Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 875

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

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

\[= 2 \bullet 3x^{2} - 2x = 6x^{2} - 2x\]

\[Равна\ нулю\ при:\]

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

\[2x \bullet (3x - 1) = 0\]

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

\[Положительна\ при:\]

\[2x \bullet (3x - 1) > 0\]

\[x < 0\ или\ x > \frac{1}{3}.\]

\[Отрицательна\ при:\]

\[2x \bullet (3x - 1) < 0\]

\[0 < x < \frac{1}{3}.\]

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

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

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

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

\[= - 9x^{2} + 4x\]

\[Равна\ нулю\ при:\]

\[- 9x^{2} + 4x = 0\]

\[x( - 9x + 4) = 0\]

\[x_{1} = 0\ и\ x_{2} = \frac{4}{9}.\]

\[Положительна\ при:\]

\[x( - 9x + 4) > 0\]

\[x(9x - 4) < 0\]

\[0 < x < \frac{4}{9}.\]

\[Отрицательна\ при:\]

\[x( - 9x + 4) < 0\]

\[x(9x - 4) > 0\]

\[x < 0\ или\ x > \frac{4}{9}.\]

\[3)\ f(x) = x^{5} - 5x^{3} - 20x\]

\[f^{'}(x) = \left( x^{5} \right)^{'} - 5\left( x^{3} \right)^{'} - (20x)^{'} =\]

\[= 5x^{4} - 5 \bullet 3x^{2} - 20 =\]

\[= 5x^{4} - 15x^{2} - 20\]

\[Равна\ нулю\ при:\]

\[5x^{4} - 15x^{2} - 20 = 0\]

\[5\left( x^{4} - 3x^{2} - 4 \right) = 0\]

\[D = 9 + 16 = 25,\]

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

\[x_{2}^{2} = \frac{3 + 5}{2} = 4\]

\[x^{2} = 4\]

\[x = \pm 2.\]

\[Положительна\ при:\]

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

\[x < - 2\ или\ x > 2.\]

\[Отрицательна\ при:\]

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

\[- 2 < x < 2.\]

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

\[= (x - 4) \bullet (x + 3)^{2} \bullet (5x - 6).\]

\[Равна\ нулю\ при:\]

\[(x - 4) \bullet (x + 3)^{2} \bullet (5x - 6) = 0\]

\[x_{1} = 4\ x_{2} = - 3\ x_{3} = 1,2.\]

\[Положительна\ при:\]

\[(x + 3)^{2} \bullet (5x - 6)(x - 4) > 0\]

\[x < - 3\text{\ \ }\]

\[- 3 < x < 1,2\]

\[x > 4.\]

\[Отрицательна\ при:\]

\[(5x - 6)(x - 4) < 0\]

\[1,2 < x < 4.\]

\[5)\ f(x) = \frac{3x + 1}{x - 2}\]

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

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

\[= \frac{3(x - 2) - (3x + 1) \bullet 1}{(x - 2)^{2}} =\]

\[= \frac{3x - 6 - 3x - 1}{(x - 2)^{2}} =\]

\[= - \frac{7}{(x - 2)^{2}}.\]

\[Производная\ не\ равна\ нулю\]

\[не\ положительна.\]

\[Отрицательна\ при:\]

\[x - 2 \neq 0\]

\[x \neq 2.\]

\[6)\ f(x) = x^{2} + \frac{2}{x}\]

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

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

\[Имеет\ смысл\ при:\]

\[x \neq 0.\]

\[Равна\ нулю\ при:\]

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

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

\[x^{3} = 1\]

\[x = 1.\]

\[Положительна\ при:\]

\[x^{3} - 1 > 0\]

\[x^{3} > 1\]

\[x > 1.\]

\[Отрицательна\ при:\]

\[x^{3} - 1 < 0\]

\[x^{3} < 1\]

\[x < 0\ или\ 0 < x < 1.\]