# Can there exist local maxima or minima at inflection point

Assuming that 1st and 2nd derivative exist, can there be local maxima or minima at inflection point?

What I think is that it cant have maxima or minima because at inflection points the original curve changes its concavity. That why it can either be increasing or decreasing at that point but it cant have any extremum at that point.

If $f(x)$ has a local max or min at $x=0$, then $f'(0)=0$. Now suppose $f(x)$ has an inflection point at $x=0$.
(case 1) $f'(x)$ is increasing on $(-\epsilon,0]$ and decreasing on $[0,\epsilon)$. Then that means $f'(x) \leq 0$ on $(-\epsilon,\epsilon)$. So $f(x)$ is strictly decreasing on $(-\epsilon,\epsilon)$ so there can't possibly be a local max/min at $x=0$.
(case 2) $f'(x)$ is decreasing on $(-\epsilon,0]$ and increasing on $[0,\epsilon)$. Then that means $f'(x) \geq 0$ on $(-\epsilon,\epsilon)$. So $f(x)$ is strictly increasing on $(-\epsilon,\epsilon)$ so there can't possibly be a local max/min at $x=0$.
• I accidently deleted the comment, can I get it back? The thing is that $$f(x)=\begin{cases} 0 \qquad \text{if } x=0\\\exp{-1/x^2}\sin^2(1/x^2)\end{cases}$$ is indeed differentiable. Commented Apr 6, 2016 at 9:05
• But our reasoning can be made correct. You just have to look at $f'$. An inflection point means that $f'$ has a maximum/minimum there. Now apply your reasoning. Commented Apr 6, 2016 at 9:06
• You say that an inflection point means that $f'$ has a max/min there. But $f(x)=x^3$ has an inflection point at $x=0$ but doesn't have a max or min. Commented Apr 6, 2016 at 9:08
• It's just not true that $f$ differentiable and $f$ has a maximum on $x_0$ implies that $f'$ is increasing/decreasing around $x_0$. Commented Apr 6, 2016 at 9:08
• The derivative of $x^3$ indeed has a minimum at $0$. Commented Apr 6, 2016 at 9:09