ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 182

\[1)\ f(x) = 3x^{4} - 4x^{3} - 12x^{2};\]

\[f^{'}(x) = 3 \bullet 4x^{3} - 4 \bullet 3x^{2} - 12 \bullet 2x = 0;\]

\[12x^{3} - 12x^{2} - 24x = 0\]

\[12x\left( x^{2} - x - 2 \right) = 0\]

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

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

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

\[Ответ:\ - 1;\ 0;\ 2.\]

\[2)\ f(x) = x^{4} + 4x^{3} - 8x^{2} - 5;\]

\[f^{'}(x) = 4x^{3} + 4 \bullet 3x^{2} - 8 \bullet 2x = 0;\]

\[4x^{3} + 12x^{2} - 16x = 0\]

\[4x\left( x^{2} + 3x - 4 \right) = 0\]

\[D = 3^{2} + 4 \bullet 4 = 9 + 16 = 25\]

\[x_{1} = \frac{- 3 - 5}{2} = - 4;\text{\ \ }\]

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

\[Ответ:\ - 4;\ 0;\ 1.\]

\[3)\ f(x) = \left( x^{2} + 3 \right)\left( 2x^{2} + 5 \right);\]

\[f^{'}(x) = 2x\left( 2x^{2} + 5 \right) + 4x\left( x^{2} + 3 \right) = 0;\]

\[4x^{3} + 10x + 4x^{3} + 12x = 0\]

\[8x^{3} + 22x = 0\]

\[2x\left( 4x^{2} + 11 \right) = 0\]

\[x = 0.\]

\[Ответ:\ \ 0.\]

\[4)\ f(x) = x + \frac{1}{x};\]

\[f^{'}(x) = 1 - \frac{1}{x^{2}} = 0;\]

\[\frac{1}{x^{2}} = 1\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[Ответ:\ - 1;\ 1.\]

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

\[f^{'}(x) = 2(x - 1)x\sqrt{x} + (x - 1)^{2} \bullet \frac{3}{2}x^{\frac{1}{2}} = 0;\]

\[2(x - 1)x\sqrt{x} + 1,5(x - 1)^{2}\sqrt{x} = 0\]

\[(x - 1)\sqrt{x} \bullet (2x + 1,5x - 1,5) = 0\]

\[(x - 1)\sqrt{x} \bullet (3,5x - 1,5) = 0\]

\[x_{1} = 0;\ \ \ x_{2} = \frac{3}{7};\ \ \ x_{3} = 1.\]

\[Ответ:\ \ 0;\ \frac{3}{7};\ 1.\]

\[6)\ f(x) = 3x^{4} - 4x^{3} + 6x^{2} - 12x;\]

\[f^{'}(x) = 3 \bullet 4x^{3} - 4 \bullet 3x^{2} + 6 \bullet 2x - 12 = 0;\]

\[12x^{3} - 12x^{2} + 12x - 12 = 0\]

\[12x^{2}(x - 1) + 12(x - 1) = 0\]

\[\left( 12x^{2} + 12 \right)(x - 1) = 0\]

\[x = 1.\]

\[Ответ:\ \ 1.\]