[tex]f(x)=ax+b \Rightarrow ax+b=y \Rightarrow x=\frac{y-b}{a}\\ \\ Inlcouind \ x\ cu \ y \ obtinem \ inversa \ functiei:\\ \\ \Rightarrow f^{-1}(x)=\frac{x-b}{a}\\ \\ f(1-x)+f(x)+f^{-1}(1+x)=x\\ \\ a(1-x)+b+a\cdot x+b+\frac{1+x-b}{a}=x\\ \\ a-ax+ax+2b+\frac{1+x-b}{a}=x[/tex]
[tex] a+2b+\frac{1+x-b}{a}=x \ |\cdot a\\ \\ a^2+2ab+1+x-b=ax\ |+b\ \\ 1\cdot x+(a^2+2ab+1)=ax+b\\ \\\Leftrightarrow \left \{ {{a=1} \atop {a^2+2ab+1=b}} \right. \Leftrightarrow\\ \\ \Leftrightarrow \left \{ {{a=1} \atop {1^2+2\cdot 1\cdot b+1=b}} \right. \\ \\ \Leftrightarrow \left \{ {{a=1} \atop {2b-b=-2}} \right. \\ \\ \Leftrightarrow \left \{ {{a=1} \atop {b= -2}} \right.\\ \\ \boxed{f(x)=x-2}[/tex]
