Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 937

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

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

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

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

\[Точки\ экстремума:\]

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

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

\[D = 1^{2} + 4 \bullet 6 = 1 + 24 = 25\]

\[x_{1} = \frac{- 1 - 5}{2} = - 3;\text{\ \ }\]

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

\[1)\ \lbrack - 4;\ 3\rbrack:\]

\[y(2) = 2 \bullet 2^{3} + 3 \bullet 2^{2} - 36 \bullet 2 =\]

\[= 16 + 12 - 72 = - 44;\]

\[y(3) = 3 \bullet 3^{3} + 3 \bullet 3^{2} - 36 \bullet 3 =\]

\[= 81 + 27 - 108 = 0.\]

\[\ y_{\min} = - 44;\ \ y_{\max} = 81.\]

\[2)\ \lbrack - 2;\ 1\rbrack:\]

\[y(1) = 2 \bullet 1^{3} + 3 \bullet 1^{2} - 36 \bullet 1 =\]

\[= 2 + 3 - 36 = - 31.\]

\[y_{\min} = - 31;\ \ y_{\max} = 68.\]