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.