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bisection method.

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ruth okoh
ruth okoh on 10 Dec 2011
Answered: Vidhi Agarwal on 2 Jun 2023
x^3 - 1.6x^2 - 2.4x + 0.3 finding the midpoint through bisection method, using both matcad & mathlab.
  2 Comments
Walter Roberson
Walter Roberson on 10 Dec 2011
http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer
Jan
Jan on 10 Dec 2011
Dear ruth okoh, do you have a question?

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Answers (1)

Vidhi Agarwal
Vidhi Agarwal on 2 Jun 2023
Please find the attched Code for the following question
% Define the function f(x)
f = @(x) x^3 - 1.6*x^2 - 2.4*x + 0.3;
% Define the interval [a,b]
a = -1;
b = 3;
% Define the tolerance (the maximum error allowed)
tol = 1e-6;
% Set the maximum number of iterations
max_iter = 1000;
% Initialize the variables
iter = 0;
midpoint = (a + b) / 2;
fa = f(a);
fb = f(b);
% Use a while loop to iteratively refine the midpoint
while abs(f(midpoint)) > tol && iter < max_iter
midpoint = (a + b) / 2;
fm = f(midpoint);
if fm == 0
break;
elseif sign(fm) == sign(fa)
a = midpoint;
fa = fm;
else
b = midpoint;
fb = fm;
end
iter = iter + 1;
end
% Display the results
fprintf('The midpoint of the function f(x) = x^3 - 1.6x^2 - 2.4x + 0.3 is: %f\n', midpoint);
The midpoint of the function f(x) = x^3 - 1.6x^2 - 2.4x + 0.3 is: 0.116597
This code provide the solution of equation f by bisection method.
if you want to find olution of more equation by bisection method you can just change the function f.

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