Here's the question you clicked on:
estudier
f(x) = x^2 +2x -1 Solve f(f(x)) = f(x)
we have f(f(x)) - f(x) =0 or (x^2 + 2x-1)^2 + (x^2 + 2x -1) - 1 =0 its a quadratic in x^2 + 2x-1 it'll give 2 real vales of that,, then we'll solve for x 4 roots shall be obtained..
is 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+x-1=0
Can'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...
if 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