# Is $x_0$ local minimum/maximum or inflection point if $f'(x_0)=0$?

Is the following statement true or false? (I think it is true, and I tried proving it by contradiction and failed.)

Let $f:(a,b)\to \mathbb{R}$ be differentiable, and let $x_0\in (a,b)$ such that $f'(x_0)=0$. Then $x_0$ is a local minimum/maximum or inflection point.

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What precisely do you mean by "inflection point"? – Chris Eagle Mar 9 '13 at 16:02
@ChrisEagle: en.wikipedia.org/wiki/Inflection_point – 17SI.34SA Mar 9 '13 at 16:05
If you don't assume that $f''$ exists, then I think what you mean by inflection point (or saddle point) is that for any $\epsilon > 0$, there exist $x_1$ and $x_2$ in $(x_0 - \epsilon, x_0 + \epsilon) \cap (a,b)$ such that $f(x_1) < f(x_0) < f(x_2)$. In this sense, your claim is correct. – Dominique Mar 9 '13 at 21:12
@Dominique: Defining "inflection point" to mean "point which is not a local extremum" makes the question rather uninteresting. – Chris Eagle Mar 9 '13 at 22:36
@ChrisEagle When $f'(x)=0$, that's exactly what we mean in optimization by saddle point: a stationary point that isn't a min or a max. – Dominique Mar 9 '13 at 22:43

This is false. Let $$f(x)=\begin{cases}x^2 \sin (1/x) & x \neq 0 \\ 0 & x=0\end{cases}$$ Then $f$ is differentiable and $f'(0)=0$, but $0$ is neither a local extremum nor an inflection point.
So, $f''(x)$ isn't exists, hence we cannot talk about inflection point, right? But why $x_0=0$ isn't local minima? – 17SI.34SA Mar 9 '13 at 16:49
Because if for example $x=1/((2n-1/2)\pi)$ then $f(x)<0$. – Chris Eagle Mar 9 '13 at 17:59
If $f'(x_0) = 0$, then the tangent to the curve $y = f(x)$ at $x = x_0$ is horizontal. Thus it is as you say.
If the function is differentiable, where it is decreasing $f'(x) < 0$, where it is increasing $f'(x) > 0$. If $f'(x_0) = 0$, it isn't increasing nor decreasing there. It either reached a highest point and will start going down, it reached a lowest point and will start growing, or it is a turn (inflection point). Just plot $y = x^2$ or $y = -x^2$, or $y = x^3$ around $x_0 = 0$. – vonbrand Mar 9 '13 at 16:29