Let's go step by step with solving the problems visible in your image.

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.