Given that, f(x)=x3 and g(x)=x3−4x Since, f (x) and g(x) are both continuous at [−2,2] and differentiable at [−2,2] So,f(x) and g(x) satisfy mean value theorem. Now,f(−2)=−8,f(2)=8 So, f(−2)≠f(2)
g(2)=(2)3−4(2)=0=g(−2)=(−2)3−4(−2)=0
Therefore, f(x) does not satisfy Rolle’s theorem but g(x) satisfy Rolle’s theorem.