ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 229

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

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

\[f^{'}(0) = 0 - e^{0} = - 1;\]

\[a = arctg( - 1) = - \frac{\pi}{2};\]

\[b = \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{4}.\]

\[Ответ:\ \ \frac{3\pi}{4}.\]

\[2)\ f(x) = \cos x,\ \ \ x_{0} = 0:\]

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

\[f^{'}(0) = - \sin 0 = 0;\]

\[a = arctg\ 0 = 0;\]

\[b = \frac{\pi}{2} - 0 = \frac{\pi}{2}.\]

\[Ответ:\ \ \frac{\pi}{2}.\]

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

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

\[f^{'}(0) = \frac{1}{2\sqrt{1}} + \frac{1}{2}e^{0} = 1;\]

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

\[b = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}.\]

\[Ответ:\ \ \frac{\pi}{4}.\]

\[4)\ f(x) = x^{2} + 3x + \frac{2}{2x + 1};\text{\ \ \ }\]

\[x_{0} = 0:\]

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

\[= 2x + 3 + 2 \bullet 2 \bullet \left( - \frac{1}{(2x + 1)^{2}} \right);\]

\[f^{'}(0) = 0 + 3 - \frac{4}{1^{2}} = - 1;\]

\[a = arctg( - 1) = - \frac{\pi}{2};\]

\[b = \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{4}.\]

\[Ответ:\ \ \frac{3\pi}{4}.\]

\[5)\ f(x) = \ln(2x + 1) + \frac{3}{x + 1};\text{\ \ \ }\]

\[x_{0} = 0:\]

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

\[= 2 \bullet \frac{1}{2x + 1} + 3 \bullet \left( - \frac{1}{(x + 1)^{2}} \right);\]

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

\[a = arctg( - 1) = - \frac{\pi}{2};\]

\[b = \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{4}.\]

\[Ответ:\ \ \frac{3\pi}{4}.\]

\[6)\ f(x) = \frac{2}{3}(x + 3)\sqrt{x + 3};\text{\ \ \ }\]

\[x_{0} = 0:\]

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

\[f^{'}(0) = 3^{\frac{1}{2}} = \sqrt{3};\]

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

\[b = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}.\]

\[Ответ:\ \ \frac{\pi}{6}.\]