Finding inverse of the rational fucntion I am trying to find the inverse of $f(x)=\frac{2x-1}{x(x-1)}$. So far I figured out that the domain should be (0,1) and the range is R itself. I found that by drawing a graph after I turned the function $f(x)=\frac{1}{x}+\frac{1}{x-1}$. since function is 1-1 and onto between this domain and range I know this function must have an inverse. But i could not go further after the following point.
$f(x)=y=\frac{1}{x}+\frac{1}{x}-1$ and I exchange the $y$ and $x$ and get $x=\frac{1}{y}+\frac{1}{y}-1$ and after that point I know I should solve it with respect to $y$ but I could not any help will be appreciated.
thanks

If
$y = \frac{2x-1}{x(x-1)}
$,
$yx(x-1) = 2x-1$
or
$yx^2-yx-(2x-1) = 0$
or
$yx^2-(y+2)x+1 = 0
$.
Applying the quadratic formula,
the discriminant
$D^2
=(y+2)^2-4y
=y^2+4y+4-4y
=y^2+4
$
so the roots are
$x
= \frac{y+2\pm D}{2y}
= \frac{y+2\pm \sqrt{y^2+4}}{2y}
$.
