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Question 1089842: Let f(x) = x^2 - 2x. Find all real numbers x such that f(x) = f(f(x)).
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! f(x) = x^2 - 2x
f(f(x)) = (f(x))^2 - 2*(f(x)) ... replace every x with f(x)
f(f(x)) = (x^2 - 2x)^2 - 2*(x^2 - 2x) ... replace the f(x) on the right side with x^2-2x
f(f(x)) = x^4 - 4x^3 + 4x^2 - 2*(x^2 - 2x)
f(f(x)) = x^4 - 4x^3 + 4x^2 - 2x^2 + 4x
f(f(x)) = x^4 - 4x^3 + 2x^2 + 4x
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f(x) = f(f(x))
x^2 - 2x = x^4 - 4x^3 + 2x^2 + 4x
0 = x^4 - 4x^3 + x^2 + 6x
x^4 - 4x^3 + x^2 + 6x = 0
x(x^3 - 4x^2 + x + 6) = 0
x = 0 or x^3 - 4x^2 + x + 6 = 0
Solve x^3 - 4x^2 + x + 6 = 0 using a graphing calculator to get the following results: x = -1, x = 2, x = 3 (note how these values are roots, or x intercepts)
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So overall, the four solutions are:
x = 0
x = -1
x = 2
x = 3
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