If f(x) = sin(x/2), then there exists a number c in the interval pie/2 < x < 3pie/2 that satisfies the conclusion of the mean value theorem. what could be c?

For the answer to the question above, 
The mean value theorem states the if f is a continuous function on an interval [a,b], then there is a c in [a,b] such that: 
f ' (c) = [f(b) - f(a)] / (b - a) 

So [f(a) - f(b)] ( b - a ) = [sin(3pi/4) - sin(pi/4)]/pi 
= [sqrt(2)/2 - sqrt(2)/2]/pi = 0 
So for some c in [pi/2, 3pi/2] we must have f ' (c) = 0 

In general f ' (x) = (1/2) cos (x/2) 
We ask ourselves for what values x in [pi/2, 3pi/2] does the above equation equal 0. 
0 = (1/2) cos (x/2) 
0 = cos (x/2) 
x/2 = ..., -5pi/2, -3pi/2, -pi/2, pi/2, 3pi/2, 5pi/2,... 
x = ..., -5pi, -3pi, -pi, pi. 3pi, 5pi,  
and x = pi is the only solution in our interval. 

So c = pi is a solution that satisfies the conclusion of the MVT