How do I solve the following functional equation: $$f(x)+12f^{-1}(x)=\frac{1}{x}f(x)$$ I've been doing a lot of functional equations, but I haven't done one yet that has the function and its inverse together. All I've done so far is figure out that $f(x)$ has a fixed point at $x=\frac{1}{13}$ and that $f(0)$ starts a cycle of orbit $2$.
Thanks! All help is appreciated!
On
This is not a full answer but an alternate approach in addition to hypfco's. We have $y = f(x)$ and $x = f^{-1}(y)$.
Writing your equation as $(\frac{1}{x} - 1)f(x) = 12f^{-1}(x)$ we get
$$(\frac{1}{x} - 1)y = 12f^{-1}(f^{-1}(y))$$ $$(\frac{1}{f^{-1}(y)} - 1)y = 12f^{-1}(f^{-1}(y))$$ $$y = \frac{12}{\frac{1}{f^{-1}(y)} - 1}f^{-1}(f^{-1}(y))$$ $$y = -12\frac{f^{-1}(y)}{f^{-1}(y) - 1}f^{-1}(f^{-1}(y))$$
If we just write $g(x) = f^{-1}(x)$, we can solve the problem by finding a $g$ such that:
$$-12\frac{g(x)}{g(x)-1}g(g(x)) = x$$
We have
$$ y=f(x) $$
and:
$$ x=f^{-1}(y)=g(y) $$
Thus
$$ (\frac{1}{x}-1)f(x)=12f^{-1}(x)\\ x=f(f^{-1}(x))=f(\frac{1}{12}(\frac{1}{x}-1)f(x))\\ $$
Thus, we can formulate a recursive numeric search on $f$:
$$ min_{df}\left|x-f\left(\frac{1}{12}(\frac{1}{x}-1)f(x)+df(x)\right)+df\left(\frac{1}{12}(\frac{1}{x}-1)f(x)\right)\right|\\ $$