Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1508

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\[y = \sin x + 2\sqrt{2} \bullet \cos x;\ на\left\lbrack 0;\ \frac{\pi}{2} \right\rbrack:\]

\[y^{'}(x) = \left( \sin x \right)^{'} + 2\sqrt{2} \bullet \left( \cos x \right)^{'} =\]

\[= \cos x - 2\sqrt{2} \bullet \sin x.\]

\[Стационарные\ точки:\]

\[\cos x - 2\sqrt{2} \bullet \sin x = 0\ \ \ |\ :\cos x\]

\[1 - 2\sqrt{2} \bullet tg\ x = 0\]

\[2\sqrt{2} \bullet tg\ x = 1\]

\[tg\ x = \frac{\sqrt{2}}{4}.\]

\[\cos(x) = \sqrt{\frac{1}{tg^{2}\ x + 1}} =\]

\[= \sqrt{\frac{1}{\left( \frac{\sqrt{2}}{4} \right)^{2} + 1}} = \sqrt{\frac{1}{\frac{2}{16} + \frac{8}{8}}} =\]

\[= \sqrt{1\ :\frac{9}{8}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3};\]

\[\sin(x) = \sqrt{1 - \cos^{2}(x)} =\]

\[= \sqrt{1 - \frac{8}{9}} = \sqrt{\frac{9}{9} - \frac{8}{9}} = \sqrt{\frac{1}{9}} = \frac{1}{3}.\]

\[y(0) = \sin 0 + 2\sqrt{2} \bullet \cos 0 =\]

\[= 0 + 2\sqrt{2} \bullet 1 = 2\sqrt{2};\]

\[y(x) = \frac{1}{3} + 2\sqrt{2} \bullet \frac{2\sqrt{2}}{3} =\]

\[= \frac{1}{3} + \frac{4 \bullet 2}{3} = \frac{1 + 8}{3} = \frac{9}{3} = 3;\]

\[y\left( \frac{\pi}{2} \right) = \sin\frac{\pi}{2} + 2\sqrt{2} \bullet \cos\frac{\pi}{2} =\]

\[= 1 + 2\sqrt{2} \bullet 0 = 1.\]

\(Ответ:\ \ y_{\min} = 1;\ \ y_{\max} = 3.\)