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.
 A: 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 $x\in[0,1]$.
A: By Jensen $$\left(\frac{\sqrt[3]{1+x}+\sqrt[3]{1-x}}{2}\right)^3\leq\frac{1+x+1-x}{2}=1,$$
which gives $\sqrt[3]{1+x}+\sqrt[3]{1-x}\leq2$.
The equality occurs for $x=0$.
Id est, the answer is $2$.
We used the following inequality.
For non-negatives $a$ and $b$ we have:
$$\left(\frac{a+b}{2}\right)^3\leq\frac{a^3+b^3}{2},$$
which is $(a+b)(a-b)^2\geq0$.
Also we can use P_M.
A: Differentiating the mentioned function gives:
$$\frac{d}{dx}f(x)=\frac{1}{3(x+1)^{\frac{2}{3}}}-\frac{1}{3(x-1)^{\frac{2}{3}}}
$$
Since we are trying to find the maximum value of a function $(0,1)$, First we set the differential of the function equal to zero, Which is the definition of the maximum, as well as minimum (we'll see which is the case). Doing this we get the following equation:
$$\frac{1}{3(x_1+1)^{\frac{2}{3}}}=\frac{1}{3(x_1-1)^{\frac{2}{3}}}
$$  
Now we can simplify this equation in a simpler form:
$$\frac{1}{(x_1+1)^{2}}=\frac{1}{(x_1-1)^{2}}
$$
$$(x_1+1)^2=(x_1-1)^2
$$
Thus we can see that $x_1=0$, and is the only solution because this is a first order equation, Thus the function $f(x)$ has only one point at which differential of it equals 0.
Plugging in $x=0$ we get:
$$(1)^2=(-1)^2
$$
The only global maximum of the function $f(x)$ is located at $x=0$, Thus there is no local maximum on the interval $(0,1)$
We can see see that $x_1=0$ is the global maximum of the function from the graph of $f(x)$:

By using Desmos graphing calculator.
A: You can recheck : For 
$$f(x)=(x+1)^{\frac{1}{3}}-(x-1)^{\frac{1}{3}}$$
$$f'(x)=\frac{1}{3}(x+1)^{-\frac{2}{3}}-\frac{1}{3}(x-1)^{-\frac{2}{3}}$$
You won't find a point here. Before $x=1$, the second term is a negative number, which raised to fractional exponent gives a complex number. Even $f(x)$ before $x=1$ will evaluate to a complex number. So, as Andrews says, "There needn't be a maximum value" on $(0,1)$
A: The derivative should be 
$$ f'(x) = \frac{1}{3}\left( \frac{1}{(x+1)^{2/3}} - \frac{1}{(x-1)^{2/3}} \right)$$
This is always negative because $(x+1)^{2/3} > (x-1)^{2/3} \ge 0$
Therefore $f(x)$ is always decreasing. If the endpoints were included, the maximum would be $f(0) = 2$, but as it stands there is no maximum.
