## Rolle's Theorem

**Let f be continuous on [a,b] and differentiable on (a,b).**

Then f(a) = f(b) ==> 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) ≥ f(x) > f(a) = f(b).

Since f(c) ≠ f(a), and f(c) ≠
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)

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Graeme McRae.