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) $2x - 6$.

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.