Let $g(x) = \sqrt[3]{x}$

Asked Apr 21 '16 at 19:45

Active Apr 21 '16 at 19:51

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a) Prove that $g$ is continuous at $0$ This is fairly straight forward $ \forall \epsilon >0$ $\exists \delta >0$ s.t. when $|x-c|<\delta$, it follows that $|f(x)-f(c)|<\epsilon$

At $c=0$, $|f(x)-0|<\epsilon$

$\sqrt[3]{x} < \epsilon$

$x < \epsilon^3$

Letting $\delta = \epsilon^3$

$|x-c|<\delta \rightarrow |x-c|<\epsilon^3$, $c=0$

$\sqrt[3]{x-0}<\epsilon$

$|f(x)-c|<\epsilon$

b) Prove that g is continuous at a point $c \neq 0$ I am having a bit more trouble with this one but somehow I'm supposed to use $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ Would I let $a=\sqrt[3]{x} $ and $b=\sqrt[3]{c}$? Do I have to use this or is there a different way? Exactly how do I use that identity?

Thanks!

edited Apr 21 '16 at 19:51

asked Apr 21 '16 at 19:45

Jabernet

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It should be $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ – Nicholas Stull Apr 21 '16 at 19:50
@NicholasStull thanks for catching that, I fixed it. – Jabernet Apr 21 '16 at 19:51
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See here. – Mhenni Benghorbal Apr 21 '16 at 19:54
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I believe that you would use the substitutions you propose. Note that the quadratic polynomial has no zeroes, so that factor always has one sign. – colormegone Apr 21 '16 at 19:55
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One thing that might help: If $a>0$, $b>0$, $\min{a^2,b^2} < a^2+b^2+ab \leq 3\max{a^2,b^2}$. A similar statement holds if $a<0,b<0$. – Nicholas Stull Apr 21 '16 at 20:03
@NicholasStull, what I have now is that as $f(c)$ varies by $\epsilon$, i.e. $\sqrt[3]{c} + \epsilon, \sqrt[3]{c} - \epsilon$ , we have $x$ that varies by $(\sqrt[3]{c} - \epsilon)^3, (\sqrt[3]{c}+\epsilon)^3$ and I want the min of those as my $\delta$? Or am I missing something here? – Jabernet Apr 27 '16 at 21:38

Let $g(x) = \sqrt[3]{x}$

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