Question

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

Answer 1

`x inRR,x!=-1`

`y inRR,y!=3`

Explanation:

`"let " y=1/(x+1)+3`

`"expressing y as a single rational function"`

`y=1/(x+1)+(3(x+1))/(x+1)=(3x+4)/(x+1)`

The denominator of y cannot be zero as this would make y undefined. 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 value in the range, rearrange the function making x the subject.

`y(x+1)=3x+4larr" cross-multiplying"`

`xy+y=3x+4`

`rArrxy-3x=4-y`

`rArrx(y-3)=4-y`

`rArrx=(4-y)/(y-3)`

`"The denominator cannot be zero."`

`"solve "y-3=0rArry=3larrcolor(red)" excluded value"`

`"range is " y inRR,y!=3`