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МатематикаРешебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 35
Задание 35
\[\boxed{\mathbf{35}\mathbf{.}}\]
\[\textbf{а)}\ f(x) = \frac{4x}{x^{2} + 1};\ \ f^{'}(x) = 0;\]
\[f^{'}(x) = \left( \frac{4x}{x^{2} + 1} \right)^{2} =\]
\[= \frac{4x^{'} \cdot \left( x^{2} + 1 \right) - 4x \cdot \left( x^{2} + 1 \right)^{'}}{\left( x^{2} + 1 \right)^{2}} =\]
\[= \frac{4 \cdot \left( x^{2} + 1 \right) - 4x \cdot 2x}{\left( x^{2} + 1 \right)^{2}} =\]
\[= \frac{4x^{2} + 4 - 8x^{2}}{\left( x^{2} + 1 \right)^{2}} = \frac{4 - 4x^{2}}{\left( x^{2} + 1 \right)^{2}};\]
\[\frac{4 - 4x^{2}}{\left( x^{2} + 1 \right)^{2}} = 0\]
\[4 - 4x^{2} = 0\]
\[4x^{2} = 4\]
\[x^{2} = 1\]
\[x = \pm 1.\]
\[\textbf{б)}\ f(x) = \frac{4x}{x^{2} + 1};\ \ f^{'}(x) > 0;\]
\[f^{'}(x) = \left( \frac{4x}{x^{2} + 1} \right)^{2} =\]
\[= \frac{4x^{'} \cdot \left( x^{2} + 1 \right) - 4x \cdot \left( x^{2} + 1 \right)^{'}}{\left( x^{2} + 1 \right)^{2}} =\]
\[= \frac{4 \cdot \left( x^{2} + 1 \right) - 4x \cdot 2x}{\left( x^{2} + 1 \right)^{2}} =\]
\[= \frac{4x^{2} + 4 - 8x^{2}}{\left( x^{2} + 1 \right)^{2}} = \frac{4 - 4x^{2}}{\left( x^{2} + 1 \right)^{2}};\]
\[\frac{4 - 4x^{2}}{\left( x^{2} + 1 \right)^{2}} > 0\]
\[4 - 4x^{2} > 0\]
\[4x^{2} - 4 < 0\]
\[4 \cdot (x + 1)(x - 1) < 0\]
\[- 1 < x < 1.\]
\[\textbf{в)}\ (x) = \frac{4x}{x^{2} + 1};\ \ f^{'}(x) < 0;\]
\[f^{'}(x) = \left( \frac{4x}{x^{2} + 1} \right)^{2} =\]
\[= \frac{4x^{'} \cdot \left( x^{2} + 1 \right) - 4x \cdot \left( x^{2} + 1 \right)^{'}}{\left( x^{2} + 1 \right)^{2}} =\]
\[= \frac{4 \cdot \left( x^{2} + 1 \right) - 4x \cdot 2x}{\left( x^{2} + 1 \right)^{2}} =\]
\[= \frac{4x^{2} + 4 - 8x^{2}}{\left( x^{2} + 1 \right)^{2}} = \frac{4 - 4x^{2}}{\left( x^{2} + 1 \right)^{2}};\]
\[\frac{4 - 4x^{2}}{\left( x^{2} + 1 \right)^{2}} < 0\]
\[4 - 4x^{2} < 0\]
\[4x^{2} - 4 > 0\]
\[4 \cdot (x + 1)(x - 1) > 0\]
\[x < - 1;\ \ x > 1.\]
