On extrema of real functions

If $f:\mathbb{R}\to \mathbb{R}$ is a sufficiently smooth function and $f'(a) = f''(a) = \dots = f^{(k-1)}(a) = 0$ and $f^{(k)}(a) \ne 0$, then we have the following result:

1. If $k$ is even and $f^{(k)}(a)<0$, then $a$ is a (local) maximun.

2. If $k$ is even and $f^{(k)}(a)>0$, then $a$ is a (local) minimum.

3. If $k$ is odd, then $a$ is not a local extremum.

Can you give some idea/intuition regarding the difference between the case $k$ odd and $k$ even?

$f^{(k)}$ is the derivative of $f^{(k-1)}$, so if $f^{(k)}(a)<0$ this implies that in an $\varepsilon$ set around $a$ the function $f^{(k-1)}$ is decreasing. Now since $f^{(k-1)}(a)=0$, in this epsion area $f^{(k-1)}(x)>0$ on the left of $a$, i.e. for all $x<a$ and $f^{(k-1)}(x)<0$ on the right of $a$ where $x>a$. But this implies that $f^{(k-2)}$ has a maximum in $a$ since $f^{(k-2)}$ is increasing in an epsilon area left of $a$ and decreasing right of $a$ in that epsilon area.

Now the assertions follow per iterative application of this reasoning.