Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 926

\[\boxed{\mathbf{926}\mathbf{.}}\]

\[1)\ y = x^{3} - 3x^{2} + 4\]

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

\[\textbf{б)}\ y^{'}(x) =\]

\[= \left( x^{3} \right)^{'} - 3 \bullet \left( x^{2} \right)^{'} + (4)^{'};\]

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

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

\[\textbf{в)}\ 3x^{2} - 6x = 0\]

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

\[x_{1} = 0\ и\ x_{2} = 2.\]

\[\textbf{г)}\ f(2) = 2^{3} - 3 \bullet 2^{2} + 4 =\]

\[= 8 - 12 + 4 = 0;\]

\[f(0) = 0^{2} - 3 \bullet 0^{2} + 4 = 4.\]

\[\textbf{д)}\ Возрастает\ \]

\[на\ ( - \infty;\ 0) \cup (2;\ + \infty)\ и\ \]

\[убывает\ на\ (0;\ 2);\]

\[x = 2 - точка\ минимума;\ \ \]

\[x = 0 - точка\ максимума.\]

\[\textbf{е)}\]

\[2)\ y = 2 + 3x - x^{3}\]

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

\[\textbf{б)}\ y^{'}(x) =\]

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

\[\textbf{в)}\ 3 - 3x^{2} = 0\]

\[1 - x^{2} = 0\]

\[(1 - x)(1 + x) = 0\]

\[x_{1} = 1\ и\ x_{2} = - 1.\]

\[\textbf{г)}\ f( - 1) =\]

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

\[= 2 - 3 + 1 = 0;\]

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

\[= 2 + 3 - 1 = 4.\]

\[\textbf{д)}\ Возрастает\ на\ ( - 1;\ 1)\ и\ \]

\[убывает\ на\ \]

\[( - \infty;\ - 1) \cup (1;\ + \infty);\]

\[x = - 1 - точка\ минимума;\ \ \]

\[x = 1 - точка\ максимума.\]

\[\textbf{е)}\ \]

\[3)\ y = - x^{3} + 4x^{2} - 4x\]

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

\[\textbf{б)}\ y^{'}(x) =\]

\[= - \left( x^{3} \right)^{'} + 4 \bullet \left( x^{2} \right)^{'} - (4x)^{'};\]

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

\[= - 3x^{2} + 8x - 4.\]

\[\textbf{в)}\ Стационарные\ точки:\]

\[- 3x^{2} + 8x - 4 = 0\]

\[D = 8^{2} - 4 \bullet 3 \bullet 4 =\]

\[= 64 - 48 = 16\]

\[x_{1} = \frac{- 8 - 4}{2 \bullet ( - 3)} = \frac{12}{6} = 2\ \ и\ \]

\[x_{2} = \frac{- 8 + 4}{2 \bullet ( - 3)} = \frac{4}{6} = \frac{2}{3}.\]

\[- \left( x - \frac{2}{3} \right)(x - 2) = 0.\]

\[\textbf{г)}\ f\left( \frac{2}{3} \right) =\]

\[= - \left( \frac{2}{3} \right)^{3} + 4 \bullet \left( \frac{2}{3} \right)^{2} - 4 \bullet \frac{2}{3} =\]

\[= - \frac{8}{27} + \frac{16}{9} - \frac{8}{3} = - \frac{32}{27} =\]

\[= - 1\frac{5}{27};\]

\[f(2) = - 2^{3} + 4 \bullet 2^{2} - 4 \bullet 2 =\]

\[= - 8 + 16 - 8 = 0.\]

\[\textbf{д)}\ Возрастает\ на\ \left( \frac{2}{3};\ 2 \right)\ и\ \]

\[убывает\ на\ \left( - \infty;\ \frac{2}{3} \right) \cup (2;\ + \infty);\]

\[x = \frac{2}{3} - точка\ минимума;\ \ \]

\[x = 2 - точка\ максимума.\]

\[\textbf{е)}\ \]

\[4)\ y = x^{3} + 6x^{2} + 9x;\]

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

\[\textbf{б)}\ y^{'}(x) =\]

\[= \left( x^{3} \right)^{'} + 6 \bullet \left( x^{2} \right)^{'} + (9x)';\]

\[y^{'}(x) = 3x^{2} + 6 \bullet 2x + 9 =\]

\[= 3x^{2} + 12x + 9.\]

\[\textbf{в)}\ Стационарные\ точки:\]

\[3x^{2} + 12x + 9 = 0\]

\[x^{2} + 4x + 3 = 0\]

\[D = 4^{2} - 4 \bullet 3 = 16 - 12 = 4\]

\[x_{1} = \frac{- 4 - 2}{2} = - 3\ \ и\ \ \]

\[x_{2} = \frac{- 4 + 2}{2} = - 1;\]

\[(x + 3)(x + 1) = 0.\]

\[\textbf{г)}\ f( - 3) =\]

\[= ( - 3)^{3} + 6 \bullet ( - 3)^{2} + 9 \bullet ( - 3) =\]

\[= - 27 + 54 - 27 = 0;\]

\[f( - 1) =\]

\[= ( - 1)^{3} + 6 \bullet ( - 1)^{2} + 9 \bullet ( - 1) =\]

\[= - 1 + 6 - 9 = - 4.\]

\[\textbf{д)}\ Возрастает\ \]

\[на\ ( - \infty;\ - 3) \cup ( - 1;\ + \infty)\ и\ \]

\[убывает\ на\ ( - 3;\ - 1);\]

\[x = - 1 - точка\ минимума;\ \ \]

\[x = - 3 - точка\ максимума.\]

\[\textbf{е)}\ \]