Classify the following functions for being even or odd:
` sin [ log (x + sqrt(x^2 +1))]`

Answer

To classify the function \( f(x) = \sin \left( \log \left( x + \sqrt{x^2 + 1} \right) \right) \) as even or odd, we will follow these steps:

### Step 1: Define the function
Let \( f(x) = \sin \left( \log \left( x + \sqrt{x^2 + 1} \right) \right) \).

### Step 2: Find \( f(-x) \)
To check if the function is even or odd, we need to compute \( f(-x) \):
\[ f(-x) = \sin \left( \log \left( -x + \sqrt{(-x)^2 + 1} \right) \right) \]
Since \( (-x)^2 = x^2 \), we can simplify:
\[ f(-x) = \sin \left( \log \left( -x + \sqrt{x^2 + 1} \right) \right) \]

### Step 3: Simplify \( f(-x) \)
Next, we simplify \( \log \left( -x + \sqrt{x^2 + 1} \right) \). We can use the property of logarithms:
\[ \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \]
We know that:
\[ -x + \sqrt{x^2 + 1} = \frac{1}{x + \sqrt{x^2 + 1}} \]
Thus:
\[ \log \left( -x + \sqrt{x^2 + 1} \right) = \log \left( \frac{1}{x + \sqrt{x^2 + 1}} \right) = -\log \left( x + \sqrt{x^2 + 1} \right) \]
So we have:
\[ f(-x) = \sin \left( -\log \left( x + \sqrt{x^2 + 1} \right) \right) \]

### Step 4: Use the sine property
Using the property of sine, \( \sin(-\theta) = -\sin(\theta) \):
\[ f(-x) = -\sin \left( \log \left( x + \sqrt{x^2 + 1} \right) \right) = -f(x) \]

### Step 5: Conclusion
Since \( f(-x) = -f(x) \), the function \( f(x) \) is classified as an **odd function**.

### Summary
The function \( f(x) = \sin \left( \log \left( x + \sqrt{x^2 + 1} \right) \right) \) is an odd function.

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