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Let \[f(x) = \frac{2x + 3}{kx - 2}.\]Find all real numbers $k$ so that $f^{-1}(x) = f(x).$

 Dec 17, 2020
 #1
avatar+117724 
+1

y  =  (2x + 3) / (kx - 2)       get x by itself

 

y ( kx - 2)   =  2x + 3

 

ykx - 2y    =  2x + 3

 

x (ky)  - 2y  =  2x  + 3

 

-2y - 3  =  2x  - x (ky)

 

-2y - 3  =  x  ( 2 - ky)

 

-(2y + 3)  / ( 2 - ky )  =  x       swap x  and  y

 

(2x + 3) / ( kx - 2)   = y    =  the inverse

 

The function is its own inverse

 

We just  can't have  kx - 2  =  0

 

So....  kx  = 2  will make the denominator  =   0

 

k = 2/x     

 

k can be  any value  except  2/x

 

 

cool cool cool

 Dec 17, 2020
 #2
avatar
+1

Hi Cphill, how would I express this in interval notation, my steps were very similar but idk how to put in interval

 Dec 17, 2020
 #3
avatar+117724 
+1

Try this  ?????

 

(-inf, 2/x)  U (2/x, infinity )

 

 

cool cool cool

CPhill  Dec 17, 2020

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