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We are given the functions:

- \( f(x) = \frac{x}{x + 9} \)
- \( g(x) = 7x - 9 \)

We need to compute the following:

### 1. Find \( (f \circ g)(x) \):
This means applying \( g(x) \) inside \( f(x) \), i.e., \( f(g(x)) \).

\[
(f \circ g)(x) = f(g(x)) = f(7x - 9)
\]

Substitute \( g(x) = 7x - 9 \) into \( f(x) \):

\[
f(g(x)) = \frac{7x - 9}{(7x - 9) + 9} = \frac{7x - 9}{7x}
\]

Simplify:

\[
f(g(x)) = 1 - \frac{9}{7x}
\]

### 2. Domain of \( (f \circ g)(x) \):
The domain of \( f(g(x)) \) depends on where the denominator is not zero. So, we solve:

\[
7x \neq 0 \implies x \neq 0
\]

Therefore, the domain of \( f \circ g \) is \( (-\infty, 0) \cup (0, \infty) \).

### 3. Find \( (g \circ f)(x) \):
This means applying \( f(x) \) inside \( g(x) \), i.e., \( g(f(x)) \).

\[
(g \circ f)(x) = g(f(x)) = g\left(\frac{x}{x + 9}\right)
\]

Substitute \( f(x) = \frac{x}{x + 9} \) into \( g(x) \):

\[
g\left(\frac{x}{x + 9}\right) = 7 \times \frac{x}{x + 9} - 9 = \frac{7x}{x + 9} - 9
\]

To simplify:

\[
g(f(x)) = \frac{7x}{x + 9} - \frac{9(x + 9)}{x + 9} = \frac{7x - 9(x + 9)}{x + 9} = \frac{7x - 9x - 81}{x + 9} = \frac{-2x - 81}{x + 9}
\]

### 4. Domain of \( (g \circ f)(x) \):
The domain of \( g(f(x)) \) depends on the domain of \( f(x) \) where \( f(x) \)'s denominator is not zero:

\[
x + 9 \neq 0 \implies x \neq -9
\]

Thus, the domain of \( g \circ f \) is \( (-\infty, -9) \cup (-9, \infty) \).

### 5. Find \( (f \circ f)(x) \):
This means applying \( f(x) \) inside \( f(x) \), i.e., \( f(f(x)) \).

\[
(f \circ f)(x) = f\left(\frac{x}{x + 9}\right)
\]

Substitute \( f(x) \) into itself:

\[
f\left(\frac{x}{x + 9}\right) = \frac{\frac{x}{x + 9}}{\frac{x}{x + 9} + 9}
\]

Simplify:

\[
f(f(x)) = \frac{\frac{x}{x + 9}}{\frac{x + 9( x + 9)}{x + 9}} = \frac{x}{x + 9 + 9(x + 9)} = \frac{x}{10x + 81}
\]

### 6. Domain of \( (f \circ f)(x) \):
The domain of \( f(f(x)) \) is defined where the denominator is not zero:

\[
10x + 81 \neq 0 \implies x \neq -\frac{81}{10}
\]

Thus, the domain of \( f \circ f \) is \( (-\infty, -\frac{81}{10}) \cup (-\frac{81}{10}, \infty) \).

---

Would you like further clarification on any of these steps?

### Here are 5 related questions:
1. What happens when we swap the order of composition between two functions?
2. How would you compute the inverse of \( f(x) \) or \( g(x) \)?
3. Can we verify the result of a composition by substituting specific values of \( x \)?
4. What is the significance of domain restrictions in function compositions?
5. How does the complexity of the function composition affect its graph?

### Tip:
Always check the domain restrictions of each individual function before performing composition to avoid undefined operations.

Find the functions f ∘ g, g ∘ f, f ∘ f, and g ∘ g and their domains. f(x) = x / (x + 9), g(x) = 7x - 9

Learn how to solve function compositions such as f ∘ g, g ∘ f, f ∘ f, and g ∘ g and find their domains. The problem involves rational functions and algebraic manipulations to determine domain restrictions. Suitable for students in Grades 10-12.

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