Question

Determine the domain of the function left-parenthesis lower f circle lower g right-parenthesis left-parenthesis x right-parenthesis where lower f left-parenthesis x right-parenthesis equals Start Fraction 3 x minus 1 over x minus 4 End Fraction and lower g left-parenthesis x right-parenthesis equals Start Fraction x plus 1 over x End Fraction.
(1 point)
Responses

left-parenthesis negative infinity comma one-third right-parenthesis union left-parenthesis one-third comma infinity right-parenthesis
Image with alt text: left-parenthesis negative infinity comma one-third right-parenthesis union left-parenthesis one-third comma infinity right-parenthesis

left-parenthesis negative infinity comma 0 right-parenthesis union left-parenthesis 0 comma one-third right-parenthesis union left-parenthesis one-third comma infinity right-parenthesis
Image with alt text: left-parenthesis negative infinity comma 0 right-parenthesis union left-parenthesis 0 comma one-third right-parenthesis union left-parenthesis one-third comma infinity right-parenthesis

left-parenthesis negative infinity comma 4 right-parenthesis union left-parenthesis 4 comma infinity right-parenthesis
Image with alt text: left-parenthesis negative infinity comma 4 right-parenthesis union left-parenthesis 4 comma infinity right-parenthesis

left-parenthesis negative infinity comma 0 right-parenthesis union left-parenthesis one-third comma infinity right-parenthesis

Answer 1

To determine the domain of the function \((f \circ g)(x)\), we need to analyze both functions \(f(x)\) and \(g(x)\).

1. **Identify the domains of \(f(x)\) and \(g(x)\)**:

- **Function \(f(x) = \frac{3x - 1}{x - 4}\)**:
- The denominator \(x - 4\) cannot be zero.
- Therefore, \(x \neq 4\).
- The domain of \(f(x)\) is \( (-\infty, 4) \cup (4, \infty) \).

- **Function \(g(x) = \frac{x + 1}{x}\)**:
- The denominator \(x\) cannot be zero.
- Therefore, \(x \neq 0\).
- The domain of \(g(x)\) is \( (-\infty, 0) \cup (0, \infty) \).

2. **Determine the input for \(f(g(x))\)**:
- For \(f(g(x))\) to be defined, \(g(x)\) must be in the domain of \(f\).
- Therefore, we need to check where \(g(x) \neq 4\) (since the domain of \(f\) prohibits \(x = 4\)).

3. **Find where \(g(x) = 4\)**:
\[
g(x) = \frac{x + 1}{x} = 4 \Rightarrow x + 1 = 4x \Rightarrow 1 = 3x \Rightarrow x = \frac{1}{3}
\]
- So, \(g(x) = 4\) when \(x = \frac{1}{3}\).

4. **Combine all restrictions**:
- The domain of \(g(x)\) is \( (-\infty, 0) \cup (0, \infty) \).
- We exclude \(x = \frac{1}{3}\) from the valid inputs since this makes \(g(x) = 4\), which is not in the domain of \(f\).
- Thus, the domain for \(f(g(x))\) will be:
- From \( (-\infty, 0) \), no exclusions.
- From \( (0, \infty) \), we exclude \(x = \frac{1}{3}\).

Hence, the final domain of \((f \circ g)(x)\) is:
\[
(-\infty, 0) \cup \left(0, \frac{1}{3}\right) \cup \left(\frac{1}{3}, \infty\right)
\]

The answer is:
\[
\boxed{(-\infty, 0) \cup (0, \frac{1}{3}) \cup (\frac{1}{3}, \infty)}
\]