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

Jika x^2+3x+2 <0 dan f(x) = x^2-3x+2 maka

${x}^{2 } + 3x + 2 = 0 \ (x + 1)(x + 2) = 0 \ x = - 1 \ x = - 2 \ - 2 < x < - 1 \ \ f( - 1) = { (- 1)}^{2} - 3( - 1) + 2 \ = 6 \ f( - 2) = ( - {2})^{2 } - 3( - 2) + 2 \ = 12 \ 6 < f(x) < 12$