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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.

 Mar 10, 2021
 #1
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Since we want k(x) to be its own inverse, k(x) = 2 + (x - 2)^2 = x^2 - 4x + 6.

 Mar 21, 2021

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