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

Apr 11, 2017

$\left(- \infty , 2\right) \cup \left(6 , \infty\right)$

#### Explanation:

Given:

$\frac{x - 2}{\sqrt{{x}^{2} - 8 x + 12}}$

This function is well defined when the radicand is positive.

We find:

${x}^{2} - 8 x + 12 = \left(x - 2\right) \left(x - 6\right)$

which is $0$ when $x = 2$ or $x = 6$ and positive outside $\left[2 , 6\right]$.

So the domain is $\left(- \infty , 2\right) \cup \left(6 , \infty\right)$.

Apr 11, 2017

The domain is $\left\{x | x < 2 \cup x > 6 , x \in \mathbb{R}\right\}$.

#### 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

${x}^{2} - 8 x + 12 = 0$

$\left(x - 6\right) \left(x - 2\right) = 0$

$x = 6 \mathmr{and} 2$

We now select test points.

Test point $1 : x = 1$

1^2 - 8(1) + 12 ≥ 0 color(green)(√)

Therefore, the intervals that work are $\left(- \infty , 2\right]$ and $\left[6 , \infty\right)$. 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 $\left\{x | x < 2 \cup x > 6 , x \in \mathbb{R}\right\}$.

Hopefully this helps!