f(x)=(x-1)^2+1
當(dāng)t+1<1,即t<0時(shí),x在[t,t+1]上遞減,
此時(shí)g(t)=f(x)min=f(t+1)=t^2+1
當(dāng)t≤1≤t+1,即0≤t≤1時(shí),
g(t)=f(x)min=f(1)=1
當(dāng)t>1時(shí),x在[t,t+1]上遞增,
此時(shí)g(t)=f(x)min=f(t)=(t-1)^2+1
因?yàn)閠^2+1>=1,(t-1)^2+1>=1
所以g（t）min=1