$$ 1 \leftarrow $$ Find $$ f+g, f-g, f g $$ and $$ \frac{f}{g} $$. Determine the domain for each function. $$ f(x)=5 x+2, g(x)=x-6 $$ What is the domain of $$ f+g $$ ? B. The domain of $$ f+g $$ is $$ (-\infty, \infty) $$. (Type your answer in interval notation.) C. The domain of $$ f+g $$ is $$ \{ $$ is $$ \varnothing $$. (Use a corma to separate answers as needed.) $$ (f-g)(x)=\square $$ (Simplify your answer.)

Question

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Let's solve the problem step by step.

**Given:**
- $$ f(x) = 5x + 2 $$
- $$ g(x) = x - 6 $$

### 1. Find $$ f + g $$, $$ f - g $$, $$ fg $$, and $$ \frac{f}{g} $$

**a. $$ f + g $$**
$$
f(x) + g(x) = (5x + 2) + (x - 6) = 5x + x + 2 - 6 = 6x - 4
$$
$$
\boxed{f + g = 6x - 4}
$$

**b. $$ f - g $$**
$$
f(x) - g(x) = (5x + 2) - (x - 6) = 5x - x + 2 + 6 = 4x + 8
$$
$$
\boxed{f - g = 4x + 8}
$$

**c. $$ fg $$**
$$
f(x) \cdot g(x) = (5x + 2)(x - 6) = 5x \cdot x + 5x \cdot (-6) + 2 \cdot x + 2 \cdot (-6) = 5x^2 - 30x + 2x - 12 = 5x^2 - 28x - 12
$$
$$
\boxed{fg = 5x^2 - 28x - 12}
$$

**d. $$ \frac{f}{g} $$**
$$
\frac{f(x)}{g(x)} = \frac{5x + 2}{x - 6}
$$
$$
\boxed{\frac{f}{g} = \frac{5x + 2}{x - 6}}
$$

### 2. Determine the Domain for Each Function

**a. Domain of $$ f + g $$**
- Both $$ f(x) $$ and $$ g(x) $$ are defined for all real numbers.
- Therefore, $$ f + g $$ is also defined for all real numbers.

$$
\boxed{(-\infty, \infty)}
$$

**b. Domain of $$ f - g $$**
- Similar to $$ f + g $$, both functions are defined for all real numbers.

$$
\boxed{(-\infty, \infty)}
$$

**c. Domain of $$ fg $$**
- Product of two functions defined for all real numbers is also defined for all real numbers.

$$
\boxed{(-\infty, \infty)}
$$

**d. Domain of $$ \frac{f}{g} $$**
- The denominator $$ g(x) = x - 6 $$ cannot be zero.
- So, $$ x \neq 6 $$.

$$
\boxed{(-\infty, 6) \cup (6, \infty)}
$$

### 3. Specific Questions

**B. What is the domain of $$ f + g $$?**

The domain of $$ f + g $$ is all real numbers.

$$
\boxed{(-\infty, \infty)}
$$

**C. The domain of $$ f + g $$ is $$ \varnothing $$ (empty set).**

This statement is **incorrect** because $$ f + g $$ is defined for all real numbers. The correct domain is $$ (-\infty, \infty) $$, not the empty set.

**$$ f - g $$ Simplified**

$$
(f - g)(x) = 4x + 8
$$
$$
\boxed{f - g = 4x + 8}
$$

### Summary

- **$$ f + g = 6x - 4 $$** with domain $$ (-\infty, \infty) $$
- **$$ f - g = 4x + 8 $$** with domain $$ (-\infty, \infty) $$
- **$$ fg = 5x^2 - 28x - 12 $$** with domain $$ (-\infty, \infty) $$
- **$$ \frac{f}{g} = \frac{5x + 2}{x - 6} $$** with domain $$ (-\infty, 6) \cup (6, \infty) $$

For **question B**, the correct domain of $$ f + g $$ is $$ (-\infty, \infty) $$.

For **question C**, the statement that the domain is $$ \varnothing $$ is incorrect.

Finally, **$$ f - g $$** simplified is $$ 4x + 8 $$.
$$ f + g = 6x - 4 $$, $$ f - g = 4x + 8 $$, $$ fg = 5x^2 - 28x - 12 $$, and $$ \frac{f}{g} = \frac{5x + 2}{x - 6} $$. The domain of $$ f + g $$ is all real numbers, $$ (-\infty, \infty) $$.

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