Fermat’s Theorem

Proof. Since \(x_{0}\) is a local extremum point of \(f\), it’s either a local maximum point or a local minimum point. Let us assume \(x_{0}\) is a local maximum point. Then since it’s also in particular an interior point, \[\exists\delta>0\quad s.t.\quad\forall x\in\left(x_{0}-\delta,x_{0}+\delta\right)\subseteq\left(a,b\right),f(x)\le f(x_{0})\]

Since \(f\) is differentiable at \(x_{0}\) it’s in particular differentiable from the right and from the left while maintaining \[f'_{-}(x_{0})=f'(x_{0})=f'_{+}(x_{0})\]

We shall examine each one-sided derivative of \(f\) at \(x_{0}\) in turn.

For every \(x\in\left(x_{0}-\delta,x_{0}\right)\) we get \(f(x)\le f(x_{0})\) and \(x<x_{0}\), which gives \[\forall x\in\left(x_{0}-\delta,x_{0}\right),\frac{f(x)-f(x_{0})}{x-x_{0}}\ge0\]

Let us recall that limits weakly preserve inequalities. Hence

\[f'_{-}\left(x_{0}\right)=\underset{x\rightarrow x_{0}^{-}}{\lim}\frac{f(x)-f(x_{0})}{x-x_{0}}\ge0\] For every \(x\in\left(x_{0},x_{0}+\delta\right)\) we get \(f(x)\le f(x_{0})\) and \(x>x_{0}\), which gives

\[\forall x\in\left(x_{0},x_{0}+\delta\right),\frac{f(x)-f(x_{0})}{x-x_{0}}\le0\]

And again since limits weakly preserve inequalities,

\[f'_{+}\left(x_{0}\right)=\underset{x\rightarrow x_{0}^{+}}{\lim}\frac{f(x)-f(x_{0})}{x-x_{0}}\le0\] This gives us \[0\le f'(x_{0})\le0\] Hence \[f'(x_{0})=0\] As required.

Now let us assume \(x_{0}\) is a local minimum of \(f\).

Let us recall that \(c\) is a local minimum of \(f\) if and only if \(c\) is a local maximum of \(-f\).

Let us also recall that by the differentiable functions arithmetic theorem, \(f\) is differentiable at point \(c\) and \(f'(c)=\alpha\) if and only if \(-f\) is differentiable at point \(c\) and \(\left(-f\right)'(c)=-\alpha\).

Let us notice that \(\left(-f\right)\) fulfills the requirements for the case we just proved, hence the derivative of \(-f\) at point \(x_{0}\) is zero, which in turn means \(f'(x_{0})=0\), as required. ∎