Question

|2x+1| + |3x| + |4x-1| ≥ x-2
imi explica si mie cineva cum se face?
in ce mod?

Answer 1

Cazul (1) x < -1/2:

-2x-1 - 3x -4x+1 ≥ x -2

-9x ≥ x -2

-10x ≥ -2

x ≤ 1/5

=> x ∈ (-ꝏ, -1/2)

Cazul (2) x ∈ [-1/2, 0):

2x+1 - 3x - (4x-1) ≥ x-2

2x-3x-4x +1+1 ≥ x-2

-6x ≥ 0

x ≤ 0

=> x ∈ [-1/2, 0)

Cazul (3) x ∈ [0, 1/4):

(2x+1) + 3x - (4x-1) ≥ x-2

x+2 ≥ x-2

2 ≥ -2

=> x ∈ [0, 1/3)

Cazul (4) x ≥ 1/4:

2x+1 + 3x + 4x-1 ≥ x-2

9x ≥ x -2

8x ≥ -2

x ≥ -1/4

=> x ∈ [1/4, +ꝏ)

Din (1) ∪ (2) ∪ (3) ∪ (4):

=> S = ℝ