F (x) = 3x^2-12x+5, find the absolute max and min value of f (x) on the interval [0,1].

F (0) = 3 (0) ^2 - 12 (0) + 5 = 5

f (1) = 3 (1) ^2 - 12 (1) + 5 = 3 - 12 + 5 = - 4

The absolute maximum in the interval [0, 1] is at y = 5 which occurs at x = 0.

The absolute minimum in the interval [0, 1] is at y = - 4 which occurs at x = 1.