Rolle's Theorem

Theorem

Let $f$ be a real function which is:

continuous on the closed interval $\closedint a b$

and:

differentiable on the open interval $\openint a b$.

Let $\map f a = \map f b$.

Then:

$\exists \xi \in \openint a b: \map {f'} \xi = 0$

Proof 1

We have that $f$ is continuous on $\closedint a b$.

It follows from Continuous Image of Closed Interval is Closed Interval that $f$ attains:

a maximum $M$ at some $\xi_1 \in \closedint a b$

and:

a minimum $m$ at some $\xi_2 \in \closedint a b$.

Suppose $\xi_1$ and $\xi_2$ are both end points of $\closedint a b$.

Because $\map f a = \map f b$ it follows that $m = M$ and so $f$ is constant on $\closedint a b$.

Then, by Derivative of Constant, $\map {f'} \xi = 0$ for all $\xi \in \openint a b$.

Suppose $\xi_1$ is not an end point of $\closedint a b$.

Then $\xi_1 \in \openint a b$ and $f$ has a local maximum at $\xi_1$.

Hence the result follows from Derivative at Maximum or Minimum‎.

Similarly, suppose $\xi_2$ is not an end point of $\closedint a b$.

Then $\xi_2 \in \openint a b$ and $f$ has a local minimum at $\xi_2$.

Hence the result follows from Derivative at Maximum or Minimum‎.

$\blacksquare$

Proof 2

First take the case where:

$\forall x \in \openint a b: \map f x = 0$

Then:

$\forall x \in \openint a b: \map {f'} x = 0$

Otherwise:

$\exists c \in \openint a b: \map f c \ne 0$

Let $\map f c > 0$.

Then there exists an absolute maximum at a point $\xi \in \openint a b$.

Hence:

 $\ds \dfrac {\map f {\xi + h} - \map f \xi} h$ $\le$ $\ds 0$ for $\xi < \xi + h < b$ $\ds \dfrac {\map f {\xi + h} - \map f \xi} h$ $\ge$ $\ds 0$ for $a < \xi + h < \xi$

As $h \to 0$, we see that both of the above approach $\map {f'} \xi$, which is then both non-negative and non-positive.

That is:

$\map {f'} \xi = 0$

Similarly, let $\map f c < 0$.

Then there exists an absolute minimum at a point $\xi \in \openint a b$.

Hence:

 $\ds \dfrac {\map f {\xi + h} - \map f \xi} h$ $\ge$ $\ds 0$ for $\xi < \xi + h < b$ $\ds \dfrac {\map f {\xi + h} - \map f \xi} h$ $\le$ $\ds 0$ for $a < \xi + h < \xi$

Again, as $h \to 0$, we see that both of the above approach $\map {f'} \xi$, which is then both non-negative and non-positive.

That is:

$\map {f'} \xi = 0$

Hence the result.

$\blacksquare$

Source of Name

This entry was named for Michel Rolle.