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To determine the intervals of concavity and the points of inflection for \( f(x) = 2 + \frac{11}{12}x - 2x^2 + \frac{1}{12}x^4 \), we will analyze the function's second derivative, \( f''(x) \).

### Steps:
1. **Find \( f'(x) \)** (the first derivative).
2. **Find \( f''(x) \)** (the second derivative).
3. **Solve \( f''(x) = 0 \)** to find possible points of inflection.
4. **Determine the concavity** by analyzing the sign of \( f''(x) \) around these points.

I'll go through these steps now. 

### 1. First Derivative \( f'(x) \):
Given:
\[
f(x) = 2 + \frac{11}{12}x - 2x^2 + \frac{1}{12}x^4
\]

The derivative \( f'(x) \) is:
\[
f'(x) = \frac{11}{12} - 4x + \frac{1}{3}x^3
\]

### 2. Second Derivative \( f''(x) \):
The second derivative \( f''(x) \) is:
\[
f''(x) = -4 + x^2
\]

### 3. Solve \( f''(x) = 0 \):
Set \( f''(x) = -4 + x^2 = 0 \) to find the points where concavity might change:
\[
x^2 = 4
\]
\[
x = \pm 2
\]

So, \( x = -2 \) and \( x = 2 \) are potential points of inflection.

### 4. Determine the Concavity:
To determine intervals of concavity, we check the sign of \( f''(x) \) on intervals around \( x = -2 \) and \( x = 2 \).

- For \( x < -2 \), choose a point like \( x = -3 \):
  \[
  f''(-3) = -4 + (-3)^2 = -4 + 9 = 5 > 0
  \]
  So, \( f(x) \) is **concave up** on \( (-\infty, -2) \).

- For \( -2 < x < 2 \), choose a point like \( x = 0 \):
  \[
  f''(0) = -4 + 0^2 = -4 < 0
  \]
  So, \( f(x) \) is **concave down** on \( (-2, 2) \).

- For \( x > 2 \), choose a point like \( x = 3 \):
  \[
  f''(3) = -4 + 3^2 = -4 + 9 = 5 > 0
  \]
  So, \( f(x) \) is **concave up** on \( (2, \infty) \).

### Summary
- The function is **concave up** on \( (-\infty, -2) \) and \( (2, \infty) \).
- The function is **concave down** on \( (-2, 2) \).
- It changes concavity at \( x = -2 \) and \( x = 2 \).

### Answer
- **Concave up on:** \((- \infty, -2)\), \((2, \infty)\)
- **Concave down on:** \((-2, 2)\)
- **Changes concavity at:** \(x = -2\), \(x = 2\)

Would you like further details, or do you have any questions?

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Here are 5 related questions for further exploration:
1. How do you interpret concavity in terms of a function's shape?
2. What is the significance of points where \( f''(x) = 0 \)?
3. How would you find concavity if \( f(x) \) had a different polynomial degree?
4. Can there be cases where \( f''(x) = 0 \) but the function doesn’t change concavity?
5. What is the relationship between the second derivative and inflection points?

**Tip:** For concavity questions, always check the sign of the second derivative around critical points where it equals zero.

Consider the function f(x) = 2 + (11/12)x - 2x^2 + (1/12)x^4. Determine the intervals where the function is concave up, concave down, and where it changes concavity.

Learn how to find intervals of concavity and points of inflection for the function f(x) = 2 + (11/12)x - 2x^2 + (1/12)x^4. This calculus problem involves analyzing the second derivative to determine where the function is concave up, concave down, and where it changes concavity.

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