Question

l. Determine the real root(s) off(x)--5xs + 14x3 + 20x2 + 10x a. Graphically on a graph paper. b. Using Bisection method c. Using False Position method to determine the root, employing initial guesses of x-2 d. Using the Newton Raphson methods to determine the root, employing initial guess to determine the root, employing initial guesses ofxn-2 and Xu-4 and Es= 18%. and r 5.0 andas answer. 1%. was this method the best for these initial guesses? Explain your xo--l and Es-1%. ES-6%.. and xo=5, Perform 4 iterations and calculate Ea for each iteration f. Using the secant method to determine the root employing initial guesses x.i-3

Answer 1

`Hey,

Note: Brother in case of any queries, just comment in box I would be very happy to assist all your queries

Bisection method

clear all
clc
f = @(x) -5*x^5+14*x^3+20*x^2+10*x;
e=0.18;
a=-2;
b=4;
iter = 0;
      
if f(a)*f(b)>=0

                disp('No Root')

else

                prev = (a+b)/2;
                p=a;
                while (abs(f(p))>e)
                    prev=p;

                                 iter =iter+ 1;

                                p = (a+b)/2;

                                if f(p) == 0
                                    p
q=1
                                                break;

                                end

                                if f(a)*f(p)<0

                                                b = p;

                                else

                                                a = p;

                                end
    fprintf('Iteration %d, root=%2.8f\n',iter,p);
    if(iter==100)
        disp('the required accuracy is not reached in 50 iterations');
    end
                end

end
fplot(f,[-2,4]);
hold on;
plot(p,f(p),'*r');

FALSE POSITION METHOD

clc

f = @(x) -5*x^5+14*x^3+20*x^2+10*x;
e=1e-2;
x0=2;
x1=5;
i=0;
imax=200;
x2=x1+1;
flag=1;
while (abs(x2-x1)>e)
    if(i~=0)
    x1=x2;
    end
    x2=x1-((f(x1)/(f(x1)-f(x0)))*(x1-x0));
    x0=x1;
    i=i+1;
     if(i==200)
       
        disp('the required accuracy is not reached in 100 iterations');
        flag=0;
    end
end
if(flag==1)
root=x2
else
    root=[];
end

NEWTON METHOD

clc
clear all
close all

f = @(x) -5*x^5+14*x^3+20*x^2+10*x;

g=@(x) - 25*x^4 + 42*x^2 + 40*x + 10;
x0=-1;
e=0.01;
maxit=100;
x1=x0;
x0=0;
N=0;
err=[];
while 1

if abs(f(x1))>e
x0=x1;
x1=x0-(f(x0)/g(x0));
N=N+1;

else

break;

end
if(N==maxit)
    break;
end

end
root=x1;
iter=N;
disp('So, final approximation is')
(root)
fprintf('Number of iteration is %d\n',N);
fprintf('Approximate absolute relative error is %f\n',abs(x1-x0)/abs(x1));

Note: Brother according to HOMEWORKLIB RULES we are only allowed to answer 3 part if there are many. So, I request you to post other part as separate posts.

Kindly revert for any queries

Thanks.