If the function f :`[1,oo) to [1,oo)` is defined by `f (x) = 2^(x (x-1)`, then `f^(-1)` (x) is

Options

A. `(1/2)^(x(x-1))`
B. `1/2 {1 +sqrt(1 +4 log_2 x)}`
C. `1/2 {1 - sqrt(1+ 4 log_2 x)}`
D. Not defined

Answer

To find the inverse of the function \( f(x) = 2^{x(x-1)} \) defined on the interval \([1, \infty)\), we will follow these steps:

### Step-by-Step Solution

1. **Set the function equal to \( y \)**:
\[ y = f(x) = 2^{x(x-1)} \]

2. **Take the logarithm of both sides**:
\[ \log_2(y) = x(x-1) \]

3. **Rearrange the equation**:
\[ x^2 - x - \log_2(y) = 0 \]
This is a quadratic equation in terms of \( x \).

4. **Use the quadratic formula to solve for \( x \)**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -1 \), and \( c = -\log_2(y) \).
Substituting these values into the formula gives:
\[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-\log_2(y))}}{2 \cdot 1} \]
Simplifying this:
\[ x = \frac{1 \pm \sqrt{1 + 4\log_2(y)}}{2} \]

5. **Determine the correct sign**:
Since \( f(x) \) is increasing on \([1, \infty)\), we take the positive root:
\[ x = \frac{1 + \sqrt{1 + 4\log_2(y)}}{2} \]

6. **Replace \( y \) with \( x \) to express the inverse function**:
Thus, the inverse function \( f^{-1}(x) \) is:
\[ f^{-1}(x) = \frac{1 + \sqrt{1 + 4\log_2(x)}}{2} \]

### Final Answer
\[ f^{-1}(x) = \frac{1 + \sqrt{1 + 4\log_2(x)}}{2} \]