|x−3|+|2x+4|+|x|≤11
If x≥3,
x−3+2x+4+x≤11
⇒4x≤10
⇒x≤2.5 No solution
If 0≤x<3,
−x+3+2x+4+x≤11⇒x≤2⇒x∈[0,2]
If −2≤x<0,
−x+3+2x+4−x≤11
⇒7≤11 which is true
⇒x∈[−2,0)
If x<−2,
−x+3−2x−4−x≤11
⇒−4x−1≤11
⇒x≥−3⇒x∈[−3,−2)
Hence, x∈[0,2]∪[−2,0)∪[−3,−2)
i.e., x∈[−3,2]