Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 11

\[\boxed{\mathbf{11}\mathbf{.}}\]

\[\textbf{а)}\ y = 2x^{3} - x^{2};\ \ \lbrack - 1;1\rbrack\]

\[f^{'}(x) = 2 \cdot 3x^{2} - 2x = 6x^{2} - 2x;\]

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

\[6x\left( x - \frac{1}{3} \right) = 0\]

\[x = 0;\ \ x = \frac{1}{3}.\]

\[f( - 1) = - 2 - 1 = - 3;\]

\[f(0) = 0;\]

\[f\left( \frac{1}{3} \right) = 2 \cdot \frac{1}{27} - \frac{1}{9} = \frac{2}{27} - \frac{3}{27} =\]

\[= - \frac{1}{27};\]

\[f(1) = 2 - 1 = 1.\]

\[\max{f(x)} = 1;\]

\[\min{f(x)} = - 3.\]

\[\textbf{б)}\ y = 2x^{3} + x;\ \ \lbrack - 1;1\rbrack\]

\[f^{'}(x) = 2 \cdot 3x^{2} + 1 = 6x^{2} + 1;\]

\[6x^{2} = - 1\]

\[нет\ корней.\]

\[f( - 1) = - 2 - 1 = - 3;\]

\[f(1) = 2 + 1 = 3.\]

\[\max{f(x)} = 3;\]

\[\min{f(x)} = - 3.\]

\[\textbf{в)}\ y = 2x^{3} + 6x^{2} + 8;\ \ \lbrack - 3;2\rbrack\]

\[f^{'}(x) = 2 \cdot 3x^{2} + 6 \cdot 2x + 0 =\]

\[= 6x^{2} + 12x;\]

\[6x^{2} + 12x = 0\]

\[6x(x + 2) = 0\]

\[x = 0;\ \ x = - 2.\]

\[f( - 3) = 2 \cdot ( - 27) + 6 \cdot 9 + 8 =\]

\[= - 54 + 54 + 8 = 8;\]

\[f( - 2) = 2 \cdot ( - 8) + 6 \cdot 4 + 8 =\]

\[= - 16 + 24 + 8 = 16;\]

\[f(0) = 8;\]

\[f(2) = 2 \cdot 8 + 6 \cdot 4 + 8 =\]

\[= 16 + 24 + 8 = 48.\]

\[\max{f(x)} = 48;\]

\[\min{f(x)} = 8.\]

\[\textbf{г)}\ y = x^{3} + 6x;\ \ \lbrack - 2;1\rbrack\]

\[f^{'}(x) = 3x^{2} + 6;\]

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

\[3x^{2} = - 6\]

\[нет\ корней.\]

\[f( - 2) = - 8 - 12 = - 20;\]

\[f(1) = 1 + 6 = 7.\]

\[\max{f(x)} = 7;\]

\[\min{f(x)} = - 20.\]