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I found this theorem on Wikipedia without a name being mentioned:

Say $f$ is a function that is differentiable $n$ times and $f^{'}(x_0)=f^{''}=(x_0)=f^(x_0){'''}=....=f^{n-1}(x_0)=0$ and $f^{n}\ne0.$ If $n$ is even, there is an extremum at $x_0$, else an inflection point.

I would like to see a proof on that, can anyone tell me what this theorem is called or even explain why this theorem is true?

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    Hint: it's an inflection point iff $f''$ changes sign. – 2017-01-20
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    Apply Taylor-Young theorem to $f$ and look at the sign of $\Delta(x)=f(x)-f(x_0)$ for $x$ close to $x_0$. – 2017-01-20

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Proof:

We have
$f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac {f''(x_0)}{2!}(x-x_0)^2+\ldots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+o((x-x_0)^n) \quad(x\to x_0)$

Because$f′(x_0)=f′′(x_0)=f'''(x_0)=....=f^{(n-1)}(x_0)=0$ and $f^{n}\not=0$
We have
$f(x)=f(x_0)+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+o((x-x_0)^n) \quad(x\to x_0)$
And without loss of generality, we assume $f^n(x_0)>0$.
If $n$ is even, so $(x-x_0)^n\geq0$.
Furthermore,
$\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\geq0$. And when $x\not=x_0,\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n>0$ holds.
According to
$f(x)-f(x_0)=\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+o((x-x_0)^n) \quad(x\to x_0)$

And $\qquad\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+o((x-x_0)^n)\geq0 \quad\forall x\in O(x_0,\delta)$

It's
$f(x)\geq f(x_0)\quad\forall x\in O(x_0,\delta)$,$x_0$ is the extremun point.

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    What does $o((x-x_0)^n) \quad(x\to x_0)$ mean? – 2017-01-20
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    It means $\lim\limits_{x\rightarrow x_0}\frac{o((x-x_0)^n)}{(x-x_0)^n}=0$ – 2017-01-20
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    ok, and what is $o$ in $o(x - x_n)$? – 2017-01-20
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    Hm, are those $o$ and $O$ Landau notation? If so I don't understand the connection. And I still am not sure if I understand your notation for limits correctly. Could you maybe edit this to more standard notation, as understanding your answer really already fails for me at parsing and interpreting your equations. – 2017-01-21