Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 218

\[f(x) = 0.\]

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

\[e^{2x} \bullet \ln(2x - 1) = 0\]

\[\ln(2x - 1) = 0\]

\[2x - 1 = 1\]

\[2x = 2\]

\[x = 1.\]

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

\[= 2e^{2x}\ln(2x - 1) + e^{2x} \bullet \frac{2}{2x - 1};\]

\[f^{'}(1) = 2e^{2}\ln 1 + e^{2} \bullet \frac{2}{1} =\]

\[= 0 + 2e^{2} = 2e^{2}.\]

\[Ответ:\ \ 2e^{2}.\]

\[2)\ f(x) =\]

\[= \frac{\sin\text{\ x} - \cos x}{\sin x} = 1 - ctg\ x;\]

\[1 - ctg\ x = 0\]

\[\text{ctg\ x} = 1\]

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

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

\[f^{'}\left( \frac{\pi}{4} + 2\pi n \right) = \frac{1}{\sin^{2}\frac{\pi}{4}} =\]

\[= 1\ :\left( \frac{\sqrt{2}}{2} \right)^{2} = 2;\]

\[f^{'}\left( \frac{3\pi}{4} + 2\pi n \right) = \frac{1}{\sin^{2}\frac{3\pi}{4}} =\]

\[= 1\ :\left( - \frac{\sqrt{2}}{2} \right)^{2} = 2;\]

\(Ответ:\ \ 2.\)