To find the domain of the function \((f/g)(x)\), we need to consider the domains of both \(f(x)\) and \(g(x)\), but more importantly, we need to ensure that the denominator \(g(x)\) does not equal zero.
Given:
- \(f(x) = 6x^2 - 2x\)
- \(g(x) = 2x\)
The function \((f/g)(x)\) can be expressed as:
\[ \frac{f(x)}{g(x)} = \frac{6x^2 - 2x}{2x} \]
First, we identify when the denominator \(g(x) = 2x\) is zero. This occurs when:
\[ 2x = 0 \implies x = 0 \]
Next, we need to express the function \((f/g)(x)\):
\[ \frac{f(x)}{g(x)} = \frac{6x^2 - 2x}{2x} = \frac{2(3x^2 - x)}{2x} = \frac{3x^2 - x}{x} \]
This expression simplifies further except for the point where \(x = 0\), but it also reveals that \(x = 0\) is a point of discontinuity since division by zero is not allowed.
Now, we simplify the function:
\[ \frac{3x^2 - x}{x} = 3x - 1 \quad \text{for } x \neq 0 \]
Thus, \((f/g)(x) = 3x - 1\), but we cannot evaluate this at \(x = 0\).
Now we analyze the entire function:
- Domain of \(f(x)\): All real numbers (since it's a polynomial).
- Domain of \(g(x)\): All real numbers but \(x = 0\) (since \(g(x) = 2x\) cannot be zero).
The only issue in the domain arises from \(g(x) = 2x\), which means we must exclude \(x = 0\) from the domain.
Therefore, the domain of \((f/g)(x)\) is all real numbers except \(x = 0\):
The correct answer is: The set of all real numbers except \(x = 0\).
It appears that there is no option that matches this exactly in the original question's phrasing, but it is closest to the second option with \(x = 1/3\) being an irrelevant consideration.
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