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

May 1, 2016

Domain: $\left[- 4 , \infty\right)$
Range: $\left[0 , \infty\right)$

#### Explanation:

Assuming we are talking about Real square roots, we have:

• The square root $\sqrt{\text{expression}}$ is only defined if $\text{expression} \ge 0$
• The notation $\sqrt{\text{expression}}$ denotes the principal, non-negative square root so is always $\ge 0$

So we require $3 t + 12 \ge 0$ for $t$ to be in the domain.

Dividing through by $3$, then subtracting $4$ from both sides this becomes:

$t \ge - 4$

So the domain is $t \in \left[- 4 , \infty\right)$

If $y \in \left[0 , \infty\right)$, then if we let $t = \frac{{y}^{2} - 12}{3}$, we find:

$\sqrt{3 t + 12} = y$

So the range includes $\left[0 , \infty\right)$

Since sqrt is always non-negative, there are no negative values in the range.

So $\left[0 , \infty\right)$ is the whole of the range.