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

2 Answers
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)#.

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!