Let $f: (-1,1) \longrightarrow \mathbb{R}$ such that $f', f''$ and $f'''$ exist and be continuous in $(-1, 1)$. Assume that $f'(0) = f''(0) = 0$ and $f'''(0) \neq 0$. Show that $f(0)$ isn't a local minimum point.
$\bf{Solution:}$ Using the Taylor's Formula:
$$f(h) - f(0) = \left[\frac{f'''(0)}{3!} + \rho(h)\right]h^{3}$$
where $\displaystyle\rho(h) = \frac{r(h)}{h^{3}}$. Thus, $\lim \rho(h) = 0$.
The signal of $h^{3}$ is the same as $h$. Therefore in a sufficiently small interval around $0$, $f(h) - f(0)$ has no constant signal, independet of the signal of $f'''(0)$. Thus, there is no local minimum in $f(0)$. $\blacksquare$
We can generalize this result to $n$ odd and $a$ arbitrary. The domain of $f$ must be a symmetric interval around $a$?
Is there another way to solve this exercise?