ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1509

\[\boxed{\mathbf{1509}\mathbf{.}}\]

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

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

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

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

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

\[Стационарные\ точки:\]

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

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

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

\[x = \pm \frac{1}{\sqrt{2}}.\]

\[Имеет\ смысл\ при:\]

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

\[x^{2} \leq 1\]

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

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

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

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

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

\[= 1 \bullet \sqrt{1 - 1} = 1 \bullet \sqrt{0} = 0.\]

\[Ответ:\ \ y_{\min} = 0;\ \ y_{\max} = 0,5.\]