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

2 Answers
George C.
Apr 11, 2017

(-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).

Noah G
Apr 11, 2017

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!

Related questions
See all questions in Domain
Impact of this question
1909 views around the world
You can reuse this answer
Creative Commons License