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

algebra-precalculus inverse-function

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 $f(x)=y$.

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