How do you find the domain and range of # 1/(3t+12)#?

1 Answer
May 22, 2018

Answer:

#"domain "(-oo,-4)uu(-4,oo)#
#"range "(-oo,0)uu(0,oo)#

Explanation:

#"let " y=1/(3t+12)#

#"the denominator cannot be zero as this would make y"#
#"undefined. Equating the denominator to zero and solving"#
#"gives the value that t cannot be"#

#"solve "3t+12=0rArrt=-4larrcolor(red)"excluded value"#

#rArr"domain "(-oo,-4)uu(-4,oo)#

#"rearrange making t the subject"#

#y(3t+12)=1#

#rArr3ty+12y=1#

#rArr3ty=1-12y#

#rArrt=(1-12y)/(3y)#

#"solve "3y=0rArry=0larrcolor(red)"excluded value"#

#rArr"range "(-oo,0)uu(0,oo)#
graph{1/(3x+12) [-10, 10, -5, 5]}