$$ 1 \leftarrow $$ For the functions $$ f(x)=6-x^{2} $$ and $$ g(x)=x^{2}+4 x-32 $$, find $$ f+g, f-g, f g $$, and $$ \frac{f}{g} $$. Determine the domain for each function. $$ (f \cdot g)(x)=-x^{4}-4 x^{3}+38 x^{2}+24 x-192 $$ (Simplify your answer.) What is the domain of fg ? Select the correct choice below and, 1 necessary, fill in the answer box to complete your choice. A. The domain of $$ f g $$ is $$ \} $$. (Use a comma to separate answers as needed.) B. The domain of $$ f g $$ is (Type your answer in interval notation.) C. The domain of $$ f g $$ is $$ \varnothing $$.

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Let's analyze the functions $$ f(x) = 6 - x^{2} $$ and $$ g(x) = x^{2} + 4x - 32 $$ to find $$ f + g $$, $$ f - g $$, $$ f \cdot g $$, and $$ \frac{f}{g} $$, along with their respective domains.

### 1. $$ f + g $$
$$
f(x) + g(x) = (6 - x^{2}) + (x^{2} + 4x - 32) = 6 - x^{2} + x^{2} + 4x - 32 = 4x - 26
$$
**Domain:** Since both $$ f $$ and $$ g $$ are polynomials, their sum is also a polynomial. Polynomials are defined for all real numbers.
$$
\text{Domain of } f + g: \quad (-\infty, \infty)
$$

### 2. $$ f - g $$
$$
f(x) - g(x) = (6 - x^{2}) - (x^{2} + 4x - 32) = 6 - x^{2} - x^{2} - 4x + 32 = -2x^{2} - 4x + 38
$$
**Domain:** Similar to addition, subtraction of polynomials results in a polynomial defined for all real numbers.
$$
\text{Domain of } f - g: \quad (-\infty, \infty)
$$

### 3. $$ f \cdot g $$
$$
f(x) \cdot g(x) = (6 - x^{2})(x^{2} + 4x - 32)
$$
Expanding:
$$
= 6(x^{2} + 4x - 32) - x^{2}(x^{2} + 4x - 32) = 6x^{2} + 24x - 192 - x^{4} - 4x^{3} + 32x^{2}
$$
Combining like terms:
$$
= -x^{4} - 4x^{3} + (6x^{2} + 32x^{2}) + 24x - 192 = -x^{4} - 4x^{3} + 38x^{2} + 24x - 192
$$
**Simplified Form:**
$$
(f \cdot g)(x) = -x^{4} - 4x^{3} + 38x^{2} + 24x - 192
$$
**Domain:** The product of two polynomials is also a polynomial, which is defined for all real numbers.
$$
\text{Domain of } f \cdot g: \quad (-\infty, \infty)
$$

### 4. $$ \frac{f}{g} $$
$$
\frac{f(x)}{g(x)} = \frac{6 - x^{2}}{x^{2} + 4x - 32}
$$
**Domain:** The division is defined when the denominator $$ g(x) \neq 0 $$.
$$
x^{2} + 4x - 32 = 0
$$
Solving for $$ x $$:
$$
x = \frac{-4 \pm \sqrt{16 + 128}}{2} = \frac{-4 \pm \sqrt{144}}{2} = \frac{-4 \pm 12}{2}
$$
$$
x = 4 \quad \text{or} \quad x = -8
$$
Therefore, the domain excludes $$ x = 4 $$ and $$ x = -8 $$:
$$
\text{Domain of } \frac{f}{g}: \quad (-\infty, -8) \cup (-8, 4) \cup (4, \infty)
$$

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### Determining the Domain of $$ f \cdot g $$

Given the multiple-choice options:

**A.** The domain of $$ f \cdot g $$ is $$ \{ \} $$.  
**B.** The domain of $$ f \cdot g $$ is $$\text{(Type your answer in interval notation.)}$$  
**C.** The domain of $$ f \cdot g $$ is $$ \varnothing $$.

**Correct Choice:**  
**B. The domain of $$ f \cdot g $$ is $$ (-\infty, \infty) $$.**

**Explanation:**  
Since both $$ f(x) $$ and $$ g(x) $$ are polynomials, their product $$ f(x) \cdot g(x) $$ is also a polynomial. Polynomials are defined for all real numbers, so the domain of $$ f \cdot g $$ is all real numbers.

$$
\boxed{\text{B. The domain of } f \cdot g \text{ is } (-\infty, \infty)}
$$
The domain of $$ f \cdot g $$ is all real numbers, so the correct choice is: **B. The domain of $$ f \cdot g $$ is $$ (-\infty, \infty) $$.**

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