The function given is:

$f(x) = \frac{x + 5}{x - 3}$

Domain:

The domain of the function consists of all real numbers except where the denominator is zero, as division by zero is undefined.

So, we set the denominator equal to zero and solve for $x$:

$x - 3 = 0$ $x = 3$

Thus, the domain is all real numbers except $x = 3$. In interval notation, the domain is:

$\text{Domain: } (-\infty, 3) \cup (3, \infty)$

Range:

To find the range, we analyze the behavior of the function. A rational function like this one typically has a horizontal asymptote that affects its range.

For large values of $x$ (both positive and negative), the function behaves like:

$f(x) \approx \frac{x}{x} = 1$

Thus, the horizontal asymptote is $y = 1$.

However, since $f(x)$ never actually equals $1$ (it only approaches it as $x$ becomes very large or very small), $y = 1$ is not included in the range.

To confirm, let's solve for $x$ in terms of $y$:

$y = \frac{x + 5}{x - 3}$ $y(x - 3) = x + 5$ $yx - 3y = x + 5$ $yx - x = 3y + 5$ $x(y - 1) = 3y + 5$ $x = \frac{3y + 5}{y - 1}$

For $x$ to be defined, $y - 1$ must not be zero, so $y \neq 1$.

Thus, the range is all real numbers except $y = 1$. In interval notation:

$\text{Range: } (-\infty, 1) \cup (1, \infty)$

Would you like to discuss any specific part in more detail, or do you have any questions?

Here are some related questions for further exploration:

1. How would you find the vertical asymptote of this function?
2. What is the significance of the horizontal asymptote in rational functions?
3. How do you determine the oblique asymptote, if it exists, for a rational function?
4. What happens to the function values as $x$ approaches the vertical asymptote?
5. How would you graph this function and label all the asymptotes?

Tip: When solving for the range of a rational function, finding where the function's denominator equals zero can help identify the vertical asymptotes, which often define the boundaries of the range.