Question

If f(x) = |(x2 - 4)(x2 + 2)|, how many numbers in the interval [0, 1] satisfy the conclusion of the Mean Value Theorem?

Answer 1

f(x) = |(x^2-4)(x^2+2)|
f is continuous and differentiable on (-2,2), so also on [0,1].

Since |u| = -u if u<0,
on [0,1], f(x) = (4-x^2)(x^2+2)

f'(x) = 4x(1-x^2)

f(0) = 8
f(1) = 9
(f(1)-f(0))/(1-0)=1
so, where is f'(x)=1?
only at 2 points in [0,1]