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

1 Answer
May 24, 2017

Answer:

#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#