The inverse of the functions `f(x)=log_(2)(x+sqrt(x^(2)+1))` is

Options

A. `2^(x)+2^(-x)`
B. `(2^(x)+2^(-x))/(2)`
C. `(2^(-x)-2^(x))/(2)`
D. `(2^(x)-2^(-x))/(2)`

Answer

To find the inverse of the function \( f(x) = \log_2(x + \sqrt{x^2 + 1}) \), we will follow these steps:

### Step 1: Set the function equal to \( y \)
Let \( y = f(x) \):
\[ y = \log_2(x + \sqrt{x^2 + 1}) \]

### Step 2: Rewrite the logarithmic equation in exponential form
Using the property of logarithms, we can rewrite the equation:
\[ x + \sqrt{x^2 + 1} = 2^y \]

### Step 3: Isolate the square root
Now, we will isolate the square root term:
\[ \sqrt{x^2 + 1} = 2^y - x \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ x^2 + 1 = (2^y - x)^2 \]

### Step 5: Expand the right side
Expanding the right side:
\[ x^2 + 1 = 2^{2y} - 2 \cdot 2^y \cdot x + x^2 \]

### Step 6: Simplify the equation
Subtract \( x^2 \) from both sides:
\[ 1 = 2^{2y} - 2 \cdot 2^y \cdot x \]

### Step 7: Rearrange to solve for \( x \)
Rearranging gives:
\[ 2 \cdot 2^y \cdot x = 2^{2y} - 1 \]
\[ x = \frac{2^{2y} - 1}{2 \cdot 2^y} \]

### Step 8: Replace \( y \) with \( x \) to find the inverse
Now, replace \( y \) with \( x \) to express the inverse function:
\[ f^{-1}(x) = \frac{2^{2x} - 1}{2 \cdot 2^x} \]

### Step 9: Simplify the inverse function
We can simplify this further:
\[ f^{-1}(x) = \frac{2^{2x} - 1}{2^{x+1}} = \frac{2^{2x}}{2^{x+1}} - \frac{1}{2^{x+1}} = 2^{x} - 2^{-x} \]

Thus, the final result for the inverse function is:
\[ f^{-1}(x) = 2^x - 2^{-x} \]