Let `f:[4,infty) to [1,infty)` be defined by
`f(x) = 11^(x(4-x)) AA x ge 4`
Then `f^(-1)(x)` is given by

Options

A. `2-sqrt(4+ 1 log_(11)(x)|) AA x ge 1`
B. `2 + sqrt(4+log_(11)(x)) AA x ge 1`
C. `2+ sqrt(4 - log_(11)(x)) AA x ge 1`
D. `sqrt(4+ log_(11)(x)-2) AA x ge 1`

Answer

To find the inverse of the function \( f(x) = 11^{x(4-x)} \) defined on the interval \( [4, \infty) \) with a codomain of \( [1, \infty) \), we will follow these steps:

### Step 1: Set up the equation for the inverse
Let \( y = f(x) \). Then we have:
\[ y = 11^{x(4-x)} \]

### Step 2: Take the logarithm of both sides
To solve for \( x \), we take the logarithm base 11 of both sides:
\[ \log_{11}(y) = x(4 - x) \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ x(4 - x) = \log_{11}(y) \]
This can be rewritten as:
\[ -x^2 + 4x - \log_{11}(y) = 0 \]

### Step 4: Identify the quadratic equation
This is a quadratic equation in the standard form:
\[ -x^2 + 4x - \log_{11}(y) = 0 \]
We can multiply through by -1 to make it easier to work with:
\[ x^2 - 4x + \log_{11}(y) = 0 \]

### Step 5: Apply the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -4, c = \log_{11}(y) \):
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot \log_{11}(y)}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{4 \pm \sqrt{16 - 4\log_{11}(y)}}{2} \]
\[ x = 2 \pm \sqrt{4 - \log_{11}(y)} \]

### Step 6: Determine the correct sign
Since \( f \) is defined for \( x \geq 4 \), we need to choose the positive root:
\[ x = 2 + \sqrt{4 - \log_{11}(y)} \]

### Step 7: Write the inverse function
Thus, the inverse function \( f^{-1}(y) \) is given by:
\[ f^{-1}(y) = 2 + \sqrt{4 - \log_{11}(y)} \]

### Final Result
The final expression for the inverse function is:
\[ f^{-1}(x) = 2 + \sqrt{4 - \log_{11}(x)} \]