[tex]\displaystyle \\
f(x)=\left \{ {{(a+b)x+1~~~\text{daca} ~~x \leq 1}\atop {(2a-3b)x~~\text{daca}~~x\ \textgreater \ 1}~~~} \right\\\\\\
\text{Rescriem functia inlocuind pe }f(x) \text{ cu } y:\\\\
y=\left \{ {{(a+b)x+1~~~\text{daca} ~~x \leq 1} \atop {(2a-3b)x~~\text{daca}~~x\ \textgreater \ 1}~~~} \right\\\\\\
\text{Rescriem functia sub forma de 2 functii:}\\\\
\left \{ {{y=(a+b)x+1~~~\text{daca} ~~x \leq 1} ~~~(f_1)\atop {y=(2a-3b)x~~\text{daca}~~x\ \textgreater \ 1}~~~~~~(f_2)} \right\\\\\\
A(-1,2)\in f_1;~~~B(2,6) \in f_2[/tex]
[tex] \text{Inlocuim fiecare punct in functia corespunzatoare si obtinem}\\
\text{un sistem de 2 ecuatii cu 2 necunoscute, a si b.}\\\\
\displaystyle\\
\left \{ {{2=(a+b)\cdot(-1)+1\atop {6=(2a-3b)\cdot2~~~~~~~\right \\\\\\
\left \{ {{2=-a-b+1\atop {6=4a-6b~~~~\right \\\\\\
\left \{ {{a+b=1-2\atop {4a-6b=6~~~~\right \\\\\\
\left \{ {{a+b=-1~~|\cdot (6)\atop {4a-6b=6~~~~\right \\\\\\
\left \{ {{6a+6b=-6\atop {4a-6b=6~~~~\right \\\\
\text{Adunam ecuatiile:}\\
10a~~~~/~~=0\\
[/tex]
[tex]\displaystyle \\
\Longrightarrow~~\boxed{a = 0}\\ \\
\text{Din ecuatia: } ~~~a+b=-1 \\
b=-1-a = -1-0 = -1 \\ \\
\Longrightarrow~~\boxed{b = -1}\\ \\ \\
\text{Refacem functia data:} \\ \\
f(x)=\left \{ {{(a+b)x+1~~~\text{daca} ~~x \leq 1}\atop {(2a-3b)x~~\text{daca}~~x\ \textgreater \ 1}~~~} \right\\\\
\text{unde: }~ a=0~~\text{ si }~~ b=-1 \\ \\
f(x)=\left \{ {{(0-1)x+1~~~\text{daca} ~~x \leq 1}\atop {(2\cdot 0-3\cdot(-1))x~~\text{daca}~~x\ \textgreater \ 1}~~~} \right
[/tex]
[tex]\displaystyle \\
\Longrightarrow~~\boxed{f(x)=\left \{ {{-x+1~~~\text{daca} ~~x \leq 1}\atop {3x~~\text{daca}~~x\ \textgreater \ 1}~~~} \right \\ \\ }[/tex]
