Question

Let $f:R\to R$ be defined by $f\left(x\right)=\frac{(e^x-e^{-x})}{2}$.
The inverse of the given function is:

• A. $f^{-1}\left(x\right)={\log }_e(x+\sqrt{x^2+1})$
• B. $f^{-1}\left(x\right)={\log }_e(x-\sqrt{x^2-1})$
• C. $f^{-1}\left(x\right)={\log }_e(x-\sqrt{x^2+1})$
• D. $f^{-1}\left(x\right)={\log }_e(x+\sqrt{x^2-1})$

Answer 1

The correct option is A $f^{-1}\left(x\right)={\log }_e(x+\sqrt{x^2+1})$

Let $y=f\left(x\right)=(e^x-e^{-x})/2$
$\Rightarrow 2y=e^x-e^{-x}$
$\Rightarrow e^{2x}-2y{e}^x-1=0$
$\Rightarrow e^x=\frac{2y\pm \sqrt{4y^2+4}}{2}=y\pm \sqrt{y^2+1}$
$\Rightarrow e^x=y+\sqrt{y^2+1}$
$(\because y-\sqrt{y^2+1}<0$ for all $y$ but $e^x$ is always positive$)$
$\Rightarrow x={\log }_e(y+\sqrt{y^2+1})$
$\therefore f^{-1}\left(x\right)={\log }_e(x+\sqrt{x^2+1})$