Question

(1) Use the Bisection method to find solutions accurate to within 10-2 for x3 – 7x2 + 14x – 6 = 0 on the interval [3.2, 4]. Using 4-digit rounding arithmatic. (2) Consider the function f(x) = cos X – X. (a). Approximate a root of f(x) using Fixed- point method accurate to within 10-2 . (b). Approximate a root of f(x) using Newton's method accurate to within 10-2.

Answer 1

soin ① 043- 742 +14% -6 = 0 [3.2.4 ] Let F(X) = 73-722+147-6 ist iteration : f(3-2) = - -0.112 (3.2)3-7(3.2)2 +14 (3.2) - 6 F13.2) -0.112 20 & F14) = (4)3- 764)2 +14 (4)-6 = 2 FC4) = 270 Root lies between 3.2 and 4 3.2+4 -1.3.6 Ho = 3.274 f(20) = f(3-6) = (3-673 - 7 (3:672 +14(3-6) -6 = 0.33670 and iteration f(3.2) = (3.2)3-713.212 +14(3.2) - 6 = -0.112 20 & F13.6) = (3.613-7 (36)2+14(3-6) - = 0.33670 Root lies between 3.2 and 3.6 - 3.2 +3.6 3.4 H = 0.33670 flai) F (3.4) - (3-4)3 - 7 (3.4)2 +14 (3.4) – 6 = -0.016 <0 3rd iteration : f13.4) -0.016 0 & F(3.6) = (3-6)3-7 (3-6)2 +14 (3.61-6 Root lies between 3.4 and 3.6 H2 = 3.4+ 3.6 3.4+3:6 f(7/2) = f(3.5) = (3.5)3- 7 (3.5)2 + 14 (3.5) -6 3.5 = 0.125 70

4th iteration : f(3.4) -0.016 20 & F (3.5) = 0.125 70 Root lies between 3.4 and 3.5 H3 3.4 +3.5 3.45 2 f (x3) = f(3:45) = (3-45)3 - 7 (3:48)2 +14 (3:45) – 6 = 0.0461 f(3:45) 0.0461 70 5th iteration : f (3.4) -0.016 to F 13.45) = 0.0461 70 & Root lies between 3.4 and 3.45 H4 = 3.4 +3:45 3.425 2 F194) = f(3-425) = (3.425)3 - 7 (3:425)2 +14.(3-425) - 6 F13.425) = 0.013 70 6th iteration : f 134) = -0.016 to df (3.425) = 0.01370 Root lies between 3.4 and 3.425 Us = 3.4 +3.425 3.4125 2 Flas) = f (3:4125) = (3.4125)3 - 7 (3.4128)2+14 13:4125) - 6 f(3-4125) = -0.002 20

7th iteration : F(3.4125) = & F 13.425) -0.002 20 0.013 70 Root lies between 3.4125 and 3.425 16 = 3.4125 + 3.425 3.4188 2 f (016) F (3-4188) = 0.0094 70 The solutions accurate to within 102 is x = 3.41

Answer 2

x³-7x+2=0, in 0,1.using a) Bisection method and find the minimum number of iteration required to approximate the root with an absolute error less than 10^-3.  b) fixed point iteration (FPI) method by choosing g(x) = 1
7 (x3 + 2) & x0 = 0 and justify why this g(x)
is a right choice in [0, 1].