f(x)=ax²+2bx+c
a=-1 c=0
f(x)=-x²+2bx=-(x-b)²+b²
开口向下,对称轴x=b
b<-1 区间x∈[-1,3]位于对称轴右侧,f(x)单调递减
g(b)=f(-1)=-2b-1
b>3 区间x∈[-1,3]位于对称轴左侧,f(x)单调递增
g(b)=f(3)=6b-9
-1≤b≤3 区间x∈[-1,3]包含对称轴,顶点为最大值
g(b)=b²
f(x)=ax²+2bx+c=a(x+b/a)²-b²/a+c
f(x)与x轴相切,-b²/a+c=0→c=b²/a
x∈[-8,-2]f(x)不单调,对称轴-b/a∈[-8,-2]→b/a∈[2,8]
f(1)/(b-2a)=(a+2b+b²/a)/(b-2a)=[a/a+2b/a+(b/a)²]/(b/a-2)
=[1+2b/a+(b/a)²]/(b/a-2)
令f(t)=f(b/a)=[1+2t+t²]/(t-2)=(1+t)²/(t-2) t∈[2,8]
f'(t)=[2(t+1)(t-2)-(1+t)²]/(t-2)=(t+1)(t-5)/(t-2)
t∈[2,8]区间内驻点t=5,经判断为极小值点
∴f(1)/(b-2a)的最小值=f(5)=36/3=12