$f$ differentiable $5$ times around $x=a,\ f'(a)=f''(a)=f'''(a)=0,\ f^{(4)}(x) <0 \Rightarrow x=a$ is either a local minimum or a local maximum point So I've been trying to prove the following statement:

Let $f$ be a function such that it is differentiable $5$ times around $x=a$. Prove or disprove, that if $f'(a)=f''(a)=f'''(a)=0$ and $f^{(4)}(x)<0$ then $x=a$ is  either a local maximum or a local minimum point. 

Since I couldn't find a counterexample, I believe this statement is false, and $x=a$ must be a maximum point, but I'm not certainly sure since I have no idea how to prove this. I thought using Taylor but it doesn't seem to help that much.
Thank you very much!.

The following theorem was proved by Colin Maclaurin in 1742.
Let $f$ be a real-valued function defined on an open interval $J$ which is $(n-1)$-times continuously differentiable in a neighborhood of a point $a \in J $ and for which moreover $f^{(n)}(a)$ exists.
Assume $f'(a) = f''(a) = \dots f^{(n-1)}(a) = 0$ and $f^{(n)}(a) \ne 0$. Then:
1) If $n$ is even, then $f$ has a local maximum [resp. local minimum] at $a$ if $f^{(n)}(a) < 0$ [resp. $f^{(n)}(a) > 0$].
2) If $n$ is odd, then $f$ has an inflection point at $a$.
So you see that your $f$ has a local maximum at $a$.
Edited:
The proof is based on Taylor's theorem. Under the above assumptions it is a bit tedious, so let us restrict to the special case that $f$ is $n$-times continuously differentiable in a neighborhood of $a$. This covers your question since you assume that $f$ is $5$-times differentiable.
We can write $f(a + h) = f(a) + \dfrac{h^n}{n!}f^{(n)}(a + \theta h)$, $0 < \theta < 1$. That means
$$f(a + h) - f(a) = \dfrac{h^n}{n!}f^{(n)}(a + \theta h) .$$
Since $f^{(n)}$ is continuous in $a$ and $f^{(n)}(a) \ne 0$, we see for $\lvert h \rvert < \epsilon$ the sign of $f^{(n)}(a + \theta h)$ agrees with the sign of $f^{(n)}(a)$. This shows that for $n$ even $f(a + h) - f(a)$ has the same sign in a neigborhood of $a$. For $n$ odd the sign is different on both sides of $a$.
