1. Given function

We have $f(x)=\left|2-|x-3|\right|.$

We need:

• the set $S$ of points where $f$ is not differentiable,
• then compute $\sum_{x\in S} f(f(x)).$

2. Find where $f(x)$ is not differentiable

The function contains absolute value expressions, so nondifferentiability can occur where the inside of an absolute value becomes zero.

Step 1: Inner absolute value

The inner expression is $|x-3|.$ This is not differentiable at $x=3.$

Step 2: Outer absolute value

The outer absolute value is applied to $2-|x-3|.$ This may be nondifferentiable where $2-|x-3|=0.$ So, $|x-3|=2 \implies x-3=\pm 2 \implies x=1,5.$

Thus, possible nondifferentiable points are $S=\{1,3,5\}.$

These are indeed all nondifferentiable points.

3. Compute $f(f(x))$ for each $x\in S$

First compute $f(1), f(3), f(5)$.

At $x=1$

$f(1)=\left|2-|1-3|\right|=|2-2|=0.$ Then $f(f(1))=f(0)=\left|2-|0-3|\right|=|2-3|=1.$

At $x=3$

$f(3)=\left|2-|3-3|\right|=|2-0|=2.$ Then $f(f(3))=f(2)=\left|2-|2-3|\right|=|2-1|=1.$

At $x=5$

$f(5)=\left|2-|5-3|\right|=|2-2|=0.$ Then $f(f(5))=f(0)=1.$

4. Sum

Therefore, $\sum_{x\in S} f(f(x))=1+1+1=3.$

5. Comparison with stored answer

Our derived answer is $3$, which matches the stored correct answer.