M106 Practice Exam 3

[Note: These problems appeared on my M106 exams in the fall of 1995.]

Instructions: Show all of your work and clearly explain your answers. This is particularly important on problems with a numerical answer, to allow the possibility of partial credit. No books are allowed during the exam, but you may use your calculator.

Problem [1] Suppose that f(x) is a differentiable function such that f(2)=3, f'(2)=5 and f'(3)=7.
(a) Differentiate the equation: y'f(x) + yf'(x) + 1y3 + 3xy2y' = 0. Now plug in x = 2 and y = -1: y'3 + (-1)5 - 1 + 6y' = 0 so y' = 2/3.
(b) y' = 35 and y = -40 when x = 0, so the tangent line is: y = (35)(x - 0) + -40

Problem [2] Suppose the graph below gives the derivative g(x) of some function y = G(x). Graph goes here.

Problem [3] Consider the limit as h approaches 0 of [ln(e+h)-1]/h.
(a) This is the derivative of ln x at x = e; i.e., f(x) = ln x, a = e.
(b) Since f(x) = ln x, we know f'(x) = 1/x, so f'(e) = 1/e, hence the value of the limit is exactly 1/e.
(c) By L'Hopital, the limit as x goes to 0 of [ln(e+x)-1]/x is the same as the limit as x goes to 0 of [ln(e+x)-1]'/x', i.e., of [1/(e+x)]/1, which is just [1/(e+0)]/1, or 1/e.

Problem [4] Let h(x) = x2x.
(a) From the graph of h(x) we expect there to be only one critical point, between -3 and -1, and that the global minimum of h(x) occurs there. So we solve h'(x) = 0 to check that there realy is only one critical point, and to find it exactly. Now h'(x) = 2x + x(ln 2)2x, and solving 2x + x(ln 2)2x = 0 we get x = -1/ln 2. So there really is only one critical point, so from the graph of h(x) that is where h(x) is as small as possible.
(b) Since the interval -3 <= x <= -1 is closed, we know that the extrema occur either at the critical point or the endpoints. We want a maximum, and we know from (a) that the critical point gives a minimum, so we only need to check h(x) at x = -2 and x = -1. But h(-3) = -3/8, and h(-1) = -1/2, so the maximum is -3/8, and it occurs at x = -3.

Problem [5] Farmer Brown wants to fence off three sides of a rectangular area; the fourth side runs along a river and does not need a fence. She wants there to be 5000 square yards of land in the rectangle. What are the dimensions of the rectangle needing the least amount of fencing?

Answer: (1) The quantity Q to be optimized is the length of fence.
(2) Q = 2w + l, if w is the width of the rectangle and l is the length.
(3) Since lw = 5000, we have l = 5000/w.
(4) Thus Q = 2w + 5000/w.
(5) The range of values of w which make sense here is: w > 0.
(6) From graphing Q(w) on the calculatore we expect that there is only one citical point and that the minimum occurs there. We now check this and find that critical point by solving Q' = 0: Q' = 2 - 5000/w2, so (for w > 0) Q' = 0 occurs only for w = 50.
(7) The question to be answered is: what are the dimensions of the rectangle? We now know that w = 50, and l is thus 5000/50 = 100.