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We are given three functions:

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

The task is to check if the results of the following compositions are the same:

1. \( f \circ (h \circ g)(x) = (f \circ h) \circ g(x) \)
2. \( h \circ (f \circ g)(x) = (h \circ f) \circ g(x) \)
3. \( g \circ (f \circ h)(x) = (g \circ f) \circ h(x) \)

Let's break them down one by one:

### 1. \( f \circ (h \circ g)(x) = (f \circ h) \circ g(x) \)
- \( h \circ g(x) = h(g(x)) = \frac{1}{g(x) + 1} = \frac{1}{x^2 + 4 + 1} = \frac{1}{x^2 + 5} \)
- \( f \circ (h \circ g)(x) = f(h(g(x))) = f\left(\frac{1}{x^2 + 5}\right) = 2\left(\frac{1}{x^2 + 5}\right) + 1 = \frac{2}{x^2 + 5} + 1 \)

Now, for the right-hand side:
- \( f \circ h(x) = f(h(x)) = f\left(\frac{1}{x + 1}\right) = 2\left(\frac{1}{x + 1}\right) + 1 = \frac{2}{x + 1} + 1 \)
- \( (f \circ h) \circ g(x) = f(h(g(x))) = f\left(\frac{1}{x^2 + 5}\right) \)

So the results are the same for the first composition.

### 2. \( h \circ (f \circ g)(x) = (h \circ f) \circ g(x) \)
- \( f \circ g(x) = f(g(x)) = f(x^2 + 4) = 2(x^2 + 4) + 1 = 2x^2 + 8 + 1 = 2x^2 + 9 \)
- \( h \circ (f \circ g)(x) = h(f(g(x))) = \frac{1}{(2x^2 + 9) + 1} = \frac{1}{2x^2 + 10} \)

Now, for the right-hand side:
- \( h \circ f(x) = h(f(x)) = h(2x + 1) = \frac{1}{(2x + 1) + 1} = \frac{1}{2x + 2} = \frac{1}{2(x + 1)} \)
- \( (h \circ f) \circ g(x) = h(f(g(x))) = \frac{1}{2(x^2 + 4 + 1)} = \frac{1}{2(x^2 + 5)} \)

Thus, the results are not the same for the second composition.

### 3. \( g \circ (f \circ h)(x) = (g \circ f) \circ h(x) \)
- \( f \circ h(x) = f(h(x)) = f\left(\frac{1}{x + 1}\right) = 2\left(\frac{1}{x + 1}\right) + 1 = \frac{2}{x + 1} + 1 \)
- \( g \circ (f \circ h)(x) = g\left(\frac{2}{x + 1} + 1\right) = \left(\frac{2}{x + 1} + 1\right)^2 + 4 \)

Now, for the right-hand side:
- \( g \circ f(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2 + 4 = 4x^2 + 4x + 1 + 4 = 4x^2 + 4x + 5 \)
- \( (g \circ f) \circ h(x) = g(f(h(x))) = (4(h(x))^2 + 4h(x) + 5) \)

The results differ, so the third composition is also not equal.

### Conclusion:
- Composition 1: Equal
- Composition 2: Not equal
- Composition 3: Not equal

Do you want any more details or clarification?

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Here are five related questions to expand on this:
1. How do you find the inverse of a function like \( f(x) = 2x + 1 \)?
2. What are the conditions for two functions to commute under composition?
3. Can you describe how to simplify complex function compositions?
4. What role does domain and range play in function composition?
5. How do you verify if a function is bijective?

**Tip:** When composing functions, be mindful of the order as \( f(g(x)) \neq g(f(x)) \) in most cases.

Diketahui fungsi f(x) = 2x + 1, g(x) = x^2 + 4, dan h(x) = 1/(x+1). Cek apakah hasil dari komposisi fungsi berikut sama:
1. (f ∘ (h ∘ g))(x) = ((f ∘ h) ∘ g)(x)
2. (h ∘ (f ∘ g))(x) = ((h ∘ f) ∘ g)(x)
3. (g ∘ (f ∘ h))(x) = ((g ∘ f) ∘ h)(x)

Cek apakah hasil komposisi fungsi f(x) = 2x + 1, g(x) = x^2 + 4, dan h(x) = 1/(x+1) menghasilkan hasil yang sama. Analisis dan perhitungan untuk tiga komposisi fungsi berbeda. Cocok untuk siswa di kelas 10-12.

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