# How do you find the domain and range of t/sqrt(t^2-25) ?

Jan 22, 2018

$\text{see explanation}$

#### Explanation:

$f \left(t\right) = \frac{t}{\sqrt{{t}^{2} - 25}}$

$\text{the denominator cannot equal zero as this would make}$
$\text{f(t) undefined}$

$\Rightarrow {t}^{2} - 25 \ne 0 \Rightarrow t \ne \pm 5$

$\text{also } {t}^{2} - 25 > 0$

$\Rightarrow \left(t - 5\right) \left(t + 5\right) > 0$

$\Rightarrow t < - 5 \text{ or } t > 5$

$\Rightarrow \text{domain is } \left(- \infty , - 5\right) \cup \left(5 , + \infty\right)$

f(t)=t/(sqrt(t^2(1-25/t^2)))=t/(tsqrt(1-25/t^2)

$\textcolor{w h i t e}{f \left(t\right)} = {\cancel{t}}^{1} / \left({\cancel{t}}^{1} \sqrt{1 - \frac{25}{t} ^ 2}\right)$

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

$\Rightarrow y = - 1 , y = 1 \leftarrow \textcolor{b l u e}{\text{excluded values}}$

$\Rightarrow \text{range is } \left(- 1 , - \infty\right) \cup \left(1 , + \infty\right)$
graph{x/(sqrt(x^2-25)) [-10, 10, -5, 5]}