introduccion

1. CHAPTER 0 Preliminaries
0.1 Concepts Review 1 ⎡ 2 1 ⎛ 1 1 ⎞⎤ 1
8. − ⎢ − ⎜ − ⎟ ⎥ = −
3 ⎣ 5 2 ⎝ 3 5 ⎠⎦ ⎡ 2 1 ⎛ 5 3 ⎞⎤
1. rational numbers 3 ⎢ − ⎜ − ⎟⎥
⎣ 5 2 ⎝ 15 15 ⎠ ⎦
2. dense 1 ⎡ 2 1 ⎛ 2 ⎞⎤ 1 ⎡2 1 ⎤
= − ⎢ − ⎜ ⎟⎥ = − ⎢ − ⎥
3. If not Q then not P. 3 ⎣ 5 2 ⎝ 15 ⎠ ⎦ 3 ⎣ 5 15 ⎦
1⎛ 6 1 ⎞ 1⎛ 5 ⎞ 1
4. theorems =− ⎜ − ⎟=− ⎜ ⎟=−
3 ⎝ 15 15 ⎠ 3 ⎝ 15 ⎠ 9
Problem Set 0.1 2 2
14 ⎛ 2 ⎞ 14 ⎛ 2 ⎞
2
⎟ = ⎜ ⎟ = ⎛ ⎞
14 6
9. ⎜ ⎜ ⎟
1. 4 − 2(8 − 11) + 6 = 4 − 2(−3) + 6 21 ⎜ 5 − 1 ⎟ 21 ⎜ 14 ⎟ 21 ⎝ 14 ⎠
⎝ 3⎠ ⎝ 3 ⎠
= 4 + 6 + 6 = 16 2
14 ⎛ 3 ⎞ 2⎛ 9 ⎞ 6
= ⎜ ⎟ = ⎜ ⎟=
21 ⎝ 7 ⎠ 3 ⎝ 49 ⎠ 49
2. 3 ⎡ 2 − 4 ( 7 − 12 ) ⎤ = 3[ 2 − 4(−5) ]
⎣ ⎦
= 3[ 2 + 20] = 3(22) = 66 ⎛2 ⎞ ⎛ 2 35 ⎞ ⎛ 33 ⎞
⎜ − 5⎟ ⎜ − ⎟ ⎜ − ⎟
10. ⎝
7 ⎠ = ⎝ 7 7 ⎠ = ⎝ 7 ⎠ = − 33 = − 11
3. –4[5(–3 + 12 – 4) + 2(13 – 7)] ⎛ 1⎞ ⎛7 1⎞ ⎛6⎞ 6 2
= –4[5(5) + 2(6)] = –4[25 + 12] ⎜1 − ⎟ ⎜ − ⎟ ⎜ ⎟
⎝ 7⎠ ⎝7 7⎠ ⎝7⎠
= –4(37) = –148
11 – 12 11 – 4 7
4. 5 [ −1(7 + 12 − 16) + 4] + 2 11. 7 21 = 7 7 = 7 = 7
11 + 12 11 + 4 15 15
= 5 [ −1(3) + 4] + 2 = 5 ( −3 + 4 ) + 2 7 21 7 7 7
= 5 (1) + 2 = 5 + 2 = 7
1 3 7 4 6 7 5
− + − +
5
5 1 65 7 58 12. 2 4 8 = 8 8 8 = 8 =
5. – = – = 1 3 7 4 6 7 3 3
7 13 91 91 91 + − + −
2 4 8 8 8 8 8
3 3 1 3 3 1 1 1 2 3 2 1
6. + − = + − 13. 1 – =1– =1– = – =
4 − 7 21 6 −3 21 6 1
1+ 2 3 3 3 3 3
42 6 7 43 2
=− + − =−
42 42 42 42
3 3 3
14. 2 + = 2+ = 2+
5 2 5 7
1 ⎡1 ⎛ 1 1 ⎞ 1⎤ 1 ⎡1 ⎛ 3 – 4 ⎞ 1⎤ 1+ −
⎜ – ⎟+ = ⎜ ⎟+
3 ⎢ 2 ⎝ 4 3 ⎠ 6 ⎥ 3 ⎢ 2 ⎝ 12 ⎠ 6 ⎥
7. 2 2 2 2
⎣ ⎦ ⎣ ⎦
6 14 6 20
1 ⎡1 ⎛ 1 ⎞ 1⎤ = 2+ = + =
= ⎢ ⎜– ⎟+ ⎥ 7 7 7 7
3 ⎣ 2 ⎝ 12 ⎠ 6 ⎦
( )( ) ( 5) – ( 3)
2 2
1⎡ 1 4⎤ 15. 5+ 3 5– 3 =
= ⎢– + ⎥
3 ⎣ 24 24 ⎦
=5–3= 2
1⎛ 3 ⎞ 1
= ⎜ ⎟=
3 ⎝ 24 ⎠ 24
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2. ( ) = ( 5) ( 5 )( 3 ) + ( 3 )
2 2 2 12 4 2
16. 5− 3 −2 27. +
+
x + 2x x x + 2
2
= 5 − 2 15 + 3 = 8 − 2 15 12 4( x + 2) 2x
= + +
x( x + 2) x( x + 2) x( x + 2)
17. (3x − 4)( x + 1) = 3 x 2 + 3 x − 4 x − 4 12 + 4 x + 8 + 2 x 6 x + 20
= =
= 3x2 − x − 4 x( x + 2) x( x + 2)
2(3 x + 10)
18. (2 x − 3)2 = (2 x − 3)(2 x − 3) =
x( x + 2)
= 4 x2 − 6 x − 6 x + 9
2 y
= 4 x 2 − 12 x + 9 28. +
6 y − 2 9 y2 −1
(3x – 9)(2 x + 1) = 6 x 2 + 3 x –18 x – 9 2 y
19. = +
2
= 6 x –15 x – 9 2(3 y − 1) (3 y + 1)(3 y − 1)
2(3 y + 1) 2y
= +
20. (4 x − 11)(3x − 7) = 12 x 2 − 28 x − 33 x + 77 2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1)
= 12 x 2 − 61x + 77 6y + 2 + 2y 8y + 2
= =
2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1)
21. (3t 2 − t + 1) 2 = (3t 2 − t + 1)(3t 2 − t + 1) 2(4 y + 1) 4y +1
= =
4 3 2 3
= 9t − 3t + 3t − 3t + t − t + 3t − t + 1 2 2 2(3 y + 1)(3 y − 1) (3 y + 1)(3 y − 1)
= 9t 4 − 6t 3 + 7t 2 − 2t + 1 0
29. a. 0⋅0 = 0 b. is undefined.
0
22. (2t + 3)3 = (2t + 3)(2t + 3)(2t + 3)
= (4t 2 + 12t + 9)(2t + 3) c.
0
=0 d.
3
is undefined.
17 0
= 8t 3 + 12t 2 + 24t 2 + 36t + 18t + 27
= 8t 3 + 36t 2 + 54t + 27 e. 05 = 0 f. 170 = 1
x 2 – 4 ( x – 2)( x + 2) 0
23. = = x+2, x ≠ 2 30. If = a , then 0 = 0 ⋅ a , but this is meaningless
x–2 x–2 0
because a could be any real number. No
x 2 − x − 6 ( x − 3)( x + 2) 0
24. = = x+2, x ≠3 single value satisfies = a .
x−3 ( x − 3) 0
t 2 – 4t – 21 (t + 3)(t – 7) 31. .083
25. = = t – 7 , t ≠ −3 12 1.000
t +3 t +3
96
2
2x − 2x 2 x(1 − x) 40
26. =
3 2 2
x − 2x + x x( x − 2 x + 1) 36
−2 x( x − 1) 4
=
x( x − 1)( x − 1)
2
=−
x −1
2 Section 0.1 Instructor’s Resource Manual
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3. 32. .285714 35. 3.6
7 2.000000 3 11.0
14 9
60 20
56 18
40 2
35
36. .846153
50
13 11.000000
49
10 4
10
60
7
52
30
80
28
78
2
20
33. .142857 13
21 3.000000 70
21 65
90 50
84 39
60 11
42
37. x = 0.123123123...
180 1000 x = 123.123123...
168 x = 0.123123...
120 999 x = 123
105 123 41
150 x= =
999 333
147
3 38. x = 0.217171717 …
1000 x = 217.171717...
34. .294117... 10 x = 2.171717...
17 5.000000... → 0.2941176470588235 990 x = 215
34 215 43
x= =
160 990 198
153
39. x = 2.56565656...
70 100 x = 256.565656...
68 x = 2.565656...
20 99 x = 254
17 254
30 x=
99
17
130 40. x = 3.929292…
119 100 x = 392.929292...
11 x = 3.929292...
99 x = 389
389
x=
99
Instructor’s Resource Manual Section 0.1 3
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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...
( )
4
52. 2− 3 ≈ 0.0102051443
100 x = 19.99999...
10 x = 1.99999...
53. 4 1.123 – 3 1.09 ≈ 0.00028307388
90 x = 18
18 1
x= = 54. ( 3.1415 )−1/ 2 ≈ 0.5641979034
90 5
42. x = 0.399999… 55. 8.9π2 + 1 – 3π ≈ 0.000691744752
100 x = 39.99999...
10 x = 3.99999... 56. 4 (6π 2 − 2)π ≈ 3.661591807
90 x = 36
36 2 57. Let a and b be real numbers with a < b . Let n
x= = be a natural number that satisfies
90 5
1 / n < b − a . Let S = {k : k n > b} . Since
43. Those rational numbers that can be expressed a nonempty set of integers that is bounded
by a terminating decimal followed by zeros. below contains a least element, there is a
k 0 ∈ S such that k 0 / n > b but
p ⎛1⎞ 1
44. = p ⎜ ⎟ , so we only need to look at . If (k 0 − 1) / n ≤ b . Then
q ⎝q⎠ q
k0 − 1 k0 1 1
q = 2n ⋅ 5m , then = − >b− > a
n m n n n n
1 ⎛1⎞ ⎛1⎞ k 0 −1 k 0 −1
= ⎜ ⎟ ⋅ ⎜ ⎟ = (0.5)n (0.2)m . The product Thus, a < n ≤ b . If n < b , then choose
q ⎝ 2⎠ ⎝5⎠
k 0 −1 k0 − 2
of any number of terminating decimals is also a r= n . Otherwise, choose r = n .
n m
terminating decimal, so (0.5) and (0.2) , 1
Note that a < b − <r.
1 n
and hence their product, , is a terminating
q Given a < b , choose r so that a < r1 < b . Then
p choose r2 , r3 so that a < r2 < r1 < r3 < b , and so
decimal. Thus has a terminating decimal
q on.
expansion.
58. Answers will vary. Possible answer: ≈ 120 in 3
45. Answers will vary. Possible answer: 0.000001,
1 ft
≈ 0.0000010819... 59. r = 4000 mi × 5280 = 21,120, 000 ft
π 12 mi
equator = 2π r = 2π (21,120, 000)
46. Smallest positive integer: 1; There is no ≈ 132, 700,874 ft
smallest positive rational or irrational number.
60. Answers will vary. Possible answer:
47. Answers will vary. Possible answer: beats min hr day
3.14159101001... 70 × 60 × 24 × 365 × 20 yr
min hr day year
48. There is no real number between 0.9999… = 735,840, 000 beats
(repeating 9's) and 1. 0.9999… and 1 represent 2
⎛ 16 ⎞
the same real number. 61. V = πr 2 h = π ⎜ ⋅12 ⎟ (270 ⋅12)
⎝ 2 ⎠
49. Irrational ≈ 93,807, 453.98 in.3
volume of one board foot (in inches):
50. Answers will vary. Possible answers: 1× 12 × 12 = 144 in.3
−π and π , − 2 and 2 number of board feet:
93,807, 453.98
≈ 651, 441 board ft
51. ( 3 + 1)3 ≈ 20.39230485 144
4 Section 0.1 Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

5. 62. V = π (8.004) 2 (270) − π (8)2 (270) ≈ 54.3 ft.3 b. Every circle has area less than or equal to
9π. The original statement is true.
63. a. If I stay home from work today then it
c. Some real number is less than or equal to
rains. If I do not stay home from work, its square. The negation is true.
then it does not rain.
b. If the candidate will be hired then she 71. a. True; If x is positive, then x 2 is positive.
meets all the qualifications. If the
candidate will not be hired then she does b. False; Take x = −2 . Then x 2 > 0 but
not meet all the qualifications. x<0.
64. a. If I pass the course, then I got an A on the 1
. Then x = <x
2 1
final exam. If I did not pass the course, c. False; Take x = 4
2
thn I did not get an A on the final exam.
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 y = x 2 + 1 . Then y > x 2 .
week, then I did not finish my research
paper. e. True; Let y be any positive number. Take
y
65. a. If a triangle is a right triangle, then x = . Then 0 < x < y .
2
2 2 2
a + b = c . If a triangle is not a right
2 2 2 72. a. True; x + ( − x ) < x + 1 + ( − x ) : 0 < 1
triangle, then a + b ≠ c .
b. If the measure of angle ABC is greater than b. False; There are infinitely many prime
0o and less than 90o, it is acute. If the numbers.
measure of angle ABC is less than 0o or
c. True; Let x be any number. Take
greater than 90o, then it is not acute.
1 1
y = + 1 . Then y > .
66. a. If angle ABC is an acute angle, then its x x
measure is 45o. If angle ABC is not an
acute angle, then its measure is not 45o. d. True; 1/ n can be made arbitrarily close
to 0.
2 2 2 2
b. If a < b then a < b. If a ≥ b then
a ≥ b. e. True; 1/ 2n can be made arbitrarily close
to 0.
67. a. The statement, converse, and
contrapositive are all true. 73. a. If n is odd, then there is an integer k such
that n = 2k + 1. Then
b. The statement, converse, and n 2 = (2k + 1) 2 = 4k 2 + 4k + 1
contrapositive are all true.
= 2(2k 2 + 2k ) + 1
68. a. The statement and contrapositive are true.
The converse is false. b. Prove the contrapositive. Suppose n is
even. Then there is an integer k such that
b. The statement, converse, and
contrapositive are all false. n = 2k . Then n 2 = (2k )2 = 4k 2 = 2(2k 2 ) .
Thus n 2 is even.
69. a. Some isosceles triangles are not
equilateral. The negation is true. 74. Parts (a) and (b) prove that n is odd if and
b. All real numbers are integers. The original only if n 2 is odd.
statement is true.
75. a. 243 = 3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3
c. Some natural number is larger than its
square. The original statement is true. b. 124 = 4 ⋅ 31 = 2 ⋅ 2 ⋅ 31 or 22 ⋅ 31
70. a. Some natural number is not rational. The
original statement is true.
Instructor’s Resource Manual Section 0.1 5
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6. c. 5100 = 2 ⋅ 2550 = 2 ⋅ 2 ⋅1275 82. a. –2 b. –2
= 2 ⋅ 2 ⋅ 3 ⋅ 425 = 2 ⋅ 2 ⋅ 3 ⋅ 5 ⋅ 85
c. x = 2.4444...;
= 2 ⋅ 2 ⋅ 3 ⋅ 5 ⋅ 5 ⋅17 or 22 ⋅ 3 ⋅ 52 ⋅17 10 x = 24.4444...
x = 2.4444...
76. For example, let A = b ⋅ c 2 ⋅ d 3 ; then
9 x = 22
A2 = b 2 ⋅ c 4 ⋅ d 6 , so the square of the number 22
is the product of primes which occur an even x=
9
number of times.
d. 1
p p2
77. 2= ;2 = ; 2q 2 = p 2 ; Since the prime
q 2 3 2
q e. n = 1: x = 0, n = 2: x = , n = 3: x = – ,
factors of p 2 must occur an even number of 2 3
p 5
times, 2q2 would not be valid and = 2 n = 4: x =
q 4
must be irrational. 3
The upper bound is .
2
p p2
78. 3= ; 3= ; 3q 2 = p 2 ; Since the prime f. 2
q q2
factors of p 2 must occur an even number of 83. a. Answers will vary. Possible answer: An
p example
times, 3q 2 would not be valid and = 3 is S = {x : x 2 < 5, x a rational number}.
q
must be irrational. Here the least upper bound is 5, which is
real but irrational.
a
79. Let a, b, p, and q be natural numbers, so b. True
b
p a p aq + bp
and are rational. + = This 0.2 Concepts Review
q b q bq
sum is the quotient of natural numbers, so it is 1. [−1,5); (−∞, −2]
also rational.
2. b > 0; b < 0
p
80. Assume a is irrational, ≠ 0 is rational, and 3. (b) and (c)
q
p r q⋅r 4. −1 ≤ x ≤ 5
a⋅ = is rational. Then a = is
q s p⋅s
rational, which is a contradiction.
Problem Set 0.2
81. a. – 9 = –3; rational 1. a.
3
b. 0.375 = ; rational
8
b.
c. (3 2)(5 2) = 15 4 = 30; rational
c.
d. (1 + 3)2 = 1 + 2 3 + 3 = 4 + 2 3;
irrational
d.
6 Section 0.2 Instructor’s Resource Manual
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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. e. –3 < 1 – 6 x ≤ 4
9. –4 < –6 x ≤ 3
2 1 ⎡ 1 2⎞
f. > x ≥ – ; ⎢– , ⎟
3 2 ⎣ 2 3⎠
2. a. (2, 7) b. [−3, 4)
c. (−∞, −2] d. [−1, 3]
10. 4 < 5 − 3x < 7
3. x − 7 < 2 x − 5 −1 < −3x < 2
−2 < x;( − 2, ∞) 1 2 ⎛ 2 1⎞
> x > − ; ⎜− , ⎟
3 3 ⎝ 3 3⎠
4. 3x − 5 < 4 x − 6
1 < x; (1, ∞ )
11. x2 + 2x – 12 < 0;
–2 ± (2)2 – 4(1)(–12) –2 ± 52
x= =
2(1) 2
7 x – 2 ≤ 9x + 3
= –1 ± 13
5. –5 ≤ 2 x
5 ⎡ 5 ⎞ ( ) (
⎡ x – –1 + 13 ⎤ ⎡ x – –1 – 13 ⎤ < 0;
⎣ ⎦⎣ ⎦ )
x ≥ – ; ⎢– , ∞ ⎟
2 ⎣ 2 ⎠ ( –1 – 13, – 1 + 13 )
6. 5 x − 3 > 6 x − 4
1 > x;(−∞,1) 12. x 2 − 5 x − 6 > 0
( x + 1)( x − 6) > 0;
(−∞, −1) ∪ (6, ∞)
7. −4 < 3 x + 2 < 5
−6 < 3 x < 3
−2 < x < 1; (−2, −1) 13. 2x2 + 5x – 3 > 0; (2x – 1)(x + 3) > 0;
⎛1 ⎞
(−∞, −3) ∪ ⎜ , ∞ ⎟
⎝2 ⎠
8. −3 < 4 x − 9 < 11
6 < 4 x < 20
14. 4 x2 − 5x − 6 < 0
3 ⎛3 ⎞
< x < 5; ⎜ ,5 ⎟
2 ⎝2 ⎠ ⎛ 3 ⎞
(4 x + 3)( x − 2) < 0; ⎜ − , 2 ⎟
⎝ 4 ⎠
x+4
15. ≤ 0; [–4, 3)
x–3
Instructor’s Resource Manual Section 0.2 7
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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. 3x − 2 ⎛ 2⎤ 3
16. ≥ 0; ⎜ −∞, ⎥ ∪ (1, ∞) 20. >2
x −1 ⎝ 3⎦ x+5
3
−2 > 0
x+5
2 3 − 2( x + 5)
<5 >0
17.
x x+5
2 −2 x − 7 ⎛ 7⎞
−5 < 0 > 0; ⎜ −5, − ⎟
x x+5 ⎝ 2⎠
2 − 5x
< 0;
x
⎛2 ⎞
(– ∞, 0) ∪ ⎜ , ∞ ⎟
⎝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 ⎟
7 ⎝ 2⎠ ⎝3 ⎠
18. ≤7
4x
7
−7 ≤ 0
4x
7 − 28 x ⎛ 3⎤
≤ 0; 23. (2 x - 3)( x -1)2 ( x - 3) ≥ 0; ⎜ – ∞, ⎥ ∪ [3, ∞ )
4x ⎝ 2⎦
( −∞, 0 ) ∪ ⎡ , ∞ ⎞
1
⎢4 ⎟
⎣ ⎠
24. (2 x − 3)( x − 1) 2 ( x − 3) > 0;
( −∞,1) ∪ ⎛1,
3⎞
1 ⎜ ⎟ ∪ ( 3, ∞ )
19. ≤4 ⎝ 2⎠
3x − 2
1
−4≤ 0
3x − 2
25. x3 – 5 x 2 – 6 x < 0
1 − 4(3 x − 2)
≤0
3x − 2 x( x 2 – 5 x – 6) < 0
x( x + 1)( x – 6) < 0;
9 − 12 x ⎛ 2 ⎞ ⎡3 ⎞
≤ 0; ⎜ −∞, ⎟ ∪ ⎢ , ∞ ⎟ (−∞, −1) ∪ (0, 6)
3x − 2 ⎝ 3 ⎠ ⎣4 ⎠
26. x3 − x 2 − x + 1 > 0
( x 2 − 1)( x − 1) > 0
( x + 1)( x − 1) 2 > 0;
(−1,1) ∪ (1, ∞)
27. a. False. b. True.
c. False.
8 Section 0.2 Instructor’s Resource Manual
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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. b. True. 33. a. ( x + 1)( x 2 + 2 x – 7) ≥ x 2 – 1
c. False. x3 + 3 x 2 – 5 x – 7 ≥ x 2 – 1
x3 + 2 x 2 – 5 x – 6 ≥ 0
29. a. ⇒ Let a < b , so ab < b 2 . Also, a 2 < ab .
( x + 3)( x + 1)( x – 2) ≥ 0
Thus, a 2 < ab < b 2 and a 2 < b 2 . ⇐ Let
[−3, −1] ∪ [2, ∞)
a 2 < b 2 , so a ≠ b Then
0 < ( a − b ) = a 2 − 2ab + b 2
2
b. x4 − 2 x2 ≥ 8
< b 2 − 2ab + b 2 = 2b ( b − a ) x4 − 2 x2 − 8 ≥ 0
Since b > 0 , we can divide by 2b to get ( x 2 − 4)( x 2 + 2) ≥ 0
b−a > 0.
( x 2 + 2)( x + 2)( x − 2) ≥ 0
b. We can divide or multiply an inequality by (−∞, −2] ∪ [2, ∞)
any positive number.
a 1 1
a < b ⇔ <1⇔ < . c. ( x 2 + 1)2 − 7( x 2 + 1) + 10 < 0
b b a
[( x 2 + 1) − 5][( x 2 + 1) − 2] < 0
30. (b) and (c) are true.
( x 2 − 4)( x 2 − 1) < 0
(a) is false: Take a = −1, b = 1 .
( x + 2)( x + 1)( x − 1)( x − 2) < 0
(d) is false: if a ≤ b , then −a ≥ −b .
(−2, −1) ∪ (1, 2)
31. a. 3x + 7 > 1 and 2x + 1 < 3
3x > –6 and 2x < 2 34. a. 1
1.99 < < 2.01
x > –2 and x < 1; (–2, 1) x
1.99 x < 1 < 2.01x
b. 3x + 7 > 1 and 2x + 1 > –4 1.99 x < 1 and 1 < 2.01x
3x > –6 and 2x > –5
x<
1 and x > 1
5
x > –2 and x > – ; ( −2, ∞ ) 1.99 2.01
2 1 1
<x<
2.01 1.99
c. 3x + 7 > 1 and 2x + 1 < –4
⎛ 1 1 ⎞
5 ⎜ , ⎟
x > –2 and x < – ; ∅ ⎝ 2.01 1.99 ⎠
2
32. a. 2 x − 7 > 1 or 2 x + 1 < 3
2 x > 8 or 2 x < 2
b. 1
x > 4 or x < 1 2.99 < < 3.01
x+2
(−∞,1) ∪ (4, ∞) 2.99( x + 2) < 1 < 3.01( x + 2)
2.99 x + 5.98 < 1 and 1 < 3.01x + 6.02
b. 2 x − 7 ≤ 1 or 2 x + 1 < 3 − 4.98 and − 5.02
x< x>
2 x ≤ 8 or 2 x < 2 2.99 3.01
x ≤ 4 or x < 1 5.02 4.98
− <x<−
3.01 2.99
(−∞, 4]
⎛ 5.02 4.98 ⎞
⎜− ,− ⎟
c. 2 x − 7 ≤ 1 or 2 x + 1 > 3 ⎝ 3.01 2.99 ⎠
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, ∞)
Instructor’s Resource Manual Section 0.2 9
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10. 36. x + 2 < 1; 1
43. − 3 > 6;
–1 < x + 2 < 1 x
–3 < x < –1 1 1
− 3 < −6 or − 3 > 6
(–3, –1) x x
1 1
37. 4 x + 5 ≤ 10; + 3 < 0 or − 9 > 0
x x
−10 ≤ 4 x + 5 ≤ 10 1 + 3x 1− 9x
< 0 or > 0;
−15 ≤ 4 x ≤ 5 x x
15 5 ⎡ 15 5 ⎤ ⎛ 1 ⎞ ⎛ 1⎞
− ≤ x ≤ ; ⎢− , ⎥ ⎜ − , 0 ⎟ ∪ ⎜ 0, ⎟
4 4 ⎣ 4 4⎦ ⎝ 3 ⎠ ⎝ 9⎠
38. 2 x – 1 > 2; 5
44. 2+ > 1;
2x – 1 < –2 or 2x – 1 > 2 x
2x < –1 or 2x > 3; 5 5
2 + < –1 or 2 + > 1
1 3 ⎛ 1⎞ ⎛3 ⎞ x x
x < – or x > , ⎜ – ∞, – ⎟ ∪ ⎜ , ∞ ⎟
2 2 ⎝ 2⎠ ⎝2 ⎠ 5 5
3 + < 0 or 1 + > 0
x x
2x 3x + 5 x+5
39. −5 ≥ 7 < 0 or > 0;
7 x x
2x 2x ⎛ 5 ⎞
− 5 ≤ −7 or −5 ≥ 7 (– ∞, – 5) ∪ ⎜ – , 0 ⎟ ∪ (0, ∞)
7 7 ⎝ 3 ⎠
2x 2x
≤ −2 or ≥ 12
7 7 45. x 2 − 3x − 4 ≥ 0;
x ≤ −7 or x ≥ 42;
3 ± (–3)2 – 4(1)(–4) 3 ± 5
(−∞, −7] ∪ [42, ∞) x= = = –1, 4
2(1) 2
x ( x + 1)( x − 4) = 0; (−∞, −1] ∪ [4, ∞)
40. +1 < 1
4
x 4 ± (−4)2 − 4(1)(4)
−1 < + 1 < 1 46. x 2 − 4 x + 4 ≤ 0; x = =2
4 2(1)
x ( x − 2)( x − 2) ≤ 0; x = 2
−2 < < 0;
4
–8 < x < 0; (–8, 0) 47. 3x2 + 17x – 6 > 0;
–17 ± (17) 2 – 4(3)(–6) –17 ± 19 1
41. 5 x − 6 > 1; x= = = –6,
2(3) 6 3
5 x − 6 < −1 or 5 x − 6 > 1
⎛1 ⎞
5 x < 5 or 5 x > 7 (3x – 1)(x + 6) > 0; (– ∞, – 6) ∪ ⎜ , ∞ ⎟
⎝3 ⎠
7 ⎛7 ⎞
x < 1 or x > ;(−∞,1) ∪ ⎜ , ∞ ⎟
5 ⎝5 ⎠ 48. 14 x 2 + 11x − 15 ≤ 0;
42. 2 x – 7 > 3; −11 ± (11) 2 − 4(14)(−15) −11 ± 31
x= =
2x – 7 < –3 or 2x – 7 > 3 2(14) 28
2x < 4 or 2x > 10 3 5
x=− ,
x < 2 or x > 5; (−∞, 2) ∪ (5, ∞) 2 7
⎛ 3 ⎞⎛ 5⎞ ⎡ 3 5⎤
⎜ x + ⎟ ⎜ x − ⎟ ≤ 0; ⎢ − , ⎥
⎝ 2 ⎠⎝ 7⎠ ⎣ 2 7⎦
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 Instructor’s Resource Manual
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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. ε 59. x –1 < 2 x – 3
51. x−2 < ⇒ 6 x − 2 < ε ⇒ 6 x − 12 < ε
6 x –1 < 2 x – 6
ε ( x –1) 2 < (2 x – 6)2
52. x+4 < ⇒ 2 x + 4 < ε ⇒ 2x + 8 < ε
2 x 2 – 2 x + 1 < 4 x 2 – 24 x + 36
53. 3x − 15 < ε ⇒ 3( x − 5) < ε 3x 2 – 22 x + 35 > 0
⇒ 3 x−5 < ε (3x – 7)( x – 5) > 0;
ε ε ⎛ 7⎞
⇒ x−5 < ;δ = ⎜ – ∞, ⎟ ∪ (5, ∞)
⎝ 3⎠
3 3
54. 4 x − 8 < ε ⇒ 4( x − 2) < ε 60. 2x −1 ≥ x + 1
2
⇒ 4 x−2 <ε (2 x − 1)2 ≥ ( x + 1)
ε ε 4 x2 − 4 x + 1 ≥ x2 + 2 x + 1
⇒ x−2 < ;δ =
4 4
3x2 − 6 x ≥ 0
55. 6 x + 36 < ε ⇒ 6( x + 6) < ε 3 x( x − 2) ≥ 0
⇒ 6 x+6 <ε (−∞, 0] ∪ [2, ∞)
ε ε 2 2 x − 3 < x + 10
⇒ x+6 < ;δ = 61.
6 6
4 x − 6 < x + 10
56. 5 x + 25 < ε ⇒ 5( x + 5) < ε (4 x − 6) 2 < ( x + 10)2
⇒ 5 x+5 <ε 16 x 2 − 48 x + 36 < x 2 + 20 x + 100
ε ε
⇒ x+5 < ;δ = 15 x 2 − 68 x − 64 < 0
5 5
(5 x + 4)(3 x − 16) < 0;
57. C = π d ⎛ 4 16 ⎞
⎜– , ⎟
C – 10 ≤ 0.02 ⎝ 5 3⎠
πd – 10 ≤ 0.02
62. 3x − 1 < 2 x + 6
⎛ 10 ⎞
π ⎜ d – ⎟ ≤ 0.02 3x − 1 < 2 x + 12
⎝ π⎠
10 0.02 (3x − 1) 2 < (2 x + 12)2
d– ≤ ≈ 0.0064
π π 9 x 2 − 6 x + 1 < 4 x 2 + 48 x + 144
We must measure the diameter to an accuracy 5 x 2 − 54 x − 143 < 0
of 0.0064 in.
( 5 x + 11)( x − 13) < 0
5 ⎛ 11 ⎞
58. C − 50 ≤ 1.5, ( F − 32 ) − 50 ≤ 1.5; ⎜ − ,13 ⎟
9 ⎝ 5 ⎠
5
( F − 32 ) − 90 ≤ 1.5
9
F − 122 ≤ 2.7
We are allowed an error of 2.7 F.
Instructor’s Resource Manual Section 0.2 11
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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 Order property: x < y ⇔ xz < yz when z is positive.
2 2
⇒ x < y Transitivity
⇒ x2 < y 2 (x 2
= x2 )
Conversely,
x2 < y 2 ⇒ x < y
2 2
(x 2
= x
2
)
2 2 2
⇒ x – y <0 Subtract y from each side.
⇒ ( x – y )( x + y ) < 0 Factor the difference of two squares.
⇒ x – y <0 This is the only factor that can be negative.
⇒ x < y Add y to each side.
( a) ( b)
2 2 x–2 x + (–2)
64. 0 < a < b ⇒ a = and b = , so 67. =
2
x +9 x2 + 9
( a) < ( b)
2 2
, and, by Problem 63, x–2 x –2
≤ +
2 2 2
a < b ⇒ a< b. x +9 x +9 x +9
x–2 x 2 x +2
≤ + =
65. a. a – b = a + (–b) ≤ a + –b = a + b x + 9 x + 9 x + 9 x2 + 9
2 2 2
1 1
Since x 2 + 9 ≥ 9, ≤
a – b ≥ a – b ≥ a – b Use Property 4 2
b. x +9 9
of absolute values. x +2 x +2
≤
x2 + 9 9
c. a + b + c = ( a + b) + c ≤ a + b + c x +2
x–2
≤
≤ a+b+c x2 + 9 9
1 1 1 ⎛ 1 ⎞ 68. x ≤ 2 ⇒ x2 + 2 x + 7 ≤ x2 + 2 x + 7
− = +⎜−
⎜ x +2⎟
66.
x2 + 3 x +2 2
x +3 ⎝ ⎟
⎠ ≤ 4 + 4 + 7 = 15
1 1 1
≤ +− and x 2 + 1 ≥ 1 so ≤ 1.
2 x +2 2
x +3 x +1
1 1 x2 + 2 x + 7 1
= + Thus, = x2 + 2 x + 7
2 2
2
x +3 x +2 x +1 x +1
1 1 ≤ 15 ⋅1 = 15
= +
x +3 2 x +2
1 3 1 2 1 1
by the Triangular Inequality, and since 69. x4 + x + x + x+
2 4 8 16
1 1
x 2 + 3 > 0, x + 2 > 0 ⇒ > 0, > 0. 1 3 1 2 1 1
2
x +3 x +2 ≤ x4 + x + x + x +
2 4 8 16
x 2 + 3 ≥ 3 and x + 2 ≥ 2, so 1 1 1 1
≤ 1+ + + + since x ≤ 1.
1 1 1 1 2 4 8 16
≤ and ≤ , thus,
x +3 2 3 x +2 2 1 1 1 1
So x 4 + x3 + x 2 + x + ≤ 1.9375 < 2.
1 1 1 1 2 4 8 16
+ ≤ +
x +3 2 x +2 3 2
12 Section 0.2 Instructor’s Resource Manual
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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. 70. a. x < x2 77.
1 1 1
≤ + +
1
R 10 20 30
x − x2 < 0
1 6+3+ 2
x(1 − x) < 0 ≤
R 60
x < 0 or x > 1
1 11
≤
2 R 60
b. x <x
2 60
x −x<0 R≥
x( x − 1) < 0 11
0 < x <1 1 1 1 1
≥ + +
R 20 30 40
71. a ≠ 0 ⇒ 1 6+4+3
2 ≥
⎛ 1⎞ 1 R 120
0 ≤ ⎜ a – ⎟ = a2 – 2 +
⎝ a⎠ a2 120
R≤
1 1 13
so, 2 ≤ a 2 + or a 2 + ≥2.
2
a a2 60 120
Thus, ≤R≤
11 13
72. a < b
a + a < a + b and a + b < b + b 78. A = 4π r 2 ; A = 4π (10)2 = 400π
2a < a + b < 2b
4π r 2 − 400π < 0.01
a+b
a< <b
2 4π r 2 − 100 < 0.01
73. 0 < a < b 0.01
r 2 − 100 <
4π
a 2 < ab and ab < b 2
0.01 2 0.01
− < r − 100 <
a 2 < ab < b 2 4π 4π
a < ab < b 0.01 0.01
100 − < r < 100 +
4π 4π
74.
1
2
1
(
( a + b ) ⇔ ab ≤ a 2 + 2ab + b2
ab ≤
4
) δ ≈ 0.00004 in
1 2 1 1 2 1 2
⇔ 0 ≤ a − ab + b = a − 2ab + b 2
4 2 4 4
( )
1 2 0.3 Concepts Review
⇔ 0 ≤ (a − b) which is always true.
4
1. ( x + 2)2 + ( y − 3)2
75. For a rectangle the area is ab, while for a
2
⎛ a+b⎞ 2. (x + 4)2 + (y – 2)2 = 25
square the area is a 2 = ⎜ ⎟ . From
⎝ 2 ⎠
1 ⎛ a+b⎞
2 ⎛ −2 + 5 3 + 7 ⎞
Problem 74, ab ≤
(a + b) ⇔ ab ≤ ⎜ ⎟
3. ⎜ , ⎟ = (1.5,5)
2 ⎝ 2 ⎠ ⎝ 2 2 ⎠
so the square has the largest area.
d −b
4.
76. 1 + x + x 2 + x3 + … + x99 ≤ 0; c−a
(−∞, −1]
Instructor’s Resource Manual Section 0.3 13
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

14. Problem Set 0.3 5. d1 = (5 + 2) 2 + (3 – 4)2 = 49 + 1 = 50
1. d 2 = (5 − 10)2 + (3 − 8)2 = 25 + 25 = 50
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 )
⎛ –2 + 4 –2 + 3 ⎞ ⎛ 1 ⎞
9. ⎜ , ⎟ = ⎜1, ⎟ ;
⎝ 2 2 ⎠ ⎝ 2⎠
d = (−3 − 2)2 + (5 + 2)2 = 74 ≈ 8.60
2
⎛1 ⎞ 25
d = (1 + 2)2 + ⎜ – 3 ⎟ = 9 + ≈ 3.91
3. ⎝ 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 2
⎛ 3 7 ⎞ ⎛ 9 11 ⎞
d = ⎜ − ⎟ +⎜ − ⎟
⎝2 2⎠ ⎝2 2 ⎠
= 4 + 1 = 5 ≈ 2.24
d = (4 – 5)2 + (5 + 8) 2 = 170 ≈ 13.04
11 (x – 1)2 + (y – 1)2 = 1
4.
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 Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

15. 14. ( x − 4) 2 + ( y − 3) 2 = r 2 21. 4 x 2 + 16 x + 15 + 4 y 2 + 6 y = 0
(6 − 4) 2 + (2 − 3) 2 = r 2 ⎛ 3 9⎞
4( x 2 + 4 x + 4) + 4 ⎜ y 2 + y + ⎟ = −15 + 16 +
9
2 ⎝ 2 16 ⎠ 4
r = 4 +1 = 5
2
⎛ 3⎞ 13
( x − 4)2 + ( y − 3)2 = 5 4( x + 2)2 + 4 ⎜ y + ⎟ =
⎝ 4⎠ 4
2
⎛ 1+ 3 3 + 7 ⎞ ⎛ 3⎞ 13
15. center = ⎜ , ⎟ = (2, 5) ( x + 2)2 + ⎜ y + ⎟ =
⎝ 2 2 ⎠ ⎝ 4⎠ 16
1 1 ⎛ 3⎞ 13
radius = (1 – 3)2 + (3 – 7)2 = 4 + 16 center = ⎜ −2, − ⎟ ; radius =
2 2 ⎝ 4⎠ 4
1
= 20 = 5
2 105
22. 4 x 2 + 16 x + + 4 y2 + 3 y = 0
2 2 16
( x – 2) + ( y – 5) = 5
⎛ 3 9 ⎞
4( x + 4 x + 4) + 4 ⎜ y 2 + y + ⎟
2
16. Since the circle is tangent to the x-axis, r = 4. ⎝ 4 64 ⎠
( x − 3)2 + ( y − 4) 2 = 16 105 9
=− + 16 +
16 16
17. x 2 + 2 x + 10 + y 2 – 6 y –10 = 0 ⎛ 3⎞
2
4( x + 2)2 + 4 ⎜ y + ⎟ = 10
x2 + 2 x + y 2 – 6 y = 0 ⎝ 8⎠
2
( x 2 + 2 x + 1) + ( y 2 – 6 y + 9) = 1 + 9 ⎛ 3⎞ 5
( x + 2)2 + ⎜ y + ⎟ =
( x + 1) 2 + ( y – 3) 2 = 10 ⎝ 8⎠ 2
center = (–1, 3); radius = 10 ⎛ 3⎞ 5 10
center = ⎜ −2, − ⎟ ; radius = =
⎝ 8⎠ 2 2
18. x 2 + y 2 − 6 y = 16
2 –1 7−5
23. =1 24. =2
x 2 + ( y 2 − 6 y + 9) = 16 + 9 2 –1 4−3
x 2 + ( y − 3) 2 = 25
–6 – 3 9 −6 + 4
center = (0, 3); radius = 5 25. = 26. =1
–5 – 2 7 0−2
19. x 2 + y 2 –12 x + 35 = 0 5–0 5 6−0
27. =– 28. =1
x 2 –12 x + y 2 = –35 0–3 3 0+6
( x 2 –12 x + 36) + y 2 = –35 + 36 29. y − 2 = −1( x − 2)
( x – 6) 2 + y 2 = 1 y − 2 = −x + 2
center = (6, 0); radius = 1 x+ y−4 = 0
20. x 2 + y 2 − 10 x + 10 y = 0 30. y − 4 = −1( x − 3)
2 2
( x − 10 x + 25) + ( y + 10 y + 25) = 25 + 25 y − 4 = −x + 3
x+ y−7 = 0
( x − 5) 2 + ( y + 5)2 = 50
center = ( 5, −5 ) ; radius = 50 = 5 2 31. y = 2x + 3
2x – y + 3 = 0
32. y = 0x + 5
0x + y − 5 = 0
Instructor’s Resource Manual Section 0.3 15
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.