Let f(x)=(9/x)+9 and g(x)=(x+3)/(x−8), find the formula for fog, find the domain of fog, find gof, and find the domain of gof, for the domain answer in interval notation

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!

Learn how to find the formulas for f(g(x)) and g(f(x)) when given rational functions f(x) = (9/x) + 9 and g(x) = (x+3)/(x-8). This detailed guide covers function composition, simplification, and finding domain restrictions. Suitable for high school students in Grades 11-12.

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