What is the absolute maximum for 2x^3-3x^2-5?
Let f(x) = 2x^3 - 3x^2 - 5
f' = 6x^2 - 6x
Setting f' = 0,
6x^2 - 6x = 0
6x(x - 1) = 0
x = 0, x = 1
f(0) = 0 - 0 - 5 = -5
f(1) = 2 - 3 - 5 = -6
|f(0) = 5 and |f(1)| = 6
So, absolute maximum of f(x) is 6