Страница 422 номер 1508, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

\(y = \sin x + 2\sqrt{2}\cos x,\)
\(y'(x) = \cos x - 2\sqrt{2}\sin x,\)
\(1^{2} + \left( 2\sqrt{2} \right)^{2} = 9,\)
\(\cos^{2}\alpha + \sin^{2}\alpha = 1,\)
\(\alpha = \arccos\dfrac{1}{3},\)
\(y'(x) = 3\left( \dfrac{1}{3}\cos x - \dfrac{2\sqrt{2}}{3}\sin x \right) = 3\cos(\alpha + x),\)
\(y'(x) = 0,\)
\(\cos{(\alpha + x)} = 0,\)
\(\alpha + x = \dfrac{\pi}{2} + \pi n, n \in Z\)
\(x = \frac{\pi}{2} - \alpha\) при \(n = 0\) (\(0 < \alpha < \frac{\pi}{2}\)), при других \(n\) – вне отрезка;
\(\cos(\alpha + x) > 0\) при
\(0 < \alpha + x < \dfrac{\pi}{2},\)
\(- \alpha < x < \dfrac{\pi}{2} - \alpha;\)
\(\cos(\alpha + x) < 0\) при
\(\dfrac{\pi}{2} - \alpha < x < \dfrac{\pi}{2};\)
\(y_{\text{наиб}} = y\left( \dfrac{\pi}{2} - \alpha \right) = \cos\alpha + 2\sqrt{2}\sin\alpha = \dfrac{1}{3} + 2\sqrt{2} \cdot \dfrac{2\sqrt{2}}{3} = 3;\)
\(y(0) = 2\sqrt{2},\)
\(y_{\text{наим}} = y\left( \dfrac{\pi}{2} \right) = 1.\)