The maximum value of `[x(x-1)+1]^(1/3), 0 le x le 1` is:

Options

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

Explanation

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

### Step 1: Define the function
We start with the function:
\[ f(x) = [x(x-1) + 1]^{1/3} \]
This can be simplified to:
\[ f(x) = [x^2 - x + 1]^{1/3} \]

### Step 2: Find the derivative
Next, we find the derivative \( f'(x) \) using the chain rule:
\[ f'(x) = \frac{1}{3}[x^2 - x + 1]^{-2/3} \cdot (2x - 1) \]
This derivative is set to zero to find critical points:
\[ \frac{1}{3}[x^2 - x + 1]^{-2/3} \cdot (2x - 1) = 0 \]
Since \( [x^2 - x + 1]^{-2/3} \) is never zero, we can focus on:
\[ 2x - 1 = 0 \]
Solving for \( x \):
\[ x = \frac{1}{2} \]

### Step 3: Evaluate the function at critical points and endpoints
Now we evaluate \( f(x) \) at the critical point \( x = \frac{1}{2} \) and 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 \( x = \frac{1}{2} \)**:
\[ f\left(\frac{1}{2}\right) = \left[\left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1\right]^{1/3} = \left[\frac{1}{4} - \frac{1}{2} + 1\right]^{1/3} = \left[\frac{1}{4} - \frac{2}{4} + \frac{4}{4}\right]^{1/3} = \left[\frac{3}{4}\right]^{1/3} \]

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

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

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