Problem

For the functions $f(x)=\frac{x}{x-1}$ and $g(x)=\frac{11}{x}$, find the composition $f \circ g$ and simplify your answer as much as possible. Write the domain using interval notation.
\[
\begin{array}{l}
(f \circ g)(x)=\square \\
\text { Domain of } f \circ g:[ \\
\end{array}
\]

Answer

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Answer

\(\boxed{\text { Domain of } f \circ g: (-\infty, 11) \cup (11, \infty)}\)

Steps

Step 1 :Given the functions $f(x)=\frac{x}{x-1}$ and $g(x)=\frac{11}{x}$, we are asked to find the composition $f \circ g$ and simplify the answer as much as possible. We also need to write the domain using interval notation.

Step 2 :The composition of two functions, $f \circ g$, is defined as $(f \circ g)(x) = f(g(x))$. So, to find $f \circ g$, we substitute $g(x)$ into $f(x)$, which gives us $f(g(x)) = \frac{g(x)}{g(x)-1}$.

Step 3 :Substituting $g(x) = \frac{11}{x}$ into the equation, we get $f(g(x)) = \frac{\frac{11}{x}}{\frac{11}{x}-1} = \frac{11}{x*(-1 + 11/x)}$.

Step 4 :Simplifying the above expression, we get $f(g(x)) = -\frac{11}{x-11}$.

Step 5 :Now, we need to find the domain of this function. The domain of a rational function is all real numbers except for those that make the denominator equal to zero. So, we need to find the x-values that make $x - 11 = 0$.

Step 6 :Solving the equation $x - 11 = 0$, we find that $x = 11$.

Step 7 :Therefore, the domain of $f \circ g$ is all real numbers except 11. In interval notation, this is $(-\infty, 11) \cup (11, \infty)$.

Step 8 :\(\boxed{(f \circ g)(x)=-\frac{11}{x-11}}\)

Step 9 :\(\boxed{\text { Domain of } f \circ g: (-\infty, 11) \cup (11, \infty)}\)

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