<s>[INST] Greatest value of $f(x)= (x+1)^{1/3}-(x-1)^{1/3}$ on $(0,1)$ Greatest value of $f(x)= (x+1)^{1/3}-(x-1)^{1/3}$ on $(0,1)$
Please guide me to solve this problem. I have differentiated it with respect to $x$ and make equal to zero, but couldn't get any point.
 [/INST] HINT: Here we can use the trigonometric substitution $x=\cos 2\theta$ where $\theta\in[0,\frac{\pi}{4}]$
Now $$f(x)=(x+1)^{1/3}-(x-1)^{1/3}=(2\cos^2 2\theta)^{1/3}-(2\sin^2 2\theta)^{1/3}.$$ Then we can use the parametric differentiation for find the minimum and maximums.
OR:
Using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ we can obtain that, $$(x+1)^{1/3}-(x-1)^{1/3}=\dfrac{2}{(x+1)^{2/3}+(x+1)^{1/3}(x-1)^{1/3}+(x-1)^{1/3}}.$$ Therefore it is enough, find the minimum value of $$a^2+ab+b^2=\Big(a+\dfrac{b}{2}\Big)^2+\dfrac{3b^2}{4}$$ in which $a=-\dfrac{b}2,$ where $a=(x+1)^{1/3},\,\,\,b=(x-1)^{1/3}$ and $x\in[0,1]$.
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