Rolle's Theorem and a Proof

A special case of the Mean Value Theorem, known as Rolle's Theorem, is:

Rolle's Theorem

If f(x) is continuous an [a,b] and differentiable on (a,b) and if f(a) = f(b) then there is some c in the interval (a,b) such that f '(c) = 0.

The reason that this is a special case is that under the stated hypothesis the MVT guarantees the existence of a point c with

f(b) - f(a) f(b) - f(a)
f '(c) =
= 0.
b - a b - a

Proof of Rolle's Theorem!

Most proofs in CalculusQuestTM are done on enrichment pages. This is one exception, simply because the proof consists of putting together two facts we have used quite a few times already.

In the statement of Rolle's theorem, f(x) is a continuous function on the closed interval [a,b]. Hence by the Intermediate Value Theorem it achieves a maximum and a minimum on [a,b]. Either

  1. One of these occurs at a point c with a < c < b,

    Since f(x) is differentiable on (a,b) and c is an extremum we then conclude that f '(c) = 0.


  2. Both the maximum and minimum occur at endpoints.

    Since f(a) = f(b), this means that the function is never larger or smaller than f(a). In other words, the function f(x) is constant on the interval [a,b] and its derivative is therefore 0 at every point in (a,b).

That's the proof! Truth in Proofs Statement

Proof of the Mean Value Theorem

Rolle's theorem is a special case of the MVT, but the Mean Value Theorem is also a consequence of Rolle's Theorem.

Why is THAT true?.

If two mathematical statements are each consequences of each other, they are called equivalent. Thus Rolle's Theorem is equivalent to the Mean Value Theorem.



© CalculusQuestTM
Version 1
All rights reserved---1996
William A. Bogley
Robby Robson