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 $y = 1$:

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

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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 $y = 1$:

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

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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!