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$f(x)=x^3+1$ and $g(x)=\sqrt[3]{x-1}$

What is the domain of $(f\circ g)(x)$

I thought that since the root function must be greater than or equal to 0 in order to be a real number, I would calculate domain by determining where the radicand is greater than or equal zero and then excluding it:

$x-1\geqslant 0 \Longrightarrow x=1$

So, I thought the domain would therefore be $[1,\infty)$

However, my textbook solutions section says the domain is actually $(-\infty, \infty)$.

Why is that?

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You don't need $x-1\geqslant 0$. It is $\sqrt[3]{x-1}$(which range and domain are $\mathbb R$), not $\sqrt{x-1}$(which domain is $[1,\infty)$ and range is $[0,\infty)$).

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  • 2
    $\begingroup$ This is correct, but it should be mentioned that taking a negative value for a cube root of a negative conflicts with the usual choice of principal value for general powers. $\endgroup$ – K B Dave Sep 5 '19 at 14:21
  • $\begingroup$ As a follow up, is this true for any odd number nth root? 3, 5 7? $\endgroup$ – Doug Fir Sep 5 '19 at 14:37
  • $\begingroup$ @Doug Fir Yeah. $\endgroup$ – FFjet Sep 5 '19 at 15:00
  • $\begingroup$ @K B Dave That's a good complement. $\endgroup$ – FFjet Sep 5 '19 at 15:02

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