How do you find the domain and range of `f(x)=x/(x^2 - 2x -3)`?

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`