Let's analyze the problem provided in the image.

Problem Statement

We are given two functions:

• $f(x) = x^2$
• $g(x) = \sqrt{16 - x} - 2$

We are asked to find the composite function $\left(\frac{f}{g}\right)(x)$ and state its domain in interval notation.

Solution

Step 1: Determine the Composite Function $\left(\frac{f}{g}\right)(x)$

The composite function $\left(\frac{f}{g}\right)(x)$ is defined as: $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$ Substituting the given functions $f(x)$ and $g(x)$: $\left(\frac{f}{g}\right)(x) = \frac{x^2}{\sqrt{16 - x} - 2}$

Step 2: Determine the Domain

The domain of the function $\left(\frac{f}{g}\right)(x)$ is determined by the following conditions:

1. The denominator $g(x)$ must not be zero: $g(x) = \sqrt{16 - x} - 2 \neq 0$ Solving $\sqrt{16 - x} = 2$: $\sqrt{16 - x} = 2 \implies 16 - x = 4 \implies x = 12$ Therefore, $x \neq 12$.

2. The expression under the square root in $g(x)$ must be non-negative: $16 - x \geq 0 \implies x \leq 16$ This gives $x$ must be in the interval $(-\infty, 16]$.

Combining these two conditions, the domain is: $(-\infty, 12) \cup (12, 16]$

Final Answer

The composite function $\left(\frac{f}{g}\right)(x) = \frac{x^2}{\sqrt{16 - x} - 2}$ has the domain $(-\infty, 12) \cup (12, 16]$.

Would you like more details or have any questions about this solution?

Here are five related questions to explore further:

1. How would the domain change if $f(x)$ was $x^3$ instead of $x^2$?
2. Can you find the domain of $g(x) = \sqrt{16 - x} - 2$ alone?
3. How does the behavior of $\left(\frac{f}{g}\right)(x)$ change near $x = 12$?
4. What would happen to the domain if the square root in $g(x)$ was replaced with a cube root?
5. How do you determine the domain of a function involving logarithms?

Tip: Always check the conditions under which the denominator of a fraction and the expression under a square root are valid to ensure the correct domain is determined.