Question

1.Use Newton's method to find all solutions of the equation correct to six decimal places. (Enter your answers as a comma-separated list.)

ln(x) = 1/(x-3)

2. Use Newton's method to find all solutions of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations. (Enter your answers as a comma-separated list.)

6e^−x2 sin(x) = x² − x + 1

Answer 1

(Question 1) MATLAB Script:

close all
clear
clc

syms x
f = log(x) * (x - 3) - 1; % f(x) = 0
df = diff(f); % f'(x)
tol = 1e-6; % Error tolerance (6 decimal places)

% Plot f(x) for locating approximate solutions
x_vals = 0:0.01:5;
plot(x_vals, subs(f, x_vals))
xlabel('x'), ylabel('f(x)')
title('f(x) = ln(x) * (x - 3) - 1')
grid on

fprintf('From the graph, the approximate roots are: 0.65 and 3.76\n\n')

x0 = 0.65; % Initial guess 1
result = newtraph(f,df,x0,tol);
fprintf('Solution (using initial guess = %.2f): %.6f\n', x0, result)

x0 = 3.76; % Initial guess 2
result = newtraph(f,df,x0,tol);
fprintf('Solution (using initial guess = %.2f): %.6f\n', x0, result)

% Newton's Method
function x = newtraph(f,df,x0,tol)
x = x0;
while true
x_ = x; % Backup previous iteration's result
x = double(x - subs(f, x)/subs(df, x)); % Newton Update Rule
if abs(x_ - x) < tol % Termination condition
break
end
end
end

Plot:

Output:

From the graph, the approximate roots are: 0.65 and 3.76

Solution (using initial guess = 0.65): 0.653060
Solution (using initial guess = 3.76): 3.755701