ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 860

\[\boxed{\mathbf{860}\mathbf{.}}\]

\[y = f\left( x_{0} \right) + f^{'}\left( x_{0} \right)\left( x - x_{0} \right).\]

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

\[f^{'}(x) = \left( x^{2} \right)^{'} + (x + 1)^{'} = 2x + 1\]

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

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

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

\[= 3 + 3x - 3 = 3x.\]

\[Ответ:\ \ y = 3x.\]

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

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

\[= 1 - 3 \bullet 2x = 1 - 6x\]

\[f^{'}(2) = 1 - 6 \bullet 2 = 1 - 12 = - 11\]

\[f(2) = 2 - 3 \bullet 2^{2} = 2 - 3 \bullet 4 =\]

\[= 2 - 12 = - 10\]

\[y = - 10 - 11(x - 2) =\]

\[= - 10 - 11x + 22 = 12 - 11x.\]

\[Ответ:\ \ y = 12 - 11x.\]

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

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

\[f^{'}(3) = - \frac{1}{3^{2}} = - \frac{1}{9}\]

\[f(3) = \frac{1}{3}\]

\[y = \frac{1}{3} - \frac{1}{9}(x - 3) =\]

\[= \frac{1}{3} - \frac{1}{9}x + \frac{1}{3} = \frac{2}{3} - \frac{1}{9}x.\]

\[Ответ:\ \ y = \frac{2}{3} - \frac{1}{9}\text{x.}\]

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

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

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

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

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

\[= - \frac{1}{2} - \frac{1}{4}x - \frac{1}{2} = - \frac{1}{4}x - 1.\]

\[Ответ:\ \ y = - \frac{1}{4}x - 1.\]

\[5)\ f(x) = \sin x\text{\ \ }и\ \ x_{0} = \frac{\pi}{4}:\]

\[f^{'}(x) = \left( \sin x \right)^{'} = \cos x\]

\[f^{'}\left( \frac{\pi}{4} \right) = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}\]

\[f\left( \frac{\pi}{4} \right) = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}\]

\[y = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\left( x - \frac{\pi}{4} \right) =\]

\[= \frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2} - \frac{\pi\sqrt{2}}{8}.\]

\[Ответ:\ \ y = \frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2} - \frac{\pi\sqrt{2}}{8}.\]

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

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

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

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

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

\[Ответ:\ \ y = x + 1.\]

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

\[f^{'}(x) = \left( \ln x \right)^{'} = \frac{1}{x}\]

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

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

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

\[Ответ:\ \ y = x - 1.\]

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

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

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

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

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

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

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