# M106 Practice Exam 2 Solutions

[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 the graph shown is the graph of y = f(x):
• (a) For what values of x is f '(x) < 0? f '(x) < 0 for all x < 0 and for all x > 2
• (b) For what values of x is f ''(x) > 0? f ''(x) > 0 for all x < 1 ( i.e., where the graph of f(x) is concave up)
• (c) Sketch the graph of f '(x). shown in bold

Problem [2] Suppose now that the graph above gives the graph of g '(x), where g(x) is some differentiable function. Although what you are given is the graph of the derivative, pay attention to the fact that the questions below refer to g(x) itself.
• (a) For what values of x is g(x) decreasing? for x > - 1, since that is where g ' is not positive
• (b) For what values of x is g(x) concave up? for 0 < x < 2, since that is where g ' is increasing and hence g '' is positive
• (c) For what value of x is g(x) biggest? Justify your answer. At x = - 1: since g ' is positive to the left of - 1 and nonpositive to the right of - 1, we see g increases up to where x = - 1, then g decreases, so g must be biggest at x = - 1

Problem [3] Suppose that f(t) is a differentiable function such that f(3.99)=251.1, f(4)=251, and whose graph is concave up.
• (a) What is the average rate of change of f(t) over the interval 3.99 <= t <= 4? [f(4) - f(3.99)]/(4 - 3.99) = (251 - 251.1) / .01 = - 10
• (b) Is the answer to (a) bigger or smaller than f '(4)? Explain. Because f is concave up, f ' (4) must be bigger
• (c) Suppose that f(5) = 245 and f '(5) = -2. Use this information to estimate f(5.001); indicate whether the estimate should be bigger or smaller than the actual value of f(5.001), and explain why. We'll use the tangent line to approximate f. Since f is concave up, the tangent line is under the graph of f, so our estimate will be low. We get y = - 2 (x - 5) + 245 for the tangent line, so f(5.001) is about - 2 (5.001 - 5) + 245 = 244.998.

Problem [4] The temperature one winter day in Lincoln was 42 degrees Fahrenheit at noon and decreased all day until at 8:00 pm it was only 10 degrees Fahrenheit. Let T = f(t) be the temperature at time t, where t is the number of hours past noon and T is in degrees Fahrenheit.
• (a) Express the average temperature between noon and 8:00 pm in terms of a definite integral. It's inconvenient to use an actual integral sign on the web, so I'll use calculator notation (but on the exam I'll expect correct mathematical notation): [1/(8-0)]fnInt(f(t),t,0,8)
• (b) Suppose you took the temperature every two hours, obtaining the following table of data:
```    t    (hrs)    0  2  4  6  8
f(t) (deg F) 42 26 18 13 10
```
Using this table and appropriate Riemann sums, give upper and lower estimates for the average temperature between noon and 8:00 pm:
• Upper estimate: 2[42+26+18+13]/8 = 24.75
• Lower estimate: 2[26+18+13+10]/8 = 16.75
• Would taking the temperature every 5 minutes be often enough to ensure that the Left Hand Riemann Sum estimate for the average temperature between noon and 8:00 pm is within 0.1 degrees Fahrenheit of the actual value? Why or why not. If we take a reading every 5 minutes, the difference between the LHS and the RHS is (1/8)(42 - 10)(1/12) = 1/3, so the LHS estimate could be off by as much as 1/3 degree. Thus this is not accurate to 0.1 degrees.

Problem [5] This problem refers to the function T = f(t) discussed in the preceding problem.
• (a) Use the table in the preceding problem to estimate the derivative of f(t) at t = 5 (show your work), and specify the units of your answer. f ' (5) is about [f(6) - f(4)]/(6-4) = - 2.5 degrees Fahrenheit per hour.
• (b) Which is a possible value of the derivative of f -1(T) at T = 18, 0.32 or -0.32? Justify your answer. If it were 0.32, this would mean that it takes about 0.32 hours for the temperature to increase 1 degree, which means that the temperature would be increasing, which it is not. Hence the correct answer is - 0.32.
• (c) What are the units to your answer to (b), and what is its practical meaning? The units are hours per degree. The practical meaning of the derivative of f -1(T) at T = 18 being - 0.32 is that when the temperature is 18 degrees Fahrenheit, it takes about 0.32 hours for the temperature to fall another degree.

Problem [6] The graph of the derivative F '(t) of a differentiable function F(t) is given below:
• (a) Given that F(1) = 0, determine the values of F(-1) and F(3). Use the fact that fnInt(F '(t), t, a, b) = F(b) - F(a) (this is the fundamental theorem of calculus). Thus fnInt(F '(t), t, 1, 3) = F(3) - F(1) = F(3) - 0; now compute fnInt(F '(t), t, 1, 3) by using the fact that fnInt(F '(t), t, 1, 3) is the signed area between the graph and the x-axis from x = 1 to x = 3. Thus fnInt(F '(t), t, 1, 3) = 0, so F(3) = 0. Similarly, fnInt(F '(t), t, - 1, 1) = F(1) - F(- 1) = 0 - F(- 1) = - F(- 1), but fnInt(F '(t), t, - 1, 1) = -5/4, so F(- 1) = 5/4.
• (b) Evaluate the definite integral of F '(t) from t = 0 to t = 3, and indicate how you obtain your answer. fnInt(F '(t), t, 0, 3) = the signed area between the graph of F ' and the x-axis from x = 0 to x = 3, which is - 3/4.

Problem [7] Suppose that f(x) is a differentiable function such that f(2) = 3, f '(2) = 5 and f '(3) = 7.
• (a) Find the derivative of (f(x))2 at x = 2. f '(2) = 2f(2) f '(2) = 30.
• (b) Find the derivative of f(f(x)) at x = 2. (f(f(x))) ' at x = 2 is f '(f(2))f '(2) = 35.
• (c) Find the derivative of x/f(x) at x = 2. Use the quotient rule: (x/f(x)) ' = [1f(x) - x f '(x)]/(f(x))2 which gives (3 - 10)/9 = - 7/9.
• (d) Find the line tangent to the graph of g(x) = x7 - 8x2 + 35x - 40 at x = 0. g ' (0) = 7(06) - 16(0) + 35 = 35, so the tangent line is y = 35 (x - 0) + g(0) or y = 35x - 40.

Problem [8] Consider the limit limh->0 (ln(e+h)-1)/h. This limit is actually the definition of the derivative of some function f(x) at some value x = a. Determine f(x) and a. This is just the derivative of ln(x) at x = e. Thus f(x) = ln(x) and and a = e.