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

1 Answer
Mar 7, 2018

Answer:

The domain is #x in RR-{-3}#. The range is #RR-{3}#

Explanation:

Let

#y=1/(x+3)+3=(1+3(x+3))/(x+3)=(1+3x+9)/(x+3)#

#=(3x+10)/(x+3)#

As you cannot divide by #0#, the

#"denominator "!= 0#

#x+3!=0#

#x!=-3#

Therefore,

The domain is #x in RR-{-3}#

To find the range, proceed as follows

#y=(3x+10)/(x+3)#

#y(x+3)=3x+10#

#yx+3y=3x+10#

#yx-3x=10-3y#

#x(y-3)=(10-3y)#

#x=(10-3y)/(y-3)#

AS you cannot divide by #0#,

#y-3!=0#

#y!=3#

The range is #RR-{3}#

graph{(3x+10)/(x+3) [-18.67, 17.36, -5.59, 12.43]}