Let S be the set of points where the function f(x)=|2-|x-3||, xR, is not differentiable.

Question

Let $S$ be the set of points where the function $f(x)=|2-|x-3||,\ x\in\mathbb{R}$, is not differentiable. Then $\displaystyle\sum_{x\in S}f(f(x))$ is equal to \_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

$f$ is non-differentiable at $x=1,3,5$ (corners of the nested absolute values). $f(1)=|2-2|=0\Rightarrow f(f(1))=f(0)=|2-3|=1$. $f(3)=|2-0|=2\Rightarrow f(f(3))=f(2)=|2-1|=1$. $f(5)=|2-2|=0\Rightarrow f(f(5))=f(0)=1$.

Sum $=1+1+1=3$.

Solutions · 1