Math graph: f(x) = 3x^(2/3) - x^2?

Question

i need the critical numbers, x intercepts , and the derivative

YO isnt the derivative : 2x^(-1/3) - 2x ?

Answer 1

f(x) = 3x^(2/3) - x^2 = (x^2)(3x^(-4/3) - 1) = 0

3x^(-4/3) - 1 = 0
3x^(-4/3) = 1
x^(-4/3) = (1/3)
1/(x^(4/3)) = (1/3)
1/(1/3) = x^(4/3)
x^(4/3) = 3
x^4 = 27
x = 4thrt(27)

x-intercepts are 0 and about 2.27951

f(x) = 3x^(2/3) - x^2
f'(x) = 2x^((2/3) - 1) - 2x
f'(x) = 2x^(-1/3) - 2x
f'(x) = (2x)(x^(-4/3) - 1)

x^(-4/3) - 1 = 0
x^(-4/3) = 1
x = 1

Critical Points are 0 and 1
x-intercepts are 0 and about 2.27951
Derivative is 2x^(-1/3) - 2x or (2 - 2x^(1/3))/(x^(1/3)) or 2((1 - x^(1/3))/(x^(1/3)))

Answer 2

x intercepts:
0=3x^2/3-x^2
3x^2/3=x^2
x^4=27
x=0 or 2.2795 or -2795

derivative:
f'(x)=2x^(1/3)-2x

critical numbers:
0=2x^1/3-2x
x=x^3
x=1 or -1 or 0