7- Encontre o polinomio interpolador de bagrang da funcaio of que relaciona os seguintes valores: x & -1 & 0 & 1 & 2 f(x) & 0,5 & 2 & -0,9 & 3

## Step 1: Understand the Problem
### We need to find the Lagrange interpolating polynomial for the given set of points. The table provides us with four data points: $(-1, 0.5)$, $(0, 2)$, $(1, -0.8)$, and $(2, 3)$.

## Step 2: Write the General Form of the Lagrange Polynomial
### The Lagrange interpolating polynomial $P(x)$ for a set of points $(x_0, y_0), (x_1, y_1), \ldots, (x_n, y_n)$ is given by:
$$P(x) = \sum_{i=0}^{n} y_i L_i(x)$$
where $L_i(x)$ are the Lagrange basis polynomials defined as:
$$L_i(x) = \prod_{\substack{0 \leq j \leq n \\ j \neq i}} \frac{x - x_j}{x_i - x_j}$$

## Step 3: Calculate Each Lagrange Basis Polynomial
### For each point, calculate $L_i(x)$:

- $L_0(x)$ for $x_0 = -1$:
$$L_0(x) = \frac{(x - 0)(x - 1)(x - 2)}{(-1 - 0)(-1 - 1)(-1 - 2)} = \frac{(x)(x - 1)(x - 2)}{-1 \cdot -2 \cdot -3} = \frac{x(x - 1)(x - 2)}{-6}$$

- $L_1(x)$ for $x_1 = 0$:
$$L_1(x) = \frac{(x + 1)(x - 1)(x - 2)}{(0 + 1)(0 - 1)(0 - 2)} = \frac{(x + 1)(x - 1)(x - 2)}{1 \cdot -1 \cdot -2} = \frac{(x + 1)(x - 1)(x - 2)}{2}$$

- $L_2(x)$ for $x_2 = 1$:
$$L_2(x) = \frac{(x + 1)(x)(x - 2)}{(1 + 1)(1 - 0)(1 - 2)} = \frac{(x + 1)(x)(x - 2)}{2 \cdot 1 \cdot -1} = \frac{(x + 1)(x)(x - 2)}{-2}$$

- $L_3(x)$ for $x_3 = 2$:
$$L_3(x) = \frac{(x + 1)(x)(x - 1)}{(2 + 1)(2 - 0)(2 - 1)} = \frac{(x + 1)(x)(x - 1)}{3 \cdot 2 \cdot 1} = \frac{(x + 1)(x)(x - 1)}{6}$$

## Step 4: Construct the Lagrange Interpolating Polynomial
### Substitute the values of $f(x)$ into the polynomial:
$$P(x) = 0.5 \cdot L_0(x) + 2 \cdot L_1(x) - 0.8 \cdot L_2(x) + 3 \cdot L_3(x)$$

Substitute the expressions for $L_i(x)$:
$$P(x) = 0.5 \cdot \frac{x(x - 1)(x - 2)}{-6} + 2 \cdot \frac{(x + 1)(x - 1)(x - 2)}{2} - 0.8 \cdot \frac{(x + 1)(x)(x - 2)}{-2} + 3 \cdot \frac{(x + 1)(x)(x - 1)}{6}$$

Simplify each term and combine them to get the final polynomial.