Using higher order derivatives to analyze local extrema 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?
 A: Here are some steps you can follow:


*

*Use Taylor's theorem to deduce that $$f(x)=f(x_0)+\frac{f^{(n)}(c)}{n!} (x-x_0)^n$$ for some $c$ between $x_0$ and $x$.

*What is the relationship between the sign of $f^{(n)}(x_0)$ and the sign of $f^{(n)}(x)$ for $x$ sufficiently close to $x_0$? What do you know about $f^{(n)}$ that allows you to prove this? How about the sign of $f^{(n)}(c)$?

*What happens to the sign of $(x-x_0)^n$ for $x$ near $x_0$ when $n$ is even, and what happens when $n$ is odd? (When $n$ is odd, can $f$ sometimes have a local minimum/maximum at $x_0$?)

*Finish your proof.

A: These two facts are useful,
1) if $x_0$ is a local minimum, then in a neighbourhood of $x_0$, $f(x)-f(x_0)>0,$ 
2) if $x_0$ is a local maximum, then in a neighbourhood of $x_0$, $f(x)-f(x_0)<0.$ 
A: Here is the sketch of a  proof without using the Taylor formula. We will only consider one special case of the theorem, because the proof will be essentially the same in other cases. Furthermore, it is okay to drop the condition "$n+1$ -  times - continuously differentiable" and replace this by a weaker condition. Under these considerations let us write the exact statement:
Given a function $f$ having derivatives of order $n$ at the point $x_0$. If $n$ is even and $f^{(1)}(x_0) = f^{(2)}(x_0) = ... = f^{(n-1)}(x_0) = 0 $ and $f^{(n)}(x_0) > 0$ then $x_0$ is a local minimum of the function $f$.
Put $g(x) = f^{(n-1)}(x)$. Since $f^{(n)}(x)$ is defined at $x_0$, $g(x)$ is defined in some neighborhood of $x_0$. We have $g'(x_0) = \lim_{h \to 0} \frac{g(x_0 +h) - g(x_0)}{h}$ and $g'(x_0) > 0$, so for $h$ sufficiently close to 0, we get $g(x_0 + h) > g(x_0)$ if $h>0$ and $g(x_0 + h) < g(x_0)$ if $h<0$.
This means that $f^{(n-1)}(x)$ is negative in some left neighborhood of $x_0$ (LN) and positive in some right neighborhood of $x_0$ (RN) ($*$). So $f^{(n-2)}(x)$ is decreasing in LN and increasing in RN. But $f^{(n-2)}(x_0) = 0$, so $f^{(n-2)}(x)$ is positive in both LN and RN. That means $f^{(n-3)}(x)$ is increasing in both LN and RN, so $f^{(n-3)}(x)$ is negative in LN and positive in RN (**).
Repeating the way of argument from ($*$) to ($**$) and notice that $n$ is odd, we get a statement analogous to ($*$) about $f'(x)$: $f'(x)$ is negative in LN and positive in RN. Therefore $x_0$ is a local minimum.
