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.)

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 \( (-\infty, \infty) \).**