Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.1 Inverse Functions - 6.1 Exercises - Page 407: 27


The inverse function is given by $$f^{-1}(x)=\left(\frac{1-x}{1+x}\right)^2.$$

Work Step by Step

We have to express $x$ in terms of $y$. Then $x=f^{-1}(y)$. $$y=\frac{1-\sqrt{x}}{1+\sqrt{x}}\Rightarrow(1+\sqrt{x})y=1-\sqrt{x}\Rightarrow y+y\sqrt{x}=1-\sqrt{x}.$$ Rearranging terms $$y\sqrt{x}+\sqrt{x}=1-y\Rightarrow \sqrt{x}(y+1)=1-y\Rightarrow \sqrt{x}=\frac{1-y}{y+1},$$ which finally gives $$x=\left(\frac{1-y}{1+y}\right)^2$$ so $$f^{-1}(y)=\left(\frac{1-y}{1+y}\right)^2.$$ Renaming $y$ back to $x$ we have $$f^{-1}(x)=\left(\frac{1-x}{1+x}\right)^2.$$
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