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?

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 $g$.