ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 947

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\[1)\ f(x) = x \bullet \sqrt[4]{5 - x};\ \ (0;\ 5)\]

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

\[= (x)^{'} \bullet \sqrt[4]{5 - x} + x \bullet {(5 - x)^{\frac{1}{4}}}^{'};\]

\[f^{'}(x) = \sqrt[4]{5 - x} - \frac{x}{4\sqrt[4]{(5 - x)^{3}}};\]

\[f^{'}(x) = \frac{4(5 - x) - x}{4\sqrt[4]{(5 - x)^{3}}} =\]

\[= \frac{20 - 4x - x}{4\sqrt[4]{(5 - x)^{3}}} = \frac{20 - 5x}{4\sqrt[4]{(5 - x)^{3}}}.\]

\[Промежуток\ возрастания:\]

\[20 - 5x > 0\]

\[5x < 20\]

\[x < 4.\]

\[f(4) = 4 \bullet \sqrt[4]{5 - 4} = 4 \bullet \sqrt[4]{1} = 4.\]

\[Ответ:\ \ y_{\max} = 4.\]

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

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

\[= (x)^{'} \bullet \ \sqrt[3]{4 - x} + x \bullet {(4 - x)^{\frac{1}{3}}}^{'};\]

\[f^{'}(x) = \sqrt[3]{4 - x} - \frac{x}{3\sqrt[3]{(4 - x)^{2}}};\]

\[f^{'}(x) = \frac{3(4 - x) - x}{3\sqrt[3]{(4 - x)^{2}}} =\]

\[= \frac{12 - 3x - x}{3\sqrt[3]{(4 - x)^{2}}} = \frac{12 - 4x}{3\sqrt[3]{(4 - x)^{2}}}.\]

\[Промежуток\ возрастания:\]

\[12 - 4x > 0\]

\[4 \bullet (3 - x) > 0\]

\[3 - x > 0\]

\[x < 3.\]

\[f(3) = 3 \bullet \sqrt[3]{4 - 3} = 3 \bullet \sqrt[3]{1} = 3.\]

\[Ответ:\ \ y_{\max} = 3.\]

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

\[Пусть\ u = x^{2} - x^{3};\ \ \text{\ f}(u) = u^{\frac{1}{3}}:\]

\[f^{'}(x) = \left( x^{2} - x^{3} \right)^{'} \bullet \left( u^{\frac{1}{3}} \right)^{'};\]

\[f^{'}(x) = \left( 2x - 3x^{2} \right) \bullet \frac{1}{3} \bullet u^{- \frac{2}{3}} =\]

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

\[Промежуток\ возрастания:\]

\[2x - 3x^{2} > 0\]

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

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

\[0 < x < \frac{2}{3}.\]

\[f\left( \frac{2}{3} \right) = \sqrt[3]{\left( \frac{2}{3} \right)^{2} - \left( \frac{2}{3} \right)^{3}} =\]

\[= \sqrt[3]{\frac{4 \bullet 3 - 8}{27}} = \sqrt[3]{\frac{4}{27}} = \frac{\sqrt[3]{4}}{3}.\]

\[Ответ:\ \ y_{\max} = \frac{\sqrt[3]{4}}{3}.\]

\[4)\ f(x) = \sqrt[3]{\left( x^{2} - 4x + 5 \right)^{- 1}};\ \]

\[( - 1;\ 5)\]

\[Пусть\ u = x^{2} - 4x + 5;\ \]

\[f(u) = u^{- \frac{1}{3}}:\]

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

\[f^{'}(x) = (2x - 4) \bullet \left( - \frac{1}{3} \right) \bullet u^{- \frac{4}{3}};\]

\[f^{'}(x) = - \frac{2x - 4}{3 \bullet \sqrt[3]{\left( x^{2} - 4x + 5 \right)^{4}}}.\]

\[Промежуток\ возрастания:\]

\[- (2x - 4) > 0\]

\[2x - 4 < 0\]

\[x - 2 < 0\]

\[x < 2.\]

\[f(2) = \sqrt[3]{\left( 2^{2} - 4 \bullet 2 + 5 \right)^{- 1}} =\]

\[= \frac{1}{\sqrt[3]{4 - 8 + 5}} = \frac{1}{\sqrt[3]{4 - 8 + 5}} =\]

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

\[Ответ:\ \ y_{\max} = 1.\]