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Let's analyze the five functions and find their domain and range. Here's how we compute both for each function:

### 1. \( f(x) = \frac{4}{x} \)
- **Domain**: The denominator cannot be zero, so \( x \neq 0 \). The domain is all real numbers except 0: 
  \[
  \text{Domain} = (-\infty, 0) \cup (0, \infty)
  \]
- **Range**: For \( f(x) = \frac{4}{x} \), the function can take any value except 0 (as there is no \( x \) such that \( f(x) = 0 \)). Hence, the range is all real numbers except 0:
  \[
  \text{Range} = (-\infty, 0) \cup (0, \infty)
  \]

### 2. \( f(x) = \frac{3}{x + 2} \)
- **Domain**: The denominator cannot be zero, so \( x + 2 \neq 0 \), i.e., \( x \neq -2 \). The domain is:
  \[
  \text{Domain} = (-\infty, -2) \cup (-2, \infty)
  \]
- **Range**: Similar to the previous function, \( f(x) \) will never be zero because the numerator is a constant. The range is all real numbers except 0:
  \[
  \text{Range} = (-\infty, 0) \cup (0, \infty)
  \]

### 3. \( f(x) = \frac{x + 5}{7} \)
- **Domain**: The function is defined for all real \( x \) since there is no restriction (no denominator that could become zero). Therefore, the domain is:
  \[
  \text{Domain} = (-\infty, \infty)
  \]
- **Range**: Simplifying, \( f(x) = \frac{1}{7}x + \frac{5}{7} \), which is a linear function. The range of a linear function is all real numbers:
  \[
  \text{Range} = (-\infty, \infty)
  \]

### 4. \( f(x) = \frac{x - 6}{3} \)
- **Domain**: This is another linear function with no restrictions on \( x \), so the domain is:
  \[
  \text{Domain} = (-\infty, \infty)
  \]
- **Range**: Simplifying, \( f(x) = \frac{1}{3}x - 2 \), which is also a linear function. The range of a linear function is all real numbers:
  \[
  \text{Range} = (-\infty, \infty)
  \]

### 5. \( f(x) = \frac{x - 2}{x + 4} \)
- **Domain**: The denominator cannot be zero, so \( x + 4 \neq 0 \), i.e., \( x \neq -4 \). The domain is:
  \[
  \text{Domain} = (-\infty, -4) \cup (-4, \infty)
  \]
- **Range**: The horizontal asymptote of this function is \( y = 1 \) (as \( x \to \infty \), \( f(x) \approx 1 \)). Thus, the function can take any value except 1:
  \[
  \text{Range} = (-\infty, 1) \cup (1, \infty)
  \]

Let me know if you'd like more detailed explanations!

Here are 5 related questions for further practice:
1. How do vertical asymptotes affect the domain of a rational function?
2. What conditions create holes in the graph of a rational function?
3. How do you determine the horizontal asymptote of a rational function?
4. Can a rational function have more than one vertical asymptote? Why or why not?
5. How do you find the x- and y-intercepts of a rational function?

**Tip**: Always check the denominator of a rational function to avoid undefined points in the domain!

Compute the domain and range of the following rational functions.

Explore the computation of domain and range for five distinct rational functions, highlighting conditions for undefined points and asymptotic behavior. Suitable for high school students studying advanced algebra.

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