Art of Problem Solving
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

Rolle's Theorem

(Redirected from Rolle's theorem)

Rolle's theorem is an important theorem among the class of results regarding the value of the derivative on an interval.

Statement

Let $f:[a,b]\rightarrow\mathbb{R}$

Let $f$ be continous on $[a,b]$ and differentiable on $(a,b)$

Let $f(a)=f(b)$

Then $\exists$ $c\in (a,b)$ such that $f'(c)=0$

Proof

The result is trivial for the case $f([a,b])=\{f(a)\}$. Hence, let us assume that $f$ is a non-constant function.

Let $M=\sup\{f([a,b])\}$ and $m=\inf\{f([a,b])\}$ Without loss of generality, we can assume that $M\neq f(a)$

By the Maximum-minimum theorem, $\exists c\in (a,b)$ such that $f(c)=M$

Assume if possible $f'(c)>0$

Let $\epsilon=\frac{f'(c)}{2}$

Hence, $\exists \delta>0$ such that $x\in V_{\delta}(c)\implies |\frac{f(x)-f(c)}{x-c}-f'(c)|<\epsilon$

i.e. $\forall x\in V_{\delta}(c)$, $\frac{f(x)-f(c)}{x-c}>0$

Thus we have that $f(x)>f(c)$ if $x\in (c,c+\delta)$, contradicting the assumption that $f(c)$ is a maximum.

Similarly we can show that $f'(c)<0$ leads to contradiction.

Therefore, $f'(c)=0$

QED

See Also