Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 305

\[1)\ f(x) = \cos x;\ - \pi < x < \pi:\]

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

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

\[- \cos x = 0\]

\[\cos x = 0\]

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

\[Ответ:\ - \frac{\pi}{2};\ \frac{\pi}{2}.\]

\[2)\ f(x) = x^{5} - 80x^{2};\]

\[f^{'}(x) = 5x^{4} - 80 \bullet 2x =\]

\[= 5x^{4} - 160x;\]

\[f^{''}(x) = 5 \bullet 4x^{3} - 160 =\]

\[= 20x^{3} - 160.\]

\[20x^{3} - 160 = 0\]

\[20x^{3} = 160\]

\[x^{3} = 8\]

\[x = 2.\]

\[Ответ:\ \ 2.\]

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

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

\[= 3x^{2} - 4x + 1;\]

\[f^{''}(x) = 3 \bullet 2x - 4 + 0 =\]

\[= 6x - 4.\]

\[6x - 4 = 0\]

\[6x = 4\]

\[x = \frac{2}{3}.\]

\[Ответ:\ \ \frac{2}{3}.\]

\[4)\ f(x) = \sin x - \frac{1}{4}\sin{2x};\]

\[- \pi < x < \pi:\]

\[f^{'}(x) = \cos x - \frac{1}{4} \bullet 2\cos{2x} =\]

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

\[f^{''}(x) = - \sin x + \frac{1}{2} \bullet 2\sin{2x} =\]

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

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

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

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

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

\[x = \pi n.\text{\ \ }\]

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

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

\[Ответ:\ - \frac{\pi}{3};\ 0;\ \frac{\pi}{3}.\]