### Instruction: Let  \[f(x) =
\begin{cases}
k(x) &\text{if }x>2, \\
2+(x-2)^2&\text{if }x\leq2.
\end{cases}
\]Find the function $k(x)$ such that $f$ is its own inverse.### Response: To find the inverse of a function, I usually switch the roles of $x$ and $y$ and solve for $y$ in terms of $x$.
In this case, I have to consider two cases: when $x>2$ and when $x\leq2$.
When $x>2$, I have $y=k(x)$, so $k$ is already in terms of $x$.
To make $f$ its own inverse, I need $k(y)=x$ for this case, or in other words, $k$ is symmetric about the line $y=x$.
When $x\leq2$, I have $y=2+(x-2)^2$, so I need to solve for $x$ in terms of $y$.
Subtracting $2$ from both sides and taking the square root, I get $x-2=\pm\sqrt{y-2}$.
Adding $2$ to both sides, I get $x=2\pm\sqrt{y-2}$.
To make $f$ its own inverse, I need $y=2\pm\sqrt{x-2}$ for this case, or in other words, $f$ is symmetric about the line $y=x$.
This means that $k(x)=2\pm\sqrt{x-2}$ for $x>2$, but I have to choose the sign carefully.
If I choose the plus sign, then $k(3)=3+\sqrt{1}=4$, which means that $f(4)=k(4)=4+\sqrt{2}$, which is not equal to $3$.
If I choose the minus sign, then $k(3)=3-\sqrt{1}=2$, which means that $f(2)=2+(3-2)^2=3$, which is equal to $k(f(2))$.
So the minus sign works, and I have found the function $k(x)$.