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To solve this problem, let's go through the following steps:

1. **Identify Critical Points**:
   The function given is:
   \[
   f(x) = -2x^3 + 27x^2 - 108x + 8
   \]
   Critical points occur where the derivative, \( f'(x) \), is zero or undefined. We’ll start by finding \( f'(x) \).

2. **Compute \( f'(x) \)**:
   \[
   f(x) = -2x^3 + 27x^2 - 108x + 8
   \]
   Using the power rule, the derivative \( f'(x) \) is:
   \[
   f'(x) = -6x^2 + 54x - 108
   \]

3. **Set \( f'(x) = 0 \) to Find Critical Points**:
   Solve the equation:
   \[
   -6x^2 + 54x - 108 = 0
   \]
   This can be simplified by dividing each term by -6:
   \[
   x^2 - 9x + 18 = 0
   \]
   Factor the quadratic:
   \[
   (x - 3)(x - 6) = 0
   \]
   So, the critical points are:
   \[
   x = 3 \quad \text{and} \quad x = 6
   \]
   Therefore, \( A = 3 \) and \( B = 6 \).

4. **Determine Increasing or Decreasing Intervals**:
   To determine whether \( f(x) \) is increasing or decreasing on each interval, we’ll examine the sign of \( f'(x) \) in the intervals:
   - \( (-\infty, A] = (-\infty, 3] \)
   - \( [A, B] = [3, 6] \)
   - \( [B, \infty) = [6, \infty) \)

   Let's test values within each interval in \( f'(x) = -6x^2 + 54x - 108 \):
   
   - For \( x \in (-\infty, 3) \): Choose \( x = 1 \).
     \[
     f'(1) = -6(1)^2 + 54(1) - 108 = -6 + 54 - 108 = -60 \quad (\text{negative, so } f(x) \text{ is decreasing})
     \]
   - For \( x \in (3, 6) \): Choose \( x = 4 \).
     \[
     f'(4) = -6(4)^2 + 54(4) - 108 = -96 + 216 - 108 = 12 \quad (\text{positive, so } f(x) \text{ is increasing})
     \]
   - For \( x \in (6, \infty) \): Choose \( x = 7 \).
     \[
     f'(7) = -6(7)^2 + 54(7) - 108 = -294 + 378 - 108 = -24 \quad (\text{negative, so } f(x) \text{ is decreasing})
     \]

### Summary of Results:
- **Critical Points**: \( A = 3 \), \( B = 6 \)
- **Intervals**:
  - \( (-\infty, 3] \): Decreasing (DEC)
  - \( [3, 6] \): Increasing (INC)
  - \( [6, \infty) \): Decreasing (DEC)

Would you like further details on any step? Here are some related questions:

1. How do we determine the critical points of a polynomial function?
2. Why do we use test points to check increasing or decreasing intervals?
3. How does the sign of the derivative relate to the function's behavior?
4. Can this method be applied to any polynomial function?
5. What would change if we had additional critical points?

**Tip**: For polynomial functions, critical points are found by setting the derivative to zero and solving for \( x \).

Consider the function f(x) = -2x^3 + 27x^2 - 108x + 8. Find the critical points A and B, and determine whether f(x) is increasing or decreasing in the intervals (-∞, A], [A, B], and [B, ∞).

Find the critical points and determine where the function f(x) = -2x^3 + 27x^2 - 108x + 8 is increasing or decreasing. This solution covers key calculus concepts such as critical points and intervals of increase and decrease. Suitable for students in Grades 11-12.

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