Why odd derivative of a function doesn't gives maxima or minima? Why odd derivative of a function doesn't gives maxima or minima for example the third derivative of x^3 gives 6 in third derivative test at x=0 . Can anyone help me visualise this.
 A: I think that this theorem may be what you are trying to say:
Let $f$ be a funcition differentiable $n$ times at $x_0$, and suppose that $f^{(k)}(x_0)=0$ for $1\le k<n$ and $f^{(n)}(x_0)\neq 0$. Then:

*

*if $n$ is even and $f^{(n)}(x_0)>0$, $x_0$ is a local maximum.

*if $n$ is even and $f^{(n)}(x_0)<0$, $x_0$ is a local minimum.

*if $n$ is odd, $x_0$ is neither a maximum nor a minimum.

The reason why this theorem holds is the Taylor polynomial. More specifically, for $x\rightarrow x_0$ you have that $f(x)=f(x_0)+\frac{1}{n!}f^{(n)}(x_0)(x-x_0)^n+o((x-x_0)^n)$. To get rid of the little-o notation we can say there exists a function $w(x)$ such that $\lim_{x\rightarrow x_0}w(x)=0$ and $f(x)=f(x_0)+(x-x_o)^n\left(\frac{1}{n!}f^{(n)}(x_0)+w(x)\right)$. Now since $w(x)$ goes to zero we get that the sign of $f(x)-f(x_0)$ near $x_0$ is completely determined by the one of $(x-x_0)^n$ and $f^{(n)}(x_0)$. Now it is easy to see that if $n$ is odd the difference $f(x)-f(x_0)$ will be positive for $x$ slightly larger than $x_0$ and negative for $x$ sightly smaller than $x_0$ or viceversa. In either case, $x_0$ is not a maximum or a minimum. If on the other hand $n$ is even then $f(x)-f(x_0)$ ha a constant sign near $x_0$ so it is a maximum or a minum and it depends on the sign of the $n$-th derivative.
