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Find the domain of the function. \( f(x)=\frac{1}{\sqrt{x-9}} \) What is the domain of \( f \) ? A. \( (-\infty, 9) \cup(9, \infty) \) B. \( [9, \infty) \) C. \( (9, \infty) \) D. \( [0, \infty) \)

Ask by Bush Rogers. in the United States
Mar 16,2025

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Answer

The domain of \( f(x) = \frac{1}{\sqrt{x-9}} \) is all real numbers greater than 9, which is \( (9, \infty) \).

Solution

1. The function is given by \[ f(x)=\frac{1}{\sqrt{x-9}}. \] 2. The expression under the square root, \(x-9\), must be positive to ensure two things: - The radicand is non-negative (a requirement for real square roots). - The denominator is not zero (since division by zero is undefined). 3. Therefore, we require \[ x - 9 > 0. \] 4. Solving the inequality: \[ x > 9. \] 5. Thus, the domain of \(f\) is all \(x\) such that \(x > 9\), which in interval notation is \[ (9, \infty). \] The correct answer is C. \( (9, \infty) \).

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To find the domain of the function \( f(x) = \frac{1}{\sqrt{x-9}} \), we need to consider two main conditions. First, the expression under the square root, \( x - 9 \), must be greater than zero since we cannot take the square root of a negative number, and the square root itself cannot be zero because it is in the denominator. This leads us to the inequality \( x - 9 > 0 \), which simplifies to \( x > 9 \). Thus, the domain of \( f \) is \( (9, \infty) \), which corresponds to option C. The process of determining the domain can often be intuitive once you understand the constraints placed by operations like division and square roots. A common mistake is to overlook the fact that the radicand must be greater than zero, leading some to incorrectly include the boundary point! Keep practicing with different function types to sharpen your skills!

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