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Let's solve each part of the problem by finding \((f \circ g)(x)\), \((g \circ f)(x)\), and \((f \circ g)(2)\) for all the given functions.

### 1. \( f(x) = 2x \) and \( g(x) = x + 5 \)

- **\( (f \circ g)(x) \)**: 
  \[
  f(g(x)) = f(x + 5) = 2(x + 5) = 2x + 10
  \]
- **\( (g \circ f)(x) \)**: 
  \[
  g(f(x)) = g(2x) = 2x + 5
  \]
- **\( (f \circ g)(2) \)**: 
  \[
  f(g(2)) = f(2 + 5) = f(7) = 2(7) = 14
  \]

---

### 2. \( f(x) = 4x \) and \( g(x) = x - 3 \)

- **\( (f \circ g)(x) \)**: 
  \[
  f(g(x)) = f(x - 3) = 4(x - 3) = 4x - 12
  \]
- **\( (g \circ f)(x) \)**: 
  \[
  g(f(x)) = g(4x) = 4x - 3
  \]
- **\( (f \circ g)(2) \)**: 
  \[
  f(g(2)) = f(2 - 3) = f(-1) = 4(-1) = -4
  \]

---

### 3. \( f(x) = 2x + 1 \) and \( g(x) = 3x - 2 \)

- **\( (f \circ g)(x) \)**: 
  \[
  f(g(x)) = f(3x - 2) = 2(3x - 2) + 1 = 6x - 4 + 1 = 6x - 3
  \]
- **\( (g \circ f)(x) \)**: 
  \[
  g(f(x)) = g(2x + 1) = 3(2x + 1) - 2 = 6x + 3 - 2 = 6x + 1
  \]
- **\( (f \circ g)(2) \)**: 
  \[
  f(g(2)) = f(3(2) - 2) = f(6 - 2) = f(4) = 2(4) + 1 = 9
  \]

---

### 4. \( f(x) = 5x - 3 \) and \( g(x) = x^2 - 3 \)

- **\( (f \circ g)(x) \)**: 
  \[
  f(g(x)) = f(x^2 - 3) = 5(x^2 - 3) - 3 = 5x^2 - 15 - 3 = 5x^2 - 18
  \]
- **\( (g \circ f)(x) \)**: 
  \[
  g(f(x)) = g(5x - 3) = (5x - 3)^2 - 3 = 25x^2 - 30x + 9 - 3 = 25x^2 - 30x + 6
  \]
- **\( (f \circ g)(2) \)**: 
  \[
  f(g(2)) = f(2^2 - 3) = f(4 - 3) = f(1) = 5(1) - 3 = 2
  \]

---

### 5. \( f(x) = x^2 + 3 \) and \( g(x) = x^2 - 1 \)

- **\( (f \circ g)(x) \)**: 
  \[
  f(g(x)) = f(x^2 - 1) = (x^2 - 1)^2 + 3 = x^4 - 2x^2 + 1 + 3 = x^4 - 2x^2 + 4
  \]
- **\( (g \circ f)(x) \)**: 
  \[
  g(f(x)) = g(x^2 + 3) = (x^2 + 3)^2 - 1 = x^4 + 6x^2 + 9 - 1 = x^4 + 6x^2 + 8
  \]
- **\( (f \circ g)(2) \)**: 
  \[
  f(g(2)) = f(2^2 - 1) = f(4 - 1) = f(3) = 3^2 + 3 = 9 + 3 = 12
  \]

---

Would you like further clarification on any of these calculations?

Here are five related questions you might find helpful:
1. How do you calculate compositions of functions in general?
2. What are the common mistakes to avoid when composing functions?
3. Can we derive the domain and range for \( f \circ g \) in these cases?
4. How would the results change if \( f(x) \) and \( g(x) \) were swapped?
5. What are some real-life applications of function compositions?

**Tip**: Always carefully substitute the functions when composing them to avoid errors in algebraic manipulation!

Find (f∘g)(x), (g∘f)(x), and (f∘g)(2) for the given pairs of functions.

Learn how to solve function composition problems, including finding (f∘g)(x), (g∘f)(x), and (f∘g)(2). Explore key algebraic techniques and detailed explanations for function pairs like f(x) = 2x and g(x) = x + 5.

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