Question

The function $f(x) = \frac{x}{x^2 - 6x - 16}$, $x \in \mathbb{R} - \\{-2, 8\\}$

• A. decreases in $(-2, 8)$ and increases in $(-\infty, -2) \cup (8, \infty)$
• B. decreases in $(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$
• C. decreases in $(-\infty, -2)$ and increases in $(8, \infty)$
• D. increases in $(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$

Answer 1

Answer: (B)

decreases in $(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$

Answer 2

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:$(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$