find the absolute minimum and maximum values of f(x)= x^(2/3) (3+x) on the interval [-2,1] i did a mistake on the other one sorry.
f ' (x) = (2/3)x^(-1/3) (3+x) + x^(2/3)
= (1/3)x^(-1/3) [ 2(3+x) + 3x]
= (5x+6)/(3x^(1/3) )
for a max/min,
(5x+6)/(3x^(1/3) ) = 0
5x+6 = 0
x = -6/5
f(-6/5) = (-6/5)^(2/3) (3-6/5) = appr 2.03
f(-2) = (-2)^(2/3) (1) = 1.59
f(1) = 4
max is 4, min is appr 1.59