Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1581

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\[1)\ y = \frac{2}{(x - 1)(x - 3)} =\]

\[= \frac{2}{x^{2} - 3x - x + 3} =\]

\[= \frac{2}{x^{2} - 4x + 3};\]

\[x - 1 \neq 0 \rightarrow x \neq 1;\]

\[x - 3 \neq 0 \rightarrow x \neq 3;\]

\[D(x) = ( - \infty;\ 1) \cup (1;\ 3) \cup (3;\ + \infty).\]

\[f^{'}(x) = 2 \bullet {\left( x^{2} - 4x + 3 \right)^{- 1}}^{'} =\]

\[= 2 \bullet (2x - 4) \bullet ( - 1) \bullet \left( x^{2} - 4x + 3 \right)^{- 2} =\]

\[= \frac{- 2(2x - 4)}{\left( x^{2} - 4x + 3 \right)^{2}} =\]

\[= \frac{8 - 4x}{\left( x^{2} - 4x + 3 \right)^{2}}.\]

\[Промежуток\ возрастания:\]

\[8 - 4x > 0\]

\[2 - x > 0\]

\[x < 2.\]

\[x = 2 - точка\ максимума.\]

\[y(2) = \frac{2}{(2 - 1)(2 - 3)} =\]

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

\[\lim_{x \rightarrow \infty}\frac{2}{x^{2} - 4x + 3} =\]

\[= \lim_{x \rightarrow \infty}\frac{\frac{2}{x^{2}}}{1 - \frac{4}{x} + \frac{3}{x^{2}}} =\]

\[= \frac{0}{1 - 0 + 0} = 0.\]

\[2)\ y = \frac{1}{\cos x}\]

\[\cos x \neq 0;\]

\[x \neq \arccos 0 + \pi n \neq \frac{\pi}{2} + \pi n.\]

\[y^{'}(x) = \left( \cos x \right)^{- 1} =\]

\[= ( - 1) \bullet \left( - \sin x \right) \bullet \left( \cos x \right)^{- 2} =\]

\[= \frac{\sin x}{\cos^{2}x}.\]

\[Промежуток\ возрастания:\]

\[\sin x > 0\]

\[\arcsin 0 + 2\pi n < x < \pi - \arcsin 0 + 2\pi n\]

\[2\pi n < x < \pi + 2\pi n.\]

\[x = \pi + 2\pi n - точки\ максимума;\]

\[x = 2\pi n - точки\ минимума.\]

\[Максимум\ и\ минимум:\]

\[y(\pi + 2\pi n) = \frac{1}{\cos(\pi + 2\pi n)} =\]

\[= \frac{1}{\cos\pi} = \frac{1}{- 1} = - 1;\]

\[y(2\pi n) = \frac{1}{\cos(2\pi n)} =\]

\[= \frac{1}{\cos 0} = \frac{1}{1} = 1.\]

\[3)\ y = \frac{1}{\ln x}\]

\[x > 0\]

\[\ln x \neq 0\]

\[x \neq 1;\]

\[D(x) = (0;\ 1) \cup (1;\ + \infty).\]

\[y^{'}(x) = \left( \ln x \right)^{- 1} =\]

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

\[Промежуток\ убывания:\]

\[- x < 0\]

\[x > 0.\]

\[\lim_{x \rightarrow \infty}\frac{1}{\ln x} = 0.\]

\[Функция\ отрицательна:\]

\[\frac{1}{\ln x} < 0\]

\[\ln x < 0\]

\[x < 1.\]