To check the monotonicity of the function \( f(x) = x^3 - 3x + 1 \) at the points \( x = -1 \) and \( x = 2 \), we will follow these steps:
### Step 1: Find the first derivative of the function
We start by differentiating the function with respect to \( x \).
\[
f'(x) = \frac{d}{dx}(x^3 - 3x + 1) = 3x^2 - 3
\]
### Step 2: Set the first derivative to zero to find critical points
Next, we need to find the critical points by setting the first derivative equal to zero.
\[
3x^2 - 3 = 0
\]
Dividing both sides by 3:
\[
x^2 - 1 = 0
\]
Factoring the equation:
\[
(x - 1)(x + 1) = 0
\]
Thus, the critical points are:
\[
x = 1 \quad \text{and} \quad x = -1
\]
### Step 3: Analyze the intervals around the critical points
We will analyze the sign of \( f'(x) \) in the intervals determined by the critical points \( -1 \) and \( 1 \). The intervals are \( (-\infty, -1) \), \( (-1, 1) \), and \( (1, \infty) \).
1. **Interval \( (-\infty, -1) \)**:
Choose a test point, for example \( x = -2 \):
\[
f'(-2) = 3(-2)^2 - 3 = 12 - 3 = 9 \quad (> 0)
\]
So, \( f(x) \) is increasing in this interval.
2. **Interval \( (-1, 1) \)**:
Choose a test point, for example \( x = 0 \):
\[
f'(0) = 3(0)^2 - 3 = -3 \quad (< 0)
\]
So, \( f(x) \) is decreasing in this interval.
3. **Interval \( (1, \infty) \)**:
Choose a test point, for example \( x = 2 \):
\[
f'(2) = 3(2)^2 - 3 = 12 - 3 = 9 \quad (> 0)
\]
So, \( f(x) \) is increasing in this interval.
### Step 4: Check the points of interest
Now we check the points \( x = -1 \) and \( x = 2 \):
- At \( x = -1 \):
\[
f'(-1) = 3(-1)^2 - 3 = 3 - 3 = 0
\]
Since \( f'(-1) = 0 \) and the sign changes from positive to negative, \( f(x) \) is neither increasing nor decreasing at \( x = -1 \).
- At \( x = 2 \):
\[
f'(2) = 3(2)^2 - 3 = 12 - 3 = 9 \quad (> 0)
\]
Since \( f'(2) > 0 \), \( f(x) \) is increasing at \( x = 2 \).
### Final Conclusion
- At \( x = -1 \): \( f(x) \) is neither increasing nor decreasing.
- At \( x = 2 \): \( f(x) \) is increasing.
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