Question

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

Answer 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.