Let $f\colon (a,b)\to\mathbb{R}$ be an $(n+1)$-time continuously differentiable function on $(a,b)$ with $n\geq2$. Assume $f'(x_0)=\dotsc=f^{(n-1)}(x_0)=0$ and $f^{(n)}(x_0)\neq 0$ at some $x_0\in (a,b)$. Show if $n$ is even then $f$ has a local extremum at $x_0$. When is it a maximum? If $n$ is odd, does $f$ have a local minimum at $x_0$?
First I tried looking at a couple of examples to convince myself that the statement is true. Just in the easiest case, I immediately get a counterexample for the last question. Consider $f(x)=x^3$ on $(-1,1)$ at $x_0=0$. Then $f'(x)=3x^2, f''(x)=6x$, and $f'''(x)=6$. We also have that $f'(0)=f''(0)=0$ and $f'''(0)=6\neq 0$ so we satisfy all of the appropriate hypotheses, but $x_0=0$ is not a local minimum.
To answer the first question, I'm not really sure where to proceed. This looks like a place to use the second derivative test, but I'm not sure how to apply it in the sense of these higher derivatives. Based on the simple example of $f(x)=x^2$, it looks like $x_0$ will be a local maximum when $f^{(n)}(x_0)<0$. I'm just not sure how to convince myself that looking at higher derivatives will convey the same sort of information as the first and second derivative tests. Can anyone clear this up for me?