Chapter 1: Answers 1 Jack K. Cohen Colorado School of Mines

  1. (1.56) A = P2/16.
    Figure: Problem 1: Area as a function of perimeter .
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  2. (1.58) V = ${\frac{{S^{2/3}}}{{6 \sqrt{\pi}}}}$.

    Figure: Problem 2: Volume as a function of surface area .
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  3. For a > 0, | a| = a. Moreover | - a| = a too. Note that | - a| = | a| all the time!

  4. For a < 0, | a| = - a; | - a| = - a.

  5. For xa, the inequality reads bx - ac. Adding a everywhere gives a + bxa + c. Since cb, this inequality can be satisfied by a non-empty set and since b≥ 0 the lower limit is compatible with the assumption xa. 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.

    1. The solution set is empty.
    2. The solution set is the same as for b = 0.

    1. The solution set of b≤| x - a| < c is (a - c, a - b]∪[a + b, a + c).
    2. The solution set of b < | x - a|≤c is [a - c, a - b)∪(a + b, a + c].
    3. The solution set of b < | x - a|  < c is (a - c, a - b)∪(a + b, a + c).

  6. (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].

  7. (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, + ∞).

  8. (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.