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МатематикаРешебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 8
Задание 8
\[\boxed{\mathbf{8}\mathbf{.}}\]
\[\textbf{а)}\ y = \left( e^{x} \right) - x;\ \ \lbrack - 3;2\rbrack\]
\[f^{'}(x) = e^{x} - 1;\]
\[e^{x} - 1 = 0\]
\[e^{x} = 1\]
\[x = 0.\]
\[0 \in \lbrack - 3;2\rbrack - критическая\ точка.\]
\[Ответ:0.\]
\[\textbf{б)}\ y = e^{x} - xe;\ \ \lbrack - 2;2\rbrack\]
\[f^{'}(x) = e^{x} - e;\]
\[e^{x} - e = 0\]
\[e^{x} = e\]
\[x = 1.\]
\[1 \in \lbrack - 2;2\rbrack - критическая\ точка.\]
\[Ответ:1.\]
\[\textbf{в)}\ y = \sin{2x} - x;\ \ \lbrack - \pi;\ \pi\rbrack\]
\[f^{'}(x) = 2 \cdot \cos{2x} - 1;\]
\[2\cos{2x} - 1 = 0\]
\[2\cos{2x} = 1\]
\[\cos{2x} = \frac{1}{2}\]
\[2x = \pm \arccos\frac{1}{2} + 2\pi k\]
\[2x = \pm \frac{\pi}{3} + 2\pi k\]
\[x = \pm \frac{\pi}{6} + \pi k.\]
\[При\ k = 0:\]
\[x = - \frac{\pi}{6} + \pi \cdot 0 = - \frac{\pi}{6};\]
\[x = \frac{\pi}{6} + \pi \cdot 0 = \frac{\pi}{6}.\]
\[При\ k = 1:\]
\[x = - \frac{\pi}{6} + \pi \cdot 1 = - \frac{\pi}{6} + \frac{6\pi}{6} = \frac{5\pi}{6}.\]
\[При\ k = - 1:\]
\[x = \frac{\pi}{6} - \pi = - \frac{5\pi}{6}.\]
\[Ответ:\ - \frac{5\pi}{6};\ - \frac{\pi}{6};\frac{\pi}{6};\]
\[\frac{5\pi}{6} - критические\ точки.\]
\[\textbf{г)}\ y = \cos{2x} + x;\ \ \ \lbrack - \pi;\ \pi\rbrack\]
\[f^{'}(x) = - 2\sin{2x} + 1;\]
\[- 2\sin{2x} + 1 = 0\]
\[- 2\sin{2x} = - 1\]
\[\sin{2x} = \frac{1}{2}\]
\[2x = ( - 1)^{k} \cdot \arcsin\frac{1}{2} + \pi k\]
\[2x = ( - 1)^{k} \cdot \frac{\pi}{6} + \pi k\]
\[x = ( - 1)^{k} \cdot \frac{\pi}{12} + \frac{\text{πk}}{2}.\]
\[x = ( - 1)^{0} \cdot \frac{\pi}{12} + \frac{\pi \cdot 0}{2} = \frac{\pi}{12};\]
\[x = ( - 1)^{1} \cdot \frac{\pi}{12} + \frac{\pi \cdot 1}{2} =\]
\[= - \frac{\pi}{12} + \frac{6\pi}{12} = \frac{5\pi}{12};\]
\[x = ( - 1)^{- 1} \cdot \frac{\pi}{12} + \frac{\pi \cdot ( - 1)}{2} =\]
\[= - \frac{\pi}{12} - \frac{6\pi}{12} = - \frac{7\pi}{12};\]
\[x = ( - 1)^{- 2} \cdot \frac{\pi}{12} + \frac{\pi \cdot ( - 2)}{2} =\]
\[= \frac{\pi}{12} - \frac{12\pi}{12} = - \frac{11\pi}{12}.\]
\[Ответ:\ - \frac{11\pi}{12};\ - \frac{7\pi}{12};\frac{\pi}{12};\]
\[\frac{5\pi}{12} - критические\ точки.\ \]
