Show that the function f(x) = |x - 1| + |x + 1|, for all x ∈ R, is not differentiable at the points x = -1 and x = 1.

From the definition of absolute value we have:
f(x) = \begin{cases} -(x - 1) + -(x + 1), & x < -1 \ x - 1 + -(x + 1), & -1 ≤ x < 1 \ x - 1 + x + 1, & x ≥ 1 \end{cases} = \begin{cases} -2x, & x < -1 \ -2, & -1 ≤ x < 1 \ 2x, & x ≥ 1 \end{cases}
\implies f'(x) = \begin{cases} -2, & x < -1 \ 0, & -1 ≤ x < 1 \ 2, & x ≥ 1 \end{cases}
\implies f'(-1^-) = -2, f'(-1^+) = 0 \implies f'(1^-) = 0, f'(1^+) = 2
The above derivative calculations can be done from the first principle also.
Since the left hand and right hand derivatives are not equal at the points 1 and -1, the function is not differentiable at those points.