bisection method.

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ruth okoh 2011-12-10

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https://ww2.mathworks.cn/matlabcentral/answers/23575-bisection-method

回答： Vidhi Agarwal 2023-6-2

x^3 - 1.6x^2 - 2.4x + 0.3 finding the midpoint through bisection method, using both matcad & mathlab.

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Walter Roberson 2011-12-10

http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

Jan 2011-12-10

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Answer 1

Vidhi Agarwal 2023-6-2

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https://ww2.mathworks.cn/matlabcentral/answers/23575-bisection-method#answer_1249124

在 MATLAB Online 中打开

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