\(|2x - 1| = |x| + |x - 1|\)

holds for \(x\in [-\infty , 0] \cup [1, \infty)\) 

\(\therefore |2x - 1| = |x + (x - 1)|\)

\(\le |x| + |x - 1|\)

But equality holds for \(x\in(\infty, 0] \cup [1, \infty)\)

Alternatively:

Case I: \(x \in (-\infty, 0]\) 

Then |x| = -x

& \(|x - 1| = -(x - 1)\)

Also \(|2x - 1| = -(2x - 1)\)

\(\because -(2x - 1) = -2x + 1 = - x - x + 1 = -x -(x - 1)\)

⇒ \(|2x - 1| = |x | + |x - 1| \)

given equality holds.

Case II: \(x\in \left(0, \frac 12\right)\)

Then \(|x| = x\)

\(|x - 1| = -(x - 1) = 1 - x\)

\(|2x - 1| = -(2x - 1)\)

\(\because |2x -1| = -(2x - 1) = 1- 2x = 1 -x - x \)

\(= |x - 1| - |x|\)

given equality not hold

Case III: \(x\in\left(\frac 12 , 1\right)\) 

Then \(|x| = x\)

\(|x - 1| = -(x - 1) = 1 -x\)

\(|2x - 1| = 2x - 1\) 

\(\because |2x- 1| = 2x - 1 = x + x - 1 =|x| - (1 -x)\)

\(=|x| - |x - 1|\)

given inequality not hold.

Case IV: \(x\in[1, \infty)\)

Then \(|x| = x\) 

\(|x - 1| = x - 1\)

\(|2x| - 1 = 2x - 1\)

\(\because |2x - 1| = 2x -1 = x + x - 1 = |x | + |x - 1|\)

given equality holds.

Hence, given equality holds for \(x\in(-\infty, 0]\cup[1,\infty)\).