Solving for f(3) using Lagrange's Interpolation

The problem asks us to find the value of the function f(3) using Lagrange's interpolation formula, given two data points: f(1) = 4 and f(5) = 6. This means we have two points (1, 4) and (5, 6).

Lagrange's interpolation formula is used to find a polynomial that passes through a given set of data points. For two points $(x_0, y_0)$ and $(x_1, y_1)$, the linear interpolating polynomial $P(x)$ is given by:

$$\small P(x) = y_0 \frac{x - x_1}{x_0 - x_1} + y_1 \frac{x - x_0}{x_1 - x_0}$$

In this problem, we have:

• $(x_0, y_0) = (1, 4)$, so $x_0 = 1$ and $y_0 = 4$.
• $(x_1, y_1) = (5, 6)$, so $x_1 = 5$ and $y_1 = 6$.
• We need to find the value at $x = 3$.

Now, we substitute these values into the Lagrange's formula:

$$\small P(3) = f(3) = 4 \frac{3 - 5}{1 - 5} + 6 \frac{3 - 1}{5 - 1}$$

Let's calculate each term step by step:

• First term: $$4 \frac{3 - 5}{1 - 5} = 4 \frac{-2}{-4} = 4 \times \frac{1}{2} = 2$$
• Second term: $$6 \frac{3 - 1}{5 - 1} = 6 \frac{2}{4} = 6 \times \frac{1}{2} = 3$$

Now, add the results of the two terms to get the value of f(3):

$$\small f(3) = 2 + 3 = 5$$

So, the value of f(3) using Lagrange's interpolation with the given points is 5.

Revision Table: Key Concepts

Additional Information: Understanding Interpolation

Interpolation is a technique used to estimate values between known data points. When you have a set of data points from a function or experiment, but you don't have the exact formula for the function, interpolation helps you guess the value of the function at points not explicitly given.

Lagrange's method is one way to do this. It constructs a unique polynomial that goes through every single data point you provide. The degree of this polynomial is one less than the number of points you have. For two points, as in this problem, the polynomial is a straight line, which is also called linear interpolation.

The formula for Lagrange's interpolation for $\small n+1$ points $\small (x_0, y_0), (x_1, y_1), \dots, (x_n, y_n)$ is:

$$\small P(x) = \sum_{j=0}^{n} y_j L_j(x)$$

where $\small L_j(x)$ is the Lagrange basis polynomial, defined as:

$$\small L_j(x) = \prod_{i=0, i \neq j}^{n} \frac{x - x_i}{x_j - x_i}$$

For two points ($\small n=1$), the formula simplifies to the one we used: $\small P(x) = y_0 L_0(x) + y_1 L_1(x)$, where $\small L_0(x) = \frac{x - x_1}{x_0 - x_1}$ and $\small L_1(x) = \frac{x - x_0}{x_1 - x_0}$. This confirms the linear interpolation formula used for solving this specific problem.