Because $f$ is continuous on a compact (closed and bounded) interval $I = [a,b]$ , it attains its maximum and minimum values. In case $f(a)=f(b)$ is both the maximum and the minimum, then there is nothing more to say, for then $f$ is a constant function and $f' \equiv 0$ on the whole interval $I$ . So suppose otherwise, and $f$ attains an extremum in the open interval $(a,b)$ , and without loss of generality, let this extremum be a maximum, considering $-f$ in lieu of $f$ as necessary. We claim that at this extremum $f(c)$ we have $f'(c) = 0$ , with $a < c < b$ .

To show this, note that $f(x) - f(c) \leq 0$ for all $x \in I$ , because $f(c)$ is the maximum. By definition of the derivative, we have that $$ f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}. $$ Looking at the one-sided limits, we note that $$ R = \lim_{x \to c^+} \frac{f(x) - f(c)}{x - c} \leq 0 $$ because the numerator in the limit is nonpositive in the interval $I$ , yet $x - c > 0$ , as $x$ approaches $c$ from the right. Similarly, $$ L = \lim_{x \to c^-} \frac{f(x) - f(c)}{x - c} \geq 0. $$ Since $f$ is differentiable at $c$ , the left and right limits must coincide, so $0 \leq L = R \leq 0$ , that is to say, $f'(c) = 0$ .