# What is the domain and range of f(x) = 5/(x-9)?

Jan 13, 2016

DOMAIN: $x \in \left(- \infty , 9\right) \cup \left(9 , + \infty\right)$

RANGE: $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

#### Explanation:

$y = f \left(x\right) = \frac{k}{g} \left(x\right)$

Existence Condition is:

$g \left(x\right) \ne 0$

$\therefore x - 9 \ne 0$

$\therefore x \ne 9$

Then:

$F . E .$= Field of Existence=Domain: $x \in \left(- \infty , 9\right) \cup \left(9 , + \infty\right)$

$x = 9$ could be a vertical asymptote

To find the range we have to study the behavior for:

• $x \rightarrow \pm \infty$

${\lim}_{x \rightarrow - \infty} f \left(x\right) = {\lim}_{x \rightarrow - \infty} \frac{5}{x - 9} = \frac{5}{-} \infty = {0}^{-}$

${\lim}_{x \rightarrow + \infty} f \left(x\right) = {\lim}_{x \rightarrow + \infty} \frac{5}{x - 9} = \frac{5}{+ \infty} = {0}^{+}$

Then

$y = 0$ is a horizontal asymptote.

Indeed,

$f \left(x\right) \ne 0 \forall x \in F . E .$

• $x \rightarrow {9}^{\pm}$

${\lim}_{x \rightarrow {9}^{-}} f \left(x\right) = {\lim}_{x \rightarrow {9}^{-}} \frac{5}{x - 9} = \frac{5}{0} ^ \left(-\right) = - \infty$

${\lim}_{x \rightarrow {9}^{+}} f \left(x\right) = {\lim}_{x \rightarrow {9}^{+}} \frac{5}{x - 9} = \frac{5}{0} ^ \left(+\right) = + \infty$

Then

$x = 9$ it's a vertical asympote

$\therefore$ Range of $f \left(x\right)$: $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$