+137
$f(x) =
\begin{cases}
k(x) &\text{if }x>3, \\
x^2-6x+12&\text{if }x\leq3.
\end{cases}
$
Find the function $k(x)$ such that $f$ is its own inverse.
---
I found inverse of $x^2 - 6x + 12$ as $3 \pm \sqrt{x - 3}$, but will that make the function inverse of itself? Hints appreciated.
+137 What!? No way... I dont think that is the right answer... The inverse of a function that is itself can't be itself! :P
Do you have a better idea? :)
