What is the minimum value of `[x(x-1)+1]^(1/3)` where `0 lt x lt 1` ?

Options

A. `(3/4)^(1/3`
B. 1
C. `1/3`
D. `(3/8)^(1/3)`

Explanation

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

### Step 1: Simplify the expression inside the cube root
We start with the expression inside the cube root:
\[ f(x) = [x(x-1) + 1]^{1/3} \]
This simplifies to:
\[ f(x) = [x^2 - x + 1]^{1/3} \]

### Step 2: Define a new function
Let’s define a new function \( g(x) = x^2 - x + 1 \). We will find the minimum value of \( g(x) \) in the interval \( 0 < x < 1 \).

### Step 3: Find the vertex of the quadratic
The function \( g(x) \) is a quadratic function, and its minimum value occurs at the vertex. The x-coordinate of the vertex for a quadratic \( ax^2 + bx + c \) is given by:
\[ x = -\frac{b}{2a} \]
For \( g(x) = x^2 - x + 1 \), we have \( a = 1 \) and \( b = -1 \):
\[ x = -\frac{-1}{2 \cdot 1} = \frac{1}{2} \]

### Step 4: Evaluate \( g(x) \) at the vertex
Now we evaluate \( g(x) \) at \( x = \frac{1}{2} \):
\[ g\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \]

### Step 5: Find the minimum value of \( f(x) \)
Now we substitute this minimum value of \( g(x) \) back into \( f(x) \):
\[ f(x) = [g(x)]^{1/3} = \left(\frac{3}{4}\right)^{1/3} \]

### Conclusion
Thus, the minimum value of \( f(x) \) for \( 0 < x < 1 \) is:
\[ \left(\frac{3}{4}\right)^{1/3} \]