Question

If $f\left(x\right)=\frac{e^x+e^{-x}}{2}$, then the inverse of $f\left(x\right)$ is

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

Answer 1

The correct option is D ${\log }_e(x+\sqrt{x^2-1})$

Let $f\left(x\right)=\frac{e^x+e^{-x}}{2}=y$
$\therefore x=f^{-1}\left(y\right)$
$e^x+e^{-x}=2y$
$e^{2x}-2y{e}^x+1=0$
$\Rightarrow e^x=\frac{2y\pm \sqrt{4y^2-4}}{2}$
$e^x=\frac{y\pm \sqrt{y^2-1}}{1}$
Range of $f$ is $(-\infty ,\infty )\Rightarrow e^x=\frac{y+\sqrt{y^2-1}}{1}$
As if we take $e^x=\frac{y-\sqrt{y^2-1}}{1}$, which is always small
$\therefore x={\log }_e\left(\frac{y+\sqrt{y^2-1}}{1}\right)$$\therefore f^{-1}\left(x\right)={\log }_e\left(\frac{x+\sqrt{x^2-1}}{1}\right)$