find critical numbers?

Question

G(x) = cube root (x^2-x)

Answer 1

critical numbers mean f'=0 or f' DNE

rewrite G(x) = (x^2-x)^(1/3), then ready to derive:
G'(x) = (1/3) (x^2-x)^(-2/3) (2x-1) (power rule and chain rule)
set =0 and see where undefined.
G'(x)=0 when 2x-1=0 or x = 1/2
G'(x) DNE when x^2-x = 0 or x=1 or x=0

three critical points (0,0) (1,0) and (1/2,cube root -1/4)

Answer 2

G(x) = (x^2-x)^1/3.The derivative of this function exists at all points where x^2-x is NOT zero
x(x-1)=0 x=0 and x= 1
G´(x) = 1/(x^2-x)^2/3 * (2x-1)
G´(x)= 0 x=1/2 is the critical point. (x=0 and x=1 are singular points where the derivative does nor exist)