Question

f(x)=2x^3-3x^2+7
demonstrati că f(x) ≧ 6, pentru orice Xㅌ [0,+⑅)

Answer 1

$\displaystyle 2x^3-3x^2+7= \\ \\ =2x^3-3x^2+1+6= \\ \\ =2x^3-2x^2-x^2+1+6= \\ \\ =2x^2(x-1)-(x^2-1)+6= \\ \\ =2x^2(x-1)-(x-1)(x+1)+6= \\ \\ =(x-1)(2x^2-x-1)+6=$

$\displaystyle =(x-1)(x^2-x+x^2-1)+6= \\ \\ =(x-1) \Big( x(x-1)+(x+1)(x-1) \Big)+6= \\ \\ =(x-1)(x-1)(x+x+1)+6= \\ \\ = \underbrace{(x-1)^2}_{ \ge 0} \underbrace{(2x+1)}_{>0}+6 \ge 0+6=6.$