MATH 185 EXAM #2
1. The theorem of the mean policeman. Consider the following situation.
“Jimmy is hungry and decides he wants food at Tim’s drive-in which is in a town exactly
30 miles away (in a straight line), we will call this town X. The only road between these
two towns has a speed limit of 50mph. He gets in his ’57 mustang convertible and drives
to town X, picks up lunch at the drive-in and then drives home. The local sheriff notices
Jimmy leave town at 12:30pm. He notices Jimmy return at 1:30pm, this time carrying a
bag with a half-eaten hamburger from Tim’s drive-in. As soon as Jimmy parks his car, the
sheriff walks over to Jimmy and writes him a ticket for speeding by 10mph.”
Give a mathematically rigorous explanation which proves that the sheriff is correct in
assuming that Jimmy was speeding by at least 10mph at some instant. You may assume
that the function J (t), which gives the total distance that Jimmy has driven since leaving
town, is differentiable. (5 points)
Proof. We assume t is in hours. Note that J(0) = 0 and that J(1) = 60. Thus
By the “mean” value theorem , there exists a t 0∈ (0, 1) such that
This completes the proof. If one wants to argue something about Jimmy’s average velocity,
and one wants to make it rigorous, one needs to make some kind of argument about integrals
or at least areas under curves .
2. Prove that the function
is differentiable at x = 1. (5 points)
Hint: You may use the fact that a limit exists whenever the left and right limits exist and
Proof. There are a number of correct ways to do this problem. We’ll do a short one here.
Note that regardless of anything else, we know that f' is defined for x ≠ 1. We see that
But then and because polynomials and constant functions
are continuous. By a Theorem at the end of section 3-21, we see that f is differentiable
at x = 1.
3. Let f : R → R be a function with a third derivative and
suppose that there exists x 0∈ R
so that and . Show that there exists so that f is
increasing on (10 points)
Hint: What does the graph of f' look like at x = x0 (a graph is not a rigorous proof
Proof. Note that f' (x) is concave up at x0. Thus there exists such that
for all Plugging values in, we see that
for all Since we see that on the interval
and furthermore that except at a finite number of points.
By a theorem in section 3-21, this proves that f (x) is increasing.
4. Consider the parametric equation
(a) Calculate the value of for t = t0. (3 points)
Thus d At t0 we get
Solving for b we get
and so our final equation is
(c) What value of t0 in the interval [0, 3] minimizes the y-intercept of tangent line you just
wrote down. Justify your answer (5 points).
Hint: Recall that the y- intercept of a line written in the form y = m x + b is simply the
value of b.
We view b as a function of t0 and so we write (Note that now t0 is actually
a variable , and not a constant). , so that b is concave up. and so
we see the only critical point is at t0 = 1. This is in the interval [0, 3] and by the second
derivative test, it must be a local min. Therefore, it is the absolute minimum by a theorem
from chapter 6.
(EC) Suppose that f : R → R is twice differentiable and
one-to-one. Further suppose that
f'' (x) > 0 for all x ∈ R and that f is a monotonically increasing function and that
Show that f -1 is concave down. (5 points)
Proof. Since f is twice differentiable, f' is differentiable, and thus continuous. Therefore
(because if it was < 0, it would be < 0 on some interval, and thus it would be
decreasing). But this implies, because that f' (x) > 0.
Now we know that
Taking another derivative (using the chain rule) we see that
Plugging in our formula for yet again, we see that
Now, f'' of anything is always positive , as is f', so
and thus is negative for all x, so that f -1 is concave down.
(EC) Suppose that f : R→ R is twice differentiable.
Further suppose that f'' (x) < 0 for all
x ∈ R. Prove that for each real number L, the equation L = f (x) has at most two solutions.
Hint: Analyze the solutions to the equation f' (x) = 0.
Proof. Suppose that L = f(x) has at least three different solutions, we will prove a contradiction.
Let us label x1 < x2 < x3 as three of the solutions. Since f(x1) = f(x2), by
Rolle’s theorem, we see that there exists a x4 ∈ (x1, x2) such that Likewise,
there exists such that Now, since f'' (x) < 0, f' (x) is monotonically
decreasing and thus f' is one to one. However, (since they are both zero )
and which demonstrates that , which is a contradiction.
Remark 0.1. It is possible to do a careful and correct by first showing that f' can cross zero
at most once. Furthermore, the sign of f' changes only at that point (if it exists) because
f' is continuous (essentially by the intermediate value theorem). Then one can analyze the
cases where f' < 0 and f' > 0 individually, and make the same conclusion.