and find its domain and range.

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

1. (f + g)(x) and D f + x

We are given two functions: 

\[
f(x) = \frac{x - 5}{x + 3} \quad \text{and} \quad g(x) = \frac{x - 1}{x}
\]

### Part 1: Finding \( (f + g)(x) \)

The sum of two functions \( f(x) + g(x) \) is found by adding them together:

\[
(f + g)(x) = f(x) + g(x)
= \frac{x - 5}{x + 3} + \frac{x - 1}{x}
\]

To add these fractions, we need to find a common denominator, which is \( (x + 3)x \):

\[
(f + g)(x) = \frac{(x - 5)x + (x - 1)(x + 3)}{(x + 3)x}
\]

Now simplify the numerator:
\[
= \frac{(x^2 - 5x) + (x^2 + 3x - x - 3)}{(x + 3)x}
\]
\[
= \frac{x^2 - 5x + x^2 + 2x - 3}{(x + 3)x}
\]
\[
= \frac{2x^2 - 3x - 3}{(x + 3)x}
\]

Thus, 
\[
(f + g)(x) = \frac{2x^2 - 3x - 3}{(x + 3)x}
\]

### Part 2: Domain of \( f(x) + g(x) \)

The domain of a function is the set of all \( x \)-values for which the function is defined. In this case, \( f(x) + g(x) \) involves two denominators: \( x + 3 \) and \( x \), which must not be zero.

- For \( f(x) \), the denominator \( x + 3 \neq 0 \) implies \( x \neq -3 \).
- For \( g(x) \), the denominator \( x \neq 0 \).

So, the domain of \( (f + g)(x) \) is all real numbers except \( x = -3 \) and \( x = 0 \).

In interval notation, the domain is:

\[
\text{Domain of } (f + g)(x): \quad (-\infty, -3) \cup (-3, 0) \cup (0, \infty)
\]

### Part 3: Range of \( f(x) + g(x) \)

To find the range of \( f(x) + g(x) \), we need to consider the output of the function as \( x \) takes values from its domain. Since \( (f + g)(x) \) is a rational function, determining its exact range algebraically involves solving \( y = (f + g)(x) \) for \( x \), which is not trivial. However, in most cases, rational functions like this can take almost all real values except for potential restrictions from vertical and horizontal asymptotes.

The behavior at vertical asymptotes (where the denominator is zero) suggests the range excludes some values near \( x = -3 \) and \( x = 0 \), but the exact range typically includes all real numbers depending on the asymptotes.

### Part 4: Domain of \( f(x) \) and \( g(x) \)

- The domain of \( f(x) = \frac{x - 5}{x + 3} \) is all real numbers except \( x = -3 \), since the denominator cannot be zero.
  \[
  \text{Domain of } f(x): \quad (-\infty, -3) \cup (-3, \infty)
  \]

- The domain of \( g(x) = \frac{x - 1}{x} \) is all real numbers except \( x = 0 \), since the denominator cannot be zero.
  \[
  \text{Domain of } g(x): \quad (-\infty, 0) \cup (0, \infty)
  \]

Would you like to dive deeper into the range of \( f(x) + g(x) \) or explore another concept?

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### Related Questions:
1. How do you find the product \( (f \cdot g)(x) \)?
2. How do vertical asymptotes affect the range of a function?
3. What is the behavior of \( f(x) \) and \( g(x) \) as \( x \to \infty \)?
4. Can we simplify the difference \( (f - g)(x) \)?
5. How do you determine if two functions are inverses?

### Tip:
When combining functions, always simplify by finding a common denominator before adding or subtracting!

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

Learn how to add the rational functions f(x) = (x - 5)/(x + 3) and g(x) = (x - 1)/x and find their domain. This problem involves function operations, algebraic manipulation, and domain determination. Suitable for high school students in Grades 10-12.

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