To write $ f(x)=|x+1|-|2x-1| $ as a piecewise linear function, we split at the points where the absolute values change sign:

• $x+1=0 \Rightarrow x=-1$
• $2x-1=0 \Rightarrow x=\tfrac12$

So we consider three intervals.

1. For (x < -1)

• $x+1 < 0 \Rightarrow |x+1| = -(x+1)$
• $2x-1 < 0 \Rightarrow |2x-1| = -(2x-1)$

$$f(x)=-(x+1)-\big(-(2x-1)\big) = -x -1 +2x -1 = x -2$$

2. For $-1 \le x < \frac12)$

• $x+1 \ge 0 \Rightarrow |x+1| = x+1$
• $2x-1 < 0 \Rightarrow |2x-1| = -(2x-1)$

$$f(x)= (x+1)-\big(-(2x-1)\big) = x+1 +2x -1 = 3x$$

3. For $x \ge \tfrac12$

• $x+1 \ge 0 \Rightarrow |x+1| = x+1$
• $2x-1 \ge 0 \Rightarrow |2x-1| = 2x-1$

$$f(x)= (x+1)-(2x-1) = x+1 -2x +1 = -x+2$$

Final Piecewise Function

$$f(x)= \begin{cases} x-2, & x < -1,\\ 3x, & -1 \le x < \tfrac12,\\ -x+2, & x \ge \tfrac12 . \end{cases}$$