Why isn't the domain of $y=x^{1/3}$ just $x \in \Bbb R$?
Could someone explain it to me? I got the answer, and went to check if i was right on a graph plotter, and found out that I got the domain wrong(alongside with the range, because of my domain, of course). My logic was: If I input x$=-1$, I'll get $-1$ as the answer right? Because $(-1)(-1)(-1)=-1=(-1)^3$, so the domain should be able to go to the negatives, right?
What am I doing wrong?
(the actual question was a somewhat harder function, however the problem broke down to this simpler one).
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$\begingroup$You are correct. Most precalculus and calculus text interpret $f(x)=x^{1/3}$ in the way you interpret it. However many computer algebra systems interpret it as the principal third root.
If you want Wolfram alpha to plot the function you have in mind use the "surd" command.
$\endgroup$ $\begingroup$There are three (complex) cube roots of each real number (apart from $0$), and only one of these is real. Depending on your software, the definition it uses of $x^{1/3}$ will, in general, use the real root for positive $x$, but not necessarily for negative $x$. (The reason for choosing a non-real root is to ensure continuity as you pass from the positive to the negative real axis by paths in the complex plane that do not just go straight through the origin.)
$\endgroup$ $\begingroup$I think the domain of $y = \sqrt[3]{x}$ should be $(-\infty,\infty)$, but the domain of $y = x^{\tfrac{1}{3}}$ should be $[0,\infty)$.
This is because we end up with a contradiction if we define $(-1)^{\tfrac{1}{3}}$ to be $\sqrt[3]{-1} = -1$ and still want to keep other index laws.
We have$$ -1 = (-1)^{\tfrac{1}{3}} = (-1)^{2\cdot\tfrac{1}{6}} = \big((-1)^2\big)^{\tfrac{1}{6}} = 1^{\tfrac{1}{6}}, $$and similarly$$ -1 = (-1)^{\tfrac{1}{3}} = (-1)^{4\cdot\tfrac{1}{12}} = \big((-1)^4\big)^{\tfrac{1}{12}} = 1^{\tfrac{1}{12}}. $$
But now$$ -1 = 1^{\tfrac{1}{6}} = 1^{\tfrac{1}{12}\cdot 2} = \big(1^{\tfrac{1}{12}}\big)^2 = (-1)^2 = 1, $$which is a contradiction.
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