The function \( f(x) = \frac{x}{x^2 - 6x - 16} \), \( x \in \mathbb{R} - \{-2, 8\} \)
- decreases in \( (-2, 8) \) and increases in \( (-\infty, -2) \cup (8, \infty) \)
- decreases in \( (-\infty, -2) \cup (-2, 8) \cup (8, \infty) \)
- decreases in \( (-\infty, -2) \) and increases in \( (8, \infty) \)
- increases in \( (-\infty, -2) \cup (-2, 8) \cup (8, \infty) \)
The Correct Option is B
Solution and Explanation
The given function is: \(f(x) = \frac{x}{x^2 - 6x - 16}\)
Step 1. Calculate the derivative \( f'(x) \):\(f'(x) = \frac{-(x^2 + 16)}{(x^2 - 6x - 16)^2}\)
Step 2. Since \( f'(x) < 0 \), the function \( f(x) \) is decreasing in all intervals where it is defined.**
Step 3. Therefore, \( f(x) \) is decreasing in \( (-\infty, -2) \cup (-2, 8) \cup (8, \infty) \).**
The Correct Answer is:\( (-\infty, -2) \cup (-2, 8) \cup (8, \infty) \)
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