Find the range of the following function Find the range of the following function:
$$f(x)=\frac{x-1}{x^2-5x-6}$$
I know how to find the domain only
Please provide an explanation.
EDIT: Another question: find the range of
$$f(x)=\frac{1}{(1-x)(5-x)}$$
How would we do that?
 A: Suppose $\;r\in\Bbb R\;$ , so we want to check when
$$\frac{x-1}{x^2-5x-6}=r\iff rx^2-(5r+1)x+1-6r=0$$
The above quadratic's discriminant is :
$$\Delta=25r^2+10r+1+24r^2-4r=49r^2+6r+1$$
and certainly $\;49r^2+6r+1>0\;,\;\;\forall\;r\in\Bbb R\;$ (why? Look again at this new quadratic's discriminant)  and from here that the quadratic's a solution for any $\;r\in\Bbb R\;$ , and thus the function is onto (or what's the same: the range is $\;\Bbb R\;$ )
A: the range of $f(x) = \dfrac{x-1}{(x+1)(x-6)}$ is $(-\infty, \infty).$ here is the reason. show that $\lim_{x \to -1+}f(x) = \infty$ and $\lim_{x \to 6-} f(x)= -\infty$ and $f$ is continuous in $(-1, 6).$  these three conditions imply the conclusion that the range of $f$ is $(-\infty, \infty).$
A: From partial fractions decomposition,
$$\frac{x-1}{x^2-5x-6}=\frac{x-1}{(x+1)(x-6)} =\frac{2}{7(x+1)}+\frac{5}{7(x-6)}.$$
Both fractions are decreasing continuous functions on the domain: $(-\infty,-1)\cup (-1,6)\cup (6,+\infty)$. 
In particular, it is continuous decreasing on $(-1,6)$. On another hand, $$\lim_{x\to -1_+}f(x)=+\infty, \quad\lim_{x\to 6_-}f(x)=-\infty, $$
Hence, by the intermediate value theorem, the range of $f$ is $(-\infty,+\infty)$
A: If we factor, we have $\dfrac{x-1}{(x+1)(x-6)}$, and since there are no holes in the graph of the function. There is a horizontal asymptote at $y=0$, but this function has a zero at $x=1$, and thus the value $y=0$ is reached. Therefore the function has range $(-\infty,\infty)$.
A: 
Pure Algebraic Way of Doing This Without any Graphic Calculator

Let y = $\frac{x-1}{x^2-5x-6}$
Using Cross Multiplication
x$^{2}$y -x$\left(5y+1\right)$-6y+1=0$\Longrightarrow$x=$\frac{\left(5y+1\right)\pm\sqrt{\left\{ 5y+1\right\} ^{2}-4y\left(1-6y\right)}}{2y}$
$\Longrightarrow$$\left\{ 5y+1\right\} ^{2}-4y\left(1-6y\right)$>
0
=$\left\{ 5y+1\right\} ^{2}-4y\left(1-6y\right)$= $\left(7y+\frac{3}{7}\right)^{2}+\frac{40}{49}$>0
which is always true $\forall$ y $\in.$$\mathbb{R}$

Hence the Range is= $\mathbb{R}$

