Key Concepts and Formulas

• First Derivative Test for Monotonicity: If $f'(x) > 0$ on an interval, then $f(x)$ is strictly increasing on that interval. If $f'(x) < 0$ on an interval, then $f(x)$ is strictly decreasing on that interval.
• Quotient Rule: If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.
• Domain of a Rational Function: A rational function is undefined where the denominator is zero.

Step 1: Understanding the Function and its Domain

The given function is: $f(x)=\frac{x}{x^2-6 x-16}$ The domain is $x \in \mathbb{R}-\{-2,8\}$. We need to determine why these points are excluded by factoring the denominator: $x^2-6x-16 = (x-8)(x+2)$ Thus, the function can be written as: $f(x)=\frac{x}{(x-8)(x+2)}$ The function is undefined when $x = 8$ or $x = -2$. Therefore, the domain is correctly specified.

Step 2: Calculating the First Derivative $f'(x)$

We will use the quotient rule to find $f'(x)$. Let $u(x) = x$, so $u'(x) = 1$. Let $v(x) = x^2 - 6x - 16$, so $v'(x) = 2x - 6$. Applying the quotient rule: $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} = \frac{(1)(x^2-6x-16) - (x)(2x-6)}{(x^2-6x-16)^2}$ Simplify the numerator: $f'(x) = \frac{x^2-6x-16 - 2x^2+6x}{(x^2-6x-16)^2} = \frac{-x^2-16}{(x^2-6x-16)^2}$ We can factor out a negative sign from the numerator: $f'(x) = \frac{-(x^2+16)}{(x^2-6x-16)^2}$

Step 3: Analyzing the Sign of $f'(x)$

We need to analyze the sign of $f'(x)$ for $x \in \mathbb{R}-\{-2,8\}$.

1. Numerator: $-(x^2+16)$

   • Since $x^2 \ge 0$ for all real $x$, $x^2 + 16 \ge 16 > 0$.
   • Therefore, $-(x^2+16) < 0$ for all real $x$.

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

   • Since the expression is squared, it is always non-negative.
   • Since $x \neq -2$ and $x \neq 8$, the denominator is strictly positive: $(x^2-6x-16)^2 > 0$ for all $x \in \mathbb{R}-\{-2,8\}$.

3. Combining Signs: $f'(x) = \frac{\text{negative}}{\text{positive}} < 0$ Therefore, $f'(x) < 0$ for all $x \in \mathbb{R}-\{-2,8\}$.

Step 4: Determining Intervals of Monotonicity

Since $f'(x) < 0$ for all $x$ in its domain, the function $f(x)$ is strictly decreasing on each interval in its domain. The domain is $\mathbb{R}-\{-2,8\}$, which can be expressed as $(-\infty,-2) \cup (-2,8) \cup (8, \infty)$. Thus, $f(x)$ is decreasing on $(-\infty,-2) \cup (-2,8) \cup (8, \infty)$.

Step 5: Comparing with Options and Final Answer

Comparing our conclusion with the given options: (A) decreases in $(-\infty,-2) \cup(-2,8) \cup(8, \infty)$ (B) increases in $(-\infty,-2) \cup(-2,8) \cup(8, \infty)$ (C) decreases in $(-2,8)$ and increases in $(-\infty,-2) \cup(8, \infty)$ (D) decreases in $(-\infty,-2)$ and increases in $(8, \infty)$

Our result matches option (A).

Common Mistakes & Tips:

• Domain Awareness: Always remember to consider the domain of the function. Monotonicity analysis is only valid within the function's domain.
• Algebraic Errors: Be extremely careful with algebraic manipulations, especially when applying the quotient rule.
• Sign Analysis: Systematically analyze the signs of the numerator and denominator of the derivative to determine the intervals of increase or decrease.

Summary

By calculating the first derivative of the function and analyzing its sign, we found that $f'(x)$ is always negative on its domain. Therefore, the function is strictly decreasing on $(-\infty,-2) \cup (-2,8) \cup (8, \infty)$, which corresponds to option (A).

The final answer is $\boxed{\text{A}}$.