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 Let f(x)=x^2 -2x. Find all real numbers x such that f(x)=f(f(x))+2x-1.

 Jun 6, 2022
 #1
avatar+124676 
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f(f(x))  + 2x - 1 = (x^2 -2x)^2 - 2 (x^2 - 2x) + 2x - 1 =   [ x^4 - 4x^3 + 4x^2 ] - 2x^2 + 4x + 2x- 1 =

 

x^4 - 4x^3 + 2x^2 + 6x - 1

 

So

 

x^2 - 2x =  x^4 - 4x^3 + 2x^2 + 6x - 1            rearrange as

 

x^4 - 4x^3 + x^2 + 8x - 1   =  0

 

See the graph here :  https://www.desmos.com/calculator/os2uyqpfze

 

The real numbers that make this true   are     x ≈ -1.209   and x ≈ .124

 

 

cool cool cool

 Jun 6, 2022

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