Classifying extremums using higher order derivatives I was reading this from my calculus book:

If $f(x)$  has enough consecutive continuous derivatives at $x_0$ and we have
$n\ge2$ where  $$f'(x_0)=\cdots=f^{(n-1)}(x_0)=0,f^{(n)}(x_0)\neq0$$
Then if $n$ is an even number, $x_0$ is local extremum point( for
$f^{(n)}>0$ it is local minimum and for $f^{(n)}(x_0)<0$ it is local
maximum). If $n$ is an odd number $f$ has not extremum at $x_0$.

Can you please explain it . I don't understand this note. in fact I don't have intuition about $f'(x_0)=\cdots=f^{(n-1)}(x_0)=0,f^{(n)}(x_0)\neq0$ and how it can be connected with local extremum?
 A: This is just  a consequence of Taylor's formula. You can decide if $x_0$ is a local extrema observing how the values of $f$ vary when you move away from $x_0$. Taylor's formula states that
$$
f(x_0+h) = f(x_0) + f'(x_0) h + \cdots  +\frac{f^{(n-1)}(x_0)}{(n-1)!} h^{p-1} + \frac{f^{(n)}(\xi)}{n!} h^n
$$
if $f'(x_0) = \cdots =f^{(n-1)}(x_0) = 0$, the formula becomes
$$
f(x_0+h) -f(x_0) = \frac{f^{(n)}(\xi)}{n!} h^n.
$$
Also, assuming that $f^{(n)}$ is continuous, if $h$ is sufficiently small, the sign of $f^{(n)}(\xi)$ is the same as $f^{(n)}(x_0)$. So,

*

*If $n$ is odd, the sign of $h^n$ changes passing through 0, and so the sign of $f(x_0+h)-f(x_0)$ is not fixed ($x_0$ is not an extrema).


*If $n$ is even, $h^n \ge 0$ and so the sign of $f(x_0+h)-f(x_0)$ is (for small $h$) determined by the sign of $f^{(n)}(x_0)$. If it is positive, $f(x_0+h) \ge f(x_0)$ and you get a local minimum and, if it is negative, $f(x_0+h)\leq f(x_0)$ we you have a maximum.
Intuition works mainly for the case $n=2$, relating the sign of $f''(x_0)$ with local convexity/concavity.
A: Hint
Use Taylor formula and notice that $x \mapsto x^n$ has a local minimum at zero when $n$ is even while it is not the case for $n$ odd.
A: In the given situation there are two cases:
1-If the order of first non-zero derivative at $x=x_0$ is odd, there is no local max/ min. e.g. $f(x)=(x-3)^5$ has no local max/min at $x=3$ as $f^{v}(3)\ne 0.$
2-If the order of first non-zero derivative at $x=x_0$ is even:
(a) there is  local min  if its value at $x=x_0$ is positive  e.g., $f(x)=(x-2)^4$ has local min at $x=2$ as $f^{iv}(2)>0$
(b) there is local max if its value at $x=x_0$ is negative e.g., $f(x)=1-(x-4)^6$ has local max at $x=4$ as $f^{vi}(4)<0.$
