If a function $f$ is $L$-periodic then $f'$ has $2$ zeros in $[0,L)$?

Question

Let $f: \mathbb{R} \longrightarrow \mathbb{R} $ be a differentiable and odd function. If $f$ is periodic and the (minimal) period $L>0$, then $f'$ has $2$ zeros in $[0,L)$?

For example, this occurs if we consider $f(x)=\sin(x)$, for all $x \in \mathbb{R} $, since in this case $L=2\pi$.

This is in general true? ${}$

Answer 1

Yes; follows immediately from Rolle's Theorem.

To elaborate: $f(0)=0$. $f$ has a root at $a$, with $0<a<L$. Then $f$ has a root at $-a$ as well, so it also has a root at $-a+L$. By Rolle's Theorem, $f'$ has a root on $(0,a)$ and on $(a,L)$.