Prove that there are $0<c<\frac{\pi }{2}$ such that $f'(c)=(f(\frac{\pi }{2})-f(0))\cos (c)$

Question

Let $f(x)$ be a differentiable function in $[0,\frac{\pi }{2}]$. Prove that there is $0<c<\frac{\pi }{2}$ such that $$f'(c)=\left(f\left(\frac{\pi }{2}\right)-f(0)\right)\cos(c).$$

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$f(x)$ is a differentiable function in $[0,\frac{\pi }{2}]$ so it is also continuous in $[0,\frac{\pi }{2}]$.

To prove this I need to use the mean value theorem.

Is this correct and how to continue from here?

thanks

Answer 1

Hint: Put $g(x)=f(x)-(f(\frac{\pi}{2})-f(0))\sin(x)$, and compute $g(0)$ and $g(\frac{\pi}{2})$.

Answer 2

Cauchy's mean value theorem is the key here. Just use $g(x) =\sin x$ so that $g'$ does not vanish in $(0,\pi/2)$ and thus we have $$\frac{f(\pi/2)-f(0)}{g(\pi/2)-g(0)}=\frac{f'(c)}{g'(c)}$$ for some $c\in(0,\pi/2)$ and we are done.