Let
\(f(x) = \begin{cases} k(x) &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases} \)
Find the function k(x) such that f is its own inverse.
Thanks in advance
y= x^2 - 6x + 12
The vertex of the graph is ( 3,3)
Two other points on this graph are (0,12) and ( 2,4)
So.. the points (3,3) ( 12,0) and (4,2) are on the inverse
y - 12 + 9 = x^2 - 6x + 9
(y - 3) = (x -3)^2
±sqrt (y - 3) = x - 3
±sqrt (y- 3) + 3 = x
y = ± (x - 3) + 3
If we take the negative root we have
y = -sqrt (x - 3) + 3
And (3,3) (12,0) and ( 4,2) will be on this graph
So
k(x) = - √(x - 3) + 3, x > 3
See the graph here : https://www.desmos.com/calculator/8oxnmjor5o