Proof of Rolle's Theorem

If $f$ is a function continuous on $[a,b]$ and differentiable on $(a,b)$, with $f(a) = f(b) = 0$, then there exists some $c$ in $(a,b)$ where $f'(c)=0$.


Consider the two cases that could occur:

Case 1:

$f(x) = 0$ for all $x$ in $[a,b]$.

In this case, any value between $a$ and $b$ can serve as the $c$ guaranteed by the theorem, as the function is constant on $[a,b]$ and the derivatives of constant functions are zero.

Case 2:

$f(x) \ne 0$ for some $x$ in $(a,b)$.

We know by the Extreme Value Theorem, that $f$ attains both its absolute maximum and absolute minimum values somewhere on $[a,b]$.

Recall by our hypothesis, $f(a)=f(b)=0$, and that in this case, $f(x)$ is not zero for some $x$ in $(a,b)$. Thus, $f$ will have either a positive absolute maximum value at some $c_{max}$ in $(a,b)$ or a negative absolute minimum value at some $c_{min}$ in $(a,b)$ or both.

Take $c$ to be either $c_{min}$ or $c_{max}$, depending on which you have.

Note then, the open interval $(a,b)$ contains $c$, and either:

  1. $f(c) \ge f(x)$ for all $x$ in $(a,b)$; or
  2. $f(c) \le f(x)$ for all $x$ in $(a,b)$.
Either way, this means $f$ has a local extremum at $c$.

As $f$ is also differentiable at $c$, Fermat's Theorem applies, and concludes that $f'(c)=0$.