Question

|x−1|+|x−2|+|x−3|≥6 solve for x

Answer 1

Step-by-step explanation:

Case 1:  x≥3  or  [3,∞)  

Here all the bars will open with a positive sign.

x−1+x−2+x−3≥6  

3x≥12  

x≥4  

Case 2:  2≤x<3  or  [2,3)  

Here, only  |x−3|  will open with a negative sign.

x−1+x−2−x+3≥6  

x≥6  

But this lies out of  [2,3)  

So no solution

Case 3:  1≤x<2  or  [1,2)  

Here only  |x−1|  will open with a positive sign.

x−1−x+2−x+3≥6  

−x≥2  

x≤−2  

Case 4:  x<1(−∞,1)  

Here all bars will open with a negative sign.

−x+1−x+2−x+3≥6  

x≤0  

Eliminating the redundant solutions, we get the final answer as:

x∈(−∞,0]∪[4,∞)  

or  x∈R−(0,4)

Hope it is helpful, please mark as brainliest