How to solve the question related to differentiation. The question is :
Find the point of differentiability of the function $f : \mathbb {R} \longrightarrow \mathbb {R}$ defined by $f(x) = |x - x^3|$.
Please help me.Thank you in advance.
 A: Note that $x-x^3=x(1-x^2)=0$ for $x=0,\pm1$.  
The derivative fails to exist at $x=0,\pm1$.
For $x>1$, we find $f(x)=x^3-x$ and $f'(x)=3x^2-1$.
For $0<x<1$, $f(x)=x-x^3$ and $f'(x) =1-3x^2$.
For $-1<x<0$, $f(x)=x^3-x$ and $f'(x)=3x^2-1$.
For $x<-1$, $f(x)=x-x^3$ and $f'(x)=1 -3x^2$.
Therefore, we have
$$f'(x)=\begin{cases}3x^2-1&,x>1\\\\
1-3x^2&,0<x<1\\\\
3x^2-1&,-1<x<0\\\\
1-3x^2&,x<-1\end{cases}$$
To see clearly the failure of existence of $f'(0)$ and $f'(\pm1)$, note that 
$$\lim_{h\to 0}\frac{|h-h^3|}{h}=\lim_{h\to 0}\frac{|h|}{h}|1-h^2|$$
does not exist since the limits from the right-hand side and left-hand side are not equal.
A: We know the function $|y|$ is differentiable everywhere except at $0$, where the left hand derivative is not equal to the right hand derivative.
The function $x-x^3$ is a polynomial and hence is differentiable everywhere. 
Thus, in order to see where $|x-x^3|$ is not differentiable, it suffices to find at which points $x-x^3$ becomes zero. This happens when $x$ is $0$, $1$ or $-1$. At these points, we can find the right hand and left hand derivatives of $f$ and see that they are unequal.
Thus, $f$ is differentiable everywhere except $0$, $1$ and $-1$.
A: Hint: $$f\left( x \right) =\left| x \right| $$ is not differentiable at the point $x=0$
A: Get intuition by drawing a graphical representation (see below): the values of $x$ for which there is no differentiability are the values of x with "angular points" on the curve (left derivative $\neq$ right derivative).

