Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 261

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

\[f^{'}(x) = 2x - 4 = 0;\]

\[2x = 4\]

\[x = 2.\]

\[f(2) = 4 - 8 = - 4;\]

\[y = - 4.\]

\[Ответ:\ \ y = - 4.\]

\[2)\ f(x) = (x - 1)(x - 2);\]

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

\[2x - 3 = 0\]

\[2x = 3\]

\[x = \frac{3}{2}.\]

\[f\left( \frac{3}{2} \right) = \frac{1}{2} \bullet \left( - \frac{1}{2} \right) = - \frac{1}{4}.\]

\[Ответ:\ \ y = - \frac{1}{4}.\]

\[3)\ f(x) = x^{4} + 32x - 3;\]

\[f^{'}(x) = 4x^{3} + 32 = 0;\]

\[4x^{3} = - 32\]

\[x^{3} = - 8\]

\[x = - 2.\]

\[f(2) = 16 - 64 - 3 = - 51.\]

\[Ответ:\ \ y = - 51.\]

\[4)\ f(x) = x^{6} + 6x - 2;\]

\[f^{'}(x) = 6x^{5} + 6 = 0;\]

\[6x^{5} = - 6\]

\[x^{5} = - 1\]

\[x = - 1.\]

\[f( - 1) = 1 - 6 - 2 = - 7.\]

\[Ответ:\ - 7.\]