Plotting of different nonlinear equations in different ranges segment wise in a single graph

Can anyone help me to plot different nonlinear equations in different ranges segmentwise in a single graph? (using fsolve, for loop, if else)

For Example:

for the range 0 < x < 0.3, plot y^2 - 2*(x+y)

for the range 0.3 < x < 0.7, plot y^2 - 2*(x+y) + 4*x*y

for the range 0.7 < x < 2. plot y^2 - 2*(x+y) + 4*x*x

I think there is some mistake in the following code: (plot (A,Y).. A:X axis, Y : Y axis)

A = 0:0.01:2;
b = 0.3;
c = 0.7;
y0 = 0;
for i = numel(A)
    x = A(i);
    if  (A(i)<=c)
        if (A(i)<b)
            fun = @(y) y^2 - 2*(x+y);
            Y(i) = fsolve(fun,y0);
            y0 = Y(i);
        elseif (A(i)>b)
            fun = @(y) y^2 - 2*(x+y) + 4*x*y;
            Y(i) = fsolve(fun,y0);
            y0 = Y(i);  
        end
    else if (A(i)>c)
            fun = @(y) y^2 - 2*(x+y) + 4*x*x;
            Y(i) = fsolve(fun,y0);
            y0 = Y(i);   
    end
    end
end
plot (A,Y);

Answer 1

[VBBV]

A = 0:0.1:2;
b = 0.3;
c = 0.7;
y0 = 0;
% y = 1:10;
for i = 1: numel(A)
   x = A(i);
    if  (A(i)<=c) & (A(i)<b)
         
            fun = @(y) y^2 - 2*(x+y);
            Y(i) = fsolve(fun,y0);
            y0 = Y(i);
        elseif (A(i)>b)
             
            fun = @(y) y^2 - 2*(x+y) + 4*x*y;
            Y(i) = fsolve(fun,y0);
            y0 = Y(i); 
        end
    if (A(i)>c)
         
            fun = @(y) y^2 - 2*(x+y) + 4*x*x;
            Y(i) = fsolve(fun,y0);
            y0 = Y(i);   
    end
   
end

Equation solved at initial point.

fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

Equation solved at initial point.

fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

plot(A,Y)

%plot (A,Y);

[Supriya Khatoniar]

Got my mistake. Many thanks.

Answer 2

[Torsten]

x = 0:0.01:2;
b = 0.3;
c = 0.7;
y = zeros(size(x));
idx = x>=0 & x<=0.3;
y(idx) = 1 + sqrt(1+2*x(idx));                            % or y(idx) = 1-sqrt(1+2*x(idx));
idx = x>0.3 & x<0.7;
y(idx) = (1-2*x(idx)) + sqrt(4*(x(idx)).^2-2*x(idx)+1);   % or y(idx) = (1-2*x(idx))-sqrt(4*(x(idx)).^2-2*x(idx)+1);
idx = x>=0.7 & x<=2;          
y(idx) = 1 + sqrt(1+2*x(idx)-4*(x(idx)).^2);              % or y(idx) = 1 - sqrt(1+2*x(idx)-4*(x(idx)).^2);
                                                          % sqrt has no real
                                                          % roots in this
                                                          % interval for x
plot(x,y)

Choose the branches of the sqrt you want to plot. The argument of the sqrt in the last expression is negative and thus gives a complex root.

[Supriya Khatoniar]

Sure, many thanks!