How do you find the domain and range of $h(x) = sqrt(x-3)$ ?

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Answer 1 · 2016-07-23T18:14:26

Domain is $[3,oo)$ .

Range is $[0,oo)$ .

In $RR$ , negative nos. do not have square-root.

So, $(x-3)>=0, i.e., x>=3$ . Hence, Domain is $[3,oo)$ .

Also, square-root means $+ve$ square-root, so, $h(x)>=0$

Hence, the Range is $[0,oo)$ .

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