Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 225

\[1)\ f(x) = \frac{1}{3}x^{2},\ \ \ x_{0} = 1:\]

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

\[f^{'}(1) = \frac{2}{3};\]

\[a = arctg\frac{2}{3}.\]

\[Ответ:\ \ arctg\frac{2}{3}.\]

\[2)\ f(x) = \frac{1}{3}x^{3},\ \ \ x_{0} = 1:\]

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

\[f^{'}(1) = 1^{2} = 1;\]

\[a = arctg\ 1 = \frac{\pi}{4}.\]

\[Ответ:\ \ \frac{\pi}{4}.\]

\[3)\ f(x) = \frac{1}{x},\ \ \ x_{0} = 1:\]

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

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

\[a = arctg( - 1) = \frac{3\pi}{4}.\]

\[Ответ:\ \ \frac{3\pi}{4}.\]

\[4)\ f(x) = 2\sqrt{x},\ \ \ x_{0} = 3:\]

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

\[f^{'}(3) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3};\]

\[a = arctg\frac{\sqrt{3}}{3} = \frac{\pi}{6}.\]

\[Ответ:\ \ \frac{\pi}{6}.\]

\[5)\ f(x) = e^{\frac{3x + 1}{2}},\ \ \ x_{0} = 1:\]

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

\[f^{'}(1) = \frac{3}{2} \bullet e^{\frac{4}{2}} = \frac{3}{2}e^{2};\]

\[a = arctg\left( \frac{3}{2}e^{2} \right).\]

\[Ответ:\ \ arctg\left( \frac{3}{2}e^{2} \right).\]

\[6)\ f(x) = \ln(2x + 1);x_{0} = \frac{1}{2}:\]

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

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

\[a = arctg\ 1 = \frac{\pi}{4}.\]

\[Ответ:\ \ \frac{\pi}{4}.\]