How do you find the domain and range of f(t) = 3sqrt(t + 4)?

Jun 21, 2016

Assuming we are restricted to Real values:
Domain: $t \in \left[- 4 , + \infty\right)$
Range:$f \left(t\right) \in \left[0 , + \infty\right)$

Explanation:

$3 \sqrt{t + 4}$ is only defined (in Real values) if $t + 4 \ge 0$
$\Rightarrow t \ge - 4$
however, it is defined for all $t \ge - 4$;
therefore the Domain is $t \ge - 4 \mathmr{and} t \in \left[- 4 , + \infty\right)$

$\sqrt{t + 4} \ge 0$ (by definition of the square root function)
It is equal to $0$ when $t = - 4$
As $t \rightarrow + \infty$
$\textcolor{w h i t e}{\text{XXX}} \sqrt{t + 4} \rightarrow + \infty$ (and so does $3 \sqrt{t + 4}$)
Therefore the Range is $f \left(t\right) > 0 \mathmr{and} f \left(t\right) \in \left[0 , + \infty\right)$