Answer
1) Domain: $(-\infty, \infty)$
2) Range: $(-\infty,\infty )$
Work Step by Step
Given \begin{equation}
f(x)=\sqrt[3]{-x+9}.
\end{equation} a) This is an odd root function because the index, $n=3$, is odd. Both the domain and range are all real numbers.
1) The domain is $(-\infty, \infty)$.
2) The range is $(-\infty,\infty )$.
See the graph for proof.
