[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  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  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  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  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):
(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  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
Problem  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  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  Consider the limit
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.