@vv

You write
 "First note that there is not such thing as "global inverse" of f, unless f (supposed to be C^1) is strictly monotonic."
Thank you vv for this quick math reminder but I know this perfectly.

Maybe I should have written
I want to construct some kind of pseudo global inverse of f over R by putting "side by side" local inverse functions
instead of
I want to construct the global inverse of f over R by putting "side by side" local inverse functions
to be clearer ?

I had thought that I was clear enough when saying
     The idea is to define the global inverse g of f over R by
     g := y ->  piecewise(y < f(a__1), g__0(y), ..., y < f(a__n), g__(n-1)(y))
    where g__p(y), is the inverse function of the restriction of f to ] a__p, a__(p+1) [

I realize it was not the case ...  or maybe you where too scandalized (with good reason) by reading the first lines (find the inverse of a non monotonic function) that you did not keep reading the rest of my question ?
I do not hold this against you: "This function is not strictly monotonic over R [and] I want to construct [its] global inverse" is really disturbing and I guess that hearing this would make me hit the roof too.

@Carl Love 

Thank you Carl.
My today problem concerns a polynomial function f, then your answer will be very valuable.

By the way: I have always been surprised that  solve(sin(x)=y, x)  returns arcsin(y), just as if the inverse of "sin" was defined everywhere.
If it is not too much to ask could you say a little more about this ?

Thank for all

@Kitonum 

Thank you Kitonum.

I'm often puzzled with the differences of behavior between  -> and unapply ... I guess I have to read more carefully the dedicated helps pages ...

@acer

Thanks acer
Loading 2015.2 here is not easily due to the very strict safety policy 

Nevertheless I will test this tomorrow with (some) version 2016.
Maybe I will join you later on from home, where I use 2015.2 on a imac.

Thanks again

@acer 

Here it is

Issue.mw

@Daniel Skoog 

Thanks Daniel,

I have just gave some info to acer.

About point 1 : this is a very interesting way to customize the DataSummary procedure
Concerning  your point 2 : right, a screenshot is a roundabout way to answer the question.

Thanks also for your last line, I appreciate that

@acer 

Thank you acer.

I had aleady tried the plotsetup(jpeg, ...) command but I had got an error saying the embeded summary was not of plot type (did not tried textplot : I'm going to see righr now).
What do I mean by "save this table" : you are right, this is not clear. I mean "ideally" save the appearance of the table as an image ... just like some image capture tool could do.
Of course I could use fprintf to save these informations in a text file in a "smart" form, but I found that the appearance of
DataSummarize(...summarize=embed) was realy very smart.


 

@vv 

It looks like our replies have crossed

@vv 

I found myself the solution :-)
I just  declared  use plots  within the procedure and it  works correctly now.

(I thought that loading the package plots in the worksheet itself would be sufficient)

Sorry for the inconvenience

 

@vv 

This works well but your answer made me realize that  I oversimplified my problem.
If it's not too much to ask, could you answer this one please ?


 f := proc(DrawThis)
     plotsetup(jpeg, plotoutput=SomeJpegFile);
     print(DrawThis)
     plotsetup(default);
end proc;

MyPlot := plot(x, x=0..1)
f(MyPlot);   # That's fine, thank you,

MyPlot := display(plot(x, x=0..1), plot(x^2, x=0..1))
f(MyPlot);   # No file is created

Even this simplified procedure
 f := proc(DrawThis)
     print(DrawThis)
end proc;

returns what I would have obtained by writting
MyPlot := display(plot(x, x=0..1), plot(x^2, x=0..1)) ;


Could you help me please ?

@acer 
Thanks a lot for your fruitful help.
I tested tou .mw file and it works beond my expectations.

 

@tomleslie 

I am going to implement these two solutions and look what happens on my machine.

Thanks again

@tomleslie 

It works perfectly well

@tomleslie 

Thanks for the suggestion.


To make it more concrete ...

The unknown is a function r(x) which represents tjhe radius of a revolution object at abscissa. r(x) verifies a nonlinear first order ODE and I'm interested in the radius of curvarure of "solution shap" in some range of x. Assessing this radius by divided differences is not accurate, hence the idea to work with a piecewise solution.

@tomleslie 

Thank you for the information.

I had already used the work-around you refet to but the result is less nice.

Thanks again