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shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.2we have f(f(x))  f(x) =0 or (x^2 + 2x1)^2 + (x^2 + 2x 1)  1 =0 its a quadratic in x^2 + 2x1 it'll give 2 real vales of that,, then we'll solve for x 4 roots shall be obtained..

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0is this correct ? f(f(x))=f(x) f(x)=f^1 (f(x)) f(x) = x to get x^2 +2x 1 = x so solving this : x^2+x1=0

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0Can't see the wood for the trees... The equation is f(x)^2 + 2f(x) 1 = f(x) > f(x)^2 + f(x) 1 = 0 as shubhamsrg said...

harsh314
 2 years ago
Best ResponseYou've already chosen the best response.0if we do it like this \[f(x)=(x+1)^{2}2 \implies f(f(x))=(f(x)+1)^{2}2\] using it in the given equation we have \[f(x)=(f(x)+1)^{2}2\] putting the value of f(x) \[(x+1)^{2}2=((x+1)^{2}2+1)^{2}2\] solving that we get \[x+1=(x+1)^{2}1\] and then solve again
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