ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 228

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

\[x_{0} = 0:\]

\[f(0) = 0 + 0 - 0 + 2 = 2;\]

\[f^{'}(x) = 4x^{3} + 3 \bullet 2x - 4;\]

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

\[y = 2 - 4(x - 0) = 2 - 4x.\]

\[Ответ:\ \ y = 2 - 4x.\]

\[2)\ f(x) = \sqrt[3]{x + 1};\text{\ \ \ }x_{0} = 0:\]

\[f(0) = \sqrt[3]{1} = 1;\]

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

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

\[y = 1 + \frac{1}{3}(x - 0) = \frac{1}{3}x + 1.\]

\[Ответ:\ \ y = \frac{1}{3}x + 1.\]

\[3)\ f(x) = 2x - \sqrt{x + 1};\ x_{0} = 0:\]

\[f(0) = 0 - \sqrt{1} = - 1;\]

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

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

\[y = - 1 + \frac{3}{2}(x - 0) = \frac{3}{2}x - 1.\]

\[Ответ:\ \ \ y = \frac{3}{2}x - 1.\]

\[4)\ f(x) = x + \frac{1}{x + 1},\ \ \ x_{0} = 0:\]

\[f(0) = 0 + \frac{1}{1} = 1;\]

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

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

\[y = 1 + 0(x - 0) = 1.\]

\[Ответ:\ \ y = 1.\]

\[5)\ f(x) = e^{3x} + \cos x;\ x_{0} = 0:\]

\[f(0) = e^{0} + \cos 0 = 2;\]

\[f^{'}(x) = 3e^{3x} - \sin x;\]

\[f^{'}(0) = 3e^{0} - \sin 0 = 3;\]

\[y = 2 + 3(x - 0) = 3x + 2.\]

\[Ответ:\ \ y = 3x + 2.\]

\[6)\ f(x) = \sin{2x} - \ln(x + 1);\text{\ \ \ }\]

\[x_{0} = 0:\]

\[f(0) = \sin 0 - \ln 1 = 0;\]

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

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

\[y = 0 + 1(x - 0) = x.\]

\[Ответ:\ \ y = x.\]