Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 957

\[\boxed{\mathbf{957}\mathbf{.}}\]

\[1)\ y = x^{4} - 4x^{3} - 8x^{2} + 1\]

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

\[= 4x^{3} - 4 \bullet 3x^{2} - 8 \bullet 2x + 0 =\]

\[= 4x^{3} - 12x^{2} - 16x.\]

\[Стационарные\ точки:\]

\[4x^{3} - 12x^{2} - 16x = 0\]

\[4x \bullet \left( x^{2} - 3x - 4 \right) = 0\]

\[D = 3^{2} + 4 \bullet 4 = 9 + 16 = 25\]

\[x_{1} = \frac{3 - 5}{2} = - 1;\ \ \text{\ \ }\]

\[x_{2} = \frac{3 + 5}{2} = 4.\]

\[(x + 1) \bullet 4x \bullet (x - 4) = 0.\]

\[x_{1} = - 1;\ \ x_{2} = 0;\ \ x_{3} = 4.\]

\[Ответ:\ \ x_{1} = - 1;\ \ x_{2} = 0;\ \ \]

\[x_{3} = 4.\]

\[2)\ y = 4x^{4} - 2x^{2} + 3\]

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

\[= 4 \bullet \left( x^{4} \right)^{'} - 2 \bullet \left( x^{2} \right)^{'} + (3)^{'};\]

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

\[= 16x^{3} - 4x.\]

\[Стационарные\ точки:\]

\[16x^{3} - 4x = 0\]

\[4x \bullet \left( 4x^{2} - 1 \right) = 0\]

\[(2x + 1) \bullet 4x \bullet (2x - 1) = 0\]

\[Ответ:\ \ x_{1} = - 0,5;\ \ x_{2} = 0;\ \ \]

\[x = 0,5.\]

\[3)\ y = \frac{x}{3} - \frac{12}{x}\]

\[y^{'}(x) = \frac{1}{3} \bullet (x)^{'} - 12 \bullet \left( \frac{1}{x} \right)^{'};\]

\[y^{'}(x) = \frac{1}{3} - 12 \bullet \left( - \frac{1}{x^{2}} \right) = \frac{1}{3} + \frac{12}{x^{2}}.\]

\[Стационарные\ точки:\]

\[\frac{1}{3} + \frac{12}{x^{2}} = 0\]

\[x^{2} + 12 \bullet 3 = 0\]

\[x^{2} = - 36 - нет\ корней.\]

\[Ответ:\ \ нет\ таких\ точек.\]

\[4)\ y = \cos{2x} + 2\cos x\]

\[y^{'}(x) = \left( \cos{2x} \right)^{'} + 2 \bullet \left( \cos x \right)^{'};\]

\[y^{'}(x) = - 2\sin{2x} - 2\sin x =\]

\[= - 2 \bullet \left( \sin{2x} + \sin x \right).\]

\[Стационарные\ точки:\]

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

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

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

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

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

\[x = \pi n.\]

\[2)\ 2\cos x + 1 = 0\]

\[2\cos x = - 1\]

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

\[x = \pm \left( \pi - \arccos\frac{1}{2} \right) + 2\pi n\]

\[x = \pm \left( \pi - \frac{\pi}{3} \right) + 2\pi n\]

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

\[Ответ:\ \ x_{1} = \pi n;\ \ \]

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