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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?

zd_
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    You must specify the domain of your function, and for each input value $x$ in the domain you must specify the output value $f(x)$. So, what is your domain for $f(x) = x^{\frac{4}{5}}$ (it is your responsibility to specify the domain)? And for each input value $x$ in the domain, what is the output value $f(x)$ (it is your responsibility to specify a unique output)? Once you specify that, then the question becomes answerable. – Lee Mosher Jan 31 '20 at 03:05
  • $x^{\frac{4}{5}}$ is a multi-valued function so its values are sets, not numbers; if you want to make it single-valued you need to specify a domain and a branch and then it may or may not be even depending on your choices – Conrad Jan 31 '20 at 03:05

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When you define a function you need to define the domain. When we talk of even and odd functions we are normally talking of functions $\Bbb {R \to R}$. The behavior of the complex values (not roots) of $x^{1/5}$ does not matter. As real functions $x^{1/5}$ is odd and $x^{4/5}$ is even.

Ross Millikan
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The function is even. It seems like what's really in question is $(-1)^{\frac{p}{q}}$ with $p$ even, $q$ odd. This equals $1$. To see this, consider that $(-1)^{\frac{1}{q}}=-1$ is the unique real solution to $x^q=-1$. Then $(-1)^{\frac{p}{q}}=((-1)^{\frac{1}{q}})^p=(-1)^p=1$ since $p$ is even.

This is also true if you do the order in reverse in that $(-1)^p=1$ and $1^\frac{1}{q}=1$ is the unique real solution to $x^q=1$.

In conclusion, $(-x)^\frac{p}{q}=x^\frac{p}{q}(-1)^\frac{p}{q}=x^\frac{p}{q}$.

ProfOak
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Split up the fraction. Either $$f(x)=(x^4)^\frac15$$ Where $(x^4)=(-x)^4$ so these are equivalent, Or: $$f(x)=(x^\frac15)^4$$ Where $x^\frac15=-(-x)^\frac15$ (take the fifth power on both sides to see this), but using $(x)^4=(-x)^4$ these are equivalent as well. It is understood that when selecting which root to take, you take the same root on both sides.

For example, if you were calculating values of $f(x) =x^\frac13+(-x)^\frac13$, to say that $f(8)=2+(-1+i\sqrt3)=1+i\sqrt3$ is totally wrong.

Rhys Hughes
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