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Find the domain of $f(x)=x^{2/3}$. Now, $0^2=0$ and $\sqrt[3]{0}=0$, all positive and negative numbers squared will give positive answer, these numbers can give us cuberoot, so entire real line is the domain, am I correct? (one of the online help provided the answer as $\{x \in \mathbb{R} : x \geq 0\}$ (all non-negative real numbers), why negative numbers are not included in the domain, where am I going wrong?

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What online tool gave $[0,\infty)$ as the answer, and what was the input? (On the other hand, the domain of $x^{3/2}$ is the nonnegative real numbers...) – Arturo Magidin Feb 14 '12 at 16:39
Are you talking about the range of $f(x)$? – user21436 Feb 14 '12 at 16:41
can I disclose the website name here? – Vikram Feb 14 '12 at 17:04
@Vikram: Why not? – Arturo Magidin Feb 14 '12 at 17:12… – Vikram Feb 14 '12 at 17:13
up vote 5 down vote accepted

By definition, $x^{2/3} = \sqrt[3]{x^2}$.

Since we can compute $x^2$ for any real number $x$, and since we can compute a cubic root for any real number, whether it be positive, negative, or zero, the domain is all real numbers.

On the other hand, as Kannappan hints, the range of this function is $[0,\infty)$. So you should really check the input of that on-line tool.

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One online tool that claims the domain is $\[0,\infty)$ is Wolfram Alpha. It chooses a complex branch for negative arguments, possibly in an attempt to have only a single cut in the complex $x$-plane. – Henning Makholm Feb 14 '12 at 17:02
thanx Henning, you are correct about the online tool name – Vikram Feb 14 '12 at 17:10
and I am not interested in complex numbers, so my question is answered by Arturo – Vikram Feb 14 '12 at 17:11
And yet, the plot given by Wolfram Alpha clearly includes negative numbers. Yet another reason for me not to like Wolfram Alpha all that much... – Arturo Magidin Feb 14 '12 at 17:12
@lhf: Most of the time, you can't define $\exp$ until you've developed "enough" exponentials (unless you want to go the route of Taylor series; or as the inverse of $\log$, with $\log$ defined via integrals), so the definition for rational exponents often precedes that of $\exp$ (making it difficult to define $x^{a/b}$ as $\exp(b\log(x)/a)$). One is limited to $x\gt 0$ for arbitrary rational exponents, but one can extend the rational exponents definition when $b$ is odd to negative exponents. – Arturo Magidin Feb 14 '12 at 21:31

The answer is unfortunately context-dependent. That's a fancy way of saying that on a test the answer is what your instructor says it is. For whatever it is worth, Wolfram Alpha thinks that the domain is the non-negative reals.

I assume we are working in the reals. We are concerned with the domain of $x^y$, where $y$ is a positive real number. (I am taking $y$ positive to avoid division by $0$ issues, and $0$-th power issues.)

There is general agreement that $x^y$ is defined when $x\ge 0$. But there are disagreements in the case $x<0$.

If $y$ is irrational, there is general agreement that $x^y$ is not defined at negative $x$.

If $y$ is rational, and $y$ can be expressed as $a/b$, where $a$ is an odd integer, and $b$ is an even integer, there is general agreement that $x^y$ is not defined at negative $x$.

When $y$ is a rational, which, when put in lowest terms, has an odd denominator, there is disagreement.

In high school, generally the convention is that in that case, $x^y$ is defined when $x$ is negative. And that convention is not confined to high school.

But when we are dealing with exponential functions, with $y$ thought of as variable, there are good arguments for considering, for example, $(-2)^{2/3}$ to be undefined.

The issue is this. Consider the function $(-2)^y$. Arbitrarily close to $2/3$, there are irrational $y$ at which $(-2)^y$ is definitely not defined. So even if we consider $(-2)^{2/3}$ to have a meaning, the function $(-2)^y$ is not continuous at $y=2/3$.

Another argument is that a common definition of $x^y$ is that $x^y=\exp(y\ln x)$. Note that $\ln x$ is not defined when $x$ is negative. Of course, $\ln x$ is not defined at $x=0$, but that generally does not stop people from considering $x^{2/3}$ defined at $x=0$.

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The problem with second argument is that if $y=1$ for $x^{y}$ then $x=e^{\ln(x)}$. At $x=-1$ this is not the case. It is possible when $x$ is positive but $y=x$ must still have a negative domain. For your first argument is there a theorem stating a function must be continuous at point to be definable? – Arbuja Dec 28 '15 at 17:06
Certainly there is no theorem, or convention, that a function must be continuous at $a$ to be defined at $a$. However, if certain "natural" manipulations are correct for positive $a$ but incorrect for negative $a$, it may be useful to restrict the domain. That said, we certainly would not want $(-1)^n$ to be taken away from us! – André Nicolas Dec 28 '15 at 17:14

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