For the functions $f(x)=\frac{x}{x+3}$ and $g(x)=\frac{13}{x}$, find the composition $f \circ g$ and simplify your answer as much as possible. Write the domain using interval notation. \[ (f-g)(x)=\square \] Domain of $f=g: \square$ \begin{tabular}{|c|c|c|} \hline$\frac{\mathrm{D}}{\mathrm{D}}$ & $\mathrm{c}^{\mathrm{a}}$ & $\sqrt{\square}$ \\ \hline 미미 & $(\square, \square)$ & {$[0,0]$} \\ \hline qua & $(0,0]$ & {$[0, \square)$} \\ \hline$\varnothing$ & $\infty$ & $-\infty$ \\ \hline$x$ & & 5 \\ \hline \end{tabular}

Step 1 ：The composition of functions is a function that applies one function to the results of another. In this case, we need to substitute $g(x)$ into $f(x)$ to get $f(g(x))$.

Step 2 ：The domain of a function is the set of all possible input values (x-values) which will produce a valid output. To find the domain of the composition $f(g(x))$, we need to consider the domains of both $f(x)$ and $g(x)$, and exclude any x-values that would make either function undefined.

Step 3 ：For the function $f(x)=\frac{x}{x+3}$, the denominator cannot be zero, so $x \neq -3$. For the function $g(x)=\frac{13}{x}$, the denominator cannot be zero, so $x \neq 0$.

Step 4 ：When we substitute $g(x)$ into $f(x)$, we get $f(g(x)) = \frac{13}{x*(3 + 13/x)}$.

Step 5 ：Simplifying the above expression, we get $f(g(x)) = \frac{13}{3x + 13}$.

Step 6 ：The denominator of the function $f(g(x))$ is zero when $x = -\frac{13}{3}$. So, the domain of $f(g(x))$ is all real numbers except $x = 0$ (from the domain of $g(x)$) and $x = -\frac{13}{3}$ (from the domain of $f(g(x))$).

Step 7 ：In interval notation, this is $(-\infty, -\frac{13}{3}) \cup (-\frac{13}{3}, 0) \cup (0, \infty)$.

Step 8 ：So, the composition $f(g(x))$ is $\frac{13}{3x + 13}$ and its domain is $(-\infty, -\frac{13}{3}) \cup (-\frac{13}{3}, 0) \cup (0, \infty)$.

Step 9 ：Final Answer: $(f-g)(x)=\boxed{\frac{13}{3x + 13}}$ and Domain of $f=g: \boxed{(-\infty, -\frac{13}{3}) \cup (-\frac{13}{3}, 0) \cup (0, \infty)}$.