Question

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

• A. $\log {\left(\frac{x-2}{x-1}\right)}^{1\setminus 2}$
• B. $\frac{1}{2}{\log }_e\left(\frac{x-1}{3-x}\right)$
• C. $\begin{array}{cc}& \log {\left(\frac{x}{2-x}\right)}^{1\setminus 2}\\ & \end{array}$
• D. $\log {\left(\frac{x-1}{3-x}\right)}^{1\setminus 2}$

Answer 1

The correct option is B $\frac{1}{2}{\log }_e\left(\frac{x-1}{3-x}\right)$

$f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}+2$

Let $f\left(x\right)=y$

Then, $y=\frac{e^x-e^{-x}}{e^x-e^{-x}}+2$

$y=\frac{e^x-e^{-x}+2e^x+2e^{-x}}{e^x-e^{-x}}$

$\Rightarrow y{e}^x+y{e}^{-x}=3e^x+e^{-x}$

$\Rightarrow y{e}^x+\frac{y}{e^x}=3e^x+\frac{1}{e^x}$

$\Rightarrow y.e^{2x}+y=3e^{2x}+1$

$\Rightarrow y.e^{2x}-3e^{2x}=1-y$

$\Rightarrow e^{2x}(y-3)=1-y$

$\Rightarrow e^{2x}=\frac{1-y}{y-3}$

$2x={\log }_e\left(\frac{1-y}{y-3}\right)$

$\therefore x=\frac{1}{2}{\log }_e\left(\frac{1-y}{y-3}\right)$

$y=f\left(x\right)\Rightarrow x=f^{-1}\left(y\right)$

$\therefore f^{-1}\left(y\right)=\frac{1}{2}{\log }_e\left(\frac{1-y}{y-3}\right)$

$\therefore f^{-1}\left(x\right)=\frac{1}{2}{\log }_e\left(\frac{1-x}{x-3}\right)$

$=\frac{1}{2}{\log }_e\left(\frac{x-1}{3-x}\right)$