What is the range of the function g(x) = (x-3)/(x+1)?

May 30, 2017

$x \in \mathbb{R} , x \ne - 1$
$y \in \mathbb{R} , y \ne 1$

Explanation:

$g \left(x\right) \text{ is defined for all real values of x except the value}$
$\text{that makes the denominator equal to zero}$

$\text{equating the denominator to zero and solving gives the }$
$\text{value that x cannot be}$

$\text{solve " x+1=0rArrx=-1larrcolor(red)" excluded value}$

$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne - 1$

$\text{to find any excluded values in the range, rearrange y = g(x)}$
$\text{making x the subject}$

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

$\Rightarrow x y + y = x - 3$

$\Rightarrow x y - x = - 3 - y$

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

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

$\text{the denominator cannot equal zero}$

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

$\Rightarrow \text{range is } y \in \mathbb{R} , y \ne 1$