function analysis question

[quinyu]

No, I want the calculator to tell me where it can interpret the function. That's the domain. I want the calculator to tell the domain of an arbitrary function f(x). What I showed was an example. Also, I want the calculator to tell the range - that is, the set of all the values that f(x) has, on the domain. How do I do this?

For your info, on the TI-nspire CAS, the command domain(asin(sqrt(x)),x) results in the output 0<=x<=1, which is what I'm looking for. (This is not the precise input but close enough to it.) While it cannot directly find the range (or at least I know of no method to do so), if I set x=asin(sqrt(y)), and have it solve for y, gives y=(sin(x))^2 (that is the inverse function), with the requirements that sin(x)>=0 and -pi/2<=x<=pi/2. From that point on, I can get a domain again for (sin(x))^2, which is obviously any real number, so that bit is irrelevant. If I solve for the other criteria, it gives max(-pi/2,2*n1*pi)<=x<=min(pi/2,2*n1*pi+pi) as a result. For the principal value (that is, n1=0) of the expression, that means that 0<=x<=pi/2. This is therefore our function range.

So, how do I execute the same sort of procedure on the HP Prime?