Let `g (x)` be the inverse of `f (x) =(2 ^(x+1)-2^(1-x))/(2 ^(x)+2 ^(-x))` then g (x) be :

Options

A. `1/2 log _(2)((2+x)/(2-x))`
B. `-1/2 log _(2)((2+x)/(2-x))`
C. ` log _(2)((2+x)/(2-x))`
D. ` log _(2)((2-x)/(2+x))`

Answer

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

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

### Step 2: Rewrite the equation
To simplify the equation, we can rewrite it:
\[ y = \frac{2 \cdot 2^x - \frac{2}{2^x}}{2^x + \frac{1}{2^x}} \]
This can be expressed as:
\[ y = \frac{2^{x+1} - 2^{1-x}}{2^x + 2^{-x}} = \frac{2^{x+1} - 2 \cdot 2^{-x}}{2^x + 2^{-x}} \]

### Step 3: Cross-multiply
Cross-multiplying gives us:
\[ y(2^x + 2^{-x}) = 2^{x+1} - 2^{1-x} \]

### Step 4: Expand and rearrange
Expanding both sides:
\[ y \cdot 2^x + y \cdot 2^{-x} = 2^{x+1} - 2^{1-x} \]
Rearranging gives:
\[ y \cdot 2^x + y \cdot 2^{-x} - 2^{x+1} + 2^{1-x} = 0 \]

### Step 5: Substitute \( 2^x = z \)
Let \( z = 2^x \), then \( 2^{-x} = \frac{1}{z} \). The equation becomes:
\[ y \cdot z + \frac{y}{z} - 2z + 2/z = 0 \]
Multiplying through by \( z \) to eliminate the fraction:
\[ y z^2 + y - 2z^2 + 2 = 0 \]

### Step 6: Rearranging the quadratic equation
This is a quadratic equation in \( z \):
\[ y z^2 - 2z^2 + y + 2 = 0 \implies (y - 2)z^2 + y + 2 = 0 \]

### Step 7: Solve for \( z \)
Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = y - 2 \), \( b = y + 2 \), and \( c = 0 \):
\[ z = \frac{-(y + 2) \pm \sqrt{(y + 2)^2 - 4(y - 2)(2)}}{2(y - 2)} \]

### Step 8: Substitute back for \( x \)
Since \( z = 2^x \), we can take the logarithm:
\[ x = \log_2(z) \]

### Step 9: Final expression for \( g(x) \)
After simplifying, we find:
\[ g(x) = \frac{1}{2} \log_2 \left(\frac{y + 2}{2 - y}\right) \]

### Conclusion
Thus, the inverse function \( g(x) \) is:
\[ g(x) = \frac{1}{2} \log_2 \left(\frac{x + 2}{2 - x}\right) \]