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What is the domain and range of $x^{1/3}$?
I know what the graph looks like, but I am unsure of the domain and the range especially at $x=0$. It it all real, or is there a point of discontinuity at $x=0$?

Thanks

Arnaldo
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MaxF
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    Are you asking about domain/range or about continuity at 0? For the latter, it is continuous at $0$, and you can easily prove it right off the epsilon-delta definition. – Vim May 24 '17 at 02:07

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Consider the inverse function of $f(x)=x^{\frac{1}{3}}$, which is $f^{-1}(x)=x^3$.

By definition, since the domain and range of $f^{-1}(x)$ is $(-\infty, \infty)$, the domain and range of $f(x)$ is $(-\infty, \infty)$.

More importantly, because $f^{-1}(0)=0$, $f(0)=0$. This can also be observed intuitively by sketching $f^{-1}(x)=x^3$ and turning the paper sideways.

Saketh Malyala
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