Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 227

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

\[f(1) = 1 + 1 + 1 = 3;\]

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

\[f^{'}(1) = 3 + 2 = 5;\]

\[y = 3 + 5(x - 1) = 5x - 2.\]

\[Ответ:\ \ y = 5x - 2.\]

\[2)\ f(x) = 6x - 3x^{2};\text{\ \ \ }x_{0} = 2:\]

\[f(2) = 12 - 12 = 0;\]

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

\[f^{'}(2) = 6 - 12 = - 6;\]

\[y = 0 - 6(x - 2) = 12 - 6x.\]

\[Ответ:\ \ y = 12 - 6x.\]

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

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

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

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

\[y = 1 - 3(x - 1) = 4 - 3x.\]

\[Ответ:\ \ y = 4 - 3x.\]

\[4)\ f(x) = \frac{1}{x^{2}},\ \ \ x_{0} = - 2:\]

\[f( - 2) = \frac{1}{4};\]

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

\[f^{'}( - 2) = - \frac{2}{- 8} = \frac{1}{4};\]

\[y = \frac{1}{4} + \frac{1}{4}(x + 2) = \frac{1}{4}x + \frac{3}{4}.\]

\[Ответ:\ \ y = \frac{1}{4}x + \frac{3}{4}.\]

\[5)\ f(x) = \cos x,\ \ \ x_{0} = \frac{\pi}{3}:\]

\[f\left( \frac{\pi}{3} \right) = \cos\frac{\pi}{3} = \frac{1}{2};\]

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

\[f^{'}\left( \frac{\pi}{3} \right) = - \sin\frac{\pi}{3} = - \frac{\sqrt{3}}{2};\]

\[y = \frac{1}{2} - \frac{\sqrt{3}}{2}\left( x - \frac{\pi}{3} \right) =\]

\[= - \frac{\sqrt{3}}{2}x + \frac{1}{2} + \frac{\pi\sqrt{3}}{6}.\]

\[Ответ:\ \ y = - \frac{\sqrt{3}}{2}x + \frac{1}{2} + \frac{\pi\sqrt{3}}{6}.\]

\[6)\ f(x) = e^{x},\ \ \ x_{0} = 0:\]

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

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

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

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

\[Ответ:\ \ y = x + 1.\]

\[7)\ f(x) = \ln x,\ \ \ x_{0} = 1:\]

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

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

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

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

\[Ответ:\ \ y = x - 1.\]

\[8)\ f(x) = \sqrt{x},\ \ \ x_{0} = 1:\]

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

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

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

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

\[Ответ:\ \ y = \frac{1}{2}x + \frac{1}{2}.\]