**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#