To find the domain of $h(x) = (f \times g)(x)$, we need to determine the domains of $f(x)$ and $g(x)$ individually, then find the intersection of those domains because the product $h(x) = f(x) \cdot g(x)$ will only be valid where both functions are defined.

Step 1: Domain of $f(x) = \sqrt{x}$

For $f(x)$, the square root function is only defined when $x \geq 0$ because the square root of a negative number is not a real number. Therefore, the domain of $f(x)$ is: $D_f = [0, \infty)$

Step 2: Domain of $g(x) = x^2 - 64$

The function $g(x) = x^2 - 64$ is a quadratic function, and quadratic functions are defined for all real values of $x$. Thus, the domain of $g(x)$ is: $D_g = (-\infty, \infty)$

Step 3: Domain of $h(x) = f(x) \cdot g(x)$

The function $h(x) = f(x) \cdot g(x)$ will only be defined where both $f(x)$ and $g(x)$ are defined. Since $f(x)$ restricts the domain to $x \geq 0$, the domain of $h(x)$ is the intersection of $D_f$ and $D_g$, which is: $D_h = [0, \infty)$

Final Answer (in interval notation):

The domain of $h(x)$ is: $[0, \infty)$

Do you want more details or have any questions? Here are five related questions:

1. How would the domain change if $f(x) = \sqrt{x-1}$ instead?
2. What is the range of $g(x) = x^2 - 64$?
3. How would you graph $h(x) = f(x) \cdot g(x)$?
4. How does the product of two functions affect the domain of the resulting function?
5. What happens to the domain if $g(x) = \frac{1}{x^2 - 64}$?

Tip: When combining functions, always check the individual domains to ensure the result is well-defined!