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 \(k,l,m,\text{ and }n\) are integers.
\(\begin{array}{|rcll|} \hline f\left( f^{-1}(c) \right) &=& c \quad | \quad \mathbf{f(c)=\dfrac{3}{2c-3}} \\ \dfrac{3}{2f^{-1}(c)-3} &=& c \\ 3 &=& c\left( 2f^{-1}(c)-3 \right) \\ 3 &=& 2cf^{-1}(c)-3c \\ 2cf^{-1}(c) &=& 3+3c \\ 2cf^{-1}(c) &=& 3(1+c) \\ \mathbf{ f^{-1}(c) } &=& \mathbf{ \dfrac{3(1+c)}{2c} } \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline && \mathbf{ f^{-1}(c)\times c \times f(c) } \\\\ &=& \dfrac{3(1+c)}{2c}\times c \times \dfrac{3}{2c-3} \\ \\ &=& \dfrac{3(1+c)}{2} \times \dfrac{3}{2c-3} \\ \\ &=& \dfrac{9(1+c)}{2(2c-3)} \\\\ &=& \mathbf{ \dfrac{9c+9}{4c-6} } \quad | \quad \text{compare }~\dfrac{kc+l}{mc+n} \\ \hline k&=& 9 \\ l &=& 9 \\ m &=& 4 \\ n &=& -6 \\\\ && \mathbf{ \dfrac{kn^2}{lm} } \\\\ &=& \dfrac{9*(-6)^2}{9*4} \\\\ &=& \dfrac{(-6)^2}{4} \\\\ &=& \dfrac{36}{4} \\\\ &=& \mathbf{9} \\ \hline \end{array}\)
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