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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.

 

Thanks in advance wink

 Jan 10, 2021
 #1
avatar+114361 
+1

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

 

 

cool cool cool

 

 

 

 

 


 

 Jan 10, 2021
edited by CPhill  Jan 10, 2021
 #2
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+1

Thanks for your clear and concise explanation CPhill!

 

Love the profile pic (MAD fan here)

smiley

Guest Jan 10, 2021
 #3
avatar+114361 
0

HAHAHA!!!!....thanks.....many  on here never heard of "MAD"  magazine   !!!!

 

 

cool cool cool

CPhill  Jan 10, 2021

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