+585 The function
\[f(x) = \frac{cx}{(4x - 5)(3x + 2)}\]
satisfies $f(f(x))=x$ for all real numbers $x\neq \frac{5}{4}$ and $x \neq -\frac{2}{3}$. Find $c$.
0 we need:
\(f(f(x))=x\\ f(x)=f^{-1}(x)\)
The only way this is possible is if \(f(x)=f^{-1}(x)=x\):
So:
\(\cfrac{cx}{(4x-5)(3x+2)}=x\\ \boxed{\displaystyle{c=(4x-5)(3x+2)}}\)
