How do you find the domain and range of # 1/(3t+12)#?
1 Answer
May 22, 2018
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]}