Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 863

\[\boxed{\mathbf{863}\mathbf{.}}\]

\[k = tg\ a\ \]

\[a = arctg\ k:\]

\[\beta = \frac{\pi}{2} - arctg\ k.\]

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

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

\[= 1 + ( - 1) \bullet e^{- x} = 1 - e^{- x}\]

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

\[\beta = \frac{\pi}{2} - arctg\ 0 = \frac{\pi}{2} - 0 = \frac{\pi}{2}.\]

\[Ответ:\ \ \beta = \frac{\pi}{2}.\]

\[2)\ f(x) = \cos x\ и\ \ x_{0} = 0:\]

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

\[k = f^{'}(0) = - \sin 0 = 0\]

\[\beta = \frac{\pi}{2} - arctg\ 0 = \frac{\pi}{2} - 0 = \frac{\pi}{2}.\]

\[Ответ:\ \ \beta = \frac{\pi}{2}.\]

\[3)\ f(x) = \sqrt{x + 1} + e^{\frac{x}{2}}\text{\ \ }и\ \ x_{0} = 0:\]

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

\[= \frac{1}{2} \bullet (x + 1)^{- \frac{1}{2}} + \frac{1}{2} \bullet e^{\frac{x}{2}} =\]

\[= \frac{1}{2\sqrt{x + 1}} + \frac{e^{\frac{x}{2}}}{2}\]

\[k = f^{'}(0) = \frac{1}{2\sqrt{0 + 1}} + \frac{e^{0}}{2} =\]

\[= \frac{1}{2\sqrt{1}} + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1\]

\[\beta = \frac{\pi}{2} - arctg\ 1 = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}.\]

\[Ответ:\ \ \beta = \frac{\pi}{4}.\]