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To solve this problem, let’s break it down step-by-step:

1. **Identify Points of Inflection**:
   - Points of inflection occur where the second derivative \( f''(x) \) changes sign, indicating a shift in concavity.

2. **Find \( f'(x) \)** (the first derivative of \( f(x) = \frac{1}{2}x^4 + 2x^3 \)):
   \[
   f(x) = \frac{1}{2}x^4 + 2x^3
   \]
   Differentiate \( f(x) \) with respect to \( x \):
   \[
   f'(x) = 2x^3 + 6x^2
   \]

3. **Find \( f''(x) \)** (the second derivative):
   Differentiate \( f'(x) \) with respect to \( x \):
   \[
   f''(x) = 6x^2 + 12x = 6x(x + 2)
   \]

4. **Set \( f''(x) = 0 \)** to find potential inflection points:
   \[
   6x(x + 2) = 0
   \]
   This equation gives:
   \[
   x = 0 \quad \text{or} \quad x = -2
   \]

5. **Determine the Concavity on Intervals Around \( x = -2 \) and \( x = 0 \)**:
   - To understand where the graph is concave up or down, test values in the intervals created by these points: \( (-\infty, -2) \), \( (-2, 0) \), and \( (0, \infty) \).
   - For concavity:
     - If \( f''(x) > 0 \), the function is concave upward.
     - If \( f''(x) < 0 \), the function is concave downward.

6. **Evaluate \( f(x) \) at \( x = -2 \) and \( x = 0 \) to find the corresponding \( y \)-values**:
   - For \( x = -2 \):
     \[
     f(-2) = \frac{1}{2}(-2)^4 + 2(-2)^3 = \frac{1}{2}(16) + 2(-8) = 8 - 16 = -8
     \]
     So, the point \( (-2, -8) \) is a potential inflection point.
   - For \( x = 0 \):
     \[
     f(0) = \frac{1}{2}(0)^4 + 2(0)^3 = 0
     \]
     So, the point \( (0, 0) \) is another potential inflection point.

### Solution Summary:

- **Inflection Points**: 
  - Smaller \( x \)-value: \( (-2, -8) \)
  - Larger \( x \)-value: \( (0, 0) \)

- **Concavity Intervals**:
  - **Concave Upward**: \( (-\infty, -2) \cup (0, \infty) \)
  - **Concave Downward**: \( (-2, 0) \)

Would you like a detailed explanation of any specific step? Here are some questions for further exploration:

1. What are the practical applications of identifying points of inflection?
2. How does the sign of the second derivative relate to concavity?
3. How can the first derivative test help verify inflection points?
4. Can inflection points exist where the second derivative does not exist?
5. How would this approach change if the function were more complex (e.g., involving trigonometric functions)?

**Tip**: Always verify points of inflection by checking if the concavity changes across the identified points.

Find the points of inflection of the graph of the function f(x) = (1/2)x^4 + 2x^3. Discuss the concavity of the graph of the function.

Find the inflection points and analyze the concavity of the function f(x) = (1/2)x^4 + 2x^3. This solution covers key calculus concepts, including finding the second derivative to identify inflection points and testing intervals for concavity. Suitable for students in Grades 11-12.

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