ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 933

\[\boxed{\mathbf{933}\mathbf{.}}\]

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

\[\textbf{а)}\ D(x) = ( - \infty;\ 2) \cup (2;\ + \infty);\]

\[\textbf{б)}\ y^{'}(x) =\]

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

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

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

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

\[\textbf{в)}\ Стационарные\ точки:\]

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

\[x \bullet (x - 4) = 0\]

\[x_{1} = 0\ и\ x_{2} = 4.\]

\[\textbf{г)}\ f(0) = \frac{0^{2}}{0 - 2} = 0;\]

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

\[\textbf{д)}\ Возрастает\ \]

\[на\ ( - \infty;\ 0) \cup (4;\ + \infty)\ и\ убывает\ \]

\[на\ (0;\ 2) \cup (2;\ 4);\]

\[x = 4 - точка\ минимума;\text{\ \ }\]

\[x = 0 - точка\ максимума.\]

\[\textbf{е)}\ \]

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

\[= - x + 3 - \frac{1}{x}\]

\[\textbf{а)}\ D(x) = ( - \infty;\ 0) \cup (0;\ + \infty);\]

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

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

\[\textbf{в)}\ Стационарные\ точки:\]

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

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

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

\[x_{1} = - 1\ и\ x_{2} = 1.\]

\[\textbf{г)}\ f( - 1) = 1 + 3 + \frac{1}{1} = 5;\]

\[f(1) = - 1 + 3 - \frac{1}{1} = 1;\]

\[\textbf{д)}\ Возрастает\ \]

\[на\ ( - 1;\ 0) \cup (0;\ 1)\ и\ убывает\ \]

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

\[x = - 1 - точка\ минимума;\text{\ \ }\]

\[x = 1 - точка\ максимума.\]

\[\textbf{е)}\ \]

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

\[\textbf{а)}\ D(x) = ( - \infty;\ 2) \cup (2;\ + \infty);\]

\[= \frac{7x^{2} - 24x + 20}{(x - 4)^{4}}.\]

\[\textbf{в)}\ Стационарные\ точки:\]

\[7x^{2} - 24x + 20 = 0\]

\[D = 24^{2} - 4 \bullet 7 \bullet 20 =\]

\[= 576 - 560 = 16\]

\[x_{1} = \frac{24 - 4}{2 \bullet 7} = \frac{20}{14} = \frac{10}{7};\ \text{\ \ }\]

\[x_{2} = \frac{24 + 4}{2 \bullet 7} = \frac{28}{14} = 2.\]

\[\left( x - 1\frac{3}{7} \right)(x - 2) = 0.\]

\[\textbf{г)}\ f\left( \frac{10}{7} \right) = \frac{4 + \frac{10}{7} - 2 \bullet \left( \frac{10}{7} \right)^{2}}{\left( \frac{10}{7} - 2 \right)^{2}} =\]

\[= \frac{\frac{38}{7} - \frac{200}{49}}{\left( - \frac{4}{7} \right)^{2}} = \frac{66}{49}\ :\frac{16}{49} = \frac{33}{8} =\]

\[= 4\frac{1}{8}.\]

\[\textbf{д)}\ Возрастает\ \]

\[на\left( - \infty;\ 1\frac{3}{7} \right) \cup (2;\ + \infty)\ и\ \]

\[убывает\ на\ \left( 1\frac{3}{7};\ 2 \right);\]

\[x = 1\frac{3}{7} - точка\ максимума.\]

\[\textbf{е)}\ \]