What is the domain and range of # f(x) = (x+3) /( x^2 + 8x + 15)#?

2 Answers
Jul 15, 2018

Answer:

The domain is #x in (-oo, -5)uu(-5,+oo)#. The range is #y in (-oo,0)uu(0,+oo)#

Explanation:

The function is

#f(x)=(x+3)/(x^2+8x+15)=(x+3)/((x+3)(x+5))=1/(x+5)#

The denominator must be #!=0#

Therefore,

#x+5!=0#

#x!=-5#

The domain is #x in (-oo, -5)uu(-5,+oo)#

To calculate the range, let

#y=(1)/(x+5)#

#y(x+5)=1#

#yx+5y=1#

#yx=1-5y#

#x=(1-5y)/y#

The denominator must be #!=0#

#y!=0#

The range is #y in (-oo,0)uu(0,+oo)#

graph{1/(x+5) [-16.14, 9.17, -6.22, 6.44]}

Jul 15, 2018

Answer:

Domain: #x inRR, x!=-5#

Range: #y inRR, y!=0#

Explanation:

We can factor the denominator as #(x+3)(x+5)#, since #3+5=8#, and #3*5=15#. This leaves us with

#(x+3)/((x+3)(x+5))#

We can cancel out common factors to get

#cancel(x+3)/(cancel(x+3)(x+5))=>1/(x+5)#

The only value that will make our function undefined is if the denominator is zero. We can set it equal to zero to get

#x+5=0=>x=-5#

Therefore, we can say the domain is

#x inRR, x!=-5#

To think about our range, let's go back to our original function

#(x+3)/((x+3)(x+5))#

Let's think about the horizontal asymptote. Since we have a higher degree on the bottom, we know we have a HA at #y=0#. We can show this graphically:

graph{(x+3)/((x+3)(x+8)) [-17.87, 2.13, -4.76, 5.24]}

Notice, our graph never touches the #x#-axis, which is consistent with having a horizontal asymptote at #y=0#.

We can say our range is

#y inRR, y!=0#

Hope this helps!