Exista functii \( f:\mathbb{R}\to\mathbb{R} \) continue astfel incat
\( f(1)=2 \) si \( f(f(x))=x^2-2x+2,\forall x\in\mathbb{R} \)?
Cristinel Mortici, Olimpiada locala Constanta, 1999
Ecuatie functionala cu functii continue
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Tudorel Lupu
- Euclid
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Presupunem ca exista functii \( f \) cu aceasta proprietate.
\( f(1)=2 \Rightarrow f(2)=1. \)
Fie \( g:\mathbb{R} \to \mathbb{R}, g(x)=f(x)-x \), continua.
\( g(1)=1, g(2)=-1 \) si cum e continua \( \Rightarrow \) exista \( x_0 \in [1;2] \) a.i. \( g(x_0)=0 \Leftrightarrow f(x_0)=x_0, f(f(x_0))=x_0. \)
Ajungem la \( x_0^2-2x_0+2=x_0 \Rightarrow x_0=1 \) sau \( x_0=2 \).
Contradictie deoarece \( f \) nu are ca pct fixe pe \( 1, 2 \).
\( f(1)=2 \Rightarrow f(2)=1. \)
Fie \( g:\mathbb{R} \to \mathbb{R}, g(x)=f(x)-x \), continua.
\( g(1)=1, g(2)=-1 \) si cum e continua \( \Rightarrow \) exista \( x_0 \in [1;2] \) a.i. \( g(x_0)=0 \Leftrightarrow f(x_0)=x_0, f(f(x_0))=x_0. \)
Ajungem la \( x_0^2-2x_0+2=x_0 \Rightarrow x_0=1 \) sau \( x_0=2 \).
Contradictie deoarece \( f \) nu are ca pct fixe pe \( 1, 2 \).
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