Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises: 26

Answer

$y=\ln\frac{1+x}{1-x}$

Work Step by Step

$y=\frac{1-e^{-x}}{1+e^{-x}}$ $y(1+e^{-x})=1-e^{-x}$ $y+ye^{-x}=1-e^{-x}$ $ye^{-x}+e^{-x}=1-y$ $e^{-x}(y+1)=1-y$ $e^{-x}=\frac{1-y}{1+y}$ $-x=\ln\frac{1-y}{1+y}$ $x=-\ln\frac{1-y}{1+y}=\ln(\frac{1-y}{1+y})^{-1}=\ln\frac{1+y}{1-y}$ $y\rightleftharpoons x$ $f^{-1}(x)=y=\ln\frac{1+x}{1-x}$
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