解:y=x²-2x-3
对y求导,再令导数=0
y'=2x-2,y'=0,得x=1
当x=1时,y=-4,函数最小值
如果令t=1,t≤x≤t+1
则g(t)=-4,h(t)=-3
对称轴x=1,顶点(1,-4)
(1)
t+1≤1即t≤0时
y=f(x)在[t,t+1]单调递减
g(t)=f(t)=t²-2t-3
h(t)=f(t+1)=t²-4
(2)
t≥1时
y=f(x)在[t,t+1]单调递增
g(t)=f(t+1)=t²-4
h(t)=f(t)=t²-2t-3
(3)
t<1,t+1>1即0
f(x)min=f(1)=-4
f(t)=t²-2t-3
f(t+1)=t²-4
f(t)
f(t)≥f(t+1)即0≤t≤1/2时 g(t)=t²-2t-3 h(t)=1
很简单,分为三个情况,
1.当t+1<1,即t<0,g(t)=Y(t),h(t)=y(t+1)
2.当t>1,g(t)=y(t+1),h(t)=y(t)
3.当0