Precalculus (10th Edition)

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: 63

Answer

(a) $ f^{-1}(x)=\frac{2}{x}-3$. (b) $f$: $(-\infty,-3)U(-3,\infty)$ and $(-\infty,0)U(0,\infty)$. $f^{-1}$: $(-\infty,0)U(0,\infty)$ and $(-\infty,-3)U(-3,\infty)$.

Work Step by Step

(a) $f(x)=\frac{2}{3+x} \longrightarrow y=\frac{2}{3+x}\longrightarrow x=\frac{2}{3+y}\longrightarrow y=\frac{2}{x}-3\longrightarrow f^{-1}(x)=\frac{2}{x}-3$. Check $f(f^{-1})=\frac{2}{3+\frac{2}{x}-3}=x$ and $f^{-1}(f)=\frac{2}{\frac{2}{3+x}}-3=x$ (b) Domain and range of $f$: $(-\infty,-3)U(-3,\infty)$ and $(-\infty,0)U(0,\infty)$. Domain and range of $f^{-1}$: $(-\infty,0)U(0,\infty)$ and $(-\infty,-3)U(-3,\infty)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.