ГДЗ по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 104

\[\boxed{\mathbf{104}\mathbf{.}}\]

\[\textbf{а)}\ y = \frac{2x^{2} + 1}{x}\]

\[k = \lim_{x \rightarrow + \infty}\frac{2x^{2} + 1}{x} =\]

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

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

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

\[Наклонная\ асимптота\ \]

\[при\ x \rightarrow + \infty:\]

\[y = 2x.\]

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

\[b = \lim_{x \rightarrow - \infty}\left( f(x) - kx \right) = 0.\]

\[Наклонная\ асимптота\ \]

\[при\ x \rightarrow - \infty:\]

\[y = 2x.\]

\[Вертикальная\ асимптота:\]

\[x = 0.\]

\[Ответ:k = 2;b = 0;y = 2x.\]

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

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

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

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

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

\[Наклонная\ асимптота\ \]

\[при\ x \rightarrow + \infty:\]

\[y = 2x.\]

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

\[b = \lim_{x \rightarrow - \infty}\left( f(x) - kx \right) = 0.\]

\[Наклонная\ асимптота\ \]

\[при\ x \rightarrow - \infty:\]

\[y = 2x.\]

\[Вертикальная\ асимптота:\]

\[x = 0.\]

\[Ответ:k = 2;b = 0;y = 2x.\]

\[\textbf{в)}\ y = \frac{2x^{2} - 5x + 5}{x - 2}\]

\[k = \lim_{x \rightarrow + \infty}\frac{2x^{2} - 5x + 5}{x - 2} =\]

\[= \lim_{x \rightarrow + \infty}\left( \frac{2 - \frac{5}{x} + \frac{5}{x^{2}}}{1 - \frac{2}{x}} \right) = 2;\]

\[b = \lim_{x \rightarrow + \infty}\left( \frac{2x^{2} - 5x + 5}{x - 2} - 2x \right) =\]

\[= \lim_{x \rightarrow + \infty}\frac{2x^{2} - 5x + 5 - 2x^{2} + 4x}{x - 2} =\]

\[= - 1.\]

\[Наклонная\ асимптота\ \]

\[при\ x \rightarrow + \infty:\]

\[y = 2x - 1.\]

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

\[b = \lim_{x \rightarrow - \infty}\left( f(x) - kx \right) = - 1.\]

\[Наклонная\ асимптота\ \]

\[при\ x \rightarrow - \infty:\]

\[y = 2x - 1.\]

\[Вертикальная\ асимптота:\]

\[x = 2.\]

\[Ответ:k = 2;b = - 1;\]

\[y = 2x - 1.\]

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

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

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

\[b =\]

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

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

\[= - 1.\]

\[Наклонная\ асимптота\ \]

\[при\ x \rightarrow + \infty:\]

\[y = 3x - 1.\]

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

\[b = \lim_{x \rightarrow - \infty}\left( f(x) - kx \right) = - 1.\]

\[Наклонная\ асимптота\ \]

\[при\ x \rightarrow - \infty:\]

\[y = 3x - 1.\]

\[Ответ:k = 3;b = - 1;\]

\[y = 3x - 1.\]

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

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

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

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

\[= \lim_{x \rightarrow + \infty}\frac{3x^{2} - 6x + 3 - 3x^{2} - 1}{3x^{2} + 1} =\]

\[= 0.\]

\[Наклонная\ асимптота\ \]

\[при\ x \rightarrow + \infty:\]

\[y = \frac{1}{3}\text{x.}\]

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

\[b = \lim_{x \rightarrow - \infty}\left( f(x) - kx \right) = 0.\]

\[Наклонная\ асимптота\ \]

\[при\ x \rightarrow - \infty:\]

\[y = \frac{1}{3}\text{x.}\]

\[Ответ:k = \frac{1}{3};b = 0;y = \frac{1}{3}\text{x.}\]

\[\textbf{е)}\ y = \frac{x^{2} + 2x + 1}{2x - 1}\]

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

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

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

\[= \frac{5}{2}.\]

\[Наклонная\ асимптота\ \]

\[при\ x \rightarrow + \infty:\]

\[y = \frac{1}{2}x + \frac{5}{2}.\]

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

\[b = \lim_{x \rightarrow - \infty}\left( f(x) - kx \right) = \frac{5}{2}.\]

\[Наклонная\ асимптота\ \]

\[при\ x \rightarrow - \infty:\]

\[y = \frac{1}{2}x + \frac{5}{2}.\]

\[Ответ:k = \frac{1}{2};b = \frac{5}{2};\]

\[y = \frac{1}{2}x + \frac{5}{2}.\]