Find the inverse function of $f(x)=\frac{1-x}{-x}$

Question

When I tried solving for the inverse of $f(x)=\frac{1-x}{-x}$, I got this:

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

I know that the way to check my answer would be to take the inverse of the inverse I just found, but this is what I get:

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

The last part, when multiplied by $\frac{-1}{-1}$ is indeed my original function, but am I allowed to do that? And why does that get lost when taking the inverse?

Answer 1

A better way to check your answer is;

the composite functions of both should be equal to x.

$f(f^{-1}(x))=x$

and

$f^{-1}(f(x))=x$

Answer 2

It really helps to simplify $f(x)$ so that $x$ only appears once in its definition.

$$f(x)=\frac{1-x}{-x} = 1-\frac 1x $$

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

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