Given the functions $f(x) = \sqrt{x}$ and $g(x) = \sqrt{x^2 - 16}$, we need to find $\left( \frac{f}{g} \right)(x)$ and determine its domain.

Step 1: Define $\left( \frac{f}{g} \right)(x)$

The function $\left( \frac{f}{g} \right)(x)$ is defined as: $\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{\sqrt{x^2 - 16}}$ This expression is valid as long as both the numerator and denominator are defined and the denominator is not zero.

Step 2: Determine the domain

Domain restrictions:

1. For $f(x) = \sqrt{x}$:

   • The square root function $\sqrt{x}$ requires $x \geq 0$, because the square root of a negative number is not defined in the real number system.
   • So, $x \geq 0$.

2. For $g(x) = \sqrt{x^2 - 16}$:

   • The square root function $\sqrt{x^2 - 16}$ requires $x^2 - 16 \geq 0$, because we cannot take the square root of a negative number.
   • Solving $x^2 - 16 \geq 0$: $x^2 \geq 16$ $x \geq 4 \quad \text{or} \quad x \leq -4$ So, $x \in (-\infty, -4] \cup [4, \infty)$.

3. Denominator cannot be zero:

   • $g(x) = \sqrt{x^2 - 16}$ must not be zero, so we require $x^2 - 16 \neq 0$, or $x^2 \neq 16$, which means $x \neq \pm 4$.

Combine restrictions:

• From $f(x)$, we know $x \geq 0$.
• From $g(x)$, we have $x \in (-\infty, -4] \cup [4, \infty)$, but combining this with the restriction $x \geq 0$ gives $x \in [4, \infty)$.
• Finally, we must exclude $x = 4$, because $g(4) = 0$ would make the denominator zero.

Domain of $\left( \frac{f}{g} \right)(x)$:

The domain is $(4, \infty)$.

Final Answer:

$\left( \frac{f}{g} \right)(x) = \frac{\sqrt{x}}{\sqrt{x^2 - 16}}, \quad \text{with domain } (4, \infty)$

Would you like more details or have any questions?

Here are five related questions:

1. What happens if the function $f(x)$ is replaced by a cubic root function?
2. How would the domain change if $g(x) = \sqrt{x^2 - 25}$?
3. Can we simplify $\left( \frac{f}{g} \right)(x)$ any further?
4. What is the behavior of $\left( \frac{f}{g} \right)(x)$ as $x \to 4^+$?
5. How does the function behave as $x \to \infty$?

Tip: Always check both the numerator and denominator for restrictions when finding the domain of a quotient function.