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МатематикаРешебник по алгебре и начала математического анализа 10 класс Алимов Задание 931
Задание 931
\[\boxed{\mathbf{931}\mathbf{.}}\]
\[1)\ y = 3x + \frac{1}{3x}\]
\[\textbf{а)}\ D(x) = ( - \infty;\ 0) \cup (0;\ + \infty);\]
\[\textbf{б)}\ y^{'}(x) = (3x)^{'} + \frac{1}{3} \bullet \left( \frac{1}{x} \right)^{'} =\]
\[= 3 - \frac{1}{3x^{2}};\]
\[\textbf{в)}\ Стационарные\ точки:\]
\[3 - \frac{1}{3x^{2}} = 0\]
\[9x^{2} - 1 = 0\]
\[(3x + 1)(3x - 1) = 0\]
\[x_{1} = - \frac{1}{3}\ и\ x_{2} = \frac{1}{3}.\]
\[\textbf{г)}\ f\left( - \frac{1}{3} \right) =\]
\[= 3 \bullet \left( - \frac{1}{3} \right) + \frac{1}{3} \bullet ( - 3) =\]
\[= - 1 - 1 = - 2;\]
\[f\left( \frac{1}{3} \right) = 3 \bullet \frac{1}{3} + \frac{1}{3} \bullet 3 = 1 + 1 = 2;\]
\[f( - 2) = 3 \bullet ( - 2) - \frac{1}{3 \bullet 2} =\]
\[= - 6 - \frac{1}{6} = - 6\frac{1}{6};\]
\[f(2) = 3 \bullet 2 + \frac{1}{3 \bullet 2} = 6 + \frac{1}{6} =\]
\[= 6\frac{1}{6}.\]
\[\textbf{д)}\ Возрастает\ \]
\[на\ \left( - \infty;\ - \frac{1}{3} \right) \cup \left( \frac{1}{3};\ + \infty \right)\ и\ \]
\[убывает\ на\ ( - \frac{1}{3};\ 0) \cup \left( 0;\ \frac{1}{3} \right);\]
\[x = - \frac{1}{3} - точка\ минимума;\ \ \]
\[x = \frac{1}{3} - точка\ максимума.\]
\[\textbf{е)}\ \]
| \[x\] | \[x < - \frac{1}{3}\] | \[- \frac{1}{3}\] | \[- \frac{1}{3} < x < 0\] | \[0 < x < \frac{1}{3}\] | \[\frac{1}{3}\] | \[x > \frac{1}{3}\] |
|---|---|---|---|---|---|---|
| \[f^{'}(x)\] | \[+\] | \[0\] | \[-\] | \[-\] | \[0\] | \[+\] |
| \[f(x)\] | \[\nearrow\] | \[- 2\] | \[\searrow\] | \[\searrow\] | \[2\] | \[\nearrow\] |
\[2)\ y = \frac{4}{x} - x\]
\[\textbf{а)}\ D(x) = ( - \infty;\ 0) \cup (0;\ + \infty);\]
\[\textbf{б)}\ y^{'}(x) = 4 \bullet \left( \frac{1}{x} \right)^{'} - (x)^{'} =\]
\[= - \frac{4}{x^{2}} - 1;\]
\[\textbf{в)}\ Стационарные\ точки:\]
\[- \frac{4}{x^{2}} - 1 = 0\]
\[- 4 - x^{2} = 0\]
\[x^{2} = - 4 - корней\ нет.\]
\[\textbf{г)}\ \frac{4}{x} - x = 0\]
\[4 - x^{2} = 0\]
\[x^{2} = 4\ \]
\[x = \pm 2.\]
\[\textbf{д)}\ f( - 4) = - \frac{4}{4} + 4 =\]
\[= - 1 + 4 = 3;\]
\[f(4) = \frac{4}{4} - 4 = 1 - 4 = - 3.\]
\[\textbf{е)}\ Убывает\ \]
\[на\ ( - \infty;\ 0) \cup (0;\ + \infty).\]
\[\textbf{ж)}\ \]
| \[x\] | \[- 2\] | \[x < 0\] | \[x > 0\] | \[2\] | |
|---|---|---|---|---|---|
| \[f^{'}(x)\] | \[-\] | \[-\] | \[-\] | \[-\] | |
| \[f(x)\] | \[0\] | \[\searrow\] | \[\searrow\] | \[0\] |
\[3)\ y = x - \frac{1}{\sqrt{x}}\]
\[\textbf{а)}\ D(x) = (0;\ + \infty);\]
\[\textbf{б)}\ y^{'}(x) = (x)^{'} - \left( x^{- \frac{1}{2}} \right)^{'} =\]
\[= 1 + \frac{1}{2} \bullet x^{- \frac{3}{2}} = 1 + \frac{1}{2x\sqrt{x}};\]
\[\textbf{в)}\ Стационарные\ точки:\]
\[1 + \frac{1}{2x\sqrt{x}} = 0\]
\[2x\sqrt{x} + 1 = 0\]
\[x\sqrt{x} = - \frac{1}{2} - корней\ нет.\]
\[\textbf{г)}\ x - \frac{1}{\sqrt{x}} = 0\]
\[x\sqrt{x} - 1 = 0\]
\[x\sqrt{x} = 1\]
\[x = 1.\]
\[\textbf{г)}\ f(4) = 4 - \frac{1}{\sqrt{4}} = 4 - \frac{1}{2} = 3,5.\]
\[\textbf{д)}\ Возрастает\ на\ (0;\ + \infty).\]
\[\textbf{е)}\ \]
| \[x\] | \[x > 0\] | \[1\] |
|---|---|---|
| \[f^{'}(x)\] | \[+\] | \[+\] |
| \[f(x)\] | \[\nearrow\] | \[0\] |
