Question

What is the domain of the function `(x-2)/sqrt(x^2-8x+12)` ?

Answer 1

`(-oo, 2) uu (6, oo)`

Explanation:

Given:

`(x-2)/sqrt(x^2-8x+12)`

This function is well defined when the radicand is positive.

We find:

`x^2-8x+12 = (x-2)(x-6)`

which is `0` when `x=2` or `x=6` and positive outside `[2, 6]`.

So the domain is `(-oo, 2) uu (6, oo)`.

Answer 2

The domain is `{x| x < 2 uu x > 6, x in RR}`.

Explanation:

We have a couple of conditions that need to be addressed:

•When will the value under the `√` be inferior to `0`?
•When will the denominator equal `0`?

For the function to be defined on `x`, the following inequality must hold true

`sqrt(x^2 - 8x + 12) ≥ 0`

Solve as an equation

`x^2 - 8x + 12 =0`

`(x - 6)(x - 2) =0`

`x= 6 or 2`

We now select test points.

Test point `1: x = 1`

`1^2 - 8(1) + 12 ≥ 0 color(green)(√)`

Therefore, the intervals that work are `(-oo, 2]` and `[6, oo)`. However, the problem here is that when the `√` becomes `0`, the entire function becomes undefined. Therefore, we must exclude the points `x= 2` and `x= 6` from the domain

Our domain becomes `{x| x < 2 uu x > 6, x in RR}`.

Hopefully this helps!