A Computational Functional Equation Problem: $ f ( x ) f \big( f ( x ) \big) = 1 $

Question

How do I approach the following contest problem:

Let $ f : \mathbb R \to \mathbb R $ be a continuous function such that $ f ( x ) f \big( f ( x ) \big) = 1 $. Given that $ f ( 1000 ) = 999 $, compute $ f ( 500 ) $.

Answer 1

Letting $y = f(x)$ in the equation $f(x)f(f(x)) = 1$ we get that for all $y$ in the range of $f(x)$ we have $$f(y) = {1 \over y}$$ The condition $f(1000) = 999$ implies that $999$ is in the range of $f(x)$, so that $$f(999) = {1 \over 999}$$ This implies that ${1 \over 999}$ is in the range too. Since $f(x)$ is continuous, the intermediate value theorem applies and the range of $f(x)$ contains $[1/999, 999]$. In particular it contains $500$, so we have $$f(500) = {1 \over 500}$$

Answer 2

$f(500)=\frac1{500}$.

I substituted $1000$ for $x$ into the original equation and found that $999*f(999)=1$, so most logically $f(x)=\frac1x$. In more formal terms, $f(y) = \frac1y$ for all $y \in R.$ To address continuity issues:

$$f(999) = 1/999$$ $$f(1/999) = 999$$

Since $f$ is continuous, it must take on all values between $1/999$ and $999$. So $500 \in f(R).$ $$f(500)=\frac1{500}$$