Monotonicity of function $f(x)=\sqrt[3]{(x+1)^2}-\sqrt[3]{x^2}$ I'm examining the properties of the function
$$f(x)=\sqrt[3]{(x+1)^2}-\sqrt[3]{x^2}$$
and I'm stuck at monotonicity.
I know that monotonicity depends from first derivative (if first derivative is grater, equal or lower than zero).
So I found the first derivative and got
$$f'(x)=\frac{2}{3}\left(\frac{1}{\sqrt[3]{x+1}}-\frac{1}{\sqrt[3]{x}}\right)$$ 
and I know it's correct.  
Now I need the zeroes of the derivative so I can see important points where monotonicty potentially changes.
When I try to find zeroes I get:
$$\frac{2}{3}\left(\frac{1}{\sqrt[3]{x+1}}-\frac{1}{\sqrt[3]{x}}\right)=0$$
$$\frac{1}{\sqrt[3]{x+1}}-\frac{1}{\sqrt[3]{x}}=0$$
$$\frac{1}{\sqrt[3]{x+1}}=\frac{1}{\sqrt[3]{x}}$$
$$x+1=x$$
which makes no sense...  
What's the problem here?
How can I find monotonicity of $f(x)$ when I can't find the zeroes? Is my method of finding zeroes wrong? I did this with other similar functions and was able to find zeroes without a problem and monotonicity was correct after.  
Edit: To clarify, $\sqrt[3]{x}$ is defined on $\Bbb R$, and the whole function is defined on $\Bbb R$.
 A: Hint:
Your derivative
$$f'(x)=\frac{2}{3}\left(\frac{1}{\sqrt[3]{x+1}}-\frac{1}{\sqrt[3]{x}}\right)$$
does not exists for $x=-1$ and $x=0$ ( in these points the function has a cusp). It is easy to see that  the derivative is positive for $-1<x<0$ and negative outside this interval.
A: The function  $$f(x)=\sqrt[3]{(x+1)^2}-\sqrt[3]{x^2}$$ is decreasing on $(-\infty, -1)\cup (0, \infty)$ and increasing on $(-1, 0)$
It changes course at critical points $x=-1$ and $x=0$ from decreasing to increasing and again from increasing to decreasing. 
You may verify these results by testing the derivative on each interval.  
A: You need to study the sign of $f'(x)$ that is
$$\frac{1}{\sqrt[3]{x+1}}-\frac{1}{\sqrt[3]{x}}=\frac{\sqrt[3]{x}-\sqrt[3]{x+1}}{\sqrt[3]{x+1}\cdot\sqrt[3]{x}}$$
note that since $\sqrt[3]{.}$ is strictly increasing and odd $$\sqrt[3]{x}-\sqrt[3]{x+1}<0$$
and
$$\sqrt[3]{x+1}\cdot\sqrt[3]{x}>0\iff (x+1)\cdot x>0$$
$$\sqrt[3]{x+1}\cdot\sqrt[3]{x}<0\iff (x+1)\cdot x<0$$
thus
$$f'(x)\begin{cases}>0\quad x\in(-1,0)\implies \text{f strictly increasing}\\ \\<0\quad x\in(-\infty,-1)\cup(0,+\infty)\implies \text{f strictly decreasing}\end{cases}$$
A: $g(x)=x^{2/3}$ is blatantly a concave function on $\mathbb{R}^+$. It follows that
$$ \Delta(a,b) = \frac{g(a)-g(b)}{a-b}\text{ if }a\neq b,\qquad g'(a)\text{ if } a=b $$
is decreasing with respect to both $a$ and $b\in\mathbb{R}^+$. By considering $a=x+1$ and $b=x$ we have that $f(x)$ is decreasing on $\mathbb{R}^+$.
