Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1468

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

\[1)\ \ y(x) < 0:\]

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

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

\[D = 9 + 16 = 25\]

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

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

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

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

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

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

\[= - 2 \bullet 2x + 3 = 3 - 4x;\]

\[3 - 4x \leq 0\]

\[4x \geq 3\]

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

\[Что\ и\ требовалось\ доказать.\]

\[3)\ Наибольшее\ значение:\]

\[x = \frac{3}{4} - точка\ максимума;\]

\[y\left( \frac{3}{4} \right) = - 2 \bullet \frac{9}{16} + 3 \bullet \frac{3}{4} + 2 =\]

\[= - \frac{9}{8} + \frac{18}{8} + 2 = 1\frac{1}{8} + 2 = 3\frac{1}{8}.\]

\[4)\ Ниже\ графика\ функции\ \]

\[y = 3x + 2:\]

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

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

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

\[x \neq 0.\]

\[5)\ Уравнение\ касательной\ к\ \]

\[y = 3:\]

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

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

\[D = 9 - 8 = 1\]

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

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

\[1)\ f^{'}(x) = - 2\left( x^{2} \right)^{'} + (3x + 2)^{'} =\]

\[= - 2 \bullet 2x + 3 = 3 - 4x;\]

\[f^{'}\left( \frac{1}{2} \right) = 3 - 4 \bullet \frac{1}{2} = 3 - 2 = 1;\]

\[y = 3 + 1\left( x - \frac{1}{2} \right) =\]

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

\[2)\ f^{'}(1) = 3 - 4 \bullet 1 = 3 - 4 = - 1;\]

\[y = 3 - 1(x - 1) = 3 - x + 1 =\]

\[= 4 - x.\]