Question

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

Answer 1

`x inRR,x!=-1`
`y inRR,y!=1`

Explanation:

`g(x)" is defined for all real values of x except the value"`
`"that makes the denominator equal to zero"`

`"equating the denominator to zero and solving gives the "`
`"value that x cannot be"`

`"solve " x+1=0rArrx=-1larrcolor(red)" excluded value"`

`rArr"domain is " x inRR,x!=-1`

`"to find any excluded values in the range, rearrange y = g(x)"`
`"making x the subject"`

`rArry(x+1)=x-3`

`rArrxy+y=x-3`

`rArrxy-x=-3-y`

`rArrx(y-1)=-(3+y)`

`rArrx=-(3+y)/(y-1)`

`"the denominator cannot equal zero"`

`"solve " y-1=0rArry=1larrcolor(red)" excluded value"`

`rArr"range is " y inRR,y!=1`