The positive difference between the local maximum value and the local minimum value of the function `f(x)=x^(3)-3x-1, AA x in [-2, 3]` is

Options

A. 20
B. 4
C. 14
D. 22

Answer

To solve the problem of finding the positive difference between the local maximum and local minimum values of the function \( f(x) = x^3 - 3x - 1 \) on the interval \([-2, 3]\), we can follow these steps:

### Step 1: Find the derivative of the function
To find the local maxima and minima, we first need to find the derivative of the function:
\[ f'(x) = \frac{d}{dx}(x^3 - 3x - 1) = 3x^2 - 3 \]

### Step 2: Set the derivative equal to zero
Next, we set the derivative equal to zero to find the critical points:
\[ 3x^2 - 3 = 0 \]
\[ x^2 = 1 \]
\[ x = 1 \quad \text{or} \quad x = -1 \]

### Step 3: Determine the second derivative
Now we find the second derivative to classify the critical points:
\[ f''(x) = \frac{d}{dx}(3x^2 - 3) = 6x \]

### Step 4: Evaluate the second derivative at the critical points
We evaluate the second derivative at \( x = -1 \) and \( x = 1 \):
- For \( x = -1 \):
\[ f''(-1) = 6(-1) = -6 \quad (\text{local maximum}) \]
- For \( x = 1 \):
\[ f''(1) = 6(1) = 6 \quad (\text{local minimum}) \]

### Step 5: Calculate the function values at the critical points
Now we find the values of the function at these critical points:
- For \( x = -1 \):
\[ f(-1) = (-1)^3 - 3(-1) - 1 = -1 + 3 - 1 = 1 \quad (\text{local maximum}) \]
- For \( x = 1 \):
\[ f(1) = (1)^3 - 3(1) - 1 = 1 - 3 - 1 = -3 \quad (\text{local minimum}) \]

### Step 6: Calculate the positive difference
Finally, we calculate the positive difference between the local maximum and local minimum values:
\[ \text{Positive difference} = |f(-1) - f(1)| = |1 - (-3)| = |1 + 3| = 4 \]

### Final Answer
The positive difference between the local maximum value and the local minimum value of the function is:
\[ \boxed{4} \]