Step-by-step Solution:

\[Step 1:\] Rewrite the Function The function \( f(x) = ||x| - 1| \) can be rewritten as: \[ f(x) = \begin{cases} | -x - 1 | = x + 1 & \text{if } x < -1, \\ | -x - 1 | = -x - 1 & \text{if } -1 \leq x < 0, \\ | x - 1 | = 1 - x & \text{if } 0 \leq x < 1, \\ | x - 1 | = x - 1 & \text{if } x \geq 1. \end{cases} \] \[Step 2:\] Identify Critical Points The function \( f(x) \) involves absolute values, which introduce potential points of non-differentiability at: - \( x = -1 \), - \( x = 0 \), - \( x = 1 \). \[Step 3:\] Check Differentiability at Critical Points 1. At \( x = -1 \): - Left derivative: \( \lim_{h \to 0^-} \frac{f(-1 + h) - f(-1)}{h} = \lim_{h \to 0^-} \frac{(-(-1 + h) - 1) - 0}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1 \). - Right derivative: \( \lim_{h \to 0^+} \frac{f(-1 + h) - f(-1)}{h} = \lim_{h \to 0^+} \frac{(-(-1 + h) - 1) - 0}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1 \). - Since the left and right derivatives are not equal, \( f(x) \) is not differentiable at \( x = -1 \). 2. At \( x = 0 \): - Left derivative: \( \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{(-h - 1) - 1}{h} = \lim_{h \to 0^-} \frac{-h - 2}{h} = \text{undefined} \). - Right derivative: \( \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{(1 - h) - 1}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1 \). - Since the left derivative is undefined, \( f(x) \) is not differentiable at \( x = 0 \). 3. At \( x = 1 \):- Left derivative: \( \lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^-} \frac{(1 - (1 + h)) - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1 \). - Right derivative: \( \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^+} \frac{((1 + h) - 1) - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 \). - Since the left and right derivatives are not equal, \( f(x) \) is not differentiable at \( x = 1 \). \[Step 4:\] Conclusion The function \( f(x) = ||x| - 1| \) is not differentiable at \( x = -1 \), \( x = 0 \), and \( x = 1 \). Correct Answer: \[ \boxed{A} \]