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Solution Manual for Applied Calculus for The Managerial, Life and Social Sciences, 1st Edition
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Solution Manual for Applied Calculus for The Managerial, Life and Social Sciences, 1st Edition
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CHAPTER 1EXERCISES 1.1, page 9
1. The statement is false because −3 is greater than −20. (See the number line that
follows).
2. The statement is true because −5 is equal to −5.
3. The statement is false because 2/3 [which is equal to (4/6)] is less than 5/6.
4. The statement is false because −5/6 (which is −10/12) is greater than −11/12.
5. The interval (3,6) is shown on the number line that follows. Note that this is an
open interval indicated by ( and ).
x
0 3 6
6. The interval (−2,5] is shown on the number line that follows.
7. The interval [−1,4) is shown on the number line that follows. Note that this is a
half-open interval indicated by [ (closed) and ) (open).
x
–1 4
8. The closed interval [−6/5, −1/2] is shown on the number line that follows.
9. The infinite interval (0,∞) is shown on the number line that follows.
x
0
1 1 Preliminaries
,10. The infinite interval (−∞,5] is shown in the number line that follows.
11. First, 2x + 4 < 8 (Add −4 to each side of the inequality.)
Next, 2x < 4 (Multiply each side of the inequality by 1/2)
and x < 2.
We write this in interval notation as ( −∞,2).
12. −6 > 4 + 5x ⇒ −6 − 4 > 5x ⇒ −10 > 5x ⇒ −2 > x or x < −2.
We write this in interval notation as (−∞,−2).
13. We are given the inequality −4x > 20.
Then x < −5. (Multiply both sides of the inequality by −1/4
and reverse the sign of the inequality.)
We write this in interval notation as (−∞,−5].
14. −12 < −3x ⇒ 4 > x, or x < 4. We write this in interval notation as (−∞,4].
15. We are given the inequality −6 < x − 2 < 4.
First −6 + 2 < x < 4 + 2 (Add +2 to each member of the inequality.)
and −4 < x < 6,
so the solution set is the open interval (−4,6).
16. We add −1 to each member of the given double inequality 0 < x + 1 < 4 to
obtain
−1 < x < 3,
and the solution set is [−1,3].
17. We want to find the values of x that satisfy the inequalities
x + 1 > 4 or x + 2 < −1.
Adding −1 to both sides of the first inequality, we obtain
x + 1 − 1 > 4 − 1,
or x > 3.
Similarly, adding −2 to both sides of the second inequality, we obtain
x + 2 − 2 < −1 − 2,
or x < −3.
Therefore, the solution set is (−∞,−3) ∪ (3,∞).
18. We want to find the values of x that satisfy the inequalities
x + 1 > 2 or x − 1 < −2.
Solving these inequalities, we find that
1 Preliminaries 2
, x > 1 or x < −1,
and the solution set is (−∞,−1) ∪ (1,∞).
19. We want to find the values of x that satisfy the inequalities
x + 3 > 1 and x − 2 < 1.
Adding −3 to both sides of the first inequality, we obtain
x + 3 − 3 > 1 − 3,
or x > −2.
Similarly, adding 2 to each side of the second inequality, we obtain
x < 3,
and the solution set is (−2,3).
20. We want to find the values of x that satisfy the inequalities
x − 4 < 1 and x + 3 > 2.
Solving these inequalities, we find that
x < 5 and x > −1, and the solution set is (−1,5].
21. −6 + 2 = 4. 22. 4 + −4 = 4 + 4 = 8.
−12 + 4 −8 0.2 − 14
. −12
.
23. = = 2. 24. = = 15
..
16 − 12 4 . − 2.4 −0.8
16
25. 3 −2 + 3 − 3 = 3 ( 2 ) + 3 3 = 5 3 . 26. −1 + 2 −2 = 1 + 2 2 .
27. π − 1 + 2 = π − 1 + 2 = π + 1. 28. π − 6 − 3 = 6 − π − 3 = 3 − π .
29. 2 − 1 + 3 − 2 = 2 − 1 + 3 − 2 = 2.
30. 2 3 − 3 − 3 − 4 = 2 3 − 3 − (4 − 3 ) = 3 3 − 7.
31. False. If a > b, then −a < − b, −a + b < −b + b, and b − a < 0.
32. False. Let a = −2 and b = −3. Then a/b = −2/−3 < 1.
33. False. Let a = −2 and b = −3. Then a2 = 4 and b2 = 9, and 4 < 9. Note that we
only need to provide a counterexample to show that the statement is not always
true.
34. False. Let a = −2 and b = −3. Then 1/a = −1/2 and 1/b = −1/3 and −1/2 < −1/3.
3 1 Preliminaries
, 35. True. There are three possible cases.
Case 1 If a > 0, b > 0, then a3 > b3, since a3 − b3 = (a − b)(a2 + ab + b2) > 0.
Case 2 If a > 0, b < 0, then a3 > 0 and b3 < 0 and it follows that a3 > b3.
Case 3 If a < 0 and b < 0, then a3 − b3 = (a − b)(a2 + ab + b2) > 0, and we see
that a3 > b3. (Note that (a − b) > 0 and ab > 0.)
36. True. If a > b, then it follows that −a < −b because an inequality symbol is
reversed when both sides of the inequality are multiplied by a negative number.
37. False. Take a = −2, then − a = − ( −2) = 2 = 2 ≠ a.
38. True. If b < 0, then b2 > 0, and b 2 = b 2 .
39. True. If a − 4 < 0, then a − 4 = 4 − a = 4 − a . If a − 4 > 0, then
4 − a = a − 4 = a − 4.
40. False. Let a = −2, then a + 1 = −2 + 1 = −1 = 1 ≠ −2 + 1 = 3.
41. False. Take a = 3, b = −1. Then a + b = 3 − 1 = 2 ≠ a + b = 3 + 1 = 4.
42. False. Take a = 3, b = −1. Then a − b = 4 ≠ a − b = 3 − (1) = 2.
43. 272/3 = (33)2/3 = 32 = 9.
FG 1 IJ = 1 = 1 .
44. 8− =
H 8 K 2 16
4/3 4
FG 1 IJ 0
45.
H 3K = 1. Recall that any number raised to the zero power is 1.
46. (71/2)4 = 74/2 = 72 = 49.
LMFG 1IJ OP
1/ 3 −2
F 1I
=G J
−2
47.
MNH 8K PQ H 2K = (2 2 ) = 4.
LMFG − 1IJ OP
2 −3
FG 1 IJ −3
48.
MNH 3K PQ =
H 9K = (9) 3 = 729.
1 Preliminaries 4
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