Let `f :[2,oo)to {1,oo)` defined by `f (x)=2^(x ^(4)-4x ^(2))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(4+sqrt(4+log _(2)x))`
D. `sqrt(4- sqrt(2+ log _(2)x))`

Answer

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

### Step 1: Set up the equation
Let \( y = f(x) \). Therefore, we have:
\[ y = 2^{x^4 - 4x^2} \]

### Step 2: Take the logarithm
To eliminate the exponent, take the logarithm base 2 of both sides:
\[ \log_2(y) = x^4 - 4x^2 \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ x^4 - 4x^2 - \log_2(y) = 0 \]

### Step 4: Substitute \( z = x^2 \)
Let \( z = x^2 \). Then the equation becomes:
\[ z^2 - 4z - \log_2(y) = 0 \]

### Step 5: Apply the quadratic formula
Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = -4 \), and \( c = -\log_2(y) \).
\[ z = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-\log_2(y))}}{2 \cdot 1} \]
\[ z = \frac{4 \pm \sqrt{16 + 4\log_2(y)}}{2} \]
\[ z = 2 \pm \sqrt{4 + \log_2(y)} \]

### Step 6: Solve for \( x \)
Since \( z = x^2 \), we take the positive root (as \( x \) must be non-negative):
\[ x^2 = 2 + \sqrt{4 + \log_2(y)} \]
\[ x = \sqrt{2 + \sqrt{4 + \log_2(y)}} \]

### Step 7: Substitute back to find \( f^{-1}(x) \)
Now, replace \( y \) with \( x \) to express the inverse function:
\[ f^{-1}(x) = \sqrt{2 + \sqrt{4 + \log_2(x)}} \]

Thus, the final answer is:
\[ f^{-1}(x) = \sqrt{2 + \sqrt{4 + \log_2(x)}} \]