Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1064

\[1)\ y = \ln x - x\ на\ \lbrack 0,5;\ 4\rbrack:\]

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

\[\frac{1 - x}{x} \geq 0\]

\[\frac{x - 1}{x} \leq 0\]

\[0 < x \leq 1.\]

\[y(0,5) = \ln{0,5} - 0,5 = \ln\frac{1}{2} - \frac{1}{2};\]

\[y(1) = \ln 1 - 1 = 0 - 1 = - 1;\]

\[y(4) = \ln 4 - 4.\]

\[Ответ:\ \ \ln 4 - 4;\ - 1.\]

\[2)\ y = x\sqrt{1 - x^{2}}\ на\ \lbrack 0;\ 1\rbrack:\]

\[y^{'}(x) = \left( \sqrt{x^{2} - x^{4}} \right)^{'} =\]

\[= \frac{2x - 4x^{3}}{2\sqrt{x^{2} - 4x^{3}}} = 0;\]

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

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

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

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

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

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

\[= \frac{1}{\sqrt{2}} \bullet \frac{1}{\sqrt{2}} = \frac{1}{2};\]

\[y(1) = 1 \bullet \sqrt{1 - 1^{2}} = 1 \bullet \sqrt{0} = 0.\]

\(Ответ:\ \ 0;\ 0,5.\)