Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 242

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

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

\[9x^{2} - 4x < 0\]

\[x(9x - 4) < 0\]

\[0 < x < \frac{4}{9}.\]

\[f^{'}(x) = 0\ при\ x = 0;\ x = \frac{4}{9};\]

\[f^{'}(x) > 0\ при\ 0 < x < \frac{4}{9};\]

\[f^{'}(x) < 0\ при\ x < 0;\ x > \frac{4}{9}.\]

\[2)\ f(x) = (x + 3)^{3}(x - 4)^{2};\]

\[f^{'}(x) =\]

\[(x + 3)^{2}(x - 4) \bullet (5x - 6) > 0\]

\[x < \frac{6}{5};\ \ \ x > 4;\text{\ \ \ x} \neq - 3.\]

\[f^{'}(x) = 0\ при\ \]

\[x = - 3;\ x = \frac{6}{5};\ x = 4;\]

\[f^{'}(x) > 0\ при\ \]

\[x < - 3;\ - 3 < x < \frac{6}{5};\ x > 4;\]

\[f^{'}(x) < 0\ при\ \frac{6}{5} < x < 4.\]

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

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

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

\[- \frac{7}{(x - 2)^{2}} > 0\]

\[x \in \varnothing;\text{\ \ \ x} \neq 2.\]

\[f^{'}(x) < 0\ при\ x \neq 2.\]

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

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

\[x^{3} - 1 > 0\]

\[x^{3} > 1\]

\[x > 1.\]

\[f^{'}(x) = 0\ при\ x = 1;\]

\[f^{'}(x) > 0\ при\ x > 1;\]

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