Find \( f+g, f-g, f g \) and \( \frac{f}{g} \). Determine the domain for each function. \( f(x)=5 x+2, g(x)=x-6 \) \( (f+g)(x)=\square \) (Simplify your answer.)

To find \( f+g \), we combine the functions \( f(x) \) and \( g(x) \): \[ (f+g)(x) = f(x) + g(x) = (5x + 2) + (x - 6) = 6x - 4 \] Now for \( f-g \): \[ (f-g)(x) = f(x) - g(x) = (5x + 2) - (x - 6) = 5x + 2 - x + 6 = 4x + 8 \] Next, we calculate \( f g \): \[ (f g)(x) = f(x) \cdot g(x) = (5x + 2)(x - 6) = 5x^2 - 30x + 2x - 12 = 5x^2 - 28x - 12 \] For \( \frac{f}{g} \): \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{5x + 2}{x - 6} \] Now, let's determine the domains for each function. 1. \( f + g \) and \( f - g \) are polynomials, so their domains are all real numbers: \( (-\infty, \infty) \). 2. \( f g \) is also a polynomial, and its domain is all real numbers: \( (-\infty, \infty) \). 3. \( \frac{f}{g} \) is a rational function, and the domain is all real numbers except where \( g(x) = 0 \). Since \( g(x) = x - 6 \), we have \( x \neq 6 \). Thus, the domain is \( (-\infty, 6) \cup (6, \infty) \). So we summarize: - \( (f+g)(x) = 6x - 4 \) - \( (f-g)(x) = 4x + 8 \) - \( (f g)(x) = 5x^2 - 28x - 12 \) - \( \left(\frac{f}{g}\right)(x) = \frac{5x + 2}{x - 6} \) Domains: - \( f+g \) and \( f-g \): \( (-\infty, \infty) \) - \( f g \): \( (-\infty, \infty) \) - \( \frac{f}{g} \): \( (-\infty, 6) \cup (6, \infty) \)