and find its domain and range.

IIL Given f(x) = (x - 5)/(x + 3) and g(x) = (x - 1)/x

Let's solve the problem step by step to find the domain and range for both functions:

### 1. **Function \( f(x) = \frac{x - 5}{x + 3} \)**

#### **Domain of \( f(x) \):**
The domain of a function consists of all the possible input values (x-values) that do not cause the function to be undefined. For rational functions, the function is undefined where the denominator is zero.

For \( f(x) = \frac{x - 5}{x + 3} \), the denominator is \( x + 3 \). Setting the denominator equal to zero:

\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]

Thus, \( f(x) \) is undefined at \( x = -3 \). Therefore, the domain of \( f(x) \) is all real numbers except \( x = -3 \), or:

\[
\text{Domain of } f(x): \quad (-\infty, -3) \cup (-3, \infty)
\]

#### **Range of \( f(x) \):**
The range of a function is the set of all possible output values (y-values). To find the range of a rational function, it is useful to think about the behavior of the function as \( x \) approaches certain values.

1. As \( x \to \infty \) or \( x \to -\infty \), the highest degree terms dominate, so the function approaches a horizontal asymptote. Here, the function simplifies to:

   \[
   \frac{x - 5}{x + 3} \approx \frac{x}{x} = 1
   \]

   Therefore, the function has a horizontal asymptote at \( y = 1 \).

2. To check if the function can ever equal 1, set \( f(x) = 1 \):

   \[
   \frac{x - 5}{x + 3} = 1 \quad \Rightarrow \quad x - 5 = x + 3
   \]

   Simplifying, we find this is a contradiction (no solution), so \( f(x) \neq 1 \).

Thus, the range of \( f(x) \) is all real numbers except \( y = 1 \):

\[
\text{Range of } f(x): \quad (-\infty, 1) \cup (1, \infty)
\]

---

### 2. **Function \( g(x) = \frac{x - 1}{x} \)**

#### **Domain of \( g(x) \):**
For \( g(x) = \frac{x - 1}{x} \), the denominator is \( x \). Setting the denominator equal to zero:

\[
x = 0
\]

Thus, \( g(x) \) is undefined at \( x = 0 \). Therefore, the domain of \( g(x) \) is all real numbers except \( x = 0 \), or:

\[
\text{Domain of } g(x): \quad (-\infty, 0) \cup (0, \infty)
\]

#### **Range of \( g(x) \):**
Again, we analyze the function’s behavior:

1. As \( x \to \infty \) or \( x \to -\infty \), the highest degree terms dominate, so the function simplifies to:

   \[
   \frac{x - 1}{x} \approx \frac{x}{x} = 1
   \]

   So, the function approaches a horizontal asymptote at \( y = 1 \).

2. To check if the function can equal 1, set \( g(x) = 1 \):

   \[
   \frac{x - 1}{x} = 1 \quad \Rightarrow \quad x - 1 = x
   \]

   This is a contradiction, so \( g(x) \neq 1 \).

Thus, the range of \( g(x) \) is all real numbers except \( y = 1 \):

\[
\text{Range of } g(x): \quad (-\infty, 1) \cup (1, \infty)
\]

---

### **Summary:**

- **Domain of \( f(x) = \frac{x - 5}{x + 3} \):** \( (-\infty, -3) \cup (-3, \infty) \)
- **Range of \( f(x) \):** \( (-\infty, 1) \cup (1, \infty) \)
- **Domain of \( g(x) = \frac{x - 1}{x} \):** \( (-\infty, 0) \cup (0, \infty) \)
- **Range of \( g(x) \):** \( (-\infty, 1) \cup (1, \infty) \)

Would you like to go into further detail on any part of this? Let me know!

Here are 5 related questions to explore this further:
1. How do vertical asymptotes affect the domain of a rational function?
2. Why can't rational functions reach their horizontal asymptotes?
3. Can you find the x-intercept and y-intercept of \( f(x) \) and \( g(x) \)?
4. How do you find slant asymptotes for more complex rational functions?
5. What happens to the range of a function if you modify its numerator or denominator?

**Tip:** Always check for points where the denominator is zero to ensure the function remains defined when determining the domain of rational functions!

Given f(x) = (x - 5)/(x + 3) and g(x) = (x - 1)/x, find their domain and range.

This guide explains how to determine the domain and range of rational functions f(x) = (x - 5)/(x + 3) and g(x) = (x - 1)/x. Learn about vertical and horizontal asymptotes and how they influence the domain and range. Suitable for students in Grades 10-12.

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