\(Let f(x) = \begin{cases} k(x) &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases}\) Find the function k(x) such that f is its own inverse.
Since f is its own inverse, then f(f(x)) = x for all x in the domain of f.
If x > 3, then f(x) = k(x) and f(k(x)) = k(k(x)) = x.
If x <= 3, then f(x) = x^2 - 6x + 12 and f(x^2 - 6x + 12) = (x^2 - 6x + 12)^2 - 6(x^2 - 6x + 12) + 12 = x^4 - 12x^3 + 36x^2 - 36x + 144 - 6x^2 + 36x - 72 + 12 = x^4 - 18x^2 + 120 = (x^2 - 6x + 10)^2.
Therefore, we need k(k(x)) = x^2 - 6x + 10 for all x in the domain of f. This means that k(x) must be the inverse of x^2 - 6x + 10.
The inverse of x^2 - 6x + 10 is (x - 2)^2 + 5, so k(x) = (x - 2)^2 + 5.
Therefore, the function k(x) is:
k(x) = (x - 2)^2 + 5
for all x in the domain of f.
