The function f(x) = frac{x}{x^2-6x-16},where | vedclass

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(B) Given $f(x) = \frac{x}{x^2-6x-16}$.Using the quotient rule,$f'(x) = \frac{(x^2-6x-16)(1) - x(2x-6)}{(x^2-6x-16)^2}$.Simplifying the numerator: $x^

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

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(B) Given $f(x) = \frac{x}{x^2-6x-16}$.Using the quotient rule,$f'(x) = \frac{(x^2-6x-16)(1) - x(2x-6)}{(x^2-6x-16)^2}$.Simplifying the numerator: $x^2 - 6x - 16 - 2x^2 + 6x = -x^2 - 16 = -(x^2 + 16)$.Thus,$f'(x) = \frac{-(x^2+16)}{(x^2-6x-16)^2}$.Since $x^2+16 > 0$ for all $x \in \mathbb{R}$ and $(x^2-6x-16)^2 > 0$ for all $x 
eq -2, 8$,it follows that $f'(x) Therefore,the function $f(x)$ is strictly decreasing in its entire domain $(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$.

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

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