Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 841

\[\boxed{\mathbf{841}\mathbf{.}}\]

\[1)\ f(x) = x - \cos x\]

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

\[= 1 + \sin x\]

\[1 + \sin x = 0\]

\[\sin x = - 1\]

\[x = - \arcsin 1 + 2\pi n =\]

\[= - \frac{\pi}{2} + 2\pi n.\]

\[Ответ:\ \ - \frac{\pi}{2} + 2\pi n.\]

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

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

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

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

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

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

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

\[Ответ:\ \ \pm \frac{\pi}{3} + 2\pi n.\]

\[3)\ f(x) = 2\ln(x + 3) - x\]

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

\[= \frac{2}{x + 3} - 1\]

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

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

\[x + 3 = 2\]

\[x = - 1.\]

\[Ответ:\ \ - 1.\]

\[4)\ f(x) = \ln(x + 1) - 2x\]

\[f^{'}(x) = \left( \ln(x + 1) \right)^{'} - (2x)^{'} =\]

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

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

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

\[2(x + 1) = 1\]

\[2x + 2 = 1\]

\[2x = - 1\]

\[x = - 0,5.\]

\[Ответ:\ \ - 0,5.\]

\[5)\ f(x) = x^{2} + 2x - 12\ln x\]

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

\[= \left( x^{2} \right)^{'} + (2x)^{'} - 12 \bullet \left( \ln x \right)^{'} =\]

\[= 2x + 2 - \frac{12}{x}\]

\[2x + 2 - \frac{12}{x} = 0\]

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

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

\[D = 1 + 24 = 25\]

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

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

\[Ответ:\ \ - 3\ \ 2.\]

\[6)\ f(x) = x^{2} - 6x - 8\ln x\]

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

\[= \left( x^{2} \right)^{'} - (6x)^{'} - 8 \bullet \left( \ln x \right)^{'} =\]

\[= 2x - 6 - \frac{8}{x}\]

\[2x - 6 - \frac{8}{x} = 0\]

\[2x^{2} - 6x - 8 = 0\]

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

\[D = 9 + 16 = 25\]

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

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

\[Ответ:\ \ - 1\ \ 4.\]