### Instruction: Is
\[f(x) = \frac{1}{2^x - 1} + \frac{1}{2}\]an even function, odd function, or neither?

Enter "odd", "even", or "neither".### Response: To check if a function is even, I need to see if f(-x) = f(x) for any x in the domain.
To check if a function is odd, I need to see if f(-x) = -f(x) for any x in the domain.
To find f(-x), I need to replace x with -x in the function and simplify.
So f(-x) = \[\frac{1}{2^{-x} - 1} + \frac{1}{2}\].
I can rewrite 2^{-x} as \[\frac{1}{2^x}\] using the exponent rule.
So f(-x) = \[\frac{1}{\frac{1}{2^x} - 1} + \frac{1}{2}\].
I can multiply the numerator and denominator by 2^x to get rid of the fraction in the denominator.
So f(-x) = \[\frac{2^x}{1 - 2^x} + \frac{1}{2}\].
I notice that this is not equal to f(x), since the sign of the denominator is different.
So f is not an even function.
I also notice that this is not the negative of f(x), since the sign of the numerator is different.
So f is not an odd function either.
Therefore, f is neither an even nor an odd function.