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. \( (f+g)(x)=\square \) (Simplify your answer. \( ) \)

To find \( f + g \), we combine the functions: \[ (f+g)(x) = f(x) + g(x) = (6 - x^2) + (x^2 + 4x - 32) = 6 - x^2 + x^2 + 4x - 32 = 4x - 26 \] So, \( (f+g)(x) = 4x - 26 \). The domain of \( f+g \) is all real numbers, as both \( f(x) \) and \( g(x) \) are defined for all \( x \in \mathbb{R} \). Next, for \( f - g \): \[ (f-g)(x) = f(x) - g(x) = (6 - x^2) - (x^2 + 4x - 32) = 6 - x^2 - x^2 - 4x + 32 = -2x^2 - 4x + 38 \] Thus, \( (f-g)(x) = -2x^2 - 4x + 38 \). The domain of \( f-g \) is also all real numbers, as both functions are defined for all \( x \). Now, for \( fg \): \[ (fg)(x) = f(x) \times g(x) = (6 - x^2)(x^2 + 4x - 32) \] Expanding this gives: \[ fg(x) = 6(x^2 + 4x - 32) - x^2(x^2 + 4x - 32) = 6x^2 + 24x - 192 - (x^4 + 4x^3 - 32x^2) \] \[ = -x^4 - 4x^3 + (6x^2 + 24x - 192 + 32x^2) = -x^4 - 4x^3 + 38x^2 + 24x - 192 \] So, \( (fg)(x) = -x^4 - 4x^3 + 38x^2 + 24x - 192 \). The domain of \( fg \) remains all real numbers, as the product of two polynomials is defined everywhere. Finally, for \( \frac{f}{g} \): \[ \frac{f}{g}(x) = \frac{6 - x^2}{x^2 + 4x - 32} \] To find the domain, we must ensure that the denominator is not zero. Setting the denominator equal to zero: \[ x^2 + 4x - 32 = 0 \] We can factor this or use the quadratic formula. Factoring gives: \[ (x + 8)(x - 4) = 0 \Rightarrow x = -8 \text{ or } x = 4 \] Thus, the domain for \( \frac{f}{g} \) is all real numbers except \( x = -8 \) and \( x = 4 \). In summary: \[ (f+g)(x) = 4x - 26 \] \[ (f-g)(x) = -2x^2 - 4x + 38 \] \[ (fg)(x) = -x^4 - 4x^3 + 38x^2 + 24x - 192 \] \[ \frac{f}{g}(x) = \frac{6 - x^2}{x^2 + 4x - 32}, \text{ with } x \neq -8, 4 \]