Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 33

\[\boxed{\mathbf{33.}}\]

\[\textbf{а)}\ y = x^{2} + x - 2;\ \ x = 0\]

\[y(0) = - 2;\]

\[(0; - 2) - точка\ пересечения\ \]

\[графика\ с\ \text{Oy.}\]

\[f^{'}(x) = 2x + 1;\]

\[f^{'}(0) = 1 = tg\ a\]

\[a = \frac{\pi}{4};\]

\[a + \beta = \frac{\pi}{2}\]

\[\beta = \frac{\pi}{2} - a = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}.\]

\[Ответ:\ \ \frac{\pi}{4}.\]

\[\textbf{б)}\ y = 5x^{2} + 4x - 9;\]

\[y(0) = - 9;\]

\[(0; - 9) - точка\ пересечения\]

\[\ графика\ с\ \text{Oy.}\]

\[f^{'}(x) = 5x + 4;\]

\[f^{'}(0) = 4 = tg\ a\]

\[a = artg\ 4;\]

\[\beta = \frac{\pi}{2} - a = \frac{\pi}{2} - arctg\ 4.\]

\[Ответ:\ \frac{\pi}{2} - arctg\ 4.\]

\[\textbf{в)}\ y = 2x^{3} - 12x^{2} + x - 6;\]

\[y(0) = - 6;\]

\[(0; - 6) - точка\ пересечения\]

\[\ графика\ с\ \text{Oy.}\]

\[f^{'}(x) = 6x^{2} - 24x + 1;\]

\[f^{'}(0) = 1 = tg\ a\]

\[a = \frac{\pi}{4};\]

\[\beta = \frac{\pi}{2} - a = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}.\]

\[Ответ:\ \frac{\pi}{4}.\]

\[\textbf{г)}\ y = x^{3} + 6x^{2} + 5x;\]

\[y(0) = 0;\]

\[(0;0) - точка\ пересечения\]

\[\ графика\ с\ \text{Oy.}\]

\[f^{'}(x) = 3x^{2} + 12x + 5;\]

\[f^{'}(0) = 5 = tg\ a\]

\[a = artg\ 5;\]

\[\beta = \frac{\pi}{2} - a = \frac{\pi}{2} - arctg\ 5.\]

\[Ответ:\ \frac{\pi}{2} - arctg\ 5.\ \]