If x satisfies |x−1|+|x−2|+|x−3|≥6, then
x≤0 or x≥4
We have to solve
|x−1|+|x−2|+|x−3|≥6
Let |x−1|+|x−2|+|x−3|<6
Now |(x−1)+(x−2)+(x−3)|<|x−1|+|x−2|+|x−3|<6
∴|3x−6|<6∴|x−2|<2∴−2<x−2<2
or 0<x<4
Hence, for |x−1|+|x−2|+|x−3|≥6,x≤0 or x≥4
Alternative Method:
We have f(x) =|x-1|+|x-2|+|x-3|
To draw the graph of the function, we consider
f(0) =1 +2 +3 =6
f(1) =0 +1 +2 =3
f(2) =1+0+1 =2
f(3) =2 +1 +0 =3
f(4) =3 +2 +1 =6
Now, join these points with line segments as shown in the following figure.
Note: The methods to drew the graphs of such functions is explained in the book "Graphs for JEE Main and JEE Advanced".
Also, draw the line y =6
For f(x)=|x−1|+|x−2|+|x−3|≥6, graph y=f(x) must lie above the graph of y=6.
As shown in the figure it occurs for x≤0 or x≥4