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