What is the range of the function #g(x) = (x-3)/(x+1)#?

1 Answer
May 30, 2017

Answer:

#x inRR,x!=-1#
#y inRR,y!=1#

Explanation:

#g(x)" is defined for all real values of x except the value"#
#"that makes the denominator equal to zero"#

#"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 values in the range, rearrange y = g(x)"#
#"making x the subject"#

#rArry(x+1)=x-3#

#rArrxy+y=x-3#

#rArrxy-x=-3-y#

#rArrx(y-1)=-(3+y)#

#rArrx=-(3+y)/(y-1)#

#"the denominator cannot equal zero"#

#"solve " y-1=0rArry=1larrcolor(red)" excluded value"#

#rArr"range is " y inRR,y!=1#