Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1006

\[y = - 2x^{2} + 3x + 2.\]

\[1)\ x_{0} = - \frac{b}{2a} = - \frac{3}{2 \bullet ( - 2)} =\]

\[= \frac{3}{4} = 0,75;\]

\[y_{0} = - 1,125 + 2,25 + 2 = 3,125.\]

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

\[2x^{2} - 3x - 2 > 0\]

\[D = 9 + 16 = 25\]

\[x_{1} = \frac{3 - 5}{2 \bullet 2} = - \frac{1}{2};\]

\[x_{2} = \frac{3 + 5}{2 \bullet 2} = 2;\]

\[\left( x + \frac{1}{2} \right)(x - 2) > 0\]

\[x < - \frac{1}{2};\ \text{\ \ }x > 2.\]

\[2)\ Убывает\ на\ отрезке\ \lbrack 1;\ 2\rbrack:\]

\[y^{'}(x) = \left( - 2x^{2} \right)^{'} + (3x + 2)^{'} =\]

\[= - 4x + 3 \geq 0\]

\[4x \leq 3\]

\[x \leq \frac{3}{4}.\]

\[Что\ и\ требовалось\ доказать.\]

\[3)\ Наибольшее\ значение:\]

\[y^{'}(x) = \left( - 2x^{2} \right)^{'} + (3x + 2)^{'} =\]

\[= - 4x + 3 \geq 0;\]

\[4x \leq 3\]

\[x \leq \frac{3}{4}\]

\[x = 0,75.\]

\[4)\ - 2x^{2} + 3x + 2 < 3x + 2\]

\[- 2x^{2} < 0\]

\[x^{2} > 0;\]

\[x \neq 0.\]

\[5)\ y = 3:\]

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

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

\[D = 9 - 8 = 1\]

\[x_{1} = \frac{3 - 1}{2 \bullet 2} = 0,5;\]

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

\[f^{'}(x) = \left( - 2x^{2} \right)^{'} + (3x + 2)^{'} =\]

\[= - 4x + 3;\]

\[f^{'}(0,5) = - 2 + 3 = 1;\]

\[f^{'}(1) = - 4 + 3 = - 1.\]

\[Уравнение\ касательной:\]

\[y_{1} = 3 + 1(x - 0,5) = 2,5 + x;\]

\[y_{2} = 3 - 1(x - 1) = 4 - x.\]