At a Glance - I Like Abstract Stuff; Why Should I Care?
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 if it's to equal 0. Therefore
Case 3: The minimum value of f on [a, b] is less than f(a).
Finish the proof in Case 3.
This is the same as Case 2. Let c be a value of x where a minimum occurs:
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.