Find the domain of $x^{2/3}$ Find the domain of $f(x)=x^{2/3}$.  Now, $0^2=0$ and $\sqrt[3]{0}=0$, all positive and negative numbers squared will give positive answer, these numbers can give us cuberoot, so entire real line is the domain, am I correct? (one of the online help provided the answer as $\{x \in \mathbb{R} : x \geq 0\}$ (all non-negative real numbers), why negative numbers are not included in the domain, where am I going wrong?
 A: By definition, $x^{2/3} = \sqrt[3]{x^2}$.
Since we can compute $x^2$ for any real number $x$, and since we can compute a cubic root for any real number, whether it be positive, negative, or zero, the domain is all real numbers.
On the other hand, as Kannappan hints, the range of this function is $[0,\infty)$. So you should really check the input of that on-line tool.
A: The answer is unfortunately context-dependent. That's a fancy way of saying that on a test the answer is what your instructor says it is.  For whatever it is worth, Wolfram Alpha thinks that the domain is the non-negative reals.
I assume we are working in the reals.  We are concerned with the domain of $x^y$, where $y$ is a positive real number. (I am taking $y$ positive to avoid division by $0$ issues, and $0$-th power issues.)
There is general agreement that $x^y$ is defined when $x\ge 0$. But there are disagreements in the case $x<0$.
If $y$ is irrational, there is general agreement that $x^y$ is not defined at negative $x$. 
If $y$ is rational, and $y$ can be expressed as $a/b$, where $a$ is an odd integer, and $b$ is an even integer, there is general agreement that $x^y$ is not defined at negative $x$.
When $y$ is a rational, which, when put in lowest terms, has an odd denominator, there is disagreement.
In high school, generally the convention is that in that case, $x^y$ is defined when $x$ is negative.  And that convention is not confined to high school.
But when we are dealing with exponential functions, with $y$ thought of as variable, there are good arguments for considering, for example, $(-2)^{2/3}$ to be undefined.
The issue is this. Consider the function $(-2)^y$.   Arbitrarily close to $2/3$, there are irrational $y$ at which $(-2)^y$ is definitely not defined. So even if we consider $(-2)^{2/3}$ to have a meaning, the function $(-2)^y$ is not continuous at $y=2/3$.  
Another argument is that a common definition of $x^y$ is that $x^y=\exp(y\ln x)$. Note that $\ln x$ is not defined when $x$ is negative. Of course, $\ln x$ is not defined at $x=0$, but that generally does not stop people from considering $x^{2/3}$ defined at $x=0$. 
