Verified answer

I assume that you mean f(x) = x³/(x² + 1)

Rational functions don't exist if the denominator is zero.

i.e. where x² + 1 = 0

=> x² = -1....no solutions exist

so, there are no discontinuities => domain is all real values of x

:)>

When it is a fraction the denominator cannot be zero, but here x^2 + 1 is not zero for no value of x.

Then the domain is all real number x.

Your problem is difficult to read without parentheses. But (x^3)/(x^2) is just x (if that's what the problem is trying to say). So basically you have

f(x)=x +1 which gives you a domain of all reals.

It all real numbers because -1^2 still gives you 1.

I guess:

f(x) = x^3/(x^2 + 1)

Domain = ℜ

If you do x = - 1

f( - 1) = (( - 1)^3)/(( - 1)^2 + 1) = - 1/2, therefore ∃ point ( - 1; - 1/2)