## Calculus (3rd Edition)

The inverse function is $g(x)=(x^2+8)^{1/3}$; its domain is $[0,\infty)$ and range is $(2,\infty)$.
To find the inverse of $y=\sqrt{x^3-8}$, we have $$y^2=x^3-8 \Longrightarrow x^3=y^2+8\Longrightarrow x=(y^2+8)^{1/3}.$$ Hence, the inverse function is $g(x)=(x^2+8)^{1/3}$. Note that the lowest value that $x$ can take on is $0$, which gives the range of $(2,\infty)$. The domain is the range of the original function, which is $[0,\infty)$.