In which interval the function `f(x)=x/(x^2-6x-16)` is increasing

Options

A. `phi`
B. `[1, 3/7]cup(5/4,infty)
C. `(5/4,infty)`
D. `[3/4, 5/4]`

Answer

To determine the interval in which the function \( f(x) = \frac{x}{x^2 - 6x - 16} \) is increasing, we need to find the derivative \( f'(x) \) and analyze its sign.

### Step 1: Find the derivative \( f'(x) \)

Using the quotient rule, which states that if \( f(x) = \frac{g(x)}{h(x)} \), then

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

we can identify \( g(x) = x \) and \( h(x) = x^2 - 6x - 16 \).

- The derivative of \( g(x) \) is \( g'(x) = 1 \).
- The derivative of \( h(x) \) is \( h'(x) = 2x - 6 \).

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

Expanding the numerator:

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

This simplifies to:

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

### Step 3: Analyze the sign of \( f'(x) \)

The denominator \( (x^2 - 6x - 16)^2 \) is always positive (since a square is always non-negative). Therefore, the sign of \( f'(x) \) is determined by the numerator \( -x^2 - 16 \).

Since \( -x^2 - 16 \) is always negative (as \( -x^2 \) is non-positive and subtracting 16 makes it negative), we conclude that:

\[ f'(x) < 0 \quad \text{for all } x \]

### Step 4: Conclusion

Since \( f'(x) < 0 \) for all \( x \), the function \( f(x) \) is decreasing everywhere and does not have any interval where it is increasing.

Thus, the final answer is that there are no intervals in which the function is increasing.