If `f:[1, oo) rarr [1, oo)` is defined as `f(x) = 3^(x(x-2))` then `f^(-1)(x)` is equal to

Options

A. (a)`1+sqrt(1+log_(3)x)`
B. (b)`1-sqrt(1+log_(3)x)`
C. (c)`1+sqrt(1-log_(3)x)`
D. (d)Does not exist

Explanation

To find the inverse function \( f^{-1}(x) \) for the function \( f(x) = 3^{x(x-2)} \), we will follow these steps:

### Step 1: Set up the equation
We start with the equation for the function:
\[ y = f(x) = 3^{x(x-2)} \]
We want to express \( x \) in terms of \( y \).

### Step 2: Take the logarithm
To solve for \( x \), we take the logarithm of both sides:
\[ \log_3(y) = x(x-2) \]
This simplifies to:
\[ x^2 - 2x - \log_3(y) = 0 \]

### Step 3: Use the quadratic formula
Now we can use the quadratic formula to solve for \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2 \), and \( c = -\log_3(y) \). Plugging these values into the formula gives:
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-\log_3(y))}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{2 \pm \sqrt{4 + 4\log_3(y)}}{2} \]
\[ x = 1 \pm \sqrt{1 + \log_3(y)} \]

### Step 4: Determine the correct root
Since \( f(x) \) is defined from \( [1, \infty) \) to \( [1, \infty) \), we need to take the positive root to ensure \( x \) remains in the valid range:
\[ x = 1 + \sqrt{1 + \log_3(y)} \]

### Step 5: Write the inverse function
Thus, we have:
\[ f^{-1}(y) = 1 + \sqrt{1 + \log_3(y)} \]

### Final Answer
The inverse function is:
\[ f^{-1}(x) = 1 + \sqrt{1 + \log_3(x)} \]