f(x)-4=-x³+3x+2-4= -x³+3x-2
Fie g(x)= -x³+3x-2
g `(x)= -3x²+3 g`(x)=0=>x1=-1 si x2=+1
Studiem semnul pt functia de gradul 2 g(x)
x∈(-∞,-1) g `(x)≤0 => g(x) este descrescatoare
x∈[-1,1] g `(x)>0 => g(x) crescatoare=> x=-1 punct de minim
pt x>1 g `(x)<0=> g(x) crescatoare.=> x=1 punct de maxim pt x∈[-1, ∞)=>
∀x>-1 . g(x)<g(1)=0=>
- x³+3x-2≤0 <=>
-x³+3x+2-4≤0
-x³+3x-2≤4 concluzia
