lim f ( x) = 2x 1lim f ( x)
= 1x 1+lim f ( x ) = +x 15 2Section 1.165 2007 Pearson Education,
Inc., Upper Saddle River, NJ. All rights reserved. This material is
protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any
means, without permission in writing from the publisher. 66. lim f
( x) does not exist.b.lim f ( x) does not exist.b.f(3) = 1c.f(1) =
2c.f(1) = 1d.30. a.d.x 3lim f ( x) = 2x 1e.lim f ( x) = 2x
1+34.f(1) = 1f.x 1lim f ( x) does not exist.g. h. i. 31. a. b. c.
d. e. f. 32. a. b. c.x 1lim f ( x) = 1x 1lim f ( x) does not
exist.x 1+lim f ( x ) = 2x 1+a.f(3) = 2 f(3) is undefined.x 1b.
g(1) does not exist.lim f ( x) = 2c.x 3lim f ( x) = 4x 3+d.lim f (
x) does not exist.x 3lim g ( x) = 035.lim g ( x ) = 1x2lim g ( x )
= 1x 2+f ( x) = x [ x ] lim f ( x) does not exist.x 3+lim f ( x) =
2x 1lim f ( x) = 2x 1+lim f ( x) = 2x 1d. f (1) = 2 e.lim f ( x) =
0f.f (1) = 0x 1a. b.33. c.d.a.66f(0) = 0 lim f ( x) does not
exist.x 0lim f ( x ) = 1x 0 lim f ( x) =x 1 21 2lim f ( x) = 0x
0Section 1.1Instructors 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. 67. f ( x)
=36.41. lim f ( x) exists for a = 1, 0, 1.x xxa42. The changed
values will not change lim f ( x) at xaany a. As x approaches a,
the limit is still a 2 . 43. a.x 1limx 1limx 1 x 1x 1b.f (0) does
not exist.a.lim f ( x) does not exist.b.x 0limx 1limc.does not
exist.x 1 x 1 x 1= 1 and lim+x 1x 1=1= 1x2 x 1 1 x 1x 1x 1= 3lim f
( x ) = 1c.x 0 d. 1 1 lim does not exist. x 1 x 1 x 1 d.lim f ( x)
= 1x 1244. a.x2 1 37. lim does not exist. x 1 x 1 limx 1x2 1 x2 1
=2 = 2 and lim x 1 x 1+ x 1x 0= lim( x + 2 2)( x + 2 + 2)= limx 0x
0x( x + 2 + 2)x 0= limx 01x1 does not exist. x 1/ xlim x (1)2 = = =
4 0+2 + 2 2 2 x+2+ 2=0b) 01c)=01/ xx 0+45. a) 1x( x + 2 + 2)1+x
0+x( x + 2 + 2) x+22x x =0lim x(1)d.= limb.limc.x 039. a.b.x+2 2
x38. limlim x 1+d)1146. a) Does not exist c)lim f ( x) does not
exist.1b) 0 d) 0.556x 1lim f ( x) = 047. lim x does not exist since
x 0x 040.x is not definedfor x < 0. 48.lim x x = 1x 0+49. limx
0x =0 x50. lim x = 1 x 0sin 2 x 1 = x 0 4 x 251. limInstructors
Resource ManualSection 1.267 2007 Pearson Education, Inc., Upper
Saddle River, NJ. All rights reserved. This material is protected
under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means,
without permission in writing from the publisher. 68. 52. limx 07.
If x is within 0.001 of 2, then 2x is within 0.002 of 4.sin 5 x 5 =
3x 31 53. lim cos does not exist. x 0 x 1 54. lim x cos = 0 x 0 xx3
155. lim56. limx 057.=62x + 2 2x 1x sin 2 x sin( x 2 )limx 258.
lim+x 18. If x is within 0.0005 of 2, then x2 is within 0.002 of
4.=2x2 x 2 = 3 x22 1/( x 1)1+ 2=059. lim x ; The computer gives a
value of 0, but x 0limx 09. If x is within 0.0019 of 2, then 0.002
of 4.8 x is withinx does not exist.1.2 Concepts Review 1. L ; L +
2. 0 < x a < ; f ( x) L < 10. If x is within 0.001 of 2,
then 3.8 is within 0.002 xof 4.34. ma + bProblem Set 1.2 1. 0 <
t a < f (t ) M < 2. 0 < u b < g (u ) L < 11. 0 <
x 0 < (2 x 1) (1) < 2x 1+ 1 < 2x < 3. 0 < z d <
h( z ) P < 2 x 0. ) x x=39. For x < 0, x = x, thus limx 0 x
x= limx 0 x = 1 x40. For x > 0, x = x, thus limx 0 +45.x x= limx
0 +x =1 x41. As x 0 , 1 + cos x 2 while sin x 0 . 1 + cos x lim =
sin x x 0limx 2x 2 = lim = 2, x 3 x 1 3 x2x 2 lim = lim = 2, x x 3
x 1 3 xHorizontal asymptote y = 2 2x 2x lim = , lim = ; x 3+ x 3 x
3 x 3 Vertical asymptote x = 342. 1 sin x 1 for all x, and 1 sin x
lim = 0, so lim = 0. x x x xInstructors Resource ManualSection
1.581 2007 Pearson Education, Inc., Upper Saddle River, NJ. All
rights reserved. This material is protected under all copyright
laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in
writing from the publisher. 82. 46.lim3 2x 9 3= 0, lim249. f ( x )
= 2 x + 3 = 0;x 9 x x Horizontal asymptote y = 0 3 3 lim = , lim =
, 9 x2 + 9 x2 x 3 x 3 3 3 lim = , lim = ; 2 2 x 3+ 9 x x 3 9 x
Vertical asymptotes x = 3, x = 350.14x 2 x2f ( x) = 3x + 4 4x + 3
x2 + 1, thusWe say that lim f ( x) = if to each x c +negative
number M there corresponds a > 0 such that 0 < x c < f(x)
< M.14= 0, lim, thus 4x + 3 lim [ f ( x) (3 x + 4)] = lim x x x
2 + 1 4+ 3 x x2 = lim =0. 1+ 1 x 2 x The oblique asymptote is y =
3x + 4.= 0; x 2 x2 + 7 +7 Horizontal asymptote y = 0 2 Since 2x + 7
> 0 for all x, g(x) has no vertical asymptotes. limx 11 lim [ f
( x) (2 x + 3)] = lim =0 3 x x x 1 The oblique asymptote is y = 2x
+ 3.51. a.47.1 3b. We say that lim f ( x) = if to each x c positive
number M there corresponds a > 0 such that 0 < c x < f(x)
> M. We say that lim f ( x) = if to each52. a.x positive number
M there corresponds an N > 0 such that N < x f(x) > M. b.
We say that lim f ( x ) = if to each x positive number M there
corresponds an N < 0 such that x < N f(x) > M. 53. Let
> 0 be given. Since lim f ( x ) = A, there is x 48.limx limx 2x
2x +5= lim2x x2 + 5x = lim2 1+x 5 x2=2 1+5 x22 1 =a corresponding
number M1 such that= 2, 2 1x > M1 f ( x) A < . Similarly,
there is a 2= 2Since x 2 + 5 > 0 for all x, g(x) has no vertical
asymptotes.number M2 such that x > M 2 g ( x) B < . 2 Let M =
max{M1 , M 2 } , then x > M f ( x) + g ( x) ( A + B) = f ( x) A
+ g ( x) B f ( x) A + g ( x) B= 2 2 Thus, lim [ f ( x) + g ( x)] =
A + B 0. Thus, there is at least one number c between 0 and 1 such
that x 3 + 3x 2 = 0. 53. Because the function is continuous on [
0,2 ] and (cos 0)03 + 6sin 5 0 3 = 3 < 0,Cost $0.60(cos 2)(2)3 +
6sin 5 (2) 3 = 83 3 > 0, there is at least one number c between
0 and 2 such0.48that (cos t )t 3 + 6sin 5 t 3 = 0.0.720.3654. Let f
( x ) = x 7 x + 14 x 8 . f(x) is continuous at all values of x.
f(0) = 8, f(5) = 12 Because 0 is between 8 and 12, there is at
least one number c between 0 and 5 such that 30.24 0.12 1 3 5 2 4 6
Length of call in minutes50. The function is continuous on the
intervals [0, 200], (200,300], (300, 400], 2f ( x ) = x 3 7 x 2 +
14 x 8 = 0 .This equation has three solutions (x = 1,2,4)Cost $ 80
60 4055. Let f ( x ) = x cos x. . f(x) is continuous at all20100
200 300 400 500 Miles Driven51. The function is continuous on the
intervals (0, 0.25], (0.25, 0.375], (0.375, 0.5], values of x 0.
f(0) = 1, f(/2) = / 2 Because 0 is between 1 and / 2 , there is at
least one number c between 0 and /2 such that f ( x ) = x cos x =
0. The interval [0.6,0.7] contains the solution.Cost $ 4 3 2
10.250.5 0.75 Miles Driven156. Let f ( x) = x5 + 4 x3 7 x + 14 f(x)
is continuous at all values of x. f(2) = 36, f(0) = 14 Because 0 is
between 36 and 14, there is at least one number c between 2 and 0
such that f ( x) = x5 + 4 x3 7 x + 14 = 0.86Section 1.6Instructors
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. 87. 57. Suppose that f is continuous
at c, so lim f ( x) = f (c). Let x = c + t, so t = x c, then x cas
x c , t 0 and the statement lim f ( x) = f (c) becomes lim f (t + c
) = f (c). x ct 0Suppose that lim f (t + c) = f (c) and let x = t +
t 0c, so t = x c. Since c is fixed, t 0 means that x c and the
statement lim f (t + c) = f (c) t 0becomes lim f ( x) = f (c) , so
f is continuous at x cc. 58. Since f(x) is continuous at c, lim f (
x) = f (c) > 0. Choose = f ( c ) , then x cthere exists a > 0
such that 0 < x c < f ( x) f (c) < . Thus, f ( x ) f ( c )
> = f ( c ) , or f ( x ) > 0 . Since also f ( c ) > 0 , f
( x ) > 0 for all x in (c , c + ).59. Let g(x) = x f(x). Then,
g(0) = 0 f(0) = f(0) 0 and g(1) = 1 f(1) 0 since 0 f(x) 1 on [0, 1]
. If g(0) = 0, then f(0) = 0 and c = 0 is a fixed point of f. If
g(1) = 0, then f(1) = 1 and c = 1 is a fixed point of f. If neither
g(0) = 0 nor g(1) = 0, then g(0) < 0 and g(1) > 0 so there is
some c in [0, 1] such that g(c) = 0. If g(c) = 0 then c f(c) = 0 or
f(c) = c and c is a fixed point of f. 60. F