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

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To determine the domain of $$ \frac{1}{g} $$, where $$ g(x) = x^2 + 4x - 32 $$, we need to identify all real numbers $$ x $$ for which $$ g(x) $$ is not equal to zero (since division by zero is undefined).

1. **Find the roots of $$ g(x) $$:**
   $$
   g(x) = x^2 + 4x - 32 = 0
   $$
   Solving this quadratic equation:
   $$
   x = \frac{-4 \pm \sqrt{(4)^2 - 4(1)(-32)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 128}}{2} = \frac{-4 \pm \sqrt{144}}{2} = \frac{-4 \pm 12}{2}
   $$
   This gives two solutions:
   $$
   x = \frac{8}{2} = 4 \quad \text{and} \quad x = \frac{-16}{2} = -8
   $$

2. **Determine the domain of $$ \frac{1}{g} $$:**
   The function $$ \frac{1}{g} $$ is defined for all real numbers except where $$ g(x) = 0 $$. Therefore, the domain excludes $$ x = 4 $$ and $$ x = -8 $$.

In interval notation, the domain is:
   $$
   (-\infty, -8) \cup (-8, 4) \cup (4, \infty)
   $$

**Correct Choice:**

**B. The domain of $$ \frac{1}{g} $$ is $$ (-\infty, -8) \cup (-8, 4) \cup (4, \infty) $$.**
The domain of $$ \frac{1}{g} $$ is all real numbers except $$ x = -8 $$ and $$ x = 4 $$. In interval notation, this is $$ (-\infty, -8) \cup (-8, 4) \cup (4, \infty) $$.

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