已知函数f(x)=x2+4x+3,x∈[t,t+1],t∈R,g(t)表示f(x)在区间[t,t+1]上的最小值,求g(

当t+1≤-2,即t≤-3时,
f(x)在[t,t+1]上单调递减,
所以f(x)的最小值g(t)=f(t+1)=(t+1)^2+4(t+1)+3=t^2+6t+8
当t＜-2＜t+1,即-3＜t＜-2时,
f(x)的最小值g(t)=f(x)|x=-2=f(-2)=4-8+3=-1
当-2≤t,即t≥-2时,
f(x)在[t,t+1]上单调递增,
所以f(x)的最小值g(t)=f(t)=t^2+4t+3
综上所述:
g(t)=t^2+6t+8,t≤-3
g(t)=-1,-3＜t＜-2
g(t)=t^2+4t+3,t≥-2
其中t∈R