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

2 Answers
Jul 5, 2018

Answer:

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]}

Jul 5, 2018

Answer:

#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]}