How do you find the domain and range of #y = 3 sqrt (x-2) #?

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Answer 1 · 2015-07-17T19:45:04

For #x-2# to have a real square root, we require #x-2 >= 0#, hence #x >= 2#.

Given #x >= 2# we find #y# can have any positive value.

So the domain is #[2, oo)# and range is #[0, oo)#

For #sqrt(x-2)# to have a value in #RR#, we require #x-2 >= 0#

Add #2# to both sides of this inequality to get:

#x >= 2#.

If #x >= 2#, then #sqrt(x-2) >= 0# is well defined and hence #y = 3sqrt(x-2)# is well defined.

So the domain is #[2, oo)#

#sqrt(x - 2) >= 0# so #y = 3sqrt(x-2) >= 0#

In fact, for any #y >= 0#, let #x = (y/3)^2+2#.

Then #3sqrt(x-2) = 3sqrt((y/3)^2) = 3(y/3) = y#

So the range is the whole of #[0, oo)#