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Let \(f(x) = x^2-3x\). For what values of \(x\) is \(f(f(x)) = f(x)\)?

 May 10, 2019
 #1
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 May 10, 2019
edited by Guest  May 10, 2019
 #2
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f(x) = x^2 - 3x

f(f(x) )  = (x^2 - 3x)^2 - 3(x^2 - 3x)

 

So

 

(x^2 - 3x)^2 - 3(x^2 - 3x)  = x^2 - 3x         let   x^2 - 3x =  m          and we have that

 

m^2 - 3m  = m

 

m^2 - 4m  =   0

 

m(m - 4)  = 0

 

So     m =  0       so    x^2 - 3x = 0      so   x(x - 3) = 0      so x = 0   or x = 3

 

And

 

m - 4  = 0         so     x^2 - 3x - 4 = 0     so   (x - 4)(x + 1)  = 0

 

Setting both of the last two factors to  0  and solving for x gives that   x = 4  and x = -1

 

So....x =  { -1, 0, 3, 4 }

 

 

cool cool cool

 May 10, 2019

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