The function `f(x)=log_(10)(x+sqrt(x^(2))+1)` is

Options

A. an even function
B. an odd function
C. periodic function
D. none of these

Answer

To determine the nature of the function \( f(x) = \log_{10}(x + \sqrt{x^2 + 1}) \), we need to check if it is even, odd, or periodic.

### Step-by-Step Solution:

1. **Define the function:**
\[ f(x) = \log_{10}(x + \sqrt{x^2 + 1}) \]

2. **Find \( f(-x) \):**
We need to substitute \(-x\) into the function:
\[ f(-x) = \log_{10}(-x + \sqrt{(-x)^2 + 1}) = \log_{10}(-x + \sqrt{x^2 + 1}) \]

3. **Simplify \( f(-x) \):**
To simplify \( f(-x) \), we can rationalize the expression:
\[ f(-x) = \log_{10}(-x + \sqrt{x^2 + 1}) \]
We can multiply and divide by the conjugate:
\[ f(-x) = \log_{10}\left(\frac{(-x + \sqrt{x^2 + 1})(-x + \sqrt{x^2 + 1})}{(-x + \sqrt{x^2 + 1})(-x + \sqrt{x^2 + 1})}\right) \]
This simplifies to:
\[ f(-x) = \log_{10}\left(\frac{(x^2 + 1) - x^2}{\sqrt{x^2 + 1} + x}\right) = \log_{10}\left(\frac{1}{\sqrt{x^2 + 1} + x}\right) \]

4. **Rewrite \( f(-x) \):**
Using the property of logarithms:
\[ f(-x) = -\log_{10}(\sqrt{x^2 + 1} + x) \]

5. **Compare \( f(-x) \) and \( -f(x) \):**
Now, we compare \( f(-x) \) with \(-f(x)\):
\[ -f(x) = -\log_{10}(x + \sqrt{x^2 + 1}) \]
We see that:
\[ f(-x) = -f(x) \]

6. **Conclusion:**
Since \( f(-x) = -f(x) \), the function \( f(x) \) is an **odd function**.

### Final Answer:
The function \( f(x) = \log_{10}(x + \sqrt{x^2 + 1}) \) is an odd function.

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