The inverse of the function f(x)=ex−e−xex+e−x+2 is given by

let y=f(x) ⇒y-2=ex-e-xex+e-x apply compodendo divedendo y-1y-3=2ex-2e-x ⇒e2x=y-13-y ⇒x=12log(y-13-y) ⇒f-1(x)=log(x-13-x)12

The inverse of the function f(x)=ex−e−xex+e−x+2 is given by

let y=f(x) ⇒y-2=ex-e-xex+e-x apply compodendo divedendo y-1y-3=2ex-2e-x ⇒e2x=y-13-y ⇒x=12log(y-13-y) ⇒f-1(x)=log(x-13-x)12

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The inverse of the function $f (x) = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} + 2$ is given by

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$\log_{e} {(\frac{x - 1}{x + 1})}^{- 2}$

$\log_{e} {(\frac{x - 2}{x - 1})}^{1 / 2}$

$\log_{e} {(\frac{x}{2 - x})}^{1 / 2}$

$\log_{e} {(\frac{x - 1}{3 - x})}^{1 / 2}$

answer is D.

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Detailed Solution

$l e t y = f (x) \Rightarrow y - 2 = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} a p p l y c o m p o d e n d o d i v e d e n d o \frac{y - 1}{y - 3} = \frac{2 e^{x}}{- 2 e^{- x}} \Rightarrow e^{2 x} = \frac{y - 1}{3 - y} \Rightarrow x = \frac{1}{2} \log (\frac{y - 1}{3 - y}) \Rightarrow f^{- 1} (x) = \log {(\frac{x - 1}{3 - x})}^{\frac{1}{2}}$