ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 166

\[1)\ f(x) = x^{3} - 2x;\]

\[f^{'}(x) = 3x^{2} - 2 = 0\]

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

\[x^{2} = \frac{2}{3}\]

\[x = \pm \frac{\sqrt{2}}{\sqrt{3}} = \pm \frac{\sqrt{6}}{3}.\]

\[Ответ:\ - \frac{\sqrt{6}}{3};\ \frac{\sqrt{6}}{3}.\]

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

\[f^{'}(x) = - 2x + 3 = 0\]

\[2x = 3\]

\[x = 1,5.\]

\[Ответ:\ \ 1,5.\]

\[3)\ f(x) = 2x^{3} + 3x^{2} - 12x - 3;\]

\[f^{'}(x) = 2 \bullet 3x^{2} + 3 \bullet 2x - 12 = 0\]

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

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

\[D = 1 + 8 = 9\]

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

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

\[Ответ:\ - 2;\ 1.\]

\[4)\ f(x) = (x - 3)(x + 4);\]

\[f^{'}(x) = (x + 4) + (x - 3) = 0\]

\[2x + 1 = 0\]

\[2x = - 1\]

\[x = - 0,5.\]

\[Ответ:\ - 0,5.\]

\[5)\ f(x) = (x - 2)^{2}(x + 1);\]

\[f^{'}(x) =\]

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

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

\[3x(x - 2) = 0\]

\[x_{1} = 0;\ \ \ x_{2} = 2.\]

\[Ответ:\ \ 0;\ 2.\]

\[6)\ f(x) = (x + 1)^{3};\]

\[f^{'}(x) = 3(x + 1)^{2} = 0\]

\[x + 1 = 0\]

\[x = - 1.\]

\[Ответ:\ - 1.\]