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\(f(x) = \begin{cases} k(x) &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases}\)


Find k(x) such that f is its own inverse.

 Jan 16, 2019

\(f(x) = \begin{cases} k(x) &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases}\)


The LaTex for this question is really cool.  I've not seen this code before.

f(x) = \begin{cases} k(x) &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases}





\(y=x^2-6x+12\\ y=x^2-6x+9+3\\ y=(x-3)^2+3\\ y-3=(x-3)^2\\ \pm \sqrt{y-3}=x-3 \qquad where\;\;y\ge3\\ x=\pm \sqrt{y-3}+3\\ \text{The desired inverse will be one of these.}\\ y=\pm \sqrt{x-3}+3\\ \text{BUT is it the plus or the minus?}\)


Consider f(0)=12  

The reflection of (0,12) in the line y=x (which is the inverse) is  (12,0)

This works when the sign is negative.



so this underneath will be its own inverse.


\(f(x) = \begin{cases}3 -\sqrt{x-3} &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases} \)



And here is the graph:

 Jan 16, 2019

I did not put that latex in the latex box. It was supposed to just display as code.

The latex just keeps showing me things on this post!

Melody  Jan 16, 2019

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