How do you find the domain and range of $\sqrt{3 t + 12}$ $sqrt(3t+12)$ ?

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Answer 1 · 2016-05-01T21:12:42

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

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

The square root $\sqrt{expression}$ $sqrt("expression")$ is only defined if $expression \geq 0$ $"expression" >= 0$
The notation $\sqrt{expression}$ $sqrt("expression")$ denotes the principal, non-negative square root so is always $\geq 0$ $>= 0$

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

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

$t \geq - 4$ $t >= -4$

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

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

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

So the range includes $[0, \infty)$ $[0, oo)$

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

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

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