Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 955

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

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

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

\[Точки\ перегиба:\]

\[- \cos x = 0\]

\[\cos x = 0\]

\[x = \arccos 0 + \pi n\]

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

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

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

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

\[= 5x^{4} - 80 \bullet 2x = 5x^{4} - 160x;\]

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

\[= 5 \bullet 4x^{3} - 160 = 20x^{3} - 160.\]

\[Точки\ перегиба:\]

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

\[x^{3} - 8 = 0\]

\[x^{3} = 8\]

\[x = 2.\]

\[Ответ:\ \ 2.\]

\[3)\ f(x) = 12x^{3} - 24x^{2} + 12x\]

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

\[= 12 \bullet 3x^{2} - 24 \bullet 2x + 12 =\]

\[= 36x^{2} - 48x + 12.\]

\[f^{''}(x) = 36 \bullet \left( x^{2} \right)^{'} - (48x - 12)^{'};\]

\[f^{''}(x) = 36 \bullet 2x - 48 =\]

\[= 72x - 48.\]

\[Точки\ перегиба:\]

\[72x - 48 = 0\]

\[3x - 2 = 0\]

\[3x = 2\]

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

\[Ответ:\ \ \frac{2}{3}.\]

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

\[- \pi < x < \pi;\]

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

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

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

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

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

\[Точки\ перегиба:\]

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

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

\[\sin x \bullet \left( 4\cos x - 1 \right) = 0.\]

\[1)\ \sin x = 0;\]

\[x = \arcsin 0 + \pi n\]

\[x = \pi n.\]

\[2)\ 4\cos x - 1 = 0\]

\[4\cos x = 1\]

\[\cos x = \frac{1}{4}\]

\[x = \pm \arccos\frac{1}{4} + 2\pi n.\]

\[Ответ:\ x = \pm \arccos\frac{1}{4}.\]