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 function. $$ (1+g)(x)=4 x-\angle 0 $$ (Impiry your answer.) What is the domain of $$ f+g $$ ? Select the correct choice below and, if necessary, fill in the answer box to co your choice. A. The domain of $$ f+g $$ is $$ \} $$. (Use a comma to separate answers as needed.) B. The domain of $$ f+g $$ is $$ (-\infty, \infty) $$. (Type your answer in interval notation.) C. The domain of $$ f+g $$ is $$ \varnothing $$. (f-g)(x)= (Simplify your answer.)

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Let's analyze the given functions and perform the required operations step by step.

### Given Functions:
- $$ f(x) = 6 - x^2 $$
- $$ g(x) = x^2 + 4x - 32 $$

### 1. **Sum: $$ f + g $$**
$$
\begin{align*}
f + g &= (6 - x^2) + (x^2 + 4x - 32) \
&= 6 - x^2 + x^2 + 4x - 32 \
&= 4x - 26
\end{align*}
$$
**Simplified Form:**
$$
f + g = 4x - 26
$$

### 2. **Difference: $$ f - g $$**
$$
\begin{align*}
f - g &= (6 - x^2) - (x^2 + 4x - 32) \
&= 6 - x^2 - x^2 - 4x + 32 \
&= -2x^2 - 4x + 38
\end{align*}
$$
**Simplified Form:**
$$
f - g = -2x^2 - 4x + 38
$$

### 3. **Product: $$ f \cdot g $$**
$$
\begin{align*}
f \cdot g &= (6 - x^2)(x^2 + 4x - 32) \
&= 6(x^2) + 6(4x) + 6(-32) - x^2(x^2) - x^2(4x) - x^2(-32) \
&= 6x^2 + 24x - 192 - x^4 - 4x^3 + 32x^2 \
&= -x^4 - 4x^3 + 38x^2 + 24x - 192
\end{align*}
$$
**Simplified Form:**
$$
f \cdot g = -x^4 - 4x^3 + 38x^2 + 24x - 192
$$

### 4. **Quotient: $$ \frac{f}{g} $$**
$$
\frac{f}{g} = \frac{6 - x^2}{x^2 + 4x - 32}
$$
**Factorization:**
- **Numerator:** $$ 6 - x^2 = -(x^2 - 6) $$ (cannot be factored further over real numbers)
- **Denominator:** 
  $$
  x^2 + 4x - 32 = (x - 4)(x + 8)
  $$
  
**Simplified Form:**
$$
\frac{f}{g} = \frac{6 - x^2}{(x - 4)(x + 8)}
$$
(No further simplification is possible.)

### 5. **Determining the Domain**

- **Domain of $$ f(x) $$:** All real numbers, since it's a polynomial.
- **Domain of $$ g(x) $$:** All real numbers, since it's a polynomial.
- **Domain of $$ f + g $$, $$ f - g $$, and $$ f \cdot g $$:** All real numbers, as these operations on polynomials result in polynomials.
- **Domain of $$ \frac{f}{g} $$:** All real numbers except where $$ g(x) = 0 $$.

**Finding where $$ g(x) = 0 $$:**
  $$
  x^2 + 4x - 32 = 0 \
  x = \frac{-4 \pm \sqrt{16 + 128}}{2} = \frac{-4 \pm 12}{2} \
  x = 4 \quad \text{or} \quad x = -8
  $$
  
  **Thus, the domain of $$ \frac{f}{g} $$ is:**
  $$
  \mathbb{R} \setminus \{ -8, 4 \}
  $$

### 6. **Multiple Choice Question:**

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

**Options:**
- **A.** The domain of $$ f + g $$ is $$ \} $$.  
  *(This seems to be a typographical error and is not a valid interval.)*
  
- **B.** The domain of $$ f + g $$ is $$ (-\infty, \infty) $$.  
  *(Correct)*

- **C.** The domain of $$ f + g $$ is $$ \varnothing $$.  
  *(Incorrect, as $$ f + g $$ is defined for all real numbers.)*

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

### 7. **Simplifying $$ f - g $$:**
$$
f - g = -2x^2 - 4x + 38
$$
**This is already in its simplest form.**

---

**Summary of Results:**
- $$ f + g = 4x - 26 $$ with domain $$ (-\infty, \infty) $$
- $$ f - g = -2x^2 - 4x + 38 $$ with domain $$ (-\infty, \infty) $$
- $$ f \cdot g = -x^4 - 4x^3 + 38x^2 + 24x - 192 $$ with domain $$ (-\infty, \infty) $$
- $$ \frac{f}{g} = \frac{6 - x^2}{(x - 4)(x + 8)} $$ with domain $$ \mathbb{R} \setminus \{ -8, 4 \} $$
- **The correct domain for $$ f + g $$ is $$ (-\infty, \infty) $$.**
The domain of $$ f + g $$ is all real numbers, which can be written as $$ (-\infty, \infty) $$.

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