I want to use Reduce to my own function, but it generates the error message: This system cannot be solved with the methods available to Reduce. Here is a simplified example.
test[x_] := x^2 - 3 /; 0<x<10; test[x_] := -1 /; 10<x Reduce[test[x]<1, 0<x<5] I think /; makes the problem. However, how can I define the domain of my own function without using /;?
To summarize insightful comments, Piecewise is the appropriate construct to express the mathematical idea of a piecewise function, as Natas suggested. Condition (i.e. /;) is a programming construct. As Szabolcs mentioned, "it is not meant to represent mathematical ideas; it is meant to implement algorithms". In the following, note also the corrected syntax for Reduce.
Clear[test] test[x_] := Piecewise[{{x^2 - 3, 0 < x < 10}, {-1, x > 10}}] Reduce[test[x] < 1 && 0 < x < 5, x] (* Out: 0 < x < 2 *) As an alternative approach, Bill proposed using Boole:
Clear[test] test[x_] := ((x^2 - 3)*Boole[0 < x < 10]) + (-1*Boole[10 < x]); Reduce[test[x] < 1 && 0 < x < 5, x] (* Out: 0 < x < 2 *) 
test and test2? Does it make a difference if you add the variable to be solved for explicitly, as we have been suggesting? Reduce[{test[x] > test2[x], 0 < x < 2}, x] $\endgroup$
Piecewiselets you define such functions. Also it probably should beReduce[test[x] && 0<x<5, x]. $\endgroup$/;is a programming construct. It is not meant to represent mathematical ideas. It is meant to implement algorithms. $\endgroup$Boolehad been incorporated into many functions in Mathematica.test[x_]:=((x^2-3)*Boole[0<x<10])+(-1*Boole[10<x]); Reduce[test[x]<1&&0<x<5,x]$\endgroup$