Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 283

\[1)\ f(x) = 2\sin x + \cos{2x}\ на\ \lbrack 0;\ 2\pi\rbrack:\]

\[f^{'}(x) = 2\cos x - 2\sin{2x}.\]

\[Стационарные\ точки:\]

\[2\cos x - 2\sin{2x} = 0\]

\[2\cos x - 4\sin x \bullet \cos x = 0\]

\[2\cos x \bullet \left( 1 - 2\sin x \right) = 0;\]

\[\cos x = 0\]

\[x = \frac{\pi}{2} + \pi n.\text{\ \ }\]

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

\[x = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]

\[f(0) = 2\sin 0 + \cos 0 = 1;\]

\[f\left( \frac{\pi}{6} \right) = 2\sin\frac{\pi}{6} + \cos\frac{\pi}{3} =\]

\[= 1 + \frac{1}{2} = 1,5;\]

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

\[= 2 - 1 = 1;\]

\[f\left( \frac{5\pi}{6} \right) = 2\sin\frac{5\pi}{6} + \cos\frac{5\pi}{3} =\]

\[= 1 + \frac{1}{2} = 1,5;\]

\[f\left( \frac{3\pi}{2} \right) = 2\sin\frac{3\pi}{2} + \cos{3\pi} =\]

\[= - 2 - 1 = - 3;\]

\[f(2\pi) = 2\sin{2\pi} + \cos{4\pi} =\]

\[= 0 + 1 = 1.\]

\[Ответ:\ - 3;\ 1,5.\]

\[2)\ f(x) = 2\cos x - \cos{2x}\ на\ \lbrack 0;\ \pi\rbrack:\]

\[f^{'}(x) = - 2\sin x + 2\sin{2x}.\]

\[Стационарные\ точки:\]

\[2\sin{2x} - 2\sin x = 0\]

\[4\sin x \bullet \cos x - 2\sin x = 0\]

\[2\sin x \bullet \left( 2\cos x - 1 \right) = 0\]

\[\sin x = 0\text{\ \ }\]

\[x = \pi n.\]

\[\cos x = \frac{1}{2}\text{\ \ }\]

\[x = \pm \frac{\pi}{3} + 2\pi n.\]

\[f(0) = 2\cos 0 - \cos 0 = 2 - 1 = 1;\]

\[f\left( \frac{\pi}{3} \right) = 2\cos\frac{\pi}{3} - \cos\frac{2\pi}{3} =\]

\[= 1 + \frac{1}{2} = 1,5;\]

\[f(\pi) = 2\cos\pi - \cos{2\pi} =\]

\[= - 2 - 1 = - 3.\]

\[Ответ:\ - 3;\ 1,5.\]