Question:
Let $S$ be the set of real numbers strictly greater than $-1 .$ Find all functions $f: S \rightarrow S$ satisfying the two conditions:$$$$ 1. $f(x+f(y)+x f(y))=y+f(x)+y f(x)$ for all $x$ and $y$ in $S$$$$$ 2. $\frac{f(x)}{x}$ is strictly increasing on each of the intervals $-1<x<0$ and $0<x$
My doubt -
I am having trouble with first few lines of the solution that was given in my book...
Let f be a function of desired type.since $f(x)/x$ is strictly increasing on (-1,0), equation $f(x)=x$ can have at most one solution in (-1,0)....
I didn't able to understand why this is true,is there a reason without calculus of this...if there is not then I will also manage to understand with calculus...
Thankyou
If $\frac{f(x)}{x}$ is strictly increasing in $(-1,0)$, then it can only cut the line $y = 1$ at most once (as otherwise, it has to "come back down" and cut it again. Whenever it cuts the line $y = 1$, we have $\frac{f(x)}{x} = 1 \implies f(x) = x$. Conversely, whenever $f(x) = x$, we have $\frac{f(x)}{x} = 1$, so it only has at most one solution.