To Find the domain
Equate the denominator(#x^2-2x-3#) to zero, then solve the equation for #x#
#rarr x^2-2x-3=0#
#rarr x=(-(-2)+-sqrt((-2)^2-4*(1)*(-3)))/(2*1)#
#rarr x= 1+-2#
#=> x= -1# and # x=3#
This means that, when #x=-1# or #3#, we have the # x^2-2x-3=0#
Implying that #f(x)=color(red)(x/0)# which is undefined.
Hence, the domain is all real numbers except #-1#
and #3#
Also written as #D_f=(-oo,-1)uu(-1,3)uu(3,+oo)#
To Find the Range
Step 1
say #f(x)=y# and rearrange the function as a quadratic equation
#y=x/(x^2-2x-3)#
#rarr y(x^2-2x-3)=x #
#rarr yx^2-2yx-3y-x= 0#
#rarr yx^2+(-2y-1)x-3y= 0#
Step 2
We know from the quadratic formula,
#x=(-b+-sqrt(b^2-4ac))/(2a)#
that the solutions of #x# are real when #b^2-4ac>=0#
So likewise, we say, #(-2y-1)^2-4*(y)*(-3y)>=0#
#color(white)" "color(red)b##color(white)" "color(red)a##color(white)" "color(red)c#
Step 3
We solve the inequality for the values set of values of #y#
#=>4y^2+4y+1+12y^2>=0#
#rarr16y^2+4y+1>=0#
#rarr16[y^2+1/4y+1/16]>=0#
#rarr16[(y+1/8)^2-1/64+1/16]>=0#
#rarr16[(y+1/8)^2+3/64]>=0#
Notice that for all values of #y# the left hand side of the inequality be greater than (but not equal) to zero.
We then conclude that, #y# can take all real values.
#y in RR <=> f(x) in RR#
So the Range is #RR#
