I've completely confused myself as to determining the domain of polynomial expressions that have rational powers, e.g. $$y = x^{2/3} \qquad \text{or} \qquad y_2 = (x^2-1)^{2/ 3}.$$
A calculus textbook I've consulted asserts that the domain of $y_2$ is all real values of $x$; however, when I plot the function using Grapher or Wolframalpha, it excludes the values for $-1 \leq x \leq 1$:
Similarly, when I try to use Wolfram to evaluate $(-4)^{2/ 3}$ it returns a complex number, whereas I would've expected a real value (because I thought we could think of this as $(-4)^{2/ 3} = \big((-4)^2\big)^{1/ 3} = (16)^{1/ 3} = \sqrt[3]{16}$, which I thought was the same as $\big( (-4)^{1/ 3} \big)^{2} = (-\sqrt[3]{4})^{2} = \sqrt[3]{4}^2$
I consulted this question, which clarified that the property $a^{bc} = (a^b)^c$ only applies to all $a$, if $b$ $c$ are integers (otherwise, we must assume $a > 0$), but I'm still unclear as to how I can determine the domain of these functions, given the disagreement between my textbook and what I'm finding with graphing apps.
Any clarification would be greatly appreciated!
ps. I could not determine which of the "domain" tags was appropriate, I did not see anything like "domain of a function," and most say not to use for this purpose.
Edit
I just consulted a 3rd textbook that includes the following statement:
For all real numbers $a$ for which the indicated roots exist, and for any rational number $m/n$, $$a^{m/n} = (a^{1/n})^m$$.
This prompted me to try graphing the function as $y_2 = \big((x^2-1)^{1 /3}\big)^2$; however, this yielded the same result as before (ie excluding -1 < x < 1).
On Wolfram, I tried y = (cube-root(x^2-1))^2 and this yielded the same graph as the one provided by my textbook.
I must admit, I'm a bit at a loss as to why $y_2 = \big((x^2-1)^{1 /3}\big)^2$ didn't resolve the issue. If that had worked, it would've made (some) sense to me.
Best Answer
I believe the issue here is the definition of negative numbers raised to rational powers that's used by Wolfram or your graphing software. A robust definition of exponentials that's used by most software because it generalises to complex numbers is: $$a^b := \exp({b\ln a}),$$ where $\exp x=e^x$, and in the case where $a$ is complex, $\ln$ is the principal logarithm. The issue now, is when you try to evaluate something like $(-1)^{2/3}$. If we use the definition we have $$(-1)^{2/3}=\exp\left(\frac23\ln(-1)\right),$$ which clearly makes no sense in $\mathbb R$ (the $\ln$ of a negative number is undefined). So to answer your question, the reason wolfram thinks $(-1)^{2/3}$ is undefined is because of the robust definition for exponentiation, which gives $(-1)^{2/3}$ as a complex number.