Barrons AP Calculus - David Bock

(A) Note that f (0) = = 0 and that f ′(x) exists on the given interval. By the MVT, there is
a number, c, in the interval such that f ′(c) = 0. If c = 1, then 6c^2 − 6 = 0. (−1 is not in the
interval.)