1.简答题(10分)已知f(-1)=3,f(0)=2,f(2)=6,求f(x)的二次Lagrange插值多项式.

为了找到给定数据点 $ f(-1) = 3 $, $ f(0) = 2 $, 和 $ f(2) = 6 $ 的二次Lagrange插值多项式，我们使用Lagrange插值公式： \[ f(x) = f(x_0) L_0(x) + f(x_1) L_1(x) + f(x_2) L_2(x) \] 其中 $ L_0(x) $, $ L_1(x) $, 和 $ L_2(x) $ 是Lagrange基多项式，定义为： \[ L_0(x) = \frac{(x - x_1)(x - x_2)}{(x_0 - x_1)(x_0 - x_2)} \] \[ L_1(x) = \frac{(x - x_0)(x - x_2)}{(x_1 - x_0)(x_1 - x_2)} \] \[ L_2(x) = \frac{(x - x_0)(x - x_1)}{(x_2 - x_0)(x_2 - x_1)} \] 这里，数据点是 $ (x_0, f(x_0)) = (-1, 3) $, $ (x_1, f(x_1)) = (0, 2) $, 和 $ (x_2, f(x_2)) = (2, 6) $。让我们计算每个Lagrange基多项式。 1. 计算 $ L_0(x) $: \[ L_0(x) = \frac{(x - 0)(x - 2)}{(-1 - 0)(-1 - 2)} = \frac{x(x - 2)}{1 \cdot -3} = \frac{x(x - 2)}{-3} = -\frac{x(x - 2)}{3} = -\frac{x^2 - 2x}{3} = -\frac{x^2}{3} + \frac{2x}{3} \] 2. 计算 $ L_1(x) $: \[ L_1(x) = \frac{(x + 1)(x - 2)}{(0 + 1)(0 - 2)} = \frac{(x + 1)(x - 2)}{1 \cdot -2} = \frac{(x + 1)(x - 2)}{-2} = -\frac{(x + 1)(x - 2)}{2} = -\frac{x^2 - 2x + x - 2}{2} = -\frac{x^2 - x - 2}{2} = -\frac{x^2}{2} + \frac{x}{2} + 1 \] 3. 计算 $ L_2(x) $: \[ L_2(x) = \frac{(x + 1)(x - 0)}{(2 + 1)(2 - 0)} = \frac{x(x + 1)}{3 \cdot 2} = \frac{x(x + 1)}{6} = \frac{x^2 + x}{6} = \frac{x^2}{6} + \frac{x}{6} \] 现在，将这些基多项式代入Lagrange插值公式： \[ f(x) = f(-1) L_0(x) + f(0) L_1(x) + f(2) L_2(x) \] \[ f(x) = 3 \left( -\frac{x^2}{3} + \frac{2x}{3} \right) + 2 \left( -\frac{x^2}{2} + \frac{x}{2} + 1 \right) + 6 \left( \frac{x^2}{6} + \frac{x}{6} \right) \] \[ f(x) = -x^2 + 2x + (-x^2 + x + 2) + (x^2 + x) \] \[ f(x) = -x^2 + 2x - x^2 + x + 2 + x^2 + x \] \[ f(x) = -x^2 + 4x + 2 \] 因此，二次Lagrange插值多项式是： \[ \boxed{-x^2 + 4x + 2} \]