Finding minimum value of function (interval mathematics)

Questions
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1

Hi! I must find minimum of function as follows.
For example, I have such a graphic

k1.jpg
k1.jpg1231×608 43.7 KB
I saw that there are two points of minimum and some of rectangles are exactly higher than others, so I need discard them and continue only with the remaining rectangles.
k2.jpg
k2.jpg1231×608 27.6 KB

If I can’t see exactly which rectangles are higher I must divide my interval on more rectangles.
Thanks,
Viktoriya.
P.S. I know how draw rectangles on a function:

gr[f_, xmin_, xmax_, slices_] := 
 Module[{span = (xmax - xmin)/slices, ext}, 
  ext[op_, r_] := First@Quiet@op[{f[x], r <= x <= r + span}, x]; 
  Plot[f[x], {x, xmin, xmax}, 
   Prolog -> {FaceForm[Gray], EdgeForm, 
     Rectangle @@@ ({{#, ext[Minimize, #]}, {# + span, 
           ext[Maximize, #]}} & /@ Most@Range[xmin, xmax, span])}]]; 
f[x_] := x + 3*Sin[x + 2]; gr[f, -5, 5, 10]
2

You should first find all the stationary points of the function.

FindRoot[f'[x]==0,x]

Then check each x to identify maxima and minima and identify largest maximum. Can use

Max[x1,x2,...]
3

The notation might be a bit confusing. I suppose that x1, x2 etc are the x-coordinates of the extrema. If that is so to determine the global maximum one might want to calculate

Max[f[x1],f[x2],...] (*This is not Mathematica notation*)

Perhaps this is better:

f[x_] := x + 3*Sin[x + 2];
Solve[D[f[x], x] == 0, x]

and then

f[x] /. {{x -> -2 - ArcCos[-(1/3)]}, {x -> -2 + ArcCos[-(1/3)]}}f[x] /. {{x -> -2 - ArcCos[-(1/3)]}, {x -> -2 + ArcCos[-(1/3)]}}

(*{-2 - 2 Sqrt[2] - ArcCos[-(1/3)], -2 + 2 Sqrt[2] + ArcCos[-(1/3)]}{-2 - 2 Sqrt[2] - ArcCos[-(1/3)], -2 + 2 Sqrt[2] + ArcCos[-(1/3)]}*)

But I don’t think that this answers Victoriya’s questions.

M.

4

Stan Wagon has written global optimization code that works via intervals. Below are a couple of references.

http://www-sbras.nsc.ru/interval/Library/Surveys/ThinkGlobally.pdf?PHPSESSID=df71c15e5c2bcf30c47aeb0f975ed2cc

http://www.mathematica-journal.com/issue/v9i3/newpublications/ISBN089871561X.html

5

I have found that NMinimize and NMaximize are reasonably good at finding the global extremum if the method used is “SimulatedAnnealing” or “RandomSearch” but not so good with the default method.

In[20]:= f = (x - 1)^2 + 5 Sin[x];

In[21]:= NMinimize[f, x, Method -> "SimulatedAnnealing"]

Out[21]= {-0.348, {x -> -0.779015}}