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We are tasked with solving the composition of three functions, \( (f \circ g \circ h)(x) \), where:

- \( f(x) = 2x - 2 \)
- \( g(x) = \sqrt{4 - x} \)
- \( h(x) = 4x - x^2 \)

To find \( (f \circ g \circ h)(x) \), we need to first calculate \( g(h(x)) \), then substitute that result into \( f(x) \).

### Step 1: Find \( g(h(x)) \)

Since \( h(x) = 4x - x^2 \), we substitute it into \( g(x) = \sqrt{4 - x} \):

\[
g(h(x)) = \sqrt{4 - (4x - x^2)} = \sqrt{4 - 4x + x^2} = \sqrt{(x - 2)^2} = |x - 2|
\]

Thus, \( g(h(x)) = |x - 2| \).

### Step 2: Find \( f(g(h(x))) \)

Now, we substitute \( g(h(x)) = |x - 2| \) into \( f(x) = 2x - 2 \):

\[
f(g(h(x))) = f(|x - 2|) = 2|x - 2| - 2
\]

For the possible answers, we assume \( x \geq 2 \), which simplifies \( |x - 2| = x - 2 \). Therefore:

\[
f(g(h(x))) = 2(x - 2) - 2 = 2x - 4 - 2 = 2x - 6
\]

### Final Answer:

The correct answer is **e) \( 2x - 6 \)**.

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Would you like further details on any part of this solution? Here are five related questions to consider:

1. What happens to \( f(g(h(x))) \) if \( x < 2 \)?
2. How do we handle function compositions with different domains?
3. What is the general rule for function composition \( (f \circ g)(x) \)?
4. Can function compositions be commutative?
5. How would the composition change if one of the functions was modified?

**Tip:** Always verify the domain of each function in compositions, especially when dealing with square roots or absolute values.

Given the functions f(x) = 2x - 2, g(x) = sqrt(4 - x), and h(x) = 4x - x^2, find the composition (f o g o h)(x).

This problem explores the composition of three functions including quadratic and square root functions to solve (f o g o h)(x). The detailed solution demonstrates how to handle compositions involving absolute values and different function types. This is suitable for high school students studying advanced algebra.

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