I am confused if the function $f(x) = x^{4\over5}$ is an even function. If we only consider the real roots, then the function is even in the sense that $f(-x) = f(x)$. 

However, since $x^{1\over 5}$ is likely to have some non-real roots for $x \in \mathbb{R}$, and hence it may be the case when $x^{1\over 5} \neq (-x)^{1\over 5}$, would this $f(x)$ still fit into the definition of an even function?