Let `f :[2,oo)to {1,oo)` defined by `f (x)=2^(x ^(4)-4x ^(3))and g : [(pi)/(2), pi] to A ` defined by `g (x) = (sin x+4)/(sin x-2)` be two invertible functions, then
` f ^(-1) (x)` is equal to

Options

A. `sqrt(2+sqrt(4-log_(2)x))`
B. `sqrt(2+sqrt(4+log_(2)x))`
C. `sqrt(2-sqrt(4+log_(2)x))`
D. None of these

Answer

To find the inverse of the function \( f(x) = 2^{x^4 - 4x^3} \), we need to follow these steps:

### Step 1: Set up the equation
We start with the equation for the inverse function:
\[ f(f^{-1}(x)) = x \]
This means we need to express \( f^{-1}(x) \) in terms of \( x \).

### Step 2: Substitute \( f^{-1}(x) \) with \( t \)
Let \( t = f^{-1}(x) \). Then we have:
\[ f(t) = 2^{t^4 - 4t^3} = x \]

### Step 3: Rewrite the equation
From the above, we can rewrite it as:
\[ t^4 - 4t^3 = \log_2(x) \]
This simplifies to:
\[ t^4 - 4t^3 - \log_2(x) = 0 \]

### Step 4: Recognize the quadratic form
Notice that this is a polynomial equation in \( t^2 \). We can let \( u = t^2 \), transforming our equation into:
\[ u^2 - 4u - \log_2(x) = 0 \]

### Step 5: Apply the quadratic formula
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -4, c = -\log_2(x) \):
\[ u = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-\log_2(x))}}{2 \cdot 1} \]
This simplifies to:
\[ u = \frac{4 \pm \sqrt{16 + 4\log_2(x)}}{2} \]
\[ u = 2 \pm \sqrt{4 + \log_2(x)} \]

### Step 6: Solve for \( t \)
Since \( u = t^2 \), we take the positive root (as \( t^2 \) cannot be negative):
\[ t^2 = 2 + \sqrt{4 + \log_2(x)} \]
Taking the square root gives us:
\[ t = \sqrt{2 + \sqrt{4 + \log_2(x)}} \]

### Step 7: Write the final expression for \( f^{-1}(x) \)
Thus, we have:
\[ f^{-1}(x) = \sqrt{2 + \sqrt{4 + \log_2(x)}} \]

### Final Answer
The inverse function is:
\[ f^{-1}(x) = \sqrt{2 + \sqrt{4 + \log_2(x)}} \]