ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 280

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

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

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

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

\[Промежуток\ возрастания:\]

\[2x^{3} + 6x^{2} \geq 0\]

\[2x^{2} \bullet (x + 3) \geq 0\]

\[x + 3 \geq 0\]

\[x \geq - 3.\]

\[Ответ:\ \ \]

\[x = - 3 - точка\ минимума.\]

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

\[y^{'} =\]

\[= \frac{\left( x^{5} \right)^{'} \bullet (5 - x) - x^{5} \bullet (5 - x)^{'}}{(5 - x)^{2}} =\]

\[= \frac{5x^{4} \bullet (5 - x) - x^{5} \bullet ( - 1)}{(x - 5)^{2}} =\]

\[= \frac{25x^{4} - 5x^{5} + x^{5}}{(x - 5)^{2}} = \frac{25x^{4} - 4x^{5}}{(x - 5)^{2}}.\]

\[Промежуток\ возрастания:\]

\[25x^{4} - 4x^{5} \geq 0\]

\[x^{4} \bullet (25 - 4x) \geq 0\]

\[4x - 25 \leq 0\]

\[x \leq \frac{25}{4}.\]

\[Ответ:\ \ \]

\[x = \frac{25}{4} - точка\ максимума.\]

\[3)\ y = |x - 5|(x - 3)^{3};\]

\[x \geq 5:\]

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

\[y^{'} =\]

\[= (x - 5)^{'} \bullet (x - 3)^{3} + (x - 5) \bullet {(x - 3)^{3}}^{'} =\]

\[= (x - 3)^{3} + (x - 5) \bullet 3(x - 3)^{2} =\]

\[= (x - 3)^{2} \bullet (x - 3 + 3x - 15) =\]

\[= (x - 3)^{2} \bullet (4x - 18).\]

\[x \leq 5:\]

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

\[y^{'} = (x - 3)^{2} \bullet (18 - 4x).\]

\[Промежуток\ возрастания:\]

\[18 - 4x \geq 0.\]

\[4x \leq 18\]

\[x \leq \frac{9}{2}.\]

\[Ответ:\ \ \]

\[x = \frac{9}{2} - точка\ максимума;\]

\[x = 5 - точка\ минимума.\]

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

\[y^{'} =\]

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

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

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

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

\[Промежуток\ возрастания:\]

\[(x + 3)(x - 1) \geq 0\]

\[x \leq - 3;\ \ \ x \geq 1.\]

\[Ответ:\ \ \]

\[x = - 3 - точка\ максимума;\]

\(x = 1 - точка\ минимума.\)