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 Feb 11, 2018
 #1
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f(g(x)  =  [2x^2 + 3 ]  - 1

 

g (f(x))  = 2 [ x - 1 ] ^2  + 3  

 

So.....we are trying to solve

 

f(g(x)  = g(f(x))

 

[ 2x^2 + 3 ] - 1  =  2 [ x - 1 ]^2 + 3     simplify

 

2x^2  +  2   =   2 ( x^2 - 2x + 1) + 3

 

2x^2 + 2  = 2 x^2 - 4x + 2 + 3     subtract 2x^2 from both sides

 

2  =  -4x + 5   add 4x to both sides, subtract 2 from each side

 

4x  =  3   divide both sides by 4

 

x  = 3 / 4

 

 

cool cool cool

 Feb 11, 2018

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