Q: Does the inverse of $f(x)=x^3$ have a non-negative domain to have a real output? I'm not familiar with complex analysis. While playing with Mathematica (a mathematics software), I found that it keeps spitting out unexpected results, and the reason was that it considers differently from me of the inverse of certain polynomial functions such as $f(x)=x^3$.
Mathematica evaluates the following line
N[InverseFunction[Function[x,x^3]][-1]]

as 0.5 + 0.866025 I. The above means to evaluate numerically the value of $f^{-1}(-1)$ where $f(x)=x^3$.
I was thinking naively that an inverse function is plotted as symmetric to $y=x$ from its original form, so that the result above ought to be $-1$. What did I miss?

A: It depends which root of the equation the software considers by default. The point is that for real $y$, the equation $x^3-y=0$ always has $3$ complex roots. If $y>0$, there is no controversy about taking the positive real root, which we normally call $\sqrt[3]{y}$. On the other hand, if $y<0$, then there are no positive real roots; instead the roots are
$$ -\sqrt[3]{-y}, \quad -\omega \sqrt[3]{-y}, \quad -\omega^2 \sqrt[3]{-y}, $$
where $\omega$ is a complex number which satisfies $\omega^3=1$.
Mathematica's convention is to take the root with the smallest complex argument; this happens to be $-\omega^2 \sqrt[3]{-y}$ when $y$ is negative, for a particular choice of $\omega$. Other calculators will use $-\sqrt[3]{-y}$, but they normally only have to deal with real numbers: the point of Mathematica's convention is that it is consistent and continuous in most of the complex plane for any power of $x$, with the result that the odd integers do not behave quite as you expect. 
On the other hand, if you plot the same thing multiplied by $i\sqrt{3}-1$, you will find the negative part reappears where you expect, so it is where it ought to be with Mathematica's definition.

A: For $f^{-1}$ to exist $f$ must be injective. This means that if $x \neq y$ then $f(x) \neq f(y)$.
But $f(x) = x^3$ isn't injective since
$$
f\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -1,\ 
f\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -1, \ 
f(-1) = -1
$$
So $f(x) = x^3$ doesn't have an inverse! If you try to define one you realize that $f^{-1}(-1)$ actually has three possible values. Mathematica only gave you one, but not the one you were expecting: $$\frac{1}{2} + i\frac{\sqrt{3}}{2} \approx 0.5 + i0.866025$$
$f(x) = x^3$ can have an inverse so that $f^{-1}(-1) = -1$, but only if $f$ is restricted to $\mathbb{R}$.
