, 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 $x_1 \neq x_0$. 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 $x \neq 1$, $f$ 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 $x > 0$.