ГДЗ по алгебре и начала математического анализа 11 класс Колягин Проверь себя III

\[\mathbf{1.}\ \]

\[1)\ y = 2x^{2} - 5x;\]

\[y^{'}(x) = 2 \bullet 2x - 5 \geq 0;\]

\[4x \geq 5\]

\[x \geq 1\frac{1}{4}.\]

\[Ответ:\ \ \]

\[возрастает\ на\ \left\lbrack 1\frac{1}{4};\ + \infty \right);\]

\[убывает\ на\ \left( - \infty;\ 1\frac{1}{4} \right\rbrack.\]

\[2)\ y = - \sqrt{x + 4};\]

\[y^{'}(x) = - \frac{1}{2\sqrt{x + 4}} < 0;\]

\[x + 4 > 0\]

\[x > - 4.\]

\[Ответ:\ \ убывает\ на\ \lbrack - 4;\ + \infty).\]

\[\mathbf{2}.\]

\[y = x^{4} - 4x^{3} + 20;\]

\[y^{'}(x) = 4x^{3} - 4 \bullet 3x^{2} \geq 0;\]

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

\[x \geq 3;\]

\[y(3) = 81 - 108 + 20 = - 7.\]

\[Ответ:\ \ x_{\min} = 3;\ y(3) = - 7.\]

\[\mathbf{3.}\text{\ y}\]

\[= x^{3} + 3x^{2} - 4;\]

\[y^{'}(x) = 3x^{2} + 3 \bullet 2x \geq 0;\]

\[3x(x + 2) \geq 0\]

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

\[y( - 2) = - 8 + 12 - 4 = 0;\]

\[y(0) = 0 + 0 - 4 = - 4.\]

\[\mathbf{4}\text{.\ }\]

\[f(x) = x + \frac{9}{x};\ \ \ x \in \lbrack 1;\ 4\rbrack:\]

\[f^{'}(x) = 1 - \frac{9}{x^{2}} \geq 0;\]

\[\frac{x^{2} - 9}{x^{2}} \geq 0\]

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

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

\[f(1) = 1 + 9 = 10;\]

\[f(3) = 3 + 3 = 6;\]

\[f(4) = 4 + 2,25 = 6,25.\]

\[Ответ:\ \ 10;\ 6.\]

\[\mathbf{5}\text{.\ }\]

\[a,\ b,\ c\ дм - стороны\ отливки:\]

\[a\ :b = 1\ :2;\]

\[b = 2a;\]

\[\text{abc} = 72;\]

\[c = \frac{72}{\text{ab}} = \frac{72}{2a^{2}} = \frac{36}{a^{2}}.\]

\[Площадь\ поверхности:\]

\[S(a) = 2ab + 2ac + 2bc =\]

\[= 4a^{2} + \frac{72}{a} + \frac{144}{a} =\]

\[= 4a^{2} + \frac{216}{a};\]

\[S^{'}(a) = 4 \bullet 2a - \frac{216}{a^{2}} \geq 0.\]

\[8a^{3} - 216 \geq 0\]

\[8a^{3} \geq 216\]

\[a^{3} \geq 27\]

\[a \geq 3.\]

\[Точка\ минимума:\]

\[a = 3;\]

\[b = 2 \bullet 3 = 6;\]

\[c = \frac{36}{3^{2}} = \frac{36}{9} = 4.\]

\[Ответ:\ \ 3\ дм;\ 4\ дм;\ 6\ дм.\]