Question

Sa se rezolve Inecuatia pe intervalul mentionat:
(2x+1)^3 - 3(1-2x)^2 <= 8(x^3 - 2) , x apartine [-2,4]

Answer 1

Scriem sub forma: (2x+1)³-8x³-3(1-4x+4x²)+16≤0,
(2x+1-2x)[(2x+1)²+(2x+1)*2x+(2x)²]-3+12x-12x²+16≤0,
4x²+4x+1+4x²+2x+4x²-3+12x-12x²+16≤0 ⇒ 18x ≤-14 ⇒ x≤$-\frac{7}{9}$
 intersectat cu conditia initiala avem x∈[-2;4]∩(-∞;$-\frac{7}{9}$] ⇒
x∈[-1; -7/9].

Answer 2

(2x+1)^3 - 3(1-2x)^2 ≤ 8(x^3 - 2)
8x^3+18x−2≤8x^3−16
8x^3+18x−2-8x^3+16≤0
18x+14≤0 => x≤-14/18 => x≤-7/9

x≤-7/9 dar x ∈ [-2,4] => x ∈ [-2; -7/9]