`f(x) = -3xsqrt(x + 1)` Find the two x-intercepts of the function `f` and show that `f'(x) = 0` at some point between the two x-intercepts.
f(x) = 0 for x=0 and x=-1 (obvious).
f is continuous on [-1, 0] and is differentiable on (-1, 0) (as an elementary function). So we can aplly Rolle's Theorem and conclude that there is at least one point in (-1, 0) where f'(x)=0.
Actually, we can find this point:
`f'(x) = -3sqrt(x+1) - 3x/(2sqrt(x+1)) = (-3/2)*(2(x+1) + x)/sqrt(x+1) = (-3/2)*(3x+2)/sqrt(x+1),`