$\sqrt[3]{-x}$ for x>0

Question

I am sure this topic must have come up before, but all posts i could find, did not answer all my questions, my math background is very week so sorry for this dumb question.

If i have a function with a cubic root, is it defined on all x? I know that the cubic root of lets say (-8) has the real solution -2 and 2 imaginary solutions.

Let’s say the function $\sqrt[3]{x^2-1}$, will it be defined on all x? Can i just „forget“ about the imaginary part when i am only interested in the real numbers?

Answer 1

Well of course, if you define $f: \mathbb{R} \rightarrow \mathbb{R}$ with the formula $f(x) = \sqrt[3]{x^2-1}$ then it's defined for every real number $x$, just because the remark you made; for negative values, that is for $x \in (-1,1)$, cubic root is well defined and is a negative number.

Answer 2

Yes, the domain of the function $f(x)=\sqrt[3]x$ is $\mathbb R$ and $\sqrt[3]{-8}=-2.$

Answer 3

I suppose that the problem happened to you typing in Wolfram Alpha

Plot (-x)^(1/3) from x=-2 to x=2

which will produce the left branch and nothing of the right side. Try instead

Plot  cuberoot(-x) from x=-2 to x=2