Anyone could explain this matlab code step by step and clearly, im new in matlab

Question

Anyone could explain this matlab code step by step and clearly, im new in matlab, thanks in advance!

%% D1

p = [1 3 -87 -307 822 2160];
x = roots(p);
y = polyval(p, x);

%% D2

p = poly([-5 -4 2 8])
x = linspace(-50, 50)

plot(x, polyval(p, x))
axis([-50, 50, -750, 750])

%% D3

f = @(t) 4*sin(3*t) + 3*exp(1).^(-0.5*t);%. mul
S = integral(f, 1.4, 3.1)

%% D4

h2 = @(x) cos(exp(1).^x)./(1 - x)
min = fminbnd(h2, 2, 3)
y = h2(min)

fprintf('Min: x = %d g(x) = %d ', min, y)

x = linspace(2, 3);
plot(x, h2(x))
grid on;

%% D5

h = @(x) 3*exp(1).^(-x.^2) - 9*x.^2 + x;
h2 = @(x) -h(x);

%fplot(h, [-10 10])

x_roots = [fzero(h, 0) fzero(h, 0.2)];
y_roots = h2(x_roots);

max = fminbnd(h2, -50, 50);
ymax = h(max);

%% D6

rows = 7;
M = eye(rows) + 5*diag(ones(rows-1, 1), 1) + 5*diag(ones(rows-1, 1), -1)

Explanation / Answer

%% D1

p = [1 3 -87 -307 822 2160];
x = roots(p);%this command returns the roots of a polynomial whose coefficients are described in descending order of power in p
%so for given vector p, the polynomial will be x^5+3*x^4-87*x^3-307x^2+822*x+2160
y = polyval(p, x);%polyval evaluates the polynomial whose coefficients are in p, at points specified in x
%since x stores the roots of this polynomial, the result in y should be 0
%% D2

p = poly([-5 -4 2 8])%poly returns the coefficients of a polynomial whose roots are in the vector passed to it
%so p will hold the coefficients of a polynomial whose roots are -5 -4 2 8
x = linspace(-50, 50)
%linspace creates 100 values from-50 to 50
plot(x, polyval(p, x))%this plots the x vector on x axis and the values of polynomial p evealuated at points described in x
axis([-50, 50, -750, 750])%this command controls the range to be displayed on the plot
%x axis will be from -50 to 50, y axis will be from -750 to 750
%% D3

f = @(t) 4*sin(3*t) + 3*exp(1).^(-0.5*t);%this creates a function f according to the equation provided
%if we want to evaluate it at any point x, we type f(x) and the value is calculated
S = integral(f, 1.4, 3.1)%this finds the integral of function f from 1.4 to 3.1

%% D4

h2 = @(x) cos(exp(1).^x)./(1 - x)%a function created, like above
min = fminbnd(h2, 2, 3)% this returns the x value at which the function returns minimum value between 2 and 3
y = h2(min)%the value of function at that point

fprintf('Min: x = %d g(x) = %d ', min, y)%this line displays the minimum point and the value at that point
%the %d and %d are placeholders for variables, which are then specified to
%be min and y, so main and y will be displayed in place of them, rest all
%will be printed as shown
x = linspace(2, 3);%100 points between 2 and 3 created
plot(x, h2(x))%function h2 plotted from 2 to 3
grid on;% this command shows gridlines on plot

%% D5

h = @(x) 3*exp(1).^(-x.^2) - 9*x.^2 + x;%function created
h2 = @(x) -h(x);%this function is the negatve of above function

%fplot(h, [-10 10])%this line plots the function h from x=-10 to 10

x_roots = [fzero(h, 0) fzero(h, 0.2)];%this is a 2 element vector of roots of h around 0 and around .2
y_roots = h2(x_roots);%value of function h at points in x_roots

max = fminbnd(h2, -50, 50);%minimum value between -50 to 50, since it is of -h, for h this will be maximum value
ymax = h(max);%value at that point

%% D6

rows = 7;
M = eye(rows) + 5*diag(ones(rows-1, 1), 1) + 5*diag(ones(rows-1, 1), -1)
%eye(rows) returns an identity matrix of rows rows
%diag(r) returns a diagonal matrix with the elements at diagonal those specified in r
%if 1 is passed like diag(r,1) then the diagonal above the main diagonal is
%filled, and for -1 the diagonal below the main diagonal is filled

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