Question

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Answer 1

   
[tex]\displaystyle\\ I)\\ I.1)\\ f(-2)=2\cdot(-2)-3=-4-3=-7~\Longrightarrow~(-2,~-7)~\text{ este corect.}\\\\ f(0)=2\cdot0-3=0-3=-3~\Longrightarrow~(0,~-3)~\text{ este corect.}\\\\ f(2)=2\cdot2-3=4-3=1~\Longrightarrow~(2,~1)~\text{ este corect.}\\\\ \Longrightarrow~G_f=\{(-2,~-7);~(0,~-3);~(2,~1)\}~\text{ este corect.}\\\\\\ I.2)\\ \text{Verificam daca punctul }~N(-3,~1)\in G_f\\\\ f(x)=-\frac{1}{3}x+1\\ f(-3)=-\frac{1}{3}\cdot(-3)+1=-\frac{-3}{3}+1=\frac{3}{3}+1=1+1=2\\ 2\neq1\\ Concluzie:~~N(-3,~1)\notin G_f[/tex]