Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 799

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

\[f^{'}(x) = 2 \bullet 2 \bullet (2x - 1)^{2 - 1} =\]

\[= 4(2x - 1)\]

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

\[(2x - 1)^{2} - 4(2x - 1) = 0\]

\[(2x - 1)(2x - 1 - 4) = 0\]

\[(2x - 1)(2x - 5) = 0\]

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

\[x_{1} = \frac{1}{2}\]

\[2x_{2} - 5 = 0\]

\[x_{2} = \frac{5}{2}.\]

\[Ответ:\ \ \frac{1}{2}\ \frac{5}{2}.\]

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

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

\[= 9(3x + 2)^{2}\]

\[(3x + 2)^{3} = 9(3x + 2)^{2}\]

\[(3x + 2)^{3} - 9(3x + 2)^{2} = 0\]

\[(3x + 2)^{2} \bullet (3x + 2 - 9) = 0\]

\[(3x + 2)^{2} \bullet (3x - 7) = 0\]

\[3x_{1} + 2 = 0\]

\[x_{1} = - \frac{2}{3}\]

\[3x_{2} - 7 = 0\]

\[x_{2} = \frac{7}{3}.\]

\[Ответ:\ - \frac{2}{3}\ \frac{7}{3}.\]