For a function $f: R \rightarrow R$, for any numbers $x, y$,

$$
|f(x)-f(y)|=|x-y| \text {. }
$$

What can be the value of $f(2)$ if $f(1)=3$?
Solution. Let's write 1 in place of $x$ and 2 in place of $y$. Thus, $|f(1)-f(2)|=|1-2|=1$. Since $f(1)=3$, we have $|3-f(2)|=1$, so $f(2)$ is either 2 or 4.

In the first case, $f(x)=-x+4$ could be possible. For this, the problem's condition is satisfied for all real $x, y$:

$$
|f(x)-f(y)|=|-x+4-(-y+4)|=|-x+y|=|x-y|
$$

In the second case, $f(x)=x+2$ could be possible, then

$$
|f(x)-f(y)|=|x+2-y-2|=|x-y|
$$

No other value can come into play for $f(2)$.