Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1516

\[\boxed{\mathbf{1516}\mathbf{.}}\]

\[\mathbf{Н}а\ отрезке\ \left\lbrack e^{\frac{3}{4}};\ e^{3} \right\rbrack;\]

\[f(x) = 2\ln^{3}x - 9\ln^{2}x + 12\ln x;\]

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

\[= 2\left( \ln^{3}x \right)^{'} - 9\left( \ln^{2}x \right)^{'} + \left( 12\ln x \right) =\]

\[= 2 \bullet \frac{1}{x} \bullet 3\ln^{2}x - 9 \bullet \frac{1}{x} \bullet 2\ln x + 12 \bullet \frac{1}{x} =\]

\[= \frac{6}{x} \bullet \left( \ln^{2}x - 3\ln x + 2 \right).\]

\[Стационарные\ точки:\]

\[\ln^{2}x - 3\ln x + 2 = 0\]

\[D = 9 - 8 = 1\]

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

\[\ln x_{2} = \frac{3 + 1}{2} = 2;\]

\[x_{1} = e^{1}\text{\ \ }и\ \ x_{2} = e^{2}.\]

\[f\left( e^{\frac{3}{4}} \right) =\]

\[= 2 \bullet \left( \frac{3}{4} \right)^{3} - 9 \bullet \left( \frac{3}{4} \right)^{2} + 12 \bullet \frac{3}{4} =\]

\[= \frac{54 - 324 + 576}{64} = \frac{306}{64} = 4\frac{25}{32};\]

\[f\left( e^{1} \right) = 2 \bullet 1^{3} - 9 \bullet 1^{2} + 12 \bullet 1 =\]

\[= 2 - 9 + 12 = 5;\]

\[f\left( e^{2} \right) = 2 \bullet 2^{3} - 9 \bullet 2^{2} + 12 \bullet 2 =\]

\[= 16 - 36 + 24 = 4;\]

\[f\left( e^{3} \right) = 2 \bullet 3^{3} - 9 \bullet 3^{2} + 12 \bullet 3 =\]

\[= 54 - 81 + 36 = 9.\]

\[Ответ:\ \ y_{\min} = 4;\ \ y_{\max} = 9.\]