ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1063

\[1)\ y = \sqrt{x + 5}\ на\ \lbrack - 1;\ 4\rbrack:\]

\[y^{'}(x) = \frac{1}{2\sqrt{x + 5}} > 0;\]

\[y( - 1) = \sqrt{- 1 + 5} = \sqrt{4} = 2;\]

\[y(4) = \sqrt{4 + 5} = \sqrt{9} = 3.\]

\[Ответ:\ \ 2;\ 3.\]

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

\[y^{'}(x) = \cos x - 2\sqrt{2}\sin x = 0\]

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

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

\[tg\ x = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}.\]

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

\[= \sqrt{\frac{1}{\frac{2}{16} + \frac{8}{8}}} = \sqrt{1\ :\frac{9}{8}} =\]

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

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

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

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

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

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

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

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

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

\[Ответ:\ \ 1;\ 3.\]