The difference between the maximum and minimum values of the function `f(x)=x^(3)-3x+4, AA x in {0, 1]` is

Answer

To find the difference between the maximum and minimum values of the function \( f(x) = x^3 - 3x + 4 \) for \( x \) in the interval \([0, 1]\), we will follow these steps:

### Step 1: Find the derivative of the function
To analyze the behavior of the function, we first need to find its derivative:
\[ f'(x) = \frac{d}{dx}(x^3 - 3x + 4) = 3x^2 - 3 \]

### Step 2: Set the derivative to zero to find critical points
Next, we set the derivative equal to zero to find the critical points:
\[ 3x^2 - 3 = 0 \]
\[ 3(x^2 - 1) = 0 \]
\[ x^2 - 1 = 0 \]
\[ x = \pm 1 \]
Since we are only interested in the interval \([0, 1]\), we consider \( x = 1 \) as a critical point.

### Step 3: Evaluate the function at the endpoints and critical points
Now, we evaluate the function at the endpoints of the interval and at the critical point:
1. At \( x = 0 \):
\[ f(0) = 0^3 - 3(0) + 4 = 4 \]
2. At \( x = 1 \):
\[ f(1) = 1^3 - 3(1) + 4 = 1 - 3 + 4 = 2 \]

### Step 4: Determine the maximum and minimum values
From the evaluations:
- The maximum value of \( f(x) \) in the interval \([0, 1]\) is \( f(0) = 4 \).
- The minimum value of \( f(x) \) in the interval \([0, 1]\) is \( f(1) = 2 \).

### Step 5: Calculate the difference between the maximum and minimum values
Now, we find the difference between the maximum and minimum values:
\[ \text{Difference} = \text{Maximum} - \text{Minimum} = 4 - 2 = 2 \]

### Final Answer
The difference between the maximum and minimum values of the function \( f(x) \) in the interval \([0, 1]\) is \( \boxed{2} \).

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