Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.1 Inverse Functions - 4.1 Exercises - Page 418: 70

Answer

(a) $f^{-1}(x)=\frac{1}{x}-2$ (b) see graph (c) $f(x)$ domain $(-\infty,-2)U(-2,\infty)$ and range $(-\infty,0)U(0,\infty)$, $f^{-1}(x)$ domain $(-\infty,0)U(0,\infty)$ and range $(-\infty,-2)U(-2,\infty)$.

Work Step by Step

(a) This function $f(x)=\frac{1}{x+2}, x\ne-2$ is one-to-one. Find the inverse as the following: $y=\frac{1}{x+2}\longrightarrow x=\frac{1}{y}-2\longrightarrow f^{-1}(x)=\frac{1}{x}-2$ (b) see graph (c) $f(x)$ domain $(-\infty,-2)U(-2,\infty)$ and range $(-\infty,0)U(0,\infty)$, $f^{-1}(x)$ domain $(-\infty,0)U(0,\infty)$ and range $(-\infty,-2)U(-2,\infty)$.
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