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Let's analyze and solve the given problems step by step.

### Functions given:
- \( f(x) = \frac{1}{x} \)
- \( g(x) = x - 7 \)

### Part (a): Find the domain of the following functions:

1. **Domain of \( f(x) = \frac{1}{x} \)**:
   - \( f(x) \) is undefined when \( x = 0 \), because division by zero is undefined.
   - Therefore, the domain of \( f(x) \) is \( x \in (-\infty, 0) \cup (0, \infty) \).

2. **Domain of \( g(x) = x - 7 \)**:
   - \( g(x) \) is defined for all real numbers, so the domain is \( (-\infty, \infty) \).

3. **Domain of \( f + g \)**:
   - \( f + g \) is the sum of \( f(x) \) and \( g(x) \), i.e., \( f(x) + g(x) = \frac{1}{x} + (x - 7) \).
   - The domain of \( f + g \) is restricted by the domain of \( f(x) \), which excludes \( x = 0 \).
   - So, the domain of \( f + g \) is \( x \in (-\infty, 0) \cup (0, \infty) \).

4. **Domain of \( f - g \)**:
   - \( f - g \) is the difference of \( f(x) \) and \( g(x) \), i.e., \( f(x) - g(x) = \frac{1}{x} - (x - 7) \).
   - The domain is the same as for \( f + g \), which is \( x \in (-\infty, 0) \cup (0, \infty) \).

5. **Domain of \( f \cdot g \) (product \( fg \))**:
   - \( f(x) \cdot g(x) = \frac{1}{x} \cdot (x - 7) = \frac{x - 7}{x} \).
   - Again, this function is undefined at \( x = 0 \), so the domain is \( x \in (-\infty, 0) \cup (0, \infty) \).

6. **Domain of \( f \circ f \) (composition \( ff \))**:
   - \( f(f(x)) = f\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} = x \).
   - Since \( f(x) \) is only defined for \( x \neq 0 \), the domain of \( f \circ f \) is \( x \in (-\infty, 0) \cup (0, \infty) \).

7. **Domain of \( \frac{f}{g} \)**:
   - \( \frac{f(x)}{g(x)} = \frac{\frac{1}{x}}{x - 7} = \frac{1}{x(x - 7)} \).
   - This function is undefined when \( x = 0 \) or \( x = 7 \) (since \( g(x) = 0 \) when \( x = 7 \)).
   - Therefore, the domain is \( x \in (-\infty, 0) \cup (0, 7) \cup (7, \infty) \).

8. **Domain of \( \frac{g}{f} \)**:
   - \( \frac{g(x)}{f(x)} = \frac{x - 7}{\frac{1}{x}} = x(x - 7) \).
   - This function is undefined when \( x = 0 \).
   - Therefore, the domain is \( x \in (-\infty, 0) \cup (0, \infty) \).

### Part (b): Find the following:

1. **\( (f + g)(x) \)**:
   - \( (f + g)(x) = \frac{1}{x} + (x - 7) = \frac{1 + x(x - 7)}{x} = \frac{1 + x^2 - 7x}{x} = \frac{x^2 - 7x + 1}{x} \).

2. **\( (f - g)(x) \)**:
   - \( (f - g)(x) = \frac{1}{x} - (x - 7) = \frac{1 - x(x - 7)}{x} = \frac{1 - x^2 + 7x}{x} = \frac{-x^2 + 7x + 1}{x} \).

3. **\( (fg)(x) \)**:
   - \( (fg)(x) = f(x) \cdot g(x) = \frac{1}{x} \cdot (x - 7) = \frac{x - 7}{x} \).

4. **\( (ff)(x) \)**:
   - \( (ff)(x) = f(f(x)) = f\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} = x \).

5. **\( \left(\frac{f}{g}\right)(x) \)**:
   - \( \left(\frac{f}{g}\right)(x) = \frac{\frac{1}{x}}{x - 7} = \frac{1}{x(x - 7)} \).

6. **\( \left(\frac{g}{f}\right)(x) \)**:
   - \( \left(\frac{g}{f}\right)(x) = \frac{x - 7}{\frac{1}{x}} = x(x - 7) = x^2 - 7x \).

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Let me know if you want further clarification on any of these steps!

Here are five related questions to extend the problem:

1. What is the range of \( f(x) = \frac{1}{x} \) and \( g(x) = x - 7 \)?
2. How would you find the inverse functions of \( f(x) \) and \( g(x) \)?
3. What is the composition \( g(f(x)) \), and what is its domain?
4. What is the behavior of \( f(x) + g(x) \) as \( x \to \infty \)?
5. How do you simplify \( f(x) \cdot f(g(x)) \)?

**Tip**: Always check for restrictions in the domain when dealing with fractions and compositions of functions to avoid undefined expressions.

The functions f and g are defined as f(x) = 1/x and g(x) = x - 7. a) Find the domain of f, g, f + g, f - g, fg, ff, f/g, and g/f. b) Find (f + g)(x), (f - g)(x), (fg)(x), (ff)(x), (f/g)(x), and (g/f)(x).

Explore the domain and operations of functions f(x) = 1/x and g(x) = x - 7. Learn to find the domain of f, g, f + g, f - g, fg, ff, f/g, and g/f, as well as perform operations like addition, subtraction, multiplication, division, and composition of functions. Suitable for students in Grades 11-12.

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