What is the domain of #h(x)=sqrt(x-2)#?

1 Answer
Jun 17, 2018

Answer:

#x in [2, infty)#

Explanation:

For radical functions, we cannot have a quantity less than #0# inside the square root.

In this case, we know that #h(2) = 0#, but if #x# is decreased any more than this, the radical will be undefined.

So we know that #x = 2# is the minimum value of the domain. As we increase #x#, we have no issues as the radical always contains a positive number. So #x -> infty#.

So the domain would be all values of #x >= 2#, or

#x in [2, infty)#