ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 970

\(\boxed{\mathbf{970}\mathbf{.}}\)

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

\[\textbf{б)}\ y^{'}(x) = \left( x^{2} - 4 \right)^{'} \bullet \left( \frac{2}{u} \right)^{'} =\]

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

\[\textbf{в)}\ Стационарные\ точки:\]

\[- 4x = 0\ \]

\[x = 0.\]

\[\textbf{г)}\ y(0) = \frac{2}{0^{2} - 4} = - \frac{2}{4} = - \frac{1}{2}.\]

\[\textbf{д)}\ Уравнение\ горизонтальной\ \]

\[асимптоты:\]

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

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

\[\textbf{е)}\ Возрастает\ \]

\[на\ ( - \infty;\ - 2) \cup ( - 2;\ 0)\ и\ \]

\[убывает\ на\ (0;\ 2) \cup (2;\ + \infty);\]

\[x = 0 - точка\ максимума.\]

\[\textbf{ж)}\ \]

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

\[\textbf{а)}\ D(x) = ( - \infty;\ + \infty);\]

\[\textbf{б)}\ y^{'}(x) = \left( x^{2} + 4 \right)^{'} \bullet \left( \frac{2}{u} \right)^{'};\]

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

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

\[\textbf{в)}\ Стационарные\ точки:\]

\[- 4x = 0\ \]

\[x = 0.\]

\[\textbf{г)}\ y(0) = \frac{2}{0^{2} + 4} = \frac{2}{4} = \frac{1}{2}.\]

\[\textbf{д)}\ Уравнение\ горизонтальной\ \]

\[асимптоты:\]

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

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

\[\textbf{е)}\ Возрастает\ на\ ( - \infty;\ 0)\ и\ \]

\[убывает\ на\ (0;\ + \infty);\]

\[x = 0 - точка\ максимума.\]

\[\textbf{ж)}\ \]

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

\[\textbf{а)}\ D(x) = ( - \infty;\ + \infty);\]

\[\textbf{в)}\ Стационарные\ точки:\]

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

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

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

\[x_{1} = - 1\ и\ x_{2} = 1.\]

\[\textbf{г)}\ y( - 1) =\]

\[= ( - 1 - 1)^{2} \bullet ( - 1 + 2) =\]

\[= ( - 2)^{2} \bullet 1 = 4;\]

\[y(1) = (1 - 1)^{2} \bullet (1 + 2) =\]

\[= 0^{2} \bullet 3 = 0.\]

\[\textbf{д)}\ Возрастает\ \]

\[на\ ( - \infty;\ - 1) \cup (1;\ + \infty)\ и\ \]

\[убывает\ на\ ( - 1;\ 1);\]

\[x = 1 - точка\ минимума;\ \]

\[x = - 1 - точка\ максимума.\]

\[\textbf{е)}\ \]

\[4)\ y = x \bullet (x - 1)^{3}\ \]

\[\textbf{а)}\ D(x) = ( - \infty;\ + \infty);\]

\[\textbf{б)}\ y^{'}(x) =\]

\[= (x)^{'} \bullet (x - 1)^{3} + x \bullet {(x - 1)^{3}}^{'};\]

\[y^{'}(x) =\]

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

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

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

\[\textbf{в)}\ Стационарные\ точки:\]

\[(x - 1)^{2} \bullet (4x - 1) = 0\]

\[x_{1} = 1\ и\ x_{2} = \frac{1}{4}.\]

\[\textbf{г)}\ y\left( \frac{1}{4} \right) = \frac{1}{4} \bullet \left( \frac{1}{4} - 1 \right)^{3} =\]

\[= \frac{1}{4} \bullet \left( - \frac{3}{4} \right)^{3} = - \frac{27}{256};\]

\[y(1) = 1 \bullet (1 - 1)^{3} = 1 \bullet 0^{3} = 0.\]

\[\textbf{д)}\ Возрастает\ на\ \left( \frac{1}{4};\ + \infty \right)\ и\ \]

\[убывает\ на\ \left( - \infty;\ \frac{1}{4} \right);\]

\[x = \frac{1}{4} - точка\ минимума.\]

\[\textbf{е)}\ \]