Let's solve the problem step-by-step.

1. Finding $f(g(x))$ or $f \circ g(x)$

Given:

• $f(x) = \frac{9}{x} + 9$
• $g(x) = \frac{(x+3)}{(x-8)}$

To find $f(g(x))$, substitute $g(x)$ into $f(x)$: $f(g(x)) = f\left( \frac{x+3}{x-8} \right)$ Now, replace $x$ in $f(x)$ with $\frac{x+3}{x-8}$: $f\left( \frac{x+3}{x-8} \right) = \frac{9}{\frac{x+3}{x-8}} + 9$ Simplify $\frac{9}{\frac{x+3}{x-8}}$: $\frac{9 \cdot (x-8)}{x+3}$ So, $f(g(x)) = \frac{9(x-8)}{x+3} + 9$

Next, find a common denominator to simplify further: $f(g(x)) = \frac{9(x-8) + 9(x+3)}{x+3}$ $f(g(x)) = \frac{9(x-8+x+3)}{x+3}$ $f(g(x)) = \frac{9(2x - 5)}{x+3}$

So, the formula for $f \circ g(x)$ is: $f(g(x)) = \frac{18x - 45}{x+3}$

2. Domain of $f(g(x))$

For $f(g(x))$, we need to consider two things:

1. The domain restrictions of $g(x)$.
2. The domain restrictions of $f(g(x))$.

Domain restrictions of $g(x)$:

• $g(x) = \frac{x+3}{x-8}$
• $x-8 \neq 0 \Rightarrow x \neq 8$

Domain restrictions of $f(g(x))$:

• The formula $f(g(x)) = \frac{18x - 45}{x+3}$ implies $x+3 \neq 0 \Rightarrow x \neq -3$.

Combining both restrictions:

• $x \neq 8$
• $x \neq -3$

So, the domain of $f(g(x))$ is: $(-\infty, -3) \cup (-3, 8) \cup (8, \infty)$

3. Finding $g(f(x))$ or $g \circ f(x)$

To find $g(f(x))$, substitute $f(x)$ into $g(x)$: $g(f(x)) = g\left( \frac{9}{x} + 9 \right)$ Now, replace $x$ in $g(x)$ with $\frac{9}{x} + 9$: $g\left( \frac{9}{x} + 9 \right) = \frac{\left( \frac{9}{x} + 9 + 3 \right)}{\left( \frac{9}{x} + 9 - 8 \right)}$ Simplify the numerator: $\frac{9}{x} + 9 + 3 = \frac{9}{x} + 12$ Simplify the denominator: $\frac{9}{x} + 9 - 8 = \frac{9}{x} + 1$ So, $g(f(x)) = \frac{\frac{9}{x} + 12}{\frac{9}{x} + 1}$

To simplify, multiply numerator and denominator by $x$: $g(f(x)) = \frac{9 + 12x}{9 + x}$

So, the formula for $g \circ f(x)$ is: $g(f(x)) = \frac{12x + 9}{x + 9}$

4. Domain of $g(f(x))$

For $g(f(x))$, we need to consider two things:

1. The domain restrictions of $f(x)$.
2. The domain restrictions of $g(f(x))$.

Domain restrictions of $f(x)$:

• $f(x) = \frac{9}{x} + 9$
• $x \neq 0$

Domain restrictions of $g(f(x))$:

• The formula $g(f(x)) = \frac{12x + 9}{x + 9}$ implies $x+9 \neq 0 \Rightarrow x \neq -9$.

Combining both restrictions:

• $x \neq 0$
• $x \neq -9$

So, the domain of $g(f(x))$ is: $(-\infty, -9) \cup (-9, 0) \cup (0, \infty)$

Summary:

1. Formula for $f(g(x))$: $f(g(x)) = \frac{18x - 45}{x+3}$
2. Domain of $f(g(x))$: $(-\infty, -3) \cup (-3, 8) \cup (8, \infty)$
3. Formula for $g(f(x))$: $g(f(x)) = \frac{12x + 9}{x + 9}$
4. Domain of $g(f(x))$: $(-\infty, -9) \cup (-9, 0) \cup (0, \infty)$

Do you want more details on any of these steps, or have any other questions?

Related Questions:

1. How do we find the domain of a composition of functions in general?
2. What are some other methods to simplify complex rational functions?
3. How can we find the inverse of a composition of two functions?
4. What is the graphical interpretation of composing two functions?
5. How do we determine whether a composite function is increasing or decreasing?

Tip:

Always check the domain restrictions of both the inner and outer functions when finding the domain of a composite function!