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Let's review the problem and solution provided for part (A).

You are given:
- \( h(x) = \sqrt{\frac{1 + \sqrt{x}}{2 + \sqrt{x}}} \)
- \( g(x) = \sqrt{x} \)
  
You need to find a function \( f(x) \) such that \( h(x) = f(g(x)) \). This means:
\[
h(x) = f(g(x)) = f(\sqrt{x})
\]

From the given \( h(x) = \sqrt{\frac{1 + \sqrt{x}}{2 + \sqrt{x}}} \), we set:
\[
f(\sqrt{x}) = \sqrt{\frac{1 + \sqrt{x}}{2 + \sqrt{x}}}
\]

To isolate \( f(x) \), replace \( \sqrt{x} \) with a new variable, say \( u = \sqrt{x} \). Then \( x = u^2 \), and:
\[
f(u) = \sqrt{\frac{1 + u}{2 + u}}
\]

Thus, \( f(x) \) is:
\[
f(x) = \sqrt{\frac{1 + x}{2 + x}}
\]

This matches the solution you have written for part (A), so **yes, part A is correct**.

Would you like more details or have any other questions?

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Here are some related questions:
1. How can we verify that \( f(x) \) is correctly derived from \( h(x) \)?
2. What happens if we switch the order of \( g(x) \) and \( f(x) \)?
3. What are the domain restrictions for \( f(x) \) and \( h(x) \)?
4. Can this technique of substitution be used for any composite function?
5. How does this process change if \( h(x) \) were not a square root?

**Tip**: When solving for \( f(x) \) in a composite function, always check for domain compatibility between \( f \) and \( g \).

Learn how to find a function f(x) such that h(x) = f(g(x)) for the given functions h(x) = √((1 + √x) / (2 + √x)) and g(x) = √x. This detailed solution explores the process of function composition and substitution, making it suitable for students in Grades 11-12.

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