Determining range of a function I was trying to determine the domain and range of a function. The function is: $$y = \frac{x}{x^2 + 1}$$
I determined the domain which is $\mathbb{R}$. In this equation, when the value of $x$ is 0, the value of $y$ is $0$.  Then, I tried to determine the range by this equation: 
$$x = \frac{1 \pm \sqrt{1 - 4y^2}}{2y}$$ 
I derived the equation from previous equation using quadratic equation formula. In this method we can see, the range is $\{-0.5 \leq y < 0\}$ and $\{0 < y \leq 0.5\}$. Zero cannot be a range because the denominator cannot be zero. 
But if $0$ is not in the range, then we do not find any value for $x$ when $x$ is $0$ which is a contradiction of the definition of a function. Where is the fault?
 A: A reason you get a contradiction is that you got too far ahead before looking to see where you were going. Indeed if we want to solve for $x$ in
$$y = \frac{x}{x^2 + 1},$$
the first steps could be
$$y (x^2 + 1) = x,$$
$$y x^2 + - x + y = 0.$$
Now if $y \neq 0$ this is a quadratic equation that can be solved by the quadratic formula, but if $y = 0$ then it is a linear equation that can be solved much more simply:
$$ -x = 0, $$
$$ x = 0.$$
Indeed,  you should simply take note of the obvious fact that $0$ is in the range of the function (since $y= 0$ when $x = 0$) and look for additional possible non-zero values of $y,$
that is, to look for any possible ways that $y \neq 0.$
Your quadratic formula then finds the rest of the range for you.
A: An idea what to do with this kind of questions: assume $\;a\in\Bbb R\;$ belongs to the function's range, then there exists $\;x\in\Bbb R\;$ (the function's domain) s.t.
$$a=f(x)=\frac x{x^2+1}\implies ax^2-x+a=0\implies \Delta=1-4a^2\ge0\implies |a|\le\frac12$$
and thus you get that it must be that $\;-\frac12\le a\le\frac12\;$ , and the range of the function is waaaay different from what you thought.
