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What is the inverse of the function $f(x)=\large\frac{e^x-e^{-x}}{e^x+e^{-x}}$

$\begin{array}{1 1}f^{-1}(x)=\log\large\big(\frac{1-x}{1+x}\big) \\f^{-1}(x)=\frac{1}{2}\log\large\big(\frac{1+x}{1-x}\big) \\ f^{-1}(x)=\frac{1}{2}\log\large\big(\frac{1-x}{1+x}\big) \\f^{-1}(x)=\log\large\big(\frac{1+x}{1-x}\big)\end{array}$

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Toolbox:
  • Express x in terms of y to get the inverse.
$y=\large\frac{e^x-e^{-x}}{e^x+e^{-x}}$
$\Rightarrow y=\large\frac{e^{2x}-1}{e^{2x}+1}$
Using componendo and dividendo we get
$\large\frac{y+1}{y-1}=\frac{2e^{2x}}{-2}$
$\Rightarrow\:e^{2x}=\large\frac{1+y}{1-y}$
Taking log on both sides $\Rightarrow 2x=\log\bigg(\frac{1+y}{1-y}\bigg)$
$\Rightarrow x=f^{-1}(y)=\frac{1}{2}\log\large\big(\frac{1+y}{1-y}\big)$
$f^{-1}(x)=\frac{1}{2}\log\large\big(\frac{1+x}{1-x}\big)$
answered May 2, 2013 by rvidyagovindarajan_1
 

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