f(x)=x⁴-2x², f'(x)=4x³-4x=0=4x(x-1)(x+1), so c=-1,0,1 all in (-2,2).
f(-2)=f(2)=16-8=8, so Rolle’s Theorem says there must a value c between -2 and 2 where the gradient f' must be zero (flat). Also f(0)=0 so between -2 and 0, 0 and 2 there must also be zero gradients f(-√2)=f(√2)=f(0)=0.
The three intercepts for the gradient concur with Rolle’s Theorem since -1 lies between -√2 and 0, 0 lies between -2 and 2, and 1 lies between 0 and √2.