# What is the domain and range of f(x)=(x^2-9)/(x^2-25)?

Jun 28, 2017

#### Answer:

$x \in \mathbb{R} , x \ne \pm 5$
$y \in \mathbb{R} , y \ne 1$

#### 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.

$\text{solve } {x}^{2} - 25 = 0 \Rightarrow \left(x - 5\right) \left(x + 5\right) = 0$

$\Rightarrow x = \pm 5 \leftarrow \textcolor{red}{\text{ are excluded values}}$

$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne \pm 5$

$\text{to find any excluded value in the range we can use the}$
$\text{horizontal asymptote}$

$\text{horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

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

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} - \frac{9}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{25}{x} ^ 2} = \frac{1 - \frac{9}{x} ^ 2}{1 - \frac{25}{x} ^ 2}$

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

$\Rightarrow y = 1 \text{ is the asymptote and thus excluded value}$

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