Question

question 3 please

The first 5 questions refer to finding solutions to the equation exp(w) = 3.8 ln(1+x). You will need to write it in the form f(x)-0, and use various root finding methods. 1. (10 pts) Plot the curves y- exp(Vx), and y 3.8 ln(1+x) on the same graph in the range 0 x 6. Read off intervals in which there are roots of the equation exp(k)- 3.8 In(1+x) Now find the roots to 6 decimal places using the bisection method (write a program). Print the root found, the value of the function at the root, and the number of iterations required. Use double precision 2. (10 pts) Write a C or Java program to use Newton's method to find a root to an accuracy of 10Use starting values 0.5, 1.6, 1.8, and 3.2, and print all iterations, the roots found, the value of the function at the root, and the number of iterations required. Use double precision 3. (20 pts) We want to compare inverse quadratic interpolation with the results of questions 1 and 2. Using the same function, use inverse quadratic interpolation to find a root between 1.6 and 3.2, printing each iteration. Use starting values x11.6 and x2 3.2. Inverse Quadratic Interpolation x3 0.5* (xi +x2) fit a quadratic x-x(y) through (yl,xl ), (y2.x2), and (y3x3) Use the Lagrange polynomial set y-0 to get x4 discard the xi and xj with the largest fx) I rename the remaining values as x1 and x2 and repeat

Answer 1

clc%clears screen
clear all%clears history
close all%closes all open files
format long
fplot(@(x) exp(sqrt(x)),[0 6],'b');
hold on;
fplot(@(x) 3.8*log(1+x),[0 6],'r');
legend('exp(sqrt(x))','3.8*log(1+x)');

f=@(x) exp(sqrt(x))-3.8*log(1+x);
e=1e-6;
a=[1 2];
b=[2 3];
for i=1:2
    iter(i) = 0;

if f(a(i))*f(b(i))>=0

                disp('No Root')

else

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

                                 iter =iter+ 1;

                                p = (a(i)+b(i))/2;

                                if f(p) == 0

                                                break;

                                end

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

                                                b(i) = p;

                                else

                                                a(i) = p;

                                end

                end

end
    fprintf('Number of Iteration %d, root= %.6f and value of function at root= %.10f\n',iter(i),p,f(p));

end

PROOF: