The function \( f(x) = \frac{x}{x^2 - 6x - 16} \), \( x \in \mathbb{R} - \{-2, 8\} \)
- decreases in \( (-2, 8) \) and increases in \( (-\infty, -2) \cup (8, \infty) \)
- decreases in \( (-\infty, -2) \cup (-2, 8) \cup (8, \infty) \)
- decreases in \( (-\infty, -2) \) and increases in \( (8, \infty) \)
- increases in \( (-\infty, -2) \cup (-2, 8) \cup (8, \infty) \)
The Correct Option is B
Approach Solution - 1
The function given is \(f(x) = \frac{x}{x^2 - 6x - 16}\). To determine where this function increases or decreases, we first need to find its derivative.
This is a quotient of two functions, so we apply the quotient rule:
\(\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2}\)
For \(f(x) = \frac{x}{x^2 - 6x - 16}\), let:
- \(u(x) = x\) (which gives \(u'(x) = 1\))
- \(v(x) = x^2 - 6x - 16\) (which gives \(v'(x) = 2x - 6\))
Now, apply the quotient rule:
\(f'(x) = \frac{1 \cdot (x^2 - 6x - 16) - x \cdot (2x - 6)}{(x^2 - 6x - 16)^2}\)
Simplify the numerator:
\(= \frac{x^2 - 6x - 16 - (2x^2 - 6x)}{(x^2 - 6x - 16)^2}\)
\(= \frac{x^2 - 6x - 16 - 2x^2 + 6x}{(x^2 - 6x - 16)^2}\)
\(= \frac{-x^2 - 16}{(x^2 - 6x - 16)^2}\)
The critical points are determined by setting the numerator equal to zero. However, \(-x^2 - 16 = 0\) has no real roots (as the equation simplifies to \(x^2 = -16\), which is not possible in real numbers).
Next, examine the sign of \(f'(x)\):
- \(-x^2 - 16\) is always negative since \(x^2 \geq 0\).
- The denominator \((x^2 - 6x - 16)^2\) is always positive except at points of discontinuity, \(x = -2\) and \(x = 8\), where the function is not defined.
Therefore, \(f'(x)\) is negative for all intervals where the function is defined, indicating that the function \(f(x)\) is decreasing in those intervals.
Thus, the correct answer is that the function decreases in \((-\infty, -2) \cup (-2, 8) \cup (8, \infty)\).
Approach Solution -2
The given function is: \(f(x) = \frac{x}{x^2 - 6x - 16}\)
Step 1. Calculate the derivative \( f'(x) \):\(f'(x) = \frac{-(x^2 + 16)}{(x^2 - 6x - 16)^2}\)
Step 2. Since \( f'(x) < 0 \), the function \( f(x) \) is decreasing in all intervals where it is defined.**
Step 3. Therefore, \( f(x) \) is decreasing in \( (-\infty, -2) \cup (-2, 8) \cup (8, \infty) \).**
The Correct Answer is:\( (-\infty, -2) \cup (-2, 8) \cup (8, \infty) \)
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