1. Let f(x) = x^2 - 2x. Find all real numbers x such that f(x) = f(f(x)).
f (f(x) = ( x^2 - 2x)^2 - 2(x^2 - 2x)
So
x^2 - 2x = ( x^2 - 2x)^2 - 2(x^2 - 2x)
We can write
( x^2 - 2x)^2 - 2(x^2 - 2x) - (x^2 - 2x) = 0
x^4 - 4x^3 + 4x^2 - 2x^2+ 4x - x^2 + 2x = 0
x^4 - 4x^3 + x^2 + 6x = 0
x ( x^3 - 4x^2 + x + 6 )
One solution is x = 0
(x^3 -3x^2) - (x^2 - x - 6) = 0
x^2 ( x - 3) - 1[ (x -3)( x + 2) ] = 0
(x - 3) (x^2 -1(x + 2) ) = 0
(x - 3) (x^2 - x - 2) = 0
(x-3) (x -2) (x+ 1) = 0
Setting the linear factors = 0 and solving for x, we have that
x = -1, 2 and 3
So....the solutions that make the original equation true are
x = -1, 0, 2 and 3
