How do you find the domain and range of $Y = g(x) = (x-3)/(x+1)$ ?

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Answer 1 · 2018-04-02T11:54:16

The domain is $x in RR-{-1}$ . The range is $y in RR-{1}$

The denominator must $!=0$

Therefore,

$x+1!=0$

$x!=-1$

The domain of $g(x)$ is $x in RR-{-1}$

To find the range, proceed as follows

$y=(x-3)/(x+1)$

$yx+y=x-3$

$yx-x=-3-y$

$x(y-1)=-(3+y)$

$x=-(3+y)/(y-1)$

The denominator is $!=0$

$y-1!=0$

$y!=1$

The range of $g(x)$ is $y in RR-{1}$

graph{(x-3)/(x+1) [-22.8, 22.83, -11.4, 11.4]}

How do you find the domain and range of Y = g(x) = (x-3)/(x+1)Y = g(x) = (x-3)/(x+1)?