Question

If $f\left(x\right)=\frac{x-1}{x}=x\ne \pm 1$ then verify $fo{f}^{-1}\left(x\right)=x$

Answer 1

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

$\Rightarrow x=\frac{1}{1-y}$

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

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

$=1-(1-x)=x$

Hence verified