Question

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Answer 1

   
[tex]\displaystyle\\ \text{II)}\\ \text{II.1)}\\ f:\{-2,~1\}\to R,~~f(x)=-x+3\\ f(-2)=-(-2)+3=2+3=5\\ f(1)=-1+3=2\\ \text{II.2)}\\ g:R\to R,~~g(x)=mx-4,~m\in R\\ g(3)=5\\ m\cdot 3 - 4 = 5\\ 3m-4=5\\ 3m=5+4\\ 3m=9\\\\ m= \frac{9}{3} =3 [/tex]


[tex]\displaystyle\\ \text{II.3)}\\ f(x)=ax+b\\ A(2,~-1)\in G_f\\ B(-1,~2)\in G_f\\ \text{Inlocuim coordonatele punctelor in functie si obtinem un sistem}\\ \text{de 2 ecuatii cu necunoscutele a si b.}\\ \begin{cases} a\cdot 2+b=-1\\ a\cdot(-1)+b=2 \end{cases}\\\\ \begin{cases} 2a+b=-1\\ -a+b=2~~\Big| \cdot (-1) \end{cases}\\\\ \begin{cases} 2a+b=-1\\ a-b=-2 \end{cases}~~~~\text{\}Adunam ecuatiile.}\\\\ 3a=-3\\\\ a=\frac{-3}{3}=\boxed{\bf -1}\\\\ -1-b=2\\ -b=2+1\\ -b=3\\ b=\boxed{\bf -3}\\ \boxed{\bf f(x)=-x-3}[/tex]