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 an even function, odd function, or neither.

### Step 1: Definition of Even and Odd Functions
A function \( f(x) \) is:
- **Even** if \( f(-x) = f(x) \) for all \( x \).
- **Odd** if \( f(-x) = -f(x) \) for all \( x \).

### Step 2: Calculate \( f(-x) \)
We will compute \( f(-x) \):
\[ f(-x) = \log_{10}(-x + \sqrt{(-x)^2 + 1}) \]
Since \( (-x)^2 = x^2 \), we can simplify:
\[ f(-x) = \log_{10}(-x + \sqrt{x^2 + 1}) \]

### Step 3: Simplify \( f(-x) \)
Now we need to analyze the expression \( -x + \sqrt{x^2 + 1} \):
\[ f(-x) = \log_{10}(-x + \sqrt{x^2 + 1}) \]
Notice that \( \sqrt{x^2 + 1} \) is always greater than \( |x| \) because \( \sqrt{x^2 + 1} > |x| \) for all \( x \). Therefore, \( -x + \sqrt{x^2 + 1} \) is always positive.

### Step 4: Compare \( f(-x) \) and \( f(x) \)
Now we can compare \( f(-x) \) with \( f(x) \):
\[ f(x) = \log_{10}(x + \sqrt{x^2 + 1}) \]
To see if \( f(-x) = f(x) \), we can manipulate \( f(-x) \):
\[ f(-x) = \log_{10}(-x + \sqrt{x^2 + 1}) = \log_{10}(\sqrt{x^2 + 1} - x) \]

### Step 5: Use Logarithmic Properties
Using the property of logarithms, we can express \( f(-x) \):
\[ f(-x) = \log_{10}(\sqrt{x^2 + 1} - x) = \log_{10}\left(\frac{1}{x + \sqrt{x^2 + 1}}\right) \]
This is because:
\[ \sqrt{x^2 + 1} - x = \frac{1}{x + \sqrt{x^2 + 1}} \]

### Step 6: Final Comparison
Thus, we can express \( f(-x) \) as:
\[ f(-x) = -\log_{10}(x + \sqrt{x^2 + 1}) = -f(x) \]
This shows that:
\[ f(-x) = -f(x) \]
indicating that \( f(x) \) is an **odd function**.

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

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