The given problem involves decomposing the function $f(x) = \ln\left(\sin^{-1}\left(\frac{x}{3}\right) + 2\right)$ into two simpler functions $g(x)$ and $h(x)$, such that $f(x) = g(h(x))$.

Solution:

Part (a) - Decomposing $f(x)$

We rewrite $f(x)$ as $f(x) = g(h(x))$. Here's one way to do it:

1. Let $h(x) = \sin^{-1}\left(\frac{x}{3}\right) + 2$, which isolates the inner composition.
2. Let $g(x) = \ln(x)$, which is the outer function.

Thus: $f(x) = g(h(x)) = \ln\left(\sin^{-1}\left(\frac{x}{3}\right) + 2\right).$

Part (b) - Writing $f(x)$ in the form $g(h(x))$

Using the decomposition above: $h(x) = \sin^{-1}\left(\frac{x}{3}\right) + 2, \quad g(x) = \ln(x).$

Therefore: $f(x) = g(h(x)) = \ln(h(x)) \quad \text{with} \quad h(x) = \sin^{-1}\left(\frac{x}{3}\right) + 2.$

Would you like further details or an explanation of any step?

Related Questions:

1. What are the domain restrictions for $h(x) = \sin^{-1}\left(\frac{x}{3}\right) + 2$?
2. How would you decompose $f(x)$ differently if we defined $g(x)$ as a logarithmic transformation of $x + 2$?
3. What is the derivative of $f(x) = \ln\left(\sin^{-1}\left(\frac{x}{3}\right) + 2\right)$?
4. Can $h(x) = \sin^{-1}\left(\frac{x}{3}\right) + 2$ be inverted? If so, what is its inverse?
5. How does the value of $f(x)$ change as $x$ approaches 3?

Tip:

When decomposing functions, start by identifying the innermost operation as $h(x)$ and the final transformation as $g(x)$. This makes the composition clear and easier to verify.