Domain and Range of a rational function

Question

given the rational function $\frac{1}{x^{2} - \frac{x}{2}-3 }$ and asked for the domain and range, I multiplied thru by 2 and got $\frac{2}{2x^{2} - x-6 }$ I understand the domain includes all real numbers except for (vertical asymptotes) $\frac{-3}{2}$ and 2. i thought there were no horizontal asymptotes until I graphed it on a calculator. according to the calculator the range is $(-\infty, \frac{-16}{49}] \cup (0,\infty)$. this makes sense to me looking at the graph but how can I obtain these results for the range using algebra? thank you

Answer 1

First, we need to find the roots of the denominator and the range of the denominator.

We denote the denominator as $f$.

$f=x^2-\frac12 x-3=0$ has roots $x_1=2$ and $x_2=-\frac32$. So the domain is the real line excluding the two roots.

Next we need to find the range.

$-\frac {49}{16}<x^2-\frac12 x-3$.

$-\frac {49}{16}<f$.

To find the range of the rational function is to find the range of $\frac 1f$.

I'll leave it to you.

Answer 2

You may use a following property of a continuous function:

If a function $f(x) : D(f(x)) \rightarrow R(f(x))$ is continuous and monotonic on interval $(a,b) \subseteq D(f(x))$, then it takes all values from interval $(f(a), f(b)) \subseteq R(f(x))$

Here you have four intervals of monotony: $D_1 = (- \infty, -\frac32 - 0)$, $D_2 = (-\frac32 + 0, \frac14)$, $D_3 = [\frac14, 2 - 0)$ and $D_4 = (2 + 0, + \infty)$. $R_1 = (0, + \infty)$, $R_2 = (- \infty, -\frac{16}{49})$, $R_3 = [-\frac{16}{49}, -\infty)$, $R_4 = (+ \infty, 0)$. I wrote some intervals with left limit greater that right one for simpler correspondence between $R_i(f(x))$ and $D_i(f(x))$.
$R(f(x)) = \cup_iR_i(f(x)) = (- \infty, - \frac{16}{49}] \cup (0, + \infty)$.