Interior extremum and Fermat's lemma

Local extremum

Let $f$ be defined on an interval $I$. We say that $f$ has a local maximum at $c$ if there is a neighborhood of $c$ on which $f(x) \leq f(c)$; a local minimum is defined in the same way. The word local is important: nothing is required far from $c$.

An extremum is a maximum or a minimum. It is called interior when $c$ is not an endpoint of $I$.

Fermat's lemma

If $f$ is differentiable at $c$, a point in the interior of $I$, and if $f$ has a local extremum at $c$, then $f'(c) = 0$.

Proof, in the case of a local maximum. For $h > 0$ small enough, $f(c + h) \leq f(c)$, so the difference quotient is negative:

$$\frac{f(c + h) - f(c)}{h} \leq 0 \qquad \text{hence} \qquad f'(c) = \lim_{h \to 0^+} \frac{f(c + h) - f(c)}{h} \leq 0$$

For $h < 0$ small enough, the numerator is still negative but the denominator changes sign, so the quotient is positive, and taking the limit from the left, $f'(c) \geq 0$.

The two inequalities give $f'(c) = 0$. The case of a minimum follows by applying the above to $-f$.

Three counterexamples to keep in mind

Situation	Function	What goes wrong
Extremum at an endpoint	$f(x) = x$ on $[0 \,;\, 1]$	maximum at $1$, but $f'(1) = 1 \neq 0$
Non-differentiable function	$f(x) = \lvert x \rvert$ on $[-1 \,;\, 1]$	minimum at $0$, no derivative there
False converse	$f(x) = x^3$ on $\mathbb{R}$	$f'(0) = 0$ with no extremum at $0$

The third row is worth remembering as is: $f'(c) = 0$ is a necessary condition for an interior extremum, never a sufficient one. That is why studying a function does not stop at where the derivative vanishes, but examines its change of sign.

Classic pitfall

Applying Fermat's lemma at an endpoint of the interval. On $[0 \,;\, 1]$, the function $x \mapsto x$ does reach its maximum, at $1$, without the derivative vanishing there: the proof uses increments $h$ of both signs, which no longer makes sense at an endpoint. Always check that the point is interior before writing $f'(c) = 0$.