Does $f'(x)=0$ implies Max/min or point of inflection

Question

Let $f$ be a differentiable function on $(a,b)$ such that $f'(c)=0$ for some $c\in (a,b)$. Also there is no maximum or minimum at $c$ .Does this means $c$ is point of inflection?

Consider $f(x)=x^3$ , this is true. But I want to know if it is true in general. I try to find counter example but can't get any. I hope this is true. If it is please give a hint to begin a proof

Edit : hint this question by maxima / minima, I mean strict maxima or minima

Answer 1

Consider the function $$g(x)=x\sin(\frac{1}{x})$$ with $g(0)=0$. This function is continuous on $\mathbb R$.

Let $F(x)$ be any antiderivative of $g(x)$. Then $F'(0)=0$.

Now, since $g(x)$ is an even function, $F$ is odd, and hence it cannot have a local max/min at $x=0$.

Moreover, $F(x)$ cannot have an inflection point at $x=0$, since this would imply that for some $(0,a)$ the function $g(x)$ would be monotonic.

Added: if $n\geq 2$, $f$ is $n$ times differentiable, $f^{(n)}$ is continuous at $c$ and $$f'(c)=...=f^{(n-1)}(c)=0 \\ f^{(n)}(c) \neq 0$$ then

• If $n$ is odd $c$ is an inflection point
• If $n$ is even and $f^{(n)}(c) >0$ then $c$ is a local min.
  • If $n$ is even and $f^{(n)}(c) <0$ then $c$ is a local max.

Sketch Proof: Since $f^{(n)}(c) \neq 0$ it is either positive or negative.

If $f^{(n)}(c)<0$ then replace $f$ by $-f$. Note that in the case $n$ even thiwill change local max to min.

By continuity, there exists some $a>0$ such that $f^{(n)}>0$ on $(c-a, c+a)$.

Now you do an inductive argument. $f^{(n)}>0$ on $(c-a, c+a)$ means $f^{(n-1)}$ is strictly increasing on $(c-a, c+a)$ and thus, since $f^{(n-1)}(c)=0$ you get $$f^{n-1}(x) <0 \forall x \in (c-a,c) \\ f^{n-1}(x) >0 \forall x \in (c,c-a) (*)\\$$

This gives that $f^{n-2}$ is decreasing on (c-a,c)$ and increasing on $(c,c+a)$. Therefore $$f^{n-2}(x) >0 \forall x \in (c-a,c) \ f^{n-2}(x) >0 \forall x \in (c,c-a) (**)\$$

Then, same argument shows that $$f^{n-3}(x) <0 \forall x \in (c-a,c) \\ f^{n-3}(x) >0 \forall x \in (c,c-a) (*)\\$$

and so on, with (*) and (**)alternating. Based of $n$ being odd or even, either $f'$ or $f''$ will satisfy $(*)$ [this is why we need $n \geq 2$].

Now, if $f'$ satisfies $(*)$ then it is easy to see that $c$ is a local max.

If $f''$ satisfies $(*)$ theb it is easy to see that $c$ is an inflection point.

QED

P.P.S. The key for the proof is the existence and continuity of the first derivative which doesn't vanish at $c$. The above counterexample fails this :)

Answer 2

The simplest counter example is the constant function $$ f\equiv 0 $$ on any interval $(a,b)$. Clearly $f'(c)=0$ for any $c\in(a,b)$ but it's not a point of inflection.

Edit: I interpreted the words max/min to mean strict minima/maxima, if the question refers to nonstrict ones then this is not a counter example.