## Task 3 - 241243

Determine all functions $f$ that are defined for all real numbers $x$ with $x \neq 0$ and satisfy the following conditions (1), (2), (3):

(1) For all real numbers $x$ with $x \neq 0$, $f\left(\frac{1}{x}\right)=x \cdot f(x)$.

(2) For all real numbers $x$ and $y$ with $x \neq 0, y \neq 0$ and $x+y \neq 0$, $f\left(\frac{1}{x}\right)+f\left(\frac{1}{y}\right)=1+f\left(\frac{1}{x+y}\right)$.

(3) $f(1)=2$.
I. If a function $f$ defined for all $x \neq 0$ satisfies conditions (1), (2), (3), then for all $x \neq 0$ it follows from (2) that

$$
2 \cdot f\left(\frac{1}{x}\right)=1+f\left(\frac{1}{2 x}\right)
$$

Substituting $\frac{1}{2 x}=u$ here, we get: For all $u \neq 0$,

$$
2 \cdot f(2 u)=1+f(u)
$$

From (5), (1), (4) it follows for all $x \neq 0$ that

$$
x(1+f(x))=2 x f(2 x)=f\left(\frac{1}{2 x}\right)=2 f\left(\frac{1}{x}\right)-1=2 x f(x)-1
$$

Thus, $1+f(x)=2 f(x)-\frac{1}{x}$. Therefore, only the function $f$ defined for all $x \neq 0$ by

$$
f(x)=1+\frac{1}{x}
$$

can have the required properties.

II. This function has these properties; indeed, for all $x, y$ with $x \neq 0, y \neq 0, x+y \neq 0$:

$$
\begin{aligned}
f\left(\frac{1}{x}\right) & =1+x=x\left(1+\frac{1}{x}\right)=x f(x) \\
f\left(\frac{1}{x}\right)+f\left(\frac{1}{y}\right) & =1+x+1+y=1+(1+x+y)=1+f\left(\frac{1}{x+y}\right) \\
f(1) & =1+\frac{1}{1}=2
\end{aligned}
$$

Therefore, exactly the function given in (6) has the required properties.

## Adapted from $[5]$