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 \( g(x) \) as defined in the question.

1. **Define the functions**:
\[ f(x) = x^3 - x^2 + x + 1 \]
\[ g(x) = \begin{cases} \max(f(t), 0 \leq t \leq x) & \text{for } 0 \leq x \leq 1 \\ 3 - x & \text{for } 1 < x \leq 2 \end{cases} \]

2. **Analyze \( f(t) \)**:
We first need to find the derivative of \( f(t) \) to determine if it is increasing or decreasing:
\[ f'(t) = 3t^2 - 2t + 1 \]
To check if \( f'(t) \) is always positive, we can analyze the discriminant:
\[ D = (-2)^2 - 4 \cdot 3 \cdot 1 = 4 - 12 = -8 \]
Since the discriminant is negative, \( f'(t) \) has no real roots and is always positive. Thus, \( f(t) \) is an increasing function.

3. **Determine \( g(x) \) for \( 0 \leq x \leq 1 \)**:
Since \( f(t) \) is increasing, the maximum value of \( f(t) \) for \( 0 \leq t \leq x \) will occur at \( t = x \):
\[ g(x) = f(x) = x^3 - x^2 + x + 1 \quad \text{for } 0 \leq x \leq 1 \]

4. **Determine \( g(x) \) for \( 1 < x \leq 2 \)**:
For \( 1 < x \leq 2 \), \( g(x) \) is defined as:
\[ g(x) = 3 - x \]

5. **Check continuity at \( x = 1 \)**:
We need to check the limits from both sides at \( x = 1 \):
- From the left:
\[ \lim_{x \to 1^-} g(x) = f(1) = 1^3 - 1^2 + 1 + 1 = 2 \]
- From the right:
\[ \lim_{x \to 1^+} g(x) = 3 - 1 = 2 \]
Since both limits are equal, \( g(x) \) is continuous at \( x = 1 \).

6. **Check differentiability at \( x = 1 \)**:
We need to find the derivatives from both sides:
- For \( 0 \leq x \leq 1 \):
\[ g'(x) = f'(x) = 3x^2 - 2x + 1 \]
Evaluating at \( x = 1 \):
\[ g'(1^-) = 3(1)^2 - 2(1) + 1 = 2 \]
- For \( 1 < x \leq 2 \):
\[ g'(x) = -1 \]
Evaluating at \( x = 1 \):
\[ g'(1^+) = -1 \]

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

**Final Result**:
Thus, the function \( g(x) \) in the interval \([0, 2]\) is:
\[ g(x) = \begin{cases} x^3 - x^2 + x + 1 & \text{for } 0 \leq x \leq 1 \\ 3 - x & \text{for } 1 < x \leq 2 \end{cases} \]
and it is continuous at \( x = 1 \) but not differentiable at that point.