ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1515

\[\boxed{\mathbf{1515}\mathbf{.}}\]

\[\mathbf{Н}а\ отрезке\ \lbrack - 3;\ 6\rbrack;\]

\[f(x) = x^{2}(2x - 3) - 12(3x - 2) =\]

\[= 2x^{3} - 3x^{2} - 36x + 24;\]

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

\[= 2\left( x^{3} \right)^{'} - 3\left( x^{2} \right)^{'} - (36x - 24)^{'} =\]

\[= 2 \bullet 3x^{2} - 3 \bullet 2x - 36 =\]

\[= 6x^{2} - 6x - 36.\]

\[Стационарные\ точки:\]

\[6x^{2} - 6x - 36 = 0\]

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

\[D = 1 + 24 = 25\]

\[x_{1} = \frac{1 - 5}{2} = - 2;\]

\[x_{2} = \frac{1 + 5}{2} = 3.\]

\[f( - 3) = - 54 - 27 + 108 + 24 = 51;\]

\[f( - 2) = - 16 - 12 + 72 + 24 = 68;\]

\[f(3) = 54 - 27 - 108 + 24 = - 57;\]

\[f(6) = 432 - 108 - 216 + 24 = 132.\]

\[Ответ:\ \ \]

\[y_{\min} = - 57;\ \ y_{\max} = 132.\]