Let `f :[1/2,oo)rarr[3/4,oo),` where `f(x)=x^2-x+1.` Find the inverse of f(x). Hence or otherwise solve the equation, `x^2-x+1=1/2+sqrt(x-3/4.`

Answer

To find the inverse of the function \( f(x) = x^2 - x + 1 \) defined on the interval \([1/2, \infty)\) and then solve the equation \( x^2 - x + 1 = \frac{1}{2} + \sqrt{x - \frac{3}{4}} \), we will follow these steps:

### Step 1: Find the Inverse of \( f(x) \)

1. **Set \( y = f(x) \)**:
\[ y = x^2 - x + 1 \]

2. **Rearrange the equation to solve for \( x \)**:
\[ x^2 - x + (1 - y) = 0 \]
This is a quadratic equation in \( x \).

3. **Use the quadratic formula**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -1, c = 1 - y \). Plugging in these values:
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (1 - y)}}{2 \cdot 1} \]
\[ x = \frac{1 \pm \sqrt{1 - 4(1 - y)}}{2} \]
\[ x = \frac{1 \pm \sqrt{4y - 3}}{2} \]

4. **Select the appropriate root**:
Since \( f(x) \) is increasing on the interval \([1/2, \infty)\), we take the positive root:
\[ f^{-1}(y) = \frac{1 + \sqrt{4y - 3}}{2} \]

### Step 2: Solve the Equation

Now we need to solve the equation:
\[ x^2 - x + 1 = \frac{1}{2} + \sqrt{x - \frac{3}{4}} \]

1. **Substitute \( f(x) \)**:
From the definition of \( f(x) \):
\[ f(x) = \frac{1}{2} + \sqrt{x - \frac{3}{4}} \]

2. **Isolate the square root**:
Rearranging gives:
\[ x^2 - x + \frac{1}{2} = \sqrt{x - \frac{3}{4}} \]

3. **Square both sides**:
\[ (x^2 - x + \frac{1}{2})^2 = x - \frac{3}{4} \]

4. **Expand the left side**:
\[ (x^2 - x + \frac{1}{2})^2 = x^4 - 2x^3 + \frac{5}{2}x^2 - x + \frac{1}{4} \]

5. **Set the equation to zero**:
\[ x^4 - 2x^3 + \frac{5}{2}x^2 - x + \frac{1}{4} - x + \frac{3}{4} = 0 \]
\[ x^4 - 2x^3 + \frac{5}{2}x^2 - 2x + 1 = 0 \]

6. **Use numerical methods or graphing to find roots**:
The roots can be found using numerical methods or graphing techniques.

### Final Solution

After solving the polynomial, we find that \( x = 1 \) is a solution.