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To find the local extrema for the function \( f(x) = -x^3 + 6x^2 \) using the Second Derivative Test, we proceed as follows:

### Step 1: Find the First Derivative
Calculate \( f'(x) \) to locate the critical points:

\[
f'(x) = \frac{d}{dx}(-x^3 + 6x^2) = -3x^2 + 12x
\]

### Step 2: Set the First Derivative to Zero
To find critical points, set \( f'(x) = 0 \):

\[
-3x^2 + 12x = 0
\]

Factor out \(-3x\):

\[
-3x(x - 4) = 0
\]

This gives the critical points \( x = 0 \) and \( x = 4 \).

### Step 3: Find the Second Derivative
Now, compute the second derivative \( f''(x) \) to use in the Second Derivative Test:

\[
f''(x) = \frac{d}{dx}(-3x^2 + 12x) = -6x + 12
\]

### Step 4: Apply the Second Derivative Test
Evaluate \( f''(x) \) at each critical point:

1. **At \( x = 0 \):**
   \[
   f''(0) = -6(0) + 12 = 12
   \]
   Since \( f''(0) > 0 \), there is a **local minimum** at \( x = 0 \).

2. **At \( x = 4 \):**
   \[
   f''(4) = -6(4) + 12 = -24 + 12 = -12
   \]
   Since \( f''(4) < 0 \), there is a **local maximum** at \( x = 4 \).

### Step 5: Find the Function Values at the Critical Points
Finally, substitute these \( x \)-values back into \( f(x) \) to find the \( y \)-coordinates for the ordered pairs:

- For \( x = 0 \):
  \[
  f(0) = -(0)^3 + 6(0)^2 = 0
  \]
  So, the local minimum is at \( (0, 0) \).

- For \( x = 4 \):
  \[
  f(4) = -(4)^3 + 6(4)^2 = -64 + 96 = 32
  \]
  So, the local maximum is at \( (4, 32) \).

### Answer
The local extrema, written as ordered pairs, are:
- Local minimum at \( (0, 0) \)
- Local maximum at \( (4, 32) \)

Would you like more details on any of these steps, or have any questions?

---

1. What is the First Derivative Test, and how does it differ from the Second Derivative Test?
2. Why does the sign of \( f''(x) \) at critical points determine the type of extremum?
3. How would the process change if \( f(x) \) were a different type of polynomial?
4. What is the significance of the critical points in finding extrema?
5. Can we use the Second Derivative Test if \( f''(x) = 0 \) at a critical point?

**Tip:** The Second Derivative Test is useful but inconclusive if \( f''(x) = 0 \) at a critical point; in such cases, consider the First Derivative Test.

Use the Second Derivative Test to find any local extrema for f(x) = -x^3 + 6x^2. Write your answers as ordered pairs.

Learn how to find local extrema of the function f(x) = -x^3 + 6x^2 using the Second Derivative Test. This solution includes calculating first and second derivatives, identifying critical points, and determining local minima and maxima. Suitable for Grades 11-12 calculus students.

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