Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.2 One-to-One Functions; Inverse Functions - 4.2 Assess Your Understanding - Page 292: 69

Answer

$ f^{-1}(x)=-\frac{2x-3}{x-2}$

Work Step by Step

Step 1. $f(x)=\frac{2x+3}{x+2} \Longrightarrow y=\frac{2x+3}{x+2} \Longrightarrow x=\frac{2y+3}{y+2} \Longrightarrow y=-\frac{2x-3}{x-2} \Longrightarrow f^{-1}(x)=-\frac{2x-3}{x-2}$ Step 2. Check $f(f^{-1}(x))=\frac{2(-\frac{2x-3}{x-2})+3}{(-\frac{2x-3}{x-2})+2}=x$ and $f^{-1}(f(x))=-\frac{2(\frac{2x+3}{x+2})-3}{(\frac{2x+3}{x+2})-2}=x$
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