Question

g(x)＝f(x,f(x,x))，求g(x)的导数

Answer 1

f(x)=x^3+bx^2+d
x∈(-∞0)f'(x)>0
x∈[0,1] f'(x)<=0
b<=-3/2
求Min g(x)=[f(x)-f(1)]/(x-1)
解:
令f'(x)=3x^2+2bx=0 b<=-3/2
<==>x=0 or x=-2/3b >=1
f'(x)>0 x<0 or x>-2/3b
f'(x)<=0 0<=x<=-2/3b
又因x∈[0,1] f'(x)<=0
<==> b=-3/2
<==>f(x)=x^3 - 3/2x^2 +d
g(x)=[x^3 - 3/2x^2 +d - (1-3/2 +d)]/(x-1)
<==>g(x)=(x^3-3/2x^2+1/2)/(x-1)
g(x)=x^2-1/2x-1/2 ( x≠1 x∈R)
g'(x)=2x-1/2
令g'(x)=0
解得 x=1/4
当x=1/4 时,g(x)存小值 Min g(x)=-9/16

追问

答非所问啊。。