How to plot a function ?

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Alexandru Stefan
Alexandru Stefan on 17 Mar 2014
Commented: Alexandru Stefan on 19 Mar 2014
Helli, i'm a newbie in mathlab and i try for hours to make the program below, please would you explained me how to begin or make the program and write the code to see where i make some mistake.
Write a program that recieve as arguments two real numbers a and b with a>0 and b∈(0,1) and who represent graphic in cartesian coordinate function f:[0,2]->R.
{ (a*x)/(x^2+b), x[0,1]
f(x)= {
{ ln(x^2-3*x+3), x(1,2]
with dashed line and black color. The program display an error message if the requirements about arguments are not respected.
Thank You!!

Accepted Answer

Carlos
Carlos on 17 Mar 2014
Edited: Carlos on 17 Mar 2014
Some ideas here
function y=new(a,b)
if(a<0)
disp('a must be bigger than zero');
end
if(b<0||b>1)
disp('a must be bigger than zero and smaller than 1');
end
x=0:0.1:2;
y=zeros(1,length(x));
for k=1:length(x)
if (x(k)<1)
y(k)=(a*x(k))/(x(k)^2+b);
else
y(k)=log(x(k)^2-3*x(k)+3);
end
end
plot(x,y)
  3 Comments
Carlos
Carlos on 17 Mar 2014
My bad, you are right.
Alexandru Stefan
Alexandru Stefan on 19 Mar 2014
Thank you, both were really helpful.

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More Answers (1)

Niklas Nylén
Niklas Nylén on 17 Mar 2014
function fx = myFunction(a,b)
x=0:0.1:2; % Create the x vector with step size 0.1
fx = zeros(size(x)); % Initialize the function output fx with zeros
% Make the calculation for when x<=1
index = x <= 1; % Creates a logical vector with 0's and 1's
% Calculate f(x) for the indices of x which fulfill the previous logical statement
fx(index) = (a.*x(index))./(x(index).^2+b);
% Use the second equation when x > 1
index = x > 1;
fx(index) = log(x(index).^2-3.*x(index)+3);
figure;
plot(x,fx)
  2 Comments
Alexandru Stefan
Alexandru Stefan on 17 Mar 2014
Thank You was really helpful, one more question how to make the program show an error if the requirements about arguments are not respected.
Carlos
Carlos on 17 Mar 2014
I have edited my previous code including the imput argument control you requested

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