Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 21

\[\boxed{\mathbf{21}\mathbf{.}}\]

\[\textbf{а)}\ y = x^{2} + 6x + 5\]

\[f^{'}(x) = \left( x^{2} \right)^{'} + 6x^{'} + 5^{'} =\]

\[= 2x + 6.\]

\[f^{'}(x) = 0:\]

\[2x + 6 = 0\]

\[2x = - 6\]

\[x = - 3.\]

\[f^{'}(x) > 0:\]

\[x > - 3.\]

\[f^{'}(x) < 0:\]

\[x < - 3.\]

\[\textbf{б)}\ y = x^{3} + 3x^{2} - 17\]

\[f^{'}(x) = \left( x^{3} \right)^{'} + 3 \cdot \left( x^{2} \right)^{'} - 17^{'} =\]

\[= 3x^{2} + 3 \cdot 2x - 0 = 3x^{2} + 6x;\]

\[f^{'}(x) = 0:\]

\[3x^{2} + 6x = 0\]

\[3x(x + 2) = 0\]

\[x = 0;\ \ x = - 2.\]

\[f^{'}(x) > 0:\]

\[3x(x + 2) > 0\]

\[x < - 2;\ \ x > 0.\]

\[f^{'}(x) < 0:\]

\[3x(x + 2) < 0\]

\[- 2 < x < 0.\]

\[\textbf{в)}\ y = \frac{1}{3}x^{3} - 3x^{2} + 9x - 15\]

\[f^{'}(x) = \frac{1}{3} \cdot \left( x^{3} \right)^{'} - 3 \cdot \left( x^{2} \right)^{'} +\]

\[+ 9x^{'} - 15^{'} =\]

\[= \frac{1}{3} \cdot 3x^{2} - 3 \cdot 2x + 9 \cdot 1 - 0 =\]

\[= x^{2} - 6x + 9;\]

\[f^{'}(x) = 0\]

\[x^{2} - 6x + 9 = 0\]

\[(x - 3)^{2} = 0\]

\[x = 3.\]

\[f^{'}(x) > 0:\]

\[(x - 3)^{2} > 0\]

\[x - любое\ число.\]

\[f^{'}(x) < 0:\]

\[(x - 3)^{2} < 0\]

\[нет\ таких\ \text{x.}\]

\[\textbf{г)}\ y = x^{3} + 5x^{2} - 13x + 7\]

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

\[- 13x^{'} + 7^{'} =\]

\[= 3x^{2} - 5 \cdot 2x - 13 \cdot 1 + 0 =\]

\[= 3x^{2} - 10x - 13;\]

\[f^{'}(x) = 0:\]

\[3x^{2} - 10x - 13 = 0\]

\[D_{1} = 25 + 39 = 64\]

\[x_{1} = \frac{5 + 8}{3} = \frac{13}{3} = 4\frac{1}{3};\]

\[x_{2} = \frac{5 - 8}{3} = - \frac{3}{3} = - 1.\]

\[f^{'}(x) > 0:\]

\[3 \cdot (x + 1)\left( x - 4\frac{1}{3} \right) > 0\]

\[x < - 4\frac{1}{3};\ \ x > 1.\]

\[f^{'}(x) < 0:\]

\[3 \cdot (x + 1)\left( x - 4\frac{1}{3} \right) < 0\]

\[- 4\frac{1}{3} < x < 1.\]