Inverse Function (and WolframAlpha gives different Result)

Question

I wanted to calculate the inverse function of $$ f(x) = \frac{1}{x} + \frac{1}{x-1} $$ Quite simple I thought, put $$ y = \frac{1}{x} + \frac{1}{x-1} = \frac{2x-1}{x(x-1)} $$ rearrange and solve $$ y(x(x-1)) - 2x + 1 = 0 $$ which give the quadratic equation $$ yx^2 - (y + 2)x + 1 = 0 $$ Using the Solution Formula we got $$ x = \frac{(y+2) \pm \sqrt{y^2+4}}{2y} $$ So the inverse function is $$ f^{-1}(x) = \frac{(x+2) \pm \sqrt{x^2+4}}{2x} $$ Just to confirm I put in WolframAlpha and it gives me $$ \frac{-x-2}{2x} \pm \frac{\sqrt{x^2+4}}{2x} $$ (just click on the link to start WolframAlpha with this parameter), which is different up to a sign in the first summand, can not see an error, do you (or is WolframAlpha wrong...)?

EDIT: If the link is not working for you:

Answer 1

Nothing wrong with your answer! Actually Wolfram's answer is wrong! Just check it by $x=3/2$ in wolfram's inverse.

Answer 2

Your error is in the solution formula. You have $(y+2)^2 - 4\cdot y\cdot 1 = y^2+4 \neq (y+2)(y-2)$. It would be $y^2-4 = (y+2)(y-2)$.

Answer 3

this function is not a one by one function , so that have not inverse .
but in some interval is 1-1 function so may have inverse