APPENDIXES
A Numbers, Inequalities, and Absolute Values
1. |5 − 23| = |−18| = 18
2. |5| − |−23| = 5 − 23 = −18
3. |−| = because 0.
4. | − 2| = − 2 because − 2 0.
√ √ √ √
5. 5 − 5 = − 5 − 5 = 5 − 5 because 5 − 5 0.
6. |−2| − |−3| = |2 − 3| = |−1| = 1
7. If 2, − 2 0, so | − 2| = −( − 2) = 2 − .
8. If 2, − 2 0, so | − 2| = − 2.
+1 if + 1 ≥ 0 +1 if ≥ −1
9. | + 1| = =
−( + 1) if + 1 0 − − 1 if −1
2 − 1 if 2 − 1 ≥ 0 2 − 1 if ≥ 1
2
10. |2 − 1| = =
−(2 − 1) if 2 − 1 0 1 − 2 if 1
2
11. 2 + 1 = 2 + 1 [since 2 + 1 ≥ 0 for all ].
√
12. Determine when 1 − 22 0 ⇔ 1 22 ⇔ 2 1
2
⇔ 2 12 ⇔ || 1
2
⇔
1 − 22 if − √12 ≤ ≤ 1
√
2
− √12 or √1 .
2
Thus, 1 − 22 =
22 − 1 if − √12 or √1
2
13. 2 + 7 3 ⇔ 2 −4 ⇔ −2, so ∈ (−2 ∞).
14. 3 − 11 4 ⇔ 3 15 ⇔ 5, so ∈ (−∞ 5).
15. 1 − ≤ 2 ⇔ − ≤ 1 ⇔ ≥ −1, so ∈ [−1 ∞).
16. 4 − 3 ≥ 6 ⇔ −3 ≥ 2 ⇔ ≤ − 23 , so ∈ −∞ − 23 .
17. 2 + 1 5 − 8 ⇔ 9 3 ⇔ 3 , so ∈ (3 ∞).
1
18. 1 + 5 5 − 3 ⇔ 8 4 ⇔ 12 , so ∈ 2
∞ .
19. −1 2 − 5 7 ⇔ 4 2 12 ⇔ 2 6, so ∈ (2 6).
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
° 1097
,1098 ¤ APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES
20. 1 3 + 4 ≤ 16 ⇔ −3 3 ≤ 12 ⇔ −1 ≤ 4, so ∈ (−1 4].
21. 0 ≤ 1 − 1 ⇔ −1 ≤ − 0 ⇔ 1 ≥ 0, so ∈ (0 1].
22. −5 ≤ 3 − 2 ≤ 9 ⇔ −8 ≤ −2 ≤ 6 ⇔ 4 ≥ ≥ −3, so ∈ [−3 4].
23. 4 2 + 1 ≤ 3 + 2. So 4 2 + 1 ⇔ 2 1 ⇔ 12 , and
2 + 1 ≤ 3 + 2 ⇔ −1 ≤ . Thus, ∈ −1 12 .
24. 2 − 3 + 4 3 − 2. So 2 − 3 + 4 ⇔ 7, and
+ 4 3 − 2 ⇔ 6 2 ⇔ 3 , so ∈ (3 7).
25. ( − 1)( − 2) 0.
Case 1: (both factors are positive, so their product is positive) − 1 0 ⇔ 1,
and − 2 0 ⇔ 2, so ∈ (2 ∞).
Case 2: (both factors are negative, so their product is positive) − 1 0 ⇔ 1,
and − 2 0 ⇔ 2, so ∈ (−∞ 1).
Thus, the solution set is (−∞ 1) ∪ (2 ∞).
26. (2 + 3) ( − 1) ≥ 0.
Case 1: 2 + 3 ≥ 0 ⇔ ≥ − 32 , and − 1 ≥ 0 ⇔ ≥ 1, so ∈ [1 ∞).
Case 2: 2 + 3 ≤ 0 ⇔ ≤ − 32 , and − 1 ≤ 0 ⇔ ≤ 1, so ∈ −∞ − 32 .
Thus, the solution set is −∞ − 32 ∪ [1 ∞).
27. 22 + ≤ 1 ⇔ 22 + − 1 ≤ 0 ⇔ (2 − 1) ( + 1) ≤ 0.
Case 1: 2 − 1 ≥ 0 ⇔ ≥ 12 , and + 1 ≤ 0 ⇔ ≤ −1,
which is an impossible combination.
Case 2: 2 − 1 ≤ 0 ⇔ ≤ 12 , and + 1 ≥ 0 ⇔ ≥ −1, so ∈ −1 12 .
Thus, the solution set is −1 12 .
28. 2 2 + 8 ⇔ 2 − 2 − 8 0 ⇔ ( − 4)( + 2) 0.
Case 1: 4 and −2, which is impossible.
Case 2: 4 and −2.
Thus, the solution set is (−2 4).
2
29. 2 + + 1 0 ⇔ 2 + + 1
4
+ 3
4
0 ⇔ 0. But since
+ 12 + 3
4
2
+ 12 ≥ 0 for every real , the original inequality will be true for all real as well.
Thus, the solution set is (−∞ ∞).
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
, APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES ¤ 1099
30. 2 + 1 ⇔ 2 + − 1 0. Using the quadratic formula, we obtain
√ √
2 + − 1 = − −1 −
2
5
− −1 +2 5 0.
√ √ √
Case 1: − −1 −
2
5
0 and − −1 +
2
5
0, so that −1 +
2
5
.
√ √ √
Case 2: − −1 −
2
0 and − −1 +2 5 0, so that
5 −1 −
2
5
.
√ √
Thus, the solution set is −∞ −1 − 2
5
∪ −1 + 5
2
∞ .
√ √
31. 2 3 ⇔ 2 − 3 0 ⇔ − 3 + 3 0.
√ √
Case 1: 3 and − 3, which is impossible.
√ √
Case 2: 3 and − 3.
√ √
Thus, the solution set is − 3 3 .
√ √ √
Another method: 2 3 ⇔ || 3 ⇔ − 3 3.
√ √
32. 2 ≥ 5 ⇔ 2 − 5 ≥ 0 ⇔ − 5 + 5 ≥ 0.
√ √ √
Case 1: ≥ 5 and ≥ − 5, so ∈ 5 ∞ .
√ √ √
Case 2: ≤ 5 and ≤ − 5, so ∈ −∞ − 5 .
√ √
Thus, the solution set is −∞ − 5 ∪ 5 ∞ .
√ √ √
Another method: 2 ≥ 5 ⇔ || ≥ 5 ⇔ ≥ 5 or ≤ − 5.
33. 3 − 2 ≤ 0 ⇔ 2 ( − 1) ≤ 0. Since 2 ≥ 0 for all , the inequality is satisfied when − 1 ≤ 0 ⇔ ≤ 1.
Thus, the solution set is (−∞ 1].
34. ( + 1)( − 2)( + 3) = 0 ⇔ = −1, 2, or −3. Construct a chart:
Interval +1 −2 +3 ( + 1)( − 2)( + 3)
−3 − − − −
−3 −1 − − + +
−1 2 + − + −
2 + + + +
Thus, ( + 1)( − 2)( + 3) ≥ 0 on [−3 −1] and [2 ∞), and the solution set
is [−3 −1] ∪ [2 ∞).
35. 3 ⇔ 3 − 0 ⇔ 2 − 1 0 ⇔ ( − 1)( + 1) 0. Construct a chart:
Interval −1 +1 ( − 1)( + 1)
−1 − − − −
−1 0 − − + +
01 + − + −
1 + + + +
Since 3 when the last column is positive, the solution set is (−1 0) ∪ (1 ∞).
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
, 1100 ¤ APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES
36. 3 + 3 42 ⇔ 3 − 42 + 3 0 ⇔ 2 − 4 + 3 0 ⇔ ( − 1)( − 3) 0.
Interval −1 −3 ( − 1)( − 3)
0 − − − −
01 + − − +
13 + + − −
3 + + + +
Thus, the solution set is (−∞ 0) ∪ (1 3).
37. 1 4. This is clearly true for 0. So suppose 0. then 1 4 ⇔
1
1 4 ⇔ 1
4
. Thus, the solution set is (−∞ 0) ∪ 4 ∞ .
38. −3 1 ≤ 1. We solve the two inequalities separately and take the intersection of the solution sets. First, −3 1 is
clearly true for 0. So suppose 0. Then −3 1 ⇔ −3 1 ⇔ − 13 , so for this inequality, the solution
set is −∞ − 13 ∪ (0 ∞). Now 1 ≤ 1 is clearly true if 0. So suppose 0. Then 1 ≤ 1 ⇔ 1 ≤ , and the
solution set here is (−∞ 0) ∪ [1 ∞).
Taking the intersection of the two solution sets gives the final solution set:
−∞ − 13 ∪ [1 ∞).
39. = 5
9
( − 32) ⇒ = 95 + 32. So 50 ≤ ≤ 95 ⇒ 50 ≤ 95 + 32 ≤ 95 ⇒ 18 ≤ 95 ≤ 63 ⇒
10 ≤ ≤ 35. So the interval is [10 35].
40. Since 20 ≤ ≤ 30 and = 5
9
( − 32), we have 20 ≤ 59 ( − 32) ≤ 30 ⇒ 36 ≤ − 32 ≤ 54 ⇒ 68 ≤ ≤ 86.
So the interval is [68 86].
41. (a) Let represent the temperature in degrees Celsius and the height in km. = 20 when = 0 and decreases by 10◦ C
for every km (1◦ C for each 100-m rise). Thus, = 20 − 10 when 0 ≤ ≤ 12.
(b) From part (a), = 20 − 10 ⇒ 10 = 20 − ⇒ = 2 − 10. So 0 ≤ ≤ 5 ⇒ 0 ≤ 2 − 10 ≤ 5 ⇒
−2 ≤ − 10 ≤ 3 ⇒ −20 ≤ − ≤ 30 ⇒ 20 ≥ ≥ −30 ⇒ −30 ≤ ≤ 20. Thus, the range of
temperatures (in ◦ C) to be expected is [−30 20].
42. The ball will be at least 32 ft above the ground if ≥ 32 ⇔ 128 + 16 − 162 ≥ 32 ⇔ 162 − 16 − 96 ≤ 0 ⇔
16( − 3)( + 2) ≤ 0. = 3 and = −2 are endpoints of the interval we’re looking for, and constructing a table gives
−2 ≤ ≤ 3. But ≥ 0, so the ball will be at least 32 ft above the ground in the time interval [0 3].
43. |2| = 3 ⇔ either 2 = 3 or 2 = −3 ⇔ = 3
2
or = − 32 .
44. |3 + 5| = 1 ⇔ either 3 + 5 = 1 or −1. In the first case, 3 = −4 ⇔ = − 43 , and in the second case,
3 = −6 ⇔ = −2. So the solutions are −2 and − 43 .
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
A Numbers, Inequalities, and Absolute Values
1. |5 − 23| = |−18| = 18
2. |5| − |−23| = 5 − 23 = −18
3. |−| = because 0.
4. | − 2| = − 2 because − 2 0.
√ √ √ √
5. 5 − 5 = − 5 − 5 = 5 − 5 because 5 − 5 0.
6. |−2| − |−3| = |2 − 3| = |−1| = 1
7. If 2, − 2 0, so | − 2| = −( − 2) = 2 − .
8. If 2, − 2 0, so | − 2| = − 2.
+1 if + 1 ≥ 0 +1 if ≥ −1
9. | + 1| = =
−( + 1) if + 1 0 − − 1 if −1
2 − 1 if 2 − 1 ≥ 0 2 − 1 if ≥ 1
2
10. |2 − 1| = =
−(2 − 1) if 2 − 1 0 1 − 2 if 1
2
11. 2 + 1 = 2 + 1 [since 2 + 1 ≥ 0 for all ].
√
12. Determine when 1 − 22 0 ⇔ 1 22 ⇔ 2 1
2
⇔ 2 12 ⇔ || 1
2
⇔
1 − 22 if − √12 ≤ ≤ 1
√
2
− √12 or √1 .
2
Thus, 1 − 22 =
22 − 1 if − √12 or √1
2
13. 2 + 7 3 ⇔ 2 −4 ⇔ −2, so ∈ (−2 ∞).
14. 3 − 11 4 ⇔ 3 15 ⇔ 5, so ∈ (−∞ 5).
15. 1 − ≤ 2 ⇔ − ≤ 1 ⇔ ≥ −1, so ∈ [−1 ∞).
16. 4 − 3 ≥ 6 ⇔ −3 ≥ 2 ⇔ ≤ − 23 , so ∈ −∞ − 23 .
17. 2 + 1 5 − 8 ⇔ 9 3 ⇔ 3 , so ∈ (3 ∞).
1
18. 1 + 5 5 − 3 ⇔ 8 4 ⇔ 12 , so ∈ 2
∞ .
19. −1 2 − 5 7 ⇔ 4 2 12 ⇔ 2 6, so ∈ (2 6).
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
° 1097
,1098 ¤ APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES
20. 1 3 + 4 ≤ 16 ⇔ −3 3 ≤ 12 ⇔ −1 ≤ 4, so ∈ (−1 4].
21. 0 ≤ 1 − 1 ⇔ −1 ≤ − 0 ⇔ 1 ≥ 0, so ∈ (0 1].
22. −5 ≤ 3 − 2 ≤ 9 ⇔ −8 ≤ −2 ≤ 6 ⇔ 4 ≥ ≥ −3, so ∈ [−3 4].
23. 4 2 + 1 ≤ 3 + 2. So 4 2 + 1 ⇔ 2 1 ⇔ 12 , and
2 + 1 ≤ 3 + 2 ⇔ −1 ≤ . Thus, ∈ −1 12 .
24. 2 − 3 + 4 3 − 2. So 2 − 3 + 4 ⇔ 7, and
+ 4 3 − 2 ⇔ 6 2 ⇔ 3 , so ∈ (3 7).
25. ( − 1)( − 2) 0.
Case 1: (both factors are positive, so their product is positive) − 1 0 ⇔ 1,
and − 2 0 ⇔ 2, so ∈ (2 ∞).
Case 2: (both factors are negative, so their product is positive) − 1 0 ⇔ 1,
and − 2 0 ⇔ 2, so ∈ (−∞ 1).
Thus, the solution set is (−∞ 1) ∪ (2 ∞).
26. (2 + 3) ( − 1) ≥ 0.
Case 1: 2 + 3 ≥ 0 ⇔ ≥ − 32 , and − 1 ≥ 0 ⇔ ≥ 1, so ∈ [1 ∞).
Case 2: 2 + 3 ≤ 0 ⇔ ≤ − 32 , and − 1 ≤ 0 ⇔ ≤ 1, so ∈ −∞ − 32 .
Thus, the solution set is −∞ − 32 ∪ [1 ∞).
27. 22 + ≤ 1 ⇔ 22 + − 1 ≤ 0 ⇔ (2 − 1) ( + 1) ≤ 0.
Case 1: 2 − 1 ≥ 0 ⇔ ≥ 12 , and + 1 ≤ 0 ⇔ ≤ −1,
which is an impossible combination.
Case 2: 2 − 1 ≤ 0 ⇔ ≤ 12 , and + 1 ≥ 0 ⇔ ≥ −1, so ∈ −1 12 .
Thus, the solution set is −1 12 .
28. 2 2 + 8 ⇔ 2 − 2 − 8 0 ⇔ ( − 4)( + 2) 0.
Case 1: 4 and −2, which is impossible.
Case 2: 4 and −2.
Thus, the solution set is (−2 4).
2
29. 2 + + 1 0 ⇔ 2 + + 1
4
+ 3
4
0 ⇔ 0. But since
+ 12 + 3
4
2
+ 12 ≥ 0 for every real , the original inequality will be true for all real as well.
Thus, the solution set is (−∞ ∞).
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
, APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES ¤ 1099
30. 2 + 1 ⇔ 2 + − 1 0. Using the quadratic formula, we obtain
√ √
2 + − 1 = − −1 −
2
5
− −1 +2 5 0.
√ √ √
Case 1: − −1 −
2
5
0 and − −1 +
2
5
0, so that −1 +
2
5
.
√ √ √
Case 2: − −1 −
2
0 and − −1 +2 5 0, so that
5 −1 −
2
5
.
√ √
Thus, the solution set is −∞ −1 − 2
5
∪ −1 + 5
2
∞ .
√ √
31. 2 3 ⇔ 2 − 3 0 ⇔ − 3 + 3 0.
√ √
Case 1: 3 and − 3, which is impossible.
√ √
Case 2: 3 and − 3.
√ √
Thus, the solution set is − 3 3 .
√ √ √
Another method: 2 3 ⇔ || 3 ⇔ − 3 3.
√ √
32. 2 ≥ 5 ⇔ 2 − 5 ≥ 0 ⇔ − 5 + 5 ≥ 0.
√ √ √
Case 1: ≥ 5 and ≥ − 5, so ∈ 5 ∞ .
√ √ √
Case 2: ≤ 5 and ≤ − 5, so ∈ −∞ − 5 .
√ √
Thus, the solution set is −∞ − 5 ∪ 5 ∞ .
√ √ √
Another method: 2 ≥ 5 ⇔ || ≥ 5 ⇔ ≥ 5 or ≤ − 5.
33. 3 − 2 ≤ 0 ⇔ 2 ( − 1) ≤ 0. Since 2 ≥ 0 for all , the inequality is satisfied when − 1 ≤ 0 ⇔ ≤ 1.
Thus, the solution set is (−∞ 1].
34. ( + 1)( − 2)( + 3) = 0 ⇔ = −1, 2, or −3. Construct a chart:
Interval +1 −2 +3 ( + 1)( − 2)( + 3)
−3 − − − −
−3 −1 − − + +
−1 2 + − + −
2 + + + +
Thus, ( + 1)( − 2)( + 3) ≥ 0 on [−3 −1] and [2 ∞), and the solution set
is [−3 −1] ∪ [2 ∞).
35. 3 ⇔ 3 − 0 ⇔ 2 − 1 0 ⇔ ( − 1)( + 1) 0. Construct a chart:
Interval −1 +1 ( − 1)( + 1)
−1 − − − −
−1 0 − − + +
01 + − + −
1 + + + +
Since 3 when the last column is positive, the solution set is (−1 0) ∪ (1 ∞).
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
, 1100 ¤ APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES
36. 3 + 3 42 ⇔ 3 − 42 + 3 0 ⇔ 2 − 4 + 3 0 ⇔ ( − 1)( − 3) 0.
Interval −1 −3 ( − 1)( − 3)
0 − − − −
01 + − − +
13 + + − −
3 + + + +
Thus, the solution set is (−∞ 0) ∪ (1 3).
37. 1 4. This is clearly true for 0. So suppose 0. then 1 4 ⇔
1
1 4 ⇔ 1
4
. Thus, the solution set is (−∞ 0) ∪ 4 ∞ .
38. −3 1 ≤ 1. We solve the two inequalities separately and take the intersection of the solution sets. First, −3 1 is
clearly true for 0. So suppose 0. Then −3 1 ⇔ −3 1 ⇔ − 13 , so for this inequality, the solution
set is −∞ − 13 ∪ (0 ∞). Now 1 ≤ 1 is clearly true if 0. So suppose 0. Then 1 ≤ 1 ⇔ 1 ≤ , and the
solution set here is (−∞ 0) ∪ [1 ∞).
Taking the intersection of the two solution sets gives the final solution set:
−∞ − 13 ∪ [1 ∞).
39. = 5
9
( − 32) ⇒ = 95 + 32. So 50 ≤ ≤ 95 ⇒ 50 ≤ 95 + 32 ≤ 95 ⇒ 18 ≤ 95 ≤ 63 ⇒
10 ≤ ≤ 35. So the interval is [10 35].
40. Since 20 ≤ ≤ 30 and = 5
9
( − 32), we have 20 ≤ 59 ( − 32) ≤ 30 ⇒ 36 ≤ − 32 ≤ 54 ⇒ 68 ≤ ≤ 86.
So the interval is [68 86].
41. (a) Let represent the temperature in degrees Celsius and the height in km. = 20 when = 0 and decreases by 10◦ C
for every km (1◦ C for each 100-m rise). Thus, = 20 − 10 when 0 ≤ ≤ 12.
(b) From part (a), = 20 − 10 ⇒ 10 = 20 − ⇒ = 2 − 10. So 0 ≤ ≤ 5 ⇒ 0 ≤ 2 − 10 ≤ 5 ⇒
−2 ≤ − 10 ≤ 3 ⇒ −20 ≤ − ≤ 30 ⇒ 20 ≥ ≥ −30 ⇒ −30 ≤ ≤ 20. Thus, the range of
temperatures (in ◦ C) to be expected is [−30 20].
42. The ball will be at least 32 ft above the ground if ≥ 32 ⇔ 128 + 16 − 162 ≥ 32 ⇔ 162 − 16 − 96 ≤ 0 ⇔
16( − 3)( + 2) ≤ 0. = 3 and = −2 are endpoints of the interval we’re looking for, and constructing a table gives
−2 ≤ ≤ 3. But ≥ 0, so the ball will be at least 32 ft above the ground in the time interval [0 3].
43. |2| = 3 ⇔ either 2 = 3 or 2 = −3 ⇔ = 3
2
or = − 32 .
44. |3 + 5| = 1 ⇔ either 3 + 5 = 1 or −1. In the first case, 3 = −4 ⇔ = − 43 , and in the second case,
3 = −6 ⇔ = −2. So the solutions are −2 and − 43 .
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
