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y=(x.*exp(x)+0.3*x-0.5) ./(x-5);
y_prime=((x-5) .*(x.*exp(x)+exp(x)+0.3)-x.*exp(x)+0.3*x-0.5)) ./(x-5) .^2;
x_intial=-1;
epsilon=10^(-6);
newton(@(x)(x.*exp(x)+0.3*x-0.5) ./(x-5),@(x)((x-5).*exp(x)+exp(x)+0.3)-(x.*exp(x)+0.3*x-0.5)) ./(x-5).^2,-1,10(^6))
function [x]= newton(y,y_prime,x_initial,epsilon) 
    x(-1) =x_initial;
    i = 1;
    while abs(y(x(i))>epsilon)
        x(i+1) = x(i) - y(x(i))/y_prime(x(i));
        i = i+1;
    end
    fprintf('this answer is x=%.4f',x(i))
end

Best Answer

Hello Edward,
Looking at your code, seems like you are trying to solve an euqation using Newton's Mathod. I see many errors in your code. So i have written a new code to solve the equation, you are trying to solve. Have added comments along the lines for you to understand.
syms f(x) x; %Symbolic variables needs to be defined before using them
f(x) = (x*exp(x)+0.3*x-0.5)/(x-5); % Equation you want to solve
g = diff(f); %To find the derivative of function
epsilon=1e-6;%Max allowed deviation
x = zeros();
x(1)= -1 ;%In Matlab index starts from 1, so this is the initial value of x. 
%No need to create another variable for it.
i =1;
while(abs(f(x(i)))>epsilon) %Loop should be stopped when tolerance is reached
        y(i) = double(f(x(i))); %If you need function values too, otherwise, not needed
        x(i+1) = x(i) - f(x(i))/g(x(i));% Calculating values for x iteratively
        i = i+1;
end
Thanks,
Deepak
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