### Instruction: Let  \[f(x) =
\begin{cases}
k(x) &\text{if }x>0, \\
-\frac1{2x}&\text{if }x< 0\\
0&\text{if }x=0.
\end{cases}
\]Find the function $k(x)$ such that $f(x)$ is its own inverse function.### Response: An inverse function is a function that undoes another function.
That is, if $f(x)$ and $f^{-1}(x)$ are inverse functions, then $f(f^{-1}(x))=f^{-1}(f(x))=x$ for all $x$ in the domain of both functions.
To find the inverse function of $f(x)$, I need to switch the roles of $x$ and $y$ in the equation $y=f(x)$ and then solve for $y$ in terms of $x$.
For $x>0$, the equation is $y=k(x)$.
Switching $x$ and $y$, I get $x=k(y)$.
To solve for $k(y)$, I need to undo whatever operation $k$ does to $y$.
Since I don't know what $k$ is, I'll call it $h$ and write $k(y)=h^{-1}(y)$.
Then $x=h^{-1}(y)$ and $y=h(x)$.
This means that $k(x)=h^{-1}(x)$ is the inverse function of $h(x)$.
So I need to find a function $h(x)$ such that $h(x)=k^{-1}(x)$ and $h^{-1}(x)=k(x)$ for $x>0$.
For $x<0$, the equation is $y=-\frac1{2x}$.
Switching $x$ and $y$, I get $x=-\frac1{2y}$.
Multiplying both sides by $-2y$, I get $-2xy=1$.
Dividing by $-2x$, I get $y=\frac1{-2x}$.
This means that $f^{-1}(x)=\frac1{-2x}$ for $x<0$.
For $x=0$, the equation is $y=0$.
Switching $x$ and $y$, I get $x=0$.
This means that $f^{-1}(x)=0$ for $x=0$.
So the inverse function of $f(x)$ is \[f^{-1}(x) =
\begin{cases}
\frac1{-2x}&\text{if }x< 0, \\
0&\text{if }x=0,\\
k(x)&\text{if }x>0.
\end{cases}
\]
Since $f(x)$ is its own inverse function, I can equate $f(x)$ and $f^{-1}(x)$ for all $x$ and get \[f(x) =
\begin{cases}
k(x) &\text{if }x>0, \\
-\frac1{2x}&\text{if }x< 0\\
0&\text{if }x=0.
\end{cases}
=
\begin{cases}
\frac1{-2x}&\text{if }x< 0, \\
0&\text{if }x=0,\\
k(x)&\text{if }x>0.
\end{cases}
\]
This implies that $k(x)=-\frac1{2x}$ for $x>0$.
Therefore, the function $k(x)$ that I'm looking for is \[k(x) =
\begin{cases}
-\frac1{2x}&\text{if }x>0, \\
\text{undefined}&\text{if }x\leq 0.
\end{cases}
\]