TITLE: Prove there exists $x$ such that $f'(x)=\sin x$
QUESTION [5 upvotes]: Let $f$ be a function which is differentiable on $[0, \frac{\pi}{2}]$, such that $0\leq f'(x)\leq1$ for all $x$ in this interval. I'm asked to prove that there exists $x\in [0, \frac{\pi}{2}]$ such that $f'(x)=\sin x$.
I believe that I should use the Darboux's theorem, but I fail to do it here.

REPLY [5 votes]: Take $g(x)=f(x)+\cos{x}$. This function is derivable on $[0,\pi/2]$. Its derivative is $g'(x)=f'(x)-\sin{x}$. We have $0\leq f'(x)\leq 1$ and $-1\leq -\sin{x}\leq 0$ so $-1\leq g'(x)\leq 1$ Can you take it from there?