Increasing function $ f : \mathbb R ^ + \to \mathbb R $ with $ x f ( x ) + 2 > 0 $ and $ f ( x ) f \left( f ( x ) + \frac 4 x \right) = 1 $

Let $ f : \mathbb R ^ + \to \mathbb R $ be an increasing function, such that $ x f ( x ) + 2 > 0 $ and $ f ( x ) f \left( \frac { x f ( x ) + 4 } x \right) = 1 $, then find the value of $ \lfloor f ' ( 1 ) \rfloor $ (where $ \lfloor \cdot \rfloor $ represents greatest integer function).

My approach is as follow $ f ( x ) f ( y ) = 1 $ as $ y = \frac { x f ( x ) + 4 } x $, I am trying to use the function given as $ x f ( x ) + 2 > 0 $ in $ y $ but I'm not able to proceed.

Solution 1:

Let $g(x)=\frac{xf(x)+4}{x}=f(x)+\frac{4}{x}$. We have $$ f(x)f(g(x)) =1$$ Since $xf(x)+2>0$, we also have $g(x)>\frac{2}{x}>0$, which means that we also have $$ f(g(x))f(g(g(x))) =1$$ which means that $$ f(x) = f(g(g(x)))$$ since $f$ is an increasing function, it means that $$ x=g(g(x))=f(g(x))+\frac{4}{g(x)}= \frac{1}{f(x)}+\frac{4}{f(x)+\frac{4}{x}}$$ Solving for $f(x)$, we can find $$ f(x) = \frac{1\pm\sqrt{17}}{2x} $$ Since we know that $f$ is increasing, we must conclude that the minus sign is the correct one $$ f(x) = \frac{1-\sqrt{17}}{2x} $$ Form this, we find $$ f'(1) = \frac{\sqrt{17}-1}{2} $$ $$ \lfloor f'(1)\rfloor = 1$$