Golden Search Optimization Problem

Hello everyone! I've been trying to figure out how to code the Golden Search algorithm for the function G with an initial interval of [-2,4]. However, I've been getting odd answers when I compared it with Newton's method. I was wondering if there was anything wrong with my code.

G = 4*x - (1.8*x^2) + (1.2*x^3) - (.3*x^4)

%Golden search method

%Define the parameters
phi = (1 + sqrt(5))/2;
a = -2;
b = 4;
n = 0;
c1 = (phi - 1)*a + (2-phi)*b;
c2 = (2-phi)*a + (phi - 1)*b;
G = @(x)(4*x - (1.8*x^2) + (1.2*x^3) - (.3*x^4));
g1 = G(c1);
g2 = G(c2);

while abs(b-a) > 0.0001
  n = n+1;
    if g1 < g2
      b = c2;
      c2 = c1;
      g2 = g1;
      c1 = (phi - 1)*a + (2 - phi)*b;
      g1 = G(c1);
    else
      a = c1;
      c1 = c2;
      g1 = g2;
      c2 = (2 - phi)*a + (phi - 1)*b;
      f2 = G(c2);
    end
end

for a comparison, I wrote a simple newton's method algorithm.

%Initial guess
x = 3;

%Desired tolerance and dummy tolerance to start iteration
Tol = 0.0001;
dummytol = 100;

while dummytol > Tol
    X = x;
    G = 4*x - (1.8*x^2) + (1.2*x^3) - (.3*x^4);
    %Use derivative of the function to find the new volume
    dG = 4 - 3.6*x + 3.6*x^2 - 1.2*x^3;
    x = x - G/dG;
    %Replace dummy tolerance with new tolerance
    dummytol = abs(x - X);
end;

I got a maximum of 3.39 for newtons and -2 for the golden search. Thank you for any help!

[SERGIO RODRIGUEZ]

You implement a minimization problem with the golden search....The result its reasonable.

[SERGIO RODRIGUEZ]

And.. phi = (-1 + sqrt(5))/2;

[Aric Acius]

Might be 5 years late, but in your else statement you calculate f2 instead of g2.