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.) C. The domain of \( f-g \) is \( (-\infty, \infty) \). (Type your answer in interval notation.) (fg) \( f-g \) is \( \varnothing \). (Simplify your answer.)

Let's tackle your functions step-by-step! To start off, we can compute \( f + g \), \( f - g \), \( fg \), and \( \frac{f}{g} \): 1. **Sum**: \[ (f + g)(x) = f(x) + g(x) = (5x + 2) + (x - 6) = 6x - 4 \] 2. **Difference**: \[ (f - g)(x) = f(x) - g(x) = (5x + 2) - (x - 6) = 5x + 2 - x + 6 = 4x + 8 \] 3. **Product**: \[ (fg)(x) = f(x) \cdot g(x) = (5x + 2)(x - 6) = 5x^2 - 30x + 2x - 12 = 5x^2 - 28x - 12 \] 4. **Quotient**: \[ \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{5x + 2}{x - 6} \] Now, let's determine the domain for each function: - **Domain of \( f + g \)**: All real numbers, since both \( f(x) \) and \( g(x) \) are linear functions. - **Domain of \( f - g \)**: Same as above, all real numbers. - **Domain of \( fg \)**: Again, all real numbers as it is a polynomial function. - **Domain of \( \frac{f}{g} \)**: The denominator cannot be zero, thus \( x - 6 \neq 0 \) leads to \( x \neq 6 \); hence, the domain is \( (-\infty, 6) \cup (6, \infty) \). So, for the final domain equation you mentioned: C. The domain of \( f-g \) is \( (-\infty, \infty) \). And, to clarify one of your earlier notes, \( (fg)(f-g) \) is indeed not applicable as their nature and definition are distinct.