Consider the following functions.
\[
f(x)=\frac{1}{x} \text { and } g(x)=\frac{x+1}{3}
\]
Step 2 of 2 : Find the formula for $(g \circ f)(x)$ and simplify your answer. Then find the domain for $(g \circ f)(x)$. Round your answer to two decimal places, if necessary.
Answer
\[
(g \circ f)(x)=
\]
Domain $=$

Step 1 ：Consider the following functions. \[f(x)=\frac{1}{x} \text { and } g(x)=\frac{x+1}{3}\]

Step 2 ：The function \((g \circ f)(x)\) means that we are applying the function \(g\) to the result of the function \(f(x)\). So, we need to substitute \(f(x)\) into \(g(x)\).

Step 3 ：The domain of a function is the set of all possible input values (often the "x" variable), which produce a valid output from a particular function. The domain of \((g \circ f)(x)\) is all the values of \(x\) for which both \(f(x)\) and \(g(f(x))\) are defined.

Step 4 ：In this case, \(f(x)\) is defined for all \(x\) except \(0\), because we cannot divide by zero. After we find the formula for \((g \circ f)(x)\), we will check if there are any other restrictions on the domain.

Step 5 ：Substitute \(f(x)\) into \(g(x)\) to get \(g(f(x)) = \frac{1}{3} + \frac{1}{3x}\)

Step 6 ：Simplify \(g(f(x))\) to get \(g(f(x)) = \frac{x + 1}{3x}\)

Step 7 ：The domain of this function is all real numbers except for those that make the denominator equal to zero. In this case, the denominator is \(3x\), so the domain is all real numbers except for \(x = 0\).

Step 8 ：Final Answer: \[\boxed{(g \circ f)(x)=\frac{x + 1}{3x}}\] Domain \[\boxed{= \mathbb{R} - \{0\}}\]