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Let's analyze and solve this problem based on the provided derivative \( f'(x) = (x - 4)(x + 1)(x + 4) \).

### Step 1: Finding Critical Points
To find the critical points, we set \( f'(x) = 0 \):
\[
(x - 4)(x + 1)(x + 4) = 0
\]
This equation is satisfied when:
1. \( x - 4 = 0 \Rightarrow x = 4 \)
2. \( x + 1 = 0 \Rightarrow x = -1 \)
3. \( x + 4 = 0 \Rightarrow x = -4 \)

So, the critical points are \( x = -4 \), \( x = -1 \), and \( x = 4 \).

### Step 2: Testing Intervals for Increase and Decrease
To determine the intervals of increase and decrease, we can analyze the sign of \( f'(x) \) on intervals determined by these critical points: \( (-\infty, -4) \), \( (-4, -1) \), \( (-1, 4) \), and \( (4, \infty) \).

1. **Interval \( (-\infty, -4) \):** Choose a test point, like \( x = -5 \).
   - \( f'(-5) = (-)(-)(-) = - \)
   - So, \( f(x) \) is decreasing on \( (-\infty, -4) \).

2. **Interval \( (-4, -1) \):** Choose a test point, like \( x = -2 \).
   - \( f'(-2) = (-)(-)(+) = + \)
   - So, \( f(x) \) is increasing on \( (-4, -1) \).

3. **Interval \( (-1, 4) \):** Choose a test point, like \( x = 0 \).
   - \( f'(0) = (+)(-)(+) = - \)
   - So, \( f(x) \) is decreasing on \( (-1, 4) \).

4. **Interval \( (4, \infty) \):** Choose a test point, like \( x = 5 \).
   - \( f'(5) = (+)(+)(+) = + \)
   - So, \( f(x) \) is increasing on \( (4, \infty) \).

### Step 3: Determining Local Maxima and Minima
Using the information on intervals of increase and decrease, we can determine the nature of each critical point:

- **At \( x = -4 \):** \( f(x) \) changes from decreasing to increasing, so \( x = -4 \) is a **local minimum**.
- **At \( x = -1 \):** \( f(x) \) changes from increasing to decreasing, so \( x = -1 \) is a **local maximum**.
- **At \( x = 4 \):** \( f(x) \) changes from decreasing to increasing, so \( x = 4 \) is a **local minimum**.

### Step 4: Sketching a Possible Graph
Since \( f(x) \) is not unique, we can sketch a general shape based on the increasing and decreasing intervals and the local maxima and minima. The function will have:

- Local minima at \( x = -4 \) and \( x = 4 \).
- A local maximum at \( x = -1 \).

The graph would generally decrease towards \( x = -4 \), increase from \( x = -4 \) to \( x = -1 \), decrease again from \( x = -1 \) to \( x = 4 \), and finally increase from \( x = 4 \) onwards.

Would you like a detailed sketch of this graph or further explanation on any specific part?

### Follow-up Questions for Deeper Understanding:
1. How can we determine if a critical point is a maximum or minimum using the second derivative?
2. Can we apply this approach to a function with a different derivative form?
3. How would we adjust the intervals if there were more critical points?
4. What happens to the graph if one of the factors in \( f'(x) \) is squared?
5. How would the graph differ if the polynomial degree were higher?

**Tip:** When analyzing intervals for increase and decrease, always pick test points within each interval to determine the sign of the derivative.

Use the derivative f'(x) = (x - 4)(x + 1)(x + 4) to determine the local maxima and minima and the intervals of increase and decrease. Sketch a possible graph of f (f is not unique).

Explore how to determine local maxima, minima, and intervals of increase and decrease for a function with derivative f'(x) = (x - 4)(x + 1)(x + 4). This calculus problem uses critical points and the First Derivative Test for analysis, suitable for students in Grades 11-12.

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