Question

use the mean value theorem to show that if f(x) = sinx/x then there is a point c in (pi/4,pi/2) such that f'(c) = -8(sqrt2 -1) pi^2

Answer 1

I'm going to assume you meant divided by pi^2, not multiplied by.
The Mean Value Theorem states that for a function \(f(x)\) continuous on \([a,b]\) and differentiable on \((a,b)\) then \(\displaystyle\exists \;c \in (a,b) : f'(c)=\frac{f(b)-f(a)}{b-a} \). To answer your question:
Let \(\displaystyle f(x)=\frac{\sin x}{x}\). Then \(f(x)\) is obviously continuous on \(\left[\tfrac{\pi}{4},\tfrac{\pi}{2}\right]\) and \(\displaystyle f'(x)=\frac{\cos x}{x} - \frac{\sin x}{x^2}\) exists on \(\left(\tfrac{\pi}{4},\tfrac{\pi}{2}\right)\). Therefore the Mean Value Theorem applies. Therefore by the mean value theorem there exists \(c\) in the open interval \(\left(\tfrac{\pi}{4},\tfrac{\pi}{2}\right)\) such that \[\small f'(c) = \frac{f(b)-f(a)}{b-a} = \frac{\frac{\sin b}{b}-\frac{\sin a}{a}}{b-a} = \frac{\frac{\sin \frac{\pi}{2}}{\frac{\pi}{2}}-\frac{\sin \frac{\pi}{4}}{\frac{\pi}{4}}}{\frac{\pi}{2}-\frac{\pi}{4}} = \frac{\frac{2}{\pi}-\frac{2\sqrt{2}}{\pi}}{\frac{\pi}{4}}=\frac{8-8\sqrt{2}}{\pi^2}=-\frac{8(\sqrt{2}-1)}{\pi^2} \]
as required.