ALL CHAPTERS COVERED
,P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-01 JWDD027-Salas-v1 November 25, 2006 15:52
SECTION 1.2 1
CHAPTER 1
SECTION 1.2
1. rational 2. rational 3. rational
4. irrational 5. rational 6. irrational
7. rational 8. rational 9. rational
3 1
10. rational 11. = 0.75 12. 0.33 <
4 3
√ √ 2
13. 2 > 1.414 14. 4= 16 15. − < −0.285714
7
22
16. π< 17. |6| = 6 18. | − 4| = 4
7
19. | − 3 − 7| = 10 20. | − 5| − |8| = −3 21. | − 5| + | − 8| = 13
√ √
22. |2 − π| = π − 2 23. |5 − 5| = 5 − 5 24.
25. 26. 27.
28. 29. 30.
31. 32. 33.
34. 35. 36.
37. 38. 39.
40. 41. bounded, lower bound 0, upper bound 4
42. bounded above by 0 43. not bounded
44. bounded above by 4 45. not bounded
√
46. bounded; lower bound 0, upper bound 1 47. bounded above, upper bound 2
√ √ √
π
48. 2< 3
π<2 < π 3 < 3π
49. x0 = 2, x1 ∼
= 2.75, x2 ∼
= 2.58264, x3 ∼
= 2.57133, x4 ∼
= 2.57128, x5 ∼ = 2.57128; bounded; lower bound
∼ ∼
2, upper bound 3 (the smallest upper bound = 2.57128 · · ·); xn = 2.5712815907 (10 decimal places)
,P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-01 JWDD027-Salas-v1 November 25, 2006 15:52
2 SECTION 1.2
50. xn → 2.970...; bounded
51. x2 − 10x + 25 = (x − 5)2 52. 9(x − 23 )(x + 23 )
53. 8x6 + 64 = 8(x2 + 2)(x4 − 2x2 + 4) 54. 27(x − 23 )(x2 + 23 x + 49 )
55. 4x2 + 12x + 9 = (2x + 3)2 56. 4(x2 + 12 )2
57. x2 − x − 2 = (x − 2)(x + 1) = 0; x = 2, −1 58. −3, 3
59. x2 − 6x + 9 = (x − 3)2 ; x=3 60. − 12 , 3
61. x2 − 2x + 2 = 0; no real zeros 62. −4
63. no real zeros 64. no real zeros
5! 1 1 8! 8·7·6
65. 5! = 120 66. = = 67. = = 56
8! 8·7·6 336 3!5! 3·2·1
9! 9·8·7 7! 7!
68. = = 84 69. = =1
3!6! 3·2·1 0!7! 1 · 7!
p1 p2 p1 q2 + p2 q1
70. + = , p1 q2 + p2 q1 and q1 q2 are integers, and q1 q2 = 0
q1 q2 q1 q2
71. Let r be a rational number and s an irrational number. Suppose r + s is rational. Then (r + s) − r = s
is rational, a contradiction.
p1 p2 p 1 p2
72. = , p1 p2 and q1 q2 are integers, and q1 q2 = 0
q1 q2 q1 q 2
√
73. The product of a rational and an irrational number may either be rational or irrational; 0 · 2=0
√ √
is rational, 1 · 2 = 2 is irrational.
√ √ √
74. 2 + 3 2 = 4 2 irrational; π + (1 − π) = 1, rational.
√ √ √ √ √
( 2)( 3) = 6 irrational; ( 2)(3 2) = 6, rational.
√
75. Suppose that 2 = p/q where p and q are integers and q = 0. Assume that p and q have no common
factors (other than ±1). Then p2 = 2q 2 and p2 is even. This implies that p = 2r is even. Therefore
2q 2 = 4r2 which implies that q 2 is even, and hence q is even. It now follows that p and q are both
even, contradicting the assumption that p and q have no common factors.
√ p p2
76. Assume 3 = , where p and q have no common factors. Then 3 = 2 , so p2 = 3q 2 . Thus p2 is divisible
q q
by 3, and therefore p is divisible by 3, say p = 3a. Then 9a2 = 3q 2 , so 3a2 = q 2 , where q must also be
divisible by 3, contracting our assumption.
, P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-01 JWDD027-Salas-v1 November 25, 2006 15:52
SECTION 1.3 3
77. Let x be the length of a rectangle that has perimeter P . Then the width y of the rectangle is given by
y = (1/2)P − x and the area is
2 2
1 P P
A=x P −x = − x− .
2 4 4
It follows that the area is a maximum when x = P/4. Since y = P/4 when x = P/4, the rectangle
of perimeter P having the largest area is a square.
p p2
78. Circle: perimeter 2πr = p =⇒ r= =⇒ area = πr2 =
2π 4π
p 2 p2 p2
square: perimeter 4x = p =⇒ x = =⇒ area = x = < .
4 16 4π
p p
For an arbitrary rectangle, p = 2(x + y), so y = − x, and area = xy = x( − x). This is the
2 2
p p
equation of a parabola with vertex (hence maximum value) at x = . Thus y = and the rectangle
4 4
is a square. The circle still has larger area.
SECTION 1.3
1
1. 2 + 3x < 5 2. 2 (2x + 3) < 6 3. 16x + 64 ≤ 16
3x < 3 2x + 3 < 12 16x ≤ −48
9
x<1 x< 2 x ≤ −3
9
Ans: (−∞, 1) Ans: (−∞, 2) Ans: (−∞, −3]
4. 3x + 5 > 14 (x − 2) 5. 1
2 (1 + x) < 13 (1 − x) 6. 3x − 2 ≤ 1 + 6x
12x + 20 > x − 2 3(1 + x) < 2(1 − x) −3x ≤ 3
11x > −22 3 + 3x < 2 − 2x x ≥ −1
x > −2 5x < −1 Ans: [−1, ∞)
Ans: (−2, ∞) x< − 15
Ans: (−∞, − 15 )
7. x2 − 1 < 0 8. x2 + 9x + 20 < 0 9. x2 − x − 6 ≥ 0
(x + 1)(x − 1) < 0 (x + 5)(x + 4) < 0 (x − 3)(x + 2) ≥ 0
Ans: (−1, 1) Ans: (−5, −4) Ans: (∞, −2] ∪ [3, ∞)
10. x2 − 4x − 5 > 0 11. 2x2 + x − 1 ≤ 0 12. 3x2 + 4x − 4 ≥ 0
(x − 5)(x + 1) > 0 (2x − 1)(x + 1) ≤ 0 (3x − 2)(x + 2) ≥ 0
Ans: (−∞, −1) ∪ (5, ∞) Ans: [−1, 1/2] Ans: (−∞, −2] ∪ [2/3, ∞)
13. x(x − 1)(x − 2) > 0 14. x(2x − 1)(3x − 5) ≤ 0 15. x3 − 2x2 + x ≥ 0
x(x − 1)2 ≥ 0
Ans: (0, 1) ∪ (2, ∞) Ans: (−∞, 0] ∪ [ 12 , 53 ] Ans: [0, ∞)
,P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-01 JWDD027-Salas-v1 November 25, 2006 15:52
SECTION 1.2 1
CHAPTER 1
SECTION 1.2
1. rational 2. rational 3. rational
4. irrational 5. rational 6. irrational
7. rational 8. rational 9. rational
3 1
10. rational 11. = 0.75 12. 0.33 <
4 3
√ √ 2
13. 2 > 1.414 14. 4= 16 15. − < −0.285714
7
22
16. π< 17. |6| = 6 18. | − 4| = 4
7
19. | − 3 − 7| = 10 20. | − 5| − |8| = −3 21. | − 5| + | − 8| = 13
√ √
22. |2 − π| = π − 2 23. |5 − 5| = 5 − 5 24.
25. 26. 27.
28. 29. 30.
31. 32. 33.
34. 35. 36.
37. 38. 39.
40. 41. bounded, lower bound 0, upper bound 4
42. bounded above by 0 43. not bounded
44. bounded above by 4 45. not bounded
√
46. bounded; lower bound 0, upper bound 1 47. bounded above, upper bound 2
√ √ √
π
48. 2< 3
π<2 < π 3 < 3π
49. x0 = 2, x1 ∼
= 2.75, x2 ∼
= 2.58264, x3 ∼
= 2.57133, x4 ∼
= 2.57128, x5 ∼ = 2.57128; bounded; lower bound
∼ ∼
2, upper bound 3 (the smallest upper bound = 2.57128 · · ·); xn = 2.5712815907 (10 decimal places)
,P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-01 JWDD027-Salas-v1 November 25, 2006 15:52
2 SECTION 1.2
50. xn → 2.970...; bounded
51. x2 − 10x + 25 = (x − 5)2 52. 9(x − 23 )(x + 23 )
53. 8x6 + 64 = 8(x2 + 2)(x4 − 2x2 + 4) 54. 27(x − 23 )(x2 + 23 x + 49 )
55. 4x2 + 12x + 9 = (2x + 3)2 56. 4(x2 + 12 )2
57. x2 − x − 2 = (x − 2)(x + 1) = 0; x = 2, −1 58. −3, 3
59. x2 − 6x + 9 = (x − 3)2 ; x=3 60. − 12 , 3
61. x2 − 2x + 2 = 0; no real zeros 62. −4
63. no real zeros 64. no real zeros
5! 1 1 8! 8·7·6
65. 5! = 120 66. = = 67. = = 56
8! 8·7·6 336 3!5! 3·2·1
9! 9·8·7 7! 7!
68. = = 84 69. = =1
3!6! 3·2·1 0!7! 1 · 7!
p1 p2 p1 q2 + p2 q1
70. + = , p1 q2 + p2 q1 and q1 q2 are integers, and q1 q2 = 0
q1 q2 q1 q2
71. Let r be a rational number and s an irrational number. Suppose r + s is rational. Then (r + s) − r = s
is rational, a contradiction.
p1 p2 p 1 p2
72. = , p1 p2 and q1 q2 are integers, and q1 q2 = 0
q1 q2 q1 q 2
√
73. The product of a rational and an irrational number may either be rational or irrational; 0 · 2=0
√ √
is rational, 1 · 2 = 2 is irrational.
√ √ √
74. 2 + 3 2 = 4 2 irrational; π + (1 − π) = 1, rational.
√ √ √ √ √
( 2)( 3) = 6 irrational; ( 2)(3 2) = 6, rational.
√
75. Suppose that 2 = p/q where p and q are integers and q = 0. Assume that p and q have no common
factors (other than ±1). Then p2 = 2q 2 and p2 is even. This implies that p = 2r is even. Therefore
2q 2 = 4r2 which implies that q 2 is even, and hence q is even. It now follows that p and q are both
even, contradicting the assumption that p and q have no common factors.
√ p p2
76. Assume 3 = , where p and q have no common factors. Then 3 = 2 , so p2 = 3q 2 . Thus p2 is divisible
q q
by 3, and therefore p is divisible by 3, say p = 3a. Then 9a2 = 3q 2 , so 3a2 = q 2 , where q must also be
divisible by 3, contracting our assumption.
, P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-01 JWDD027-Salas-v1 November 25, 2006 15:52
SECTION 1.3 3
77. Let x be the length of a rectangle that has perimeter P . Then the width y of the rectangle is given by
y = (1/2)P − x and the area is
2 2
1 P P
A=x P −x = − x− .
2 4 4
It follows that the area is a maximum when x = P/4. Since y = P/4 when x = P/4, the rectangle
of perimeter P having the largest area is a square.
p p2
78. Circle: perimeter 2πr = p =⇒ r= =⇒ area = πr2 =
2π 4π
p 2 p2 p2
square: perimeter 4x = p =⇒ x = =⇒ area = x = < .
4 16 4π
p p
For an arbitrary rectangle, p = 2(x + y), so y = − x, and area = xy = x( − x). This is the
2 2
p p
equation of a parabola with vertex (hence maximum value) at x = . Thus y = and the rectangle
4 4
is a square. The circle still has larger area.
SECTION 1.3
1
1. 2 + 3x < 5 2. 2 (2x + 3) < 6 3. 16x + 64 ≤ 16
3x < 3 2x + 3 < 12 16x ≤ −48
9
x<1 x< 2 x ≤ −3
9
Ans: (−∞, 1) Ans: (−∞, 2) Ans: (−∞, −3]
4. 3x + 5 > 14 (x − 2) 5. 1
2 (1 + x) < 13 (1 − x) 6. 3x − 2 ≤ 1 + 6x
12x + 20 > x − 2 3(1 + x) < 2(1 − x) −3x ≤ 3
11x > −22 3 + 3x < 2 − 2x x ≥ −1
x > −2 5x < −1 Ans: [−1, ∞)
Ans: (−2, ∞) x< − 15
Ans: (−∞, − 15 )
7. x2 − 1 < 0 8. x2 + 9x + 20 < 0 9. x2 − x − 6 ≥ 0
(x + 1)(x − 1) < 0 (x + 5)(x + 4) < 0 (x − 3)(x + 2) ≥ 0
Ans: (−1, 1) Ans: (−5, −4) Ans: (∞, −2] ∪ [3, ∞)
10. x2 − 4x − 5 > 0 11. 2x2 + x − 1 ≤ 0 12. 3x2 + 4x − 4 ≥ 0
(x − 5)(x + 1) > 0 (2x − 1)(x + 1) ≤ 0 (3x − 2)(x + 2) ≥ 0
Ans: (−∞, −1) ∪ (5, ∞) Ans: [−1, 1/2] Ans: (−∞, −2] ∪ [2/3, ∞)
13. x(x − 1)(x − 2) > 0 14. x(2x − 1)(3x − 5) ≤ 0 15. x3 − 2x2 + x ≥ 0
x(x − 1)2 ≥ 0
Ans: (0, 1) ∪ (2, ∞) Ans: (−∞, 0] ∪ [ 12 , 53 ] Ans: [0, ∞)
