Question

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

Answer 1

The domain is `x in RR-{-3}`. The range is `RR-{3}`

Explanation:

Let

`y=1/(x+3)+3=(1+3(x+3))/(x+3)=(1+3x+9)/(x+3)`

`=(3x+10)/(x+3)`

As you cannot divide by `0`, the

`"denominator "!= 0`

`x+3!=0`

`x!=-3`

Therefore,

The domain is `x in RR-{-3}`

To find the range, proceed as follows

`y=(3x+10)/(x+3)`

`y(x+3)=3x+10`

`yx+3y=3x+10`

`yx-3x=10-3y`

`x(y-3)=(10-3y)`

`x=(10-3y)/(y-3)`

AS you cannot divide by `0`,

`y-3!=0`

`y!=3`

The range is `RR-{3}`

graph{(3x+10)/(x+3) [-18.67, 17.36, -5.59, 12.43]}