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МатематикаРешебник по алгебре и начала математического анализа 11 класс Колягин Задание 229
Задание 229
\[1)\ f(x) = x^{2} + e^{- x},\ \ \ x_{0} = 0:\]
\[f^{'}(x) = 2x - e^{- x};\]
\[f^{'}(0) = 0 - e^{0} = - 1;\]
\[a = arctg( - 1) = - \frac{\pi}{2};\]
\[b = \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{4}.\]
\[Ответ:\ \ \frac{3\pi}{4}.\]
\[2)\ f(x) = \cos x,\ \ \ x_{0} = 0:\]
\[f^{'}(x) = - \sin x;\]
\[f^{'}(0) = - \sin 0 = 0;\]
\[a = arctg\ 0 = 0;\]
\[b = \frac{\pi}{2} - 0 = \frac{\pi}{2}.\]
\[Ответ:\ \ \frac{\pi}{2}.\]
\[3)\ f(x) = \sqrt{x + 1} + e^{\frac{x}{2}},\ \ \ x_{0} = 0:\]
\[f^{'}(x) = \frac{1}{2\sqrt{x + 1}} + \frac{1}{2}e^{\frac{x}{2}};\]
\[f^{'}(0) = \frac{1}{2\sqrt{1}} + \frac{1}{2}e^{0} = 1;\]
\[a = arctg\ 1 = \frac{\pi}{4};\]
\[b = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}.\]
\[Ответ:\ \ \frac{\pi}{4}.\]
\[4)\ f(x) = x^{2} + 3x + \frac{2}{2x + 1};\text{\ \ \ }\]
\[x_{0} = 0:\]
\[f^{'}(x) =\]
\[= 2x + 3 + 2 \bullet 2 \bullet \left( - \frac{1}{(2x + 1)^{2}} \right);\]
\[f^{'}(0) = 0 + 3 - \frac{4}{1^{2}} = - 1;\]
\[a = arctg( - 1) = - \frac{\pi}{2};\]
\[b = \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{4}.\]
\[Ответ:\ \ \frac{3\pi}{4}.\]
\[5)\ f(x) = \ln(2x + 1) + \frac{3}{x + 1};\text{\ \ \ }\]
\[x_{0} = 0:\]
\[f^{'}(x) =\]
\[= 2 \bullet \frac{1}{2x + 1} + 3 \bullet \left( - \frac{1}{(x + 1)^{2}} \right);\]
\[f^{'}(0) = \frac{2}{1} - \frac{3}{1^{2}} = - 1;\]
\[a = arctg( - 1) = - \frac{\pi}{2};\]
\[b = \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{4}.\]
\[Ответ:\ \ \frac{3\pi}{4}.\]
\[6)\ f(x) = \frac{2}{3}(x + 3)\sqrt{x + 3};\text{\ \ \ }\]
\[x_{0} = 0:\]
\[f^{'}(x) = \frac{2}{3} \bullet \frac{3}{2}(x + 3)^{\frac{1}{2}};\]
\[f^{'}(0) = 3^{\frac{1}{2}} = \sqrt{3};\]
\[a = arctg\ \sqrt{3} = \frac{\pi}{3};\]
\[b = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}.\]
\[Ответ:\ \ \frac{\pi}{6}.\]
