Part 3 of 12
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For the given functions $f$ and $g$, complete parts (a)-(h). For parts (a)-(d), also find the domain.
\[
f(x)=1+\frac{6}{x} ; g(x)=\frac{6}{x}
\]
$(f+g)(x)=1+\frac{12}{x}$ (Simplify your answer.)
What is the domain of $f+g$ ? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The domain is $\{x \mid x \neq 0\}$
(Use integers or fractions for any numbers in the expression)
B. The domain is $\{x \mid x$ is any real number $\}$
(b) Find $(f-g)(x)$.
$(f-g)(x)=\square($ Simplify your answer $)$

Step 1 ：Given the functions \(f(x) = 1 + \frac{6}{x}\) and \(g(x) = \frac{6}{x}\), we are asked to find the function \((f+g)(x)\) and its domain, and the function \((f-g)(x)\).

Step 2 ：First, we find \((f+g)(x)\) by adding the functions \(f(x)\) and \(g(x)\) together. This gives us \((f+g)(x) = 1 + \frac{12}{x}\).

Step 3 ：The domain of a function is the set of all possible input values (often denoted as \(x\)) which will make the function 'work', and will output real numbers. In this case, the function \((f+g)(x)\) will not work when \(x = 0\), because division by zero is undefined. Therefore, the domain of \((f+g)(x)\) is \(\{x \mid x \neq 0\}\).

Step 4 ：Next, we find \((f-g)(x)\) by subtracting the function \(g(x)\) from \(f(x)\). This gives us \((f-g)(x) = 1\).

Step 5 ：Final Answer: The domain of \((f+g)(x)\) is \(\boxed{\{x \mid x \neq 0\}}\). The function \((f-g)(x)\) is equal to \(\boxed{1}\).