ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1118

\[1)\ f(x) = \sin{2x} - x;\]

\[f^{'}(x) = 2\cos{2x} - 1 = 0;\]

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

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

\[2x = \pm \arccos\frac{1}{2} + 2\pi n\]

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

\[x = \frac{1}{2} \bullet \left( \pm \frac{\pi}{3} + 2\pi n \right) = \pm \frac{\pi}{6} + \pi n.\]

\[Ответ:\ \pm \frac{\pi}{6} + \pi n.\]

\[2)\ f(x) = \cos{2x} + 2x;\]

\[f^{'}(x) = - 2\sin{2x} + 2 = 0;\]

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

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

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

\[x = \frac{\pi}{4} + \pi n.\]

\[Ответ:\ \ \frac{\pi}{4} + \pi n.\]

\[3)\ f(x) = (2x - 1)^{3};\]

\[f^{'}(x) = 2 \bullet 3(2x - 1)^{2} = 0;\]

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

\[2x - 1 = 0\]

\[2x = 1\]

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

\[Ответ:\ \ \frac{1}{2}.\]

\[4)\ f(x) = (1 - 3x)^{5};\]

\[f^{'}(x) = - 3 \bullet 5(1 - 3x)^{4} = 0;\]

\[(1 - 3x)^{4} = 0\]

\[1 - 3x = 0\]

\[3x = 1\]

\[x = \frac{1}{3}.\]

\[Ответ:\ \ \frac{1}{3}.\]