Step 1: Understanding the Concept:
A real-valued function $f(x)$ fails to be differentiable (meaning its derivative $f'(x)$ is undefined) at points where its geometric graph contains sharp corners, spikes, or cusps. At these specific locations, the slope of the graph changes abruptly, causing the left-hand derivative ($\text{LHD}$) and right-hand derivative ($\text{RHD}$) to have completely different values. For absolute value functions of the linear form $y = |x - c|$, a sharp corner always forms at the critical point where the expression inside the modulus brackets equals zero ($x = c$).
Step 2: Detailed Explanation:
Let's analyze the given composite absolute value function:
$$ f(x) = |x - 1| + |x - 2| + |x - 3| $$
This function is a sum of three separate modulus expressions. Each individual absolute value component has its own critical corner point where the slope flips signs:
1. The term $|x - 1|$ switches behavior at $x - 1 = 0 \implies x = 1$.
2. The term $|x - 2|$ switches behavior at $x - 2 = 0 \implies x = 2$.
3. The term $|x - 3|$ switches behavior at $x - 3 = 0 \implies x = 3$.
Let's mathematically evaluate one of these critical boundary points, for example $x = 2$, by calculating the directional derivatives to show why the derivative is undefined:
- For a region just to the left of 2 ($1<x<2$):
$$ |x-1| = (x-1), \quad |x-2| = -(x-2), \quad |x-3| = -(x-3) $$
$$ f(x) = (x - 1) - (x - 2) - (x - 3) = -x + 4 \implies \text{LHD} = f'(x) = -1 $$
- For a region just to the right of 2 ($2<x<3$):
$$ |x-1| = (x-1), \quad |x-2| = (x-2), \quad |x-3| = -(x-3) $$
$$ f(x) = (x - 1) + (x - 2) - (x - 3) = x - 2 \implies \text{RHD} = f'(x) = +1 $$
Because the left-hand slope ($-1$) does not match the right-hand slope ($+1$) at $x = 2$, the graph forms a sharp corner vertex, making the derivative undefined at that point. The same non-differentiable corner behavior occurs at $x = 1$ and $x = 3$.
Thus, the derivative is undefined at $x = 1, 2, 3$. This corresponds to option (B).
Step 3: Final Answer:
The points where the derivative $f'(x)$ is undefined are $x = 1, 2, 3$.