Check monotonocity at following points for
`f(x) =x^(3) -3x +1` at `x= -1,2`.

Answer

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 derivative of the function.
The first step is to differentiate the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^3 - 3x + 1) = 3x^2 - 3 \]

### Step 2: Factor the derivative.
Next, we can factor the derivative to make it easier to analyze.

\[ f'(x) = 3(x^2 - 1) = 3(x - 1)(x + 1) \]

### Step 3: Check the derivative at \( x = -1 \).
Now, we will evaluate the derivative at \( x = -1 \).

\[ f'(-1) = 3((-1)^2 - 1) = 3(1 - 1) = 3 \cdot 0 = 0 \]

### Step 4: Analyze the result at \( x = -1 \).
Since \( f'(-1) = 0 \), we cannot determine whether the function is increasing or decreasing at this point. Therefore, we conclude that the function is neither increasing nor decreasing at \( x = -1 \).

### Step 5: Check the derivative at \( x = 2 \).
Next, we will evaluate the derivative at \( x = 2 \).

\[ f'(2) = 3(2^2 - 1) = 3(4 - 1) = 3 \cdot 3 = 9 \]

### Step 6: Analyze the result at \( x = 2 \).
Since \( f'(2) = 9 \), which is greater than 0, we conclude that the function is monotonically increasing at \( x = 2 \).

### Final Conclusion:
- At \( x = -1 \): The function is neither increasing nor decreasing (not monotonic).
- At \( x = 2 \): The function is monotonically increasing.

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