## Task 24/73

A function $f(x)$ has the following properties:

1. For real $x$, $f(x)>0$.
2. For real $x_{i}$, $f\left(x_{1}+x_{2}\right)=2 \cdot f\left(x_{1}\right) f\left(x_{2}\right)$.
3. $f(2)=2$.

Determine the function values $f(1), f(0), f(3)$, and $f(6)$! What function is this?
Due to $2=f(2)=f(1+1)=2 \cdot f(1) \cdot f(1)$, it immediately follows that $[f(1)]^{2}=1$, from which, due to $f(x)>0$, it follows that $f(1)=1$.

In an analogous way, $1=f(1)=f(1+0)=2 \cdot f(1) \cdot f(0)$, so $f(0)=0.5$. Further, $f(3)=f(2+1)=2 \cdot f(2) \cdot f(1)=4, f(6)=f(3+3)=2 \cdot f(3) \cdot f(3)=32$.

Sketching the curve from the calculated values, one might conjecture that the function is an exponential function of the form $a \cdot b^{x}$.

Indeed, for such a function, $f(x)>0$ for real $x$ (Property 1) always holds. Furthermore, for such a function,

$$
f\left(x_{1}+x_{2}\right)=a \cdot b^{x_{1}+x_{2}}=a \cdot b^{x_{1}} \cdot b^{x_{2}}=\frac{1}{2} \cdot a \cdot b^{x_{1}} \cdot a \cdot b^{x_{2}}=c \cdot f\left(x_{1}\right) \cdot f\left(x_{2}\right)
$$

If we set $c=\frac{1}{a}=2$, so $a=0.5$, then Property 2 is also satisfied. It remains to check whether Property 3 can occur for such a function:

$0.5 \cdot b^{2}=2$ immediately leads to $b=2$. The sought function is thus $f(x)=0.5 \cdot 2^{x}$. Substituting the values 0, 1, 3, and 6 for $x$ confirms the correctness.