jika x^2+3x+2 <0 dan f(x) = x^2-3x+2 maka A. 0 < f(x) < 6 B. f(x) ≥ 3/2 C. f(x) < 12 D. f(x) > 0 E. 6 < f(x) < 12 Tolong disertai caranya ya terimakasih

[tex] {x}^{2 } + 3x + 2 = 0 \\ (x + 1)(x + 2) = 0 \\ x = - 1 \\ x = - 2 \\ - 2 < x < - 1 \\ \\ f( - 1) = { (- 1)}^{2} - 3( - 1) + 2 \\ = 6 \\ f( - 2) = ( - {2})^{2 } - 3( - 2) + 2 \\ = 12 \\ 6 < f(x) < 12[/tex]