Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 83

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\[\textbf{а)}\ f(x) = \frac{1}{2}x + \sin x;\ \ \lbrack 0;\ \pi\rbrack\]

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

\[\frac{1}{2} + \cos x = 0\]

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

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

\[x = \frac{2\pi}{3} - критическая\ точка.\]

\[f^{''}(x) = - \sin x < 0\ при\ \]

\[x \in \lbrack 0;\ \pi\rbrack;\]

\[x = \frac{2\pi}{3} - точка\ максимума.\]

\[f_{\max} = f\left( \frac{2\pi}{3} \right) = \frac{1}{2} \cdot \frac{2\pi}{3} +\]

\[+ \sin\frac{2\pi}{3} = \frac{\pi}{3} + \frac{\sqrt{3}}{2};\]

\[f(0) = \frac{1}{2} \cdot 0 + \sin 0 = 0;\]

\[f(\pi) = \frac{1}{2} \cdot \pi + \sin\pi = \frac{\pi}{2};\]

\[f_{\min} = 0.\]

\[\textbf{б)}\ f(x) = \frac{1}{2}x + \sin x;\ \ \lbrack\pi;2\pi\rbrack\]

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

\[\frac{1}{2} + \cos x = 0\]

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

\[x = \frac{2\pi}{3} + 2\pi k;\]

\[x = \frac{4\pi}{3} + 2\pi k.\]

\[x = \frac{4\pi}{3} - критическая\ точка.\]

\[f^{''}(x) = - \sin x > 0\ при\ \]

\[x \in \lbrack\pi;2\ \pi\rbrack;\]

\[x = \frac{4\pi}{3} - точка\ минимума.\]

\[f_{\min} = f\left( \frac{4\pi}{3} \right) = \frac{1}{2} \cdot \frac{4\pi}{3} +\]

\[+ \sin\frac{4\pi}{3} = \frac{2\pi}{3} - \frac{\sqrt{3}}{2};\]

\[f(\pi) = \frac{1}{2} \cdot \pi + \sin\pi = \frac{\pi}{2};\]

\[f(2\pi) = \frac{1}{2} \cdot 2\pi + \sin{2\pi} = \pi;\]

\[f_{\max} = \pi.\]

\[\textbf{в)}\ y = - \frac{1}{2}x + \cos x;\ \ \left\lbrack - \frac{\pi}{2};\ \frac{\pi}{2} \right\rbrack.\]

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

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

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

\[x = - \frac{\pi}{6} + 2\pi k;\]

\[x = \frac{7\pi}{6} + 2\pi k.\]

\[x = - \frac{\pi}{6} \rightarrow критическая\ точка.\]

\[f^{''}(x) = - \cos x < 0\ при\ \]

\[x \in \left\lbrack - \frac{\pi}{2};\frac{\pi}{2} \right\rbrack.\]

\[x = - \frac{\pi}{6} \rightarrow точка\ максимума:\]

\[f_{\max} = f\left( - \frac{\pi}{6} \right) = - \frac{1}{2} \cdot \left( - \frac{\pi}{6} \right) +\]

\[+ \cos\left( - \frac{\pi}{6} \right) = \frac{\pi}{12} + \frac{\sqrt{3}}{2}.\]

\[f\left( - \frac{\pi}{2} \right) = - \frac{1}{2} \cdot \left( - \frac{\pi}{2} \right) +\]

\[+ \cos\left( - \frac{\pi}{2} \right) = \frac{\pi}{4};\]

\[f\left( \frac{\pi}{2} \right) = - \frac{1}{2} \cdot \frac{\pi}{2} + \cos\frac{\pi}{2} = - \frac{\pi}{4}.\]

\[f_{\min} = - \frac{\pi}{4}.\]

\[\textbf{г)}\ y = - \frac{1}{2}x + \cos x;\ \ \left\lbrack \frac{\pi}{2};\ \frac{3\pi}{2} \right\rbrack.\]

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

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

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

\[x = - \frac{\pi}{6} + 2\pi k;\]

\[x = \frac{7\pi}{6} + 2\pi k.\]

\[x = \frac{7\pi}{6} \rightarrow критическая\ точка.\]

\[f^{''}(x) = - \cos x > 0\ при\ \]

\[x \in \left\lbrack \frac{\pi}{2};\frac{3\pi}{2} \right\rbrack.\]

\[x = \frac{7\pi}{6} \rightarrow точка\ минимума:\]

\[f_{m\text{in}} = f\left( \frac{7\pi}{6} \right) = - \frac{1}{2} \cdot \left( \frac{7\pi}{6} \right) +\]

\[+ \cos\left( \frac{7\pi}{6} \right) = \frac{- 7\pi}{12} - \frac{\sqrt{3}}{2}.\]

\[f\left( \frac{\pi}{2} \right) = - \frac{1}{2} \cdot \frac{\pi}{2} + \cos\frac{\pi}{2} = - \frac{\pi}{4};\]

\[f\left( \frac{3\pi}{2} \right) = - \frac{1}{2} \cdot \frac{3\pi}{2} + \cos\frac{3\pi}{2} =\]

\[= - \frac{3\pi}{4}.\]

\[f_{\max} = - \frac{\pi}{4}.\]