What are the critical numbers of the function: f (x)= (2x^3-2)^1/4

Question

anonymous · Answer

f'[x]=1/4(2x^3-2)^(-3/4)*(6x^2)
then set it equal to zero to solve for x, whatever x u found will be your critical values.

anonymous · Answer

I get x=0 but the answer key says it's x=1 because 0 is not in the domain. But how do I know it is 1, if you get that part.

anonymous · Answer

that's weird, i'm using mathematica and still got x=0, unless you copy the question wrong.

anonymous · Answer

1. Find all critical numbers for the function:  f (x)= (2x^3-2)^1/4
 That is exactly what it says and the answer key says: 1. x = 1, zero is not part of the domain.

anonymous · Answer

it is true that 0 is not part of the domain, but I don't know why 1 is part of the answer, Mathematica only give zero as answer.

anonymous · Answer

Oh ok I'll ask my teacher tomorrow but thanx for trying anyway.