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The problem asks to find all relative extrema of the function \( f(x) = 3x^2 - x^3 \) using the Second-Derivative Test where applicable.

### Step 1: Find the first derivative
The first derivative of the function will help identify critical points.

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

### Step 2: Find critical points
Set the first derivative equal to zero to find critical points:

\[
6x - 3x^2 = 0
\]

Factor the equation:

\[
3x(2 - x) = 0
\]

This gives two critical points:

\[
x = 0 \quad \text{or} \quad x = 2
\]

### Step 3: Second-Derivative Test
To classify the critical points, we use the second derivative.

Find the second derivative of \( f(x) \):

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

Now evaluate \( f''(x) \) at each critical point:

- At \( x = 0 \):

\[
f''(0) = 6 - 6(0) = 6
\]

Since \( f''(0) > 0 \), the function is concave up at \( x = 0 \), indicating a **relative minimum** at \( x = 0 \).

- At \( x = 2 \):

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

Since \( f''(2) < 0 \), the function is concave down at \( x = 2 \), indicating a **relative maximum** at \( x = 2 \).

### Step 4: Conclusion
- There is a **relative minimum** at \( x = 0 \).
- There is a **relative maximum** at \( x = 2 \).

Would you like more details or have any questions?

Here are 5 related questions to explore:
1. How do you determine concavity using the second derivative?
2. What are the conditions for using the Second-Derivative Test?
3. Can you explain the significance of critical points in finding extrema?
4. What if the second derivative at a critical point is zero? How do you proceed?
5. How can you find absolute extrema over a specific interval?

**Tip:** If the second derivative test is inconclusive, use the First-Derivative Test for classification of critical points.

Find all relative extrema of the function. Use the Second-Derivative Test when applicable: f(x) = 3x^2 - x^3

Learn how to find relative extrema for the function f(x) = 3x^2 - x^3 using the Second-Derivative Test. This detailed solution involves finding critical points, determining concavity, and classifying them as maxima or minima. Suitable for Grades 11-12 or introductory college calculus.

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