Handwritten solutions to function operations and compositions problems on a math worksheet.

Problem Analysis and Solution

The image contains a series of function operations and compositions. Let's break down each part systematically.

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#### Part 1: Function Operations for \( f(x) = 4x + 8 \) and \( g(x) = 2x - 12 \)

We are tasked with performing the following operations:

1. Addition: \( (f + g)(x) \)
2. Subtraction: \( (f - g)(x) \)
3. Multiplication: \( (f \cdot g)(x) \)
4. Division: \( \left( \frac{f}{g} \right)(x) \)

##### Step 1: Addition
\[
(f + g)(x) = f(x) + g(x)
\]
\[
= (4x + 8) + (2x - 12)
\]
\[
= 4x + 2x + 8 - 12
\]
\[
= 6x - 4
\]

##### Step 2: Subtraction
\[
(f - g)(x) = f(x) - g(x)
\]
\[
= (4x + 8) - (2x - 12)
\]
\[
= 4x + 8 - 2x + 12
\]
\[
= 2x + 20
\]

##### Step 3: Multiplication
\[
(f \cdot g)(x) = f(x) \cdot g(x)
\]
\[
= (4x + 8)(2x - 12)
\]
Using the distributive property (FOIL method):
\[
= 4x \cdot 2x + 4x \cdot (-12) + 8 \cdot 2x + 8 \cdot (-12)
\]
\[
= 8x^2 - 48x + 16x - 96
\]
\[
= 8x^2 - 32x - 96
\]

##### Step 4: Division
\[
\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)}
\]
\[
= \frac{4x + 8}{2x - 12}
\]
Factor both the numerator and the denominator:
\[
= \frac{4(x + 2)}{2(x - 6)}
\]
Simplify by canceling the common factor of 2:
\[
= \frac{2(x + 2)}{x - 6}
\]
Restriction: The denominator cannot be zero, so \( x \neq 6 \).

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#### Part 2: Function Operations for \( f(x) = x - 2 \) and \( g(x) = x^2 \)

We are tasked with finding the following values:

1. \( (f + g)(-2) \)
2. \( (f - g)(-2) \)
3. \( (f \cdot g)(-2) \)
4. \( \left( \frac{f}{g} \right)(-2) \)

##### Step 1: Evaluate \( f(-2) \) and \( g(-2) \)
\[
f(x) = x - 2 \quad \Rightarrow \quad f(-2) = -2 - 2 = -4
\]
\[
g(x) = x^2 \quad \Rightarrow \quad g(-2) = (-2)^2 = 4
\]

##### Step 2: Addition
\[
(f + g)(-2) = f(-2) + g(-2)
\]
\[
= -4 + 4
\]
\[
= 0
\]

##### Step 3: Subtraction
\[
(f - g)(-2) = f(-2) - g(-2)
\]
\[
= -4 - 4
\]
\[
= -8
\]

##### Step 4: Multiplication
\[
(f \cdot g)(-2) = f(-2) \cdot g(-2)
\]
\[
= -4 \cdot 4
\]
\[
= -16
\]

##### Step 5: Division
\[
\left( \frac{f}{g} \right)(-2) = \frac{f(-2)}{g(-2)}
\]
\[
= \frac{-4}{4}
\]
\[
= -1
\]

---

#### Part 3: Function Operations for \( f(x) = 3x \) and \( g(x) = x + 17 \)

We are tasked with finding the following values:

1. \( f(g(2)) \)
2. \( f(g(-3)) \)
3. \( g(f(4)) \)

##### Step 1: Evaluate \( g(2) \) and \( g(-3) \)
\[
g(x) = x + 17 \quad \Rightarrow \quad g(2) = 2 + 17 = 19
\]
\[
g(x) = x + 17 \quad \Rightarrow \quad g(-3) = -3 + 17 = 14
\]

##### Step 2: Evaluate \( f(g(2)) \) and \( f(g(-3)) \)
\[
f(x) = 3x \quad \Rightarrow \quad f(g(2)) = f(19) = 3 \cdot 19 = 57
\]
\[
f(x) = 3x \quad \Rightarrow \quad f(g(-3)) = f(14) = 3 \cdot 14 = 42
\]

##### Step 3: Evaluate \( g(f(4)) \)
\[
f(x) = 3x \quad \Rightarrow \quad f(4) = 3 \cdot 4 = 12
\]
\[
g(x) = x + 17 \quad \Rightarrow \quad g(f(4)) = g(12) = 12 + 17 = 29
\]

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#### Part 4: Function Operations for \( f(x) = 2x^2 - 3 \) and \( g(x) = \frac{x + 1}{x} \)

We are tasked with finding the following values:

1. \( f(g(2)) \)
2. \( g(f(2)) \)
3. \( f(g(x)) \)

##### Step 1: Evaluate \( g(2) \)
\[
g(x) = \frac{x + 1}{x} \quad \Rightarrow \quad g(2) = \frac{2 + 1}{2} = \frac{3}{2}
\]

##### Step 2: Evaluate \( f(g(2)) \)
\[
f(x) = 2x^2 - 3 \quad \Rightarrow \quad f(g(2)) = f\left( \frac{3}{2} \right)
\]
\[
= 2\left( \frac{3}{2} \right)^2 - 3
\]
\[
= 2 \cdot \frac{9}{4} - 3
\]
\[
= \frac{18}{4} - 3
\]
\[
= \frac{18}{4} - \frac{12}{4}
\]
\[
= \frac{6}{4}
\]
\[
= \frac{3}{2}
\]

##### Step 3: Evaluate \( f(2) \)
\[
f(x) = 2x^2 - 3 \quad \Rightarrow \quad f(2) = 2(2)^2 - 3
\]
\[
= 2 \cdot 4 - 3
\]
\[
= 8 - 3
\]
\[
= 5
\]

##### Step 4: Evaluate \( g(f(2)) \)
\[
g(x) = \frac{x + 1}{x} \quad \Rightarrow \quad g(f(2)) = g(5)
\]
\[
= \frac{5 + 1}{5}
\]
\[
= \frac{6}{5}
\]

##### Step 5: Find \( f(g(x)) \)
\[
g(x) = \frac{x + 1}{x}
\]
\[
f(x) = 2x^2 - 3 \quad \Rightarrow \quad f(g(x)) = f\left( \frac{x + 1}{x} \right)
\]
\[
= 2\left( \frac{x + 1}{x} \right)^2 - 3
\]
\[
= 2 \cdot \frac{(x + 1)^2}{x^2} - 3
\]
\[
= \frac{2(x + 1)^2}{x^2} - 3
\]
\[
= \frac{2(x^2 + 2x + 1)}{x^2} - 3
\]
\[
= \frac{2x^2 + 4x + 2}{x^2} - 3
\]
\[
= \frac{2x^2}{x^2} + \frac{4x}{x^2} + \frac{2}{x^2} - 3
\]
\[
= 2 + \frac{4}{x} + \frac{2}{x^2} - 3
\]
\[
= -1 + \frac{4}{x} + \frac{2}{x^2}
\]

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Final Answers

1. Part 1:
- \( (f + g)(x) = 6x - 4 \)
- \( (f - g)(x) = 2x + 20 \)
- \( (f \cdot g)(x) = 8x^2 - 32x - 96 \)
- \( \left( \frac{f}{g} \right)(x) = \frac{2(x + 2)}{x - 6} \), \( x \neq 6 \)

2. Part 2:
- \( (f + g)(-2) = 0 \)
- \( (f - g)(-2) = -8 \)
- \( (f \cdot g)(-2) = -16 \)
- \( \left( \frac{f}{g} \right)(-2) = -1 \)

3. Part 3:
- \( f(g(2)) = 57 \)
- \( f(g(-3)) = 42 \)
- \( g(f(4)) = 29 \)

4. Part 4:
- \( f(g(2)) = \frac{3}{2} \)
- \( g(f(2)) = \frac{6}{5} \)
- \( f(g(x)) = -1 + \frac{4}{x} + \frac{2}{x^2} \)

\boxed{\text{See detailed solutions above}}