ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 315

\[1)\ y = \frac{x^{2}}{x - 2};\]

\[y^{'} = \frac{2x(x - 2) - x^{2}}{(x - 2)^{2}} = \frac{x^{2} - 4x}{(x - 2)^{2}};\]

\[y^{''} =\]

\[= \frac{8x - 16}{(x - 2)^{4}} = \frac{8(x - 2)}{(x - 2)^{4}} = \frac{8}{(x - 2)^{3}}.\]

\[Промежуток\ возрастания:\]

\[x^{2} - 4x \geq 0\]

\[x(x - 4) \geq 0\]

\[x \leq 0;\ \ \ x \geq 4.\]

\[Выпукла\ вниз:\]

\[x - 2 \geq 0\]

\[x \geq 2.\]

\[x \neq 2.\]

\[\lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\left( \frac{x}{x - 2} \right) =\]

\[= \frac{1}{1 - 0} = 1;\]

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]

\[= \lim_{x \rightarrow \infty}\left( \frac{x^{2}}{x - 2} - x \right) =\]

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

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) = \lim_{x \rightarrow \infty}\frac{2x}{x - 2} =\]

\[= \frac{2}{1 - 0} = 2;\]

\[x = 2;\ \ \ y = x + 2.\]

\[2)\ y = \frac{- x^{2} + 3x - 1}{x};\]

\[y^{'} = - 1 + 0 + \frac{1}{x^{2}} = \frac{1 - x^{2}}{x^{2}};\]

\[y^{''} = - \frac{2}{x^{3}} - 0 = - \frac{2}{x^{3}}.\]

\[Промежуток\ возрастания:\]

\[1 - x^{2} \geq 0\]

\[(x + 1)(x - 1) \leq 0\]

\[- 1 \leq x \leq 1.\]

\[Выпукла\ вниз:\]

\[- \frac{2}{x^{3}} \geq 0\]

\[x \leq 0;\]

\[x \neq 0.\]

\[\lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\left( - 1 + \frac{3}{x} - \frac{1}{x^{2}} \right) =\]

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

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]

\[= \lim_{x \rightarrow \infty}\left( - x + 3 - \frac{1}{x} + x \right) =\]

\[= \lim_{x \rightarrow \infty}\left( 3 - \frac{1}{x} \right) = 3;\]

\[x = 0;\ \ \ y = - x + 3.\]

\[3)\ y = \frac{x^{2} + x - 1}{x^{2} - 2x + 1};\]

\[y^{'} =\]

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

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

\[y^{''} =\]

\[= \frac{6x^{2} - 6x}{(x - 1)^{5}} = \frac{6x(x - 1)}{(x - 1)^{5}} =\]

\[= \frac{6x}{(x - 1)^{4}}.\]

\[Промежуток\ возрастания:\]

\[- 3x^{2} + 4x - 1 \geq 0\]

\[3x^{2} - 4x + 1 \leq 0\]

\[D = 16 - 12 = 4\]

\[x_{1} = \frac{4 - 2}{2 \bullet 3} = \frac{1}{3};\text{\ \ }\]

\[x_{2} = \frac{4 + 2}{2 \bullet 3} = 1;\]

\[\left( x - \frac{1}{3} \right)(x - 1) \leq 0\]

\[\frac{1}{3} \leq x \leq 1.\]

\[Выпукла\ вниз:\]

\[x \geq 0;\]

\[x \neq 1.\]

\[\lim_{x \rightarrow \infty}{f(x)} = \lim_{x \rightarrow \infty}\frac{x^{2} + x - 1}{x^{2} - 2x + 1} =\]

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

\[x = 1;\text{\ \ \ y} = 1.\]

\[4)\ y = \frac{4 + x - 2x^{2}}{(x - 2)^{2}};\]

\[y^{'} =\]

\[= \frac{7x - 10}{(x - 2)^{3}};\]

\[y^{''} =\]

\[= \frac{7(x - 2)^{3} - 3(7x - 10)(x - 2)^{2}}{(x - 2)^{6}} =\]

\[= \frac{(x - 2)^{2}(7x - 14 - 21x + 30)}{(x - 2)^{6}} =\]

\[= \frac{16 - 14x}{(x - 2)^{4}}.\]

\[Промежуток\ возрастания:\]

\[\frac{7x - 10}{x - 2} \geq 0\]

\[x \leq 1\frac{3}{7};\text{\ \ \ x} \geq 2.\]

\[Выпукла\ вниз:\]

\[16 - 14x \geq 0\]

\[14x \leq 16\]

\[x \leq 1\frac{1}{7};\]

\[x \neq 2.\]

\[Уравнения\ асимптот:\]

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

\[= \frac{0 + 0 - 2}{1 - 0 + 0} = - 2;\]

\[x = 2;\text{\ \ \ y} = - 2.\]

\[5)\ y = \frac{(x - 1)^{3}}{(x - 2)^{2}};\]

\[y^{'} =\]

\[= \frac{(x - 1)^{2}(x - 2)(3x - 6 - 2x + 2)}{(x - 2)^{4}} =\]

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

\[y^{''} =\]

\[= \frac{6x - 6}{(x - 2)^{4}} = \frac{6(x - 1)}{(x - 2)^{4}}.\]

\[Промежуток\ возрастания:\]

\[\frac{x - 4}{x - 2} \geq 0\]

\[x \leq 2;\text{\ \ \ x} \geq 4.\]

\[Выпукла\ вниз:\]

\[x - 1 \geq 0\]

\[x \geq 1;\]

\[x \neq 2.\]

\[\lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\frac{(x - 1)^{3}}{x(x - 2)^{2}} = \frac{1}{1} = 1;\]

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]

\[= \lim_{x \rightarrow \infty}\left( \frac{(x - 1)^{3}}{(x - 2)^{2}} - x \right) =\]

\[= \lim_{x \rightarrow \infty}\frac{(x - 1)^{3} - x(x - 2)^{2}}{(x - 2)^{2}};\]

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]

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

\[x = 2;\ \ \ y = x + 1.\]

\[6)\ y = \frac{x^{3}}{x^{2} - 1};\]

\[y^{'} = \frac{3x^{2}\left( x^{2} - 1 \right) - x^{3} \bullet 2x}{\left( x^{2} - 1 \right)^{2}} =\]

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

\[y^{''} =\]

\[= \frac{2x^{5} - 4x^{3} + 6x}{\left( x^{2} - 1 \right)^{3}}.\]

\[Промежуток\ возрастания:\]

\[x^{4} - 3x^{2} \geq 0\]

\[x^{2} - 3 \geq 0\]

\[\left( x + \sqrt{3} \right)\left( x - \sqrt{3} \right) \geq 0\]

\[x \leq - \sqrt{3};\ \ x \geq \sqrt{3};\text{\ \ \ x} = 0.\]

\[Выпукла\ вниз:\]

\[\frac{2x^{5} - 4x^{3} + 6x}{x^{2} - 1} \geq 0\]

\[\frac{x\left( 2x^{4} - 4x^{2} + 6 \right)}{(x + 1)(x - 1)} \geq 0\]

\[D = 16 - 48 = - 32\]

\[- 1 < x \leq 0;\ \ \ x > 1;\]

\[x \neq \pm 1.\]

\[\lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\frac{x^{2}}{x^{2} - 1} =\]

\[= \frac{1}{1 - 0} = 1;\]

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]

\[= \lim_{x \rightarrow \infty}\left( \frac{x^{3}}{x^{2} - 1} - x \right) =\]

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

\[\lim_{x \rightarrow \infty}\left( f(x) - kx \right) = \lim_{x \rightarrow \infty}\frac{x}{x^{2} - 1} =\]

\[= \frac{0}{1 - 0} = 0;\]

\[x = \pm 1;\ \ \ y = x.\]

\[Функция\ нечетная:\]

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

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