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# Weird inverse question :(

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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.

Jan 10, 2021

#1
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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

Jan 10, 2021
edited by CPhill  Jan 10, 2021
#2
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Thanks for your clear and concise explanation CPhill!

Love the profile pic (MAD fan here)

Guest Jan 10, 2021
#3
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HAHAHA!!!!....thanks.....many  on here never heard of "MAD"  magazine   !!!!

CPhill  Jan 10, 2021