Given the functions $f(x) = \sqrt{x}$ and $g(x) = x^2 - 64$, we are tasked with finding the domain of $h(x) = (f \cdot g)(x)$, where $h(x) = f(x) \cdot g(x)$.

Step-by-Step Solution:

1. Determine the domain of $f(x)$: The function $f(x) = \sqrt{x}$ requires that the argument under the square root is non-negative. Therefore, the domain of $f(x)$ is: $x \geq 0$ In interval notation, the domain of $f(x)$ is: $[0, \infty)$

2. Determine the domain of $g(x)$: The function $g(x) = x^2 - 64$ is a polynomial, which is defined for all real numbers. So, the domain of $g(x)$ is: $(-\infty, \infty)$

3. Determine the domain of $h(x) = f(x) \cdot g(x)$: Since $h(x) = f(x) \cdot g(x)$, the domain of $h(x)$ is restricted by the domain of $f(x)$, because the product of two functions is only defined where both functions are defined. Therefore, the domain of $h(x)$ is determined by the domain of $f(x)$, which is $[0, \infty)$.

Conclusion:

The domain of $h(x) = f(x) \cdot g(x)$ is: $[0, \infty)$ This is because $f(x) = \sqrt{x}$ restricts the domain to non-negative values of $x$.

Final Answer (Interval Notation):

$[0, \infty)$

Would you like more details or have any questions?

Here are 5 relative questions for further understanding:

1. What is the range of the function $f(x) = \sqrt{x}$?
2. How would the domain change if $f(x) = \sqrt{x - 1}$?
3. How does multiplying functions affect their domains in general?
4. Can the function $g(x) = x^2 - 64$ have any roots? If so, where?
5. How does the domain of $f(x)$ influence composite functions?

Tip: Always check for restrictions like square roots or denominators when determining the domain of a function.