Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 364

\[1)\ f(x) = 2x + 3;\text{\ \ \ M}(1;\ 2):\]

\[F(x) = 2 \bullet \frac{x^{2}}{2} + 3x + C =\]

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

\[2 = 1^{2} + 3 \bullet 1 + C\]

\[C = - 2.\]

\[Ответ:\ \ F(x) = x^{2} + 3x - 2.\]

\[2)\ f(x) = 4x - 1;\text{\ \ \ M}( - 1;\ 3):\]

\[F(x) = 4 \bullet \frac{x^{2}}{2} - x + C =\]

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

\[3 = 2 \bullet ( - 1)^{2} + 1 + C\]

\[C = 0.\]

\[Ответ:\ \ F(x) = 2x^{2} - x.\]

\[3)\ f(x) = \sqrt{x + 2};\text{\ M}(2;\ - 3):\]

\[F(x) = (x + 2)^{\frac{3}{2}}\ :\frac{3}{2} + C =\]

\[= \frac{2}{3}(x + 2)^{\frac{3}{2}} + C;\]

\[- 3 = \frac{2}{3}(2 + 2)^{\frac{3}{2}} + C\]

\[- 3 = \frac{2}{3} \bullet 2^{3} + C\]

\[- 3 = \frac{16}{3} + C\]

\[C = - \frac{25}{3}.\]

\[Ответ:\ \ F(x) = \frac{2}{3}(x + 2)^{\frac{3}{2}} - \frac{25}{3}.\]

\[4)\ f(x) = \frac{1}{\sqrt{x + 3}};\ M( - 2;\ - 1):\]

\[F(x) = (x + 3)^{\frac{1}{2}}\ :\frac{1}{2} + C =\]

\[= 2\sqrt{x + 3} + C;\]

\[- 1 = 2\sqrt{- 2 + 3} + C\]

\[- 1 = 2 + C\]

\[C = - 3.\]

\[Ответ:\ \ F(x) = 2\sqrt{x + 3} - 3.\]

\[5)\ f(x) = \sin{2x};\ \text{\ M}\left( \frac{\pi}{2};\ 5 \right):\]

\[F(x) = - \frac{1}{2}\cos{2x} + C;\]

\[5 = - \frac{1}{2}\cos\pi + C\]

\[5 = \frac{1}{2} + C\]

\[C = \frac{9}{2}.\]

\[Ответ:\ \ F(x) = - \frac{1}{2}\cos{2x} + \frac{9}{2}.\]

\[6)\ f(x) = \cos{3x};\text{\ \ \ M}(0;\ 0):\]

\[F(x) = \frac{1}{3}\sin{3x} + C;\]

\[0 = \frac{1}{3}\sin 0 + C\]

\[C = 0.\]

\[Ответ:\ \ F(x) = \frac{1}{3}\sin{3x}.\]

\[7)\ f(x) = \frac{1}{x + 3};\text{\ M}( - 2;\ 4):\]

\[F(x) = \ln|x + 3| + C\]

\[4 = \ln( - 2 + 3) + C\]

\[C = 4.\]

\[Ответ:\ \ F(x) = \ln|x + 3| + 4.\]

\[8)\ f(x) = \frac{1}{(x + 1)^{2}};\text{\ M}( - 2;\ 2):\]

\[F(x) = \frac{(x + 1)^{- 1}}{- 1} + C =\]

\[= - \frac{1}{x + 1} + C;\]

\[2 = - \frac{1}{- 2 + 1} + C\]

\[2 = 1 + C\]

\[C = 1.\]

\[Ответ:\ \ F(x) = - \frac{1}{x + 1} + 1.\]