Страница 255 номер 860, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Написать уравнение касательной к графику функции \(y = f(x)\) в точке с абсциссой \(x_0\): 1) \(f(x) = x^2 + x + 1\), \(x_0 = 1\); 2) \(f(x) = x - 3x^2\), \(x_0 = 2\); 3) \(f(x) = \frac{1}{x}\), \(x_0 = 3\); 4) \(f(x) = \frac{1}{x}\), \(x_0 = -2\); 5) \(f(x) = \sin x\), \(x_0 = \frac{\pi}{4}\); 6) \(f(x) = e^x\), \(x_0 = 0\); 7) \(f(x) = \ln x\), \(x_0 = 1\); 8) \(f(x) = \sqrt{x}\), \(x_0 = 1\).
\(y = f(x_{0}) + f'(x_{0})(x - x_{0});\)
1) \(f(x) = x^{2} + x + 1, x_{0} = 1\)
\(f(x_{0}) = 1 + 1 + 1 = 3\)
\(f'(x) = 2x + 1\)
\(f'(x_{0}) = 3,\)
\(y = 3 + 3x - 3 = 3x;\)
2) \(f(x) = x - 3x^{2}, x_{0} = 2\)
\(f(x_{0}) = 2 - 3 \cdot 4 = -10\)
\(f'(x) = 1 - 6x\)
\(f'(x_{0}) = -11,\)
\(y = -10 - 11x + 22 = 12 - 11x;\)
3) \(f(x) = \frac{1}{x}, x_{0} = 3\)
\(f(x_{0}) = \dfrac{1}{3}, f'(x) = -\dfrac{1}{x^{2}}, f'(x_{0}) = -\dfrac{1}{9},\)
\(y = \dfrac{1}{3} - \dfrac{1}{9}x + \dfrac{1}{3} = \dfrac{2}{3} - \dfrac{1}{9}x;\)
4) \(f(x) = \frac{1}{x}, x_{0} = -2\)
\(f(x_{0}) = -\dfrac{1}{2}\)
\(f'(x) = -\dfrac{1}{x^{2}}\)
\(f'(x_{0}) = -\dfrac{1}{4},\)
\(y = -\dfrac{1}{2} - \dfrac{1}{4}x - \dfrac{1}{2} = -\dfrac{1}{4}x - 1;\)
5) \(f(x) = \sin x, x_{0} = \frac{\pi}{4}\)
\(f(x_{0}) = \dfrac{\sqrt{2}}{2}\)
\(f'(x) = \cos x\)
\(f'(x_{0}) = \dfrac{\sqrt{2}}{2},\)
\(y = \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}x - \dfrac{\sqrt{2}\pi}{8} = \dfrac{\sqrt{2}}{2}\left(1 + x - \dfrac{\pi}{4}\right);\)
6) \(f(x) = e^{x}, x_{0} = 0\)
\(f(x_{0}) = 1\)
\(f'(x) = e^{x}\)
\(f'(x_{0}) = 1,\)
\(y = 1 + x - 0 = 1 + x;\)
7) \(f(x) = \ln x, x_{0} = 1\)
\(f(x_{0}) = 0\)
\(f'(x) = \dfrac{1}{x}\)
\(f'(x_{0}) = 1,\)
\(y = 0 + x - 1 = x - 1;\)
8) \(f(x) = \sqrt{x}, x_{0} = 1\)
\(f(x_{0}) = 1\)
\(f'(x) = \dfrac{1}{2\sqrt{x}}\)
\(f'(x_{0}) = \dfrac{1}{2},\)
\(y = 1 + \dfrac{1}{2}x - \dfrac{1}{2} = \dfrac{1}{2}(x + 1).\)
