Let's break down the problem based on the image you uploaded.

We are given the following functions:

• $f(x) = \frac{x}{x+1}$
• $g(x) = \frac{9}{x}$

We are asked to find:

1. $(f \circ g)(x)$, which is $f(g(x))$
2. The domain of $f \circ g$

Part (a): Finding $(f \circ g)(x)$

To compute $f(g(x))$, substitute $g(x)$ into $f(x)$:

$f(g(x)) = f\left(\frac{9}{x}\right) = \frac{\frac{9}{x}}{\frac{9}{x} + 1}$

Now, simplify the expression:

$f(g(x)) = \frac{\frac{9}{x}}{\frac{9}{x} + 1} = \frac{\frac{9}{x}}{\frac{9 + x}{x}} = \frac{9}{9 + x}$

Thus, $(f \circ g)(x) = \frac{9}{9 + x}$.

Part (b): Finding the Domain of $f \circ g$

To find the domain of $f \circ g$, we need to consider the domains of both functions:

1. Domain of $g(x)$: Since $g(x) = \frac{9}{x}$, $x \neq 0$, so the domain of $g(x)$ is $(-\infty, 0) \cup (0, \infty)$.

2. Domain of $f(g(x))$: For $f(g(x)) = \frac{9}{9 + x}$, the expression $9 + x \neq 0$, which gives $x \neq -9$.

So, the domain of $f \circ g$ is all values of $x$ where $x \neq 0$ and $x \neq -9$. Therefore, the domain is:

$(-\infty, -9) \cup (-9, 0) \cup (0, \infty)$

Final Answers:

• $(f \circ g)(x) = \frac{9}{9 + x}$
• Domain: $(-\infty, -9) \cup (-9, 0) \cup (0, \infty)$

Would you like further details or have any questions?

Here are 5 follow-up questions to expand on this:

1. What is the general process of finding the domain of a composition of functions?
2. How do you approach simplifying complex rational functions like $f(g(x))$?
3. What would happen to the domain if the form of $g(x)$ changed?
4. Can you apply this same method to other types of function compositions (e.g., logarithmic or trigonometric)?
5. How does the presence of an asymptote at $x = 0$ affect the domain?

Tip: Always check the domains of both functions when dealing with compositions to avoid missing important restrictions!