Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 190

\[1)\ f(x) = x^{4} + 8x^{3} + x - 3;\]

\[f^{'}(x) = 4x^{3} + 8 \bullet 3x^{2} + 1 = 1;\]

\[4x^{3} + 24x^{2} = 0\]

\[4x^{2}(x + 6) = 0\]

\[x_{1} = - 6;\ \ x_{2} = 0.\]

\[Ответ:\ - 6;\ 0.\]

\[2)\ f(x) = 2x^{5} + 5x^{2} + x + 4;\]

\[f^{'}(x) = 2 \bullet 5x^{4} + 5 \bullet 2x + 1 = 1;\]

\[10x^{4} + 10x = 0\]

\[10x\left( x^{3} + 1 \right) = 0\]

\[x_{1} = - 1;\ \ \ x_{2} = 0.\]

\[Ответ:\ - 1;\ 0.\]

\[3)\ f(x) = \frac{x^{3} + x^{2} + 16}{x} =\]

\[= x^{2} + x + 16x^{- 1};\]

\[f^{'}(x) = 2x + 1 + 16x^{- 2} = 1;\]

\[2x = 16x^{- 2}\]

\[x^{3} = 8\]

\[x = 2.\]

\[Ответ:\ \ 2.\]

\[4)\ f(x) = \frac{x\sqrt[3]{x} + 3x + 18}{\sqrt[3]{x}} =\]

\[= x + 3x^{\frac{2}{3}} + 18x^{- \frac{1}{3}};\]

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

\[= 1 + 3 \bullet \frac{2}{3}x^{- \frac{1}{3}} + 18 \bullet \left( - \frac{1}{3} \right)x^{- \frac{4}{3}} = 1\]

\[2x^{- \frac{1}{3}} - 6x^{- \frac{4}{3}} = 0\]

\[2x - 6 = 0\]

\[2x = 6\]

\[x = 3.\]

\[Ответ:\ \ 3.\]