ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 254

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

\[f^{'}(x) = 2^{x} \bullet \ln 2 - 2^{- x} \bullet \ln 2 > 0;\]

\[\ln 2 \bullet \left( 2^{x} - 2^{- x} \right) > 0\]

\[2^{x} > 2^{- x}\]

\[x > - x\]

\[2x > 0\]

\[x > 0.\]

\[Ответ:\ \ f^{'}(x) = 0\ при\ x = 0;\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }f^{'}(x) > 0\ при\ x > 0;\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }f^{'}(x) < 0\ при\ x < 0.\]

\[2)\ f(x) = 3^{2x} - 2x\ln 3;\]

\[f^{'}(x) = 2 \bullet 3^{2x} \bullet \ln 3 - 2\ln 3 > 0;\]

\[2\ln 3 \bullet \left( 3^{2x} - 1 \right) > 0\]

\[3^{2x} > 1\]

\[2x > 0\]

\[x > 0.\]

\[Ответ:\ \ f^{'}(x) = 0\ при\ x = 0;\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }f^{'}(x) > 0\ при\ x > 0;\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }f^{'}(x) < 0\ при\ x < 0.\]

\[3)\ f(x) = x + \ln{2x};\]

\[f^{'}(x) = 1 + 2 \bullet \frac{1}{2x} > 0;\]

\[\frac{x + 1}{x} > 0\]

\[x < - 1;\text{\ \ \ x} > 0.\]

\[Ответ:\ \ f^{'}(x) > 0\ при\ x > 0.\]

\[4)\ f(x) = x + \ln(2x + 1);\]

\[f^{'}(x) = 1 + 2 \bullet \frac{1}{2x + 1} > 0;\]

\[\frac{2x + 1 + 2}{2x + 1} > 0\]

\[\frac{2x + 3}{2x + 1} > 0\]

\[x < - \frac{3}{2};\ \ \ x > - \frac{1}{2}.\]

\[Ответ:\ \ f^{'}(x) > 0\ при\ x > - \frac{1}{2}.\]

\[5)\ f(x) = 6x - x\sqrt{x};\]

\[f^{'}(x) = 6 - \frac{3}{2}x^{\frac{1}{2}} > 0;\]

\[12 - 3\sqrt{x} > 0\]

\[3\sqrt{x} < 12\]

\[\sqrt{x} < 4\]

\[0 \leq x < 16.\]

\[Ответ:\ \ f^{'}(x) = 0\ при\ \ x = 16;\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }f^{'}(x) > 0\ при\ 0 < x < 16;\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }f^{'}(x) < 0\ при\ x > 16.\]

\[6)\ f(x) = (x + 1)\sqrt{x + 1} - 3x;\]

\[f^{'}(x) = \frac{3}{2}(x + 1)^{\frac{1}{2}} - 3 > 0;\]

\[3\sqrt{x + 1} - 6 > 0\]

\[3\sqrt{x + 1} > 6\]

\[\sqrt{x + 1} > 2\]

\[x + 1 > 4\]

\[x > 3.\]

\[Ответ:\ \ f^{'}(x) = 0\ при\ x = 3;\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }f^{'}(x) > 0\ при\ x > 3;\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }f^{'}(x) < 0\ при - 1 < x < 3.\]