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!!!
+128475 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
