How to obtain argument value of an equation

Hello, I want to know for what values of x, f(x) is equal to half of its max value. I want to use value of this argument in my future computation in MATLAB. How can I do this in MATLAB?

[Sam Chak]

Hi @moh mor

Can you define the meaning of "half of its max value"?

x = linspace(0, 10, 1001);
f = 1*sin(pi/5*x) - 1*tanh(2.3*(x - 5)) + 8;
plot(x, f, 'linewidth', 1.5), grid on, ylim([0 12])
halfmax = max(f)/2

halfmax = 5.0000

yline(max(f),  '--', 'max');
yline(halfmax, '--', 'halfmax');

[moh mor]

Ok. You are right. I I just mean how to obtain value of x in for example f(x)=constant .

Answer 1

[John D'Errico]

I wrote an allcrossings tool, that I posted on the file exchange. I've attached it here.

https://www.mathworks.com/matlabcentral/fileexchange/125585-allcrossings?s_tid=srchtitle

help allcrossings

  allcrossings: locate all intersections of a pair of functions f1 and f2, on a finite domain
  usage: Xcross = allcrossings(f1,f2,Xsupport,nsamples)
 
  allcrossings uses fzero as a search engine for the roots of f1(x) - f2(x) == 0,
  once it identifies a bracket that contains a root.
 
  If your functions have discontinuities in them, such as tan(x), then an
  identified crossing may be spurious, in the sense that f1(x)~=f2(x),
  but the two functions should still exhibit a crossing at that location.
 
  arguments: (input)
   f1,f2 - a pair of function handles. It is assumed that f1 and f2 are
       both vectorized functions, so they can be evaluated at multiple
       locations in one call.
       if you are trying to solve problems of the form f1(x) == k, then
       f2 can be passed in as a scalar constant, or you can use
       f2 = @(x) k + zeros(size(x));
       as a properly vectorized function.
   
   Xsupport - vector of length 2 that defines the lower and upper limits of
       the domain where the search will be done.
 
   nsamples - (optional) integer argument that indicates the number of
       points over that domain to test, looking for where there may be
       an intersection. At a minimum, nsamples will always be at least
       100.
 
       Default value: nsamples = 500;
 
   fzeroOptions - (optional) if supplied, it must abe a valid struct
   containing the options used by fzero. The options that fzero will look
   for are: {Display, TolX, FunValCheck, OutputFcn, and PlotFcns}. The
   values assumed will otherwise be:
     Display: 'none'
     TolX: 2.2204e-16
     FunValCheck: 'off'
     OutputFcn: []
     PlotFcns: []
 
  arguments: (output)
   Xcross - vector of locations where f1(x) == f2(x)
 
  Examples:
      Intersection points of sin(x) == cos(x), between -10 and +10.
   f1 = @(x) sin(x);
   f2 = @(x) cos(x);
  xcross = allcrossings(f1,f2,[-10,10],50)
  xcross =
       -8.6394      -5.4978      -2.3562       0.7854        3.927       7.0686
 
  Example:
      Positive solutions to the problem x + tan(x) == 1
   f1 = @(x) 2*x + tan(x);
   f2 = 1;
   xcross = allcrossings(f1,f2,[0,20],100)
   xcross =
      0.32919       1.5708       1.9113        7.854       7.9213       14.137       14.174
   f1(xcross)
   ans =
     1   1.3748e+15            1  -2.4185e+14            1  -2.0929e+14            1
  As you can see, there are crossings found at certain points that are
  not technically solutions, but due to the jump in the tan function, they
  were still identified by fzero as "crossings". These are spurious solutions.
 
  see also: fzero, fzolve, solve, vpasolve
 
  author: John D'Errico
  e-mail: [e-mail]
  Date: 2/28/2023

For example, consider the function

fun = @(x) abs(x).^1.5./(1 + (x-1.23).^2);
fplot(fun,[-10,10])
grid on
hold on

Assume we are willing to search over the interval in x of [-10,10]. (I know this function approaches zero asymptotically as x goes to infinity in either direction. So I need not worry about any other peaks.)

First, find the peak value.

[xmax,fmax] = fminbnd(@(x) -fun(x),-10,10);
xmax

xmax = 1.7786

fmax = -fmax

fmax = 1.8233

So the peak value lies around 1.7786, and the peak itself is 1.823...

Now we find the two points where the function attains the value at half the peak.

xcross = allcrossings(fun,fmax/2,[-10,10],1000)

xcross = 1×2

0.9792 4.0449

xline(xcross,'r')
yline(fmax/2,'g')

So the two points where the function attains half the peak height.

[moh mor]

Thank you very much.

Answer 2

[Vilém Frynta]

My approach:

f = @(x) x^2;   % function example
[max_val, max_idx] = fminbnd(@(x) -f(x), -10, 10);  % find maximum value of the f
half_max_val = max_val/2;   % divide maximum by two

[John D'Errico]

@Vilém Frynta

NO. You misunderstood the question. The question was NOT to compute the maximum value, and then divide by 2.

The question was to locate the point x, where the function attains half that maximum value. While you did learn the maximum, you did not solve the real problem.