To determine the domain of the function \(\frac{f}{g}(x)\) where \(f(x) = -\frac{1}{x}\) and \(g(x) = \sqrt{3x - 9}\), we need to find values of \(x\) for which both \(f(x)\) and \(g(x)\) are defined and \(g(x) \neq 0\).
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For \(f(x) = -\frac{1}{x}\):
- \(f(x)\) is defined for all \(x\) except \(x = 0\) (since division by zero is undefined).
- Therefore, the condition for \(f(x)\) is \(x \neq 0\).
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For \(g(x) = \sqrt{3x - 9}\):
- The expression inside the square root must be non-negative: \[ 3x - 9 \geq 0 \] \[ 3x \geq 9 \] \[ x \geq 3 \]
- Additionally, \(g(x)\) must be greater than zero, since we cannot have \(g(x) = 0\) in the denominator of the fraction: \[ \sqrt{3x - 9} > 0 \implies 3x - 9 > 0 \implies 3x > 9 \implies x > 3 \]
So, for \(g(x)\) to be defined and non-zero, we require \(x > 3\).
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Combining Conditions:
- We have from \(f(x)\) that \(x \neq 0\) and from \(g(x)\) that \(x > 3\).
- Therefore, the domain of \(\frac{f}{g}(x)\) is \(x > 3\).
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Final Domain:
- The domain can be expressed in interval notation as: \[ (3, \infty) \]
Thus, the answer is \((3, \infty)\).
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