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МатематикаГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 313
Задание 313
\[1)\ y = 3x + \frac{1}{3x};\]
\[y^{'} = 3 + \frac{1}{3} \bullet \left( - \frac{1}{x^{2}} \right) = \frac{9x^{2} - 1}{3x^{2}};\]
\[y^{''} = 0 - \frac{1}{3} \bullet \left( - \frac{2}{x^{3}} \right) = \frac{2}{3x^{3}}.\]
\[Промежуток\ возрастания:\]
\[9x^{2} - 1 \geq 0\]
\[(3x + 1)(3x - 1) \geq 0\]
\[x \leq - \frac{1}{3};\ \ \ x \geq \frac{1}{3}.\]
\[Выпукла\ вниз:\]
\[\frac{2}{3x^{3}} \geq 0\]
\[x \geq 0.\]
\[x \neq 0.\]
\[\lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\left( 3 + \frac{1}{3x^{2}} \right) =\]
\[= 3 + 0 = 3;\]
\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]
\[= \lim_{x \rightarrow \infty}\left( 3x + \frac{1}{3x} - 3x \right) =\]
\[= \lim_{x \rightarrow \infty}\frac{1}{3x} = 0;\]
\[x = 0;\ \ \ y = 3x.\]
\[Функция\ нечетная:\]
\[y( - x) = 3( - x) + \frac{1}{3( - x)} =\]
\[= - 3x - \frac{1}{3x} = - y(x).\]
\[y\left( \frac{1}{3} \right) = 1 + 1 = 2.\]
\[2)\ y = x - \frac{9}{x};\]
\[y^{'} = 1 - 9 \bullet \left( - \frac{1}{x^{2}} \right) = \frac{x^{2} + 9}{x^{2}} > 0;\]
\[y^{''} = 0 + 9 \bullet \left( - \frac{2}{x^{3}} \right) = - \frac{18}{x^{3}}.\]
\[Выпукла\ вниз:\]
\[- \frac{18}{x^{3}} \geq 0\]
\[x \leq 0.\]
\[x \neq 0.\]
\[\lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\left( 1 - \frac{9}{x^{2}} \right) =\]
\[= 1 - 0 = 1;\]
\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]
\[= \lim_{x \rightarrow \infty}\left( x - \frac{9}{x} - x \right) =\]
\[= \lim_{x \rightarrow \infty}\left( - \frac{9}{x} \right) = 0;\]
\[x = 0;\ \ \ y = x.\]
\[Функция\ нечетная:\]
\[y( - x) = ( - x) - \frac{9}{( - x)} =\]
\[= - x + \frac{9}{x} = - y(x).\]
\[3)\ y = \frac{4}{x} - x;\]
\[y^{'} = 4 \bullet \left( - \frac{1}{x^{2}} \right) - 1 = \frac{- 4 - x^{2}}{x^{2}} < 0;\]
\[y^{''} = - 4 \bullet \left( - \frac{2}{x^{3}} \right) - 0 = \frac{8}{x^{3}}.\]
\[Выпукла\ вниз:\]
\[\frac{8}{x^{3}} \geq 0\]
\[x \geq 0.\]
\[x \neq 0.\]
\[Уравнения\ асимптот:\]
\[\lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\left( \frac{4}{x^{2}} - 1 \right) =\]
\[= 0 - 1 = - 1;\]
\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]
\[= \lim_{x \rightarrow \infty}\left( \frac{4}{x} - x + x \right) = \lim_{x \rightarrow \infty}\frac{4}{x} = 0;\]
\[x = 0;\ \ \ y = - x.\]
\[Функция\ нечетная:\]
\[y( - x) = \frac{4}{( - x)} - ( - x) =\]
\[= - \frac{4}{x} + x = - y(x).\]
\[4)\ y = x - \frac{1}{\sqrt{x}};\]
\[y^{'} = 1 - \left( - \frac{1}{2}x^{- \frac{3}{2}} \right) =\]
\[= 1 + \frac{1}{2\sqrt{x^{3}}} > 0;\]
\[y^{''} = 0 + \frac{1}{2} \bullet \left( - \frac{3}{2}x^{- \frac{5}{2}} \right) =\]
\[= - \frac{3}{4\sqrt{x^{5}}} < 0;\]
\[x > 0.\]
\[\lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\left( 1 - \frac{1}{x\sqrt{x}} \right) =\]
\[= 1 - 0 = 1;\]
\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]
\[= \lim_{x \rightarrow \infty}\left( x - \frac{1}{\sqrt{x}} - x \right) =\]
\[= \lim_{x \rightarrow \infty}\left( - \frac{1}{\sqrt{x}} \right) = 0;\]
\[x = 0;\ \ \ y = x.\]
\[y(1) = 1 - \frac{1}{\sqrt{1}} = 1 - 1 = 0.\]
