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: Rewrite the function
Start with the function:
\[ f(x) = \frac{2^{x+1} - 2^{1-x}}{2^x + 2^{-x}} \]
We can simplify the numerator and the denominator.

### Step 2: Simplify the numerator
The numerator can be rewritten as:
\[ 2^{x+1} - 2^{1-x} = 2 \cdot 2^x - \frac{2}{2^x} = 2^{x+1} - \frac{2}{2^x} \]

### Step 3: Simplify the denominator
The denominator can be rewritten as:
\[ 2^x + 2^{-x} = 2^x + \frac{1}{2^x} \]

### Step 4: Combine the terms
Now substituting back, we have:
\[ f(x) = \frac{2^{x+1} - \frac{2}{2^x}}{2^x + \frac{1}{2^x}} = \frac{2^{x+1} - 2^{1-x}}{2^x + 2^{-x}} \]

### Step 5: Set \( f(x) = y \)
To find the inverse, we set:
\[ y = f(x) \]
This gives us:
\[ y = \frac{2^{x+1} - 2^{1-x}}{2^x + 2^{-x}} \]

### Step 6: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ y(2^x + 2^{-x}) = 2^{x+1} - 2^{1-x} \]

### Step 7: Rearranging the equation
Rearranging the equation, we get:
\[ y \cdot 2^x + y \cdot 2^{-x} = 2^{x+1} - 2^{1-x} \]

### Step 8: Isolate terms involving \( x \)
We can express this as:
\[ y \cdot 2^x + \frac{y}{2^x} = 2^{x+1} - \frac{2}{2^x} \]

### Step 9: Multiply through by \( 2^x \)
To eliminate the fraction, multiply through by \( 2^x \):
\[ y \cdot (2^x)^2 + y = 2^{x+1} \cdot 2^x - 2 \]

### Step 10: Rearranging gives a quadratic in \( 2^x \)
This can be rearranged to form a quadratic equation in terms of \( 2^x \):
\[ y \cdot (2^x)^2 - 2^{x+1} \cdot 2^x + y + 2 = 0 \]

### Step 11: Solve for \( 2^x \) using the quadratic formula
Using the quadratic formula \( ax^2 + bx + c = 0 \):
\[ 2^x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = y \), \( b = -2^{x+1} \), and \( c = y + 2 \).

### Step 12: Solve for \( x \)
Taking logarithm base 2 will give us:
\[ x = \log_2\left(\frac{-(-2^{x+1}) \pm \sqrt{(-2^{x+1})^2 - 4y(y + 2)}}{2y}\right) \]

### Step 13: Replace \( x \) with \( g(y) \)
Finally, we replace \( y \) with \( x \) to express \( g(x) \):
\[ g(x) = \frac{1}{2} \log_2\left(\frac{x + 2}{2 - x}\right) \]

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