Rule: |f(x)| is non-differentiable at x = a when f(a) = 0 and f'(a) != 0.

Why? If f(a) = 0 and f'(a) > 0, then f changes sign at a (positive to the right, negative to the left near a). The absolute value creates a corner (V-shape).

When is |f(x)| differentiable at f(a) = 0? Only if f'(a) = 0 (f touches zero tangentially). Example: |$x^2$| = $x^2$ is differentiable everywhere.

Application: |P(x)| for polynomial P. Non-differentiable at each simple root of P (where P' != 0), but differentiable at repeated roots (where P' = 0).

Example: |(x-1)^2(x-3)| is non-differentiable at x = 3 (simple root, P'(3) = 4 != 0) but differentiable at x = 1 (double root, P'(1) = 0).