discrete fixed point and their stability

What is the rong with my code?

there is an error message "Conversion to logical from sym is not possible."

clc;
clear all;
syms xn xe
assume(0<xn<=1)
r=input('Enter the value of production rate','s')
f(xn)=r*xn*(1-xn);
eq=xe==subs(f,xn,xe)

display('The equilibrium points are:')
xe=solve(eq)

df=diff(r*xn*(1-xn),xn)
eq1=subs(df,xn,xe);
t=length(eq1);
eq1=double(eq1);

for i=1:t
if eq(i)>1;
    fprintf('The equilibrium point %g is unstable',eq(i))
elseif eq(i)<-1;
    fprintf('The equilibrium point %g is unstable',eq(i))
elseif 0<eq(i)&&eq(i)<1
      fprintf('The equilibrium point %g is stable',eq(i))
else -1<eq(i)&&eq(i)<0
      fprintf('The equilibrium point %g is stable',eq(i))
end
end

Answer 1

[madhan ravi]

eq1(i)

Answer 2

[Walter Roberson]

for r = -2:0.25:5
    syms xn xe
    assume(0<xn<=1)
    
    f(xn)=r*xn*(1-xn);
    eq=xe==subs(f,xn,xe);
    
    fprintf('r = %g, The equilibrium points are:\n', r)
    xe=solve(eq);
    disp(char(xe))
    
    df = diff(r*xn*(1-xn),xn);
    eq1 = simplify(subs(df,xn,xe));
    t = length(eq1);
    eq1 = double(eq1);
  
    for i=1:t
        if eq1(i)>1
            fprintf('r = %g, The equilibrium point %g is unstable',r,eq1(i));
        elseif eq1(i)<-1
            fprintf('r = %g, The equilibrium point %g is unstable',r,eq1(i));
        elseif 0<eq1(i)&&eq1(i)<1
            fprintf('r = %g, The equilibrium point %g is stable',r,eq1(i));
        elseif -1<eq1(i)&&eq1(i)<0
            fprintf('r = %g, The equilibrium point %g is stable',r,eq1(i));
        else
            fprintf('r = %g, failed to classify %g\n', r, eq1(i));
        end
    end
    fprintf('\n-----\n');
end

r = -2, The equilibrium points are:

[0; 3/2]

r = -2, The equilibrium point -2 is unstable

r = -2, The equilibrium point 4 is unstable

-----

r = -1.75, The equilibrium points are:

[0; 11/7]

r = -1.75, The equilibrium point -1.75 is unstable

r = -1.75, The equilibrium point 3.75 is unstable

-----

r = -1.5, The equilibrium points are:

[0; 5/3]

r = -1.5, The equilibrium point -1.5 is unstable

r = -1.5, The equilibrium point 3.5 is unstable

-----

r = -1.25, The equilibrium points are:

[0; 9/5]

r = -1.25, The equilibrium point -1.25 is unstable

r = -1.25, The equilibrium point 3.25 is unstable

-----

r = -1, The equilibrium points are:

[0; 2]

r = -1, failed to classify -1

r = -1, The equilibrium point 3 is unstable

-----

r = -0.75, The equilibrium points are:

[0; 7/3]

r = -0.75, The equilibrium point -0.75 is stable

r = -0.75, The equilibrium point 2.75 is unstable

-----

r = -0.5, The equilibrium points are:

[0; 3]

r = -0.5, The equilibrium point -0.5 is stable

r = -0.5, The equilibrium point 2.5 is unstable

-----

r = -0.25, The equilibrium points are:

[0; 5]

r = -0.25, The equilibrium point -0.25 is stable

r = -0.25, The equilibrium point 2.25 is unstable

-----

r = 0, The equilibrium points are:

0

r = 0, failed to classify 0

-----

r = 0.25, The equilibrium points are:

[-3; 0]

r = 0.25, The equilibrium point 1.75 is unstable

r = 0.25, The equilibrium point 0.25 is stable

-----

r = 0.5, The equilibrium points are:

[-1; 0]

r = 0.5, The equilibrium point 1.5 is unstable

r = 0.5, The equilibrium point 0.5 is stable

-----

r = 0.75, The equilibrium points are:

[-1/3; 0]

r = 0.75, The equilibrium point 1.25 is unstable

r = 0.75, The equilibrium point 0.75 is stable

-----

r = 1, The equilibrium points are:

[0; 0]

r = 1, failed to classify 1
r = 1, failed to classify 1

-----

r = 1.25, The equilibrium points are:

[0; 1/5]

r = 1.25, The equilibrium point 1.25 is unstable

r = 1.25, The equilibrium point 0.75 is stable

-----

r = 1.5, The equilibrium points are:

[0; 1/3]

r = 1.5, The equilibrium point 1.5 is unstable

r = 1.5, The equilibrium point 0.5 is stable

-----

r = 1.75, The equilibrium points are:

[0; 3/7]

r = 1.75, The equilibrium point 1.75 is unstable

r = 1.75, The equilibrium point 0.25 is stable

-----

r = 2, The equilibrium points are:

[0; 1/2]

r = 2, The equilibrium point 2 is unstable

r = 2, failed to classify 0

-----

r = 2.25, The equilibrium points are:

[0; 5/9]

r = 2.25, The equilibrium point 2.25 is unstable

r = 2.25, The equilibrium point -0.25 is stable

-----

r = 2.5, The equilibrium points are:

[0; 3/5]

r = 2.5, The equilibrium point 2.5 is unstable

r = 2.5, The equilibrium point -0.5 is stable

-----

r = 2.75, The equilibrium points are:

[0; 7/11]

r = 2.75, The equilibrium point 2.75 is unstable

r = 2.75, The equilibrium point -0.75 is stable

-----

r = 3, The equilibrium points are:

[0; 2/3]

r = 3, The equilibrium point 3 is unstable

r = 3, failed to classify -1

-----

r = 3.25, The equilibrium points are:

[0; 9/13]

r = 3.25, The equilibrium point 3.25 is unstable

r = 3.25, The equilibrium point -1.25 is unstable

-----

r = 3.5, The equilibrium points are:

[0; 5/7]

r = 3.5, The equilibrium point 3.5 is unstable

r = 3.5, The equilibrium point -1.5 is unstable

-----

r = 3.75, The equilibrium points are:

[0; 11/15]

r = 3.75, The equilibrium point 3.75 is unstable

r = 3.75, The equilibrium point -1.75 is unstable

-----

r = 4, The equilibrium points are:

[0; 3/4]

r = 4, The equilibrium point 4 is unstable

r = 4, The equilibrium point -2 is unstable

-----

r = 4.25, The equilibrium points are:

[0; 13/17]

r = 4.25, The equilibrium point 4.25 is unstable

r = 4.25, The equilibrium point -2.25 is unstable

-----

r = 4.5, The equilibrium points are:

[0; 7/9]

r = 4.5, The equilibrium point 4.5 is unstable

r = 4.5, The equilibrium point -2.5 is unstable

-----

r = 4.75, The equilibrium points are:

[0; 15/19]

r = 4.75, The equilibrium point 4.75 is unstable

r = 4.75, The equilibrium point -2.75 is unstable

-----

r = 5, The equilibrium points are:

[0; 4/5]

r = 5, The equilibrium point 5 is unstable

r = 5, The equilibrium point -3 is unstable

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