1. Given function

$f(x)=\frac{x}{x^2-6x-16}$

First, factor the denominator:

$x^2-6x-16=(x-8)(x+2)$

So the domain is:

$x\in \mathbb{R}\setminus\{-2,8\}$

Thus we study monotonicity on the intervals:

$(-\infty,-2),\quad (-2,8),\quad (8,\infty)$

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2. Find the derivative

Using the quotient rule,

$f'(x)=\frac{(x^2-6x-16)\cdot 1-x(2x-6)}{(x^2-6x-16)^2}$

Now simplify the numerator:

$f'(x)=\frac{x^2-6x-16-2x^2+6x}{(x^2-6x-16)^2}$

$f'(x)=\frac{-x^2-16}{(x^2-6x-16)^2}$

$f'(x)= -\frac{x^2+16}{(x^2-6x-16)^2}$

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3. Determine the sign of $f'(x)$

Observe:

• $x^2+16>0$ for all real $x$
• $(x^2-6x-16)^2>0$ for all $x\neq -2,8$$

Therefore,

$f'(x)=-\frac{\text{positive}}{\text{positive}}<0$

for every $x\in \mathbb{R}\setminus\{-2,8\}$.

So the function is strictly decreasing on each interval of its domain:

$(-\infty,-2),\quad (-2,8),\quad (8,\infty)$

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4. Check the options

• A: decreases in $(-\infty,-2)\cup(-2,8)\cup(8,\infty)$ ✅
• B: increases in all intervals ❌
• C: decreases in $(-2,8)$ and increases elsewhere ❌
• D: decreases in $(-\infty,-2)$ and increases in $(8,\infty)$ ❌

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5. Final answer

The correct option is:

$\boxed{A}$

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6. Comparison with stored answer

Stored correct answer: A

My derived answer also is A, so they agree.