To determine the nature of the function \(f(x) = \log \left(x + \sqrt{x^2 + 1}\right)\), we need to check if it satisfies the properties of even, odd, or periodic functions.

Analyzing Function Properties

Recall the definitions:

• Even Function: A function \(f(x)\) is even if \(f(-x) = f(x)\) for all \(x\) in its domain.
• Odd Function: A function \(f(x)\) is odd if \(f(-x) = -f(x)\) for all \(x\) in its domain.
• Periodic Function: A function \(f(x)\) is periodic if there exists a positive number \(T\) (the period) such that \(f(x+T) = f(x)\) for all \(x\) in its domain.

Evaluating f(-x)

Let's find the expression for \(f(-x)\) by substituting \(-x\) for \(x\) in the given function:

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

Simplify the expression inside the logarithm:

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

Relating f(-x) to f(x)

To see if this relates to \(f(x)\), we can manipulate the term \(\left(-x + \sqrt{x^2 + 1}\right)\). Let's multiply and divide by the conjugate of the expression inside the log, which is \(\left(\sqrt{x^2 + 1} + x\right)\):

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

Using the difference of squares formula \( (a-b)(a+b) = a^2 - b^2 \), where \(a = \sqrt{x^2 + 1}\) and \(b = x\):

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

Now substitute this back into the expression for \(f(-x)\):

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

Using the logarithm property \(\log \left(\frac{1}{a}\right) = -\log(a)\):

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

Notice that \(\log \left(\sqrt{x^2 + 1} + x\right)\) is the original function \(f(x)\). Therefore:

$$f(-x) = -f(x)$$\

Conclusion on Function Type

Since \(f(-x) = -f(x)\) holds true for all \(x\) in the domain of the function, the function \(f(x) = \log \left(x + \sqrt{x^2 + 1}\right)\) is an odd function.

It cannot be an even function because \(f(-x) = -f(x)\), which is only equal to \(f(x)\) if \(f(x)=0\), not for all \(x\). The function is also not periodic, as logarithmic functions do not exhibit periodic behavior.

Interestingly, the function \(f(x) = \log \left(x + \sqrt{x^2 + 1}\right)\) is also known as the inverse hyperbolic sine function, denoted as \(\rm \sinh^{-1}(x)\), which is a standard example of an odd function in mathematics.