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.
y = 2 + ( x - 2)^2
y - 2 = (x - 2)^2 take both roots
±√[ y - 2 ] = x - 2 add 2 to both sides
±√[ y - 2 ] + 2 = x exchange x and y
±√[ x - 2 ] + 2 = y = possible choices for k(x)
Since +√[ x - 2 ] + 2 > 0 for all x >2, then -√[ x - 2 ] + 2 is the correct function
[ To see this, note that (-1,11) is a point on 2 + ( x - 2)^2......but ( 11, -1) is not on +√[ x - 2 ] + 2 .....however, (11 , -1) is on -√[ x - 2 ] + 2 ]
So....k(x) = -√[ x - 2 ] + 2
See the inverses here : https://www.desmos.com/calculator/iiz2jtxdok