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We are tasked with finding the Lagrange interpolation polynomial for the points:

\[
(1, -3), (3, 0), (4, 30), (6, 132)
\]

and then using this polynomial to determine \( y(5) \).

The general formula for the Lagrange interpolation polynomial is given by:

\[
P(x) = \sum_{i=0}^{n} y_i \cdot L_i(x)
\]

where \( L_i(x) \) is the Lagrange basis polynomial defined as:

\[
L_i(x) = \prod_{j=0, j \neq i}^{n} \frac{x - x_j}{x_i - x_j}
\]

We have four data points, so the polynomial will be of degree 3. Let's proceed by constructing the Lagrange basis polynomials and then the full polynomial. Finally, we will compute \( y(5) \). I will now perform these calculations.The Lagrange interpolation polynomial is:

\[
P(x) = -\frac{1}{2}x^3 + \frac{27}{2}x^2 - 46x + 30
\]

Using this polynomial, the value of \( y(5) \) is:

\[
y(5) = 75
\]

Would you like further details or have any questions? Here are five related questions you might find interesting:

1. How does the Lagrange interpolation polynomial change with additional points?
2. What are some applications of the Lagrange interpolation in numerical analysis?
3. How is Lagrange interpolation different from Newton's interpolation method?
4. How can errors be minimized in Lagrange interpolation?
5. What is the significance of the degree of the polynomial in interpolation?

**Tip:** The Lagrange interpolation method is useful when you know the function's values at specific points, but it becomes less reliable when approximating outside of the given data range.

Find Lagrange's interpolation polynomial fitting the points y(1) = -3, y(3) = 0, y(4) = 30, y(6) = 132. Hence, find y(5).

This page explains how to construct the Lagrange interpolation polynomial for the points y(1) = -3, y(3) = 0, y(4) = 30, and y(6) = 132. It also demonstrates how to use this polynomial to find the value of y(5), which results in y(5) = 75.

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