We are integrating f "(x) twice here to arrive at our function. We are given a point on the graph of f ' and on f, which will allow us to solve for C in both cases:
f ' (x) = ∫(6x - 6)dx = 3x2 - 6x + C with f'(2) = 0 , so f'(2) = 12 - 12 + C = 0 and C = 0. f'(x) = 3x2 - 6x
f(x) = ∫(3x2 - 6x)dx = x3 - 3x2 + C with f(2) = 0 so f(2) = 8 - 12 + C = 0 and C = 4
f(x) = x3 - 3x2 + 4. You can verify that f has an x-intercept and a critical pt (local minimum) at x = 2, as well as the correct double derivative. (f satisfies the first two conditions because it is divisible by (x-2)2.)
