Inverse of function is itself?

Question

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.

I thought that that inverse would just be the inverse of $x^2-6x+12$ would be $3+\sqrt{x-3}$, however, after graphical analysis, I found the answer to be the conjugate: $3-\sqrt{x-3}$. Is this the right answer, and if so, why is it negative instead of positive?

Answer 1

$$f^{-1}-3=\pm\sqrt{x-3}$$ You have to choose '$-$' because $x^2-6x+12=(x-3)^2+3\ge3$ and hence $f^{-1}$ must be $\le 3$.

Answer 2

If $y=x^2-6x+12$, then $x=3\pm\sqrt{y-3}$; we take $x=3-\sqrt{y-3}$ to get $x\le3$.

Answer 3

I believe you were very close.

Missing k(x) is this: k(x) = 3 - (sqrt( 4x - 12 ))/2

Plot it and you will see, that now f(x) = inv f(x).

Have a nice day.