Range of the function $f(x) = \frac{2x}{x^{2}+1}$ Plotting the function $f(x) = \frac{2x}{x^{2}+1}$ we can see the range is $[-1,1]$.
Now I was told to do $f(x)=y$ and isolate $x$ doing this I have, using the quadratic formula $x=\frac{1\pm \sqrt{1-y^2}}{y}$ then $$range f(x)= domain \frac{1\pm \sqrt{1-y^2}}{y}$$.
But this domain does not include $0$ and $0=f(0)$ so $0$ must be in the range, what am I missing?

$y = \frac {2x}{x^2+1}\\
yx^2 - 2x + y = 0$
And then you use the quadratic formula to find values of $y,$ such that $x$ can be a real root of the polynomial.
All good.
What happens at $y = 0$?
$yx^2 - 2x + y = 0$ ceases to be a quadratic and the quadratic formula is not applicable.
