I have some problems with Domain of the following real function: $$y=x^2\ (2 - x)^{2/3}$$ Since $t^{2/3}=\sqrt[3]{t^2}$, we can rewrite function as $$y=x^2\ \sqrt[3]{(2 - x)^{2}}$$ thus Domain is $D=\mathbb R$. But, by Wolfram: $$D=\{x \in\mathbb R : x\leq 2\}$$ Why? Maybe is $t^{2/3}=\sqrt[3]{t^2}$ not always possible?
Thanks in advance.
For the set of complex numbers, the answer from Wolfram alpha is correct
But for the set of real numbers, your answer is correct.
By WolframAlpha $$\sqrt[3]{x^2}\not\equiv x^{2/3}$$
It has to do with complex numbers. For $x<0$ values of these two functions differ, as for complex numbers the rule $$(a^b)^c=a^{bc}$$ is not always true. It comes from the fact that complex roots are multivalued.
Domain of your function depends on how you define $x^{m/n}$. In school, teachers define it as $$x^{m/n}=\sqrt[n]{x^m}$$
This is pretty intuitive definition but, unfortunately, not consistent for negative $x$. When dealing with complex numbers, this expression is multivalued and its principal value is defined differently, as
$$x^{m/n}=|x|^{m/n}(\cos(\tfrac{m}n\cdot \operatorname{Arg}(x))+i\sin(\tfrac{m}n\cdot\operatorname{Arg}(x))=|x|^{m/n}\exp(\tfrac{m}n\cdot i\operatorname{Arg}(x))$$
When dealing with first definition, your domain is correct, it is $$d=\mathbb R$$
whereas when dealing with the second definition, for $x<0$ the expression gives complex numbers, so domain is $$D:x\geqslant 0$$
As per kamil09875's answer, if you're dealing with complex numbers the two functions are not equivalent. If you're dealing with real numbers and live in a world where $\sqrt[3]{-1} = -1$, then both functions are defined over all $\mathbb{R}$ and equivalent. See this discussion on $(a^b)^c = a^{bc}$.
