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f(x)+f^2(x)+f^3(x)+...= 1/(x+2), D=R-{-2}
niech f(x)=y [dla prostszego zapisu]
q=y
|q|<1
|y|<1
y/(1-y)=1/(x+2)
xy+2y=1-y
xy+3y=1
y(x+3)=1
y=1/(x+3), x≠-3
pamiętamy, że |y|<1.
| 1/(x+3) | < 1
1/|x+3| < 1
1<|x+3|
x∈(-∞,-4)U(-2,∞)
f(x)=1/(x+3), D=(-∞,-4)U(-2,∞)
f(x)∈(-1,0)U(0,1)