How do you find the domain of y=4/(2(x-2)^2-1 )?

Mar 18, 2017

The domain is ${D}_{y} = \mathbb{R} - \left\{2 - \frac{1}{\sqrt{2}} , 2 + \frac{1}{\sqrt{2}}\right\}$

Explanation:

$y = \frac{4}{2 {\left(x - 2\right)}^{2} - 1}$

As we cannot divide by $0$,

$2 {\left(x - 2\right)}^{2} - 1 \ne 0$

$2 {\left(x - 2\right)}^{2} \ne 1$

${\left(x - 2\right)}^{2} \ne \frac{1}{2}$

$\left(x - 2\right) \ne \pm \frac{1}{\sqrt{2}}$

$x \ne 2 \pm \frac{1}{\sqrt{2}}$

The domain of $y$ is ${D}_{y} = \mathbb{R} - \left\{2 - \frac{1}{\sqrt{2}} , 2 + \frac{1}{\sqrt{2}}\right\}$

Mar 18, 2017

$x \in \mathbb{R} - \left\{2 - \frac{1}{\sqrt{2}} , 2 + \frac{1}{\sqrt{2}}\right\}$

Explanation:

$y = \frac{4}{2 {\left(x - 2\right)}^{2} - 1}$ is defined for all values of $x$ for which

$\textcolor{w h i t e}{\text{XXX}} 2 {\left(x - 2\right)}^{2} - 1 \ne 0$

That is for all values except
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 2\right)}^{2} = \frac{1}{2}$

$\textcolor{w h i t e}{\text{XXX}} x - 2 = \pm \frac{1}{\sqrt{2}}$

$\textcolor{w h i t e}{\text{XXX}} x = 2 \pm \frac{1}{\sqrt{2}}$