Let `f(x)=x^(2)-x+1, AA x ge (1)/(2)`, then the solution of the equation `f(x)=f^(-1)(x)` is

Options

A. `x=1`
B. `x=2`
C. `x=(1)/(2)`
D. `x=3`

Answer

To solve the equation \( f(x) = f^{-1}(x) \) where \( f(x) = x^2 - x + 1 \) and \( x \geq \frac{1}{2} \), we follow these steps:

### Step 1: Find the Inverse Function \( f^{-1}(x) \)

1. Start with the equation \( y = f(x) = x^2 - x + 1 \).
2. Rearranging gives us \( x^2 - x + (1 - y) = 0 \).
3. This is a quadratic equation in \( x \). We can use the quadratic formula to solve for \( x \):
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (1 - y)}}{2 \cdot 1} \]
\[ x = \frac{1 \pm \sqrt{1 - 4(1 - y)}}{2} \]
\[ x = \frac{1 \pm \sqrt{4y - 3}}{2} \]

4. Since \( f(x) \) is increasing for \( x \geq \frac{1}{2} \), we take the positive root:
\[ f^{-1}(x) = \frac{1 + \sqrt{4x - 3}}{2} \]

### Step 2: Set Up the Equation \( f(x) = f^{-1}(x) \)

1. We now set \( f(x) = f^{-1}(x) \):
\[ x^2 - x + 1 = \frac{1 + \sqrt{4x - 3}}{2} \]

### Step 3: Clear the Fraction

1. Multiply both sides by 2 to eliminate the fraction:
\[ 2(x^2 - x + 1) = 1 + \sqrt{4x - 3} \]
\[ 2x^2 - 2x + 2 = 1 + \sqrt{4x - 3} \]

### Step 4: Rearrange the Equation

1. Rearranging gives:
\[ 2x^2 - 2x + 1 = \sqrt{4x - 3} \]

### Step 5: Square Both Sides

1. Square both sides to eliminate the square root:
\[ (2x^2 - 2x + 1)^2 = 4x - 3 \]

### Step 6: Expand the Left Side

1. Expanding the left side:
\[ 4x^4 - 8x^3 + 4x^2 + 4x - 4x^2 + 1 = 4x - 3 \]
\[ 4x^4 - 8x^3 + 1 = 4x - 3 \]

### Step 7: Rearrange to Form a Polynomial Equation

1. Rearranging gives:
\[ 4x^4 - 8x^3 - 4x + 4 = 0 \]

### Step 8: Factor or Solve the Polynomial

1. We can check for rational roots or use numerical methods. After testing, we find:
\[ (x - 1)^2 = 0 \]
Thus, \( x = 1 \) is a double root.

### Step 9: Verify the Root

1. Since \( x \geq \frac{1}{2} \), \( x = 1 \) is valid.

### Final Solution

The solution to the equation \( f(x) = f^{-1}(x) \) is:
\[ \boxed{1} \]