The function $f(x) = |x| + |x - 1|$ is:

The given function is $f(x) = |x| + |x - 1|$.

We can redefine the function as a piecewise function:

$f(x) = \begin{cases} -x - (x - 1) = -2x + 1, & x < 0 \\ x - (x - 1) = 1, & 0 \le x < 1 \\ x + (x - 1) = 2x - 1, & x \ge 1 \end{cases}$

Checking differentiability at $x = 0$:
Left Hand Derivative (LHD) $= \lim_{x \to 0^-} f'(x) = -2$
Right Hand Derivative (RHD) $= \lim_{x \to 0^+} f'(x) = 0$
Since LHD $\neq$ RHD, $f(x)$ is not differentiable at $x = 0$.

Checking differentiability at $x = 1$:
Left Hand Derivative (LHD) $= \lim_{x \to 1^-} f'(x) = 0$
Right Hand Derivative (RHD) $= \lim_{x \to 1^+} f'(x) = 2$
Since LHD $\neq$ RHD, $f(x)$ is not differentiable at $x = 1$.

Therefore, the function is neither differentiable at $x = 0$ nor at $x = 1$.

Answer: Neither differentiable at $x = 0$ nor $x = 1$