Problem

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}

Answer

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Answer

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

Steps

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 $f=g: \boxed{(-\infty, -\frac{13}{3}) \cup (-\frac{13}{3}, 0) \cup (0, \infty)}$.

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