What is the value $f(-4)$ in the under function such that $f(x)+f(\frac1x)=\frac{x^2-12x+1}{2x}.$

Question

Let $f$ is a function such that $$f(x)+f(\frac{1}{x})=\dfrac{x^2-12x+1}{2x}.$$ Then what is the value $f(-4)=$?

Answer 1

Rewriting would give

$f(x)+f(\frac{1}{x})=\dfrac{x^2-12x+1}{2x}=\frac{1}{2}(x+\frac{1}{x})-6$

Then, it is easy to see a possible function is

$f(x)=\frac{1}{4}(x+\frac{1}{x})-3$

So, if you put $\frac{1}{x}$ in the argument, you would get the same function back.

However, as mentioned in comments, $f(x)=\frac{x}{2}-3$ is also a valid solution. Therefore , more conditions are required to find a unique function.