Let `f(x) = x^(3) - x^(2) + x + 1 and g(x) = {{:(max f(t)",", 0 le t le x,"for",0 le x le 1),(3-x",",1 lt x le 2,,):}` Then, g(x) in [0, 2] is

Options

A. continuous for `x in [0, 2] - {1}`
B. continuous for `x in [0, 2]`
C. differentiable for all `x in [0, 2]`
D. differentiable for all `x in [0, 2] - {1}`

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) defined in the question.

### Step 1: Define the functions
We have:
- \( f(x) = x^3 - x^2 + x + 1 \)
- \( g(x) = \max \{ f(t) \,|\, 0 \leq t \leq x \} \) for \( 0 \leq x \leq 1 \) and \( g(x) = 3 - x \) for \( 1 < x \leq 2 \).

### Step 2: Analyze \( f(x) \) in the interval \( [0, 1] \)
We need to find the maximum value of \( f(t) \) for \( t \) in the interval \( [0, x] \) where \( 0 \leq x \leq 1 \).

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

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

3. **Determine the behavior of \( f(t) \)**:
We need to check the derivative \( f'(x) \) to find critical points:
\[ f'(x) = 3x^2 - 2x + 1 \]
Since \( f'(x) \) is a quadratic function, we can check its discriminant:
\[ D = (-2)^2 - 4 \cdot 3 \cdot 1 = 4 - 12 = -8 \]
Since the discriminant is negative, \( f'(x) \) has no real roots, which means \( f(x) \) is either strictly increasing or strictly decreasing.

Evaluating \( f'(0) \):
\[ f'(0) = 3(0)^2 - 2(0) + 1 = 1 > 0 \]
Thus, \( f(x) \) is strictly increasing on \( [0, 1] \).

4. **Maximum of \( f(t) \) on \( [0, x] \)**:
Since \( f(x) \) is increasing, the maximum value of \( f(t) \) for \( 0 \leq t \leq x \) occurs at \( t = x \):
\[ g(x) = f(x) \text{ for } 0 \leq x \leq 1 \]

### Step 3: Define \( g(x) \) for \( 1 < x \leq 2 \)
For \( 1 < x \leq 2 \):
\[ g(x) = 3 - x \]

### Step 4: Check continuity of \( g(x) \) at \( x = 1 \)
We need to check if \( g(x) \) is continuous at \( x = 1 \):
- Left-hand limit as \( x \to 1^- \):
\[ g(1) = f(1) = 2 \]

- Right-hand limit as \( x \to 1^+ \):
\[ g(1) = 3 - 1 = 2 \]

Since both limits equal 2, \( g(x) \) is continuous at \( x = 1 \).

### Step 5: Check differentiability of \( g(x) \) at \( x = 1 \)
1. **Left-hand derivative**:
\[ g'(1^-) = f'(1) = 3(1)^2 - 2(1) + 1 = 3 - 2 + 1 = 2 \]

2. **Right-hand derivative**:
\[ g'(1^+) = \frac{d}{dx}(3 - x) = -1 \]

Since \( g'(1^-) \neq g'(1^+) \), \( g(x) \) is not differentiable at \( x = 1 \).

### Conclusion
Thus, the function \( g(x) \) is continuous on the interval \( [0, 2] \) but not differentiable at \( x = 1 \).

### Final Answer
The function \( g(x) \) is continuous on \( [0, 2] \) but not differentiable at \( x = 1 \).