ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 307

\[1)\ f(x) = 12x^{3} - 24x^{2} + x + 5;\]

\[f^{'}(x) = 12 \bullet 3x^{2} - 24 \bullet 2x + 1 + 0 =\]

\[= 36x^{2} - 48x + 1;\]

\[f^{''}(x) = 36 \bullet 2x - 48 + 0 =\]

\[= 72x - 48.\]

\[72x - 48 = 0\]

\[72x = 48\]

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

\[Ответ:\ \ \frac{2}{3}.\]

\[2)\ f(x) = x^{4} - 12x^{3} + 48x^{2} + 3;\]

\[f^{'}(x) = 4x^{3} - 12 \bullet 3x^{2} + 48 \bullet 2x + 0 =\]

\[= 4x^{3} - 36x^{2} + 96x;\]

\[f^{''}(x) = 4 \bullet 3x^{2} - 36 \bullet 2x + 96 =\]

\[= 12x^{2} - 72x + 96.\]

\[12x^{2} - 72x + 96 = 0\]

\[x^{2} - 6x + 8 = 0\]

\[D = 36 - 32 = 4\]

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

\[x_{2} = \frac{6 + 2}{2} = 4;\]

\[Ответ:\ \ 2;\ 4.\]

\[3)\ f(x) = x^{3} \bullet e^{- 4x};\]

\[f^{'}(x) = 3x^{2} \bullet e^{- 4x} - x^{3} \bullet 4e^{- 4x} =\]

\[= e^{- 4x} \bullet \left( 3x^{2} - 4x^{3} \right);\]

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

\[= e^{- 4x} \bullet \left( 16x^{3} - 12x^{2} + 6x - 12x^{2} \right) =\]

\[= e^{- 4x} \bullet \left( 16x^{3} - 24x^{2} + 6x \right).\]

\[16x^{3} - 24x^{2} + 6x = 0\]

\[2x\left( 8x^{2} - 12x + 3 \right) = 0\]

\[D = 144 - 96 = 48\]

\[x = \frac{12 \pm \sqrt{48}}{2 \bullet 8} = \frac{12 \pm 4\sqrt{3}}{16} =\]

\[= \frac{3 \pm \sqrt{3}}{4}.\]

\[Ответ:\ \ 0;\ \frac{3 - \sqrt{3}}{4};\ \frac{3 + \sqrt{3}}{4}.\]

\[4)\ f(x) = x^{2} \bullet \ln x;\]

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

\[= 2x \bullet \ln x + x;\]

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

\[= 2\ln x + 3.\]

\[2\ln x + 3 = 0\]

\[2\ln x = - 3\]

\[\ln x = - 1,5\]

\[x = e^{- 1,5}.\]

\[Ответ:\ \ e^{- 1,5}.\]