Range of function $f(x)=\frac{x}{1-x^2}$ How can we find the range of 
$$f(x)=\frac{x}{1-x^2}?$$
From the quadratic formula, I found that the range is $\mathbb{R}\setminus\{0\}.$ However, the correct one is $\mathbb{R}.$
 A: Suppose $y$ is in $\mathbb{R}$, and we wonder if $y$ is in the range. So we are wondering if there are $x$-values that solve $$y=\frac{x}{1-x^2}$$ Of course, $x$ will not be $1$ or $-1$ or the right side is undefined. The equation is equivalent to $$yx^2+x-y=0$$ Note it is still the case that $x$ could not be $1$ or $-1$.  We are still wondering if there are solutions in $x$ for our fixed value of $y$. Assuming $y\neq0$, this is a quadratic equation in $x$, and then there are solutions if and only if the discriminant is $\geq0$. The discriminant is $$1+4y^2$$ which is always $\geq1$. So no matter what nonzero value $y$ is, there is an $x$ (two actually) with $f(x)=y$. Now what if $y=0$ and the equation is not quadratic? Then trivially observe that $x=0$ solves the first equation.
In the end, nothing in $\mathbb{R}$ is excluded from the range.
A: A subtle aspect of the quadratic formula, that the solutions of $ax^2 + bx +c=0$ are $\frac {-b \pm {b^2 - 4ac}}{2a}$ are that in order to BE a quadratic, it is assumed that $a \ne 0$.
If $a$ does equal $0$ then what you have is $bx + c=0$ and that is not a quadratic; it's a linear equation and its root is $-\frac cb$.... (That is if it is assumed that $b \ne 0$.)
(To carry this idea to an extreme if $a=0$ and $b=0$ then you have $c=0$ and that's simply a statment that is true for all reals if $c =0$ and not true for any reals if $c \ne 0$.  So if $c \ne 0$ there are no "solutions".  If $c=0$ then all $x \in \mathbb R$ are "solutions" because $0=0$ for all $x$ [which have nothing to do the statement].  ... But I digress...)
So to solve for $\frac x{1-x^2} = y$ or $yx^2 + x - y = 0$ then solutions are:
$x = \frac {-1 \pm {1 + 4y^2}}{2y}$ if $y \ne 0$.  And that has  solutions for all $y\ne 0$.  
AND
if $y = 0$ then the equation is $0x^2 + x -0 = 0$ or $x = 0$ and that has solution... $x = 0$ if $y = 0$.  And that is a solution for $y=0$.
So solutions exist for all $y$.
A: Write it as:
$$f(x)=\frac x{1-x^2}=\frac12\left[\frac1{1-x}-\frac1{1+x}\right]=\frac12(g(x)-h(x)).$$
The function $g(x)$ is decreasing and ranges $(-\infty,0)\cup (0,\infty)$, $h(x)$ is increasing and ranges $(-\infty,0)\cup (0,\infty)$ and $f(0)=0$. Hence, $f(x)$ ranges $(-\infty,\infty)$. 
