# For f(x) = 1/(x-3), what is the natural domain and range?

Mar 9, 2018

$x \in \mathbb{R} , x \ne 3$
$y \in \mathbb{R} , y \ne 0$

#### Explanation:

$\text{the denominator of "f(x)" cannot be zero as this}$
$\text{would make "f(x)" undefined. Equating the }$
$\text{denominator to zero and solving gives the value}$
$\text{that x cannot be}$

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

$\text{domain is } x \in \mathbb{R} , x \ne 3$

$\left(- \infty , 3\right) \cup \left(3 , + \infty\right) \leftarrow \textcolor{b l u e}{\text{in interval notation}}$

$\text{to find the range, rearrange making x the subject}$

$f \left(x\right) = y = \frac{1}{x - 3}$

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

$\Rightarrow x y - 3 y = 1$

$\Rightarrow x y = 1 + 3 y$

$\Rightarrow x = \frac{1 + 3 y}{y}$

$\Rightarrow y = 0 \leftarrow \textcolor{red}{\text{excluded value}}$

$\text{range is } y \in \mathbb{R} , y \ne 0$

$\left(- \infty , 0\right) \cup \left(0 , + \infty\right) \leftarrow \textcolor{b l u e}{\text{in interval notation}}$
graph{1/(x-3) [-10, 10, -5, 5]}