# Closed-form solution for $f(x)/x=y$ using $f^{-1}$

I'm programming a piece of math that requires solving an equation of a form $$f(x)/x=y$$. Now I already have $$f^{-1}(z)$$ coded (efficiently, and not by me) so I'd prefer using this implementation instead of, say, coding bisection for $$f(x)/x$$. Is there a way to do that?

If it were $$f(x)=y$$, I would just use $$x=f^{-1}(y)$$. But with $$f(x)/x=y$$ I doubt it's even possible to have a closed-form solution with $$f$$ and $$f^{-1}$$ but I can't come up with a reasoning for that.

• What is formula for $f(x)$ ? Commented Jan 21, 2019 at 17:49
• @coffeemath, it's very large and clumsy and is constructed in several steps. It's not the first time I faced this problem, previously with different expressions for $f(x)$, so I'm asking a question in general, for arbitrary $f(x)$. Commented Jan 21, 2019 at 17:55

Not possible in general. For example, if $$f(x) = e^x$$ then $$f^{-1}(x) = \log x$$. But the solution $$x$$ of $$\frac{e^x}{x} = y$$ is not elementary: $$x = -W\left(\frac{-1}{y}\right)$$ where W is the Lambert W function. Knowing how to compute $$e^x$$ and $$\log x$$ does not tell you how to compute this solution.