What is the domain and range of #y= (4x^2 - 9) / ((2x+3)(x+1))#?

1 Answer
May 5, 2018

Answer:

See below.

Explanation:

Notice:

#4x^2-9# is the difference of two squares. This can be expressed as:

#4x^2-9=(2x+3)(2x-3)#

Substituting this in numerator:

#((2x+3)(2x-3))/((2x+3)(x+1))#

Canceling like factors:

#(cancel((2x+3))(2x-3))/(cancel((2x+3))(x+1))=(2x-3)/(x+1)#

We notice that for #x=-1# the denominator is zero. This is undefined, so our domain will be all real numbers #bbx# #x!=-1#

We can express this in set notation as:

#{x in RR | x != -1}#

or in interval notation:

#(-oo , -1)uu(-1,oo)#

To find the range:

We know the function is undefined for #x=-1#, therefore the line #x=-1# is a vertical asymptote. The function will go to #+-oo# at this line.

We now see what happens as #x ->+-oo#

Divide #(2x-3)/(x+1)# by #x#

#((2x)/x-3/x)/(x/x+1/x)=(2-3/x)/(1+1/x)#

as: #x->+-oo# # \ \ \ \ \ (2-3/x)/(1+1/x)=(2-0)/(1+0)=2#

This shows the line #y=2# is a horizontal asymptote. The function can't therefore ever equal 2.

so the range can be expressed as:

#{y in RR| y !=2}#

or

#(-oo,2)uu(2 , oo)#

This can be seen from the graph of the function:

graph{(2x-3)/(x+1) [-32.48, 32.44, -16.23, 16.25]}