# Derivatives

### Topics

We've started getting into "real" math in the last couple of units, with theorems, hypotheses, and counterexamples. It's about time we actually did a proof.

**Rolle's Theorem **says:

Let *f* be a function that

- is continuous on the closed interval [
*a*,*b*]

- is differentiable on the open interval (
*a*,*b*), and

- has
*f*(*a*) =*f*(*b*).

Then there is some *c* in the open interval (*a*,* b*) with *f*'(*c*) = 0.

### Proof of Rolle's Theorem

We'll prove this in a couple of different cases.

**Case 1 (the boring case):** If *f* is constant on [*a*,* b*], then *f*(*a*) = *f*(c) for every *c* in (*a*,* b*):

A horizontal line has a derivative (slope) of zero everywhere, so *f*'(*c*) = 0 for all *c* in (*a*,* b*).

If *f* is constant on any sub-interval of [*a*, *b*], the same argument applies:

Now suppose *f* is not constant on any sub-interval of [*a*, *b*], since we've already taken care of that case. The extreme value theorem says *f* must have a maximum and a minimum on [*a*, *b*]. The max and the min can't both equal *f*(*a*) since then we would be back in Case 1.

**Case 2:** The maximum value of *f* on [*a*,* b*] is greater than *f*(*a*). Let *c* be a value of *x* where this maximum occurs:

Calculate *f*'(*c*) by looking at the limit definition:

.

We know *f*(*c* + *h*) < *f*(*c*) since *f*(*c*) is the maximum value of *f* on our interval. The numerator of the quotient will be negative for any value of *h* close to 0. If *h* is less than zero we'll have a negative over a negative, so the quotient will be positive:

.

If *h* is greater than zero we'll have a negative over a positive, so the quotient will be negative:

.

We're assuming *f* is differentiable on (*a*,*b*), which means *f*'(*c*) needs to exist. We're getting positive numbers as *h* approaches 0 from the left, and negative numbers as *h* approaches 0 from the right. The only way the limit can exist is to equal 0. Therefore

**Case 3:** The minimum value of *f* on [*a*,* b*] is less than *f*(*a*).

### Sample Problem

Finish the proof in Case 3.

This is the same as Case 2. Let *c* be a value of *x* where a minimum occurs:

We know

Since *f*(*c*) is a minimum, *f*(*c* + *h*) will be larger and therefore *f*(*c* + *h*)-*f*(*c*) is positive. If *h* is negative then the quotient

is negative; if *h* is positive then the quotient is positive. As in Case 2, the only way for the limit

to exist is to have *f*'(*c*) = 0, since we find negative numbers as *h* approaches from one side and positive numbers as *h* approaches from the other side.