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Let's analyze the problem based on the graph of \( g'(x) \), the derivative of \( g(x) \), and the related questions.

### 1. **Graph Interpretation**:
The graph represents the derivative \( g'(x) \), and we need to extract key information to answer questions about \( g(x) \).

- The graph of \( g'(x) \) crosses the x-axis at \( x = -2 \), \( x = 0 \), and \( x = 1 \). These are the points where the derivative equals zero, indicating potential stationary points for \( g(x) \).
- \( g'(x) \) is positive (above the x-axis) between \( x = -2 \) and \( x = 0 \), and between \( x = 1 \) and beyond.
- \( g'(x) \) is negative (below the x-axis) between \( x = 0 \) and \( x = 1 \).

### 2. **Answers**:
#### 4.1. For which value(s) of \( x \) is \( g(x) \):
##### 4.1.1 Strictly decreasing?
- \( g(x) \) is strictly decreasing when \( g'(x) < 0 \).
- From the graph, \( g'(x) < 0 \) in the interval \( (0, 1) \).
- **Answer:** \( x \in (0, 1) \).

##### 4.1.2 Increasing?
- \( g(x) \) is increasing when \( g'(x) > 0 \).
- From the graph, \( g'(x) > 0 \) in the intervals \( (-2, 0) \) and \( (1, \infty) \).
- **Answer:** \( x \in (-2, 0) \cup (1, \infty) \).

#### 4.2. Determine the x-coordinates of the turning point(s) of \( g(x) \):
- The turning points occur where \( g'(x) = 0 \) and the derivative changes sign.
- The graph shows that \( g'(x) = 0 \) at \( x = -2 \), \( x = 0 \), and \( x = 1 \).
  - At \( x = -2 \), \( g'(x) \) does not change sign, so no turning point here.
  - At \( x = 0 \), \( g'(x) \) changes from positive to negative, indicating a **local maximum**.
  - At \( x = 1 \), \( g'(x) \) changes from negative to positive, indicating a **local minimum**.
- **Answer:** Turning points are at \( x = 0 \) and \( x = 1 \).

#### 4.3. Classify the stationary point(s):
- Stationary points are where \( g'(x) = 0 \).
  - At \( x = -2 \), \( g'(x) = 0 \), but there's no sign change, so this is an **inflection point**.
  - At \( x = 0 \), there's a **local maximum**.
  - At \( x = 1 \), there's a **local minimum**.
- **Answer:** \( x = -2 \) is an inflection point, \( x = 0 \) is a local maximum, and \( x = 1 \) is a local minimum.

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Do you have any questions or want further details on any part of this explanation? Here are 5 related questions to consider:

1. What is the geometric meaning of a turning point on a graph?
2. How can we determine whether a stationary point is a maximum, minimum, or inflection point?
3. What happens to the concavity of the function at the inflection point \( x = -2 \)?
4. Can a function have more than one inflection point, and if so, how do we identify them from the derivative graph?
5. Why is the sign of \( g'(x) \) important for understanding the behavior of \( g(x) \)?

**Tip:** When analyzing the behavior of a function, always check where its derivative changes signs to find intervals of increase, decrease, and turning points.

Consider the graph of the derivative of g(x). 4.1 For which value(s) of x is g: 4.1.1 strictly decreasing? 4.1.2 increasing? 4.2 Determine the x-coordinates of the turning point(s) of g(x). 4.3 Classify the stationary point(s).

Explore the graph of the derivative of g(x) to determine increasing, decreasing intervals, and classify stationary points. This problem involves key calculus concepts such as derivatives, turning points, and stationary points, suitable for Grades 11-12 or college-level calculus.

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