Question

Using Lagrange interpolation with the following function values, find the approximate value of \(f(2)\).

x	0	1	3
f(x)	1	2	4

• A. 2.13
• B. 3.00
• C. 3.13
• D. 3.33

Hint:

Build the degree-2 Lagrange polynomial through the three given points and evaluate it at x = 2.

Answer 1

The Correct Option is B

Step 1: The three data points are \((x_0,y_0)=(0,1)\), \((x_1,y_1)=(1,2)\), \((x_2,y_2)=(3,4)\). The Lagrange interpolating polynomial is\[f(x) \approx y_0L_0(x)+y_1L_1(x)+y_2L_2(x),\]with\[L_0(x)=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)},\ L_1(x)=\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)},\ L_2(x)=\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}.\]

Step 2: Evaluate the Lagrange basis functions at \(x=2\).\[L_0(2)=\frac{(2-1)(2-3)}{(0-1)(0-3)}=\frac{(1)(-1)}{3}=-\frac{1}{3}\]\[L_1(2)=\frac{(2-0)(2-3)}{(1-0)(1-3)}=\frac{(2)(-1)}{-2}=1\]\[L_2(2)=\frac{(2-0)(2-1)}{(3-0)(3-1)}=\frac{(2)(1)}{6}=\frac{1}{3}\]

Step 3: Substitute into the interpolation formula.\[f(2)\approx 1\left(-\frac{1}{3}\right)+2(1)+4\left(\frac{1}{3}\right)=-\frac{1}{3}+2+\frac{4}{3}=2+1=3\]

\[\boxed{f(2)\approx 3.00}\]