b=1; a=0
f(b)=f(1)=sqrt(3)*1^3+3*1=sqrt(3)+3;
f(a)=f(0)=0
f `(ξ)=sqrt(3)+3
f `(x)=(sqrt(3)x^3+3x)`=3sqrt(3)x^2+3
f `(ξ)=3sqrt(3)ξ^2+3
3sqrt(3)ξ^2+3=sqrt(3)+3
3ξ^2=1
x^2=1/3
x= ± sqrt(1/3)
-sqrt(1/3) ∉ (0;1)
О т в е т. sqrt(1/3)