To find the domain and range of the function $h(x) = \sqrt{x - 3}$, we need to consider the properties of the square root function and the conditions under which the expression inside the square root is defined.

Domain:

The expression inside the square root, $x - 3$, must be non-negative because the square root of a negative number is not defined in the set of real numbers.

$x - 3 \geq 0$ $x \geq 3$

So, the domain of $h(x)$ is all real numbers $x$ such that $x \geq 3$.

Domain in Interval Notation:

$[3, \infty)$

Range:

For the square root function $\sqrt{x - 3}$, the output is always non-negative because the square root of any non-negative number is also non-negative.

• When $x = 3$, $h(x) = \sqrt{3 - 3} = 0$.
• As $x$ increases, $h(x)$ also increases because the square root function is increasing.

Therefore, the range of $h(x)$ is all non-negative real numbers.

Range in Interval Notation:

$[0, \infty)$

Summary:

• Domain: $[3, \infty)$
• Range: $[0, \infty)$

Would you like more details or have any questions? Here are some related questions:

1. How do you determine the domain of a function involving a square root?
2. What are the steps to find the range of a function?
3. How would the domain and range change if the function were $h(x) = \sqrt{3 - x}$?
4. Can you explain why the square root function only outputs non-negative values?
5. How do you graph the function $h(x) = \sqrt{x - 3}$ and verify the domain and range?

Tip: When finding the domain of functions involving square roots, always set the expression inside the square root greater than or equal to zero and solve for the variable.