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I have two functions, $f(x) = x^{1/3}$ and $g(x)= f(x+3) − f(x)$ and I have to prove that $g(x) \leq x^{−2/3}$ for all $x>0$ Besides that, I also have to find the extreme values and the vertical and horizontal asymptotes of g.

I know this first part has to be solved using the mean value theorem, but I don't know how to apply it in this situation. I have written down that $g(x) = (x+3)^{1/3} - x^{1/3}$

Can someone give me the domain to use in the MVT? That is the main struggle

All help is appreciated

Thanks in advance :)

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  • $\begingroup$ Formatting tips here:meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Nick Oct 16 '16 at 11:33
  • $\begingroup$ Welcome to math stack exchange! $\endgroup$ – Peter Oct 16 '16 at 11:34
  • $\begingroup$ Just apply the MWT and see what gives $\endgroup$ – Hagen von Eitzen Oct 16 '16 at 11:41
  • $\begingroup$ @HagenvonEitzen well I don't know how to do that, so could you please explain it? $\endgroup$ – Amaluena Oct 16 '16 at 12:33
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From $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$, we get $a-b=\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}}$, apply it to your question, we get $g(x)=\frac{3}{(x+3)^{\frac{2}{3}}+x^{\frac{1}{3}}(x+3)^{\frac{1}{3}}+x^{\frac{2}{3}}}\leq\frac{3}{3x^{\frac{2}{3}}}=x^{−2/3}$ for all $x>0$, then the other questions may be easier to solve.

In order to apply MVT, we can write as following:$g(x)= f(x+3) − f(x)=\int_{x}^{x+3} {f^{'}(t)dt}=((x+3)-x)(\frac{1}{3}t^{−2/3})=t^{−2/3}$ and $x\leq t\leq x+3$, then we get $(x+3)^{-2/3}\leq g(x) \leq x^{−2/3}$.

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  • $\begingroup$ where in your answer do you use the MVT? $\endgroup$ – Amaluena Oct 17 '16 at 19:42
  • $\begingroup$ Well, as for your demand of using MVT, I came up with a new solution and added it to my previous answer. $\endgroup$ – Tom Oct 18 '16 at 3:31

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