The function `f(x) =frac{x}{x^2-6x-16}`, `x in`-{-2, 8}

Options

A. decreases in (–2, 8) and increases in `(-infty, -2) cup (8, infty)`
B. decreases in (–intly, -2) cup (-2, 8) cup (8, infty)`
C. decreases in (–infty, -2) and increases in `(8, infty)`
D. increases in (–intly, -2) cup (-2, 8) cup (8, infty)`

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x}{x^2 - 6x - 16} \) and determine its behavior over the interval \( x \in (-2, 8) \).

### Step 1: Find the derivative of the function
We will use the quotient rule to find the derivative \( f'(x) \). The quotient rule states that if \( f(x) = \frac{g(x)}{h(x)} \), then

\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]

Here, \( g(x) = x \) and \( h(x) = x^2 - 6x - 16 \).

Calculating the derivatives:
- \( g'(x) = 1 \)
- \( h'(x) = 2x - 6 \)

Now applying the quotient rule:

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

### Step 2: Simplify the derivative
Now we simplify the numerator:

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

This simplifies to:

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

### Step 3: Factor the numerator
The numerator can be factored:

\[ -x^2 + 16 = -(x^2 - 16) = -(x - 4)(x + 4) \]

Thus, we have:

\[ f'(x) = \frac{-(x - 4)(x + 4)}{(x^2 - 6x - 16)^2} \]

### Step 4: Analyze the critical points
To find where the function is increasing or decreasing, we need to determine where \( f'(x) = 0 \) or where it is undefined.

Setting the numerator equal to zero gives us the critical points:

\[ -(x - 4)(x + 4) = 0 \implies x = 4 \text{ or } x = -4 \]

The function is undefined where the denominator is zero:

\[ x^2 - 6x - 16 = 0 \]

Factoring gives:

\[ (x - 8)(x + 2) = 0 \implies x = 8 \text{ or } x = -2 \]

### Step 5: Determine intervals of increase and decrease
Now we analyze the sign of \( f'(x) \) in the intervals determined by the critical points and points where the function is undefined:

1. \( (-\infty, -4) \)
2. \( (-4, -2) \)
3. \( (-2, 4) \)
4. \( (4, 8) \)
5. \( (8, \infty) \)

Choose test points in each interval to determine the sign of \( f'(x) \):

- For \( x < -4 \) (e.g., \( x = -5 \)): \( f'(-5) > 0 \) (increasing)
- For \( -4 < x < -2 \) (e.g., \( x = -3 \)): \( f'(-3) < 0 \) (decreasing)
- For \( -2 < x < 4 \) (e.g., \( x = 0 \)): \( f'(0) < 0 \) (decreasing)
- For \( 4 < x < 8 \) (e.g., \( x = 5 \)): \( f'(5) < 0 \) (decreasing)
- For \( x > 8 \) (e.g., \( x = 9 \)): \( f'(9) > 0 \) (increasing)

### Conclusion
From the analysis, we conclude:
- The function is increasing on the intervals \( (-\infty, -4) \) and \( (8, \infty) \).
- The function is decreasing on the intervals \( (-4, -2) \), \( (-2, 4) \), and \( (4, 8) \).