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]}
