Question

Ajutor! Am mare nevoie de ajutor la mate deoarece nu ma pricep. F:R=>R, f(x) =(a+b)x+1,x mai mic sau egal cu 1, (2a-3b)x, x>1, a, b aparține lui R. Astfel încât (-1,2),(2,6) aparține Gf. Vă rog să mă ajutați cum puteți!

Answer 1

   
[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]