For the functions \( f(x)=6-x^{2} \) and \( g(x)=x^{2}+4 x-32 \), find \( f+g, f-g \), \( fg \), and \( \frac{f}{g} \). Determine the domain for each function.

To find \( f + g \), \( f - g \), \( fg \), and \( \frac{f}{g} \) for the functions \( f(x) = 6 - x^2 \) and \( g(x) = x^2 + 4x - 32 \), let's evaluate each operation step by step: 1. **Sum of the functions**: \[ f + g = (6 - x^2) + (x^2 + 4x - 32) = 6 - x^2 + x^2 + 4x - 32 = 4x - 26. \] 2. **Difference of the functions**: \[ f - g = (6 - x^2) - (x^2 + 4x - 32) = 6 - x^2 - x^2 - 4x + 32 = -2x^2 - 4x + 38. \] 3. **Product of the functions**: \[ fg = (6 - x^2)(x^2 + 4x - 32). \] Using the distributive property: \[ fg = 6x^2 + 24x - 192 - x^4 - 4x^3 + 32x^2 = -x^4 - 4x^3 + 38x^2 + 24x - 192. \] 4. **Quotient of the functions**: \[ \frac{f}{g} = \frac{6 - x^2}{x^2 + 4x - 32}. \] This is a rational function, so we must determine where \( g(x) \neq 0 \): Solving \( x^2 + 4x - 32 = 0 \) using the quadratic formula: \[ x = \frac{-4 \pm \sqrt{16 + 128}}{2} = \frac{-4 \pm 12}{2} \Rightarrow x = 4 \quad \text{or} \quad x = -8. \] Therefore, the quotient is undefined at \( x = 4 \) and \( x = -8 \). For determining the domain of each function: - The domain of \( f(x) = 6 - x^2 \) is all real numbers, \( \mathbb{R} \). - The domain of \( g(x) = x^2 + 4x - 32 \) is also all real numbers, \( \mathbb{R} \). - The domain of \( \frac{f}{g} \) is all real numbers except where \( g(x) = 0 \), so: \[ \text{Domain of } \frac{f}{g} = \mathbb{R} \setminus \{ -8, 4 \}. \] In summary: - \( f + g = 4x - 26 \) - \( f - g = -2x^2 - 4x + 38 \) - \( fg = -x^4 - 4x^3 + 38x^2 + 24x - 192 \) - \( \frac{f}{g} = \frac{6 - x^2}{x^2 + 4x - 32} \), with domain \( \mathbb{R} \setminus \{-8, 4\} \).