Question

Newton's Method in MATLAB During this module, we are going to use Newton's method to compute the root(s) of the function f(x) = x° + 3x² – 2x – 4 Since we need an initial approximation ('guess') of each root to use in Newton's method, let's plot the function f(x) to see many roots there are, and approximately where they lie. Exercise 1 Use MATLAB to create a plot of the function f(x) that clearly shows the locations of its real roots, then save and upload the plot: Upload PNG here Show/hide hint You can do this in the Command Window. Make sure you initialise x as a vector. Try x = linspace(-4, 2, 100);. Then don't forget the dot-operator when writing f = and your function. You can save a figure in MATLAB, by clicking File > Save As on your figure. Make sure you save as a .png, and not a .fig! Download the following M-file, then open and view its contents. This M-file contains the function newton to help us get started. Be sure to read the comments: newton.m (Right-click > Save link as...)

Answer 1

clc;
clear all;
close all;
n=10;
f=@(x)x.^3+3*x.^2-2*x-4;
fprime=@(x) 3*x.^2+6*x-2;
x=linspace(-4,2,100);
fx=f(x);
plot(x,fx);
grid on;
xlabel('x');
ylabel('function');

function -3 -2 -1 0 1 2

%%% Matlab function

function [] = newton (x)
f=@(x)x.^3+3*x.^2-2*x-4;
fprime=@(x) 3*x.^2+6*x-2;
xp=x;
x=xp-f(xp)/fprime(xp);
n=1;
while ( abs(x-xp) > 0.001 )
xp=x;
x=xp-f(xp)/fprime(xp);
n=n+1;
end
fprintf('Newton Approximates root at %1.10f after %d iteration \n',x,n);
fprintf('f(x) at approximate root x is %1.10f \n',f(x));
fprintf('|x-xp| is %1.10f \n',abs(x-xp));

end

OUTPUT:

>> newton(10)

Newton Approximates root at 1.2360679797 after 8 iteration
f(x) at approximate root x is 0.0000000221
|x-xp| is 0.0000574524

%%%

function [] = newton (x)
f=@(x)x.^3+3*x.^2-2*x-4;
fprime=@(x) 3*x.^2+6*x-2;
xp=x;
x=xp-f(xp)/fprime(xp);
n=1;
while ( abs(x-xp) > 0.001 )
xp=x;
x=xp-f(xp)/fprime(xp);
n=n+1;
if (n == 100 )
fprintf('Roots diverges \n');
break;
end
end
if (n<100)
fprintf('Newton Approximates root at %1.10f after %d iteration \n',x,n);
fprintf('f(x) at approximate root x is %1.10f \n',f(x));
fprintf('|x-xp| is %1.10f \n',abs(x-xp));
end

end

OUTPUT:

>> newton(-3)
Newton Approximates root at -3.2360679775 after 4 iteration
f(x) at approximate root x is -0.0000000000
|x-xp| is 0.0000016550
>> newton(-1)
Newton Approximates root at -1.0000000000 after 1 iteration
f(x) at approximate root x is 0.0000000000
|x-xp| is 0.0000000000
>> newton(0)
Roots diverges