Question

Problem 2 Determine the real roots of f(x)=-25 +82.x - 90x2 +44x-8x +0.7.x a) Graphically b) Using bisection method to determine root (0.5, 1.0) 6 = 0.1 c) False position method, Use initial guesses of x=0.5 and x. -1.0 Compute the estimated errore, with stopping criteria 1%

Answer 1

By the plot of f(X) , we get the root X=0.579

0.7x0 – 8r+ + 44r3 - 90x2 + 82x - 25 (0.579,0) -25 -20 -15 20 25 30

f(x) = -25+82x - 90x2 +44x38x" +0.7x5 intaval [0.591] 1ExI=0.1 a=0.5 ; b=1 f(0.5) = -10478125 f(1) = 3.7 AS (10.5) f(1) To $ root exists in (0.5,1). 8o By bisection method; 1,7 0 0 3 1 - 0 54 - L5 - 0.15 f10.75) = 2.07236328) As f(0.5) flo-75) to do soot exist in €0.59 0.75) Po next iteration - Ug = +34,05+0.75, 0.625 $10.625) - 0.6819915771 f(0.5) flo. 625) 20 8 root exist in (0.5, 0.625) Next iteration ? Xsa+X8 0.5+0.625 -0.5625 f(0.5625) = -0-2819916725 flo. 5625) f(0.625 To so sooth in (0.5625, 0.625] next iterahon ? ve 0.5625+0.625 = 0.59375 f(0.59375) = 0.22648850 f(0.5695) f(0.59375) To root in (0.56259 0.59375) next iteration - 15 - 0.5625+0.59375 -- 0-578125 f'(X) = 0.0208410305 5) If(x) < 0.1 % E=0.1 2

As at x=0.578125 å fix has desired accuracy So noot = 0.578125 for 8=0.1. f(x) = -25 +82x-goxa+uux38x4 +0.4x5 x = 0.5 g xu = 1:0 & fl0.5) = -1,478125 g fli)= 3,7 By false position method 8 X, = 28 - fixe). (Xu-ue) +1.478125) (0.5) flxw - flxo > x = 0.5 = 0.662798 3.7- (1.978125) & - H8 x = 0.64Q7 8 & f(x) = 0.9 187886 Error = 1xi-xil = 0.1427278 $60.64278). f70.5) 70 80 root exists in C0.5,0.64898) Eq > AB Q = 0.5 & xu = 0.64 278 X3 = 0.5 - 61.478125) (0-64878-0-5) (0.9184886-61.478195)] X = 0.588017 ģ fix) = 0.1379895 Error a lxi-Xiti 20.0880172 fixg). f(x) to so soot enests in (0.5, 0.588017) MQ = 0.5 xu = 0.588017 892 05 - C1.478125) (20-588019)-0.5) (0 1372895-61.478125) - 03=0.580537 fixg) - 0.0189191 Emore = 0.0805368. AS 1 Son) col desired a curracy obtained . so desired soot - [x = 0.580537) Anew