Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 10

\[\boxed{\mathbf{10}\mathbf{.}}\]

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

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

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

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

\[3x(x - 2) = 0\]

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

\[f( - 1) = - 4;\]

\[f(0) = 0;\]

\[f(2) = - 4;\]

\[f(3) = 0.\]

\[f(x) = - 4 \rightarrow точка\ минимума;\]

\[f(x) = 0 \rightarrow точка\ максимума.\]

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

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

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

\[3\left( x^{2} + 1 \right) = 0\]

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

\[нет\ корней.\]

\[f( - 1) = - 1 - 3 = - 4;\]

\[f(2) = 8 + 6 = 14.\]

\[f(x) = - 4 \rightarrow точка\ минимума;\]

\[f(x) = 14 \rightarrow точка\ максимума.\]

\[\textbf{в)}\ y = 2x^{3} - 6x^{2} + 9;\ \ \lbrack - 2;2\rbrack\]

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

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

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

\[6x \cdot (x - 2) = 0\]

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

\[f( - 2) = - 16 - 24 + 9 = - 31;\]

\[f(0) = 9;\]

\[f(2) = 16 - 24 + 9 = 1.\]

\[f(x) = - 31 \rightarrow точка\ минимума;\]

\[f(x) = 9 \rightarrow точка\ максимума.\]

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

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

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

\[3\left( x^{2} - 1 \right) = 0\]

\[x^{2} = 1\]

\[x = \pm 1.\]

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

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

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

\[f(3) = 27 - 9 = 18.\]

\[f(x) = - 2 \rightarrow точка\ минимума;\]

\[f(x) = 18 \rightarrow точка\ максимума.\]