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.

Question

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