ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 333

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

\[y^{'} = 2 \bullet \left( - \frac{1}{\left( x^{2} - 4 \right)^{2}} \bullet 2x \right) =\]

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

\[= \frac{- 4\left( x^{2} - 4 \right) + 16x^{2}}{\left( x^{2} - 4 \right)^{3}} =\]

\[= \frac{12x^{2} + 16}{\left( x^{2} - 4 \right)^{3}}.\]

\[Промежуток\ возрастания:\]

\[- 4x \geq 0\]

\[x \leq 0.\]

\[Выпукла\ вниз:\]

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

\[(x + 2)(x - 2) \geq 0\]

\[x \leq - 2;\ \ \ x \geq 2;\]

\[x \neq \pm 2.\]

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

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

\[x = \pm 2;\ \ \ y = 0.\]

\[Функция\ четная:\]

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

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

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

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

\[y^{''} =\]

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

\[= \frac{- 4\left( x^{2} + 4 \right) + 16x^{2}}{\left( x^{2} + 4 \right)^{3}} =\]

\[= \frac{12x^{2} - 16}{\left( x^{2} + 4 \right)^{3}}.\]

\[Промежуток\ возрастания:\]

\[- 4x \geq 0\]

\[x \leq 0.\]

\[Выпукла\ вниз:\]

\[12x^{2} - 16 \geq 0\]

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

\[\left( x\sqrt{3} + 2 \right)\left( x\sqrt{3} - 2 \right) \geq 0\]

\[x \leq - \frac{2}{\sqrt{3}};\ \ \ x \geq \frac{2}{\sqrt{3}}.\]

\[\lim_{x \rightarrow \infty}{f(x)} = \lim_{x \rightarrow \infty}\frac{2}{x^{2} + 4} = \frac{0}{1 + 0} = 0;\]

\[y = 0.\]

\[Функция\ четная:\]

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

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

\[y^{'} =\]

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

\[= (x - 1)(2x + 4 + x - 1) =\]

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

\[y^{''} = 3(2x - 0) = 6x.\]

\[Промежуток\ возрастания:\]

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

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

\[x \leq - 1;\text{\ \ \ x} \geq 1.\]

\[Выпукла\ вниз:\]

\[6x \geq 0\]

\[x \geq 0.\]

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

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

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

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

\[y^{''} =\]

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

\[= (x - 1)\left( 8x - 2 + 4x^{2} - 4x \right) =\]

\[= (x - 1)\left( 4x^{2} + 8x - 2 \right).\]

\[Промежуток\ возрастания:\]

\[4x - 1 \geq 0\]

\[4x \geq 1\]

\[x \geq 0,25.\]

\[Выпукла\ вниз:\]

\[(x - 1)\left( 4x^{2} + 8x - 2 \right) \geq 0\]

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

\[D = 16 + 8 = 24\]

\[x = \frac{- 4 \pm \sqrt{24}}{2 \bullet 2} = \frac{- 4 \pm 2\sqrt{6}}{4} =\]

\[= \frac{- 2 \pm \sqrt{6}}{2};\]

\[\frac{- 2 - \sqrt{6}}{2} \leq x \leq \frac{- 2 + \sqrt{6}}{2};x \geq 1.\]