Chapter 1: Answers 1 Jack K. Cohen Colorado School of Mines
- (1.56)
A = P2/16.
Figure:
Problem 1: Area as a function of perimeter .
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- (1.58) V =
.
Figure:
Problem 2: Volume as a function of surface area .
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- For a > 0, | a| = a. Moreover | - a| = a too. Note that
| - a| = | a| all the time!
- For a < 0, | a| = - a; | - a| = - a.
- For x≥a, the inequality reads
b≤x - a≤c. Adding a everywhere gives
a + b≤x≤a + c. Since c≥b, this inequality can be satisfied by a non-empty set and since b≥ 0 the lower limit is compatible with the assumption x≥a. Thus part of the solution set is the closed interval,
[a + b, a + c]. Similar analysis of the x < a range gives the other part of the solution set.
- The solution set is empty.
- The solution set is the same as for b = 0.
- The solution set of
b≤| x - a| < c is
(a - c, a - b]∪[a + b, a + c).
- The solution set of
b < | x - a|≤c is
[a - c, a - b)∪(a + b, a + c].
- The solution set of
b < | x - a| < c is
(a - c, a - b)∪(a + b, a + c).
- (1.19) After dividing by 5, we see that
a = 4/5, b = 2/5, c = 4/5. Thus the solution set
[a - c, a - b]∪[a + b, a + c] becomes
[0, 2/5]∪[6/5, 8/5].
- (1.17) If
c→ + ∞, then the solution set becomes
(- ∞, a - b]∪[a + b, + ∞). Inequality
| 1 - 3x| > 2 can be rewritten as
2/3 < | x - 1/3|, so
a = 1/3, b = 2/3. Noting the strict inequalities in this problem, the solution set becomes
(- ∞, -1/3)∪(1, + ∞).
- (1.15) If b = 0, then the solution set
[a - c, a - b]∪[a + b, a + c] becomes
[a - c, a]∪[a, a + c] or simply
[a - c, a + c]. Rewrite
| 3 - 2x| < 5 as
| x - 3/2| < 5/2 to identify
a = 3/2, c = 5/2. The solution set is then (- 1, 4). If we have strict inequality at the lower end (
0 < | 3 - 2x| < 5), then the point a = 3/2 is ``missing'' from the solution set:
(- 1, 3/2)∪(3/2, 4). Such ``punctured'' intervals play a key role in the study of limits.