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...

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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.