If $f(x)=log_{e}(x+\sqrt{1+x^{2}}),$ then $f^{-1}(x)$ is equal to :

Question

Accepted Answer

To determine the inverse function $ f^{-1}(x) $ for the given function $ f(x) = \log_e(x + \sqrt{1 + x^2}) $, we need to find a function $ y $ such that $ f(y) = x $.

### Step-by-step Explanation:

1. **Start with the given function:**
   $$
   f(x) = \log_e(x + \sqrt{1 + x^2})
   $$

2. **Set $ f(y) = x $:**
   $$
   \log_e(y + \sqrt{1 + y^2}) = x
   $$

3. **Exponentiate both sides to eliminate the logarithm:**
   $$
   y + \sqrt{1 + y^2} = e^x
   $$

4. **Isolate $ y $:**
   - Square both sides to eliminate the square root:
     $$
     (y + \sqrt{1 + y^2})^2 = (e^x)^2
     $$
     $$
     y^2 + 2y\sqrt{1 + y^2} + 1 + y^2 = e^{2x}
     $$
     $$
     2y^2 + 1 + 2y\sqrt{1 + y^2} = e^{2x}
     $$

5. **Rearrange and simplify:**
   - Isolate the square root term:
     $$
     2y\sqrt{1 + y^2} = e^{2x} - 2y^2 - 1
     $$

6. **Solve for $ y $:**
   - Recognize that the expression $ y = \frac{e^x - e^{-x}}{2} $ is a known identity for the hyperbolic sine function, $ \sinh(x) $, which is the inverse of the original function:
     $$
     y = \sinh(x) = \frac{e^x - e^{-x}}{2}
     $$

### Verification:

- **Verify that $ y = \frac{e^x - e^{-x}}{2} $ satisfies the original equation:**
  - Substitute back into the original function:
    $$
    f\left(\frac{e^x - e^{-x}}{2}\right) = \log_e\left(\frac{e^x - e^{-x}}{2} + \sqrt{1 + \left(\frac{e^x - e^{-x}}{2}\right)^2}\right)
    $$
  - Simplify the expression inside the logarithm:
    $$
    \frac{e^x - e^{-x}}{2} + \sqrt{1 + \left(\frac{e^x - e^{-x}}{2}\right)^2} = e^x
    $$
  - Thus, the inverse function is correctly given by:
    $$
    f^{-1}(x) = \frac{e^x - e^{-x}}{2}
    $$

### Explanation of Incorrect Options:

- **Option 2: $\frac{e^x + e^{-x}}{2}$**
  - This represents the hyperbolic cosine function, $ \cosh(x) $, which does not satisfy the inverse relationship for the given function.

- **Option 3: $\frac{e^x - e^{-x}}{e^x + e^{-x}}$**
  - This expression simplifies to the hyperbolic tangent function, $ \tanh(x) $, which is not the inverse of the given function.

- **Option 4: $\frac{e^x + e^{-x}}{e^x - e^{-x}}$**
  - This expression does not correspond to any known inverse relationship for the given function and does not satisfy the original equation.

Therefore, the correct inverse function is indeed given by Option 1: $\frac{e^x - e^{-x}}{2}$.