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Is the cube root of $x$: $x^{1/3}$, a continuous function? I thought $x$ cannot be negative. Therefore, the domain is the all real numbers greater than zero.

I have a continuity problem that asks if the following is continuous: $$(2x - 1)^{1/3}.$$

The solution states that is a composite of two functions that are continuous, $x^{1/3}$ and $2x - 1$. And so, the result is continuous.

Thank you.

cqfd
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sgoodman
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    The principal square root of a negative number is nonreal. The principal cube root of a negative number however is real. $\sqrt[3]{-8}$ for instance is $-2$, noting that $(-2)\times (-2)\times (-2)=-8$ – JMoravitz Sep 09 '19 at 16:49
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    Strictly speaking, the third root is defined for negative numbers: $(-8)^{1/3}= -2$. However, you are right that, in order to be able to use the 'laws of exponents', the function $f(x)= x^{1/3}$ has domain non-negative numbers. However, it does not then follow that " the domain is the all real numbers greater than zero.". The domain is "all real numbers greater than or equal to zero. When a function is defined on such an interval, in order to be continuous at boundary points, the limit only has to be taken through points in the domain. – user247327 Sep 09 '19 at 16:59

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There is not any discontinuous point in these functions, so the domain of both $$x^{\frac13}$$ and $$(2x-1)^{\frac13}$$ is $R$.

Whenever we have a $\sqrt{}$ functions with noneven roots, we check function in sqrt. You could check dominator functions, $\tan$, $\log$, and so on.

BarzanHayati
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