How do you find the domain and range of  1/(3t+12)?

May 22, 2018

$\text{domain } \left(- \infty , - 4\right) \cup \left(- 4 , \infty\right)$
$\text{range } \left(- \infty , 0\right) \cup \left(0 , \infty\right)$

Explanation:

$\text{let } y = \frac{1}{3 t + 12}$

$\text{the denominator cannot be zero as this would make y}$
$\text{undefined. Equating the denominator to zero and solving}$
$\text{gives the value that t cannot be}$

$\text{solve "3t+12=0rArrt=-4larrcolor(red)"excluded value}$

$\Rightarrow \text{domain } \left(- \infty , - 4\right) \cup \left(- 4 , \infty\right)$

$\text{rearrange making t the subject}$

$y \left(3 t + 12\right) = 1$

$\Rightarrow 3 t y + 12 y = 1$

$\Rightarrow 3 t y = 1 - 12 y$

$\Rightarrow t = \frac{1 - 12 y}{3 y}$

$\text{solve "3y=0rArry=0larrcolor(red)"excluded value}$

$\Rightarrow \text{range } \left(- \infty , 0\right) \cup \left(0 , \infty\right)$
graph{1/(3x+12) [-10, 10, -5, 5]}