# What is the domain and range of f(x) = (x+9)/(x-3)?

May 30, 2018

Domain: $\setminus m a t h \boldsymbol{R} \setminus \setminus \left\{3\right\}$
Range: $\setminus m a t h \boldsymbol{R}$

#### Explanation:

Domain

The domain of a function is the set of points in which the function is defined. With numeric function, as you probably know, some operations are not allowed - namely division by $0$, logarithms of non-positive numbers and even roots of negative numbers.

In your case, you have no logarithms nor roots, so you only have to worry about the denominator. When imposing $x - 3 \setminus \ne 0$, you'll find the solution $x \setminus \ne 3$. So, the domain is the set of all real numbers, except $3$, which you can write as $\setminus m a t h \boldsymbol{R} \setminus \setminus \left\{3\right\}$ or in the interval form $\left(- \setminus \infty , 3\right) \setminus \cup \left(3 , \setminus \infty\right)$

Range

The range is an interval whose extrema are the lowest and highest possible values reached by the function. In this case, we already notices that our function has a point of non-definition, which leads to a vertical asymptote. When approaching vertical asymptotes, functions diverge towards $- \infty$ or $\infty$. Let's study what happens around $x = 3$: if we consider the left limit we have

${\lim}_{x \setminus \to {3}^{-}} \setminus \frac{x + 9}{x - 3} = \setminus \frac{12}{{0}^{-}} = - \setminus \infty$

In fact, if $x$ approaches $3$, but is still less than $3$, $x - 3$ will be slightly less than zero (think, for example, at $x$ assuming values like $2.9 , 2.99 , 2.999 , \ldots$

By the same logic,

${\lim}_{x \setminus \to {3}^{+}} \setminus \frac{x + 9}{x - 3} = \setminus \frac{12}{{0}^{+}} = \setminus \infty$

Since the function approaches both $- \infty$ and $\infty$, the range is $\left(- \setminus \infty , \infty\right)$, which of course is equivalent to the whole real numbers set $\setminus m a t h \boldsymbol{R}$.

May 30, 2018

$x \in \left(- \infty , 3\right) \cup \left(3 , \infty\right)$
$y \in \left(- \infty , 1\right) \cup \left(1 , \infty\right)$

#### Explanation:

The denominator of f)x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.

$\text{solve "x-3=0rArrx=3larrcolor(red)"excluded value}$

$\text{domain } x \in \left(- \infty , 3\right) \cup \left(3 , \infty\right)$

$\text{let } y = \frac{x + 9}{x - 3}$

$\text{rearrange making x the subject}$

$y \left(x - 3\right) = x + 9$

$x y - 3 y = x + 9$

$x y - x = 9 + 3 y$

$x \left(y - 1\right) = 9 + 3 y$

$x = \frac{9 + 3 y}{y - 1}$

$\text{solve "y-1=0rArry=1larrcolor(red)"excluded value}$

$\text{range } y \in \left(- \infty , 1\right) \cup \left(1 , \infty\right)$
graph{(x+9)/(x-3) [-10, 10, -5, 5]}