Let $$g(x) = f(x) + f(2 + x)$$, where $$f(x) = \begin{cases}1 - \mid x \mid, & \mid x \mid \leq 1\\0, & \mid x \mid > 1\end{cases}$$ The number of points where the function g is not differentiable is

We will break the function into different parts and then plot the graph.

f(x)= 0; x$$\in\left(-\infty\ ,-1\right)\ $$

$$f\left(x\right)=1+x;\ x\in\left[-1\ ,0\right)\ $$

$$f\left(x\right)=1-x;\ x\in\left[0,1\right]$$

$$f\left(x\right)=1-x;\ x\in\left(1,\infty\ \right)$$

f(x)= 

f(x+2) is same as f(x), but shifted 2 units to the left.

So, f(x+2) can be drawn as:

We can observe that the parts where f(x+2) takes non zero value only where f(x)=0. So, The grapgh of g(x0= f(x)+ f(x+2) will be superimposed upon each other, and there won't be any change in values.

The graph of g(x) will be:

Now, g(x) will be non-differentiable at sharp corners, because at those points Left Hand limit will not equal to Right hand limit.

The graph of g(x) has 5 such sharp edges at (-3,0), (-2,1), (-1,0), (0,1) and (1,0). Therefore g(x) is non-differentiable at 5 points.