Find all functions $f:mathbb{R}^+to mathbb{R}$ such that $xf(xf(x)-4)-1=4x$

Partial answer

Let $g(x)=xf(x)-4$.

Our functional equation becomes

$$frac{gcirc g(x)+4}{g(x)}=4+frac1x$$

If there exists an invertible function $phi(x):mathbb R^+tomathbb R^+$ such that $g(x)=phi(phi^{-1}(x)+1)$, direct substitution gives $$frac{phi(z+2)+4}{phi(z+1)}=4+frac1{phi(z)}$$ or $$phi(z+2)=left[4+frac1{phi(z)}right]phi(z+1)-4$$ by substituting $z=phi^{-1}(x)$.

Clearly, this recurrence relation extends a $phi(x)$ defined arbitrarily on $(0,2)$ to the whole $mathbb R^+$. By computing $phi^{-1}(x)$ this method generates a large class of solution to the functional equation.

(Of course there are certain restrictions on $phi(x)$ on $(0,2)$ so that $phi(x)$ is invertible.)

The solution $f(x)=4+frac4x$ corresponds to $g(x)=4x$ and $phi(x)=kcdot 4^x$ where $k$ is a positive constant.

A feature of this special solution is that $phi(x)$ is continuous. In contrast, if rather arbitrary values are assigned to $phi(x)$ on $(0,2)$, a discontinuous solution may yield. Below is the graph of $phi(x)$ where $phi(x) = x+1$ on $(0,2)$.

Partial answer:

Consider the equation $xf(xf(x)-a)-1=ax$ for $a>0$ so that $$f(xf(x)-a)=a+frac1x.$$ This means that $limlimits_{xto+infty}f(xf(x)-a)=a$ so that $limlimits_{xto+infty}f(x)=a$. Further, we have $$lim_{xto0^+}f(xf(x)-a)=+infty$$ and since $f(x)>a/ximplieslimlimits_{xto0^+}f(x)=+infty$, it follows that $limlimits_{xto0^+}xf(x)=a$.

Let $m,n$ be integers such that $m<-1$ and $n>0$. Notice that $$f(x)=sumlimits_{k=m}^na_kx^k$$ implies $limlimits_{xto0^+}xf(x)=a$ so $a_{-1}=a$ and $a_i=0$ for all $mle i<-1$. Likewise we have $limlimits_{xto+infty}f(x)=a$ so $a_0=a$ and $a_j=0$ for all $0<jle n$. Thus if $f$ is a finite Laurent polynomial then the only solution to the functional equation is $$f(x)=a+frac ax.$$