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: <span>f ' (c) = [f(b) - f(a)] / (b - a) </span> <span> So [f(a) - f(b)] ( b - a ) = [sin(3pi/4) - sin(pi/4)]/pi </span> = [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
You have to work out the mean. (The simple average of those numbers) Then for each number subtract the Mean and square the result. Then work out the average of those squared differences.