Let S be the set of all real numbers greater than -1. Find all functions f :S→S such that f(x + f(y) + xf(y)) = y + f(x) + yf(x) for all x and y, and f(x)/x is strictly increasing on each of the intervals -1 < x < 0 and 0 < x.
Answer
f(x) = -x/(x+1).
Solution
Suppose f(a) = a. Then putting x = y = a in the relation given, we get f(b) = b, where b = 2a + a2. If -1 < a < 0, then -1 < b < a. But f(a)/a = f(b)/b. Contradiction. Similarly, if a > 0, then b > a, but f(a)/a = f(b)/b. Contradiction. So we must have a = 0.
But putting x = y in the relation given we get f(k) = k for k = x + f(x) + xf(x). Hence for any x we have x + f(x) + xf(x) = 0 and hence f(x) = -x/(x+1).
Finally, it is straightforward to check that f(x) = -x(x+1) satisfies the two conditions.
Thanks to Gerhard Woeginger for pointing out the error in the original solution and supplying this solution.
© John Scholes
jscholes@kalva.demon.co.uk
26 Oct 1998
Last corrected/updated 25 Aug 03