`f(x) = cos(2x), [-pi,pi]` Determine whether Rolle’s Theorem can be applied to `f` on the closed interval `[a, b]`. If Rolle’s Theorem can be applied, find all values of `c` in the open interval
Rolle's Theorem requires f to be defined and continuous on the given closed interval, differentiable on the open interval and values of f on ends to be equal.
Here all conditions are met (cos(-2pi)=cos(2pi)=1). Therefore there is at least one point c where f'(x)=0.
To find this point(s), find the derivative:
f'(x)=-2sin(2x). It is zero at 2x=k*pi, x=k*pi/2. There are three such points on (-pi, pi): -pi/2, 0 and pi/2.