1. f(x)=(x-1)^2+2,因为(x-1)^2≧0,,①若a>1,则f(x)max=[(a+2)-1]^2+2=a^2+2a+3 f(x)min=(a-1)^2+2=a^2-2a+3, f(x)∈(a^2-2a+3,a^2+2a+3) ②若a≦1≦a+2,即-1≦a≦1,此时分2种情况考虑,1若1-a≦a+2-1,即a≧0时,f(x)max=[(a+2)-1]^2+2=a^2+2a+3,f(x)min=2(此时x=1时,f(1)最小)f(x)∈(1,a^2+2a+3), 2若1-a≧a+2-1,即a≦0,f(x)max=(a-1)^2+2=a^2-2a+3,f(x)min=1,此时f(x)∈(1,a^2-2a+3).③当a+2<1时,f(x)max=(a-1)^2+2=a^2-2a+3,f(x)min=[(a+2)-1]^2+2=a^2+2a+3,所以f(x)∈(a^2+2a+3,a^2-2a+3) 综上所述 若a>1, f(x)∈(a^2-2a+3,a^2+2a+3), 0≦x≦1时,f(x)∈(1,a^2+2a+3), -1≦a≦0,f(x)∈(1,a^2-2a+3), a<-1时,f(x)∈(a^2+2a+3,a^2-2a+3)
2. y=(x-a)^2+1-a^2,①若a>3时, f(x)min=f(3)=10-6a,f(x)max=f(1)=2-2a, f(x)∈(10-6a,2-2a),②若1≦a≦3,此时还是分2种情况讨论,即1. a-1≧3-a时, f(x)min=f(a)=1-a^2,f(x)max=f(1)=2-2a,f(x)∈(1-a^2,2-2a), 2. a-1≦3-a时, f(x)min=f(a)=1-a^2,f(x)max=f(3)=10-6a,f(x)∈(1-a^2,10-6a), ③a<1时,f(x)min=f(1)=2-2a, f(x)max=f(3)=10-6a,f(x)∈(2-2a,10-6a) 综上所述,a>3,f(x)∈(10-6a,2-2a), 2≦a≦3,f(x)∈(1-a^2,2-2a), 1≦a≦2,f(x)∈(1-a^2,10-6a), a<1,f(x)∈(2-2a,10-6a).
f(x)=x^2-2x+3=(x-1)^2+2,可知对称轴x=1
1、区间【a,a+2】在1左侧,即a+2<1即a<-1,函数在该区间递减,最小值为f(a+2)=(a+1)^2+2
2、1在区间【a,a+2】端点及内部,即a≤1≤a+2,-1≤a≤1,最小值为函数顶点f(1)=2
3、区间【a,a+2】在1有侧,即a>1,函数在该区间递增,最小值为f(a)=(a-1)^2+2
所以g(a)=(a+1)^2+2,a<-1
2,-1≤a≤1
(a-1)^2+2,a>1 2.对称轴x=-b/2a=-a
(1)当a<-3时,g(a)=f(x)max=f(3)=-8-6a
(2)当-3<=a<=-1时,g(a)=f(x)max=f(-a)=a^2+1
(3)当a>-1时,g(a)=f(x)max=f(1)=-2a
g(a)的值域为R