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Let

\(f(x) = \begin{cases} k(x) &\text{if }x>2, \\ 2+(x-2)^2&\text{if }x\leq2. \end{cases} \)

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

michaelcai  Sep 24, 2017
 #1
avatar+88898 
+2

y =  2 + ( x - 2)^2

 

y - 2  =  (x - 2)^2       take both roots 

 

±√[ y - 2 ] = x - 2      add 2 to both sides

 

±√[ y - 2 ] + 2  =  x    exchange x and y

 

±√[ x - 2 ] + 2 =  y  =  possible choices for k(x)

 

Since  +√[ x - 2 ] + 2   > 0    for all x  >2, then  -√[ x - 2 ] + 2  is the correct  function

 

[ To see this, note that   (-1,11)  is a point on  2 + ( x - 2)^2......but ( 11, -1) is not on  +√[ x - 2 ] + 2 .....however,  (11 , -1)  is on  -√[ x - 2 ] + 2  ]

 

So....k(x)  =  -√[ x - 2 ] + 2 

 

See  the  inverses here : https://www.desmos.com/calculator/iiz2jtxdok

 

 

 

cool cool cool

CPhill  Sep 24, 2017
 #2
avatar+93356 
+1

Hi Chris,

This question confused me but what you have done is certainly correct. :)

I have included the graph here.

 

For the benefit of others.

The inverse of a function is the reflection of that function in the line y=x.

That is why I included y=x on my graph.

 

Melody  Sep 25, 2017

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