isection(f,[1 1.5], %3D
Not a graded question
![For example: For finding the cube root of 2, the equation to be solved is x^3 = 2
>> f=e (x) x.^3-2; % root finding function
>>
>> bisection (f, [1 1.5], 0.05)
ans =
1.2603
>> bisection (f, [0 1], 0.05)
Error using bisection (line 13)
Use a bracketing interval
Use the above examples as two sample cases and publish & upload the pdf on Blackboard.
(b) Determine the two roots of f(x) = 2x cos(2x) – (x – 2)² accurate to 4 sig figs.
Write a script that will use your bisection.m written in subpart (a). Use fprintf to print the results to
screen and plot these solution points on the graph with red star markers. Publish to generate pdf and submit on
Blackboard.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e366a85-c23b-450a-9bc7-d2ea0be52d3f%2Fc56e9d41-ff43-4c63-bad3-b75af4d3af12%2Fe2iqjc_processed.png&w=3840&q=75)
The required function is given below.
function c= bisection(f,interval,es)
a = interval(1);
b = interval(2);
if f(a)*f(b) >= 0
error('Use a bracketing interval')
end
c = mean([a,b]);
err = 1;
while err > es
if f(a)*f(c) < 0
b = c;
else
a = c;
end
c = mean([a,b]);
err = abs(c-b);
end
end
Output
>> f = @(x) x.^3-2;
>> bisection(f,[1,1.5],0.05)
ans =
1.2812
>> bisection(f,[0,1],0.05)
Error using bisection (line 5)
Use a bracketing interval
clear; clc; close all;
% Description: Calculates the root of a given function using bisection
% method
f = @(x) 2*x.*cos(2*x) - (x-2).^2;
fplot(f);
set(gca,'XAxisLocation','origin')
set(gca,'YAxisLocation','origin')
xlim([0 4])
ylim([-5 5])
xlabel('x')
ylabel('f(x)')
grid on
c = bisection(f,[2,3],1e-6);
hold on
plot(c,f(c),'rx')
fprintf('Root: %.4f\n',c)
c = bisection(f,[3,4],1e-6);
plot(c,f(c),'rx')
fprintf('Root: %.4f\n',c)
Output
Root: 2.3707
Root: 3.7221
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