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

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Answer 1 · 2016-05-01T21:12:42

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

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

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

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

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

$t >= -4$

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

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

$sqrt(3t+12) = y$

So the range includes $[0, oo)$

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

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

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