Determine whether the following function are even/odd/neither even nor odd? `f(x)=xlog(x+sqrt(x^(2)+1))`

Explanation

To determine whether the function \( f(x) = x \log(x + \sqrt{x^2 + 1}) \) is even, odd, or neither, we will follow these steps:

### Step 1: Understand the Definitions
- A function \( f(x) \) is **even** if \( f(-x) = f(x) \).
- A function \( f(x) \) is **odd** if \( f(-x) = -f(x) \).

### Step 2: Calculate \( f(-x) \)
We start by substituting \(-x\) into the function:

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

### Step 3: Simplify \( f(-x) \)
Now we simplify the expression inside the logarithm:

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

Next, we can rationalize the expression inside the logarithm:

\[ f(-x) = -x \log\left(-x + \sqrt{x^2 + 1}\right) \]

To simplify further, we multiply and divide by the conjugate:

\[ = -x \log\left(\frac{(-x + \sqrt{x^2 + 1})(\sqrt{x^2 + 1} + x)}{\sqrt{x^2 + 1} + x}\right) \]

The numerator simplifies as follows:

\[ (-x + \sqrt{x^2 + 1})(\sqrt{x^2 + 1} + x) = (\sqrt{x^2 + 1})^2 - x^2 = 1 \]

Thus, we have:

\[ f(-x) = -x \log\left(\frac{1}{\sqrt{x^2 + 1} + x}\right) \]

Using the property of logarithms, we can rewrite this as:

\[ f(-x) = -x \left(-\log(\sqrt{x^2 + 1} + x)\right) = x \log(\sqrt{x^2 + 1} + x) \]

### Step 4: Compare \( f(-x) \) with \( f(x) \)
Now we need to compare \( f(-x) \) with \( f(x) \):

\[ f(x) = x \log(x + \sqrt{x^2 + 1}) \]

### Step 5: Check for Evenness or Oddness
We have:

- \( f(-x) = x \log(\sqrt{x^2 + 1} + x) \)
- \( f(x) = x \log(x + \sqrt{x^2 + 1}) \)

Notice that:

\[ \log(\sqrt{x^2 + 1} + x) = \log(x + \sqrt{x^2 + 1}) \]

Thus, we find that:

\[ f(-x) = f(x) \]

### Conclusion
Since \( f(-x) = f(x) \), the function \( f(x) \) is **even**.