Question

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Q1 (a) Given the function f(x)= x - 5x² - 2x +10. (1) Prove that there at least a root in the interval [1,3] by using Intermediate Value Theorem. (2 marks) (b) (i) Find the root of f(x) by using Bisection method. Iterate until i = 5. (8 marks) Prove the Lagrange interpolating polynomial of second degree for data of (0,1), (1,2) and (4,2) is P2(x) = -* x2 + x + 1. (5 marks) (ii) If then, one data points (5,3) are added to the previous data, find the new Lagrange interpolating polynomials without solving it. (10 marks)

Answer 1

@l(a)a9-54? - 22 +10 S (1)=470 $(3) = -1420 hence by entermediate value theorem, I XE (1,3) Such that Caath 2 n 1 1 f(x)=0. f(b) f(c) scal 3 4 2 - 14 2. 1.5 -0.875 2 1 TS 8 1. 4 -0.875 1.25 1.64062 1.25 1.6 4062 - 0.875 1.375 10.39648 1.375 5 0.39648 15 -0.875 1.4375 1-0·23657 6 1.375 0.39648 1.4375 -0.23657 1.40625 0.08072 71.40625 0.08072 1.4375 -0.23657 | 11:42188 1-6.07775 1.40615 0.08072 1.42188 1-0.07775 1.41406 9 1.41406 0.00153 1:42 188 -0.07775 1.41797 101.41406 0.00153 1.41797 -0.0381 -0.01828 111.41406 0.00153 (2/1.41406 0.00153 13 1.41406 0.00153 1.41455 O vols 3 1-0.0381 1-41602 1.41602 -0.01828 1041504 -0.00837 1.41504 - 0100837 1:4145S 1-0.000342 - U'U00342 1.41431 0.00ogy 19 1.41406 0.00153 1.41431 Foivoo94 1.414180.00029 solution is 1.41418

(0, 1), (1, 2) and (4,2) (X+0) (2-4) .2 1127= (x-1) (x-4) (0-1) (0-4) (1+0) (1-4) + (x+o) (2-1). 2 (4+0) (4-1) 6-1) (2.4) 2(x-4).a x(x-1) 2 4 (-3) 4.3 (12-5x+4) 2(x²-un) 2(a²-n) - 3 12 + 4 2 + I X+1. Join (5,3) La- t (2-1)(2-4212-5) (0-1210-4) (0-5) + 2 (x+o) (2-4212-5) (1+0)(1-4) (1-5) + 2(2+0) (2-1212-5) (4+0) (4.1) (4-5) + 3 (x+o) (2-1) (2-4) (5+0)(5-1)(5-4) Then you