Rolle's Theorem

Let f be continuous on [a,b] and differentiable on (a,b).
Then f(a) = f(b) Þ there exists a number, c, in (a,b) such that f'(c)=0.

Proof:

If f is a constant function, then f'(c)=0 for all c in (a,b), proving this case.

If f(x) > f(a) for some x in (a,b) and c is a maximum of f on [a,b],

(c must exist, by the Extreme Value Theorem)

then f(c) ³ f(x) > f(a) = f(b).

Since f(c) ¹ f(a), and f(c) ¹ f(b), it follows that c is not an endpoint of [a,b], so it is a relative maximum.

Since relative extrema occur only at critical numbers, c is a critical number of f, which means either f is not differentiable at c or f'(c)=0.

Well, f is differentiable at c, so f'(c)=0, proving this case.

Similarly if f(x) < f(a) for some x in (a,b) then let c be a minimum of f on [a,b].  This case is proved the same as above.

How is this theorem used?

A generalization of Rolle's Theorem is the Mean Value Theorem.

Related Pages in this Website

Extreme Value Theorem -- that a continuous function on a closed interval has a maximum (and a minimum).

Mean Value Theorem -- that f'(c) = (f(a)-f(b))/(a-b) for some c in (a,b)

The webmaster and author of the Math Help site is Graeme McRae.
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