# 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) Find dy/dx when x = 2 and y = -1, given that yf(x) + xy3 = -5.
• (b) Find the line tangent to the graph of y = x7 - 8x2 + 35x - 40 at x = 0.

(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).
• (a) Circle each x-value on the x-axis in the graph for which the original function G(x) has a critical point.
• (b) Put an X (for maX) in each of your critical point circles for which the critical point is a local maximum of the original function G(x).
• (c) Put an N (for miN) in each of your critical point circles for which the critical point is a local minimum of the original function G(x).
• (d) Put an i at each point of the graph for which the original function G(x) has a point of inflection.

Problem [3] Consider the limit as h approaches 0 of [ln(e+h)-1]/h.
• (a) This limit is actually the definition of the derivative of some function f(x) at some value x = a. Determine f(x) and a.
• (b) Use your answer to (a) to determine the limit exactly.
• (c) Use L'Hopital's rule to determine the limit exactly a second way.

(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) For what exact x-value is h(x) as small as possible? Show how you find the correct x-value and how you know h(x) has a minimum there.
• (b) For what exact x-value or values in the range -3 <= x <= -1 is h(x) as large as possible? Show how you find the correct x-value and how you know for values in the given range that h(x) has a maximum there.