Find \( f+g, f-g, f g \) and \( \frac{1}{g} \). Determine the domain for each function. \( f(x)=5 x+2, g(x)=x-6 \) What is the domain of \( f g \) ? A. The domain of \( f g \) is \( (-\infty, \infty) \). (Type your answer in interval notation.) C. The domain of \( f g \) is \( \} \). (Use a commain of to separate answers as needed.) \( \left(\frac{f}{g}\right)(x)=\square \) (Simplify your answer.)

To find \( f + g \), \( f - g \), \( fg \), and \( \frac{1}{g} \) for the functions \( f(x) = 5x + 2 \) and \( g(x) = x - 6 \): 1. **Addition \( f + g \)**: \[ f(x) + g(x) = (5x + 2) + (x - 6) = 6x - 4 \] 2. **Subtraction \( f - g \)**: \[ f(x) - g(x) = (5x + 2) - (x - 6) = 4x + 8 \] 3. **Multiplication \( fg \)**: \[ fg(x) = (5x + 2)(x - 6) = 5x^2 - 30x + 2x - 12 = 5x^2 - 28x - 12 \] 4. **Inverse of g \( \frac{1}{g} \)**: \[ \frac{1}{g(x)} = \frac{1}{x - 6} \] The domain of \( f \) and \( g \) is all real numbers, but for \( \frac{1}{g} \), it is where \( g(x) \neq 0 \), so the domain is: \[ (-\infty, 6) \cup (6, \infty) \] **Domain of \( fg \)**: Since both \( f(x) \) and \( g(x) \) are defined for all \( x \), the function \( fg(x) \) is also defined for all \( x \). Therefore, the domain of \( fg \) is: \[ (-\infty, \infty) \] Finally, simplifying \( \frac{f}{g} \): \[ \left(\frac{f}{g}\right)(x) = \frac{5x + 2}{x - 6} \] In summary: - \( f + g = 6x - 4 \) - \( f - g = 4x + 8 \) - \( fg = 5x^2 - 28x - 12 \) - \( \frac{1}{g} = \frac{1}{x - 6} \) - Domain of \( fg = (-\infty, \infty) \) - \( \left(\frac{f}{g}\right)(x) = \frac{5x + 2}{x - 6} \)