ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 232

\[y = kx.\]

\[1)\ y = x^{2} - 3x + 4,\ \ \ k = 1:\]

\[y^{'}(x) = 2x - 3 = 1\]

\[2x = 4\]

\[x = 2;\]

\[y = 1 - 3 + 4 = 2.\]

\[Ответ:\ \ (2;\ 2).\]

\[2)\ f(x) = x(x + 1),\ \ \ k = 3:\]

\[y^{'}(x) = (x + 1) + x = 3\]

\[2x = 2\]

\[x = 1;\]

\[y = 1 \bullet 2 = 2.\]

\[Ответ:\ \ (1;\ 2).\]

\[3)\ f(x) = \frac{1}{3}x^{3} + x^{2} - 2x;k = 1:\]

\[f^{'}(x) = \frac{1}{3} \bullet 3x^{2} + 2x - 2 = 1;\]

\[x^{2} + 2x - 3 = 0\]

\[D = 4 + 12 = 16\]

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

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

\[y_{1} = - 9 + 9 + 6 = 6;\ \]

\[y_{2} = \frac{1}{3} + 1 - 2 = - \frac{2}{3}.\]

\[Ответ:\ \ ( - 3;\ 6);\ \left( 1;\ - \frac{2}{3} \right).\]

\[4)\ f(x) = e^{x} + e^{- x},\ \ \ k = \frac{3}{2}:\]

\[f^{'}(x) = e^{x} - e^{- x} = \frac{3}{2};\]

\[2e^{x} - 2e^{- x} = 3\]

\[2e^{2x} - 3e^{x} - 2 = 0\]

\[D = 9 + 16 = 25\]

\[e_{1}^{x} = \frac{3 - 5}{2 \bullet 2} = - \frac{1}{2};\]

\[e_{2}^{x} = \frac{3 + 5}{2 \bullet 2} = 2;\]

\[x_{1} \in \varnothing\ \ и\ \ x_{2} = \ln 2.\]

\[y = e^{\ln 2} + e^{- \ln 2} = 2 + \frac{1}{2} = \frac{5}{2}.\]

\[Ответ:\ \ \left( \ln 2;\ \frac{5}{2} \right).\]

\[5)\ f(x) = \sqrt{3x + 1},\ \ \ k = \frac{3}{4}:\]

\[f^{'}(x) = \frac{3}{2\sqrt{3x + 1}} = \frac{3}{4};\]

\[\sqrt{3x + 1} = 2\]

\[3x + 1 = 4\]

\[3x = 3\]

\[x = 1;\]

\[y = \sqrt{4} = 2.\]

\[Ответ:\ \ (1;\ 2).\]

\[6)\ f(x) = x + \sin x,\ \ \ k = 0:\]

\[f^{'}(x) = 1 + \cos x = 0;\]

\[\cos x = - 1\]

\[x = \pi + 2\pi n;\]

\[y = \pi + 2\pi n.\]

\[Ответ:\ \ (\pi + 2\pi n;\ \pi + 2\pi n).\]