Find the absolute maximum and minimum values of the function f (x) = [x^(2/3)]*[x − 1] on the

Question

closed interval [0, 4], and state where these values occur.

Answer 1

f' (x) = x^(2/3) + (2/3)x^(-1/3) (x-1)
= (1/3)x^(-1/3) [ 3x - 2(x-1) ]
= (1/3)x^(-1/3) [ x + 2]
= 0 for a max/min

x = 0 or x = -2

checking the f(x) for those and the endvalues

f(0) = 0
f(-2) = (-2)^(2/3) (-2-1) = -3(-2)^(2/3)
= appr 4.76
f(4) = 4^(2/3) (4-1) = appr 7.56

so the max is 3( 4^(2/3) ) or appr 7.56
and the min is 0

Answer 2

thank you very much