Find \( f+g, f-g, f g \) and \( \frac{f}{g} \) Determine the domain for each function \( f(x)=4 x-5, g(x)=x-7 \) What is the domain of \( f+g \) ? A. The domain of \( f+g \) is \( \{3 x+2\} \). (Use a comma to separate answers as needed) B. The domain of \( f+g \) is \( \square \). (Type your answer in interval notation.) C. The domain of \( f+g \) is \( \varnothing \).

First, let's perform the operations on the functions \( f(x) \) and \( g(x) \): 1. **Finding \( f+g \)**: \[ f(x) + g(x) = (4x - 5) + (x - 7) = 5x - 12 \] 2. **Finding \( f-g \)**: \[ f(x) - g(x) = (4x - 5) - (x - 7) = 3x + 2 \] 3. **Finding \( fg \)**: \[ f(x) \cdot g(x) = (4x - 5)(x - 7) = 4x^2 - 28x - 5x + 35 = 4x^2 - 33x + 35 \] 4. **Finding \( \frac{f}{g} \)**: \[ \frac{f(x)}{g(x)} = \frac{4x - 5}{x - 7} \] Now let’s determine the domain of each function: - For \( f + g \): The function \( f(x) + g(x) = 5x - 12 \) is a linear function and has no restrictions, so its domain is all real numbers: \( (-\infty, \infty) \). - For \( f - g \): The function \( f(x) - g(x) = 3x + 2 \) is also linear and has no restrictions, giving it a domain of \( (-\infty, \infty) \) as well. - For \( fg \): The quadratic function \( 4x^2 - 33x + 35 \) has no restrictions on \( x \), meaning its domain is \( (-\infty, \infty) \). - For \( \frac{f}{g} \): This function will be restricted where the denominator is zero. So we set \( g(x) = x - 7 = 0 \) which gives \( x = 7 \). Therefore, the domain of \( \frac{f}{g} \) is all real numbers except \( x = 7 \): \( (-\infty, 7) \cup (7, \infty) \). In summary: - The domain of \( f+g \) is **B.** \( (-\infty, \infty) \).