Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 32

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\[\textbf{а)}\ y = x^{2} + x - 2;\]

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

\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 2\]

\[x_{1} = - 2;\ \ x_{2} = 1.\]

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

\[f'(1) = 2 + 1 = 3 = tga_{1};\]

\[f^{'}( - 2) = - 4 + 1 = - 3 = tga_{2};\]

\[a_{1} = arctg\ 3;\]

\[a_{2} = - arctg\ 3.\]

\[\textbf{б)}\ y = 5x^{2} + 4x - 9;\]

\[5x^{2} + 4x - 9 = 0\]

\[D_{1} = 4 + 45 = 49\]

\[x_{1} = \frac{- 2 + 7}{5} = 5;\]

\[x_{2} = \frac{- 2 - 7}{5} = - \frac{9}{5} = - 1,8.\]

\[f^{'}(x) = 10x + 4;\]

\[f^{'}(5) = 50 + 4 = 54 = tga_{1};\]

\[f^{'}( - 1,8) = - 18 = 4 =\]

\[= - 14 = tga_{2};\]

\[a_{1} = arctg\ 54;\]

\[a_{2} = - arctg\ 14.\]

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

\[2x^{3} - 12x^{2} + x - 6 = 0\]

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

\[(x - 6)\left( 2x^{2} + 1 \right) = 0\]

\[x - 6 = 0\]

\[x_{1} = 6.\]

\[2x^{2} + 1 = 0\]

\[2x^{2} = - 1\]

\[нет\ корней.\]

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

\[f^{'}(6) = 6 \cdot 36 - 24 \cdot 6 + 1 =\]

\[= 216 - 144 + 1 = 73 = tg\ a.\]

\[a = arctg\ 73.\]

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

\[x^{3} + 6x^{2} + 5x = 0\]

\[x\left( x^{2} + 6x + 5 \right) = 0\]

\[x_{1} = 0;\]

\[x^{2} + 6x + 5 = 0\]

\[D_{1} = 9 - 5 = 4\]

\[x_{2} = - 3 + 2 = - 1;\]

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

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

\[f^{'}(0) = 5 = tg\ a_{1};\]

\[f^{'}( - 1) = 3 - 12 + 5 = - 4 =\]

\[= tg\ a_{2};\]

\[f^{'}( - 5) = 3 \cdot 25 - 60 + 5 =\]

\[= 20 = tg\ a_{3};\]

\[a_{1} = arctg\ 5;\]

\[a_{2} = - arctg\ 4;\]

\[a_{3} = arctg\ 20.\]