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

Jul 23, 2016

Domain is $\left[3 , \infty\right)$.

Range is $\left[0 , \infty\right)$.

Explanation:

In $\mathbb{R}$, negative nos. do not have square-root.

So, $\left(x - 3\right) \ge 0 , i . e . , x \ge 3$. Hence, Domain is $\left[3 , \infty\right)$.

Also, square-root means $+ v e$ square-root, so, $h \left(x\right) \ge 0$

Hence, the Range is $\left[0 , \infty\right)$.