What is the function to use to find the intersection of two graphs?

I tried browsing for the right function and came across interx and intersection, but somehow they are not available for use in my MATLAB program. Can someone help me with the issue?

f = @(x,y) y-2 - x.^2 .* (cos(x));
fimplicit(f), grid on
hold on
g = @(x,y) y-x.^4-x-1;
fimplicit(g), grid on
hold off
xlabel('x'), ylabel('y')
legend ('y = 2 + cos(x)','y = x^4 + x + 1','location','northeast')

[Sam Chak]

Hi @Lê

Have you successfully solved the problem you opened yesterday? If you have, please consider closing the issue. If you are still facing issues in MATLAB, please post your code so that I can take a look.

https://www.mathworks.com/matlabcentral/answers/2048852-how-can-i-draw-graph-of-y-2-x-2-x-3?s_tid=srchtitle

[Stephen23]

"I tried browsing for the right function and came across interx and intersection, but somehow they are not available for use in my MATLAB program."

INTERX is not an inbuilt MATLAB function, you have to download it here:

https://www.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections

[Lê]

@Stephen23 Where do I place the INTERX file if I may ask? Also @Sam Chak I figured it out on my own.

[Stephen23]

"Where do I place the INTERX file if I may ask?"

It is exactly like your own MATLAB code files: either in the current directory or somewhere on the MATLAB search path (of course avoiding the installation folders of any application, including MATLAB).

Answer 1

[Sam Chak]

Hi @Lê

It is possible to find intersections using the fsolve() command.

f = @(x,y) y - 2 - (x.^2).*cos(x);
fimplicit(f), grid on
hold on

g = @(x,y) y - x.^4 - x - 1;
fimplicit(g), grid on
hold off

xlabel('x'), ylabel('y')
legend('y = 2 + x^{2} cos(x)','y = x^4 + x + 1','location','northeast')

%% Finding intersection points
x0  = [-1, 1];  % initial guesses
for j = 1:2
    fun = @(x) x^4 + x + 1 - 2 - (x.^2).*cos(x);    % f(x) = g(x)
    x   = fsolve(fun, x0(j))
    y   = x.^4 + x + 1
end

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

x = -1.2879

y = 2.4630

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

x = 0.8843

y = 2.4956

[Lê]

I want to calculate the volume of the region bounded by these curves, so what should I do to print the value of x min and x max on the workspace, so I can use integral to calculate the volume? Because when I ran the script, x min and x max are both 0.8843, so the volume is 0, while it should be x min = -1,2879 and x max = 0.8843.

clear
%% Plotting the graphs
f = @(x,y) y-2 - x.^2 .* (cos(x));
fimplicit(f), grid on
hold on
g = @(x,y) y-x.^4-x-1;
fimplicit(g), grid on
hold off
xlabel('x'), ylabel('y')
legend ('y = 2 + cos(x)','y = x^4 + x + 1','location','northeast')
%% Finding intersection points
x0  = [-1, 1];  % initial guesses
for j = 1:2
    fun1 = @(x) x^4 + x + 1 - 2 - (x.^2).*cos(x);    % f(x) = g(x)
    x   = fsolve(fun1, x0(j))
%% Calculating the volume
pi = 3.14;
fun2 = @(x) (2 + x.^2 * cos(x)).^2 -(x.^4 + x + 1).^2;
V = pi*integral (fun2,x,x);
end

[Sam Chak]

Hi @Lê

Maybe when calculating the volume, it should be placed outside the loop.

%% Plotting the graphs
f = @(x,y) y-2 - x.^2 .* (cos(x));
fimplicit(f), grid on
hold on
g = @(x,y) y-x.^4-x-1;
fimplicit(g), grid on
hold off
xlabel('x'), ylabel('y')
legend ('y = 2 + cos(x)','y = x^4 + x + 1','location','northeast')

%% Finding intersection points
x0  = [-1, 1];  % initial guesses
for j = 1:2
    fun1 = @(x) x^4 + x + 1 - 2 - (x.^2).*cos(x);    % f(x) = g(x)
    x(j) = fsolve(fun1, x0(j))
end

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

x = -1.2879

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

x = 1×2

-1.2879 0.8843

%% Calculating the volume
fun2 = @(x) (2 + (x.^2).*cos(x)).^2 - (x.^4 + x + 1).^2;
V    = pi*integral (fun2, x(1), x(2))

V = 23.7802

[Lê]

@Sam Chak Thank you so much, you really helped me with this!