ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 930

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Год:2020-2021-2022-2023
Тип:учебник
Серия:Базовый и углубленный уровни

Задание 930

\[\boxed{\mathbf{930}\mathbf{.}}\]

\[1)\ y = 2 + 5x^{3} - 3x^{5}\]

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

\[\textbf{б)}\ y^{'}(x) =\]

\[= (2)^{'} + 5 \bullet \left( x^{3} \right)^{'} - 3 \bullet \left( x^{5} \right)^{'};\]

\[y^{'}(x) = 0 + 5 \bullet 3x^{2} - 3 \bullet 5x^{4} =\]

\[= 15x^{2} - 15x^{4}.\]

\[\textbf{в)}\ Стационарные\ точки:\]

\[15x^{2} - 15x^{4} = 0\]

\[15x^{2} \bullet \left( 1 - x^{2} \right) = 0\]

\[15x^{2} \bullet (1 + x) \bullet (1 - x) = 0\]

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

\[\textbf{г)}\ f( - 1) =\]

\[= 2 + 5 \bullet ( - 1)^{3} - 3 \bullet ( - 1)^{5} =\]

\[= 2 - 5 + 3 = 0;\]

\[f(0) = 2 + 5 \bullet 0^{3} - 3 \bullet 0^{5} = 2;\]

\[f(1) = 2 + 5 \bullet 1^{3} - 3 \bullet 1^{5} =\]

\[= 2 + 5 - 3 = 4.\]

\[\textbf{д)}\ Возрастает\ \]

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

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

\[x = - 1 - точка\ минимума;\ \ \]

\[x = 1 - точка\ максимума.\]

\[\textbf{е)}\ \]

\[x\] \[x < - 1\] \[- 1\] \[- 1 < x < 0\] \[0\] \[0 < x < 1\] \[1\] \[x > 1\]
\[f^{'}(x)\] \[-\] \[0\] \[+\] \[0\] \[+\] \[0\] \[-\]
\[f(x)\] \[\searrow\] \[0\] \[\nearrow\] \[2\] \[\nearrow\] \[4\] \[\searrow\]

\[2)\ y = 3x^{5} - 5x^{3}\]

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

\[\textbf{б)}\ y^{'}(x) = 3 \bullet \left( x^{5} \right)^{'} - 5 \bullet \left( x^{3} \right)^{'};\]

\[y^{'}(x) = 3 \bullet 5x^{4} - 5 \bullet 3x^{2} =\]

\[= 15x^{4} - 15x^{2}.\]

\[\textbf{в)}\ Стационарные\ точки:\]

\[15x^{4} - 15x^{2} = 0\]

\[15x^{2} \bullet \left( x^{2} - 1 \right) = 0\]

\[15x^{2} \bullet (x + 1) \bullet (x - 1) = 0\]

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

\[\textbf{г)}\ f( - 1) =\]

\[= 3 \bullet ( - 1)^{5} - 5 \bullet ( - 1)^{3} =\]

\[= - 3 + 5 = 2;\]

\[f(0) = 3 \bullet 0^{5} - 5 \bullet 0^{3} = 0;\]

\[f(1) = 3 \bullet 1^{5} - 5 \bullet 1^{3} = 3 - 5 =\]

\[= - 2.\]

\[\textbf{д)}\ Возрастает\ \]

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

\[убывает\ на\ ( - 1;\ 0) \cup (0;\ 1);\]

\[x = 1 - точка\ минимума;\ \ \]

\[x = - 1 - точка\ максимума.\]

\[\textbf{е)}\ \]

\[x\] \[x < - 1\] \[- 1\] \[- 1 < x < 0\] \[0\] \[0 < x < 1\] \[1\] \[x > 1\]
\[f^{'}(x)\] \[+\] \[0\] \[-\] \[0\] \[-\] \[0\] \[+\]
\[f(x)\] \[\nearrow\] \[2\] \[\searrow\] \[0\] \[\searrow\] \[- 2\] \[\nearrow\]

\[3)\ y = 4x^{5} - 5x^{4}\]

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

\[\textbf{б)}\ y^{'}(x) = 4 \bullet \left( x^{5} \right)^{'} - 5 \bullet \left( x^{4} \right)^{'};\]

\[y^{'}(x) = 4 \bullet 5x^{4} - 5 \bullet 4x^{3} =\]

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

\[\textbf{в)}\ Стационарные\ точки:\]

\[20x^{4} - 20x^{3} = 0\]

\[20x^{3} \bullet (x - 1) = 0\]

\[x_{1} = 0\ и\ x_{2} = 1.\]

\[\textbf{г)}\ f(0) = 4 \bullet 0^{5} - 5 \bullet 0^{4} = 0;\]

\[f(1) = 4 \bullet 1^{5} - 5 \bullet 1^{4} =\]

\[= 4 - 5 = - 1.\]

\[\textbf{д)}\ Возрастает\ \]

\[на\ ( - \infty;\ 0) \cup (1;\ + \infty)\ и\ \]

\[убывает\ на\ (0;\ 1);\]

\[x = 1 - точка\ минимума;\ \ \]

\[x = 0 - точка\ максимума.\]

\[\textbf{е)}\ \]

\[x\] \[x < 0\] \[0\] \[0 < x < 1\] \[1\] \[1 < x < + \infty\]
\[f^{'}(x)\] \[+\] \[0\] \[-\] \[0\] \[+\]
\[f(x)\] \[\nearrow\] \[0\] \[\searrow\] \[- 1\] \[\nearrow\]

\[4)\ y = \frac{1}{10}x^{5} - \frac{5}{6}x^{3} + 2x\]

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

\[\textbf{б)}\ y^{'}(x) =\]

\[= \frac{1}{10} \bullet \left( x^{5} \right)^{'} - \frac{5}{6} \bullet \left( x^{3} \right)^{'} + (2x)^{'};\]

\[y^{'}(x) = \frac{1}{10} \bullet 5x^{4} - \frac{5}{6} \bullet 3x^{2} + 2 =\]

\[= 0,5x^{4} - 2,5x^{2} + 2.\]

\[\textbf{в)}\ Стационарные\ точки:\]

\[0,5x^{4} - 2,5x^{2} + 2 = 0\]

\[x^{4} - 5x^{2} + 4 = 0\]

\[D = 5^{2} - 4 \bullet 4 = 25 - 16 = 9\]

\[x_{1}^{2} = \frac{5 - 3}{2} = 1\ \ и\ \ \]

\[x_{2}^{2} = \frac{5 + 3}{2} = 4.\]

\[\left( x^{2} - 1 \right)\left( x^{2} - 4 \right) = 0\]

\[(x + 2)(x + 1)(x - 1)(x - 2) =\]

\[= 0\]

\[x_{1} = - 2,\ \ \ x_{2} = - 1,\ \ \ x_{3} = 1,\ \ \]

\[x_{4} = 2.\]

\[= - \frac{32}{10} + \frac{40}{6} - 4 = - \frac{8}{15};\]

\[f( - 1) =\]

\[= - \frac{1}{10} + \frac{5}{6} - 2 = - \frac{19}{15} = - 1\frac{4}{15};\]

\[f(1) = \frac{1}{10} \bullet 1^{5} - \frac{5}{6} \bullet 1^{3} + 2 \bullet 1 =\]

\[= \frac{1}{10} - \frac{5}{6} + 2 = \frac{19}{15} = 1\frac{4}{15};\]

\[f(2) = \frac{1}{10} \bullet 2^{5} - \frac{5}{6} \bullet 2^{3} + 2 \bullet 2 =\]

\[= \frac{32}{10} - \frac{40}{6} + 4 = \frac{8}{15}.\]

\[\textbf{д)}\ Возрастает\ \]

\[на\ ( - \infty;\ - 2) \cup ( - 1;\ 1) \cup (2;\ + \infty);\]

\[убывает\ на\ ( - 2;\ - 1) \cup (1;\ 2);\]

\[x = - 1\ и\ x = 2 - точки\ \]

\[минимума;\]

\[x = - 2\ и\ x = 1 - точки\ \]

\[максимума.\]

\[\textbf{е)}\ \]

\[x\] \[x < - 2\] \[- 2\] \[- 2 < x < - 1\] \[- 1\] \[- 1 < x < 1\] \[1\] \[1 < x < 2\] \[2\] \[x > 2\]
\[f^{'}(x)\] \[+\] \[0\] \[-\] \[0\] \[+\] \[0\] \[-\] \[0\] \[+\]
\[f(x)\] \[\nearrow\] \[- \frac{8}{15}\] \[\searrow\] \[- 1\frac{4}{15}\] \[\nearrow\] \[1\frac{4}{15}\] \[\searrow\] \[\frac{8}{15}\] \[\nearrow\]

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