Find the local and global maxima and minima for the function (x+1)^2/(1+x^2) x belongs to [-5,5].Showing local min value something different than zero and how get local values

clear
clc
syms x
f(x)=(x+1)^2/(1+x^2);
I=[-5,5];
fl(x)=-f(x);
a=I(1);b=I(2);
t=linspace(a,b,10000);
g=double(f(t));
[lmax_f,loc]=findpeaks(g);
lmax_x=round(t(loc),4);
h=double(fl(t));
[lmin_f,loc]=findpeaks(h);
lmin_x=round(t(loc),4);
disp('global maximum occurs at x=')
disp(lmax_x)
disp('the local maximum value(s) of the function are')
disp(double(f(lmax_x)))
disp('global minimum occurs at x=')
disp(lmin_x)
disp('the local minimum value(s) of the function are')
disp(double(f(lmin_x)))
plot(t,f(t));hold on;
plot(lmax_x,double(f(lmax_x)),'or');
plot(lmin_x,double(f(lmin_x)),'*g');
hold off;

Answer 1

[Sam Chak]

@Aashay, your code works. So, what exactly is the question for the math problem?

syms x
f(x)=(x+1)^2/(1+x^2);
I=[-5,5];
fl(x)=-f(x);
a=I(1);b=I(2);
t=linspace(a,b,10000);
g=double(f(t));
[lmax_f,loc]=findpeaks(g);
lmax_x=round(t(loc),4);
h=double(fl(t));
[lmin_f,loc]=findpeaks(h);
lmin_x=round(t(loc),4);
disp('global maximum occurs at x=')

global maximum occurs at x=

disp(lmax_x)

    0.9996

disp('the local maximum value(s) of the function are')

the local maximum value(s) of the function are

disp(double(f(lmax_x)))

    2.0000

disp('global minimum occurs at x=')

global minimum occurs at x=

disp(lmin_x)

   -0.9996

disp('the local minimum value(s) of the function are')

the local minimum value(s) of the function are

disp(double(f(lmin_x)))

   8.0032e-08

plot(t,f(t));hold on;
plot(lmax_x,double(f(lmax_x)),'or');
plot(lmin_x,double(f(lmin_x)),'*g');
hold off;

[Aashay]

The local min value is showing wrong here

[Aashay]

8.0032e-08 why its showing this rather than 0