What is the domain of the function \( g(x)=\sqrt{\frac{x-7}{x+9}} \) ? The domain is \( \square \). (Type your answer in interval notation. Simplify your answer. Use integers or fractions for any numbers in the expression.)
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To determine the domain of the function \( g(x) = \sqrt{\frac{x-7}{x+9}} \), we need to ensure that the expression inside the square root is non-negative. This means \( \frac{x-7}{x+9} \geq 0 \). The critical points occur when \( x-7 = 0 \) or \( x+9 = 0 \), which gives us \( x = 7 \) and \( x = -9 \). Next, we analyze the intervals determined by these points: 1. **Interval \( (-\infty, -9) \)**: Choose \( x = -10 \): \( \frac{-10-7}{-10+9} = \frac{-17}{-1} = 17 \) (positive). 2. **Interval \( (-9, 7) \)**: Choose \( x = 0 \): \( \frac{0-7}{0+9} = \frac{-7}{9} \) (negative). 3. **Interval \( (7, \infty) \)**: Choose \( x = 8 \): \( \frac{8-7}{8+9} = \frac{1}{17} \) (positive). Thus, the function is non-negative in the intervals \( (-\infty, -9) \) and \( (7, \infty) \). The points \( x = -9 \) (undefined) and \( x = 7 \) (zero) must be treated accordingly: Consequently, the domain of \( g(x) \) is \( (-\infty, -9) \cup [7, \infty) \). So, the domain is \( \square (-\infty, -9) \cup [7, \infty) \).