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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.

So I don't know how to do this. What does it mean by f is its own inverse?? Somebody please explain the problem to me!

Jun 10, 2020

#1
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oh this is an aops. This problem means that when you plug in a value for x, it goes to the other um equation, and then when you plug in that value it goes back to the original equation. (someone correct me if I am wrong)

Jun 10, 2020
#2
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I know what this means, but how do I make f the inverse of itself??

Jun 10, 2020
#3
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wait. look at this question:
https://web2.0calc.com/questions/helppppp_28

Jun 10, 2020
#4
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Jun 10, 2020
#5
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Let's get the inverse of                                                             y  =  x2 - 6x + 12

Interchange x and y:                                                                 x  =  y2 - 6y + 12

x - 12  =  y2 - 6y

Complete the square:                                                   x - 12 + 9  =  y2 - 6y + 9

x - 3  =  (y - 3)2

Take the square root (choose the negative):              - sqrt(x - 3)  =  y - 3

Finish:                                                                   - sqrt(x - 3) + 3  =  y

Create k(x) to be the inverse:     k(x)  =  - sqrt(x - 3) + 3

Now, if you choose a value for x            (say, 2),

find f(2)  (it's 4)

and then find f(4)  (it's back to your original number, 2)

Jun 10, 2020