Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 276

\[1)\ y = \frac{|x|}{1 + x^{2}}\]

\[x \geq 0:\]

\[y = \frac{x}{x^{2} + 1};\]

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

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

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

\[x \leq 0:\]

\[y = - \frac{x}{x^{2} + 1};\]

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

\[Критические\ точки:\]

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

\[x^{2} = 1\]

\[x = \pm 1.\]

\[Ответ:\ \ \]

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

\[2)\ y = x^{3} - |x - 1|\]

\[x \geq 1:\]

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

\[y^{'} = 3x^{2} - 1.\]

\[x \leq 1:\]

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

\[y' = 3x^{2} + 1.\]

\[Критические\ точки:\]

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

\[3x^{2} = 1\]

\[x^{2} = \frac{1}{3}\]

\[x = \pm \frac{1}{\sqrt{3}}.\]

\[Ответ:\ \ x = 1.\]

\[3)\ y = 2x^{2} - \left| x^{2} - 1 \right|\]

\[|x| \geq 1:\]

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

\[y^{'} = 2x + 0 = 2x.\]

\[|x| \leq 1:\]

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

\[y^{'} = 3 \bullet 2x - 0 = 6x.\]

\[Критические\ точки:\]

\[2x = 0\ \ \ \]

\[6x = 0\]

\[x = 0.\]

\[Ответ:\ \ \]

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

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

\[0 \leq x \leq 3:\]

\[y = 3x + \left( 3x - x^{2} \right) = 6x - x^{2};\]

\[y^{'} = 6 - 2x.\]

\[x \leq 0;\text{\ x} \geq 3:\]

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

\[y^{'} = 2x.\]

\[Критические\ точки:\]

\[6 - 2x = 0\]

\[x = 3.\]

\[2x = 0\]

\[x = 0.\]

\[Ответ:\ \ x_{1} = 0;\ x_{2} = 3.\]