ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 890

\[\boxed{\mathbf{890}\mathbf{.}}\]

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

\[k = y^{'}(x) = \frac{1}{3} \bullet \left( x^{3} \right)^{'} - \frac{5}{2} \bullet \left( x^{2} \right)^{'} =\]

\[= \frac{1}{3} \bullet 3x^{2} - \frac{5}{2} \bullet 2x = x^{2} - 5x.\]

\[y = 6x:\]

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

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

\[D = 25 + 24 = 49\]

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

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

\[1)\ y^{'}( - 1) = ( - 1)^{2} - 5 \bullet ( - 1) =\]

\[= 1 + 5 = 6\]

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

\[= - \frac{1}{3} - \frac{5}{2} = - \frac{2}{6} - \frac{15}{6} =\]

\[= - \frac{17}{6} = - 2\frac{5}{6}\]

\[y = - 2\frac{5}{6} + 6 \bullet (x + 1) =\]

\[= - 2\frac{5}{6} + 6x + 6 = 6x + 3\frac{1}{6}.\]

\[2)\ y^{'}(6) = 6^{2} - 5 \bullet 6 =\]

\[= 36 - 30 = 6\]

\[y(6) = \frac{1}{3} \bullet 6^{3} - \frac{5}{2} \bullet \left( 6^{2} \right) =\]

\[= \frac{216}{3} - \frac{5 \bullet 36}{2} = 72 - 90 = - 18\]

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

\[= - 18 + 6x - 36 = 6x - 54.\]

\[Ответ:\ \ \]

\[y = 6x + 3\frac{1}{6}\ \ y = 6x - 54.\]