ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 335

\[1)\ x\sqrt[3]{5 - 3x^{2}};\ \left( 0;\ \sqrt{\frac{5}{3}} \right):\]

\[y = x\left( 5 - 3x^{2} \right)^{\frac{1}{3}} = \left( 5x^{3} - 3x^{5} \right)^{\frac{1}{3}};\]

\[y^{'} =\]

\[= \frac{1}{3}\left( 5x^{3} - 3x^{5} \right)^{- \frac{2}{3}} \bullet \left( 5 \bullet 3x^{2} - 3 \bullet 5x^{4} \right) =\]

\[= \frac{15x^{2} - 15x^{4}}{3\sqrt[3]{\left( 5x^{3} - 3x^{5} \right)^{2}}} =\]

\[= \frac{15x^{2}\left( 1 - x^{2} \right);}{3\sqrt[3]{\left( 5x^{3} - 3x^{5} \right)^{2}}}.\]

\[Промежуток\ возрастания:\]

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

\[(x + 1)(x - 1) \leq 0\]

\[- 1 \leq x \leq 1.\]

\[y(1) = 1 \bullet \sqrt[3]{5 - 3 \bullet 1} = \sqrt[3]{2}.\]

\[Ответ:\ \ \sqrt[3]{2}.\]

\[2)\ x\sqrt{1 - x^{2}}\ ;\ (0;\ 1):\]

\[y = \sqrt{x^{2} - x^{4}};\]

\[y^{'} = \frac{1}{2\sqrt{x^{2} - x^{4}}} \bullet \left( 2x - 4x^{3} \right).\]

\[Промежуток\ возрастания:\]

\[2x - 4x^{3} \geq 0\]

\[2x\left( 1 - 2x^{2} \right) \geq 0\]

\[\left( x\sqrt{2} + 1 \right)x\left( x\sqrt{2} - 1 \right) \leq 0\]

\[x \leq - \frac{1}{\sqrt{2}};\ \ \ 0 \leq x \leq \frac{1}{\sqrt{2}}.\]

\[y\left( \frac{1}{\sqrt{2}} \right) = \sqrt{\frac{1}{2} - \frac{1}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}.\]

\[Ответ:\ \ \frac{1}{2}.\]