Range of $ \frac{x}{(x-2)(x+1)} $ Find Range of $$ y =\frac{x}{(x-2)(x+1)} $$
Why is the range all real numbers ? 
the denominator cannot be $0$ Hence isn't range suppose to be $y$ not equals to $0$ ?
 A: The range is $\mathbb R$ because for every $y\in\mathbb R$, there exists some $x$ such that $$\frac{x}{(x-2)(x+1)}=y.$$
For example, for $y=0$, you have $$\frac{0}{(0-2)(0+1)}=\frac{0}{(-2)\cdot 1} = -\frac{0}{2}=0.$$

For a general $y$, you have to show that the equation above has a solution, which you can try to do by multiplying it by $(x-2)(x+1)$ to get
$$x=yx^2 - yx - 2y$$
This can be further simplified to a quadratic equation, and it's fairly easy to see if a quadratic equation has a solution 
(Hint: it has something to do with the discriminant).
A: Alternatively: the function can be expressed as the sum:
$$y =\frac{x}{(x-2)(x+1)}=\frac13\left(\frac{2}{x-2}+\frac{1}{x+1}\right)=\frac13\left(g(x)+h(x)\right)$$
The functions $g(x)$ and $h(x)$ are hyperbolas with the range $g\ne0$ and $h\ne0$. And the function $y$ has the range in $\mathbb{R}$, in particular, $y(0)=0$.
A: I think this range is more properly $\bar{\mathbb{R}}$ which is the Extended Real Number line ( or the Affinely Extended Real Numbers if you prefer ) and not $\mathbb{R}$.
$\mathbb{R}$ does not include $\pm\infty$ and at the two poles ( $x=-1$ and $x=2$ ) this function can be said to take on these "values".
$\bar{\mathbb{R}}$ does include these values.
