Problem Analysis:

The problem asks us to evaluate the following operations for two functions:

• $f(x) = 9x + 1$
• $g(x) = x^2 - 4$

We need to compute:

1. $(f + g)(x)$
2. $(f - g)(x)$
3. $(fg)(x)$
4. $\frac{f}{g}(x)$
5. The domain of $\frac{f}{g}$ in interval notation.

Step-by-Step Solutions:

(a) $(f + g)(x) = f(x) + g(x)$

$f(x) + g(x) = (9x + 1) + (x^2 - 4)$ $(f + g)(x) = x^2 + 9x - 3$

(b) $(f - g)(x) = f(x) - g(x)$

$f(x) - g(x) = (9x + 1) - (x^2 - 4)$ $(f - g)(x) = -x^2 + 9x + 5$

(c) $(fg)(x) = f(x) \cdot g(x)$

$f(x) \cdot g(x) = (9x + 1) \cdot (x^2 - 4)$ Using the distributive property: $(fg)(x) = (9x)(x^2 - 4) + (1)(x^2 - 4)$ $(fg)(x) = 9x^3 - 36x + x^2 - 4$ $(fg)(x) = 9x^3 + x^2 - 36x - 4$

(d) $\frac{f}{g}(x) = \frac{f(x)}{g(x)}$

$\frac{f(x)}{g(x)} = \frac{9x + 1}{x^2 - 4}$

Domain of $\frac{f}{g}$

The domain of $\frac{f}{g}$ excludes values of $x$ that make the denominator $g(x) = 0$: $g(x) = x^2 - 4 = 0$ $x^2 = 4 \quad \Rightarrow \quad x = \pm 2$

Thus, the domain is all real numbers except $x = \pm 2$. In interval notation: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$

Final Answers:

(a) $(f + g)(x) = x^2 + 9x - 3$
(b) $(f - g)(x) = -x^2 + 9x + 5$
(c) $(fg)(x) = 9x^3 + x^2 - 36x - 4$
(d) $\frac{f}{g}(x) = \frac{9x + 1}{x^2 - 4}$
Domain of $\frac{f}{g}$: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$.

Would you like further details or clarification? Let me know!
Here are 5 related questions to expand your understanding:

1. How do you determine the domain of a composite function?
2. What are the steps for simplifying rational functions?
3. Can you verify the results by substituting specific values for $x$?
4. How would the domain change if $f(x)$ or $g(x)$ had additional restrictions?
5. How do you graph $\frac{f}{g}(x)$ and its domain on a coordinate plane?

Tip: Always factorize the denominator to find excluded values for rational functions efficiently!