# Thread: Domain of the function y=x^(1/3)

1. ## Domain of the function y=x^(1/3)

Hi, I'm having trouble on finding the domain of y=x^(1/3). Mathematica is plotting the graph for x E [0,infinity). If x is a negative number, then (-x)^(1/3) is (-1)^(1/3)*(x)^(1/3) which is -(x)^(1/3) for every x and there shouldn't be any problem to graph for this domain. So why can't I take the negative numbers as a domain of this function?
Somebody help me understand this!
Thank you.

2. Yep, fair point

$\sqrt[3]{-8} = -2$

3. Originally Posted by patzer
Anyone?
Computer algebra systems treat $x^{\frac{1}{3}}$ as a expression in the complex field. MathCad does the same a Mathematica.
However, try the notation $\sqrt[3]{x}$.
Using that notation in MathCad I get the expected graph.
I do not know the command in Mathematica.

4. So, we can say that the domain of y=(x)^(1/3) is definitely for x E (-infinity, infinity), right?

Thank you again.

5. Originally Posted by patzer
So, we can say that the domain of y=(x)^(1/3) is definitely for x E (-infinity, infinity), right?
For the set real numbers.

6. I don't know which version of Mathematica you have, but I suspect that Mathematica is actually plotting

$x^{1/3}=e^{\ln(x^{1/3})}=e^{\frac{1}{3}\,\ln(x)}.$

I'm using the equality symbol loosely here, because the instant you introduce the logarithm, you've changed the domain and therefore the function. And that's precisely your problem.

One solution is to "trick" Mathematica into plotting everything for you by means of the command

Plot[Sign[x](Abs[x])^(1/3),{x,-5,5}].

So here I erase the sign information inside the cube root, and then insert it back again with the Sign function.

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# domain (x-1)^1/3

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