Given a continuous function $f : I → R$, show that if $f$ is differentiable at $c$ and $c$ is a local extremum, then $f'(c) = 0$.

Question

Let $f : I → R$ be a continuous function on a open interval which has a local extremum at some $c\in I$. If $f$ is differentiable at $c$ then $f'(c) = 0$.

Is the following proof valid?

We study the difference quotient of $f$ at $x_0$:

$$\frac {f(x)−f(c)}{x−c}$$

Since $f$ is differentiable, we know that its limit exists for $x → c$.

Since $f$ is continuous and has a local extremum at $c$ by assumption, we get:

$$f(x)−f(c) = 0 \;(\forall x \in I)$$

Thus, by the quotient rule for limits, we find:

$$f '(c) = \lim_{ x\to c} \,\frac {f(x)−f(c)}{x−c} = \lim_{ x \to c} (0) = 0$$

Answer 1

Just because $f$ is continuous and has a local extremum at $c$, does not imply that $$ f(x) - f(x) = 0\;\forall x \in I\,.$$ The latter would imply that $f(x) = f(c)$ for all $x \in I$, meaning that $f$ would be the constant function.

Assume $c$ is a local maximum. Then use the definition of a local maximum to show that $$ \frac{f(x) - f(c)}{x-c} \leq 0\,,$$ for $x \geq c$ and then pass to the limit. Similarly you can show the other inquality thus concluding that $f'(c)=0$.

Answer 2

Wrong: If $f$ has a local maximum at $c$, then $$f(x)\le f(c)$$ for $x$ near $c$. What means this for the sign of $$\frac{f(x)−f(c)}{x−c}$$ for $x$ near $c$ in the cases $x>c$, $x<c$?