Страница 255 номер 859, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Найти угол между касательной к графику функции \(y = f(x)\) в точке с абсциссой \(x_0\) и осью \(Ox\): 1) \(f(x) = \frac{1}{3} x^3\), \(x_0 = 1\); 2) \(f(x) = \frac{1}{x}\), \(x_0 = 1\); 3) \(f(x) = 2\sqrt{x}\), \(x_0 = 3\); 4) \(f(x) = \frac{18}{\sqrt{x}}\), \(x_0 = 3\); 5) \(f(x) = e^{\frac{3x + 1}{2}}\), \(x_0 = 0\); 6) \(f(x) = \ln(2x + 1)\), \(x_0 = 2\).
1) \(f(x) = \frac{1}{3}x^{3}, x_{0} = 1\)
\(f'(x) = x^{2}\)
\(\operatorname{tg} \alpha = f'(1) = 1, \alpha = \dfrac{\pi}{4};\)
2) \(f(x) = \frac{1}{x}, x_{0} = 1\)
\(f'(x) = -\dfrac{1}{x^{2}},\)
\(\operatorname{tg} \alpha = f'(1) = -1, \alpha = -\dfrac{\pi}{4};\)
3) \(f(x) = 2\sqrt{x}, x_{0} = 3\)
\(f'(x) = \dfrac{1}{\sqrt{x}}\)
\(\operatorname{tg} \alpha = f'(3) = \dfrac{1}{\sqrt{3}}, \alpha = \dfrac{\pi}{6};\)
4) \(f(x) = \frac{18}{\sqrt{x}}, x_{0} = 3\)
\(f'(x) = -\dfrac{9}{x\sqrt{x}}\)
\(\operatorname{tg} \alpha = f'(3) = -\dfrac{9}{3\sqrt{3}} = -\sqrt{3}, \alpha = -\dfrac{\pi}{3};\)
5) \(f(x) = e^{\frac{3x + 1}{2}}, x_{0} = 0\)
\(f'(x) = \dfrac{3}{2}e^{\frac{3x + 1}{2}}\)
\(\operatorname{tg} \alpha = f'(0) = \dfrac{3}{2}\sqrt{e}, \alpha = \operatorname{arctg} \left(\dfrac{3}{2}\sqrt{e}\right);\)
6) \(f(x) = \ln(2x + 1), x_{0} = 2\)
\(f'(x) = \dfrac{2}{2x + 1}\)
\(\operatorname{tg} \alpha = f'(2) = \dfrac{2}{5}, \alpha = \operatorname{arctg} \dfrac{2}{5}.\)
