Can somebody help me please to write the code to fulfill this condition in MATLAB?

I have two functions. for example,

f(x) = 3x^2 -2y +5 (plot for x>1); g(x) = 4x^3 - 3x -7;

For f(x), x is variable and y is a arbitrary fixed number. So, by fixing arbitrary 'y', I can have many graphs for f(x) (for instance: if 'y=2' then I can have one graph for f(x), and for 'y= 2.01' I will have other graph). And For g(x), x is a variable. So, I will get only one graph for g(x) for non-negative x.

My goal is to find an unique intersection point 'x_0' between g(x) and f(x) such that,

g(x_0) = c(constant) + minimum(f(x)).

here I can change y arbitrarily in f(x), to get my correct 'x_0'. So, I might need to do iteration here. But I don't know how to write this command. Please help me. Thank you for your time.

[Walter Roberson]

? You mean like

x = 0
y = 1/(12*(c/8 + ((c/8 + 3/2)^2 - 1/1728)^(1/2) + 3/2)^(1/3)) + (c/8 + ((c/8 + 3/2)^2 - 1/1728)^(1/2) + 3/2)^(1/3)

?

For c = 0 that would be 3/2, and 4*(3/2)^3-3*(3/2)-7 == 3*0^2-2*(3/2)+5

There would be as many of these are there are values for c.

[Md Abu Helal]

@Walter Roberson, Thanks for your comment. But I set here two functions as example. Originally, my functions is so complicated and not easy to get common point for that condition, I mean it's possibly a fraction value for common y. So, I think I need to write the code to iterate and get the final answer by satisfying the condition. But I don't know how to make that command. Again, appreciate for your time.

[Md Abu Helal]

I am sorry, I should mention 'c' is same constant value for all iterations. That is c is fixed for my case. like 'c= 6' for every time.

[Walter Roberson]

Your f(x) function is complicated, but is it differentiable even in theory? Differentiable in practice? If you were to work with diff(f(x),x)==0, then is there an estimate of the number of places that would occur within the range of interest ?

[Md Abu Helal]

Yes, my f(x) is C1, in fact this an integral equation and from the contents I can get the idea the range of 'y' and that range of y is (0.5, 6)

[Walter Roberson]

Maybe

syms x y
fx = f(x);
solx = solve( diff(fx, x), x );
infimum = subs(fx, x, solx);
soly = solve(c + infimum == g(y), y);

[Md Abu Helal]

@ Walter, Thank you for your generous help. I am sorry for prior mistake. I fixed my question and expecting your valuable response.

Answer 1

[Image Analyst]

So, is this what you have:

y = linspace (0, 3, 1000);
gy = 4 * y .^ 3 - 3 * y - 7;
plot(y, gy, 'k-', 'LineWidth', 2);
grid on;
hold on;
ca = {'gy'};
% Next family of equations:
x = linspace(1, 3, 1000);
y = linspace(-12, 10, 13);
for k = 1 : length(y)
  fx = 3 * x .^ 2 - 2 * y(k) + 5;
  plot(x, fx, '-', 'LineWidth', 2);
  ca{k + 1} = sprintf('y = %.2f', y(k));
end
legend(ca, 'location', 'northwest');
fontSize = 24;
xlabel('y, and x', 'FontSize', fontSize);
ylabel('g(y) and F(x)', 'FontSize', fontSize);

and you want the coordinates of the intersections of the curves?

[Md Abu Helal]

Hi, Thank you for your help. Yes, I am looking for something like this. But, How do I write the command to get correct y? so that for that "y" they satisfy the condition:

         g(y) = c(constant) + infimum(f(x)).

I hope you understand my point. Thanks again.

[Walter Roberson]

Is c given in advance? Or are you wanting to get the formula for the intersection point for arbitrary c? Or is the concern that infimum(f(x)) might not happen to intersect the g(y) formula and might need to be shifted?

[Md Abu Helal]

I should mention 'c' is the same constant value for all iterations. like c = 6.

[Md Abu Helal]

@Walter Roberson, Thank you. Yes, c is given in advance and same for all iteration. For my case I will get the infimum, so, you can say this is minimum of (f(x)). My concern is to get the correct 'y' (by changing arbitrary y in f(x)) such that for that correct "y" they satisfy the condition:

    g(y) = c(constant) + infimum(f(x)).

I hope I described this time better. Thanks again for your support.

[Image Analyst]

Not sure I understand. I thought you said

g(y) = 4y^3 - 3y -7;

I don't know what infimum is - apparently you say it's the minimum of f(x).

So then is gy just c-min(fx) where c is some predefined contant? But then this g(y) would be just a constant, and not at all the same cubic formula for g(y). In fact it would not be a functionof y at all, just a function of f(x) which is a function of x. Actually I don't see why you're making it confusing by having the independent (horizontal) variable be called x sometimes and y other times. This is not standard.

[Md Abu Helal]

@ Image Analyst, Thank you very much for your help to lead me in right direction. After your comment, I studied full contexts three times, I found that there are some confusing term, for which I didn't get the point. Now, I am explaining again: say, I have two functions:

f(x) = 3x^2 -2y +5 (plot for x>1); g(x) = 4x^3 - 3x -7;

For f(x), x is variable and y is a arbitrary fixed number. So, by fixing arbitrary 'y', I can have many graphs for f(x) (for instance: if 'y=2' then I can have one graph for f(x), and for 'y= 2.01' I will have other graph). And For g(x), x is a variable. So, I will get only one graph for g(x) for non-negative x.

My goal is to find an unique intersection point 'x_0' between g(x) and f(x) such that,

g(x_0) = c(constant) + minimum(f(x)).

here I can change y arbitrarily in f(x), to get my correct 'x_0'. So, I might need to do iteration here. But I don't know how to write this command. Please help me and sorry for misguiding before. I appreciate your kind help.