such as this function (x^3 + 5 x^2 + 4 x + 5)/Sqrt[x^2 + 2 x - 3]
In V10 there is also
In[1]:= FunctionDomain[(x^3 + 5 x^2 + 4 x + 5)/Sqrt[x^2 + 2 x - 3], x] Out[1]= x < -3 || x > 1
I want to know for what values of x there will exist real y such that y=f[x], where f is your function. Per documentation on Resolve, it does the needed quantifier elimination.
x
y
y=f[x]
f
Resolve
What is that input essentially saying?
Resolve[Exists[y, y == (x^3 + 5 x^2 + 4 x + 5)/Sqrt[x^2 + 2 x - 3]], Reals] Out[10]= x < -3 || x > 1
