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The image contains function composition problems. Let's solve each question step by step.

### First Problem
Given:
- \( f(x) = x^2 \)
- \( g(x) = \sqrt{x - 1} \)

**Questions:**
1. \( (f \circ g)(x) \)
2. \( (g \circ f)(x) \)

**Solutions:**

1. **\( (f \circ g)(x) \):**
   - Composition means applying \( g(x) \) first and then \( f(x) \).
   - \( (f \circ g)(x) = f(g(x)) \).
   - Substitute \( g(x) \): \( f(g(x)) = f(\sqrt{x - 1}) \).
   - Now apply \( f(x) = x^2 \): \( f(\sqrt{x - 1}) = (\sqrt{x - 1})^2 = x - 1 \).

   So, \( (f \circ g)(x) = x - 1 \).

2. **\( (g \circ f)(x) \):**
   - Composition means applying \( f(x) \) first and then \( g(x) \).
   - \( (g \circ f)(x) = g(f(x)) \).
   - Substitute \( f(x) \): \( g(f(x)) = g(x^2) \).
   - Now apply \( g(x) = \sqrt{x - 1} \): \( g(x^2) = \sqrt{x^2 - 1} \).

   So, \( (g \circ f)(x) = \sqrt{x^2 - 1} \).

### Second Problem
Given:
- \( f(x) = x^2 - 3x \)
- \( g(x) = 4 - \sqrt{x} + 2 \)
- \( h(x) = \frac{2x}{x - 1} \)

**Questions:**
1. \( (f \circ g)(x) \)
2. \( (g \circ f)(x) \)
3. \( (g \circ h)(x) \)
4. \( (h \circ (g \circ f))(x) \)

**Solutions:**

1. **\( (f \circ g)(x) \):**
   - \( (f \circ g)(x) = f(g(x)) \).
   - Substitute \( g(x) \): \( f(g(x)) = f(4 - \sqrt{x} + 2) = f(6 - \sqrt{x}) \).
   - Now apply \( f(x) = x^2 - 3x \): 
     \[
     f(6 - \sqrt{x}) = (6 - \sqrt{x})^2 - 3(6 - \sqrt{x}).
     \]
   - Expand and simplify:
     \[
     (6 - \sqrt{x})^2 = 36 - 12\sqrt{x} + x,
     \]
     and
     \[
     -3(6 - \sqrt{x}) = -18 + 3\sqrt{x}.
     \]
     So,
     \[
     f(6 - \sqrt{x}) = 36 - 12\sqrt{x} + x - 18 + 3\sqrt{x} = x - 9\sqrt{x} + 18.
     \]

   So, \( (f \circ g)(x) = x - 9\sqrt{x} + 18 \).

2. **\( (g \circ f)(x) \):**
   - \( (g \circ f)(x) = g(f(x)) \).
   - Substitute \( f(x) \): \( g(f(x)) = g(x^2 - 3x) \).
   - Now apply \( g(x) = 4 - \sqrt{x} + 2 \):
     \[
     g(x^2 - 3x) = 4 - \sqrt{x^2 - 3x} + 2 = 6 - \sqrt{x^2 - 3x}.
     \]

   So, \( (g \circ f)(x) = 6 - \sqrt{x^2 - 3x} \).

3. **\( (g \circ h)(x) \):**
   - \( (g \circ h)(x) = g(h(x)) \).
   - Substitute \( h(x) \): \( g(h(x)) = g\left(\frac{2x}{x - 1}\right) \).
   - Apply \( g(x) = 4 - \sqrt{x} + 2 \):
     \[
     g\left(\frac{2x}{x - 1}\right) = 6 - \sqrt{\frac{2x}{x - 1}}.
     \]

   So, \( (g \circ h)(x) = 6 - \sqrt{\frac{2x}{x - 1}} \).

4. **\( (h \circ (g \circ f))(x) \):**
   - First find \( (g \circ f)(x) \): \( 6 - \sqrt{x^2 - 3x} \).
   - Now, \( h((g \circ f)(x)) = h(6 - \sqrt{x^2 - 3x}) \).
   - Apply \( h(x) = \frac{2x}{x - 1} \):
     \[
     h(6 - \sqrt{x^2 - 3x}) = \frac{2(6 - \sqrt{x^2 - 3x})}{(6 - \sqrt{x^2 - 3x}) - 1}.
     \]

   So, \( (h \circ (g \circ f))(x) = \frac{2(6 - \sqrt{x^2 - 3x})}{(6 - \sqrt{x^2 - 3x}) - 1} \).

Would you like more details or have any questions?

### Related Questions
1. How do you simplify compositions of functions involving radicals?
2. What are the properties of function compositions?
3. How do you handle domain restrictions in function compositions?
4. How to graph the compositions of these functions?
5. How can compositions be used in real-world applications?

**Tip:** Always check the domain of each function and the composed function to ensure they are valid for all given values.

Misalkan f(x) = x^2 dan g(x) = √(x - 1). Tentukan komposisi fungsi (f ∘ g)(x) dan (g ∘ f)(x). Misalkan f(x) = x^2 - 3x, g(x) = 4 - √x + 2, dan h(x) = 2x/(x - 1). Tentukan komposisi fungsi (f ∘ g)(x), (g ∘ f)(x), (g ∘ h)(x), (h ∘ (g ∘ f))(x).

Learn how to solve function composition problems involving square roots and algebraic functions. This guide includes step-by-step solutions for (f ∘ g)(x), (g ∘ f)(x), (g ∘ h)(x), and (h ∘ (g ∘ f))(x), covering key algebraic concepts and properties of function compositions. Suitable for high school students in Grades 10-12.

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