Asked by Trish Goal
Let f(x)=x^2-2x. Find all real numbers x such that f(x)=f(f(x)). Thank you.
Answers
Answered by
Damon
if y = x^2 - 2 x
f(y) = (x^2-2x)^2 - 2(x^2-2x) = f(f(x))
so
x^2-2x = (x^2-2x)^2 - 2(x^2-2x)
of course x = zero will do :)
let z = x^2 - 2x
z = z^2 - 2 z
z^2 - 3 z = 0
z (z-3) = 0
so another solution is
x^2-2x = 0 = x (x-2
we already have x = 0 so now also have x = 2
and also
z = 3 so x^2-2x = 3 which leads to
x^2 - 2 x - 3 = 0
(x+1)(x-3) = 0
so
x = -1
and x = 3 also
so
x = 0
x = 2
x = -1
x = 3
that was fun :)
f(y) = (x^2-2x)^2 - 2(x^2-2x) = f(f(x))
so
x^2-2x = (x^2-2x)^2 - 2(x^2-2x)
of course x = zero will do :)
let z = x^2 - 2x
z = z^2 - 2 z
z^2 - 3 z = 0
z (z-3) = 0
so another solution is
x^2-2x = 0 = x (x-2
we already have x = 0 so now also have x = 2
and also
z = 3 so x^2-2x = 3 which leads to
x^2 - 2 x - 3 = 0
(x+1)(x-3) = 0
so
x = -1
and x = 3 also
so
x = 0
x = 2
x = -1
x = 3
that was fun :)