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

Question

Lê 2023-11-19

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2049237-what-is-the-function-to-use-to-find-the-intersection-of-two-graphs

评论： Lê 2023-11-21

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')

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Lê 2023-11-19

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

Stephen23 2023-11-20

"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).

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Answer 1

Sam Chak 2023-11-20

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https://ww2.mathworks.cn/matlabcentral/answers/2049237-what-is-the-function-to-use-to-find-the-intersection-of-two-graphs#answer_1356392

在 MATLAB Online 中打开

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

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Sam Chak 2023-11-21

在 MATLAB Online 中打开

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 2023-11-21