Question

MATLAB QUESTION

please include function codes inputed

Problem 3 Determine the root (highest positive) of: F(x)= 0.95x.^3-5.9x.^2+10.9x-6; Note: Remember to compute the error Epsilon-a after each iteration. Use epsilon_$=0.01%. Part A Perform (hand calculation) 3 iterations of Newton's Raphson method to solve the equation. Use an initial guess of x0=3.5. Part B Write your own Matlab function to validate your results. Part C Compare the results of question 1 to the results of question 2, what is your conclusion ?

Answer 1

Flv) = 0.98x3_s.gx²+10.94-6 Newton Raphson method - Mariann- flan) f(n)= 0.95 n3 - 5.90²+10.50 -6 $'(W) = 2.35m2 – 11.82+10.9 ng ang flam) 't' and until da Ny=3-5 f (Mo) = 0.66 $(20)=4519433657 da=0.132 2 2. ni = 3.365 f(1) = 0.068|f'(21) = 3.469 21 dna 0.019 3.345 1223.345F(^2)=0.001 f'(^2)=3:3171 22 dmg 0.000? 1 3.3446] Hence root = 3.3446 t CS Scanned with CamScanner

MATLAB FUNCTION:-

%The Newton Raphson Method
clc;
close all;
clear all;
syms x;
f=0.95*x^3-5.9*x^2+10.9*x-6; %Enter the Function here
g=diff(f); %The Derivative of the Function
n=input('Enter the number of decimal places:');
epsilon = 0.01;
x0 = input('Enter the intial approximation:');
for i=1:100
     f0=vpa(subs(f,x,x0)); %Calculating the value of function at x0
     f0_der=vpa(subs(g,x,x0)); %Calculating the value of function derivative at x0
y=x0-f0/f0_der; % The Formula
err=abs(y-x0)/abs(y)*100;
if err<epsilon %checking the amount of error at each iteration
break
end
x0=y;
end
y = y - rem(y,10^-n); %Displaying upto required decimal places
fprintf('The Root is : %0.4f \n',y);
fprintf('No. of Iterations : %d\n',i);

OUTPUT:-

Enter the number of decimal places:4
Enter the intial approximation:3.5
The Root is : 3.3446
No. of Iterations : 4

******Two result are same