# Prove there exists $x$ such that $f'(x)=\sin x$

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.

• Darboux's theorem says that $f'\left([0,\pi/2]\right)$ is an interval. As for $\sin$, it is a non-decreasing function from $[0,\pi/2]$ to $[0,1]$. See if you can show that the graphs of the both functions meet for some $x$ in $[0,\pi/2]$. Jun 22, 2015 at 13:22
• Perhaps apply Darboux's theorem to $g(x)=f(x)+\cos x$ ?
– lhf
Jun 22, 2015 at 13:23

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?
• I must find $a,b$ such that $g'(a)<0$ and $g'(b)>0$ so then I can use the intermediate value theorem to prove the existence of $c$ such that $g'(c)=0$ and then $f'(c)=\sin c$. Am I correct? Thank you. (P.S. - I don't see how to apply Darboux's theorem for $g(x)$ on $[0,\pi/2]$ because it boils down to the derivatives $f'(0)$ and $f'(\pi/2)$ which we don't know). Jun 22, 2015 at 15:09
• Yes you're right. This is why it is not done completely. You still need to prove that $g'$ changes sign so it is zero somewhere Jun 22, 2015 at 15:40
• Here's my attempt: If $f'(0)=0$ or $f'(\pi/2)=1$ then $g'(0)=0$ or $g'(\pi/2)=0$ respectively and we're done. Otherwise, $f'(0)>0$ and $f'(\pi/2)<1$ and so $g'(0)>0$ and $g'(\pi/2)<0$. Thus, applying the intermediate value theorem there exists $c\in [0,\pi/2]$ such that $g'(c)=0$. Am I correct? Isn't it strange that I didn't have to use the Darboux's theorem here (as several people have suggested above)? Thanks. Jun 22, 2015 at 16:38