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If $f(c)=\frac{3}{2c-3}$, find $\frac{kn^2}{lm}$ when $f^{-1}(c)\times c \times f(c)$ equals the simplified fraction $\frac{kc+l}{mc+n}$, where $k,l,m,\text{ and }n$ are integers.

 

can anyone help me?

thanks so much!!!

 Jan 27, 2021
 #1
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y =   3  / (2c   -3)

 

y (2c - 3)   = 3

 

y2c  - 3y    = 3

 

2yc  =  3 + 3y

 

c  = ( 3 +3y) / (2y)        "swap"  c  and y

 

y  = (3 + 3c)  / (2c)  =  the inverse  of  f(c)

 

f-1(c)   *  f(c)  *  c    =  [ (3 + 3c)  / (2c)]  [ 3  / (2c   -3) ]  * c   =  3(2c + 3)   / [ 2 (2c  -3) ]  =  (6c + 9)  /(4c - 6)

 

So

 

(6c + 9) / ( 4c  - 6)    =  (kc + l) / ( mc + n)

 

k  = 6       l =  9         m   = 4      n   =  -6

 

So

 

kn^2  / ( lm)   =    6* (-6)^2  / ( 9  * 4)  =      216  / 36    =    6

 

 

 

cool cool cool

 Jan 27, 2021
 #2
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the system says it's incorrect :(

Guest Jan 27, 2021
 #3
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Anyone??

 Jan 27, 2021
 #4
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Hello???

Guest Jan 28, 2021

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