Question

Consider the function \( f(x) = -2x^3 - 9x^2 - 12x + 5 \). Then, which of the following is/are correct?

• A. \( f(x) \) decreasing in \((0, \infty)\) and increasing in \((-\infty, 0)\)
• B. \( f(x) \) increasing in \((-\infty, 1) \cup (2, \infty)\) and decreasing in \((1, 2)\)
• C. \( f(x) \) increasing in \((0, \infty)\) and decreasing in \((-\infty, 0)\)
• D. \( f(x) \) is increasing in \((-2, -1)\) and decreasing in \((-\infty, -2) \cup (-1, \infty)\)

Hint:

To determine intervals of increasing/decreasing, the sign of the first derivative \( f'(x) \) is crucial. Always factor \( f'(x) \) to easily analyze its sign across intervals defined by its roots. Use test points within each interval to determine the sign.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks to determine the intervals where the given function \( f(x) \) is increasing or decreasing. This involves analyzing the sign of the first derivative of the function.

Step 2: Key Formula or Approach:
1. Find the first derivative of the function, \( f'(x) \).
2. Find the critical points by setting \( f'(x) = 0 \).
3. Test the sign of \( f'(x) \) in intervals determined by the critical points.
- If \( f'(x) > 0 \), the function is increasing.
- If \( f'(x) < 0 \), the function is decreasing.

Step 3: Detailed Explanation:
Given function: \( f(x) = -2x^3 - 9x^2 - 12x + 5 \).
Differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(-2x^3 - 9x^2 - 12x + 5) \]
\[ f'(x) = -6x^2 - 18x - 12 \]
To find critical points, set \( f'(x) = 0 \):
\[ -6x^2 - 18x - 12 = 0 \]
Divide by -6:
\[ x^2 + 3x + 2 = 0 \]
Factor the quadratic equation:
\[ (x+1)(x+2) = 0 \]
The critical points are \( x = -1 \) and \( x = -2 \).
These critical points divide the number line into three intervals: \((-\infty, -2)\), \((-2, -1)\), and \((-1, \infty)\).
Now, test the sign of \( f'(x) = -6(x+1)(x+2) \) in each interval:
1. Interval \((-\infty, -2)\): Choose a test value, e.g., \( x = -3 \).
\( f'(-3) = -6(-3+1)(-3+2) = -6(-2)(-1) = -12 \).
Since \( f'(x) < 0 \), \( f(x) \) is decreasing in \((-\infty, -2)\).
2. Interval \((-2, -1)\): Choose a test value, e.g., \( x = -1.5 \).
\( f'(-1.5) = -6(-1.5+1)(-1.5+2) = -6(-0.5)(0.5) = -6(-0.25) = 1.5 \).
Since \( f'(x) > 0 \), \( f(x) \) is increasing in \((-2, -1)\).
3. Interval \((-1, \infty)\): Choose a test value, e.g., \( x = 0 \).
\( f'(0) = -6(0+1)(0+2) = -6(1)(2) = -12 \).
Since \( f'(x) < 0 \), \( f(x) \) is decreasing in \((-1, \infty)\).
Combining the results:
\( f(x) \) is increasing in \((-2, -1)\).
\( f(x) \) is decreasing in \((-\infty, -2) \cup (-1, \infty)\).
This matches option (d).

Step 4: Final Answer:
The function \( f(x) \) is increasing in \((-2, -1)\) and decreasing in \((-\infty, -2) \cup (-1, \infty)\).