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To determine the domain of the function [tex]\((f \circ g)(x)\)[/tex], where [tex]\(f(x) = \frac{3x - 1}{x - 4}\)[/tex] and [tex]\(g(x) = \frac{x + 1}{x}\)[/tex], we must follow several steps.
### Step 1: Determine the domain of [tex]\(g(x)\)[/tex]
The function [tex]\(g(x) = \frac{x + 1}{x}\)[/tex] is defined for all [tex]\(x\)[/tex] except where [tex]\(x = 0\)[/tex], because division by zero is undefined. Thus, the domain of [tex]\(g(x)\)[/tex] is all real numbers except [tex]\(0\)[/tex]:
[tex]\[ \text{Domain of } g(x) = \mathbb{R} \setminus \{0\} \][/tex]
### Step 2: Evaluate [tex]\(f(g(x))\)[/tex]
Next, we need to substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex] to form the composite function [tex]\((f \circ g)(x)\)[/tex]:
[tex]\[ (f \circ g)(x) = f\left(g(x)\right) = f\left(\frac{x + 1}{x}\right) \][/tex]
So, we substitute [tex]\(\frac{x + 1}{x}\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f\left(\frac{x + 1}{x}\right) = \frac{3\left(\frac{x + 1}{x}\right) - 1}{\left(\frac{x + 1}{x}\right) - 4} \][/tex]
Now, simplify both the numerator and the denominator:
[tex]\[ \begin{align*} \text{Numerator: } & 3\left(\frac{x + 1}{x}\right) - 1 = \frac{3(x + 1) - x}{x} = \frac{3x + 3 - x}{x} = \frac{2x + 3}{x} \\ \text{Denominator: } & \frac{x + 1}{x} - 4 = \frac{x + 1 - 4x}{x} = \frac{x + 1 - 4x}{x} = \frac{1 - 3x}{x} \end{align*} \][/tex]
Combining these, we get:
[tex]\[ f\left(\frac{x + 1}{x}\right) = \frac{\frac{2x + 3}{x}}{\frac{1 - 3x}{x}} = \frac{2x + 3}{1 - 3x} \][/tex]
### Step 3: Determine when the composite function is undefined
This composite function [tex]\( \frac{2x + 3}{1 - 3x} \)[/tex] is undefined when its denominator is zero:
[tex]\[ 1 - 3x = 0 \implies x = \frac{1}{3} \][/tex]
### Step 4: Combine the domain restrictions
The composite function [tex]\((f \circ g)(x)\)[/tex] has restrictions due to both [tex]\(g(x)\)[/tex] and [tex]\(f(g(x))\)[/tex]:
- [tex]\(g(x)\)[/tex] is undefined at [tex]\(x = 0\)[/tex]
- [tex]\( (f \circ g)(x) = f\left(\frac{x + 1}{x}\right) \)[/tex] is undefined at [tex]\(x = \frac{1}{3} \)[/tex]
Thus, the domain of [tex]\((f \circ g)(x)\)[/tex] is all real numbers except [tex]\(0\)[/tex] and [tex]\(\frac{1}{3}\)[/tex]:
[tex]\[ \text{Domain of } (f \circ g)(x) = \mathbb{R} \setminus \left\{0, \frac{1}{3}\right\} \][/tex]
### Conclusion:
The domain of the composite function [tex]\((f \circ g)(x)\)[/tex] is:
[tex]\[ \boxed{(-\infty, 0) \cup (0, \frac{1}{3}) \cup (\frac{1}{3}, \infty)} \][/tex]
### Step 1: Determine the domain of [tex]\(g(x)\)[/tex]
The function [tex]\(g(x) = \frac{x + 1}{x}\)[/tex] is defined for all [tex]\(x\)[/tex] except where [tex]\(x = 0\)[/tex], because division by zero is undefined. Thus, the domain of [tex]\(g(x)\)[/tex] is all real numbers except [tex]\(0\)[/tex]:
[tex]\[ \text{Domain of } g(x) = \mathbb{R} \setminus \{0\} \][/tex]
### Step 2: Evaluate [tex]\(f(g(x))\)[/tex]
Next, we need to substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex] to form the composite function [tex]\((f \circ g)(x)\)[/tex]:
[tex]\[ (f \circ g)(x) = f\left(g(x)\right) = f\left(\frac{x + 1}{x}\right) \][/tex]
So, we substitute [tex]\(\frac{x + 1}{x}\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f\left(\frac{x + 1}{x}\right) = \frac{3\left(\frac{x + 1}{x}\right) - 1}{\left(\frac{x + 1}{x}\right) - 4} \][/tex]
Now, simplify both the numerator and the denominator:
[tex]\[ \begin{align*} \text{Numerator: } & 3\left(\frac{x + 1}{x}\right) - 1 = \frac{3(x + 1) - x}{x} = \frac{3x + 3 - x}{x} = \frac{2x + 3}{x} \\ \text{Denominator: } & \frac{x + 1}{x} - 4 = \frac{x + 1 - 4x}{x} = \frac{x + 1 - 4x}{x} = \frac{1 - 3x}{x} \end{align*} \][/tex]
Combining these, we get:
[tex]\[ f\left(\frac{x + 1}{x}\right) = \frac{\frac{2x + 3}{x}}{\frac{1 - 3x}{x}} = \frac{2x + 3}{1 - 3x} \][/tex]
### Step 3: Determine when the composite function is undefined
This composite function [tex]\( \frac{2x + 3}{1 - 3x} \)[/tex] is undefined when its denominator is zero:
[tex]\[ 1 - 3x = 0 \implies x = \frac{1}{3} \][/tex]
### Step 4: Combine the domain restrictions
The composite function [tex]\((f \circ g)(x)\)[/tex] has restrictions due to both [tex]\(g(x)\)[/tex] and [tex]\(f(g(x))\)[/tex]:
- [tex]\(g(x)\)[/tex] is undefined at [tex]\(x = 0\)[/tex]
- [tex]\( (f \circ g)(x) = f\left(\frac{x + 1}{x}\right) \)[/tex] is undefined at [tex]\(x = \frac{1}{3} \)[/tex]
Thus, the domain of [tex]\((f \circ g)(x)\)[/tex] is all real numbers except [tex]\(0\)[/tex] and [tex]\(\frac{1}{3}\)[/tex]:
[tex]\[ \text{Domain of } (f \circ g)(x) = \mathbb{R} \setminus \left\{0, \frac{1}{3}\right\} \][/tex]
### Conclusion:
The domain of the composite function [tex]\((f \circ g)(x)\)[/tex] is:
[tex]\[ \boxed{(-\infty, 0) \cup (0, \frac{1}{3}) \cup (\frac{1}{3}, \infty)} \][/tex]
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