Yes, you can use the second-order derivative to determine whether the critical points are local maxima, minima, or points of inflection. This method is known as the **Second Derivative Test**.

Here's how it works:

### Step 1: Take the first derivative (which we've already done)
We have the first derivative:

\[
f'(x) = 6x^2 - 30x + 36
\]

### Step 2: Take the second derivative
Now, take the derivative of \( f'(x) \) to get the second derivative:

\[
f''(x) = \frac{d}{dx}(6x^2 - 30x + 36) = 12x - 30
\]

### Step 3: Evaluate the second derivative at the critical points
We previously found the critical points to be \( x = 2 \) and \( x = 3 \). Now, let's plug these values into the second derivative:

1. For \( x = 2 \):

\[
f''(2) = 12(2) - 30 = 24 - 30 = -6
\]

Since \( f''(2) < 0 \), this indicates that \( f(x) \) has a **local maximum** at \( x = 2 \).

2. For \( x = 3 \):

\[
f''(3) = 12(3) - 30 = 36 - 30 = 6
\]

Since \( f''(3) > 0 \), this indicates that \( f(x) \) has a **local minimum** at \( x = 3 \).

### Step 4: Compare the critical points with the endpoints
To determine the absolute maxima and minima on the interval \([1, 5]\), we still need to compare the function values at the critical points \( x = 2 \), \( x = 3 \), and the endpoints \( x = 1 \) and \( x = 5 \) (as we did before).

- \( f(1) = 24 \)
- \( f(2) = 29 \) (local maximum)
- \( f(3) = 28 \) (local minimum)
- \( f(5) = 56 \)

### Conclusion
- The **absolute maximum** on \([1, 5]\) is at \( x = 5 \) with \( f(5) = 56 \).
- The **absolute minimum** on \([1, 5]\) is at \( x = 1 \) with \( f(1) = 24 \).

The second derivative test helped us confirm the nature of the critical points, but we still needed to evaluate the endpoints to find the absolute extrema.

Would you like to explore more details or have any further questions?

Here are 5 questions to consider:
1. What does it mean if the second derivative at a critical point is zero?
2. How would the process change if the interval was open instead of closed?
3. Can the second derivative test always determine the nature of a critical point?
4. What if the second derivative doesn't exist at a critical point?
5. How does concavity relate to the second derivative and extrema?

**Tip:** The second derivative test is a quick way to assess the nature of critical points, but don't forget to consider endpoints for absolute extrema on closed intervals.

Explore how to use the second derivative test to determine local maxima, minima, and points of inflection. This problem involves calculus and derivatives, specifically focusing on evaluating critical points using the second derivative. Learn step-by-step with detailed examples.

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