Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 275

\[1)\ y = x^{2} - 6x + 5;\]

\[y^{'} = 2x - 6 + 0 = 2x - 6.\]

\[2x - 6 = 0\]

\[2x = 6\]

\[x = 3.\]

\[Ответ:\ \ x = 3.\]

\[2)\ y = x^{2} - 14x + 15;\]

\[y^{'} = 2x - 14 + 0 = 2x - 14.\]

\[2x - 14 = 0\]

\[2x = 14\]

\[x = 7.\]

\[Ответ:\ \ x = 7.\]

\[3)\ y = \frac{x}{2} + \frac{8}{x};\]

\[y^{'} = \frac{1}{2} + 8 \bullet \left( - \frac{1}{x^{2}} \right) = \frac{1}{2} - \frac{8}{x^{2}}.\]

\[\frac{1}{2} - \frac{8}{x^{2}} = 0\]

\[\frac{8}{x^{2}} = \frac{1}{2}\]

\[x^{2} = 16\]

\[x = \pm 4.\]

\[Ответ:\ \ x_{1} = - 4;\ x_{2} = 4.\]

\[4)\ y = \frac{x}{3} + \frac{12}{x};\]

\[y^{'} = \frac{1}{3} + 12 \bullet \left( - \frac{1}{x^{2}} \right) = \frac{1}{3} - \frac{12}{x^{2}}.\]

\[\frac{1}{3} - \frac{12}{x^{2}} = 0\]

\[\frac{12}{x^{2}} = \frac{1}{3}\]

\[x^{2} = 36\]

\[x = \pm 6.\]

\[Ответ:\ \ x_{1} = - 6;\ x_{2} = 6.\]

\[5)\ y = 2x^{3} - 15x^{2} + 36x;\]

\[y^{'} = 2 \bullet 3x^{2} - 15 \bullet 2x + 36 =\]

\[= 6x^{2} - 30x + 36.\]

\[6x^{2} - 30x + 36 = 0\]

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

\[D = 25 - 24 = 1\]

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

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

\[Ответ:\ \ x_{1} = 2;\ x_{2} = 3.\]

\[6)\ y = e^{2x} - 2e^{x};\]

\[y^{'} = 2e^{2x} - 2e^{x}.\]

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

\[2e^{2x} = 2e^{x}\]

\[e^{x} = 1\]

\[x = 0.\]

\[Ответ:\ \ x = 0.\]

\[7)\ y = \sin x - \cos x;\]

\[y^{'} = \cos x + \sin x.\]

\[\cos x + \sin x = 0\ \ \ \ \ |\ :\cos x\]

\[1 + tg\ x = 0\]

\[\text{tg\ x} = - 1\]

\[x = - \frac{\pi}{4} + \pi n.\]

\[Ответ:\ \ x = - \frac{\pi}{4} + \pi n.\]

\[8)\ y = \cos{2x} + 2\cos x;\]

\[y^{'} = - 2\sin{2x} - 2\sin x.\]

\[- 2\sin{2x} - 2\sin x = 0\]

\[- 4\sin x \bullet \cos x - 2\sin x = 0\]

\[- 2\sin x\left( 2\cos x + 1 \right) = 0.\]

\[\textbf{а)}\ \sin x = 0\]

\[x = \pi n.\]

\[\textbf{б)}\ 2\cos x + 1 = 0\]

\[\cos x = - \frac{1}{2}\]

\[x = \pm \frac{2\pi}{3} + 2\pi n.\]

\[Ответ:\ \ x_{1} = \pi n;\ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x_{2} = \pm \frac{2\pi}{3} + 2\pi n.\]