+619 The function\(f(x) = \frac{cx}{2x+3}\)satisfies $f(f(x))=x$ for all real numbers \($x\ne -\frac 32$\). Find $c$.
+129852 f (f(x) ) =
c [ cx / (2x + 3) ]
________________ = x
2 [ cx / (2x + 3)] + 3
[c^2x] / (2x + 3) ]
___________________ = x
[ 2cx + 6x + 9] / (2x + 3)
c^2x = [2cx + 6x + 9 ] * x
c^2x = 2cx^2 + 6x^2 + 9x
(2c + 6)x^2 + (9 - c^2)x = 0
For this to be 0 for all real x {except x = -3/2} we need both
2c + 6 = 0 and 9 - c^2 = 0
So using the second c = 3 or c = -3
But when c = 3, 2c + 6 = 12
But when c = -3 both 2c + 6 and 9 - c^2 will = 0
So.... c = - 3
