Hello i got problem with nr 22. Because x^(1/3) is delfined for possitive and negative so idk what its domain is.

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- Feb 8th 2013, 06:31 AMPetrusIf a function is continuous
Hello i got problem with nr 22. Because x^(1/3) is delfined for possitive and negative so idk what its domain is.

- Feb 8th 2013, 06:36 AMPlatoRe: If a function is continuous
- Feb 8th 2013, 06:41 AMjakncokeRe: If a function is continuous
if $\displaystyle \sqrt[3]{a} = x $ then $\displaystyle a = x^3 $ or a solution must exist for $\displaystyle x^3 - a = 0 $ Now what do you know about cubic roots? It has 3 roots, and complex roots only come in pairs when you have real coefficients, so u could either have 2 complex roots and 1 real root or 3 real roots. Which means you always have atleast one real root, so the domain is all real numbers.