Let $f'(x_{0})=f''(x_{0})=....=f^{(n-1)}(x_{0})=0$ but $f^{(n)}(x_{0})\neq0$.
Case 1: If n is even, then at $f^{(n)}(x_{0})<0$ there is maximum at $x_{0}$ and at $f^{(n)}(x_{0})>0$ there is a minimum at $x_{0}$.
Case 2: If n is odd then there is no extremum at $x_{0}$.
I have understood Case 1 as it says in my textbook. Can someone please explain me Case 2.
According to Taylor's formula
$ f(x) = f(x_0) + \cdots + \dfrac{f^{(n)}(x_0)}{n!} (x-x_0)^n + o((x-x_0)^n),x \to x_0$
so,if $n$ is odd, as you stated, the above equation becomes
$ f(x) =\dfrac{f^{(n)}(x_0)}{n!} (x-x_0)^n + o((x-x_0)^n),x \to x_0$
Assume $f^{(n)}(x_0)>0$. So, if $x > x_0$ in a small neighbourhood of $x_0$, $f(x) > f(x_0)$(since $(x-x_0)^n > 0$), and in the left $x<x_0$, $f(x)< f(x_0)$(since $(x-x_0)^n < 0$, recall that $n$ is odd) ,which means there is no extremum at $x_0$