The function `f(x)=log(x+sqrt(x^(2)+1))` is

Options

A. an even function
B. an odd function
C. a periodic function
D. neither an even nor an odd function

Answer

To determine the nature of the function \( f(x) = \log(x + \sqrt{x^2 + 1}) \), we need to check if it is an even function, an odd function, or neither.

### Step-by-Step Solution:

1. **Definition of Even and Odd Functions**:
- A function \( f(x) \) is **even** if \( f(-x) = f(x) \).
- A function \( f(x) \) is **odd** if \( f(-x) = -f(x) \).

2. **Calculate \( f(-x) \)**:
\[ f(-x) = \log(-x + \sqrt{(-x)^2 + 1}) = \log(-x + \sqrt{x^2 + 1}) \]

3. **Simplifying \( f(-x) \)**:
- We can rewrite \( \sqrt{x^2 + 1} \) as it is since it does not change with the sign of \( x \).
- Thus, we have:
\[ f(-x) = \log(-x + \sqrt{x^2 + 1}) \]

4. **Using Conjugates**:
- To simplify \( f(-x) \), we can multiply the numerator and denominator by the conjugate:
\[ f(-x) = \log\left(\frac{(-x + \sqrt{x^2 + 1})(\sqrt{x^2 + 1} - x)}{(\sqrt{x^2 + 1} - x)}\right) \]
- This results in:
\[ f(-x) = \log\left(\frac{1}{x + \sqrt{x^2 + 1}}\right) \]

5. **Using Logarithmic Properties**:
- We can use the property of logarithms that states \( \log\left(\frac{1}{a}\right) = -\log(a) \):
\[ f(-x) = -\log(x + \sqrt{x^2 + 1}) = -f(x) \]

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

### Final Answer:
The function \( f(x) = \log(x + \sqrt{x^2 + 1}) \) is an **odd function**.

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