Proof of Fermat Stationary Point Criterion

Assume first that $f$ has a local maximum at $c$. Then there exists $\delta_0>0$ such that

$$f(x)\le f(c)$$

whenever $x\in[a,b]$ and $|x-c|<\delta_0$. Since $c\in(a,b)$, the number

$$\eta=\min\{c-a,b-c\}$$

is positive. Hence, for every $h$ with $0<h<\min\{\delta_0,\eta\}$, one has $c+h\in[a,b]$ and

$$\frac{f(c+h)-f(c)}{h}\le 0.$$

For every $h$ with $-\min\{\delta_0,\eta\}<h<0$, one has $c+h\in[a,b]$, and since the numerator is still nonpositive while $h<0$, one gets

$$\frac{f(c+h)-f(c)}{h}\ge 0.$$

Because $f$ is differentiable at $c$ in the sense of Derivative at an Interior Point, both families of difference quotients converge to $f'(c)$. Therefore $f'(c)\le 0$ and $f'(c)\ge 0$, so $f'(c)=0$.

If instead $f$ has a local minimum at $c$, then the same argument with all inequalities reversed again yields $f'(c)=0$. Since a local extremum at $c$ means that $f$ has either a local maximum or a local minimum at $c$, the conclusion follows.