How do I find the inverse of $f(x) = \frac{x}{x^2 + 1}$

Question

$f(x) = \frac{x}{x^2 + 1}$. How do you find $f^{-1}$?

Is try and solve for y and I get the following:

$y = \frac{x}{x^2 + 1}$

$y(x^2 + 1) = x$

$yx^2 + y = x$

$y = -yx^2 + x$

$y = x(-yx + 1)$

It seems whatever I do I can't isolate the x variable, I just go in circles...

Answer 1

Solving for $x$ the equation $$ yx^2-x+y=0 $$ we find $$ x=\frac{1\pm \sqrt{1-4y^2}}{2y} $$

this means that the range of the function $f(x)=\frac{x}{x^2+1}$ is the interval $(-1/2,1/2)$ and the function is not invertible because for any value of $y$ in this interval we have two corresponding values of $x$ such that $f(x)=y$.

Answer 2

You have easily (solving a quadratic equation) for all $a$ such that $0\lt a\lt \frac 12$, the equality $$f\left(\frac{1+\sqrt{1-4a^2}}{2a}\right)=f\left(\frac{1-\sqrt{1-4a^2}}{2a}\right)=a$$ This shows that $f$ is not injective so not bijective and has no inverse.