Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 928

\[\boxed{\mathbf{928}\mathbf{.}}\]

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

\[\textbf{а)}\ D(x) = ( - \infty;\ + \infty);\]

\[\textbf{б)}\ y^{'}(x) =\]

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

\[y^{'}(x) = 3x^{2} - 3 \bullet 2x + 0 =\]

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

\[\textbf{в)}\ Стационарные\ точки:\]

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

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

\[x_{1} = 0\ и\ x_{2} = 2.\]

\[\textbf{г)}\ f( - 1) =\]

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

\[= - 1 - 3 + 2 = - 2;\]

\[f(0) = 0^{3} - 3 \bullet 0^{2} + 2 = 2;\]

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

\[= 8 - 12 + 2 = - 2;\]

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

\[= 3^{3} - 3^{3} + 2 = 2.\]

\[\textbf{д)}\ Возрастает\ на\ \]

\[( - 1;\ 0) \cup (2;\ 3)\ и\ убывает\ \]

\[на\ (0;\ 2);\]

\[x = 2 - точка\ минимума;\ \ \]

\[x = 0 - точка\ максимума.\]

\[\textbf{е)}\ \]

\[2)\ y = x^{4} - 10x^{2} + 9;\ \lbrack - 3;\ 3\rbrack\]

\[\textbf{а)}\ D(x) = ( - \infty;\ + \infty);\]

\[\textbf{б)}\ y^{'}(x) =\]

\[= \left( x^{4} \right)^{'} - 10 \bullet \left( x^{2} \right)^{'} + (9)^{'};\]

\[y^{'}(x) = 4x^{3} - 10 \bullet 2x + 0 =\]

\[= 4x^{3} - 20x.\]

\[\textbf{в)}\ Стационарные\ точки:\]

\[4x^{2} - 20x = 0\]

\[4x \bullet \left( x^{2} - 5 \right) = 0\]

\[\left( x + \sqrt{5} \right) \bullet 4x \bullet \left( x - \sqrt{5} \right) = 0\]

\[x_{1} = - \sqrt{5},\ \ \ x_{2} = 0,\ \ \ x_{3} = \sqrt{5}.\]

\[\textbf{г)}\ f( \pm 3) =\]

\[= ( \pm 3)^{4} - 10 \bullet ( \pm 3)^{2} + 9 =\]

\[= 81 - 10 \bullet 9 + 9 = 0;\]

\[f\left( \pm \sqrt{5} \right) =\]

\[= \left( \pm \sqrt{5} \right)^{4} - 10 \bullet \left( \pm \sqrt{5} \right)^{2} + 9 =\]

\[= 25 - 10 \bullet 5 + 9 = - 16;\]

\[f(0) = 0^{4} - 10 \bullet 0^{2} + 9 = 9.\]

\[\textbf{д)}\ Возрастает\ на\ \]

\[\left( - \sqrt{5};\ 0 \right) \cup \left( \sqrt{5};\ 3 \right)\ и\ убывает\ \]

\[на\ \left( - 3;\ - \sqrt{5} \right) \cup \left( 0;\ \sqrt{5} \right);\]

\[x = \pm \sqrt{5} - точки\ минимума;\ \ \]

\[x = 0 - точка\ максимума.\]

\[\textbf{е)}\ \]