Finding the maxima/minima of a function.

My lab TA assigned a small project to find and plot the absolute value of the maxima and minima of a given function. I made a quick and dirty algorithm to do so, but I'm having two issues with it and I simply cannot figure out what is wrong with my code.

1.) For some reason, it isn't grabbing the actual highest value, but rather one close to it.

2.) The max/min plot dips down to nearly zero and then climbs up steadily with the Absolute valued function.

I know there are other ways of doing it, including using the derivative of the function, but I would much rather assistance in finding out what is incorrect in my algorithm, which tests surrounding points in order to find maxima and minima.

Here is the code:

________________

%This program plots the abs val of the maxima and minima of a function.
%This max/min value will then continue to be plotted until a new maxima or
%minima is found.
clear,clc;

t=[0:0.1:20];   

y=exp(-t).*sin(pi./2.*t);  %Our Function

yAbs=abs(y);               %Take the absolute value of the function.

yMaxMin=zeros(201);        %Create an array of zeros to be filled w/ data.

for(i=2:200)

%If a point is a maxima in yAbs, it will be a maxima or a minima in y.

if(yAbs(i-1)<yAbs(i)&&yAbs(i+1)>yAbs(i))

      plotPoint=yAbs(i); %If the value is a max, store it in plotPoint.

end

    yMaxMin(i)=plotPoint;  %Store the value of plotPoint into yMaxMin(i)

end    

plot(t,yAbs);

hold on

plot(t,yMaxMin);

[Muhammad Sheraz Ali]

Question: "How do we find the maxima and minima of a continuous function ff on a closed and bounded interval [a,b]?[a,b]?" • Find the derivative 𝑓′(𝑥) of �

• Find all critical values of 𝑓, let's say 𝑐1, 𝑐2, …, 𝑐𝑘. • Compute 𝑓(𝑎) and 𝑓(𝑏), together with 𝑓(𝑐1),𝑓(𝑐2),…,𝑓(𝑐𝑘). • Compare (order) the values in step 3. • Then the minimum value is global minima of the values in step 4 and the maximum is the global maxima.

[KADALI]

find 1st and 2nd order derivatives matlab of f(X) = x^3+3xy^2-15x^2-15y^2+72x

Answer 1

[Azzi Abdelmalek]

You can use findpeaks function

[Christopher]

Thank you. I didn't realize that function existed. I rewrote my code and it's now functioning properly!

Answer 2

[Abhay Chaudhary]

%This program plots the abs val of the maxima and minima of a function.

%This max/min value will then continue to be plotted until a new maxima or

%minima is found.

clear,clc;

t=[0:0.1:20];

y=exp(-t).*sin(pi./2.*t); %Our Function

yAbs=abs(y); %Take the absolute value of the function.

yMaxMin=zeros(201); %Create an array of zeros to be filled w/ data.

for(i=2:200)

%If a point is a maxima in yAbs, it will be a maxima or a minima in y.

if(yAbs(i-1)<yAbs(i)&&yAbs(i+1)>yAbs(i))

plotPoint=yAbs(i); %If the value is a max, store it in plotPoint.

end

yMaxMin(i)=plotPoint; %Store the value of plotPoint into yMaxMin(i)

end

plot(t,yAbs);

hold on

plot(t,yMaxMin);

[VEMIREDDY]

Maxima and minima of f(x)=x^4-3x^2y^2+4x+y^4

Answer 3

[Jaideep Banoth]

if true
  % code
end

Answer 4

[Ruthika j]

find the critical points maxima and maxima of the function f(x) = 2x^3-9x^2+12x-3

Answer 5

[Dinesh C]

x*4+y*4-x*2-x*2+1