How do you find the intersection points of two functions?

My problem tells me to plot and then find and print the points of intersection for x=[2:7]. I already sought help and they explained that I should use 'find' and then the '==' to find where the output match. Sounded simple enough but I'm getting "Empty Matrix: 1-by-0" as the answer. Any one know what I'm doing wrong or a better way?

Also, any idea how I'd then go about using a "for - while" loop to find every intersection? I'm taking it one problem at a time though.

Here's my script so far:

% clear all windows and variables
clc

% Declare an array for the values of x
x=[2:7];

% Input the given functions
f1=@(x) 90.*exp(-x);
f2=@(x) 3*sin(2*pi*x);

% Graph the functions in the same window
fplot(f1,[2,7],'b')
hold on
fplot(f2,[2,7],'r')
grid on
title('Finding Intersections of Functions')
xlabel('Input Values (x)')
ylabel('Ouput Values (f)')

% Find the first intersection
f1a=90.*exp(-x);
f2a=3*sin(2*pi*x);
find(f1a==f2a)

[ankit kumar]

is there anyway i can give a txt file including x and y axis value and can find how many times a curve is crossing through a threshold and also to note down the x values.

[Elina Nikolopoulou]

How do i keep the Y and the X coordinates of the intersections in separate arrays (one for the X and one for the Y coordinates) ???

Answer 1

[Andrei Bobrov]

EDIT

f1=@(x) 90.*exp(-x);
f2=@(x) 3*sin(2*pi*x);
f = @(x)f1(x)-f2(x);

xx = 2:.1:7;

t = f(xx) > 0;
i0 = find(diff(t(:))~=0);
i0 = [i0(:)';i0(:)'+1];
n = size(i0,2);
xout = zeros(n,1);
for jj = 1:n
    xout(jj) = fzero(f,xx(i0(:,jj))); 
end

[Mathijs Frenken]

Thank you for your answer. However, you realize you are not the only one that reads your code right? It's actually unreadable.

[Jaden Delgado]

lmao

Answer 2

[Alexander Efremov]

There is a fancy contribution on "File Exchange" which appeared in the Pick of the week. This should help.

[Elina Nikolopoulou]

"intersections" does not work in the version i use :(

Answer 3

[Syed Riza]

Hello Dan Teep; you can find your answer in this modified code

% Declare an array for the values of x
x=linspace(2,7,1200);
% Input the given functions
f1=@(x) 90.*exp(-x);
f2=@(x) 3*sin(2*pi*x);
% Graph the functions in the same window
fplot(f1,[2,7],'b')
hold on
fplot(f2,[2,7],'r')
grid on
title('Finding Intersections of Functions')
xlabel('Input Values (x)')
ylabel('Ouput Values (f)')
% Find the x-cordinates of intersecting points
f1a=90.*exp(-x);
f2a=3*sin(2*pi*x);
Intersections=find(abs(f1a-f2a)<=(0.05));
X_Values=x(Intersections)

[Maya Priveetra]

Hi can I know why you (abs(fla-f2a)<=(0.05))

Like what does it do especially the 0.05

[Adeel Yousuf]

He has used a tolerance value of 0.05 as the dataset contains floating point values. We can keep tolerance as small as it seems suitable (like: 1e-4). It's meant to "detect" the closeness of 2 floating point values. Moreover in simple words, we need to make MATLAB determine if 60.2745 is the same as 60.274 or not? So for that we can use threshold/tolerance of 0.001.