How do you find the domain and range of 1/(3t+12)13t+12?
1 Answer
Explanation:
"let " y=1/(3t+12)let y=13t+12
"the denominator cannot be zero as this would make y"the denominator cannot be zero as this would make y
"undefined. Equating the denominator to zero and solving"undefined. Equating the denominator to zero and solving
"gives the value that t cannot be"gives the value that t cannot be
"solve "3t+12=0rArrt=-4larrcolor(red)"excluded value"solve 3t+12=0⇒t=−4←excluded value
rArr"domain "(-oo,-4)uu(-4,oo)⇒domain (−∞,−4)∪(−4,∞)
"rearrange making t the subject"rearrange making t the subject
y(3t+12)=1y(3t+12)=1
rArr3ty+12y=1⇒3ty+12y=1
rArr3ty=1-12y⇒3ty=1−12y
rArrt=(1-12y)/(3y)⇒t=1−12y3y
"solve "3y=0rArry=0larrcolor(red)"excluded value"solve 3y=0⇒y=0←excluded value
rArr"range "(-oo,0)uu(0,oo)⇒range (−∞,0)∪(0,∞)
graph{1/(3x+12) [-10, 10, -5, 5]}
