Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1496

\[\boxed{\mathbf{1496}\mathbf{.}}\]

\[y = 4x^{3} - 9x^{2} + 6x + 1;\]

\[y^{'}(x) = 4\left( x^{3} \right)^{'} - 9\left( x^{2} \right)^{'} + (6x + 1)^{'};\]

\[k = y^{'}(x) = 4 \bullet 3x^{2} - 9 \bullet 2x + 6 =\]

\[= 12x^{2} - 18x + 6.\]

\[k = 0:\]

\[12x^{2} - 18x + 6 = 0\]

\[2x^{2} - 3x + 1 = 0\]

\[D = 9 - 8 = 1\]

\[x_{1} = \frac{3 - 1}{2 \bullet 2} = \frac{2}{4} = \frac{1}{2};\]

\[x_{2} = \frac{3 + 1}{2 \bullet 2} = \frac{4}{4} = 1;\]

\[y_{1} = 4\left( \frac{1}{2} \right)^{3} - 9\left( \frac{1}{2} \right)^{2} + \frac{6}{2} + 1 =\]

\[= \frac{4 - 18 + 24 + 8}{8} = 2,25;\]

\[y_{2} = 4 \bullet 1^{3} - 9 \bullet 1^{2} + 6 \bullet 1 + 1 =\]

\[= 4 - 9 + 6 + 1 = 2.\]

\[Ответ:\ \ (0,5;\ 2,25);\ \ (1;\ 2).\]