The maximum value of `[x(x-1)+1]^(1//3).0lexle1` is

Options

A. `((1)/(3))^((1)/(3))`
B. `(1)/(2)`
C. 1
D. 0

Answer

To find the maximum value of the function \( f(x) = [x(x-1) + 1]^{1/3} \) for \( x \) in the interval \([0, 1]\), we can follow these steps:

### Step 1: Define the function
We start with the function:
\[ f(x) = [x(x-1) + 1]^{1/3} \]

### Step 2: Find the critical points
To find the critical points, we need to differentiate \( f(x) \) and set the derivative equal to zero. Using the chain rule, we differentiate:
\[ f'(x) = \frac{1}{3}[x(x-1) + 1]^{-2/3} \cdot (2x - 1) \]
Setting \( f'(x) = 0 \):
\[ \frac{1}{3}[x(x-1) + 1]^{-2/3} \cdot (2x - 1) = 0 \]
This implies:
\[ 2x - 1 = 0 \implies x = \frac{1}{2} \]

### Step 3: Evaluate the function at critical points and endpoints
Now we need to evaluate \( f(x) \) at the critical point \( x = \frac{1}{2} \) and at the endpoints \( x = 0 \) and \( x = 1 \).

1. **At \( x = 0 \)**:
\[ f(0) = [0(0-1) + 1]^{1/3} = [0 + 1]^{1/3} = 1^{1/3} = 1 \]

2. **At \( x = 1 \)**:
\[ f(1) = [1(1-1) + 1]^{1/3} = [0 + 1]^{1/3} = 1^{1/3} = 1 \]

3. **At \( x = \frac{1}{2} \)**:
\[ f\left(\frac{1}{2}\right) = \left[\frac{1}{2}\left(\frac{1}{2}-1\right) + 1\right]^{1/3} = \left[\frac{1}{2} \cdot \left(-\frac{1}{2}\right) + 1\right]^{1/3} = \left[-\frac{1}{4} + 1\right]^{1/3} = \left[\frac{3}{4}\right]^{1/3} \]

### Step 4: Compare the values
Now we compare the values obtained:
- \( f(0) = 1 \)
- \( f(1) = 1 \)
- \( f\left(\frac{1}{2}\right) = \left(\frac{3}{4}\right)^{1/3} \)

Since \( \left(\frac{3}{4}\right)^{1/3} < 1 \), the maximum value occurs at both endpoints \( x = 0 \) and \( x = 1 \).

### Conclusion
The maximum value of \( f(x) \) in the interval \([0, 1]\) is:
\[ \boxed{1} \]