Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 209

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

\[f^{'}(x) = 1 + \sin x = 0\]

\[\sin x = - 1\]

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

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

\[f^{'}(x) = \frac{1}{2} - \cos x = 0\]

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

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

\[3)\ f(x) = \ln(x + 1) - 2x;\]

\[f^{'}(x) = \frac{1}{x + 1} - 2 = 0\]

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

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

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

\[4)\ f(x) = 2\ln(x + 3) - x;\]

\[f^{'}(x) = 2 \bullet \frac{1}{x + 3} - 1 = 0\]

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

\[x + 3 = 2\]

\[x = - 1.\]

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

\[f^{'}(x) = 2^{x} \bullet \ln 2 - 2^{- x} \bullet \ln 2 = 0\]

\[\ln 2 \bullet \left( 2^{x} - 2^{- x} \right) = 0\]

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

\[2^{x} = 2^{- x}\]

\[x = 0.\]

\[6)\ f(x) = 3^{2x} - 2x\ln 3;\]

\[f^{'}(x) = 2 \bullet 3^{2x} \bullet \ln 3 - 2\ln 3 = 0\]

\[2\ln 3 \bullet \left( 3^{2x} - 1 \right) = 0\]

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

\[3^{2x} = 1\]

\[2x = 0\]

\[x = 0.\]

\[7)\ f(x) = 2\ln(x + 3) - x;\]

\[f^{'}(x) = 2 \bullet \frac{1}{x + 3} - 1 = 0\]

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

\[x + 3 = 2\]

\[x = - 1.\]

\[8)\ f(x) = x + \ln(2x + 1);\]

\[f^{'}(x) = 1 + 2 \bullet \frac{1}{2x + 1} = 0\]

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

\[2x + 1 = - 2\]

\[x \in \varnothing.\]