Find \( f+g, f-g, f g \) and \( \frac{f}{g} \). Determine the domain for each function. \( f(x)=5 x+2, g(x)=x-6 \) \( (f-g)(x)=4 x+8 \) (Simplify your answer.) What is the domain of \( f-g \) ? 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 \( \varnothing \). (fg) \( (x)=\square \) (Simplify your answer.)

Let's find the functions \( f+g \), \( f-g \), \( fg \), and \( \frac{f}{g} \). 1. **Addition**: \[ (f+g)(x) = f(x) + g(x) = (5x + 2) + (x - 6) = 6x - 4 \] 2. **Subtraction**: \[ (f-g)(x) = f(x) - g(x) = (5x + 2) - (x - 6) = 5x + 2 - x + 6 = 4x + 8 \] 3. **Multiplication**: \[ (fg)(x) = f(x) \cdot g(x) = (5x + 2)(x - 6) = 5x^2 - 30x + 2x - 12 = 5x^2 - 28x - 12 \] 4. **Division**: \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{5x + 2}{x - 6} \] The division is defined as long as the denominator is not zero, so we need \( x - 6 \neq 0 \) which means \( x \neq 6 \). Now, let's determine the domains: - \( f(x) \) and \( g(x) \) are both linear, which are defined for all real numbers. Thus: - For \( (f+g)(x) \), the domain is \( (-\infty, \infty) \). - For \( (f-g)(x) \), the domain is also \( (-\infty, \infty) \). - For \( (fg)(x) \), the domain is \( (-\infty, \infty) \). - For \( \left(\frac{f}{g}\right)(x) \), the domain is \( (-\infty, 6) \cup (6, \infty) \) since \( x \) cannot be 6. Thus, the answers are: - \( (f-g)(x) = 4x + 8 \) - Domain of \( f-g \): B. The domain of \( f-g \) is \( (-\infty, \infty) \). - \( (fg)(x) = 5x^2 - 28x - 12 \)