# What is the domain and range of f (x) = (x^2-2)/(x^2-4)?

Apr 19, 2017

Domain and range of this function

#### Explanation:

Domain:
$x < - 2$ or $- 2 < x < 2$ or $x > 2$

Range:
$f \left(x\right) \le \frac{1}{2}$ or $f \left(x\right) > 1$

Apr 19, 2017

$x \in \mathbb{R} , x \ne \pm 2$
$y \in \mathbb{R} , y \ne 1$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be.

$\text{solve } {x}^{2} - 4 = 0 \Rightarrow {x}^{2} = 4$

$\Rightarrow x = \pm 2 \leftarrow \textcolor{red}{\text{ excluded values}}$

$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne \pm 2$

To find the value that y cannot be find ${\lim}_{x \to \pm \infty} f \left(x\right)$

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

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

as $x \to \pm \infty , f \left(x\right) \to \frac{1 - 0}{1 - 0}$

$\Rightarrow y \to 1 \leftarrow \textcolor{red}{\text{ excluded value}}$

$\Rightarrow \text{range is } y \in \mathbb{R} , y \ne 1$

The graph of f(x) illustrates this.
graph{(x^2-2)/(x^2-4) [-10, 10, -5, 5]}