\( 1 \leftarrow \) For the functions \( f(x)=6-x^{2} \) and \( g(x)=x^{2}+4 x-32 \), find \( f+g, f-g \), \( f \), and \( \frac{f}{g} \). Determine the domain for each function. \( (f+g)(x)=4 x-26 \) (Simplify your answer.). What is the domain of \( f+g \) ? Select the correct choice below and, if necessary, fill in the answer box to complete \( y \) 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 \( \varnothing \).

To find \( f+g \), \( f-g \), \( f \), and \( \frac{f}{g} \), let's start by calculating these functions: 1. **Function Definitions**: - \( f(x) = 6 - x^{2} \) - \( g(x) = x^{2} + 4x - 32 \) 2. **Calculating \( f + g \)**: \[ (f + g)(x) = f(x) + g(x) = (6 - x^2) + (x^2 + 4x - 32) = 4x - 26 \] 3. **Calculating \( f - g \)**: \[ (f - g)(x) = f(x) - g(x) = (6 - x^2) - (x^2 + 4x - 32) = 6 - x^2 - x^2 - 4x + 32 = -2x^2 - 4x + 38 \] 4. **Calculating \( \frac{f}{g} \)**: \[ \frac{f}{g}(x) = \frac{6 - x^2}{x^2 + 4x - 32} \] 5. **Finding the domain**: - For both \( f \) and \( g \), they are polynomials, which means their domain is all real numbers, \( (-\infty, \infty) \). - For \( \frac{f}{g} \), we must ensure \( g(x) \neq 0 \). Setting \( g(x) = 0 \): \[ x^2 + 4x - 32 = 0 \] Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 128}}{2} = \frac{-4 \pm 12}{2} \] This results in roots \( x = 4 \) and \( x = -8 \). Therefore, \( g(x) = 0 \) when \( x = 4 \) and \( x = -8 \). Thus, the domain for \( f+g \) is \( (-\infty, \infty) \), and the final answer for the domain of \( f+g \) is: B. The domain of \( f+g \) is \((-∞, ∞)\).