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solve questions with detailed explanation. start with the functions 

Let's begin with **Question 1** from the "Functions and Graphs" section:

---

### **Problem:**

You are given two functions:  
\[
f(x) = x - 1 \quad \text{and} \quad g(x) = \frac{2}{x-2}
\]

1. **Find** \( p(x) = (f \circ g)(x) \), and determine its domain.  
2. **Find the inverse** of \( p(x) \), and sketch the graphs of \( p(x) \) and \( p^{-1}(x) \).

---

### **Solution:**

#### 1. **Finding \( p(x) = f(g(x)) \):**

First, let's substitute \( g(x) \) into \( f(x) \):
\[
p(x) = f(g(x)) = f\left( \frac{2}{x-2} \right)
\]
Since \( f(x) = x - 1 \), we replace the \( x \) in \( f(x) \) with \( g(x) \):
\[
p(x) = \left( \frac{2}{x-2} \right) - 1
\]
Simplifying the expression:
\[
p(x) = \frac{2}{x-2} - 1 = \frac{2}{x-2} - \frac{x-2}{x-2}
\]
\[
p(x) = \frac{2 - (x-2)}{x-2} = \frac{2 - x + 2}{x-2} = \frac{4 - x}{x-2}
\]
Thus, \( p(x) = \frac{4 - x}{x-2} \).

---

#### **Domain of \( p(x) \):**

- The domain of \( g(x) \) is restricted by the fact that \( g(x) = \frac{2}{x-2} \) is undefined when \( x = 2 \). Hence, \( x \neq 2 \).
- The expression for \( p(x) = \frac{4-x}{x-2} \) does not introduce any new restrictions, so the domain remains \( x \neq 2 \).

Thus, the domain of \( p(x) \) is:  
\[
x \in (-\infty, 2) \cup (2, \infty)
\]

---

#### 2. **Finding the inverse \( p^{-1}(x) \):**

We now solve for \( x \) in terms of \( y \), where \( y = p(x) \):
\[
y = \frac{4 - x}{x - 2}
\]
Multiply both sides by \( (x-2) \) to eliminate the denominator:
\[
y(x - 2) = 4 - x
\]
Expanding both sides:
\[
yx - 2y = 4 - x
\]
Rearranging the terms to isolate \( x \):
\[
yx + x = 4 + 2y
\]
\[
x(y + 1) = 4 + 2y
\]
Solving for \( x \):
\[
x = \frac{4 + 2y}{y + 1}
\]

Thus, the inverse function is:
\[
p^{-1}(x) = \frac{4 + 2x}{x + 1}
\]

---

#### **Graphing \( p(x) \) and \( p^{-1}(x) \):**

1. **Graph of \( p(x) = \frac{4 - x}{x-2} \):**
   - Asymptote at \( x = 2 \) (since \( p(x) \) is undefined at \( x = 2 \)).
   - Horizontal asymptote: As \( x \to \infty \), \( p(x) \to -1 \), so there is a horizontal asymptote at \( y = -1 \).

2. **Graph of \( p^{-1}(x) = \frac{4 + 2x}{x + 1} \):**
   - Asymptote at \( x = -1 \) (since \( p^{-1}(x) \) is undefined at \( x = -1 \)).
   - Horizontal asymptote: As \( x \to \infty \), \( p^{-1}(x) \to 2 \), so there is a horizontal asymptote at \( y = 2 \).

Both functions are reflected over the line \( y = x \).

---

Would you like more details on any part of the solution, or perhaps some assistance in visualizing the graphs?

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### Follow-up questions:

1. How do you find the composite of two functions in general?
2. What are the rules for determining the domain of a composite function?
3. How can you verify if a function has an inverse?
4. What is the relationship between the graphs of a function and its inverse?
5. How do asymptotes help in sketching graphs of rational functions?

### Tip: 
When finding the inverse of a function, always verify by substituting the inverse into the original function to ensure it returns the identity function \( f(f^{-1}(x)) = x \).

solve questions with detailed explanation. start with the functions

Explore a detailed step-by-step solution for finding the composition of functions f(x) = x - 1 and g(x) = 2 / (x - 2), determining the domain and range of the composite function, and finding its inverse. Ideal for advanced high school or early university students.

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