Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function. Consider the following properties:
$(1)$ $\;$ $f$ is a periodic function, with (minimal) period $a>0$;
$(2)$ $\;$ $f(0)=f(a)=0$ and $f'(a)=f'(0)=0$.
Question. Is true that $(1) \Leftrightarrow (2)$?
Note that $(1) \Rightarrow (2)$. Indeed, since by definition of periodic function, $$ f(x)=f(x+a) \Rightarrow f'(x)=f'(x+a), \; \forall \; x \in \mathbb{R} $$ we have, in particular for $x=0$, $f(0)=f(a)$ and $f'(a)=f'(0)$.
And the converse, is true? It's possible to use the fact that every periodic function is bounded?
$(2) \implies (1)$ is not always true. Take $$f(x)=x^2(x-1)^2$$
it is easy to check that
$ f $ is differentiable,
$$f(0)=f(1)=0,$$ and
$$f'(0)=f'(1)=0$$
but, clearly, $ f $ is not periodic : $$f(2+1)=36\ne f(2)$$