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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.

 Jul 22, 2020
 #1
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I got 324 but I might be wrong.

 Jul 22, 2020
 #2
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Yep that's wrong

Guest Jul 22, 2020
 #3
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If

\(f(c)=\dfrac{3}{2c-3}\), find \(\dfrac{kn^2}{lm} \)

when \(f^{-1}(c)\times c \times f(c)\) equals the simplified fraction \(\dfrac{kc+l}{mc+n}\),
where \(k\), \(l\), \(m\), and \(n\) are integers.

 

My attempt:

 

\(\begin{array}{|rcll|} \hline f(c) &=& \dfrac{3}{2c-3} \quad | \quad c=f^{-1}(c) \\\\ f\Big(f^{-1}(c)\Big) &=& \dfrac{3}{2f^{-1}(c)-3} \quad | \quad f\Big(f^{-1}(c)\Big) = c \\\\ c &=& \dfrac{3}{2f^{-1}(c)-3} \\\\ c\Big(2f^{-1}(c)-3\Big) &=& 3 \\\\ 2cf^{-1}(c)-3c &=& 3 \\\\ 2cf^{-1}(c) &=& 3c+3 \\\\ cf^{-1}(c) &=& \dfrac{3c+3}{2} \\\\ \mathbf{f^{-1}(c)\times c} &=& \mathbf{\dfrac{3c+3}{2}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline f^{-1}(c)\times c \times f(c) &=& \dfrac{(3c+3)}{2}\times \dfrac{3}{(2c-3)} \\\\ f^{-1}(c)\times c \times f(c) &=& \dfrac{3(3c+3)}{2(2c-3)} \\\\ \mathbf{f^{-1}(c)\times c \times f(c)} &=& \mathbf{\dfrac{9c+9}{4c-6}} \\ \hline && \boxed{k=9,\ l=9,\ m=4,\ n=-6} \\\\ \dfrac{kn^2}{lm} &=& \dfrac{9(-6)^2}{9*4} \\\\ \dfrac{kn^2}{lm} &=& \dfrac{36}{4} \\\\ \mathbf{\dfrac{kn^2}{lm}} &=& \mathbf{9} \\ \hline \end{array}\)

 

laugh

 Jul 23, 2020
edited by heureka  Jul 23, 2020

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