Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $S$ be the set of points where the function, $f(x)=|2-|x-3||,x\in \mathbb{R}$, is not differentiable. Then, the value of ${\sum }_{x\in S}f(f(x))$ is equal to

We begin our journey at the very heart of the function: the innermost expression, $|x-3|$. Imagine you are standing on the Cartesian plane.

The function $g(x)=|x-3|$ is the classic V-shape, a fundamental geometric object. Its vertex sits precisely at $x=3$.

At $x=3$, the function changes its behavior abruptly. To the left of $3$, the slope is $-1$; to the right, it is $1$. This sudden shift in slope is the definition of a sharp corner, and in the world of calculus, a sharp corner means the function is not differentiable.

Thus, we have identified our first member of the set $S$: $x=3$.

Flipping and Shifting

Now, we layer the function. We look at $h(x)=2-|x-3|$.

Think of this as a transformation. The negative sign in front of the modulus flips our V-shape upside down, creating an inverted V.

Then, the constant $2$ shifts this entire inverted V upwards by two units. The vertex, which was at $(3,0)$, has now migrated to $(3,2)$.

The Outer Modulus

This is where the magic happens. We apply the outermost modulus: $f(x)=|2-|x-3||$.

An outer modulus acts as a mirror. It takes any part of the graph that has dipped below the x-axis and reflects it upwards, making it positive.

Our inverted V, which had tails extending downwards to negative infinity, now has those tails flipped up. This creates a W-like shape.

Where the graph crosses the x-axis, it creates new sharp corners. These are the roots of the equation:

Solving this, we find $|x-3|=2$, which leads us to two cases: $x-3=2$ (so $x=5$) and $x-3=-2$ (so $x=1$).

These are our new points of non-differentiability. Our set $S$ is now complete: $S=\{1,3,5\}$.

The Composite Dance

Now that we have our set $S$, we calculate the sum of $f(f(x))$ for each $x$ in $S$.

For $x=1$:

For $x=3$:

For $x=5$:

A Beautiful Symmetry

We have arrived at the final step. We sum these values: $1+1+1=3$.

It is truly elegant how the complexity of the function collapses into such a clean, simple integer. The final answer is 3.