First and Fourth derivative tests giving different result

Question

Consider the function $$f(x)=2x^5-5x^4-10x^3$$

we have to find Local maxima and minima

So $$f'(x)=10x^4-20x^3-30x^2=0$$

critical points are $x=0$,$x=3$ and $x=-1$

$$f''(x)=40x^3-60x^2-60x$$

Now at $x=0$ $f''(0)=0$ so second derivative test fails.

Using First derivative test we have

$$f'(0^-)<0$$ and$$f'(0^+) \lt 0$$ so no change of sign in $f'(x)$ hence $x=0$ is neither point of Local max nor Min

But

$$f''''(x)=240x-120$$

$$f''''(0) \lt 0$$ so $x=0$ should be point of Local Max right?

what is going wrong here?

Answer 1

The fourth derivative test isn't applicable here.

When doing higher-order derivative tests, you're supposed to stop as soon as you get a nonzero derivative. The third derivative is the first nonzero one at $x = 0$. $f'''(0) = -60$. Since an odd-order derivative is the first nonzero derivative at $x = 0$, then this means there is neither a max nor a min at $x = 0$.

Answer 2

$f''''(0)<0$ and $f'(0)=f''(0)=f'''(0)=0$ says that $x_{max}=0$ because $f(x)\leq0$ around $0$, which says that $f$ a concave function around $0$ because $y=0$ is a tangent to the graph of $f$ in the point $(0,0)$.

In our case we have $f'''(0)\neq0$. Thus, all this just is not relevant.