Problem in calculating inverse of function

Question

$f(x)=\frac{x}{1-x^2}$

I wanted to calculate its inverse.

I done following

$f(x)=y$ and using quadratic formula evaluate x in term of y.

i.e $x=\frac{-1+\sqrt{1-4y^2}}{2y}$ but solution is its reciprocal .I thought hard But I think I am making silly mistake. Please Help me out .

Any Help will be appreciated

Answer 1

No, you made no mistake: your solution is correct, assuming that the domain of $f$ is $(-1,1)$. Note, however, that\begin{align}\frac{-1+\sqrt{1-4y^2}}{2y}&=\frac{\left(-1+\sqrt{1-4y^2}\right)\left(-1-\sqrt{1-4y^2}\right)}{2y\left(-1-\sqrt{1-4y^2}\right)}\\&=\frac{4y^2}{2y\left(-1-\sqrt{1-4y^2}\right)}\\&=-\frac{2y}{1+\sqrt{1-4y^2}}.\end{align}

Answer 2

It should be $$x=\frac{y}{1-y^2}$$ or $$x-xy^2=y$$ or $$xy^2+y-x=0,$$ which gives $$y=\frac{-1+\sqrt{1+4x^2}}{2x}$$ or $$y=\frac{-1-\sqrt{1+4x^2}}{2x}.$$ Now, we need to choose the right answer.

On $(-\infty,-1)$ and on $(1,+\infty)$ the second function is valid,

but on $(-1,1)$ we get that $g(x)=\frac{-1+\sqrt{1+4x^2}}{2x}$ is an inverse function to $f$.

Answer 3

I am sorry but this function, as a whole, has no inverse because it is not bijective : see figure below (the curve has three branches in black).

You have, at first, to select a branch. Consider for example the central one, which corresponds to a function from $(-1,1)$ to $(-\infty,+\infty)$. The resulting function (graphical representation in red, symmetrical of the central branch with respect to 1st quadrant line bissector) has equation :

$$x=f(y)=\dfrac{-1+\sqrt{1+4y^2}}{2y} \tag{1}$$

(for $y \neq 0$) and $f(0)=0$.

which is one you have obtained.

But if you select another branch, for example the left one, which corresponds to a function with domain $(-\infty,-1)$ and range $(0,\infty)$ you would have another expression than (1) for the inverse function.