ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1109

\[y = (x - 1)^{2};\ \ \ 0 \leq x \leq 1.\]

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

\[y^{'}(a) = 2(a - 1);\]

\[y(a) = (a - 1)^{2};\]

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

\[= (a - 1)(a - 1 + 2x - 2a) =\]

\[= (a - 1)(2x - a - 1) =\]

\[2)\ y(0) = (a - 1)(2 \bullet 0 - a - 1) =\]

\[= - (a + 1)(a - 1) =\]

\[= (1 - a)(a + 1).\]

\[3)\ (a - 1)(2x - a - 1) = 0\]

\[a = 1;\text{\ \ \ }\]

\[2x = a + 1\]

\[x = \frac{a + 1}{2}.\]

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

\[= \frac{1}{4}(1 - a)(a + 1)^{2}.\]

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

\[= \frac{(a + 1)( - a - 1 + 2 - 2a)}{4} =\]

\[= \frac{(a + 1)(1 - 3a)}{4}.\]

\[6)\ Промежуток\ возрастания:\]

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

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

\[- 1 \leq a \leq \frac{1}{3}.\]

\[7)\ Точка\ максимума:\]

\[a = \frac{1}{3};\text{\ \ \ }\]

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

\[Ответ:\ \ \left( \frac{1}{3};\ \frac{4}{9} \right).\]