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

2 Answers
Narad T.
Jul 5, 2018

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]}

Jim G.
Jul 5, 2018

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]}

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