#### Answer

$c = \pi$

#### Work Step by Step

$f(x) = sin(\frac{x}{2}),[\frac{\pi}{2},\frac{3\pi}{2}]$
$f(\frac{\pi}{2}) = sin(\frac{\pi/2}{2})=\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$
$f(\frac{3\pi}{2}) = sin(\frac{3\pi/2}{2})=\sin(\frac{3\pi}{4})=\frac{\sqrt{2}}{2}$
Differentiate using chain rule: $f'(x)=\frac{1}{2}cos(\frac{x}{2})$
Since $f'(x)$ exists, $f(x)$ is differentiable which implies continuity, and because $f(\frac{\pi}{2})=f(\frac{3\pi}{2})$, Rolle's theorem's prerequisites are met and is applicable.
Find zeroes of $f'(x)$: $cos(\frac{x}{2})$ has to equal 0, since $\frac{1}{2}\times0=0$
Using the unit circle, we know that $cos(\frac{\pi}{2})$ and $cos(\frac{3\pi}{2})$ both equal 0.
Therefore $x = \pi,3\pi$
The function is continuous on $[\frac{\pi}{2},\frac{3\pi}{2}]$ and differentiable on $(\frac{\pi}{2},\frac{3\pi}{2})$.
Only $\pi$ is on the interval $(\frac{\pi}{2},\frac{3\pi}{2})$
Therefore $c=\pi$