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

Let's have some fun with the functions you provided! First, we can combine them neatly. For \( f(x) + g(x) \), we have: \[ f + g = (6 - x^2) + (x^2 + 4x - 32) = 4x - 26 \] For \( f - g \): \[ f - g = (6 - x^2) - (x^2 + 4x - 32) = -2x^2 - 4x + 38 \] The product \( fg \) is calculated as follows: \[ fg = (6 - x^2)(x^2 + 4x - 32) = -x^4 - 2x^3 + 194x - 192 \] Finally, for the quotient \( \frac{f}{g} \): \[ \frac{f}{g} = \frac{6 - x^2}{x^2 + 4x - 32} \] Now, let’s discuss the domains! The domain of each function is determined by the values of \( x \) that keep the expressions valid. For \( f(x) \), it's a quadratic function and thus defined for all real numbers, \( (-\infty, \infty) \). Similarly, \( g(x) \) is also defined for all \( x \) as it is a polynomial. Therefore, \( f - g \) is also defined for all real numbers. So, the domain of \( f-g \) is: **B. The domain of \( f-g \) is \( (-\infty, \infty) \).** Now, for the multiplication \( (fg)(x) = -x^4 - 2x^3 + 194x - 192 \). You're now equipped to tackle these functions like a pro! Keep having fun with your math!