Question

If $f:R\to \left(0,2\right)$ defined by $f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}+1$ is invertible, find $f^{-1}$

Answer 1

Let $y=f\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}+1$

$\therefore y-1=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^x-\frac{1}{e^x}}{e^x+\frac{1}{e^x}}$

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

$\therefore y{e}^{2x}+y-e^{2x}-1=e^{2x}-1$

$e^{2x}(y-2)=-y$

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

Taking log on both sides

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

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

Now, $y=f\left(x\right)$

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

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

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