Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1593

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

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

\[f(a) = \sqrt{a};\]

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

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

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

\[Пересечение\ с\ x = 3:\]

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

\[Вертикальный\ катет:\ \ \]

\[b_{1} = \frac{a + 3}{2\sqrt{a}}.\]

\[Пересечение\ с\ \ Ox\ (y = 0):\]

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

\[a + x = 0\]

\[x = - a.\]

\[Горизонтальный\ катет:\ \ \]

\[b_{2} = | - a| + 3 = a + 3.\]

\[Площадь\ треугольника:\]

\[S(a) = \frac{1}{2} \bullet b_{1} \bullet b_{2} =\]

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

\[= \frac{a^{2} + 6a + 9}{4a^{\frac{1}{2}}} =\]

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

\[S^{'}(a) =\]

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

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

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

\[Промежуток\ возрастания:\]

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

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

\[D = 4 + 12 = 16\]

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

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

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

\[a < - 3\ \ и\ \ a > 1.\]

\[a = 1 - точка\ минимума;\]

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

\[Ответ:\ \ S = 4\ при\ a = 1.\]