The inverse of the function of `f:R to R` given by `f(x)=log_(a) (x+sqrt(x^(2)+1)(a gt 0, a ne 1)` is

Answer

To find the inverse of the function \( f: \mathbb{R} \to \mathbb{R} \) given by

\[ f(x) = \log_a \left( x + \sqrt{x^2 + 1} \right) \quad (a > 0, a \neq 1), \]

we will follow these steps:

### Step 1: Set \( f(x) = y \)

Let \( y = f(x) \). Therefore, we have:

\[ y = \log_a \left( x + \sqrt{x^2 + 1} \right). \]

### Step 2: Rewrite the logarithmic equation

Using the property of logarithms, we can rewrite the equation in exponential form:

\[ a^y = x + \sqrt{x^2 + 1}. \]

### Step 3: Isolate the square root

Next, we isolate the square root term:

\[ \sqrt{x^2 + 1} = a^y - x. \]

### Step 4: Square both sides

To eliminate the square root, we square both sides:

\[ x^2 + 1 = (a^y - x)^2. \]

### Step 5: Expand the right side

Expanding the right side gives:

\[ x^2 + 1 = a^{2y} - 2a^y x + x^2. \]

### Step 6: Simplify the equation

We can simplify the equation by canceling \( x^2 \) from both sides:

\[ 1 = a^{2y} - 2a^y x. \]

### Step 7: Rearrange to solve for \( x \)

Rearranging the equation to solve for \( x \):

\[ 2a^y x = a^{2y} - 1. \]

Dividing both sides by \( 2a^y \):

\[ x = \frac{a^{2y} - 1}{2a^y}. \]

### Step 8: Write the inverse function

Since \( y \) was originally defined as \( f(x) \), we can express the inverse function \( f^{-1}(x) \) as:

\[ f^{-1}(x) = \frac{a^{2x} - 1}{2a^x}. \]

### Final Result

Thus, the inverse of the function \( f \) is:

\[ f^{-1}(x) = \frac{a^{2x} - 1}{2a^x}. \]

---