ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 260

\[1)\ y = 2\sin\frac{x}{2},\ \ \ x_{0} = \frac{3\pi}{2}:\]

\[y\left( \frac{3\pi}{2} \right) = 2\sin\frac{3\pi}{4} = \sqrt{2};\]

\[y^{'}(x) = 2 \bullet \frac{1}{2}\cos\frac{x}{2};\]

\[y^{'}\left( \frac{3\pi}{2} \right) = \cos\frac{3\pi}{4} = - \frac{\sqrt{2}}{2}.\]

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

\[= \sqrt{2} - \frac{\sqrt{2}}{2}x + \frac{3\pi\sqrt{2}}{4}.\]

\[Ответ:\ \ \]

\[y = - \frac{\sqrt{2}}{2}x + \sqrt{2} + \frac{3\pi\sqrt{2}}{4}.\]

\[2)\ y = 2^{- x} - 2^{- 2x},\ \ \ x_{0} = 2:\]

\[y(2) = 2^{- 2} - 2^{- 4} = \frac{1}{4} - \frac{1}{16} = \frac{3}{16};\]

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

\[= - 2^{- x} \bullet \ln 2 - ( - 2) \bullet 2^{- 2x} \bullet \ln 2 =\]

\[= \ln 2 \bullet \left( 2 \bullet 2^{- 2x} - 2^{- x} \right);\]

\[y^{'}(2) = \ln 2 \bullet \left( 2 \bullet 2^{- 4} - 2^{- 2} \right) =\]

\[= \ln 2 \bullet \left( \frac{1}{8} - \frac{1}{4} \right) = - \frac{1}{8}\ln 2.\]

\[y = \frac{3}{16} - \frac{1}{8}\ln 2 \bullet (x - 2) =\]

\[= \frac{3}{16} - \frac{1}{8}\ln 2 \bullet x + \frac{1}{4}\ln 2.\]

\[Ответ:\ \ \]

\[y = - \frac{x}{8}\ln 2 + \frac{3}{16} + \frac{1}{4}\ln 2.\]

\[3)\ y = \frac{x + 3}{2 - x},\ \ \ x_{0} = 2:\]

\[y(2) = \frac{2 + 3}{2 - 2} = \frac{5}{0}.\]

\[Ответ:\ \ не\ существует.\]

\[4)\ y = x + \ln x,\ \ \ x_{0} = e:\]

\[y(e) = e + \ln e = e + 1;\]

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

\[y^{'}(e) = 1 + \frac{1}{e};\]

\[y = e + 1 + \left( 1 + \frac{1}{e} \right)(x - e) =\]

\[= e + 1 + \left( 1 + \frac{1}{e} \right)x - e - 1.\]

\[Ответ:\ \ y = \left( 1 + \frac{1}{e} \right)\text{x.}\]

\[5)\ y = e^{x^{2} - 1},\ \ \ x_{0} = 1:\]

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

\[y^{'}(x) = 2x \bullet e^{x^{2} - 1};\]

\[y^{'}(1) = 2 \bullet 1 \bullet e^{0} = 2;\]

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

\[Ответ:\ \ y = 2x - 1.\]

\[6)\ y = \sin\left( \pi x^{2} \right),\ \ \ x_{0} = 1:\]

\[y(1) = \sin\pi = 0;\]

\[y^{'}(x) = 2\pi x \bullet \cos\left( \pi x^{2} \right);\]

\[y^{'}(1) = 2\pi \bullet \cos\pi = - 2\pi;\]

\[y = 0 - 2\pi(x - 1) = 2\pi - 2\pi x.\]

\[Ответ:\ \ y = 2\pi - 2\pi\text{x.}\]