If the function `f:[1,oo)->[1,oo)` is defined by `f(x)=2^(x(x-1)),` then `f^-1(x)` is (A) `(1/2)^(x(x-1))` (B) `1/2(1+ sqrt(1+4log_2x)`) (C) `1/2(1-sqrt(1+4log_2x))` (D) not defined

Options

A. `((1)/(2))^(x(x-1))`
B. `(1)/(2)(1+sqrt(1+4log_(2)x))`
C. `(1)/(2)(1-sqrt(1+4log_(2)x))`
D. not defined

Explanation

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

### Step 1: Set up the equation
We start by letting \( y = f(x) \):
\[ y = 2^{x(x-1)} \]

### Step 2: Take the logarithm
To solve for \( x \), we take the logarithm base 2 of both sides:
\[ \log_2 y = x(x-1) \]

### Step 3: Rearrange the equation
Rearranging gives us a quadratic equation in terms of \( x \):
\[ x^2 - x - \log_2 y = 0 \]

### Step 4: Apply the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = -\log_2 y \):
\[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-\log_2 y)}}{2 \cdot 1} \]
\[ x = \frac{1 \pm \sqrt{1 + 4\log_2 y}}{2} \]

### Step 5: Select the appropriate root
Since \( f(x) \) is defined for \( x \geq 1 \) and is increasing, we take the positive root:
\[ x = \frac{1 + \sqrt{1 + 4\log_2 y}}{2} \]

### Step 6: Substitute back to find \( f^{-1}(x) \)
Now, we replace \( y \) with \( x \) to express the inverse function:
\[ f^{-1}(x) = \frac{1 + \sqrt{1 + 4\log_2 x}}{2} \]

### Final Result
Thus, the inverse function is:
\[ f^{-1}(x) = \frac{1 + \sqrt{1 + 4\log_2 x}}{2} \]

### Conclusion
The correct answer is (B) \( \frac{1}{2}(1 + \sqrt{1 + 4\log_2 x}) \).

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