# How do you find the domain and range of f(x) = 1/(x + 3)?

Aug 26, 2017

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

#### Explanation:

$\text{given } y = \frac{1}{x + 3}$

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

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

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

$\text{to find any excluded values in the range rearrange}$
$\text{making x the subject}$

$\Rightarrow y \left(x + 3\right) = 1 \leftarrow \textcolor{b l u e}{\text{ cross-multiplying}}$

$\Rightarrow x y + 3 y = 1$

$\Rightarrow x y = 1 - 3 y$

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

$\text{the denominator cannot be zero}$

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

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