Question

determining the domain of a function

s=sqrt3t+12

Answer 1

can you show your work

Answer 2

The domain would be the values of t that we are able to use in the equation to have a real output for s.

I'm unsure if you mean
\[ \sqrt{3t} + 12 \]
or
\[ \sqrt{3t + 12} \]

Either way, we'll be taking the value under the square root to be greater-than or equal-to 0.

Answer 3

it is the second one you wrote.

Answer 4

\[\sqrt{3t+12}\]

Answer 5

I got that far and then I am not sure what to do next?

Answer 6

okay.
So, we'd take "3t + 12" to be greater-than or equal to 0. (If this expression is less-than 0, then we wouldn't have a real output)
3t + 12 >= 0
3t >= -12
t >= -4
-4 <= t < infinity

Interval notation: [-4, infinity)

Answer 7

thank you I have one more that is on this same line but a cube root.