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I have an exercise for home to find the domain of :

$$f(x)= {\sqrt[5]{(1/3)^x - (1/9)}\over \sqrt[3]{e^x - 1}} $$

The solution given by our teachers (and the book itself) is $(0,2]$. My problem is that the cube root and the 5th root can have negative numbers in it... So you only take the denominator $ 0$... However our teacher says this is wrong and we should take the roots $\ge 0$

Can someone explain me why?

Gerry Myerson
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George Tsomlektsis
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  • You clearly have a problem with the denominator when $x=0$. Apart from that, it is a matter of the domains of the functions $\sqrt[3]{y}$ and $\sqrt[5]{y}$ which depend on their particular definition (e.g. in the book or from your teacher). – Henry Aug 15 '16 at 12:12
  • It depends on the definition of the root. Because for any number $x$ (except $0$), there are $3$ cube root of $x$, in the sense there are three numbers $y_1,y_2,y_3$ s.t. $y_i^3 = x$ therefore the cube root is not really well defined. However there is only one real cube root, so in your case, if you consider only $\mathbb{R}$ it would make sense to say that the roots are indeed defined over the negative numbers. – Zubzub Aug 15 '16 at 12:15

1 Answers1

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The idea that $\sqrt[3]{-8}=-2$ comes from the complex plane, where the argument of $-8$ – $\pi$ principally – is interpreted as $3\pi$ so that taking one-third of the argument like is done with complex cube roots lands the cube root on the real line again. Similar arguments can be made for other odd roots like $\sqrt[5]{-32}=-2$.

Because the underlying idea of complex numbers is usually taught later in a mathematics curriculum, your teachers and your book among others require $x\ge0$ in the definitions of $\sqrt[3]x$, $\sqrt[5]x$ and so on. If these restricted domains are used, the given domain of $f(x)$ is correct. If the full domain of $\mathbb R$ for these roots is used, the domain of $f(x)$ becomes $\mathbb R\backslash\{0\}$.

Parcly Taxel
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    There is an argument on the real numbers saying that $f(x)=x^3$ is a bijective function in a way that $g(x)=x^2$ or $h(x)=x^{\pi}$ are not, and so has an inverse on the real numbers which might be written as $f^{-1}(y)=\sqrt[3]{y}$ – Henry Aug 15 '16 at 14:50
  • Yes, yes, but some people are pedants, and impose on the real cube root the same domain restriction as the real square root, even though the former is bijective across the reals. I'm definitely not a pedant. – Parcly Taxel Aug 15 '16 at 14:54