MATH 136 – Advanced Placement Calculus Solutions for Resubmitted Problems – Problem Set 4 October 26, 2009

4.3/12. (a) We have f(x) = cos^{2}(x) − 2 sin(x), so f′(x) = −2 cos(x) sin(x) − 2 cos(x) = −2 cos(x)(sin(x) + 1). So f′(x) = 0 at x = \frac{π}{2} and \frac{3π}{2} in the interval [0, 2π]. Since sin(x) + 1 ≥ 0, the sign of − cos(x) gives the sign of f′ and this is negative on [0, \frac{π}{2} ) and (\frac{3π}{2} , 2π], positive on (\frac{π}{2} , \frac{3π}{2} ).

(b) By the first derivative test, f(x) has a local maximum at x = \frac{3π}{2} and a local minimum at x = \frac{π}{2} .

(c) The second derivative is

f′′(x) = 2 sin^{2}(x) − 2 cos^{2}(x) + 2 sin(x)
= 2 sin^{2}(x) − 2(1 − sin^{2}(x)) + 2 sin(x)
= 4 sin^{2}(x) + 2 sin(x) − 2
= 2(sin(x) + 1)(2 sin(x) − 1)

Now, sin(x) = −1 at x = \frac{3π}{2} , while sin(x) = \frac{1}{2} at x = \frac{π}{6} and x = \frac{5π}{6} . Of these, only x = \frac{π}{6} and x = \frac{5π}{6} are inflection points because f′′(x) does not change sign at x = \frac{3π}{2} . (Note that f′(x) has a double root at x = \frac{3π}{2} . This means that the graph of f(x) is very “flat” around the maximum, but the concavity does not change there.)

4.4/14. The limit lim_{x→∞} \frac{(ln(x))^{2}}{x} is indeterminate of the form ∞/∞. Applying L’Hopital a first time, we examine

lim_{x→∞} \frac{2 ln(x)\frac{1}{x}}{1} = lim_{x→∞} \frac{2 ln(x)}{x}.

Since this is still indeterminate of the form ∞/∞, we apply L’Hopital’s Rule a second time:

lim_{x→∞} \frac{\frac{2}{x}}{1} = 0.

Hence the original limit is 0 also.

4.6/12. Let x be the side of the base of the box and y be the height. We have 32000 = x^{2}y, so y = \frac{32000}{x^{2}}. The surface area of the bottom and the sides (no top) is

S = x^{2} + 4xy = x^{2} + 4x\frac{32000}{x^{2}} = x^{2} + \frac{168000}{x}.

Now we differentiate with respect to x and set the derivative equal to zero to find the critical numbers:

0 = S′ = 2x − \frac{168000}{x^{2}} = 0 ⇒ x^{3} = 64000

so x = 40. Substituting back, y = \frac{32000}{1600} = 20. This is a minimum of the S function (and the material needed for the box), since S′′ = 2 + \frac{336000}{x^{3}} > 0 for all x > 0. The second derivative test shows that S has a local minimum at x = 40, and this is a global minimum since it is the only critical number for x > 0 and S is differentiable for all x > 0.

4.8/30. Given f′′(x) = 8x^{3} + 5, and f′(1) = 8, f(1) = 0. Integrating once, f′(x) = 2x^{4} +5x+c. Then substituting x = 1 gives 8 = f′(1) = 2+5+c. Hence c = 1. Integrating again, f(x) = \frac{2}{5}x^{5} + \frac{5}{2}x^{2} + x + d. Substituting x = 1 again, \frac{2}{5} + \frac{5}{2} + 1 + d = 0. Hence d = \frac{−(4+25+10)}{10} = \frac{−39}{10}. Then

f(x) = \frac{2}{5}x^{5} + \frac{5}{2}x^{2} + x − \frac{39}{10}.