### 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: Well we know that even functions are symmetric with respect to the y-axis.
Right. So we just need to check if $f(x) = f(-x)$.
And we know that odd functions are symmetric with respect to the origin.
Right. So we just need to check if $f(x) = -f(-x)$.
Let's see if $f(x) = f(-x)$.
So $f(-x) = \frac{1}{2^{-x} - 1} + \frac{1}{2}$.
And since $2^{-x} = \frac{1}{2^x}$ we can simplify that to $\frac{1}{\frac{1}{2^x} - 1} + \frac{1}{2}$.
Which is $\frac{2^x}{1 - 2^x} + \frac{1}{2}$.
And since $\frac{1}{2} + \frac{2^x}{1 - 2^x} \neq \frac{1}{2^x - 1} + \frac{1}{2}$ we see that $f(x)$ is not an even function.
Let's see if $f(x) = -f(-x)$.
We know that $f(-x) = \frac{2^x}{1 - 2^x} + \frac{1}{2}$, so $-f(-x) = -\frac{2^x}{1 - 2^x} - \frac{1}{2}$.
And since $-\frac{2^x}{1 - 2^x} - \frac{1}{2} \neq \frac{1}{2^x - 1} + \frac{1}{2}$ we see that $f(x)$ is not an odd function either.