ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1005

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

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

\[y_{0} = 1 - 2 - 3 = - 4.\]

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

\[D = 4 + 12 = 16\]

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

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

\[(x + 1)(x - 3) < 0\]

\[- 1 < x < 3.\]

\[2)\ Возрастает\ на\ отрезке\ \lbrack 1;\ 4\rbrack:\]

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

\[= 2x - 2 \geq 0;\]

\[x - 1 \geq 0\]

\[x \geq 1.\]

\[Что\ и\ требовалось\ доказать.\]

\[3)\ Наименьшее\ значение:\]

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

\[= 2x - 2 \geq 0;\]

\[x - 1 \geq 0\]

\[x \geq 1\]

\[x = 1.\]

\[4)\ x^{2} - 2x - 3 > - 2x + 1\]

\[x^{2} - 4 > 0\]

\[(x + 2)(x - 2) > 0\]

\[x < - 2;\ \ \ x > 2.\]

\[5)\ x = 2:\]

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

\[= 2x - 2;\]

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

\[f(2) = 4 - 4 - 3 = - 3.\]

\[Уравнение\ касательной:\]

\[y = - 3 + 2(x - 2) =\]

\[= - 3 + 2x - 4 = 2x - 7.\]