C H A P T E R 0
A Precalculus Review
Section 0.1
The Real Number Line and Order
Section 0.2
Absolute Value and Distance on the Real Number Line . . . . . . 3
Section 0.3
Exponents and Radicals . . . . . . . . . . . . . . . . . . . . . . . 5
Section 0.4
Factoring Polynomials . . . . . . . . . . . . . . . . . . . . . . . . 6
Section 0.5
Fractions and Rationalization . . . . . . . . . . . . . . . . . . . . 9
Practice Test
. . . . . . . . . . . . . . . . .2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
C H A P T E R 0
A Precalculus Review
Section 0.1
The Real Number Line and Order
Solutions to Odd-Numbered Exercises
1. Since 0.7 7
, it is rational.
10
3.
3
is irrational because is irrational.
2
5. 4.3451 is rational because it has a repeating decimal expansion.
3
64 4, it is rational.
7. Since 3
60 is irrational, since 60 is not the cube of a rational number.
9. 11. (a) Yes, if x 3, then 53 12 3 > 0.
13. 0 <
(b) No, if x 3, then 53 12 27 < 0.
3
3
(d) No, if x , then 5 12 2
2
0 < x2 < 8
5
5
(c) Yes, if x , then 5
12 12 > 0.
2
2
9
2
x2
< 2
4
2 < x < 10
(a) Yes, if x 4, then 2 < x < 10.
< 0.
(b) No, if x 10, then x is not less than 10.
(c) No, if x 0, then x is not greater than 2.
(d) Yes, if x 72, then 2 < x < 10.
15.
x5 ≥ 7
x55 ≥ 75
x
10
12
14
17. 4x 1 < 2x
16
x
2
0
21.
2
4 5x < 1
0
1
2
− 12
7
2
x
2
0
2
4
1 < 2x < 7
1
2 < x <
x > 1
3
1
> x1 >
4
4
1
4
3
4
25.
x
1
2
1
4 < 2x 3 < 4
4 3 < 2x 3 3 < 4 3
5x < 5
1
3
< x < 4
4
2
1
x < 2
19. 4 2x < 3x 1
1
3
> x > 4
4
x
2x < 1
x ≥ 12
23.
− 12
1
2
x
x
> 5
2 3
0
3x 2x > 30
5x > 30
x > 6
7
2
x
4
6
8
Section 0.2
2x 2 x < 6
27.
Absolute Value and Distance on the Real Number Line
3
3
−2
x
2x 2 x 6 < 0
−2
−1
0
1
2
2x 3x 2 < 0
3
3
3
Zeros of the polynomial 2x 3x 2 are x 2 and x 2. Testing the intervals , 2 , 2, 2, and 2, , you see
3
that the solution set is 2 < x < 2.
29.
Hydrochloric
acid
−2
Pure
water
31. R 115.95x and C 95x 750 and we have:
Oven
cleaner
x
0
2
4
Lemon
juice
6
8
R > C
10 12 14
Black Baking
coffee soda
115.95x > 95x 750
20.95x > 750
x >
750
35.7995 . . .
20.95
Therefore, x ≥ 36 units.
33. Let x be the number of miles driven each week. Then the
company reimburses according to the formula
35. Given a < b,
C 100 0.35x
(a) 2a < 2b is false.
(b) a 2 < b 2 is true.
According to the allocations, you have
(c) 6a < 6b is true.
200 ≤ C ≤ 250
(d)
200 ≤ 100 0.35x ≤ 250
1 1
< is false if ab > 0 and true if ab < 0.
a b
100 ≤ 0.35x ≤ 150
100
150
≤ x ≤
0.35
0.35
285.71 < x < 428.57
Thus, the minimum is approximately 285.7 miles per
week, and the maximum is approximately 428.6.
Section 0.2
Absolute Value and Distance on the Real Number Line
1. (a) The directed distance from a to b is 75 126 51.
(b) The directed distance from b to a is 126 75 51.
(c) The distance between a and b is 75 126 51.
3. (a) The directed distance from a to b is 5.65 9.34 14.99.
(b) The directed distance from b to a is 9.34 5.65 14.99.
(c) The distance between a and b is 5.65 9.34 14.99.
16
128
5. (a) The directed distance from a to b is 112
75 5 75 .
112
128
(b) The directed distance from b to a is 16
5 75 75 .
(c) The distance between a and b is
7. x ≤ 2
112
75
128
16
5 75 .
9. x > 2
11. x 4 ≤ 2
4
Chapter 0
A Precalculus Review
13. x 2 > 2
15. x 4 < 2
19. 5 < x < 5
21.
x
−6
−4
−2
2
0
4
6
17. y a ≤ 2
x
< 3 or
2
x
> 3
2
x < 6
x > 6
23. 5 < x 2 < 5
7 < x < 3
7
3
x
−6
x3
≤ 5
2
25.
or
x
8
6
x3
≥ 5
2
27. 10 x < 4
x3
2 ≤ 52
2
x3
2 ≥ 52
2
x 3 ≤ 10
x 3 ≥ 10
x 3 3 ≤ 10 3
0
0
4
10 x > 4
x < 14
x > 6
x > 14
x < 6
x
6
x 3 3 ≥ 10 3
x ≤ 7
or
4
10
14
x ≥ 13
−7
13
x
10
0
10
1 < 9 2x < 1
29.
31.
b ≤ x a ≤ b
ab ≤ x ≤ ab
10 < 2x < 8
8b a < 3x < 8b a
x
a−b
4 < x < 5
a
a+b
1
1
a 8b < x < a 8b
3
3
x
4
3x a
< 2b
4
8b < 3x a < 8b
5 > x > 4
2
2b <
33.
6
x
a − 8b
3
35. Midpoint 7 21
14
2
41. 1083.4 0.2 ≤ M ≤ 1083.4 0.2
1083.2 ≤ M ≤ 1083.6
37. Midpoint 6.85 9.35
1.25
2
43.
a
3
39. Midpoint a + 8b
3
1
12 34
1
4
2
2 8
h 68.5
≤ 1
2.7
1 ≤
M 1083.4 ≤ 0.2
h 68.5
≤ 1
2.7
2.7 ≤ h 68.5 ≤ 2.7
65.8 ≤ h ≤ 71.2
45. x 200,000 ≤ 25,000
25,000 ≤ x 200,000 ≤ 25,000
175,000 ≤ x ≤ 225,000
47. (a) E 4750 ≤ 500 ⇒ 4250 ≤ E ≤ 5250
0.054750 237.50
E 4750 ≤ 237.50 ⇒ 4512.50 ≤ E ≤ 4987.50
(b) $5116.37 is not within 5% of the specified budgeted
amount; at variance.
Section 0.3
Exponents and Radicals
49. (a) E 20,000 ≤ 500 ⇒ 19,500 ≤ E ≤ 20,500
0.0520,000 1000
E 20,000 ≤ 1000 ⇒ 19,000 ≤ E ≤ 21,000
(b) $22,718.35 is at variance with both budget restrictions.
Section 0.3
Exponents and Radicals
1
1
3. 423 48 2
1. 323 38 24
5.
1 21 1 12 32
3
21
12
12
7. 322 423 34 48 12 32 44
3 272 11. 3 27 32 9
2
9. 6100 6100 61 1 5
13. 412 1
4
1
2
15. 3225 17. 5001.0160 908.3483
7x 2
7x 5
x3
29.
3xx 3xx
3x,
x12
x
2
1
1
22 4
3 154 5.3601
19. 21. 6y22y43 6y223 y12 6
25.
1
5 32
18 y
14
3
4y14
23. 10x 22 10x 4
27.
12x y3
4
x y5,
9x y2 3
31. (a) 8 4
x > 0
xy0
2 42 22
(b) 18 9
2 92 32
3 16x5 3 8x32x 2 3 8x3 3 2x 2 2x 3 2x 2
33. (a) 4 32x 4z5 4 16x 4z 42z 4 16x 4z 4 4 2z 2 x z 4 2z
(b) [Note: Since
x4
is under the radical, x could be positive or negative. For z5 to be under the radical, z must be positive.]
3 144x9y4z5 3 1823x33 y13y1z2z3
35. (a) 3 18y1z2
2x3y1z 2x3z
y
2x3z 3
18z2y2
y2
37. 4x 3 6x 2x2x 2 3
3
18z2
y
(b) 123x 57 3223x 563x 5
23x 5333x 5
23x 539x 15
39. 2x 52 x12 x122x 3 1 2x 3 1
x12
5
6
Chapter 0
A Precalculus Review
41. 3xx 132 6x 112 3x 112xx 1 2
43.
3x 112x 2 x 2
x 1x 12 x 13 x 12
x 1 x 1
x 12
x 12
3x 1 x 1x 2
12
x 12
2
x 12
2x 12
x 12
45. x 2 12x 112 2xx 112x 2 1 x2 1x 112x2 1 2xx 1
x 2 1x 1123x 2 2x 1
x 2 13x 2 2x 1
x 112
47. x 1 is defined when x ≥ 1. Therefore, the domain is 1, .
49. x 2 3 is defined for all real numbers. Therefore, the domain is , .
51.
1
is defined for all real numbers except x 1. Therefore, the domain is , 1 1, .
3 x 1
53. The numerator is defined for x ≥ 2. The denominator is defined for all x 4. Therefore, the domain is 2, 4 4, .
55. x 1 is defined when x ≥ 1, and 5 x is defined when x ≤ 5. Therefore, the domain of x 1 5 x is 1 ≤ x ≤ 5.
57. A 10,000 1 0.065
12
120
$19,121.84
59. A 5000 1 0.055
4
60
$11,345.46
32L
4
2
32
1
2
8
61. T 2
2
1
22
2
2
2.22 seconds
2
Section 0.4
Factoring Polynomials
1. Since a 6, b 1, and c 1, we have
x
1 ± 1 24 1 ± 5
.
12
12
Thus, x 15
15 1
1
or x .
12
2
12
3
3. Since a 4, b 12, and c 9, we have
x
12 ± 144 144 12 3
.
8
8
2
Section 0.4
5. Since a 1, b 4, and c 1, we have
y
7
7. Since a 2, b 3, and c 4, we have
4 ± 16 4 4 ± 23
2 ± 3.
2
2
9. x 2 4x 4 x 22
Factoring Polynomials
x
3 ± 9 424 3 ± 41
4
4
11. 4x 2 4x 1 2x 12
13. x 2 x 2 x 2x 1
15. 3x 2 5x 2 3x 2x 1
17. x 2 4xy 4y 2 x 2y2
19. 81 y 4 9 y 29 y 2
21. x3 8 x3 23
x 2x 2 2x 4
9 y 23 y3 y
23. y3 64 y3 43
25. x3 27 x3 33
y 4 y 2 4y 16
x 3x 2 3x 9
27. x3 4x 2 x 4 x 2x 4 x 4
29. 2x3 3x 2 4x 6 x 22x 3 22x 3
x 4x 2 1
2x 3x 2 2
x 4x 1x 1
31. 2x3 4x 2 x 2 2x 2x 2 x 2
33. x4 15x2 16 x2 16x2 1
x 22x 2 1
35. x 2 5x 0
x 4x 4x2 1
37.
xx 5 0
x2 9 0
x 3, 3
x 32 9 0
x 2 6x 9 9 0
43.
x2 x 2 0
x ± 3
45.
x 2x 1 0
xx 6 0
x2 3 0
x 3 x 3 0
x 3x 3 0
x 0, 5
41.
39.
x 2 5x 6 0
x 2x 3 0
x 2, 1
x 2, 3
x 0, 6
47. x3 64 0
x3 64
3
x 64 4
49. x 4 16 0
51.
x3 x 2 4x 4 0
x 2x 1 4x 1 0
x 4 16
4
16 ± 2
x ±
x 1x 2 4 0
x 1x 2x 2 0
x 1, ± 2
53. Since x2 4 x 2x 2, the roots are x ± 2.
By testing points inside and outside the interval 2, 2,
we find that the expression is defined when x ≤ 2 or
x ≥ 2. Thus, the domain is , 2 2, .
55. Since x 2 7x 12 x 3x 4, the roots are
x 3 and x 4. By testing points inside and outside the
interval 3, 4, we find that the expression is defined when
x ≤ 3 or x ≥ 4. Thus, the domain is , 3 4, .
8
Chapter 0
57. 1
A Precalculus Review
1
3
1
6
4
2
2
1
4
2
0
59. 1
1
2
2
1
1
1
2
1
1
0
Therefore, the factorization is
Therefore, the factorization is
2x3 x 2 2x 1 x 12x 2 x 1.
x 3 3x 2 6x 2 x 1x 2 4x 2
61. Possible rational zeros: ± 8, ± 4, ± 2, ± 1
Using synthetic division for x 1, we have
1
2
1
1
1
10
2
8
8
1
2
8
0
63. Possible rational roots: ± 1, ± 2, ± 3, ± 6
Using synthetic division for x 1, we have the following.
1
Therefore,
1
6
1
11
5
6
6
1
5
6
0
Therefore, we have
10x 8 0
x3 6x 2 11x 6 0
x 1x 2 2x 8 0
x 1x 2 5x 6 0
x 1x 2x 4 0
x 1x 2x 3 0
x3
x2
x 1, 2, 4
3
1
2
1
1
65. Possible rational zeros: ± 6, ± 3, ± 2, ±1, ± 2, ± 2, ± 3, ± 3, ± 6
Using synthetic division for x 3, we have
3
6
11
18
19
21
6
6
6
7
2
0
x 1, 2, 3.
67. Possible rational roots: ± 1, ± 2, ± 4
Using synthetic division for x 4, we have the following.
4
Therefore,
3
4
3
4
4
4
1
1
1
0
Therefore, we have
19x 6 0
x3 3x 2 3x 4 0
x 36x 2 7x 2 0
x 4x 2 x 1 0.
6x 3
1
11x 2
x 33x 22x 1 0
x 3,
69. 0.0003x 2 1200 0
2
3,
Since x 2 x 1 has no real solutions, x 4 is the only
real solution.
12
1.8 105 71.
0.0003x 2 1200
x2
4,000,000
x 2000 units
x2
1.0 104 x
1.8 109 1.8 105x x 2
x 1.8 105x 1.8 109 0
2
By the Quadratic Formula:
x
1.8 105 ± 1.8 1052 4 1.8 109
2
1.8 105 ± 7.524 109
2
3.437 105H
Section 0.5
Section 0.5
Fractions and Rationalization
Fractions and Rationalization
1.
x
5x x5
5
x1 x1 x1 x1
3.
2x
1 3x 2x 1 3x 5x 1
2
2
x2 2
x 2
x2 2
x 2
5.
2
1 x 2
1
2
x2 4 x 2 x 2x 2 x 2 x 2
7.
3
5
3
5
x3 3x x3 x3
9.
11.
2 x 2
x 2x 2
x
x2 4
x
4 x2
Ax 2 x 2 Bx 2 Cx 2 2x 1
x 12x 2
Ax 2 Ax 2A Bx 2B Cx 2 2Cx C
x 12x 2
A Cx 2 A B 2Cx 2A 2B C
x 12x 2
A
Bx C Ax 2 3 Bx Cx 6
2
x6
x 3
x 6x 2 3
1
2
x 2 1 2x
13. 2
x
x 1
xx 2 1
A Bx 2 C 6Bx 3A 6C
x 6x 2 3
x 2 2x 1
xx 2 1
x 2 2x 1
xx 2 1
x 12
xx 2 1
1
x
x
1
x 2 x 2 x 2 5x 6 x 1x 2 x 2x 3
x 3 xx 1
x 1x 2x 3
x 2 3
x 1x 2x 3
17.
2
x3
A
B
C
Ax 1x 2 Bx 2 Cx 12
x 1 x 12 x 2
x 12x 2
15.
x2 3
x 1x 2x 3
x
2
x 2x 1
x 132 x 112
x 132
x2
x 132
19.
1 t
2t
2t
1 t 1
21 t
21 t
2 t 21 t
21 t
3t
21 t
21 t
21 t
9
10
21.
Chapter 0
2xx
2
A Precalculus Review
1
x3
2xx 2 1 x3
x 2 1 x 2 1
1
x 2
23.
x3 2x
1x 2 1
xx 2 2
x 2 132
x 2 212 x 2x 2 212 x 2 212x 2 2 x 2
x2
x2
x 1
25.
x 2
1
x2 1
2
x 2x 2 2
x
x 1
x 1 x
2x 1
xx 1
x
1
27.
2x 1
x
x 2x 2
x 212 2x 212
2x 212
1
2xx 132
29.
31.
35.
39.
3x 4
2x 212
x2
2x
x2
2x
2x 3
2x 332 2x 312 2x 332 2x 312 2x 3
3
6
3
6
6
6
x 2 2x2x 3
2x 332
3x 2 6x
2x 332
3xx 2
2x 332
36 6
6
2
49x 3 49x 3
x 2 9
x 2 9
x 2 9
x 2 9
33.
37.
x
x 4
5
14 2
x
x 4
49x 3x 2 9
x 3x 3
49x 2 9
,x3
x3
2x
2x
5 3 5 3
5 3
5 3
41.
1
6 5
x 4
x 4
5
14 2
xx 4
x4
14 2
14 2
514 2
14 4
14 2
2
1
6 5
2x5 3 25 3
x5 3 11
6 5
6 5
65
6 5
6 5
Section 0.5
4 x 2
43.
2
x x 2
2
x x 2
x x 2
x x 2
2 x x 2
x x 2
47. P 10,000, r 0.14, N 60
0.1412
1
1
0.1412 1
60
2
x 24 x 2
4 x 2 2x 2
4
2
4x
x 4 x 232
x x 2
M 10,000
45.
x4
Fractions and Rationalization
$232.68
4 3x 2
4 x 232
x4
11
12
Chapter 0
A Precalculus Review
Practice Test for Chapter 0
4 81
1. Determine whether is rational or irrational.
2. Determine whether the given value of x satisfies the inequality 3x 4 ≤ x2.
(a) x 2
(b) x 0
8
(c) x 5
(d) x 6
3. Solve the inequality 3x 4 ≥ 13.
4. Solve the inequality x 2 < 6x 7.
5. Determine which of the two given real numbers is greater, 19 or
13
3.
6. Given the interval 3, 7, find (a) the distance between 3 and 7 and (b) the midpoint of the interval.
7. Solve the inequality 3x 1 ≤ 10.
8. Solve the inequality 4 5x > 29.
9. Solve the inequality 3 2x
< 8.
5
10. Use absolute values to describe the interval 3, 5.
11. Simplify
12x3
.
4x2
12. Simplify
3x3
x
,
0
x 0.
3 32x 4 y 3.
13. Remove all possible factors from the radical
3
1
1
14. Complete the factorization: 2x 113 4x 123 4x 113
15. Find the domain:
1
5 x
16. Factor completely: 3x 2 19x 14
17. Factor completely: 25x 2 81
18. Factor completely: x3 8
19. Use the Quadratic Formula to find all real roots of x 2 6x 2 0.
Practice Test for Chapter 0
20. Use the Rational Zero Theorem to find all real roots of x 3 4x 2 x 6 0.
21. Combine terms and simplify:
x
1
x 2 2x 3 x 1
22. Combine terms and simplify:
3x
x 5
2x 5
x 2
23. Combine terms and simplify:
24. Rationalize the denominator:
25. Rationalize the numerator:
x
x
x 2
2x 2
3y
y 2 9
x x 7
14
Graphing Calculator Required
26. Use a graphing calculator to find the real solutions of x3 5x 2 2x 8 0 by graphing y x3 5x 2 2x 8 and
finding the x-intercepts.
13