# 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

agree.

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 = x_{0} (a graph is not a
rigorous proof

however).

Proof. Note that f' (x) is concave up at x_{0}. 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 = t_{0}. (3 points)

Thus d At
t_{0} we get

(b) Write down a formula for the tangent line to the graph
at the point t = t_{0}. (2 points)

We know y = m x + b, so The line goes through
(x(t_{0}), y(t_{0})), so

Solving for b we get

and so our final equation is

(c) What value of t_{0} 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 t_{0} and so we write
(Note that now t_{0} is actually

a variable , and not a constant). , so that b
is concave up.
and so

we see the only critical point is at t_{0} = 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

= −(positive)/(positive)

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.

(5 points)

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 x_{1} < x_{2} < x_{3} as three of the
solutions. Since f(x_{1}) = f(x_{2}), by

Rolle’s theorem, we see that there exists a x_{4} ∈ (x_{1}, x_{2})
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.

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