ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1115

\[f(x) = \sqrt{x};\text{\ \ \ }\frac{1}{2} \leq a \leq 2.\]

\[1)\ Уравнение\ касательной:\]

\[f^{'}(x) = \left( \sqrt{x} \right)^{'} = \frac{1}{2\sqrt{x}};\]

\[f^{'}(a) = \frac{1}{2\sqrt{a}},\ \ \ f(a) = \sqrt{a};\]

\[y = \sqrt{a} + \frac{1}{2\sqrt{a}}(x - a) =\]

\[= \frac{2a + x - a}{2\sqrt{a}} = \frac{a + x}{2\sqrt{a}}.\]

\[2)\ Пересечение\ с\ x = 3:\]

\[y = \frac{a + 3}{2\sqrt{a}}.\]

\[3)\ Пересечение\ с\ Ox:\]

\[\frac{a + x}{2\sqrt{a}} = 0\]

\[a + x = 0\]

\[x = - a.\]

\[4)\ S(a) = \frac{1}{2} \bullet \frac{a + 3}{2\sqrt{a}} \bullet (a + 3) =\]

\[= \frac{(a + 3)^{2}}{4\sqrt{a}} = \frac{a^{2} + 6a + 9}{4a^{\frac{1}{2}}} =\]

\[= \frac{1}{4}\left( a^{\frac{3}{2}} + 6a^{\frac{1}{2}} + 9a^{- \frac{1}{2}} \right).\]

\[= \frac{1}{4}\left( \frac{3}{2}a^{\frac{1}{2}} + 3a^{- \frac{1}{2}} - \frac{9}{2}a^{- \frac{3}{2}} \right) =\]

\[= \frac{3}{8}\left( \sqrt{a} + \frac{2}{\sqrt{a}} - \frac{3}{a\sqrt{a}} \right).\]

\[6)\ Промежуток\ возрастания:\]

\[\sqrt{a} + \frac{2}{\sqrt{a}} - \frac{3}{a\sqrt{a}} \geq 0\ \ \ \ \ | \bullet a\sqrt{a}\]

\[a^{2} + 2a - 3 \geq 0\]

\[D = 4 + 12 = 16\]

\[a_{1} = \frac{- 2 - 4}{2} = - 3;\]

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

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

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

\[7)\ Точка\ минимума:\]

\[a = 1;\]

\[S(1) = \frac{(1 + 3)^{2}}{4\sqrt{1}} = \frac{4^{2}}{4} = 4.\]

\[Ответ:\ \ a = 1;\ S = 4.\]