To find the domain and range of the function \( f(x) = \frac{x + 3}{5x - 5} \), let's proceed step by step.

### 1. **Domain**:
The domain of a function includes all possible values of \( x \) for which the function is defined. The only restriction in this function comes from the denominator, which cannot be zero. So, we need to find the values of \( x \) that make the denominator zero.

Set the denominator equal to zero and solve for \( x \):
\[
5x - 5 = 0
\]
\[
5x = 5
\]
\[
x = 1
\]

Thus, the function is undefined at \( x = 1 \). Therefore, the domain of \( f(x) \) is all real numbers except \( x = 1 \). In interval notation, the domain is:
\[
\text{Domain} = (-\infty, 1) \cup (1, \infty)
\]

### 2. **Range**:
To find the range of the function, consider the behavior of \( f(x) \) as \( x \) approaches different values, including the vertical asymptote at \( x = 1 \).

Rewriting the function in a different form might help us see the range more clearly:
\[
f(x) = \frac{x + 3}{5(x - 1)} = \frac{1}{5} + \frac{2}{5(x - 1)}
\]

As \( x \) approaches 1 from the left or right, \( f(x) \) approaches infinity or negative infinity, showing that \( f(x) \) can take on extremely large positive or negative values.

To further solidify the range, consider the behavior of the function as \( x \) approaches infinity:
\[
\lim_{x \to \infty} f(x) = \frac{x + 3}{5x - 5} \approx \frac{x}{5x} = \frac{1}{5}
\]

The function never actually reaches \( y = \frac{1}{5} \) but can get arbitrarily close to it. Since the function can take on any real value except for the horizontal asymptote, the range of \( f(x) \) is:
\[
\text{Range} = (-\infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)
\]

### Summary:
- **Domain:** \( (-\infty, 1) \cup (1, \infty) \)
- **Range:** \( (-\infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty) \)

Would you like more details, or do you have any questions? 

Here are 5 related questions to deepen your understanding:
1. How do you determine if a function has a horizontal asymptote?
2. What is the significance of the vertical asymptote in determining the domain?
3. How would the domain and range change if the function were \( f(x) = \frac{x + 3}{5x + 5} \)?
4. What are some other functions where the domain excludes a specific value of \( x \)?
5. How can you graphically determine the domain and range of a rational function?

**Tip:** When analyzing rational functions, always check where the denominator equals zero to find the domain and identify horizontal asymptotes by comparing the degrees of the numerator and denominator.

Discover the domain and range of the rational function f(x) = (x + 3) / (5x - 5) through detailed step-by-step analysis. This problem covers key concepts in rational functions, including domain restrictions due to asymptotes and the behavior of the function at infinity. Suitable for students in Grades 11-12.

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