To determine the domain of the function $h(x) = \sqrt{6 - x^2}$, we need to consider the conditions under which the expression inside the square root is non-negative, because the square root of a negative number is not a real number.

The expression inside the square root is $6 - x^2$. Therefore, we require:

$6 - x^2 \geq 0$

This inequality can be rewritten as:

$x^2 \leq 6$

Taking the square root of both sides:

$-\sqrt{6} \leq x \leq \sqrt{6}$

So, the domain of the function $h(x) = \sqrt{6 - x^2}$ is:

$x \in [-\sqrt{6}, \sqrt{6}]$

This means that $h(x)$ is defined for all $x$ values in the interval $[- \sqrt{6}, \sqrt{6}]$.

Would you like more details or have any questions?

Here are 5 related questions:

1. How would the domain change if the function was $h(x) = \sqrt{9 - x^2}$?
2. What is the range of the function $h(x) = \sqrt{6 - x^2}$?
3. How do you find the domain of a function that includes a fraction?
4. What would be the domain if the function were $h(x) = \sqrt{x^2 - 6}$ instead?
5. Can the function $h(x) = \sqrt{6 - x^2}$ have negative values?

Tip: When finding the domain of a function involving a square root, always ensure the expression inside the square root is greater than or equal to zero.