Problem 5

Let

S

be the set of all real numbers greater than

-1

. Find all functions

f:S\to S

such that

f(x+f(y)+xf(y))=y+f(x)+yf(x)

for all

x,y\in S

, and

\frac{f(x)}{x}

is strictly increasing on each of the intervals

-1<x<0

and

0<x

Step 2 of 4: Use strict increase

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In plain words

Two distinct fixed points in one interval would give equal values of a strictly increasing ratio.

\frac{f(a)}a=\frac{f(b)}b=1

Detailed analysis

If $a\in(-1,0)$ , then $a$ and $b$ are distinct points of that interval, but $\frac{f(a)}a=\frac{f(b)}b=1$ , contradicting strict increase. The same contradiction holds for $a>0$ . Therefore the only possible fixed point is $a=0$ .