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To find the absolute extrema for the function \( f(x) = 2x^3 + 12x^2 + 18x - 2 \) on the interval \([-6, 0]\), we’ll go through the following steps:

### Step 1: Find the Critical Points
1. **Compute the derivative \( f'(x) \):**

   \[
   f(x) = 2x^3 + 12x^2 + 18x - 2
   \]
   Using the power rule, we get:
   \[
   f'(x) = 6x^2 + 24x + 18
   \]

2. **Set \( f'(x) = 0 \) to find critical points:**

   \[
   6x^2 + 24x + 18 = 0
   \]

   Dividing by 6 simplifies this to:
   \[
   x^2 + 4x + 3 = 0
   \]

   Factor the quadratic:
   \[
   (x + 3)(x + 1) = 0
   \]

   So, \( x = -3 \) and \( x = -1 \) are the critical points.

### Step 2: Evaluate \( f(x) \) at the Endpoints and Critical Points
We need to evaluate \( f(x) \) at \( x = -6 \), \( x = 0 \), and the critical points \( x = -3 \) and \( x = -1 \).

1. **At \( x = -6 \):**
   \[
   f(-6) = 2(-6)^3 + 12(-6)^2 + 18(-6) - 2
   = -432 + 432 - 108 - 2 = -110
   \]

2. **At \( x = 0 \):**
   \[
   f(0) = 2(0)^3 + 12(0)^2 + 18(0) - 2 = -2
   \]

3. **At \( x = -3 \):**
   \[
   f(-3) = 2(-3)^3 + 12(-3)^2 + 18(-3) - 2
   = -54 + 108 - 54 - 2 = -2
   \]

4. **At \( x = -1 \):**
   \[
   f(-1) = 2(-1)^3 + 12(-1)^2 + 18(-1) - 2
   = -2 + 12 - 18 - 2 = -10
   \]

### Step 3: Identify the Absolute Maximum and Minimum
From these calculations:
- \( f(-6) = -110 \)
- \( f(0) = -2 \)
- \( f(-3) = -2 \)
- \( f(-1) = -10 \)

The highest value of \( f(x) \) is \(-2\), which occurs at \( x = 0 \) and \( x = -3 \). Therefore:

- **The absolute maximum is \( -2 \), occurring at \( x = 0 \) and \( x = -3 \).**
- The absolute minimum is \( -110 \), occurring at \( x = -6 \).

### Answer:
The correct choice is:
- **A. The absolute maximum is \(-2\), which occurs at \( x = 0, -3 \).**

Would you like more details on any step or have questions?

Here are some related questions to consider:
1. How would you find the absolute minimum instead of the maximum?
2. What does it mean for a point to be a critical point in this context?
3. How can you determine if a critical point is a maximum or minimum?
4. Why is it necessary to check endpoints for absolute extrema?
5. How would the approach change if the domain were unbounded?

**Tip:** Always check both critical points and endpoints when looking for absolute extrema on a closed interval.

Find the absolute extrema if they exist, as well as all values of x where they occur, for the function f(x) = 2x^3 + 12x^2 + 18x - 2 on the domain [-6, 0].

Learn how to find the absolute maximum and minimum of the function f(x) = 2x^3 + 12x^2 + 18x - 2 on the interval [-6, 0]. This detailed solution includes finding critical points and evaluating the function at endpoints. Suitable for calculus students in Grades 11-12.

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