Find \( f+g, f-g \), fg 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 \( \{\quad\} \). (Use a comma to separate answers as nee 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.)

To find \( f + g \), \( f - g \), \( fg \), and \( \frac{f}{g} \) using the functions \( f(x) = 5x + 2 \) and \( g(x) = x - 6 \), we can perform the following calculations: 1. **Addition \( f + g \)**: \[ (f + g)(x) = f(x) + g(x) = (5x + 2) + (x - 6) = 6x - 4 \] 2. **Subtraction \( f - g \)**: \[ (f - g)(x) = f(x) - g(x) = (5x + 2) - (x - 6) = 5x + 2 - x + 6 = 4x + 8 \] 3. **Multiplication \( fg \)**: \[ (fg)(x) = f(x) \cdot g(x) = (5x + 2)(x - 6) = 5x^2 - 30x + 2x - 12 = 5x^2 - 28x - 12 \] 4. **Division \( \frac{f}{g} \)**: \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{5x + 2}{x - 6} \] Note that \( g(x) = x - 6 \) cannot be zero, so \( g(x) \neq 0 \) when \( x \neq 6 \). Now, let's determine the domain of \( f - g \): B. The domain of \( f - g \) is \( (-\infty, \infty) \) since there are no restrictions on \( x \). Thus: - \( f + g \) is \( 6x - 4 \) - \( f - g \) is \( 4x + 8 \) - \( fg \) is \( 5x^2 - 28x - 12 \) - \( \frac{f}{g} \) is \( \frac{5x + 2}{x - 6} \) with a domain of \( x \neq 6 \). Have fun exploring these functions! 😊