Question

repede!!! ceea ce este incercuit cu roșu!!!​

Answer 1

$4)f:\mathbb{R}\rightarrow\mathbb{R} \: \: \: ,f(x)=ax+b$

$a)f(0) = 1 = > a \times 0 + b = 1= > b = 1$

$f(1) = 3 = > a \times 1 + b = 3$

$a + b = 3$

$a + 1 = 3$

$a = 3 - 1$

$a = 2$

$= > f(x) = 2x + 1$

$b)A(1,1)\:\in\:G_{f} < = > f(1) = 1$

$a \times 1 + b = 1$

$a + b = 1$

$f(2) = 5 = > a \times 2 + b = 5$

$2a + b = 5$

$\left\{\begin{matrix} a + b = 1 \: | \times ( - 1) \\ 2a + b = 5\end{matrix}\right.$

$\left\{\begin{matrix} - a - b = - 1\\ 2a + b = 5\end{matrix}\right.$

$- a + 2a - b + b = - 1 + 5$

$a = 4$

$a + b = 1$

$4 + b = 1$

$b = 1 - 4$

$b = - 3$

$= > f(x) = 4x - 3$

$c)A(-2,-3)\:G_{f} < = > f( - 2) = - 3$

$a \times ( - 2) + b = - 3$

$- 2a + b = - 3$

$B(0,-2)\:G_{f} < = > f(0) = - 2$

$a \times 0 + b = - 2$

$b = - 2$

$- 2a + b = - 3$

$- 2a - 2 = - 3$

$- 2a = - 3 + 2$

$- 2a = - 1 \: | \div ( - 2)$

$a = \frac{1}{2}$

$= > f(x) = \frac{1}{2} x - 2$

$d)f(0) + f(1) = 8$

$a \times 0 + b + a \times 1 + b = 8$

$2b + a = 8$

$A(-1,1)\:\in\:G_{f} < = > f( - 1) = 1$

$a \times ( - 1) + b = 1$

$- a + b = 1$

$\left\{\begin{matrix} a + 2b = 8\\ - a + b = 1\end{matrix}\right.$

$a - a + 2b + b = 8 + 1$

$3b = 9 \: | \div 3$

$b = 3$

$- a + b = 1$

$- a + 3 = 1$

$- a = 1 - 3$

$- a = - 2 \: | \times ( - 1)$

$a = 2$

$= > f(x) = 2x + 3$

$5)f(x) = \: ?$

$a)AB$

$A(-1,-1)$

$B(1,3)$

$Fie\:f:\mathbb{R}\rightarrow\mathbb{R} \: \: \: ,f(x)=ax+b$

$A(-1,-1)\:\in\:G_{f}<=>f(-1)=-1$

$a \times ( - 1) + b = - 1$

$- a + b = - 1$

$B(1,3)\:\in\:G_{f}<=>f(1)=3$

$a \times 1 + b = 3$

$a + b = 3$

$\left\{\begin{matrix} - a + b = - 1\\ a + b = 3\end{matrix}\right.$

$- a + a + b + b = - 1 + 3$

$2b = 2 \: | \div 2$

$b = 1$

$a + b = 3$

$a + 1 = 3$

$a = 3 - 1$

$a = 2$

$= > f(x) = 2x + 1$

$b)AB$

$A(-2,3)$

$B(1,0)$

$Fie \: f:\mathbb{R}\rightarrow\mathbb{R} \: \: \: ,f(x)=ax+b$

$A(-2,3)\:\in\:G_{f}<=>f(-2)=3$

$a \times ( - 2) + b = 3$

$- 2a + b = 3$

$B(1,0)\:\in\:G_{f}<=>f(1)=0$

$a \times 1 + b = 0$

$a + b = 0$

$a = - b$

$- 2a + b = 3$

$- 2 \times ( - b) + b = 3$

$2b + b = 3$

$3b = 3 \: | \div 3$

$b = 1$

$a = - b$

$a = -1$

$= > f(x) = -x +1$