How do I prove the existence of a point in an ungiven function?
The question is:
Let $f$ be derivable in $[0,\frac{\pi}{2}]$, so $0 \leq f'(x) \leq 1$ for every $x$. Prove the existence of point $x$ in $[0,\frac{\pi}{2}]$ so that $f'(x)=\sin(x)$.
I tried to find an intersection of the functions, but couldn't since nor $f(x)$ nor $f'(x)$ is given in $[0,\frac{\pi}{2}]$ and I didn't know how to deduct that point $x$ exists in both functions.
How to do that? It interests me a lot, but I don't know how to solve it yet.
Thank you a lot.
Hint. Let $F(x)=f'(x)-\sin(x)$ then $F(0)=f'(0)-0\geq 0$ and $F(\pi/2)=f'(\pi/2)-1\leq 0$. Now note that, by Darboux's theorem, derivatives have the Intemediate Value Property and $F(x)=(f(x)+\cos(x))'$.