Let f(x)=x^3-3x^2 find the absolute maximum value of f(x) on the interval [1,4]?

Question

22 hours ago - 3 days left to answer

Answer 1

f' = 3 x^2 - 6 x = 0 at max or min
x(3x-6) = 0
x = 0 or 2 but 0 is outside interval
f" = 6 x - 6
at x = 2, f" = +6
so that is a minimum, not a max
therefore look at the end points
at x = 1, f(1) = 1-3 = -2
at x = 4, f(4) = 16
so
minimum at x = 1