Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 291

\[1)\ f(x) = \sin x + \cos x\ на\left\lbrack \pi;\ \frac{3\pi}{2} \right\rbrack:\]

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

\[Стационарные\ точки:\]

\[\cos x - \sin x = 0\ \ \ \ \ |\ :\cos x\]

\[1 - tg\ x = 0\]

\[\text{tg\ x} = 1\]

\[x = \frac{\pi}{4} + \pi n.\]

\[f(\pi) = \sin\pi + \cos\pi =\]

\[= 0 - 1 = - 1;\]

\[f\left( \frac{5\pi}{4} \right) = \sin\frac{5\pi}{4} + \cos\frac{5\pi}{4} =\]

\[= - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = - \sqrt{2};\]

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

\[= - 1 + 0 = - 1.\]

\[Ответ:\ - \sqrt{2};\ - 1.\]

\[2)\ f(x) = \sin x + \cos x\ на\ \left\lbrack 0;\ \frac{\pi}{2} \right\rbrack:\]

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

\[Стационарные\ точки:\]

\[\cos x - \sin x = 0\ \ \ \ \ |\ :\cos x\]

\[1 - tg\ x = 0\]

\[\text{tg\ x} = 1\]

\[x = \frac{\pi}{4} + \pi n.\]

\[f(0) = \sin 0 + \cos 0 = 0 + 1 = 1;\]

\[f\left( \frac{\pi}{4} \right) = \sin\frac{\pi}{4} + \cos\frac{\pi}{4} =\]

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

\[f\left( \frac{\pi}{2} \right) = \sin\frac{\pi}{2} + \cos\frac{\pi}{2} = 1 + 0 = 1.\]

\[Ответ:\ \ 1;\ \sqrt{2}.\]