Calculus: Early Transcendentals, Chapter 4, 4.2, Section 4.2, Problem 4

Given the function f(x)=cos(2x) in the interval [pi/8, (7pi)/8]
We have to see whether it sattisfies the three hypotheses of Roll's theorem.
(a) f(x) is continuous in the given interval because all periodic functions are continuous.
(b) Now f'(x)=-2sin(2x) which is differentiable everywhere the given interval.
(c) Now evaluate whether f(pi/8)=f((7pi)/8)
So,f(pi/8)=cos(2.pi/8)=cos(pi/4)=1/sqrt(2)
f((7pi)/8)=cos(2.(7pi)/8)=cos((7pi)/4)=1/sqrt(2)
Therefore f(pi/8)=f((7pi)/8)
So there exists a number c such that pi/8and f'(c)=0
Therefore, f'(c)=-2sin(2c)=0
sin(2c)=0 implies 2c=npi i.e. c=npi/2 , n=1,2,...
So c=pi/2 , pi, 3pi/2 lie in the interval [pi/8, 7pi/8]