ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 944

\[\boxed{\mathbf{944}\mathbf{.}}\]

\[1)\ f(x) = \ln x - x;\ \ \ \ \ \left\lbrack \frac{1}{2};\ 3 \right\rbrack\]

\[f^{'}(x) = \left( \ln x \right)^{'} - x = \frac{1}{x} - 1.\]

\[Точки\ экстремума:\]

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

\[1 - x = 0\ \]

\[x = 1.\]

\[y\left( \frac{1}{2} \right) = \ln\frac{1}{2} - \frac{1}{2} = - \ln 2 - 0,5;\]

\[y(1) = \ln 1 - 1 = \ln e^{0} - 1 =\]

\[= 0 - 1 = - 1;\]

\[y(3) = \ln 3 - 3.\]

\[Ответ:\ \ y_{\min} = \ln 3 - 3;\ \ \]

\[y_{\max} = - 1.\]

\[2)\ f(x) = x + e^{- x};\ \ \ \lbrack - 1;\ 2\rbrack\]

\[f^{'}(x) = (x)^{'} + \left( e^{- x} \right)^{'} = 1 - e^{- x}.\]

\[Точки\ экстремума:\]

\[1 - e^{- x} = 0\]

\[e^{- x} = 1\]

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

\[x = 0.\]

\[y( - 1) = - 1 + e^{1} = e - 1;\]

\[y(0) = 0 + e^{0} = 0 + 1 = 1;\]

\[y(2) = 2 + e^{- 2} = 2 + \frac{1}{e}.\]

\[Ответ:\ \ y_{\min} = 1;\ \ y_{\max} = 2 + \frac{1}{e}.\]

\[3)\ f(x) = 2\cos x - \cos{2x};\ \]

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

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

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

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

\[Стационарные\ точки:\]

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

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

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

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

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

\[x = \pi n.\]

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

\[2\cos x = 1\]

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

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

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

\[y(0) = 2\cos 0 - \cos 0 =\]

\[= 2 \bullet 1 - 1 = 1;\]

\[y\left( \frac{\pi}{3} \right) = 2 \bullet \cos\frac{\pi}{3} - \cos\frac{2\pi}{3} =\]

\[= 2 \bullet \frac{1}{2} + \frac{1}{2} = 1 + 0,5 = 1,5;\]

\[y(\pi) = 2 \bullet \cos\pi - \cos{2\pi} =\]

\[= 2 \bullet ( - 1) - 1 = - 2 - 1 = - 3.\]

\[Ответ:\ \ y_{\min} = - 3;\ \ y_{\max} = 1,5.\]