Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 338

\[1)\ y = 3x^{2} + 2x - 5;y = 2x - 5:\]

\[y^{'} = 3 \bullet 2x + 2 - 0 = 6x + 2;\]

\[k = 6x + 2 = 2.\]

\[6x = 0\]

\[x = 0.\]

\[y(0) = 0 - 5 = - 5;\]

\[y(0) = 0 + 0 - 5 = - 5.\]

\[Ответ:\ \ (0;\ - 5).\]

\[2)\ y = 3x^{2} - 2x + 5;y = 10x - 7:\]

\[y^{'} = 3 \bullet 2x - 2 + 0 = 6x - 2;\]

\[k = 6x - 2 = 10\]

\[6x = 12\]

\[x = 2.\]

\[y(2) = 20 - 7 = 13;\]

\[y(2) = 12 - 4 + 5 = 13.\]

\[Ответ:\ \ (2;\ 13).\]

\[3)\ y = x^{3} - 5x + 8;y = 7x + 24:\]

\[y^{'} = 3x^{2} - 5 + 0 = 3x^{2} - 5;\]

\[k = 3x^{2} - 5 = 7\]

\[3x^{2} = 12\]

\[x^{2} = 4\]

\[x = \pm 2.\]

\[y( - 2) = - 14 + 24 = 10;\]

\[y( - 2) = - 8 + 10 + 8 = 10;\]

\[y(2) = 14 + 24 = 38;\]

\[y(2) = 8 - 10 + 8 = 6.\]

\[Ответ:\ \ ( - 2;\ 10).\]

\[4)\ y = x^{3} - 5x^{2} - 3x + 11;\]

\[y = 10x + 18:\]

\[y^{'} = 3x^{2} - 5 \bullet 2x - 3 + 0 =\]

\[= 3x^{2} - 10x - 3;\]

\[k = 3x^{2} - 10x - 3 = 10\]

\[3x^{2} - 10x - 13 = 0\]

\[D = 100 + 156 = 256\]

\[x_{1} = \frac{10 - 16}{2 \bullet 3} = - 1;\text{\ \ }\]

\[x_{2} = \frac{10 + 16}{2 \bullet 3} = \frac{13}{3};\]

\[y( - 1) = - 10 + 18 = 8;\]

\[y( - 1) = - 1 - 5 + 3 + 11 = 8;\]

\[y\left( \frac{13}{3} \right) = \frac{130}{3} + 18 = \frac{184}{3};\]

\[y\left( \frac{13}{3} \right) =\]

\[= \frac{2197}{27} - \frac{845}{9} - 13 + 11 =\]

\[= - \frac{392}{27}.\]

\[Ответ:\ \ ( - 1;\ 8).\]