# How do you find the domain and range of sqrt((13x)/((x^2)-1)?

May 29, 2017

The domain is $x \in \left(- 1 , 0\right] \cup \left(+ 1 , + \infty\right)$
The range is $y \in \left[0 , + \infty\right)$

#### Explanation:

What's uner the sqrt sign is $\ge 0$

So,

$\frac{13 x}{{x}^{2} - 1} \ge 0$

$\frac{13 x}{\left(x + 1\right) \left(x - 1\right)} \ge 0$

Let $f \left(x\right) = \frac{13 x}{\left(x + 1\right) \left(x - 1\right)}$

We can build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a a}$$+ 1$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left(- 1 , 0\right] \cup \left(+ 1 , + \infty\right)$

The domain is $x \in \left(- 1 , 0\right] \cup \left(+ 1 , + \infty\right)$

Let,

$y = \sqrt{\frac{13 x}{{x}^{2} - 1}}$

When $x = - 1$, $\implies$, $y = + \infty$

When $x = 0$, $\implies$, $y = 0$

The range is $y \in \left[0 , + \infty\right)$