$f$ 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!.

Here is an alternate proof which doesn't require Taylor's Theorem (although admittedly it does use a lot of the same ideas, just in a much simpler form).
Since $f^{(4)}(a) < 0$, there exists some $\delta > 0$ such that whenever $0 < |x - a| < \delta$, we have $\frac{f^{(3)}(x)}{x - a} = \frac{f^{(3)}(x) - f^{(3)}(a)}{x-a} < 0$ (for example, using $\epsilon := -\frac{1}{2} f^{(4)}(a)$ in the definition of limit).  This implies that $f^{(3)}(x) < 0$ for $a < x < a + \delta$, whereas $f^{(3)}(x) > 0$ for $a - \delta < x < a$.  Now, for $a < x < a + \delta$, we have $f''(x) = f''(x) - f''(a) = \int_a^x f^{(3)}(t)\,dt < 0$; $f'(x) = f'(x) - f'(a) = \int_a^x f''(t)\,dt < 0$; and finally, $f(x) = f(a) + \int_a^x f'(t)\,dt < f(a)$.  Similarly, for $a - \delta < x < a$, we have $f''(x) = f''(x) - f''(a) = \int_a^x f^{(3)}(t)\,dt = -\int_x^a f^{(3)}(t)\,dt < 0$; $f'(x) = f'(x) - f'(a) = \int_a^x f''(t)\,dt = -\int_x^a f''(t)\,dt > 0$; and finally, $f(x) = f(a) + \int_a^x f'(t)\,dt = f(a) - \int_x^a f'(t)\,dt < f(a)$.
Putting together these facts, we conclude that $f(a)$ is the maximum value of $f$ for $x \in (a - \delta, a + \delta)$, i.e. $f$ has a local maximum at $x=a$.
