What does the third derivative imply here? Let $f:I \subseteq \mathbb{R} \rightarrow \mathbb{R}$ be a function that is (at least) three times differentiable on the interval $I$. Suppose that $f'''$ is continuous and that $f'(a)=f''(a)=0$ for an $a \in I$ but that $f'''(a) \neq 0$. Is it possible for $f$ to reach a local extremum in $a$?
I reckon that it is not possible, seen as $f$ has an inflection point in $a$. But I'm not sure how to put this in a proof. 
 A: The Taylor expansion at $a$ becomes
$$
f(a+h)=f(a)+f'(a)h+\frac{f''(a)}{2}h^2+\frac{f'''(a+\theta h)}{6}h^3=
f(a)+\frac{f'''(a+\theta h)}{6}h^3
$$
with $\theta\in(-1,1)$. Suppose $f'''(a)>0$; by continuity, there is $\delta>0$ such that, for $-\delta<h<\delta$, $f'''(a+h)>0$, so
$$
\begin{cases}
f(a+h)<f(a) & \text{if $-\delta<h<0$} \\[6px]
f(a+h)>f(a) & \text{if $0<h<\delta$}
\end{cases}
$$
Similarly if $f'''(a)<0$. Thus $f$ has no local extremum at $a$.
Note that if, instead, the first three derivatives are zero at $a$, but the fourth derivative is nonzero (and continuous), then $f$ has a local extremum; the argument is the same, but now $h^4>0$ for $h\in(-\delta,\delta)$, $h\ne0$.
A: You can think about the graph of $f$ and $f'$. If $f$ has, without loss of generality, a local minimum at $a$, then $f'(x)<0$ for $x<a$ and $f'(x)>0$ for $x>a$.
Now, using $f''(a)=0$ and $f'''(a)\neq 0$, what can you say about the point $a$ on the graph of $f'$?
A: If $a$ is an inner point, then it is an extrema from $f$ if and only if $f'$ changes its sign at $a$.
From $f'''(a) \neq 0 $ you know that $f''$ changes the sign at $a$. This means the function $f'$ has a local minimum or maximum at $a$. This means that $f'$ does not change its sign at $a$! Thus is has no extrema there.
