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We are given the derivative \( f'(x) = \frac{x^2(x - 1)}{x + 2} \) with \( x \neq -2 \), and the questions are about finding critical points, increasing or decreasing intervals, and local maxima or minima.

### a) Critical Points of \( f \)
Critical points occur where \( f'(x) = 0 \) or where \( f'(x) \) is undefined. To solve for these:

- Set \( f'(x) = 0 \): 
\[
\frac{x^2(x - 1)}{x + 2} = 0
\]
The numerator must be 0 for the derivative to equal 0, so solve:
\[
x^2(x - 1) = 0
\]
This gives two solutions:
1. \( x^2 = 0 \Rightarrow x = 0 \)
2. \( x - 1 = 0 \Rightarrow x = 1 \)

Therefore, the critical points from the numerator are \( x = 0 \) and \( x = 1 \).

- Where \( f'(x) \) is undefined: 
This happens when the denominator \( x + 2 = 0 \), so \( x = -2 \). However, \( x = -2 \) is excluded from the domain.

Thus, the critical points are \( x = 0 \) and \( x = 1 \).

### b) Intervals of Increase and Decrease
To determine where \( f \) is increasing or decreasing, analyze the sign of \( f'(x) \) over intervals determined by the critical points \( x = 0 \), \( x = 1 \), and the excluded point \( x = -2 \).

The intervals are: 
1. \( (-\infty, -2) \)
2. \( (-2, 0) \)
3. \( (0, 1) \)
4. \( (1, \infty) \)

Now, test the sign of \( f'(x) \) in each interval:

1. For \( x \in (-\infty, -2) \), pick \( x = -3 \):
   \[
   f'(-3) = \frac{(-3)^2(-3 - 1)}{-3 + 2} = \frac{9(-4)}{-1} = 36
   \]
   So \( f'(-3) > 0 \), meaning \( f \) is **increasing** on \( (-\infty, -2) \).

2. For \( x \in (-2, 0) \), pick \( x = -1 \):
   \[
   f'(-1) = \frac{(-1)^2(-1 - 1)}{-1 + 2} = \frac{1(-2)}{1} = -2
   \]
   So \( f'(-1) < 0 \), meaning \( f \) is **decreasing** on \( (-2, 0) \).

3. For \( x \in (0, 1) \), pick \( x = 0.5 \):
   \[
   f'(0.5) = \frac{(0.5)^2(0.5 - 1)}{0.5 + 2} = \frac{0.25(-0.5)}{2.5} = -0.05
   \]
   So \( f'(0.5) < 0 \), meaning \( f \) is **decreasing** on \( (0, 1) \).

4. For \( x \in (1, \infty) \), pick \( x = 2 \):
   \[
   f'(2) = \frac{(2)^2(2 - 1)}{2 + 2} = \frac{4(1)}{4} = 1
   \]
   So \( f'(2) > 0 \), meaning \( f \) is **increasing** on \( (1, \infty) \).

### c) Local Maxima and Minima
Local extrema occur where the derivative changes sign at critical points:

- At \( x = 0 \), \( f'(x) \) changes from negative (on \( (-2, 0) \)) to negative (on \( (0, 1) \)), so no extremum occurs at \( x = 0 \).
- At \( x = 1 \), \( f'(x) \) changes from negative (on \( (0, 1) \)) to positive (on \( (1, \infty) \)), indicating a **local minimum** at \( x = 1 \).

### Summary
- **Critical points**: \( x = 0 \), \( x = 1 \)
- \( f \) is **increasing** on \( (-\infty, -2) \) and \( (1, \infty) \), and **decreasing** on \( (-2, 0) \) and \( (0, 1) \).
- \( f \) has a **local minimum** at \( x = 1 \).

Would you like further details on any of these parts? Here are some related questions to explore:

1. How would we find the second derivative \( f''(x) \) to confirm concavity?
2. What is the behavior of \( f(x) \) near \( x = -2 \)?
3. How would a sign chart help determine the nature of the critical points?
4. Can we infer anything about the global extrema from this information?
5. How would this approach change if we were given the function \( f(x) \) instead of just \( f'(x) \)?

**Tip:** When analyzing increasing and decreasing intervals, always check the sign changes of the derivative around the critical points!

Given f'(x) = (x^2(x - 1)) / (x + 2), x ≠ -2, find the critical points, intervals of increase/decrease, and local extrema.

This problem covers the identification of critical points, intervals of increase/decrease, and local extrema for the derivative f'(x) = x^2(x - 1) / (x + 2), with x ≠ -2. It involves applying the First Derivative Test and analyzing the behavior of f'(x) on different intervals. Suitable for students at a Grade 12 or early college level.

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