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МатематикаРешебник по алгебре и начала математического анализа 11 класс Алимов Задание 930
Задание 930
\[\boxed{\mathbf{930}\mathbf{.}}\]
\[1)\ y = 2 + 5x^{3} - 3x^{5}\]
\[\textbf{а)}\ D(x) = ( - \infty;\ + \infty);\]
\[\textbf{б)}\ y^{'}(x) =\]
\[= (2)^{'} + 5 \bullet \left( x^{3} \right)^{'} - 3 \bullet \left( x^{5} \right)^{'};\]
\[y^{'}(x) = 0 + 5 \bullet 3x^{2} - 3 \bullet 5x^{4} =\]
\[= 15x^{2} - 15x^{4}.\]
\[\textbf{в)}\ Стационарные\ точки:\]
\[15x^{2} - 15x^{4} = 0\]
\[15x^{2} \bullet \left( 1 - x^{2} \right) = 0\]
\[15x^{2} \bullet (1 + x) \bullet (1 - x) = 0\]
\[x_{1} = 0,\ \ \ x_{2} = - 1,\ \ \ x_{3} = 1.\]
\[\textbf{г)}\ f( - 1) =\]
\[= 2 + 5 \bullet ( - 1)^{3} - 3 \bullet ( - 1)^{5} =\]
\[= 2 - 5 + 3 = 0;\]
\[f(0) = 2 + 5 \bullet 0^{3} - 3 \bullet 0^{5} = 2;\]
\[f(1) = 2 + 5 \bullet 1^{3} - 3 \bullet 1^{5} =\]
\[= 2 + 5 - 3 = 4.\]
\[\textbf{д)}\ Возрастает\ \]
\[на\ ( - 1;\ 0) \cup (0;\ 1)\ и\ убывает\ \]
\[на\ ( - \infty;\ - 1) \cup (1;\ + \infty);\]
\[x = - 1 - точка\ минимума;\ \ \]
\[x = 1 - точка\ максимума.\]
\[\textbf{е)}\ \]
| \[x\] | \[x < - 1\] | \[- 1\] | \[- 1 < x < 0\] | \[0\] | \[0 < x < 1\] | \[1\] | \[x > 1\] |
|---|---|---|---|---|---|---|---|
| \[f^{'}(x)\] | \[-\] | \[0\] | \[+\] | \[0\] | \[+\] | \[0\] | \[-\] |
| \[f(x)\] | \[\searrow\] | \[0\] | \[\nearrow\] | \[2\] | \[\nearrow\] | \[4\] | \[\searrow\] |
\[2)\ y = 3x^{5} - 5x^{3}\]
\[\textbf{а)}\ D(x) = ( - \infty;\ + \infty);\]
\[\textbf{б)}\ y^{'}(x) = 3 \bullet \left( x^{5} \right)^{'} - 5 \bullet \left( x^{3} \right)^{'};\]
\[y^{'}(x) = 3 \bullet 5x^{4} - 5 \bullet 3x^{2} =\]
\[= 15x^{4} - 15x^{2}.\]
\[\textbf{в)}\ Стационарные\ точки:\]
\[15x^{4} - 15x^{2} = 0\]
\[15x^{2} \bullet \left( x^{2} - 1 \right) = 0\]
\[15x^{2} \bullet (x + 1) \bullet (x - 1) = 0\]
\[x_{1} = 0,\ \ \ x_{2} = - 1,\ \ \ x_{3} = 1.\]
\[\textbf{г)}\ f( - 1) =\]
\[= 3 \bullet ( - 1)^{5} - 5 \bullet ( - 1)^{3} =\]
\[= - 3 + 5 = 2;\]
\[f(0) = 3 \bullet 0^{5} - 5 \bullet 0^{3} = 0;\]
\[f(1) = 3 \bullet 1^{5} - 5 \bullet 1^{3} = 3 - 5 =\]
\[= - 2.\]
\[\textbf{д)}\ Возрастает\ \]
\[на\ ( - \infty;\ - 1) \cup (1;\ + \infty)\ и\ \]
\[убывает\ на\ ( - 1;\ 0) \cup (0;\ 1);\]
\[x = 1 - точка\ минимума;\ \ \]
\[x = - 1 - точка\ максимума.\]
\[\textbf{е)}\ \]
| \[x\] | \[x < - 1\] | \[- 1\] | \[- 1 < x < 0\] | \[0\] | \[0 < x < 1\] | \[1\] | \[x > 1\] |
|---|---|---|---|---|---|---|---|
| \[f^{'}(x)\] | \[+\] | \[0\] | \[-\] | \[0\] | \[-\] | \[0\] | \[+\] |
| \[f(x)\] | \[\nearrow\] | \[2\] | \[\searrow\] | \[0\] | \[\searrow\] | \[- 2\] | \[\nearrow\] |
\[3)\ y = 4x^{5} - 5x^{4}\]
\[\textbf{а)}\ D(x) = ( - \infty;\ + \infty);\]
\[\textbf{б)}\ y^{'}(x) = 4 \bullet \left( x^{5} \right)^{'} - 5 \bullet \left( x^{4} \right)^{'};\]
\[y^{'}(x) = 4 \bullet 5x^{4} - 5 \bullet 4x^{3} =\]
\[= 20x^{4} - 20x^{3}.\]
\[\textbf{в)}\ Стационарные\ точки:\]
\[20x^{4} - 20x^{3} = 0\]
\[20x^{3} \bullet (x - 1) = 0\]
\[x_{1} = 0\ и\ x_{2} = 1.\]
\[\textbf{г)}\ f(0) = 4 \bullet 0^{5} - 5 \bullet 0^{4} = 0;\]
\[f(1) = 4 \bullet 1^{5} - 5 \bullet 1^{4} =\]
\[= 4 - 5 = - 1.\]
\[\textbf{д)}\ Возрастает\ \]
\[на\ ( - \infty;\ 0) \cup (1;\ + \infty)\ и\ \]
\[убывает\ на\ (0;\ 1);\]
\[x = 1 - точка\ минимума;\ \ \]
\[x = 0 - точка\ максимума.\]
\[\textbf{е)}\ \]
| \[x\] | \[x < 0\] | \[0\] | \[0 < x < 1\] | \[1\] | \[1 < x < + \infty\] |
|---|---|---|---|---|---|
| \[f^{'}(x)\] | \[+\] | \[0\] | \[-\] | \[0\] | \[+\] |
| \[f(x)\] | \[\nearrow\] | \[0\] | \[\searrow\] | \[- 1\] | \[\nearrow\] |
\[4)\ y = \frac{1}{10}x^{5} - \frac{5}{6}x^{3} + 2x\]
\[\textbf{а)}\ D(x) = ( - \infty;\ + \infty);\]
\[\textbf{б)}\ y^{'}(x) =\]
\[= \frac{1}{10} \bullet \left( x^{5} \right)^{'} - \frac{5}{6} \bullet \left( x^{3} \right)^{'} + (2x)^{'};\]
\[y^{'}(x) = \frac{1}{10} \bullet 5x^{4} - \frac{5}{6} \bullet 3x^{2} + 2 =\]
\[= 0,5x^{4} - 2,5x^{2} + 2.\]
\[\textbf{в)}\ Стационарные\ точки:\]
\[0,5x^{4} - 2,5x^{2} + 2 = 0\]
\[x^{4} - 5x^{2} + 4 = 0\]
\[D = 5^{2} - 4 \bullet 4 = 25 - 16 = 9\]
\[x_{1}^{2} = \frac{5 - 3}{2} = 1\ \ и\ \ \]
\[x_{2}^{2} = \frac{5 + 3}{2} = 4.\]
\[\left( x^{2} - 1 \right)\left( x^{2} - 4 \right) = 0\]
\[(x + 2)(x + 1)(x - 1)(x - 2) =\]
\[= 0\]
\[x_{1} = - 2,\ \ \ x_{2} = - 1,\ \ \ x_{3} = 1,\ \ \]
\[x_{4} = 2.\]
\[= - \frac{32}{10} + \frac{40}{6} - 4 = - \frac{8}{15};\]
\[f( - 1) =\]
\[= - \frac{1}{10} + \frac{5}{6} - 2 = - \frac{19}{15} = - 1\frac{4}{15};\]
\[f(1) = \frac{1}{10} \bullet 1^{5} - \frac{5}{6} \bullet 1^{3} + 2 \bullet 1 =\]
\[= \frac{1}{10} - \frac{5}{6} + 2 = \frac{19}{15} = 1\frac{4}{15};\]
\[f(2) = \frac{1}{10} \bullet 2^{5} - \frac{5}{6} \bullet 2^{3} + 2 \bullet 2 =\]
\[= \frac{32}{10} - \frac{40}{6} + 4 = \frac{8}{15}.\]
\[\textbf{д)}\ Возрастает\ \]
\[на\ ( - \infty;\ - 2) \cup ( - 1;\ 1) \cup (2;\ + \infty);\]
\[убывает\ на\ ( - 2;\ - 1) \cup (1;\ 2);\]
\[x = - 1\ и\ x = 2 - точки\ \]
\[минимума;\]
\[x = - 2\ и\ x = 1 - точки\ \]
\[максимума.\]
\[\textbf{е)}\ \]
| \[x\] | \[x < - 2\] | \[- 2\] | \[- 2 < x < - 1\] | \[- 1\] | \[- 1 < x < 1\] | \[1\] | \[1 < x < 2\] | \[2\] | \[x > 2\] |
|---|---|---|---|---|---|---|---|---|---|
| \[f^{'}(x)\] | \[+\] | \[0\] | \[-\] | \[0\] | \[+\] | \[0\] | \[-\] | \[0\] | \[+\] |
| \[f(x)\] | \[\nearrow\] | \[- \frac{8}{15}\] | \[\searrow\] | \[- 1\frac{4}{15}\] | \[\nearrow\] | \[1\frac{4}{15}\] | \[\searrow\] | \[\frac{8}{15}\] | \[\nearrow\] |
