Rolle's Theorem

Theorem
Let $f$ be a real function which is continuous on the closed interval $\left[{a \,.\,.\, b}\right]$ and differentiable on the open interval $\left({a \,.\,.\, b}\right)$.

Let $f \left({a}\right) = f \left({b}\right)$.

Then $\exists \xi \in \left({a \,.\,.\, b}\right): f^{\prime} \left({\xi}\right) = 0$.

Proof
Since $f$ is continuous on $\left[{a \,.\,.\, b}\right]$, it follows from the Continuity Property that $f$ attains a maximum $M$ at some $\xi_1 \in \left[{a \,.\,.\, b}\right]$ and a minimum $m$ at some $\xi_2 \in \left[{a \,.\,.\, b}\right]$.

Suppose $\xi_1$ and $\xi_2$ are both end points of $\left[{a \,.\,.\,b}\right]$.

Because $f \left({a}\right) = f \left({b}\right)$ it follows that $m = M$ and so $f$ is constant on $\left[{a \,.\,.\, b}\right]$.

But then $f^{\prime} \left({\xi}\right) = 0$ for all $\xi \in \left({a \,.\,.\, b}\right)$.

Suppose $\xi_1$ is not an end point of $\left[{a \,.\,.\, b}\right]$.

Then $\xi_1 \in \left({a \,.\,.\, b}\right)$ 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 $\left[{a \,.\,.\, b}\right]$.

Then $\xi_2 \in \left({a \,.\,.\, b}\right)$ and $f$ has a local minimum at $\xi_2$.

Hence the result follows from Derivative at Maximum or Minimum‎.