Assume $f:\Bbb R\to \Bbb R$ is periodic and differential.
I need to show $f'(0)=f'(p)$, $p$ being the periodic componenet i.e $f(x) = f(x+p)$. and that there exist two different $x,y$ in $[0,p]$ such that $f'(x)=f'(y)=0$
How do I approach this question? I'm having trouble even getting started...
For the second part of the question (I already gave a hint for part one in the comments), we can throw out the case where $f'(0)=f'(p) = 0$, as the lemma will then be trivially true. So WLOG assume that $f'(0) = a > 0$. Because $a$ is positive, there exists some $\delta_1,\delta_2>0$ for which $f(\delta_1)-f(0)/\delta_1 > 0$ and $f(p)-f(p-\delta)/\delta_2 > 0$. Consequently, $f(\delta_1) > f(0)$ and $f(p-\delta_2)<f(p)$. Then by the Intermediate Value Theorem, there exists some $x'\in [\delta_1,p-\delta_2]$ such that $f(x')=f(0) = f(p)$. Apply Rolle's theorem to the intervals $[0,x']$ and $[x',p]$ to finish off the proof. The case for $f'(0)<0$ follows a very similar argument.