## Precalculus (10th Edition)

domain: $[-2,\infty)$ range: $[5,\infty)$
Since $f$ is one-to-one the domain of $f$ is the range of $f^{-1}$ and the range of $f$ is the domain of $f^{-1}$. Hence $f^{-1}$ has: domain: $[-2,\infty)$ range: $[5,\infty)$