Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Discrete mathematics, continuous mathematics and probability (resit) / 1 / e

Replace $x$ by $1/x$, which stays in the domain of the functional equation. Writing $u=f(x)$ and $v=f(1/x)$ gives the linear systemTwice the second equation minus the first gives $3u=2/x-x$, so the only possible function isTo check existence as well as uniqueness, substitute this expression: $f(x)+2f(1/x)=(2/x-x+4x-2/x)/3=x$.

The algebraic identitysuggests $f(t)=t^3-3t$ for every $t>0$, and substitution verifies the functional equation.

In fact $x+1/x\ge 2$, since $(x-1{)}^2/x\ge 0$. Conversely, for each $t\ge 2$, the quadratic equation $x^2-tx+1=0$ has a positive root $(t+\sqrt{t^2-4})/2$. Thus the functional equation forces $f(t)=t^3-3t$ exactly on $[2,\infty )$; its values on $(0,2)$ are arbitrary. The displayed function is one convenient choice on the full required domain.