ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1073

\[1)\ y = \frac{1}{3}x^{3} - x^{2} - 3x + 9\]

\[D(x) = ( - \infty;\ + \infty);\]

\[ни\ четная,\ ни\ нечетная:\]

\[f( - x) =\]

\[= \frac{1}{3}( - x)^{3} - ( - x)^{2} - 3( - x) + 9 =\]

\[= - \frac{1}{3}x^{3} - x^{2} + 3x + 9.\]

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

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

\[Промежуток\ возрастания:\]

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

\[D = 4 + 12 = 16\]

\[x_{1} = \frac{2 - 4}{2} = - 1;\]

\[x_{2} = \frac{2 + 4}{2} = 3;\]

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

\[x \leq - 1;\text{\ \ \ }x \geq 3.\]

\[Максимум\ и\ минимум:\]

\[y( - 1) = - \frac{1}{3} - 1 + 3 + 9 =\]

\[= 11 - \frac{1}{3} = 10\frac{2}{3};\]

\[y(3) = \frac{27}{3} - 9 - 9 + 9 =\]

\[= 9 - 9 = 0.\]

\[2)\ y = - x^{4} + 6x^{2} - 9\]

\[D(x) = ( - \infty;\ + \infty);\]

\[является\ четной:\]

\[f( - x) = - ( - x)^{4} + 6\left( x^{2} \right) - 9 =\]

\[= - x^{4} + 6x^{2} - 9 = f(x).\]

\[f^{'}(x) = - 4x^{3} + 6 \bullet 2x =\]

\[= 12x - 4x^{3}.\]

\[Промежуток\ возрастания:\]

\[12x - 4x^{3} \geq 0\]

\[4x\left( 3 - x^{2} \right) \geq 0\]

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

\[x \leq - \sqrt{3};\ \ \ 0 \leq x \leq \sqrt{3}.\]

\[Максимум\ и\ минимум:\]

\[y\left( \sqrt{3} \right) = - 9 + 18 - 9 = 0;\]

\[y(0) = - 0^{4} + 6 \bullet 0^{2} - 9 = - 9.\]

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

\[D(x) = ( - \infty;\ 0) \cup (0;\ + \infty);\]

\[является\ нечетной:\]

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

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

\[f^{'}(x) = \left( x + \frac{1}{x} \right)^{'} = 1 - \frac{1}{x^{2}}.\]

\[Промежуток\ возрастания:\]

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

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

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

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

\[Максимум\ и\ минимум:\]

\[y(1) = \frac{1^{2} + 1}{1} = 2.\]

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

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

\[\lim_{x \rightarrow + \infty}\left( f(x) - x \right) = \lim_{x \rightarrow + \infty}\frac{1}{x} = 0;\]

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

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

\[D(x) = ( - \infty;\ 0) \cup (0;\ + \infty);\]

\[является\ нечетной:\]

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

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

\[f^{'}(x) = \left( \frac{x}{2} + \frac{1}{x} \right)^{'} = \frac{1}{2} - \frac{1}{x^{2}}.\]

\[Промежуток\ возрастания:\]

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

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

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

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

\[Максимум\ и\ минимум:\]

\[y\left( \sqrt{2} \right) = \frac{2 + 2}{2\sqrt{2}} = \frac{4}{2\sqrt{2}} =\]

\[= \frac{2}{\sqrt{2}} = \sqrt{2}.\]

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

\[= \frac{1 + 0}{2} = \frac{1}{2};\]

\[\lim_{x \rightarrow + \infty}\left( f(x) - \frac{x}{2} \right) = \lim_{x \rightarrow + \infty}\frac{1}{x} = 0;\]

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