Question

若x1,x2(x1≠x2)是函数f(x)=ax^3+bx^2-a^2x(a>0)的两个极值点

Answer 1

(1)f'(x)=3ax^2+2bx-a^2

因为x1,x2是函数的两个极值点,且x1=-1/3,x2=1

所以x1+x2=-2b/3a=2/3

x1x2=-a/3=-1/3

所以a=1,b=-1

所以f(x)=x^3-x^2-x

(2) x1+x2=-2b/3a

x1x2=-a/3

(x1+x2)^2-2x1x2+2|x1||x2|=12

4b^2/9a^2+2a/3+2a/3=12

整理得b^2=-3a^3+27a^2

令f(x)=-3x^3+27x^2(x>0)

f'(x)=-9x^2+54x

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