# NSolve with numerical function

I would like to solve numerically an equation which involves a numerical function constructed by fitting some data:

data := {{0, 1}, {1, 1}, {.2, 3}, {.4, 5}, {.6, 2}};
try[x_?NumberQ, y_?NumberQ] := (auy = y;
solution = FindFit[data, A (aux - auy) + B Exp[aux - auy], {A, B}, aux];
myFit[aux_?NumberQ] = (A (aux - auy) + B Exp[aux - auy]) /.
solution; Return[myFit[x]])

NSolve[y - try[1, y] == 0, y]


I tried with this code but it does not function (various errors). Does anyone know how to help me?

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Did you even try to test call your try function? What happens when you type try[1, y] ? does not return anything? This is really strange that no body seems to do any testing of anything any more. –  Nasser Dec 5 '13 at 11:55
I tried try[1,2] and it works. try[1,y] has not to return an output since it need numerical parameters to work. What I want is a way to solve the problem I described, i.e. using NSolve to solve the equation which involve a function numerically determined by a fit. My code obviously does not work, that's why I asked help. Does anyone knows how to solve my issue? I hope I clarified my post! –  user9994 Dec 5 '13 at 12:44
I modified a little the example code above, now it should be more clear. –  user9994 Dec 5 '13 at 13:40
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data = {{0, 1}, {1, 1}, {.2, 3}, {.4, 5}, {.6, 2}};
try[x_?NumericQ, y_?NumericQ, data_] := (A (x - y) + B Exp[x - y]) /.
FindFit[data, A (xx - y) + B Exp[xx - y], {A, B}, xx]

FindRoot[y - try[1, y, data] == 0, {y, 1}]

(*
{y -> 1.78209}
*)

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ok, this helps a lot: FindRoot is what I need. But I have to scal a big range of "y", right? Maybe with a Do[ ] statement. is there someway to optimize it?. Thank you for your help belisarius! –  user9994 Dec 5 '13 at 15:17
Sorry, I don't understand your sentence But I have to scal a big range of "y", right?. Can you explain it further for me, please? –  belisarius Dec 5 '13 at 15:37
I am really sorry, I had to write "scan" and not "scal". What I meant is that I have to search for solution in a range of ys to find more than one solution. –  user9994 Dec 5 '13 at 16:34
@user9994 FindRoot should take care of that. Just try to give it a good initial guess –  belisarius Dec 5 '13 at 16:45
But if I suspect the existence of different solutions I have to repeat the FindRoot with different intial guesses right? –  user9994 Dec 5 '13 at 17:11