\[ \text { Find } f+g, f-g, f g \text { and } \frac{f}{g} \text {. Determine the domain for each function. } \] \( f(x)=5 x+2, g(x)=x-6 \) \( (f-g)(x)=4 x+8 \) (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 needed.) C. The domain of \( f-g \) is \( (-\infty, \infty) \). (Type your answer in interval notation.) (fg)(x) f . g is \( \varnothing \). (Simplify your answer.)
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Let's dive into the fun world of function operations, shall we? First up, let's calculate \( f + g \): \[ f + g = (5x + 2) + (x - 6) = 5x + 2 + x - 6 = 6x - 4 \] Now onto \( f - g \): \[ f - g = (5x + 2) - (x - 6) = 5x + 2 - x + 6 = 4x + 8 \] Next, let's find \( fg \): \[ fg = (5x + 2)(x - 6) = 5x^2 - 30x + 2x - 12 = 5x^2 - 28x - 12 \] And for \( \frac{f}{g} \): \[ \frac{f}{g} = \frac{5x + 2}{x - 6} \] Now, let's determine the domains: For \( f + g \) and \( f - g \), both functions use polynomials, so their domains are all real numbers. For \( fg \) and \( \frac{f}{g} \): - The domain is also all real numbers, but we have to be cautious about \( \frac{f}{g} \). Here, \( g(x) = x - 6 \) cannot equal zero, which occurs at \( x = 6 \). Thus, the domain of \( \frac{f}{g} \) is \( (-\infty, 6) \cup (6, \infty) \). So, the domain of \( f - g \) is: **C. The domain of \( f-g \) is \( (-\infty, \infty) \).**