Prove that a differentiable function $f$ with $f(x+1)=f(x)$ has at least two points in $[0,1]$ such that $f ' (x) =0$. Prove that a differentiable function $f$ with $f(x+1)=f(x)$ has at least two points in $[0,1]$ such that $f ' (x) =0$. 
I used Mean value theorem to obtain one such point in $[0,1]$ , but i am not sure how to go any further.
Thanks
 A: Case 1: $f'(x)$ has at least two roots in $(0,1)$. We are done.
Case 2: $f'(x)$ has exactly one root in $(0,1)$. Lets call this root $c$.
Then $f$ is strictly monotonic on $(0,c)$ and $(c,1)$, and as $f(0)=f(1)$, the monotony on the two intervals must be different.
But then $f$ has the same monotony on $(1,1+c)$ as on $(0,c)$, therefor $f(1)$ is a local extrema. This implies $f'(1)=0$.
In this case you can also prove $f'(0)=0$, but it is not needed.
Case 3: $f'(x)$ has no root in $(0,1)$. Then, $f$ must be strictly monotonic on $(0,1)$, which is impossible as $f$ continuous and $f(0)=f(1)$.
Note: The proof uses the fact that $f'$ has the intermediate value property, in the form that if $f'$ has no root on an interval, it cannot change sign on that interval.
A: Case 1: $f$ is constant. Every point is such as $f'(x)=0$.
Case 2: $\sup_{[0,1]} f > \inf_{[0,1]} f$.
Use the fact that $f$ is bounded and that $f(\Bbb R) = f([0,1])$, and using the Bolzano Weierstrass theorem, you get $(x,y)\in [0,1]^2$ such as 
$$
\max_{[0,1]} f  = f(x)
\\
\min_{[0,1]} f = f(y)
$$
Then show that $f'(x) = f'(y) = 0$.
