Step 1: Analyze the given function.1. For \( x \geq 0 \), \( |x| = x \). The function becomes:\[f(x) = |1 - x + x| = |1| = 1.\]2. For \( x<0 \), \( |x| = -x \). The function becomes:\[f(x) = |1 - x - x| = |1 - 2x|.\]Step 2: Check continuity.For \( x \geq 0 \), \( f(x) = 1 \). For \( x<0 \), \( f(x) = |1 - 2x| \). At \( x = 0 \):\[f(0^+) = 1, \quad f(0^-) = |1 - 2(0)| = 1.\]At \( x = 1 \), \( f(1^+) = 1 \) and \( f(1^-) = 1 \). Therefore, \( f(x) \) is continuous everywhere.
Final Answer: \( \boxed{{(D)}} \)