Let f(b) be the minimum value of the expression `y=x^(2)-2x+(b^(3)-3b^(2)+4)AA x in R`. Then, the maximum value of f(b) as b varies from 0 to 4 is

Options

A. 20
B. 19
C. 63
D. 64

Answer

To solve the problem, we need to find the maximum value of the function \( f(b) = b^3 - 3b^2 + 3 \) as \( b \) varies from 0 to 4. Here’s a step-by-step solution:

### Step 1: Define the function
The function we are analyzing is:
\[ f(b) = b^3 - 3b^2 + 3 \]

### Step 2: Find the derivative
To find the critical points, we first compute the derivative of \( f(b) \):
\[ f'(b) = 3b^2 - 6b \]

### Step 3: Set the derivative to zero
Next, we set the derivative equal to zero to find the critical points:
\[ 3b^2 - 6b = 0 \]
Factoring out \( 3b \):
\[ 3b(b - 2) = 0 \]
This gives us two critical points:
\[ b = 0 \quad \text{and} \quad b = 2 \]

### Step 4: Evaluate the function at critical points and endpoints
Now we evaluate \( f(b) \) at the critical points and the endpoints \( b = 0 \) and \( b = 4 \).

1. **At \( b = 0 \)**:
\[ f(0) = 0^3 - 3(0^2) + 3 = 3 \]

2. **At \( b = 2 \)**:
\[ f(2) = 2^3 - 3(2^2) + 3 = 8 - 12 + 3 = -1 \]

3. **At \( b = 4 \)**:
\[ f(4) = 4^3 - 3(4^2) + 3 = 64 - 48 + 3 = 19 \]

### Step 5: Determine the maximum value
Now we compare the values:
- \( f(0) = 3 \)
- \( f(2) = -1 \)
- \( f(4) = 19 \)

The maximum value of \( f(b) \) as \( b \) varies from 0 to 4 is:
\[ \text{Maximum value} = 19 \]

### Final Answer
Thus, the maximum value of \( f(b) \) as \( b \) varies from 0 to 4 is:
\[ \boxed{19} \]