ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 262

\[y = \frac{1}{3}x^{3} - \frac{5}{2}x^{2};\ \ y = 6x.\]

\[1)\ y^{'}(x) = \frac{1}{3} \bullet 3x^{2} - \frac{5}{2} \bullet 2x = 6;\]

\[x^{2} - 5x - 6 = 0\]

\[D = 25 + 24 = 49\]

\[x_{1} = \frac{5 - 7}{2} = - 1;\]

\[x_{2} = \frac{5 + 7}{2} = 6.\]

\[2)\ y( - 1) = - \frac{1}{3} - \frac{5}{2} = - \frac{17}{6};\]

\[y = - \frac{17}{6} + 6 \bullet (x + 1);\]

\[y = 6x + \frac{19}{6}.\]

\[3)\ y(6) = 72 - 90 = - 18;\]

\[y = - 18 + 6(x - 6);\]

\[y = 6x - 54.\]

\[Ответ:\ \ y = 6x + \frac{19}{6};\ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }y = 6x - 54.\]