Answer
domain: $[5,\infty)$
range: $[0,\infty)$
Work Step by Step
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: $[5,\infty)$
range: $[0,\infty)$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 5 - Exponential and Logarithmic Functions - 5.2 One-to-One Functions; Inverse Functions - 5.2 Assess Your Understanding - Page 267: 80
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