Let's first solve the given problem.

We are given:

• $f(x) = x - 9$
• $g(x) = 4x^2$

We need to find:

1. $f + g$
2. $f - g$
3. $f \cdot g$
4. $\frac{f}{g}$

Then, we determine the domain for each function.

1. $f + g$

$f(x) + g(x) = (x - 9) + 4x^2 = 4x^2 + x - 9$

Domain:

Both functions $f(x)$ and $g(x)$ are defined for all real values of $x$, so the domain of $f + g$ is $(-\infty, \infty)$.

2. $f - g$

$f(x) - g(x) = (x - 9) - 4x^2 = -4x^2 + x - 9$

Domain:

Similarly, both functions are defined for all real values of $x$, so the domain of $f - g$ is also $(-\infty, \infty)$.

3. $f \cdot g$

$f(x) \cdot g(x) = (x - 9) \cdot 4x^2 = 4x^2(x - 9) = 4x^3 - 36x^2$

Domain:

Again, both functions are defined for all real values of $x$, so the domain of $f \cdot g$ is $(-\infty, \infty)$.

4. $\frac{f}{g}$

$\frac{f(x)}{g(x)} = \frac{x - 9}{4x^2}$

Domain:

For $\frac{f}{g}$, we need to exclude any $x$ where $g(x) = 0$. Solving $g(x) = 4x^2 = 0$, we get $x = 0$. Thus, the domain of $\frac{f}{g}$ is $(-\infty, 0) \cup (0, \infty)$.

Summary of Domains:

• $f + g$: $(-\infty, \infty)$
• $f - g$: $(-\infty, \infty)$
• $f \cdot g$: $(-\infty, \infty)$
• $\frac{f}{g}$: $(-\infty, 0) \cup (0, \infty)$

Would you like further details or clarification on any part of this?

Here are five related questions for deeper understanding:

1. What are the general steps to find the domain of a rational function?
2. How would the process change if the denominator involved more complex expressions, like square roots?
3. Could you apply the same operations to functions with trigonometric terms?
4. How does the degree of the polynomial in $f(x) \cdot g(x)$ affect the graph of the function?
5. What happens to the domain if we take the square root of $f(x)$ or $g(x)$?

Tip: Always check for values that make the denominator zero or result in undefined expressions when determining the domain of a function.