Question

设二次函数f(x)=ax^2-4bx+c,f(x)>=0,f'(x)满足f'(0)<0,f(2)/f'(0)最大值

设二次函数f(x)=ax^2-4bx+c对任意的x属于R，恒有f(x)>=0,且其导数f'(x)满足f'(0)<0,则f(2)/f'(0)最大值

Answer 1

二次函数f(x)=ax^2-4bx+c对任意的x属于R，△>0。
16b^2-4ac≥0
若f(x)>=0，则二次函数的最小值大于等于0。由于是二次函数，a≠0。那么a大于0
根据二次函数顶点公式，得最低点在y轴上是(4ac-16b^2)/4a≥0
c-4b^2/a≥0
4b^2≤ac
则c≥0
f(x)=ax^2-4bx+c
f'(x)=2ax-4b
f'(0)=-4b
∵f'(0)<0
∴b＞0
f(2)/f'(0)=(4a-8b+c)/(-4b)=2-(a/b+c/4b)
=2-(4a+c)/4b
由4b^2≤ac
则2b≤√ac。以使得4b取到最大值
2-(4a+c)/4b=2-(4a+c)/√ac
∵(4a+c)/√ac＞0
∴(4a+c)^2/ac＞0
2+4a/c+c/4a＞0
4a/c+c/4a≥2
(4a+c)^2/ac≥4
∴(4a+c)/√ac≥2
f(2)/f'(0)最大值为2-2=0

Answer 2

由条件f(x)>=0,f'(x)满足f'(0)<0,可知: 0<b<=根号下(ac/4) 且 a > 0, c > 0
f(2)/f'(0)= (4a-8b+c) / (-4b) = (4a+c) / (-4b) + 2 <= 0

Answer 3

f(x)=ax^2-4bx+c>=0
抛物线开口向上，a>0
抛物线与x轴至多只有1个交点。
0>=Delta=(-4b)^2-4ac=4[4b^2-ac].
4b^2<=ac
c>=0

f'(x)=2ax-4b
0>f'(0)=-4b, b>0

4<=ac/b^2
2<=(ac)^(1/2)/b

f(2)=4a-8b+c
g(a,b,c)=f(2)/f'(0) = [4a-8b+c]/(-4b) = -a/b + 2 - c/(4b) = 2-[4a+c]/(4b)<=2-2[4a*c]^(1/2)/(4b)
=2-4[ac]^(1/2)/(4b)
=2-(ac)^(1/2)/b
<=2-2
=0

等号成立当且仅当ac=4b^2,且4a=c
4b^2=ac=a*4a=4a^2
a=b
c=4a
此时
f(x)=ax^2-4ax+4a=a[x^2-4x+4]=a(x-2)^2
a>0时，f(x)>=0
f'(x)=2a(x-2)
f'(0)=-4a<0
f(2)=0
f(2)/f'(0)=0

f(2)/f'(0)的最大值为0.