Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 332

\[1)\ y = x \bullet \ln x;\]

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

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

\[\ln x \geq - 1\]

\[x \geq e^{- 1};\]

\[x = e^{- 1} - точка\ минимума.\]

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

\[y^{'} = 1 \bullet e^{x} + x \bullet e^{x} = e^{x}(1 + x).\]

\[1 + x \geq 0\]

\[x \geq - 1;\]

\[x = - 1 - точка\ минимума.\]

\[3)\ y = \frac{4}{x - 3} - \frac{16}{x - 7};\]

\[y^{'} = - \frac{4}{(x - 3)^{2}} + \frac{16}{(x - 7)^{2}} =\]

\[= \frac{16(x - 3)^{2} - 4(x - 7)^{2}}{(x - 3)^{2} \bullet (x - 7)^{2}}.\]

\[Промежуток\ возрастания:\]

\[16(x - 3)^{2} - 4(x - 7)^{2} \geq 0\]

\[12x^{2} - 40x - 52 \geq 0\]

\[3x^{2} - 10x - 13 \geq 0\]

\[D = 100 + 156 = 256\]

\[x_{1} = \frac{10 - 16}{2 \bullet 3} = - 1;\]

\[x_{2} = \frac{10 + 16}{2 \bullet 3} = 4\frac{1}{3};\]

\[(x + 1)\left( x - 4\frac{1}{3} \right) \geq 0\]

\[x \leq - 1;\ \ \ x \geq 4\frac{1}{3}.\]

\[x = 4\frac{1}{3} - точка\ минимума;\]

\[x = - 1 - точка\ максимума.\]

\[4)\ y = \frac{25}{7 - x} - \frac{9}{3 - x};\]

\[y^{'} = \frac{25}{(7 - x)^{2}} - \frac{9}{(3 - x)^{2}} =\]

\[= \frac{25(3 - x)^{2} - 9(7 - x)^{2}}{(3 - x)^{2} \bullet (7 - x)^{2}}.\]

\[Промежуток\ возрастания:\]

\[25(3 - x)^{2} - 9(7 - x)^{2} \geq 0\]

\[16x^{2} - 24x - 216 \geq 0\]

\[2x^{2} - 3x - 27 \geq 0\]

\[D = 9 + 216 = 225\]

\[x_{1} = \frac{3 - 15}{2 \bullet 2} = - 3;\text{\ \ }\]

\[x_{2} = \frac{3 + 15}{2 \bullet 2} = 4,5;\]

\[(x + 3)(x - 4,5) \geq 0\]

\[x \leq - 3;\text{\ \ \ x} \geq 4,5.\]

\[x = 4,5 - точка\ минимума;\]

\[x = - 3 - точка\ максимума.\]