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 \( f(b) \) defined as the minimum value of the expression \( y = x^2 - 2x + (b^3 - 3b^2 + 4) \) for \( x \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Identify the Expression**:
The expression we need to analyze is:
\[ y = x^2 - 2x + (b^3 - 3b^2 + 4) \]
Here, \( b^3 - 3b^2 + 4 \) is a constant with respect to \( x \).

2. **Complete the Square**:
The quadratic part \( x^2 - 2x \) can be rewritten by completing the square:
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Therefore, we can rewrite \( y \) as:
\[ y = (x - 1)^2 - 1 + (b^3 - 3b^2 + 4) \]
Simplifying this gives:
\[ y = (x - 1)^2 + (b^3 - 3b^2 + 3) \]

3. **Find the Minimum Value**:
The minimum value of \( y \) occurs when \( (x - 1)^2 = 0 \), which is at \( x = 1 \). Thus, the minimum value of \( y \) is:
\[ f(b) = b^3 - 3b^2 + 3 \]

4. **Determine the Maximum Value of \( f(b) \)**:
We need to find the maximum value of \( f(b) \) as \( b \) varies from 0 to 4. We will evaluate \( f(b) \) at the endpoints and any critical points within this interval.

5. **Find the Derivative**:
To find critical points, we differentiate \( f(b) \):
\[ f'(b) = 3b^2 - 6b \]
Setting the derivative to zero:
\[ 3b^2 - 6b = 0 \implies 3b(b - 2) = 0 \]
This gives critical points \( b = 0 \) and \( b = 2 \).

6. **Evaluate \( f(b) \) at Critical Points and Endpoints**:
- \( f(0) = 0^3 - 3(0^2) + 3 = 3 \)
- \( f(2) = 2^3 - 3(2^2) + 3 = 8 - 12 + 3 = -1 \)
- \( f(4) = 4^3 - 3(4^2) + 3 = 64 - 48 + 3 = 19 \)

7. **Compare Values**:
Now we compare the values:
- \( f(0) = 3 \)
- \( f(2) = -1 \)
- \( f(4) = 19 \)

The maximum value of \( f(b) \) in the interval \( [0, 4] \) is:
\[ \max(f(0), f(2), f(4)) = \max(3, -1, 19) = 19 \]

### Conclusion:
The maximum value of \( f(b) \) as \( b \) varies from 0 to 4 is \( \boxed{19} \).