How can I code this Newton-Raphson and Secant Methods?

Determine the root of the given equation (3*x*e^x)-1=0 for x ∈ [0,1] using Newton-Raphson and Secant methods. Stop calculation after seven iterations. I've coded a code but I'm not sure it is right or not...

%f is non-linear function.
%fder is derivation of f.
%x0 is initial value of iteration.
%e is tolerance of f function.
%d is first condition tolerance value.
%y is error amount.
%max is maximum iteration number.
syms x
x0 = 0;
x1 = 1;
d = 0.000001;
e = 0.00001;
max = 7;
f = @(x) 3*x*exp(x)-1;
fder = @(x) 3*(exp(x)+exp(x))*x;
    for iteration = 0:max
        t = fder(x0); %to check the derivation is zero or not.
        if abs(t)<e
            disp('Derivation value is very close to zero, algorithm stops')
            break
        end
        x1 = x0-(f(x0)/fder(x0));
        error = abs(x1-x0);
        odderror = 2*error/(abs(x1)+d);
        x0 = x1;
        y = f(x0);
        if(error<d)||(odderror<d)||(abs(y)<e)
            break
        end
    end
    disp('Wanted root value')
    x1
    disp('Error amount')
    y
    disp('Iteration number')
    iteration

[Muhammad Raihan Ekaputra Idrisatria]

Use this function that made by me https://www.mathworks.com/matlabcentral/fileexchange/81758-newton-raphson-secant-method

I mixed Newton-Raphson and backward euler to solve the derrivation using numeric method

Answer 1

[David Hill]

The basic code is as follows. You can do other checks as desired.Your code looked ok, except you don't need symbolic variables and your fder function was missing a ')'.

x=0;
f = @(x) 3*x.*exp(x)-1;
fder = @(x) 3*(exp(x)+exp(x).*x);
for k=1:7%newton method
    x=x-f(x)/fder(x);
end
y=[0,1];
for k=1:7%secant method
    z=y(2);
    y(2)=y(2)-f(y(2))*diff(y)/diff(f(y));
    y(1)=z;
end

[Umut Berkcan Karahan]

So thanks David.

[Muhammad Raihan Ekaputra Idrisatria]

better you add an if condition, if f(x0)=0 then x=x0. To avoid a division by 0