Reduction function from A to its complement
This is true for any recursive language $A$ different from $\emptyset$ and $\Sigma^*$ for some finite alphabet $\Sigma$. Your definition is correct. In general you could define $f$ as following. Let $a\in A$ and $\overline{a} \in \overline{A}$. 

$$f(x)= 
\begin{cases} 
      \overline{a} &amp; x\in A \\
      a &amp; x\notin A
\end{cases}
$$
From this definition it follows $x \in A \iff f(x) \in \overline{A}$. Moreover, since $A$ is recursive $f$ is computable for any $x$ and hence $A \leq_T \overline{A}$.

The whole point is to decide $x \in A$ by using $f(x)$ and $\overline{A}$. So, given $x$ you can effectively say whether or not $x \in A$ only by looking at if $f(x) \in \overline{A}$. If $f(x) =\overline{a} \in \overline{A}$ then $x$ is in $A$, otherwise $x$ is not in $A$. Hence $A \leq_T \overline{A}$.