Key Concepts and Formulas

• Differentiability of Absolute Value Functions: A function of the form $|g(x)|$ is not differentiable at points where $g(x) = 0$ if $g'(x)$ is non-zero at those points. More generally, a function is not differentiable at points where its graph has a sharp corner or a cusp.
• Composition of Functions: For a composite function $f(g(x))$, the output of the inner function $g(x)$ becomes the input for the outer function $f$.
• Properties of Absolute Value: $|a|$ represents the distance of $a$ from zero.

Step-by-Step Solution

Step 1: Analyze the function $f(x) = |2-|x-3||$ to find points of non-differentiability.

The function $f(x)$ involves nested absolute values. We need to find the points where the arguments of the absolute value functions become zero, as these are potential points of non-differentiability.

First, consider the inner absolute value: $|x-3|$. This is zero when $x-3 = 0$, which means $x=3$.

Now, consider the outer absolute value: $|2 - |x-3||$. This is zero when $2 - |x-3| = 0$, which implies $|x-3| = 2$. This equation has two possibilities:

1. $x-3 = 2 \implies x = 5$
2. $x-3 = -2 \implies x = 1$

So, the potential points of non-differentiability are $x=1$, $x=3$, and $x=5$. Let's examine the behavior of the function at these points.

• At $x=3$: Let $g(x) = 2 - |x-3|$. For $x > 3$, $|x-3| = x-3$, so $g(x) = 2 - (x-3) = 5-x$. The derivative is $g'(x) = -1$. For $x < 3$, $|x-3| = -(x-3) = 3-x$, so $g(x) = 2 - (3-x) = x-1$. The derivative is $g'(x) = 1$. Since the left-hand derivative (1) and the right-hand derivative (-1) are different at $x=3$, the function $f(x)$ is not differentiable at $x=3$.

• At $x=1$: When $x=1$, $|x-3| = |1-3| = |-2| = 2$. So, $f(1) = |2 - 2| = 0$. For $x$ slightly less than 1 (e.g., $x = 1-\epsilon$ where $\epsilon > 0$ is small), $x-3$ is negative and close to -2. So, $|x-3| = -(x-3) = 3-x$. $f(x) = |2 - (3-x)| = |x-1|$. For $x<1$, $x-1 < 0$, so $f(x) = -(x-1) = 1-x$. The derivative is $-1$. For $x$ slightly greater than 1 (e.g., $x = 1+\epsilon$), $x-3$ is negative and close to -2. So, $|x-3| = -(x-3) = 3-x$. $f(x) = |2 - (3-x)| = |x-1|$. For $x>1$, $x-1 > 0$, so $f(x) = x-1$. The derivative is $1$. Since the left-hand derivative (-1) and the right-hand derivative (1) are different at $x=1$, the function $f(x)$ is not differentiable at $x=1$.

• At $x=5$: When $x=5$, $|x-3| = |5-3| = |2| = 2$. So, $f(5) = |2 - 2| = 0$. For $x$ slightly less than 5 (e.g., $x = 5-\epsilon$), $x-3$ is positive and close to 2. So, $|x-3| = x-3$. $f(x) = |2 - (x-3)| = |5-x|$. For $x<5$, $5-x > 0$, so $f(x) = 5-x$. The derivative is $-1$. For $x$ slightly greater than 5 (e.g., $x = 5+\epsilon$), $x-3$ is positive and close to 2. So, $|x-3| = x-3$. $f(x) = |2 - (x-3)| = |5-x|$. For $x>5$, $5-x < 0$, so $f(x) = -(5-x) = x-5$. The derivative is $1$. Since the left-hand derivative (-1) and the right-hand derivative (1) are different at $x=5$, the function $f(x)$ is not differentiable at $x=5$.

Therefore, the set of points where $f(x)$ is not differentiable is $S = \{1, 3, 5\}$.

Step 2: Evaluate $f(f(x))$ for each $x \in S$.

We need to calculate $f(f(1))$, $f(f(3))$, and $f(f(5))$.

• Calculate $f(1)$: From Step 1, we found that $f(1) = |2 - |1-3|| = |2 - |-2|| = |2 - 2| = 0$. Now, calculate $f(f(1)) = f(0)$. $f(0) = |2 - |0-3|| = |2 - |-3|| = |2 - 3| = |-1| = 1$. So, $f(f(1)) = 1$.

• Calculate $f(3)$: From Step 1, we found that $f(3) = |2 - |3-3|| = |2 - |0|| = |2 - 0| = 2$. Now, calculate $f(f(3)) = f(2)$. $f(2) = |2 - |2-3|| = |2 - |-1|| = |2 - 1| = |1| = 1$. So, $f(f(3)) = 1$.

• Calculate $f(5)$: From Step 1, we found that $f(5) = |2 - |5-3|| = |2 - |2|| = |2 - 2| = 0$. Now, calculate $f(f(5)) = f(0)$. We already calculated $f(0)$ in the evaluation of $f(f(1))$. $f(0) = 1$. So, $f(f(5)) = 1$.

Step 3: Calculate the sum $\sum\limits_{x \in S} {f(f(x))}$.

The sum is $f(f(1)) + f(f(3)) + f(f(5))$. From Step 2, we have: $f(f(1)) = 1$ $f(f(3)) = 1$ $f(f(5)) = 1$

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

Common Mistakes & Tips

• Incorrectly identifying points of non-differentiability: Always check the points where the argument of an absolute value function becomes zero. For nested absolute values, work from the inside out.
• Errors in function composition: Carefully substitute the output of the inner function into the outer function. Double-check calculations for $f(a)$ where $a$ is itself an output of $f$.
• Graphing the function: Visualizing the graph of $f(x) = |2-|x-3||$ can be helpful. The graph has "V" shapes at the points of non-differentiability. The graph of $|x-3|$ is a V-shape with its vertex at (3,0). The graph of $2-|x-3|$ is an inverted V-shape with its vertex at (3,2). The graph of $|2-|x-3||$ then reflects the parts below the x-axis, leading to points of non-differentiability where the function touches or crosses the x-axis (i.e., $f(x)=0$) and where the inner function's absolute value causes a change in slope.

Summary

The problem requires us to find the sum of $f(f(x))$ for all points $x$ where the function $f(x) = |2-|x-3||$ is not differentiable. First, we identified the points of non-differentiability by analyzing the arguments of the absolute value functions, which are $x=1, 3, 5$. Then, we calculated $f(x)$ at these points and subsequently $f(f(x))$ for each of them: $f(f(1))=1$, $f(f(3))=1$, and $f(f(5))=1$. Finally, we summed these values to get the result.

The final answer is $\boxed{3}$.