Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 316

\[1)\ y = (x + 3)\sqrt{x};\]

\[y^{'} = \sqrt{x} + \frac{x + 3}{2\sqrt{x}} = \frac{2x + x + 3}{2\sqrt{x}} =\]

\[= \frac{3x + 3}{2\sqrt{x}};\]

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

\[= \frac{6x - 3x - 3}{4x \bullet \sqrt{x}} = \frac{3x - 3}{4x\sqrt{x}}.\]

\[Промежуток\ возрастания:\]

\[3x + 3 \geq 0\]

\[3x \geq - 3\]

\[x \geq - 1.\]

\[Выпукла\ вниз:\]

\[\frac{3x - 3}{x} \geq 0\]

\[x < 0;\ \ \ x \geq 1;\]

\[x \geq 0.\]

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

\[y^{'} =\]

\[= \frac{3(x + 1)^{2} \bullet x^{2} - (x + 1)^{3} \bullet 2x}{x^{4}} =\]

\[= \frac{(x + 1)^{2}\left( 3x^{2} - 2x^{2} - 2x \right)}{x^{4}} =\]

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

\[= \frac{6x^{2} + 6x}{x^{5}} = \frac{6x(x + 1)}{x^{5}} =\]

\[= \frac{6(x + 1)}{x^{4}}.\]

\[Промежуток\ возрастания:\]

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

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

\[x \leq 0;\text{\ \ \ x} \geq 2.\]

\[Выпукла\ вниз:\]

\[x + 1 \geq 0\]

\[x \geq - 1;\]

\[x \neq 0.\]

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

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

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

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

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

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

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

\[= \frac{3 + 0 + 0}{1} = 3;\]

\[x = 0;\ \ \ y = x + 3.\]

\[3)\ y = x^{2} \bullet \ln x;\]

\[y^{'} = 2x \bullet \ln x + x^{2} \bullet \frac{1}{x} =\]

\[= 2x \bullet \ln x + x;\]

\[y^{''} = 2 \bullet \ln x + 2x \bullet \frac{1}{x} + 1 =\]

\[= 2\ln x + 3;\]

\[Промежуток\ возрастания:\]

\[2x \bullet \ln x + x \geq 0\]

\[2\ln x + 1 \geq 0\]

\[\ln x \geq - \frac{1}{2}\]

\[x \geq e^{- \frac{1}{2}}.\]

\[Выпукла\ вниз:\]

\[2\ln x + 3 \geq 0\]

\[\ln x \geq - \frac{3}{2}\]

\[x \geq e^{- \frac{3}{2}};\]

\[x > 0.\]

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

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

\[= \frac{(x + 2)^{2}\left( 2x^{2} + 4x - 3x^{2} \right)}{(x + 2)^{6}} =\]

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

\[= \frac{(x + 2)^{3}\left( 8 - 2x^{2} - 16x + 4x^{2} \right)}{(x + 2)^{8}} =\]

\[= \frac{2x^{2} - 16x + 8}{(x + 2)^{5}} = \frac{x^{2} - 8x + 4}{(x + 2)^{5}}.\]

\[Промежуток\ возрастания:\]

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

\[x(x - 4) \leq 0\]

\[0 \leq x \leq 4.\]

\[Выпукла\ вниз:\]

\[\frac{x^{2} - 8x + 4}{x + 2} \geq 0\]

\[D = 64 - 16 = 48\]

\[x = \frac{8 \pm \sqrt{48}}{2} = \frac{8 \pm 4\sqrt{3}}{2} =\]

\[= 4 \pm 2\sqrt{3};\]

\[- 2 < x \leq 4 - 2\sqrt{3};\text{\ \ }\]

\[x \geq 4 + 2\sqrt{3};\]

\[x \neq - 2.\]

\[\lim_{x \rightarrow \infty}{f(x)} = \lim_{x \rightarrow \infty}\frac{x^{2}}{(x + 2)^{3}} = \frac{0}{1} = 0;\]

\[x = - 2;\text{\ \ \ y} = 0.\]