Newton Raphson Method and Bisection Method

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Irem Tas
Irem Tas 2021-12-7
回答： Irem Tas 2021-12-7
f(x)=114.94253x^2-1.31705x^3-0.00436522x^4-4.72276*10^4
I need to write codes for this function by applying Newton Raphson Method and Bisection Method.
For Bisection Method: a=0 b=48 error=0.0000001
For Newton-Raphson Method: x1=24 error=0.0000001

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回答（1 个）

Irem Tas
Irem Tas 2021-12-7
% Clearing Screen
clc
% Input Section
y = 114.94253*(x^2)-1.31705*(x^3)-0.00436522*(x^4)-(4.72276*10^4);
a = 0;
b = 48;
e = 0.0000001;
% Finding Functional Value
fa = eval(subs(y,x,a));
fb = eval(subs(y,x,b));
% Implementing Bisection Method
if fa*fb > 0
disp('Given initial values do not bracket the root.');
else
c = (a+b)/2;
fc = eval(subs(y,x,c));
fprintf('\n\na\t\t\tb\t\t\tc\t\t\tf(c)\n');
while abs(fc)>e
fprintf('%f\t%f\t%f\t%f\n',a,b,c,fc);
if fa*fc< 0
b =c;
else
a =c;
end
c = (a+b)/2;
fc = eval(subs(y,x,c));
end
fprintf('\nRoot is: %f\n', c);
end
I tried this code for Bisection Method but I couldnt result.

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