Rishi Sharma
Last Activity: 4 Years ago
Dear Student,
Please find below the solution to your problem.
To get rid of the absolute value functions, we need to take 4 cases:
Case 1: x≥3or [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)
Thanks and Regards