I am currently studying for a graduate exam and I came across the following question:
Suppose that $f$ and $g$ are real-valued functions on $\mathbb{R}$ having period 1 and having continuous first derivatives. Prove that $f'(c) = g'(c)$ for some non-negative $c \in \mathbb{R}$
My initial approach to this problem was to use the mean value theorem for $f$ and $g$ to obtain certain values where both derivatives vanish. However, this did not prove effective in proving the desired result.
Any suggestions on how to approach this problem would be greatly appreciated.
$0=f(1)-f(0)=\int_0^{1}f'(t)dt$ and $0=g(1)-g(0)=\int_0^{1}g'(t)dt$. Hence $\int_0^{1}f'(t)dt=\int_0^{1}g'(t)dt$. If $f'(c) \neq g'(c)$ for any $c$ between $0$ and $1$ then either $f'(t) >g'(t)$ for all $t$ in this interval or $g'(t) >f'(t)$ for all $t$ in this interval (by Intermedite Value Property of continuous functions). Do you see a contradiction?