If $f(x)=\begin{cases} x-2 & x\le 0 \\ 4-x^2 & x>0 \end{cases}$, then number of points where $f(f(x))$ is discontinuous is $f(x)$ is discontinuous at $x=0$, so $f(f(x))$ will be discontinuous at $f(x)=0$
I got $x=2$ as the value of where $f(x)=0$.
But the answer says there are two values. What is the other value I am missing?
 A: To solve these kind of problems, especially when we have a pretty simple function, we should always attempt to graph it and see what's going on. It's quite difficult to imagine composition of functions so we need to take it slowly.
If you graph $f(x)$ it's clear that we have three regions of interest. One when $x\leq 0$, another when $0<x<2$ and finally when $x\geq 2$. Why are these regions of interest? Well because this is when the graph (the function more precisely) admits values that are either positive or negative and remember that $f(x)$ is entirely dependent on whether $x>0$ (positive) or $x\leq 0$ (negative).
So, lets consider when $x\leq 0$. From our graph we have $f(x)\leq 0$ which is sufficient information to calculate $f(f(x))$. Because $x\leq 0$ we have $f(x)=x-2$ and since $f(x)\leq 0$ we conclude that $$f(f(x))=(x-2)-2=x-4.$$
Applying this sort of logic to the regions $0<x<2$ and $x\geq 2$ we conclude that,
$$ f(f(x)) =
 \begin{cases} 
      x-4 & x\leq 0, \\
      4-(4-x^2)^2 & 0< x< 2, \\
      2-x^2 & x\geq 2. 
   \end{cases}
$$
Hence, it is obvious there are two discontinuities, one at $x=0$ and another at $x=2$.
