Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 912

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\[1)\ y = \frac{x}{2} + \frac{8}{x}\]

\[y^{'}(x) = \frac{1}{2} \bullet (x)^{'} + 8 \bullet \left( \frac{1}{x} \right)^{'} =\]

\[= \frac{1}{2} + 8 \bullet \left( - \frac{1}{x^{2}} \right) = \frac{1}{2} - \frac{8}{x^{2}}.\]

\[Стационарные\ точки:\]

\[\frac{1}{2} - \frac{8}{x^{2}} = 0\]

\[\frac{x^{2}}{2} - 8 = 0\]

\[\frac{x^{2}}{2} = 8\]

\[x^{2} = 16\ \]

\[x = \pm 4.\]

\[Ответ:\ \ x_{1} = - 4;\ \ x_{2} = 4.\]

\[2)\ y = 2x^{3} - 15x^{2} + 36x\]

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

\[= 2 \bullet \left( x^{3} \right)^{'} - 15 \bullet \left( x^{2} \right)^{'} + (36x)^{'};\]

\[y^{'}(x) = 2 \bullet 3x^{2} - 15 \bullet 2x + 36 =\]

\[= 6x^{2} - 30x + 36.\]

\[Стационарные\ точки:\]

\[6x^{2} - 30x + 36 = 0\]

\[x^{2} - 5x + 6 = 0\]

\[D = 5^{2} - 4 \bullet 6 = 25 - 24 = 1\]

\[x_{1} = \frac{5 - 1}{2} = 2\ \ и\ \ \]

\[x_{2} = \frac{5 + 1}{2} = 3.\]

\[Ответ:\ \ x_{1} = 2;\ \ x_{2} = 3.\]

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

\[y^{'}(x) = \left( e^{2x} \right)^{'} - 2 \bullet \left( e^{x} \right)^{'} =\]

\[= 2e^{2x} - 2e^{x} = 2 \bullet \left( e^{2x} - e^{x} \right).\]

\[Стационарные\ точки:\]

\[e^{2x} - e^{x} = 0\]

\[e^{2x} = e^{x}\]

\[2x = x\]

\[x = 0.\]

\[Ответ:\ \ x = 0.\]

\[4)\ y = \sin x - \cos x\]

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

\[= \cos x + \sin x.\]

\[Стационарные\ точки:\]

\[\cos x + \sin x = 0\ \ \ \ \ |\ :\cos x\]

\[1 + tg\ x = 0\]

\[tg\ x = - 1\]

\[x = - arctg\ 1 + \pi n\]

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

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