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 \( y = [x(x-1) + 1]^{1/3} \) for \( 0 \leq x \leq 1 \), we can follow these steps:

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

### Step 2: Differentiate the function
To find the critical points, we need to differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{3}[x(x-1) + 1]^{-2/3} \cdot \frac{d}{dx}[x(x-1) + 1] \]
Now, we differentiate the inside function \( x(x-1) + 1 \):
\[ \frac{d}{dx}[x(x-1)] = x \cdot 1 + (x-1) \cdot 1 = 2x - 1 \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{3}[x(x-1) + 1]^{-2/3} \cdot (2x - 1) \]

### Step 3: Set the derivative to zero
To find the critical points, we set \( \frac{dy}{dx} = 0 \):
\[ \frac{1}{3}[x(x-1) + 1]^{-2/3} \cdot (2x - 1) = 0 \]
This gives us two cases:
1. \( 2x - 1 = 0 \)
2. \( [x(x-1) + 1]^{-2/3} \) is never zero since \( x(x-1) + 1 \) is always positive in the interval.

From \( 2x - 1 = 0 \):
\[ x = \frac{1}{2} \]

### Step 4: Evaluate the function at critical points and endpoints
Now we need to evaluate \( y \) at \( x = 0 \), \( x = 1 \), and \( x = \frac{1}{2} \).

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

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

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

### Step 5: Compare values
Now we compare the values:
- \( y(0) = 1 \)
- \( y(1) = 1 \)
- \( y\left(\frac{1}{2}\right) = \left(\frac{3}{4}\right)^{1/3} \) which is less than 1.

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