# Prove $g(x) \leq x^{−2/3}$ for all $x>0$ using the given formulas

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

• Formatting tips here:meta.math.stackexchange.com/questions/5020/… – Nick Oct 16 '16 at 11:33
• Welcome to math stack exchange! – Peter Oct 16 '16 at 11:34
• Just apply the MWT and see what gives – Hagen von Eitzen Oct 16 '16 at 11:41
• @HagenvonEitzen well I don't know how to do that, so could you please explain it? – Amaluena Oct 16 '16 at 12:33

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 $(x+3)^{-2/3}\leq g(x) \leq x^{−2/3}$.