What is the minimum value off `|x(x-1)+1|^(1//3)`, where `0 le x le 1`?

Options

A. `(3/4)^(1/3)`
B. 1
C. `1/2`
D. `(3/8)^(1/3)`

Answer

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

### Step 1: Simplify the expression inside the absolute value
First, we simplify the expression inside the absolute value:
\[ f(x) = |x(x-1) + 1|^{1/3} \]
Calculating \( x(x-1) + 1 \):
\[ x(x-1) + 1 = x^2 - x + 1 \]

### Step 2: Analyze the function \( g(x) = x^2 - x + 1 \)
Next, we analyze the function \( g(x) = x^2 - x + 1 \) to find its minimum value in the interval \( [0, 1] \).

### Step 3: Find the critical points of \( g(x) \)
To find the critical points, we take the derivative of \( g(x) \):
\[ g'(x) = 2x - 1 \]
Setting the derivative equal to zero to find critical points:
\[ 2x - 1 = 0 \implies x = \frac{1}{2} \]

### Step 4: Evaluate \( g(x) \) at the endpoints and the critical point
Now we evaluate \( g(x) \) at the endpoints \( x = 0 \) and \( x = 1 \), and at the critical point \( x = \frac{1}{2} \):
- At \( x = 0 \):
\[ g(0) = 0^2 - 0 + 1 = 1 \]
- At \( x = 1 \):
\[ g(1) = 1^2 - 1 + 1 = 1 \]
- At \( x = \frac{1}{2} \):
\[ g\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \]

### Step 5: Determine the minimum value of \( g(x) \)
The values we found are:
- \( g(0) = 1 \)
- \( g(1) = 1 \)
- \( g\left(\frac{1}{2}\right) = \frac{3}{4} \)

The minimum value of \( g(x) \) in the interval \( [0, 1] \) is \( \frac{3}{4} \).

### Step 6: Find the minimum value of \( f(x) \)
Now we substitute this minimum value back into \( f(x) \):
\[ f(x) = |g(x)|^{1/3} = \left(\frac{3}{4}\right)^{1/3} \]

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