Question

What is the domain and range of `y= 1/(x+1)`?

Answer 1

The domain is `x in (-oo,-1) uu(-1,+oo)`. The range is `y in (-oo,0)uu(0,+oo)`

Explanation:

The function is

`y=1/(x+1)`

As the denominator must be `!=0`

Therefore,

`x+1!=0`

`=>`, `x!=-1`

The domain is `x in (-oo,-1) uu(-1,+oo)`

To calculate the range, proceed as follows :

`y=1/(x+1)`

Cross multiply

`y(x+1)=1`

`yx+y=1`

`yx=1-y`

`x=(1-y)/(y)`

As the denominator must be `!=0`

`y!=0`

The range is `y in (-oo,0)uu(0,+oo)`

graph{1/(x+1) [-16.02, 16.02, -8.01, 8.01]}

Answer 2

`x in(-oo,-1)uu(-1,oo)`
`y in(-oo,0)uu(0,oo)`

Explanation:

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"`

`"domain is "x in(-oo,-1)uu(-1,oo)`

`"to find range, rearrange making x the subject"`

`y(x+1)=1`

`xy+y=1`

`xy=1-y`

`x=(1-y)/y`

`y=0larrcolor(red)"excluded value"`

`"range is "y in(-oo,0)uu(0,oo)`
graph{1/(x+1) [-10, 10, -5, 5]}