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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?

hmakholm left over Monica
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Vikram
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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
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    Are you talking about the range of $f(x)$? –  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
  • http://www.wolframalpha.com/input/?i=domain+of+the+function+x%5E%282%2F3%29 – Vikram Feb 14 '12 at 17:13
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    Fractional powers aren't as simple as we think. Arturo below says that $x^{2/3}$ means $\sqrt[3]{x^2}$, and this is one interpretation. Reasonable people might want $x^{2/3}$ to mean the exact same thing that $x^{4/6}$ does. Again, we could interpret this as $\sqrt[6]{x^4}$, and all real numbers are valid input. But why shouldn't this also be a valid interpretation: $(\sqrt[6]{x})^4$? This version cannot have negative input. So there is an issue here that comes down to what the definition of raising to a fractional power really means. – 2'5 9'2 Feb 14 '12 at 17:41
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    The only situations where $a^b$ is totally unambiguous is when $a\geq0$ or $b\in\mathbb{Z}$. – 2'5 9'2 Feb 14 '12 at 17:43
  • Powers of exponents are also unambiguous if $b$ is a completely reduced fraction. Since $x^3$ is defined for all integers than its inverse, $x^{1/3}$ must also have complete domain. Remember using your rule $x^{6/2}$ is also undefined unless $6/2$ is simplified. – Arbuja Dec 13 '15 at 16:55

2 Answers2

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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.

Arturo Magidin
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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. – hmakholm left over Monica 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
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    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
  • @Arturo: To be fair, to does give the domian under the subheading "properties as a real function". – hmakholm left over Monica Feb 14 '12 at 18:10
  • @Henning: Right: WA contradicts itself, giving the full domain in the graph, but giving a truncated domain under "properties as a real function." Hence, my lack of trust. – Arturo Magidin Feb 14 '12 at 18:16
  • It's not clear to me that $x^{2/3} = \sqrt[3]{x^2}$ by definition. – lhf Feb 14 '12 at 21:17
  • @lhf: That's the standard way of extending exponents to from integers to rational numbers: define $x^{1/n}$ as the inverse of $y\mapsto y^n$; then define $x^{a/b}$ as $(x^a)^{1/b}$. Of course, one can use different definitions. "The nice thing about standards is that there are so many to choose from." – Arturo Magidin Feb 14 '12 at 21:18
  • Yeah, I've just checked baby Rudin and it's like that there. But there is the definition via $\exp$, which I guess amounts to the same thing. In both cases however one is limited to $x>0$. – lhf Feb 14 '12 at 21:26
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    @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
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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 $x=0$.

André Nicolas
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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
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    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