Question

这道题详细点的答案

Answer 1

解：设F'(x)=f(x), [F(1)-F(0)]^2-x[F(1)-F(0)]-18x^5>=0
△=(-x)^2-4*(-18x^5)=x^2+72x^5>=0; x^2(1+72x^3)>=0; x>=-1/2*9^(1/3)=-3^(1/3)/6
F(1)-F(0)=[x^2+/-√(x^2+72x^5)]/2; F(x)=∫f(t)dt; F(1)-F0)=∫(x=(0,1)f(t)dt;
F(x)<=[x^2-√(x^2+72x^5)]/2], F(x)>=[x^2+√(x^2+72x^5)]/2;
(1) F'(x)=f(x)=x-x+36*5x^4=x-180x^4/√(x^2+72x^5)<=0；不合题意，舍去。
(2)F'(x)=f(x)=x+180x^4/√(x^2+72x^5)>=0;定义域为：x>-3^(1/3)/6; 当x=0时，f(x)=x;