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Q.
The inverse of the function
$f(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}+2$ is
Relations and Functions - Part 2
Solution:
$\frac{y-2}{1}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$
Applying comp. and dividendo:
$\frac{y-1}{3-y}=\frac{2 e^{x}}{2 e^{-x}}=e^{2 x}$
$\therefore x=\frac{1}{2} \log \left(\frac{y-1}{3-y}\right)=\log \left(\frac{y-1}{3-y}\right)^{1 / 2}$
Hence, the inverse of the function
$f(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}+2$ is
$\log _{c}\left(\frac{x-1}{3-x}\right)^{1 / 2}$