Please help with this problem.

Pv(x) = do taxtax-+ a3x' We can use the Vandermonde matrix approach to find do, a1, a2, a3. In this approach we solve the linear system W XO do f(xo) X1 a1 f(x1) = We find, 1 X 2 a2 f (x2) f (x3) X3 5. do 6. al = 7. a2 8. a3 We can use the polynomial to approximate f (x) at other points in the [-1, 1]. Thus, 9. Pv(0) 10. Pv(0.5) =Another approach to finding the interpolating polynomial is to use Lagrange polynomials. Here we find the polynomial 3 II(x - XK) k# j pi(x) = >fil; (x), where 1; (x) = 3 j=0 II (x ; - XK) C=0 Then, 11. fo = 12. f1 = 13. f2 14. f3 15. 10 (x) = 019(x + 1)(x+ 1/3)(x- 1/3)]/16 O[-27(x + 1)(x+ 1/3)(x- 1)]/16 0[27(x + 1)(x - 1/3)(x - 1)]/16 O[-9(x + 1/3)(x - 1/3)(x - 1)]/1616. l1(x) = O[9(x + 1)(x + 1/3)(x 1/3)] /16 O[27(x + 1)(x + 1/3)(x 1)] /16 O[27(x + 1)(x 1/3)(x 1)] /16 O[9(x + 1/3)(x 1/3)(x 1)] I16 17. l2(x) = O[9(x + 1)(x + 1/3)(x 1/3)] /16 O[27(x + 1)(x + 1/3)(x 1)] /16 O[27(x + 1)(x 1/3)(x 1)] /16 O[9(x + 1/3)(x 1/3)(x 1)] I16 18. l3 (x) = O[9(x + 1)(x + 1/3)(x 1/3)] /16 O[27(x + 1)(x + 1/3)(x 1)] /16 O[27(x + 1)(x 1/3)(x 1)] I16 O[9(x + 1/3)(x 1/3)(x 1)] /16 We can use the resulting polynomial to approximate f (x) at other points in the [1, 1]. Thus, M0) =S M05) =: In these questions enter your answers with at least six decimal places. We want to find a polynomial approximation to f (x) = ex2 on the interval [1, 1]. Suppose we sample the function at four distinct equidistant points x0 = 1, x1 = %, x2 = %, x3 = 1. Then, 1-f(x0) 2- f(x1) DE 3- f(x2) U 4- f(x3) E The highest degree polynomial that interpolates the data points is a polynomial of degree 3 which can be represented as pv(x) = a0 + alx + (1ng + 03x3