f ' (x) = 12 - 3x^2 = 0 for a max/min
3x^2 = 12
x^2 = 4
x = ± 2
f(2) = 24 - 8 = 16
f(-2) = -24 + 8 = -16
using ends of domain
f(-3) = -36 + 27 = -9
f(5) = 60 - 125 = -65
the minimum is -65
the maximum is 16
Determine the maximum and minimum of each function on the given interval.
f(x)���� = 12x� − x^3�� , � x∈ �[−3,5]
1 answer