[tex]\displaystyle Trebuie~sa~gasim~functiile~f: \mathbb{R} \to \mathbb{R},~f(x)=ax+b~(cu~a \neq 0) \\ \\pentru~ care~f \circ f=f. \\ \\ Avem~f \circ f=f(f(x))=f(ax+b)=a(ax+b)+b=a^2x+ab+b. \\ \\ Deci~trebuie~sa~avem~a^2x+ab+b=ax+b~\forall~x \in \mathbb{R}. \\ \\ Identificand~coeficientii,~obtinem~ \left \{ {{a^2=a} \atop {ab+b=b}} \right. \Leftrightarrow \left \{ {{a^2=a} \atop {ab=0}} \right. . [/tex]
[tex]\displaystyle Din~a^2=a~si~a \neq 0~rezulta~a=1. \\ \\ \bullet a=1 \Rightarrow 0=ab=b,~deci~b=0. \\ \\ Astfel,~am~obtinut~solutia~f(x)=x.[/tex]
[tex]\displaystyle Verificare: \\ \\ f \circ f =f(f(x))=f(x)~\checkmark[/tex]
