SOLUTION: Let f(x) = x^2 - 2x. Find all real numbers x such that f(x) = f(f(x)).

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Question 1028661: Let f(x) = x^2 - 2x. Find all real numbers x such that f(x) = f(f(x)).
Found 3 solutions by robertb, richard1234, Natolino1983:
Answer by robertb(5830) About Me  (Show Source):
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This is supposed to be equal to f%28x%29+=+x%5E2+-+2x
==> x%5E4+-+4x%5E3%2B2x%5E2%2B4x+=+x%5E2+-+2x
<==> x%5E4+-+4x%5E3%2Bx%5E2%2B6x+=+0
<==> x%28x%2B1%29%28x-2%29%28x-3%29+=+0 (Verify!!)
==> The real numbers x such that f(x) = f(f(x)) are 0, -1, 2, and 3.

Answer by richard1234(7193) About Me  (Show Source):
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Let y = f(x). We want to solve y = f(y).

Solving y = f(y) gives y = y^2 - 2y or y = 0, 3. Then we want to solve f(x) = 0, f(x) = 3.

If f(x) = 0, then x^2 - 2x = 0 --> x = 0 or 2.
If f(x) = 3, then x^2 - 2x = 3 --> x = 3 or -1.

Solutions: 0, -1, 2, 3

Answer by Natolino1983(23) About Me  (Show Source):
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If f(f(x)) = f(x) then we need to find every the solution to f(x) = x
X^2 -2x = x is equivalent to x^2-2x-x = 0
X^2 -3x = 0
X(x-3) =0
So, solutions are x=0 or x=3.
1) f(f(0)) =f(0^2-2×0) =f(0) (checked)
2) f(f(3)) = f(3^3-2×3) = f(9-6) = f(3) (checked)