Step 1: Understanding the Concept:
The increasing or decreasing behavior of a function depends on the sign of its first derivative.
A function \( f(x) \) is strictly increasing where \( f'(x)>0 \) and strictly decreasing where \( f'(x)<0 \).
The roots of \( f'(x) = 0 \) are the points where the function might change its direction.
Step 2: Key Formula or Approach:
We will differentiate \( f(x) \), find its critical points, and then use the interval test method (wavy curve) to determine the signs of \( f'(x) \).
Step 3: Detailed Explanation:
Given \( f(x) = -2x^{3} - 9x^{2} - 12x + 5 \).
Find the derivative:
\[ f'(x) = \frac{d}{dx}(-2x^{3} - 9x^{2} - 12x + 5) \]
\[ f'(x) = -6x^{2} - 18x - 12 \]
To find the critical points, set \( f'(x) = 0 \):
\[ -6x^{2} - 18x - 12 = 0 \]
Divide by -6 to simplify:
\[ x^{2} + 3x + 2 = 0 \]
Factorize the quadratic:
\[ (x + 1)(x + 2) = 0 \]
The critical points are \( x = -1 \) and \( x = -2 \).
These points divide the real line into three intervals: \( (-\infty, -2) \), \( (-2, -1) \), and \( (-1, \infty) \).
Check the sign of \( f'(x) = -6(x+1)(x+2) \) in each interval:
1. For \( x \in (-\infty, -2) \): Choose \( x = -3 \). \( f'(-3) = -6(-2)(-1) = -12 \). (Negative) \( \to \) Decreasing.
2. For \( x \in (-2, -1) \): Choose \( x = -1.5 \). \( f'(-1.5) = -6(-0.5)(0.5) = +1.5 \). (Positive) \( \to \) Increasing.
3. For \( x \in (-1, \infty) \): Choose \( x = 0 \). \( f'(0) = -6(1)(2) = -12 \). (Negative) \( \to \) Decreasing.
Thus, the function increases on \( (-2, -1) \) and decreases on \( (-\infty, -2) \cup (-1, \infty) \).
This matches option (d).
Step 4: Final Answer:
The correct description is given in option (d).