Let's break down the tasks from the two images you provided.

Given Functions:

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

The tasks:

For parts (a) to (d), we are asked to find combinations of these functions and their domains.

Part (e) to (h):

We need to compute the following:

• (e) Find $(f + g)(4)$.
• (f) Find $(f - g)(2)$.
• (g) Find $(f \cdot g)(3)$.
• (h) Find $\left( \frac{f}{g} \right)(1)$.

Step-by-Step Calculations:

(e) $(f + g)(4)$:

$(f + g)(x) = f(x) + g(x) = (2x + 3) + (9x - 2) = 11x + 1$ Substituting $x = 4$: $(f + g)(4) = 11(4) + 1 = 44 + 1 = 45$

(f) $(f - g)(2)$:

$(f - g)(x) = f(x) - g(x) = (2x + 3) - (9x - 2) = 2x + 3 - 9x + 2 = -7x + 5$ Substituting $x = 2$: $(f - g)(2) = -7(2) + 5 = -14 + 5 = -9$

(g) $(f \cdot g)(3)$:

$(f \cdot g)(x) = f(x) \cdot g(x) = (2x + 3)(9x - 2)$ Expanding: $= 18x^2 - 4x + 27x - 6 = 18x^2 + 23x - 6$ Substituting $x = 3$: $(f \cdot g)(3) = 18(3)^2 + 23(3) - 6 = 18(9) + 69 - 6 = 162 + 69 - 6 = 225$

(h) $\left( \frac{f}{g} \right)(1)$:

$\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{2x + 3}{9x - 2}$ Substituting $x = 1$: $\left( \frac{f}{g} \right)(1) = \frac{2(1) + 3}{9(1) - 2} = \frac{2 + 3}{9 - 2} = \frac{5}{7}$

Domains:

• Both $f(x) = 2x + 3$ and $g(x) = 9x - 2$ are linear functions, meaning their domains are all real numbers.
• For $\left( \frac{f}{g} \right)(x)$, the domain excludes values where $g(x) = 0$. Solving $9x - 2 = 0$, we get $x = \frac{2}{9}$. Thus, the domain for $\frac{f}{g}(x)$ is all real numbers except $x = \frac{2}{9}$.

Would you like more details or clarifications?

Here are 5 related questions to expand on this topic:

1. How do you find the domain for more complex rational functions?
2. What happens when we combine different types of functions (like quadratic and rational)?
3. Can you explain the process of finding limits for these combined functions?
4. How do you determine if a function is increasing or decreasing?
5. Can the product of two linear functions ever result in a non-quadratic expression?

Tip: Always check for undefined points when working with rational functions to avoid domain issues!