Spanija 2012

Let $y=f(x)$. The equation is: $$x+\frac{1}{x}=y+\frac{1}{y}$$ Bring all terms to one side: $$x+\frac{1}{x}-y-\frac{1}{y}=0$$ Or: $$x-y+\frac{1}{x}-\frac{1}{y}=0$$ $$x-y+\frac{y-x}{xy}=0$$ $$(x-y)\left(1-\frac{1}{xy}\right)=0$$ So either $x=y$ or $xy=1$.

Thus, for each $x>0$, $f(x)$ must be either $x$ or $1/x$.

We are given that $f$ is continuous on ${\mathbb{R}}^{+}$.

Suppose there exists $x_0>0$ such that $f(x_0)=x_0$ and $f(x_1)=1/x_1$ for some $x_1\ne x_0$. Consider the set $A=\{x>0:f(x)=x\}$ and $B=\{x>0:f(x)=1/x\}$. Both $A$ and $B$ are closed (since $f$ is continuous and the preimage of a closed set under a continuous function is closed), and $A\cup B={\mathbb{R}}^{+}$.

Suppose $A$ is nonempty and $B$ is nonempty. Let $x_0\in A$, $x_1\in B$. Since $f$ is continuous, for any sequence $x_n\to x_0$ with $x_n\in B$, $f(x_n)\to f(x_0)=x_0$, but $f(x_n)=1/x_n\to 1/x_0$ as $x_n\to x_0$. Thus, $x_0=1/x_0$, so $x_0=1$. Similarly, at $x=1$, $f(1)=1$ or $f(1)=1$.

Thus, the only possible point where $f(x)$ can switch from $x$ to $1/x$ is at $x=1$. But for $x\ne 1$, $f$ must be constant on each side due to continuity.

Therefore, the only continuous functions are:

1. $f(x)=x$ for all $x>0$. 2. $f(x)=1/x$ for all $x>0$.

Both functions satisfy the given equation: - For $f(x)=x$, $x+1/x=x+1/x$. - For $f(x)=1/x$, $x+1/x=1/x+x$.

Thus, the solutions are $f(x)=x$ and $f(x)=1/x$ for all $x>0$.