How can I change the definition of x to find the value of y(3.5112)( i mean y(x1)) for this question ?

Y

Answer 1

[Torsten]

I don't know if it helps:

%%main func
clear all;
close all;
clf;
clc;
% Define the function
x = linspace(-10,10,100);
f = @Goldenfunc;
y = f(x);
figure(1)
plot(x,y,'k',LineWidth=2), hold on
xr = GoldenSection(0,10,f);
plot(xr,'bo')

function [xr] = GoldenSection(xL,xU,y)

N = 200;
GR = (sqrt(5)-1) / 2; % Golden Ratio Number
d = GR * (xU-xL);

x1 = xL + d;
x2 = xU - d;

fx1 = y(x1);           % y(x1) alamıyor çünkü x1 positive integer değil
fx2 = y(x2);
for i=1:N
    
    if fx1 > fx2        % Move lower bound to x1
        xL  = x1;
        fL = fx1;
        
        x1 = x2;        % update x1
        fx1 = fx2;
        d = GR * (xU - xL);  %update d & x2
        x2 = xL + d;
        fx2 = y(x2);  
    elseif fx1 < fx2
        xU = x2;
        fU = fx2;
        x2 = x1;
        fx2 = fx1;
        d = GR * (xU - xL);  %update d & x1
        x1 = xU - d;
        fx1 = y(x1);
    else
        xL = (x1 + x2) / 2;
        xU = xL;
    end
end
xr = (x1 + x2) / 2;
%plot(xr,'ko')
end
function y = Goldenfunc(x)
%x = linspace(0,10,100);

y = x.^2 - 6*x + 15;

%plot(x,y);
%hold on
end

[Torsten]

This would me my suggestion:

%%main func
% Define the function
x = linspace(-10,10,100);
y = f(x);
figure(1)
plot(x,y,'k',LineWidth=2)
hold on
a = 0;
b = 10;
fun = @f;
tol = 1e-8;
x = GoldenSection(a,b,fun,tol);
plot(x,f(x),'o')

function x = GoldenSection(a,b,f,tol)
  GR = (sqrt(5)+1) / 2; % Golden Ratio Number
  c = b - (b - a) / GR;
  d = a + (b - a) / GR;
  while abs(b - a) > tol
    if f(c) < f(d)  % f(c) > f(d) to find the maximum
      b = d;
    else
      a = c;
    end
    % We recompute both c and d here to avoid loss of precision which may lead to incorrect results or infinite loop
    c = b - (b - a) / GR;
    d = a + (b - a) / GR;
  end
  x = (a+b)/2;
end

function y = f(x)
  y = x.^2 - 6*x + 15;
end