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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.

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