Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 866

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

\[f^{'}(x) = \left( e^{x} \right)^{'} + \left( e^{- x} \right) = e^{x} - e^{- x}\]

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

\[e^{x} - \frac{1}{e^{x}} = \frac{3}{2}\]

\[e^{2x} - 1 = \frac{3}{2}e^{x}\]

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

\[y = e^{x}:\]

\[2y^{2} - 3y - 2 = 0\]

\[D = 9 + 16 = 25\]

\[y_{1} = \frac{3 - 5}{2 \bullet 2} = - \frac{1}{2}\]

\[y_{2} = \frac{3 + 5}{2 \bullet 2} = 2.\]

\[1)\ e^{x} = - \frac{1}{2} < 0\]

\[корней\ нет.\]

\[2)\ e^{x} = 2\]

\[e^{x} = e^{\ln 2}\]

\[x = \ln 2.\]

\[3)\ f\left( \ln 2 \right) = e^{\ln 2} + e^{- \ln 2} =\]

\[= e^{\ln 2} + e^{\ln{0,5}} = 2 + 0,5 = 2,5.\]

\[Ответ:\ \ \left( \ln 2\ 2,5 \right).\]

\[2)\ f(x) = \sqrt{3x + 1}\text{\ \ }k = \frac{3}{4}:\]

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

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

\[k = \frac{3}{2\sqrt{3x + 1}} = \frac{3}{4}\]

\[2\sqrt{3x + 1} = 4\]

\[4(3x + 1) = 16\]

\[12x + 4 = 16\]

\[12x = 12\ \]

\[x = 1.\]

\[f(1) = \sqrt{3 + 1} = \sqrt{4} = 2.\]

\[Ответ:\ \ (1\ 2).\]

\[3)\ f(x) = \sin{2x}\text{\ \ }k = 2:\]

\[f^{'}(x) = \left( \sin{2x} \right)^{'} = 2\cos{2x}\]

\[k = 2\cos{2x} = 2\]

\[\cos{2x} = 1\]

\[2x = \arccos 1 + 2\pi n = 2\pi n\]

\[x = \frac{1}{2} \bullet (2\pi n) = \pi n.\]

\[f\left( \text{πn} \right) = \sin{2\pi n} = \sin 0 = 0.\]

\[Ответ:\ \ (\pi n\ 0).\]

\[4)\ f(x) = x + \sin x\text{\ \ }k = 0:\]

\[f^{'}(x) = (x)^{'} + \left( \sin x \right)^{'} =\]

\[= 1 + \cos x\]

\[k = 1 + \cos x = 0\]

\[\cos\text{\ x} = - 1\]

\[x = \pi - \arccos 1 + 2\pi n =\]

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

\[f(\pi + \pi n) =\]

\[= \pi + 2\pi n + \sin(\pi + 2\pi n) =\]

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

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