This Does Not Function

Wrong Ones:

Let $x = -\frac{1}{168}$ .

Plugging into the equation, we get $f \left ( -\frac{1}{168} \right ) + f(2017) = -\frac{1}{168} \Leftrightarrow f(2017) = -\frac{1}{168} - f \left ( -\frac{1}{168} \right )$ , not $f(2017) = \frac{1}{168} + f \left ( \frac{1}{168} \right )$ . $\blacksquare$

By taking the implicit derivative of $f(x) + f\left (1 - \frac{12}{x} \right )= x$ , we find that $f'(x) + f' \left (1 - \frac{12}{x} \right ) \cdot \frac{12}{x^2} = 1$ .

Letting $x = \sqrt{12}$ we see that $f' (\sqrt{12}) + f'(1-\sqrt{12}) = 1$ , and not $f' (\sqrt{12}) + f'(-\sqrt{12}) = 1$ . $\blacksquare$

Let $x = 1 - \frac{12}{x}$ .

Plugging into the equation, it yields us $f \left ( 1 - \frac{12}{x} \right ) + f \left (1 - \frac{12}{1 - \frac{12}{x}} \right ) = 1 - \frac{12}{x}$ .

Subtracting this equation from the original one, we get $f(x) - f \left ( 1 - \frac{12}{1 - \frac{12}{x}} \right ) = x - 1 + \frac{12}{x}$ .

But $x - 1 + \frac{12}{x}$ can never be equal zero, so we will never have $f(x) = f \left ( 1 - \frac{12}{1 - \frac{12}{x}} \right )$ . $\blacksquare$

Right One:

Let $x = 3$ .

Plugging into the equation, we get $f(3) + f(-3) = 3$ .

If the function were odd, we would have gotten $f(3) + f(-3) = 0$ .

Thus, the function can't be odd.