Graphing $x^{2/3}$: a question of domain I'm trying to graph $x^{2/3}$. If I enter $y=x^{2/3}$, my graphing program excludes negatives from the domain:

However, if I enter it as either $y=\sqrt[3]{x^2}$ or $y=(x^{1/3})^2$, it includes the negative values of $x$:

I'd like to understand what's going on. My best guess is that while $x^{2/3}$ is well defined for negative values of $x$ (since the fractional exponent's denominator is an odd integer), the graphing program is being overly cautious in interpreting the expression. But that's just a guess. Any ideas what's happening here?
Thanks for the responses. As it seems this is a duplicate of a previously posted question, I'll vote to close this.
 A: It likely comes down to the fact that your graphing program is using an approximation for $2/3,$ and said approximation has an even denominator.
A: Can't post this as a comment, but note:
WolframAlpha: $y = x^{2/3}$:

WolframAlpha: $y = \sqrt[3]{x^2}$:

Note: the scales for graphing are different, but there is clearly a difference with respect to how Wolfram interprets each of the inputs. And the first plot adds an additional anomaly.
A: The issue is that there are three sorts of things one could mean when one writes down an exponentiation, and the differences between them become quite significant when you consider negative bases.
Here is an answer I've written to another question talking about a similar question.
A: Here is the reason I think
Softwares interpret both expressions in a different way
$$ x^{\frac{2}{3}} = \exp\left ( \frac{2}{3}  \ln \left ( x \right ) \right ) $$
Works only if $x>0$
$$ \sqrt[3]{x^{2}} = \exp\left ( \frac{1}{3} \ln \left ( x^{2} \right ) \right ) $$
Works whatever $x$ value
