How to check the differentiability of this function. Question:

Let $\text{f(x) = max{1-x, 1+x, 2}}$. Prove that $f(x)$ is continuous at all points but not differentiable at $x = 1$ and $x = -1$.

Doubt:
I checked for the continuity of this function and successfully found it to be continuous at all points.
For differentiability, at $x=1$, I calculated the right and left hand derivative using: $$f'(a)=\lim_{h\to0}\frac{f(a\pm h)-f(a)}{\pm h}.$$
For $Rf'(1)$, the value of $f(1+h)$ will be $\text{1+1+h}$ as $\text{1+x}$ is the maximum in this case. On solving, I am getting $Rf'(1)=1$.
For $Lf'(1)$, the value of $f(1-h)$ will be $\text{2}$ as it is the maximum in this case. On solving, I am getting $Lf'(1)=\lim_{h\to0}\frac{2-2}{h}$. 
The case is similar with $\text{-1}$.
But for $\text{x=0}$, I am getting both, Right and Left hand derivative similar to $\lim_{h\to0}\frac{2-2}{h}$. But according to question, the function is non-differentiable at only $\text{x=-1,1}$.
Kindly help.
 A: At $x=0$ the right hand and left hand derivatives are both equal to $0$, since $f(x)$ is identically equal to $2$ in a neighborhood of $x=0$. So the function is differentiable with derivative $0$ at $x=0$.
Edit: A more explicit calculation
By definition, $f'(0)$ exists if and only if 
$$
\lim_{h\to0}\frac{f(h)-f(0)}{h}
$$
exists. In the case that the above limit does exist, that is exactly the derivative of $f$ at $x=0$. 
Now note that we are taking $h$ to zero; in particular, we can assume that $|h|<\frac{1}{2}$. In this case, it is easy to see $1+h<2$ and $1-h<2$, so we will have $f(h)=2$ by the definition of $f$. Thus we have 
$$
f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}\frac{2-2}{h}=\lim_{h\to0}\frac{0}{h}=0
$$
A: I prefer to draw this function first, then It is easy to see when function is continuous at all points, but not differentiable in $x=\pm1$

when you are at $$x \to 1^+ \to f(x)=max(1-x.1+x,2)=1+x\\x \to 1^- \to f(x)=max(1-x.1+x,2)=2\\x \to (-1)^+ \to f(x)=max(1-x.1+x,2)=2\\
x \to (-1)^- \to f(x)=max(1-x.1+x,2)=1-x$$ 
Now it is easy to write about $x=\pm 1$ condition
