\section*{Problem 6 - 241036}

Determine for each function \(f\) that satisfies the following properties (1), (2), and (3), the function values \(f(0), f(-1)\), and \(f\left(\frac{3}{7}\right)\).

(1) The function \(f\) is defined for all real numbers.

(2) It holds that \(f(1)=2\).

(3) For all real numbers \(a\) and \(b\), it holds that \(f(a+b)=f(a) \cdot f(b)\).
}

Due to (3):

\[
f(1+0)=f(1) \cdot f(0) \quad ; \quad f(0)=\frac{2}{2}=1
\]

Also due to (3):

\[
f(0)=f(1) \cdot f(-1) \quad ; \quad f(-1)=\frac{f(0)}{f(1)}=\frac{1}{2}
\]

In general, it holds that:

\[
\begin{gathered}
f(2 x)=f(x+x)=f(x)^{2} \\
f(3 x)=f(2 x+x)=f(2 x) \cdot f(x)=f(x)^{3}
\end{gathered}
\]

In this way, it follows generally that

\[
f(n x)=f(x)^{n} \quad ; \quad f(n x)^{\frac{1}{n}}=f(x)
\]

If \(n x=y\), it also holds that

\[
f\left(\frac{y}{n}\right)=f(y)^{\frac{1}{n}}
\]

Due to (4):

\[
f(k \cdot 1)=f(1)^{k}=2^{k}
\]

Together with (5):

\[
f\left(\frac{k}{n} \cdot 1\right)=\left(2^{k}\right)^{\frac{1}{n}}=2^{\frac{k}{n}}
\]

And thus:

\[
f\left(\frac{3}{7}\right)=2^{\frac{3}{7}}
\]