Prove that `f(x)=(1//x)logsqrt(x+sqrt(x^(2)+1))` is an even function.

Answer

To prove that the function \( f(x) = \frac{1}{x} \log \sqrt{x + \sqrt{x^2 + 1}} \) is an even function, we need to show that \( f(-x) = f(x) \).

### Step-by-Step Solution:

1. **Define the Function**:
We start with the function:
\[ f(x) = \frac{1}{x} \log \sqrt{x + \sqrt{x^2 + 1}} \]

2. **Calculate \( f(-x) \)**:
We substitute \(-x\) into the function:
\[ f(-x) = \frac{1}{-x} \log \sqrt{-x + \sqrt{(-x)^2 + 1}} \]
Simplifying the expression inside the logarithm:
\[ f(-x) = \frac{1}{-x} \log \sqrt{-x + \sqrt{x^2 + 1}} \]

3. **Factor Out the Negative**:
We can factor out the negative from the logarithm:
\[ f(-x) = -\frac{1}{x} \log \sqrt{-x + \sqrt{x^2 + 1}} \]

4. **Use Logarithmic Properties**:
Using the property of logarithms, we can rewrite:
\[ f(-x) = -\frac{1}{x} \cdot \frac{1}{2} \log(-x + \sqrt{x^2 + 1}) \]
This can be expressed as:
\[ f(-x) = \frac{1}{x} \cdot \left(-\frac{1}{2} \log(-x + \sqrt{x^2 + 1})\right) \]

5. **Simplify the Logarithmic Expression**:
Notice that:
\[ -x + \sqrt{x^2 + 1} = \sqrt{x^2 + 1} - x \]
Therefore:
\[ f(-x) = \frac{1}{x} \cdot \frac{1}{2} \log \sqrt{x + \sqrt{x^2 + 1}} = \frac{1}{x} \log \sqrt{x + \sqrt{x^2 + 1}} \]
Thus:
\[ f(-x) = f(x) \]

6. **Conclusion**:
Since \( f(-x) = f(x) \), we conclude that \( f(x) \) is an even function.