Question

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

• A. $lo{g}_e{\left(\frac{x-2}{x-1}\right)}^{\frac{1}{2}}$
• B. $lo{g}_e{\left(\frac{x-1}{3-x}\right)}^{\frac{1}{2}}$
• C. $lo{g}_e{\left(\frac{x}{2-x}\right)}^{\frac{1}{2}}$
• D. $lo{g}_e{\left(\frac{x-1}{x+1}\right)}^{-2}$

Answer 1

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

$y=\frac{e^x-e^{-x}}{e^x+e^{-x}}+2\Rightarrow y=\frac{e^{2x}-1}{e^{2x}+1}+2\Rightarrow e^{2x}=\frac{1-y}{y-3}=\frac{y-1}{3-y}\Rightarrow x=\frac{1}{2}lo{g}_e\left(\frac{y-1}{3-y}\right)\Rightarrow f^{-1}\left(y\right)=lo{g}_e{\left(\frac{y-1}{3-y}\right)}^{\frac{1}{2}}\Rightarrow f^{-1}\left(x\right)=lo{g}_e{\left(\frac{x-1}{3-x}\right)}^{\frac{1}{2}}$