Solving $|x-1| + |x-2| \ge 4$

Question

Solve $$|x-1| + |x-2| \ge 4$$

My Attempt: I know that $|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end {cases}$. How to apply this definition to solve the problem. Please help me.

Answer 1

Hint:

Try breaking it into three parts: $x\lt1, 1\le x\le2$ and $x\gt2$.

Answer 2

If $x\le 1$ then $x-2<0$ so $|x-1|+|x-2|=-(x-1)-(x-2)=-2x+3$.

If $1\lt x\lt 2$ then $|x-1|+|x-2|=x-1-(x-2)=1$.

If $x\ge2$ then $x-1\gt0$ so $|x-1|+|x-2|=(x-1)+(x-2)=2x-3$.

Thus, $|x-1|+|x-2|\ge4$ when $x\le-\frac12$ or $x\ge\frac72$.