ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1112

\[y = 4x - x^{2};\ \ \ M = \left( \frac{5}{2};\ 6 \right):\]

\[1)\ Уравнение\ касательной\ \]

\[в\ точке\ a:\]

\[y^{'}(x) = (4x)^{'} - \left( x^{2} \right)^{'} = 4 - 2x;\text{\ \ }\]

\[y^{'}(a) = 4 - 2a,\ \ \ y(a) = 4a - a^{2};\]

\[y = 4a - a^{2} + (4 - 2a)(x - a) =\]

\[= 4a - a^{2} + 4x - 4a - 2ax + 2a^{2} =\]

\[= a^{2} - 2ax + 4x.\]

\[2)\ Проходящие\ через\ точку\ M:\]

\[y\left( \frac{5}{2} \right) = a^{2} - 2a \bullet \frac{5}{2} + 4 \bullet \frac{5}{2} = 6;\]

\[a^{2} - 5a + 10 = 6;\]

\[a^{2} - 5a + 4 = 0;\]

\[D = 25 - 16 = 9\]

\[a_{1} = \frac{5 - 3}{2} = 1;\]

\[a_{2} = \frac{5 + 3}{2} = 4.\]

\[3)\ Уравнения\ касательных:\]

\[y_{1} = 1 - 2x + 4x = 1 + 2x;\]

\[y_{2} = 16 - 8x + 4x = 16 - 4x.\]

\[= \int_{2,5}^{4}{(x - 4)^{2}\text{dx}} + \int_{1}^{2,5}{(x - 1)^{2}\text{dx}} =\]

\[= \left. \ \frac{(x - 4)^{3}}{3} \right|_{2,5}^{4} + \left. \ \frac{(x - 1)^{3}}{3} \right|_{1}^{2,5} =\]

\[= \frac{0^{3}}{3} - \frac{( - 1,5)^{3}}{3} + \frac{(1,5)^{3}}{3} + \frac{0^{3}}{3} =\]

\[= \frac{2}{3} \bullet \left( \frac{3}{2} \right)^{3} = \left( \frac{3}{2} \right)^{2} = \frac{9}{4} = 2,25.\]

\[Ответ:\ \ 2,25.\]