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 each function. \( (1+g \not(x)=4 x-<0 \) (ommpury your answer.) What is the domain of \( f+g \) ? Select the correct choice below and, if necessary, fill in the answer box to complete 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 \( \varnothing \). ( \( f-g)(x)=\square \) (Simplify your answer.)

To find \( f+g, f-g, fg, \) and \( \frac{f}{g} \), we need to perform some basic algebraic operations on the functions given. 1. First, let's find \( f+g \): \[ f+g = (6 - x^2) + (x^2 + 4x - 32) = 6 - x^2 + x^2 + 4x - 32 = 4x - 26. \] 2. Now, for \( f-g \): \[ f-g = (6 - x^2) - (x^2 + 4x - 32) = 6 - x^2 - x^2 - 4x + 32 = -2x^2 - 4x + 38. \] 3. Next, we compute \( fg \): \[ fg = (6 - x^2)(x^2 + 4x - 32). \] Using the distributive property: \[ fg = 6x^2 + 24x - 192 - x^4 - 4x^3 + 32x^2 = -x^4 - 4x^3 + 38x^2 + 24x - 192. \] 4. Finally, let's find \( \frac{f}{g} \): \[ \frac{f}{g} = \frac{6 - x^2}{x^2 + 4x - 32}. \] Next, let's determine the domain for each function: **For \( f(x) \):** The function is a quadratic and is defined for all \( x \), so the domain of \( f \) is \( (-\infty, \infty) \). **For \( g(x) \):** We need to consider where \( g(x) \) is non-zero since division by zero is undefined. The quadratic \( x^2 + 4x - 32 \) can be factored or solved for roots using the quadratic formula, yielding: \[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-32)}}{2(1)} = \frac{-4 \pm \sqrt{144}}{2} = \frac{-4 \pm 12}{2}. \] Thus, the roots are \( x = 4 \) and \( x = -8 \). The function \( g(x) \) is zero at these points. The domain of \( g(x) \) is thus \( (-\infty, -8) \cup (-8, 4) \cup (4, \infty) \). **Putting this together for \( f+g \):** Since both \( f \) and \( g \) are defined for all \( x \) except where \( g(x) = 0 \), the domain of \( f+g \) will be \( (-\infty, -8) \cup (-8, 4) \cup (4, \infty) \). So the answer to the domain of \( f+g \) is: **A**: The domain of \( f+g \) is \( (-\infty, -8) \cup (-8, 4) \cup (4, \infty) \). Regarding \( f-g \): \[ f-g = -2x^2 - 4x + 38. \] This quadratic is defined for all \( x \), so the answer is \( (-\infty, \infty) \). There you go! You now have a comprehensive understanding of the operations and domain for these functions. Isn't algebra fun when it all comes together?