Question

am nevoie pentru maine exercitiile 14 15 15
multumesc anticipat

Answer 1

$14)f:\mathbb{R}\rightarrow\mathbb{R} \: \: \: ,f(x)= {x}^{2} + 2x - 5$

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

$m = {1}^{2} + 2 \times 1 - 5 = - 2$

$15)f:\mathbb{R}\rightarrow\mathbb{R} \: \: \: ,f(x) = {x}^{2} + 5x - 5$

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

${m}^{2} + 5m - 5 = 1$

${m}^{2} + 5m - 5 - 1 = 0$

${m}^{2} + 5m - 6 = 0$

${m}^{2} + 6m - m - 6 = 0$

$m(m + 6) - (m + 6) = 0$

$(m + 6)(m - 1) = 0$

$m + 6 = 0 = > m_{1} = - 6$

$m - 1 = 0 = > m_{2} = 1$

$16)f,g:\mathbb{R}\rightarrow\mathbb{R}$

$a)f(x) = 2x - 5$

$g(x) = x + 3$

$2x - 5 = x + 3$

$2x - x = 3 + 5$

$x = 8$

$f(8) = 2 \times 8 - 5 = 16 - 5 = 11$

$= > A(8,11)$

$b)f(x) = x - 1$

$g(x) = {x}^{2} - 5x + 7$

$x - 1 = {x}^{2} - 5x + 7$

${x}^{2} - 5x - x + 7 + 1 = 0$

${x}^{2} - 6x + 8 = 0$

$a = 1$

$b = - 6$

$c = 8$

$\Delta = {b}^{2} - 4ac$

$\Delta = {( - 6)}^{2} - 4 \times 1 \times 8$

$\Delta = 36 - 32$

$\Delta = 4$

$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a} = \frac{ - ( - 6) \pm \sqrt{4} }{2 \times 1} = \frac{6 \pm2}{2}$

$x_{1} = \frac{6 + 2}{2} = \frac{8}{2} = 4$

$= > f(x_{1}) = 4 - 1 = 3 = > B(4,3)$

$x_{2} = \frac{6 - 2}{2} = \frac{4}{2} = 2$

$=>f(x_{2}) = 4 - 2 = 2 = > C(2,2)$